CHEMISTRY
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SYLLABUS
CHEMISTRY (Code No. 043)
COURSE STRUCTURE, CLASS XII (2021 - 22)
TERM - II
Time : 2 Hours Max Marks : 35
S. No.
Unit
No. of Periods
1
Electrochemistry
7
2
Chemical Kinetics
5
3
Surface Chemistry
5
4
d - and f - Block Elements
7
5
Coordination Compounds
8
6
Aldehydes, Ketones and Carboxylic Acids
10
7
Amines
7
Total
49
Electrochemistry
Redox reactions, EMF of a cell, standard electrode potential, Nernst
equation and its application to chemical cells, Relation between
Gibbs energy change and EMF of a cell, conductance in electrolytic
solutions, specific and molar conductivity, variations of conductivity with
concentration, Kohlrausch's Law, electrolysis.
Chemical Kinetics
Rate of a reaction (Average and instantaneous), factors affecting rate of
reaction: concentration, temperature, catalyst; order and molecularity of a
reaction, rate law and specific rate constant, integrated rate equations and
half-life (only for zero and first order reactions).
Surface Chemistry
Adsorption - physisorption and chemisorption, factors affecting adsorption
of gases on solids, colloidal state: distinction between true solutions,
colloids and suspension; lyophilic, lyophobic, multi-molecular and
macromolecular colloids; properties of colloids; Tyndall effect, Brownian
movement, electrophoresis, coagulation.
d - and f - Block Elements
General introduction, electronic configuration, occurrence and
characteristics of transition metals, general trends in properties of the first
Marks
13
9
13
35
row transition metals – metallic character, ionization enthalpy, oxidation
states, ionic radii, colour, catalytic property, magnetic properties, interstitial
compounds, alloy formation.
Lanthanoids : Electronic configuration, oxidation states and lanthanoid
contraction and its consequences.
Coordination Compounds
Coordination compounds - Introduction, ligands, coordination number,
colour, magnetic properties and shapes, IUPAC nomenclature of
mononuclear coordination compounds. Bonding, Werner's theory, VBT,
and CFT.
Aldehydes, Ketones and Carboxylic Acids
Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods
of preparation, physical and chemical properties, mechanism of nucleophilic
addition, reactivity of alpha hydrogen in aldehydes, uses.
Carboxylic Acids : Nomenclature, acidic nature, methods of preparation,
physical and chemical properties; uses.
Amines
Amines: Nomenclature, classification, structure, methods of preparation,
physical and chemical properties, uses, identification of primary, secondary
and tertiary amines.
PRACTICALS
At the end of Term II, a 15-mark Practical would be conducted under the supervision of Board appointed external examiners. This would contribute to the overall practical
marks for the subject.
OR
In case the situation of lockdown continues beyond December 2021, a Practical Based Assessment (pen-paper) of 10 marks and Viva 5 marks would be conducted at the
end of Term II jointly by the external and internal examiners and marks would be submitted by the schools to the Board. This would contribute to the overall practical marks
for the subject.
Evaluation Scheme
S. No.
Practicals
Marks
1
Volumetric Analysis
4
2
Salt Analysis
4
3
Content Based Experiment
2
4
Project Work and Viva(Internal and External Both)
5
Total
15
CONTENTS
Chapter 1
Electrochemistry
Chapter 2
Chemical Kinetics
24-46
Chapter 3
Surface Chemistry
47-68
Chapter 4
The d- and f -Block Elements
69-89
Chapter 5
Coordination Compounds
Chapter 6
Aldehydes, Ketones and Carboxylic Acids
114-143
Chapter 7
Amines
144-164
Practice Papers 1-3
165-187
1-23
90-113
*As per the Circular Issued by CBSE on July 05, 2021, Special Scheme of Assessment
for Board Examination Class XII for the Session 2021-22 is as follows :
Term II Examination/Year-end Examination :
•At the end of the second term, the Board would organize Term II or Year-end Examination based on the rationalized
syllabus of Term II only (i.e. approximately 50% of the entire syllabus).
•
This examination would be held around March-April 2022 at the examination centres fixed by the Board.
•The paper will be of 2 hours duration and have questions of different formats (case-based/ situation based, open
ended- short answer/ long answer type).
•In case the situation is not conducive for normal descriptive examination a 90 minute MCQ based exam will be
conducted at the end of the Term II also.
•
Marks of the Term II Examination would contribute to the final overall score.
To cope up with ongoing unpredictable pandemic situation, this book contains chapterwise objective as well as subjective
questions.
In Objective Section, each question carry 1 mark and in Subjective Section, each VSA carry 1 mark, SA I carry 2 marks,
SA II carry 3 marks and LA carry 5 marks.
*As per the CBSE Term-II 2021-2022 curriculum, for latest information visit www.cbse.gov.in
CHAPTER
1
Electrochemistry
Recap Notes
Electrochemistry : It is the study of
production of electricity from energy
released during spontaneous chemical
reactions and the use of electrical energy
to bring about non-spontaneous chemical
transformations.
Differences between electrochemical cell and electrolytic cell
Electrochemical cell
Electrolytic cell
(Galvanic or Voltaic cell)
1. It is a device which converts chemical energy 1. It is a device which converts electrical energy
into electrical energy.
into chemical energy.
2. It is based upon the redox reaction which is 2. The redox reaction is non-spontaneous and
spontaneous. i.e., ∆G = –ve
takes place only when electrical energy is
supplied. i.e., ∆G = +ve
3. Two electrodes are usually set up in two 3. B oth the electrodes are suspended in the
separate beakers.
solution or melt of the electrolyte in the same
beaker.
4. The electrolytes taken in the two beakers are 4. Only one electrolyte is taken.
different.
5. The electrodes taken are of different materials. 5. The electrodes taken may be of the same or
different materials.
6. The electrode on which oxidation takes place 6. The electrode which is connected to the –ve
is called the anode (or –ve pole) and the
terminal of the battery is called the cathode;
the cations migrate to it which gain electrons
electrode on which reduction takes place is
and hence, a reduction takes place, the other
called the cathode (or +ve pole).
electrode is called the anode.
7. To set up this cell, a salt bridge/porous pot 7. No salt bridge is used in this case.
is used.
Nernst equation : For a reduction reaction,
Mn+
(aq)
+ ne
–
°
Ecell = Ecell
−
M(s);
2.303 RT
1
log
+
nF
[ M(naq
)]
At 298 K,
°
Ecell = Ecell
−
0.0591
1
log
+
n
[ M(naq
)]
For concentration cell, EMF at 298 K is
given by
0.0591
C
Ecell =
log 2
n
C1
where C2 > C1
X
Applications of Nernst equation :
X To calculate electrode potential of a
cell :
ne−
→ xX + yY
aA + bB 
CBSE Board Term-II Chemistry Class-12
2
x
y
 − 0.0591 log [ X ] [Y ] (At 298 K)
Ecell = Ecell
n
[ A]a [ B]b
X
To calculate equilibrium constant :
At equilibrium, Ecell = 0
 =
Ecell
0.0591
log K c at 298 K
n
Relation between cell potential and
Gibbs energy change :
DG° = –nFE°cell ; DG° = –2.303 RT log Kc
Conductance in electrolytic solutions :
Property
Conductance (G)
Formula
Units
Effect of dilution
1 = a = κa
l
R ρl
Ohm–1 (Ω–1)/Siemens (S)
Increases as larger number
of ions are produced.
Specific conductance
(k) or conductivity
1
l
or G
ρ
a
Ohm–1 cm–1/S m–1
Decreases as number of
ions per cm3 decreases.
Equivalent
conductivity (Λeq)
κ × V or
1000
κ×
N
Ω–1 cm2 eq–1/S m2 eq–1
Increases with dilution due
to large increase in V.
Molar conductivity
(Λm)
κ × V or
1000
κ×
M
Ω–1 cm2 mol–1/S m2 mol–1 Increases with dilution due
to large increase in V.
Limiting molar conductivity : When
concentration approaches zero i.e., at infinite
dilution, the molar conductivity is known as
limiting molar conductivity (Λ°m).
Variation of molar conductivity with
concentration : For a strong electrolyte it
is shown by Debye–Huckel Onsager equation
as follows :
Λ m = Λ °m − A C
(KCl)
Strong electrolyte
like KCl
m
Weak electrolyte like
CH3COOH
C
Here,
°
Lm
=
olar conductivity at infinite
M
dilution ( L i m i t i n g m o l a r
conductivity)
Lm = Molar conductivity at V-dilution
A = Constant which depends upon nature
of solvent and temperature
C = Concentration
Plot of Lm against C1/2 is a straight line with
°
intercept equal to Lm
and slope equal to ‘–A’.
Thus, Lm decreases linearly with C , when
C = 0, Lm = L°m and L°m can be determined
experimentally.
X
X
For weak electrolytes : There is a
very large increase in conductance with
dilution especially near infinite dilution
as no. of ions increases. Lm increases as
C decreases but does not reach a constant
value even at infinite dilution. Hence, their
L°m cannot be determined experimentally.
For a strong electrolyte : There is
only a small increase in conductance
with dilution. This is because a strong
electrolyte is completely dissociated in
solution and so, the number of ions remain
constant and on dilution, interionic
attractions decreases as ions move far
apart.
Kohlrausch’s law of independent
migration of ions : It states that limiting
molar conductivity of an electrolyte can be
represented as the sum of the individual
contributions of the anion and cation of the
electrolyte.
L°m = u+ l°+ + u–l°– ; where l°+ and l°– are the
limiting molar conductivities of the cation
and anion respectively and u + and u – are
stoichiometric number of cations and anions
respectively in one formula unit of the
electrolyte.
3
Electrochemistry
Applications of Kohlrausch’s law :
X Calculation of molar conductivity of
weak electrolytes :
Λm (CH3COOH) = λ
(
)
(
CH3COO

= λH + + λCl − + λ CH
3COO
H+
−
+ λ
Na +
(
λ
)−
Na
+
AB
A+ + B–
Initial conc.
0
0
ca
ca
c
Conc. at equil. (c – ca)
+ λ
= Λm (HCl) + Λm (CH3COONa) − Λm (NaCl)
Calculation of dissociation constant
(Kc ) of weak electrolyte :

−+ λ
X
X
Cl
−
)
Calculation of degree of dissociation :
Λ
Degree of dissociation (a) = m
Λm
Molar conductivity at concentration c
=
Molar conductivity at infinite dilution
Kc =
cα × cα
cα 2
=
(c − cα) (1 − α)
Electrolytic Cells and Electrolysis
X
Electrolysis
is
the
process
of
decomposition of an electrolyte by passing
electricity through its aqueous solution or
molten state.
Products of Electrolysis
Products
Electrolyte
At
At
cathode anode
Reactions involved
At cathode
Molten NaCl Na metal Cl2 gas Na+(l) + e– → Na(l)
At anode
–
Cl (l) →
1
Cl2(g) + e–
2
Aqueous
NaCl
H2 gas
–
Cl2 gas H2O(l) + e →
1
–
1
–
H2(g) + OH (aq) Cl (aq) → Cl2(g) + e–
2
2
Dil. H2SO4
H2 gas
+
–
O2 gas H (aq) + e →
1
H2(g)
2
2H2O(l) → O2(g) + 4H+(aq) + 4e–
H+(aq) + e– →
1
H2(g)
2
2SO42–(aq) → S2O82–
Conc. H2SO4 H2 gas
S2O82–
Overvoltage/Over potential : Oxidation of
H2O is relatively slow process and thus needs
extra potential. This extra potential needed
to oxidise H 2 O is called overvoltage/over
potential. Due to overvoltage, the oxidation
of chloride ion occurs at anode in preference
to H2O.
(aq)
+ 2e–
Discharge potential : The minimum
potential that must be applied across the
electrodes to bring about the electrolysis and
thus, discharge of the ions on the electrode
is known as discharge potential. It is infact
the ability of ions to discharge first at
electrodes.
4
CBSE Board Term-II Chemistry Class-12
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. Given below are the standard electrode
potentials of few half-cells. The correct order of
these metals in increasing reducing power will
be
K+|K = –2.93 V, Ag+|Ag = 0.80 V,
Mg2+|Mg = –2.37 V, Cr3+|Cr = –0.74 V.
(a) K < Mg < Cr < Ag
(b) Ag < Cr < Mg < K
(c) Mg < K < Cr < Ag
(d) Cr < Ag < Mg < K
2. DrG° for the cell with the cell reaction:
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s)
+ 2OH–(aq)
[E°Ag2O/Ag = 0.344 V, E°Zn2+/Zn = –0.76 V]
(a) 2.13 × 105 J mol–1
(c) 1.06 × 105 J mol–1
(b) –2.13 × 105 J mol–1
(d) –1.06 × 105 J mol–1
–
3. For a cell reaction: Mn+
(aq) + ne → M(s), the
Nernst equation for electrode potential at any
concentration measured with respect to standard
hydrogen electrode is represented as
(a) E n +
= E ° n+
−
(M /M )
(M /M )
RT
1
ln
nF [ M n + ]
(b) E
= E°
−
(M / M n+ )
(M / M n+ )
RT [ M n + ]
ln
[M ]
nF
RT
1
log
nF
[M ]
RT
(d) E n +
= E° n+
−
ln[ M n + ]
(M / M )
( M / M ) nF
(c) E n +
= E° n+
−
(M / M )
(M / M )
4. Limiting molar conductivity for some ions
is given below (in S cm2 mol–1) :
Na+ - 50.1, Cl– - 76. 3, H+ - 349.6, CH3COO– - 40.9,
Ca2+ - 119.0.
What will be the limiting molar conductivities
(L°m) of CaCl2, CH3COONa and NaCl respectively?
(a) 97.65, 111.0 and 242.8 S cm2 mol–1
(b) 195.3, 182.0 and 26.2 S cm2 mol–1
(c) 271.6, 91.0 and 126.4 S cm2 mol–1
(d) 119.0, 1024.5 and 9.2 S cm2 mol–1
5. Electrical conductance through metals is
called metallic or electronic conductance and is
due to the movement of electrons. The electronic
conductance depends on
(a) the nature and structure of the metal
(b) the number of valence electrons per atom
(c) change in temperature
(d) all of these.
6. A galvanic cell has electrical potential of
1.1 V. If an opposing potential of 1.1 V is applied
to this cell, what will happen to the cell reaction
and current flowing through the cell?
(a) The reaction stops and no current flows
through the cell.
(b) The reaction continuous but current flows
in opposite direction.
(c) The concentration of reactants becomes
unity and current flows from cathode to
anode.
(d) The cell does not function as a galvanic cell
and zinc is deposited on zinc plate.
7. In a Daniell cell,
(a) the chemical energy liberated during the
redox reaction is converted to electrical
energy
(b) the electrical energy of the cell is converted
to chemical energy
(c) the energy of the cell is utilised in conduction
of the redox reaction
(d) the potential energy of the cell is converted
into electrical energy.
8. Mark the correct Nernst equation for the
given cell.
Fe(s)|Fe2+(0.001 M)||H+ (1 M) | H2(g)(1 bar)|Pt(s)
° −
(a) Ecell = Ecell
0.591 [ Fe2+ ] [ H + ]2
log
2
[ Fe][ H 2 ]
° −
(b) Ecell = Ecell
0.591
[ Fe] [ H + ]2
log
2
[ Fe2+ ][ H 2 ]
5
Electrochemistry
° −
(c) Ecell = Ecell
0.0591 [ Fe2+ ] [ H 2 ]
log
2
[ Fe][ H + ]2
° − 0.0591 log [ Fe] [ H 2 ]
(d) Ecell = Ecell
2+
+ 2
2
[ Fe
][ H ]
9. When an aqueous solution of AgNO 3 is
electrolysed between platinum electrodes, the
substances liberated at anode and cathode are
(a) silver is deposited at cathode and O 2 is
liberated at anode
(b) silver is deposited at cathode and H 2 is
liberated at anode
(c) hydrogen is liberated at cathode and O2 is
liberated at anode
(d) silver is deposited at cathode and Pt is
dissolved in electrolyte.
10. A standard hydrogen electrode has a zero
potential because
(a) hydrogen can be most easily oxidised
(b) hydrogen has only one electron
(c) the electrode potential is assumed to be zero
(d) hydrogen is the lightest element.
11. At 25°C, Nernst equation is
(a) Ecell = E °cell −
0.0591 [ion]RHS
log
n
[ion]LHS
(b) Ecell = E °cell −
0.0591 [M ]RHS
log
n
[M ]LHS
(c) Ecell = E °cell +
0.0591 [ion]RHS
log
n
[ion]LHS
(d) Ecell = E °cell −
0.0591
[ion]LHS
log
n
[ion]RHS
12. Electrode potential data of few cells is given
below. Based on the data, arrange the ions in
increasing order of their reducing power.
–
2+
Fe3+
(aq) + e → Fe (aq) ; E° = +0.77 V
–
Al3+
(aq) + 3e → Al(s) ; E° = –1.66 V
Br2(aq) + 2e– → 2Br–(aq) ; E° = +1.09 V
(a) Br– < Fe2+ < Al
(b) Fe2+ < Al < Br–
–
2+
(c) Al < Br < Fe
(d) Al < Fe2+ < Br–
13. Mark the correct relationship from the
following.
(a) Equilibrium constant is related to emf as
log K =
nFE
2.303RT
(b) EMF of a cell Zn | Zn2+(a1) || Cu2+(a2) | Cu is
E = E° −
0.591 [a2 ]
log
n
[a1]
(c) Nernst equation is Ecell = E °cell −
0.0591
[ Products]
log
n
[ Reactants]
(d) For the electrode Mn+/M at 273 K
E = E° +
0.591
log[ M n + ]
n
14. The specific conductivity of N/10 KCl solution
at 20°C is 0.0212 ohm–1 cm–1 and the resistance
of the cell containing this solution at 20°C is 55
ohm. The cell constant is
(a) 3.324 cm–1
(b) 1.166 cm–1
–1
(c) 2.372 cm
(d) 3.682 cm–1
15. Following reactions are taking place in a
Galvanic cell,
Zn → Zn2+ + 2e– ; Ag+ + e– → Ag
Which of the given representations is the correct
method of depicting the cell?
+
(a) Zn(s) | Zn2+
(aq) || Ag (aq) | Ag(s)
(b) Zn2+ | Zn || Ag | Ag+
+
(c) Zn(aq) | Zn2+
(s) || Ag (s) | Ag(aq)
(d) Zn(s) | Ag+(aq) || Zn2+
(aq) | Ag(s)
16. What will be the molar conductivity of Al3+
ions at infinite dilution if molar conductivity of
Al2(SO4)3 is 858 S cm2 mol–1 and ionic conductance
of SO42– is 160 S cm2 mol–1 at infinite dilution?
(b) 698 S cm2 mol–1
(a) 189 S cm2 mol–1
2
–1
(d) 429 S cm2 mol–1
(c) 1018 S cm mol
17. E° value of Ni2+/ Ni is –0.25 V and Ag+ /Ag
is +0.80 V. If a cell is made by taking the two
electrodes what is the feasibility of the reaction?
(a) Since E° value for the cell will be positive,
redox reaction is feasible.
(b) Since E° value for the cell will be negative,
redox reaction is not feasible.
(c) Ni cannot reduce Ag+ to Ag hence reaction
is not feasible.
(d) Ag can reduce Ni2+ to Ni hence reaction is
feasible.
18. A cell is set up as shown in the figure. It
is observed that EMF of the cell comes out to
be 2.36 V. Which of the given statements is not
correct about the cell?
CBSE Board Term-II Chemistry Class-12
6
19. Limiting molar conductivity of NaBr is
(a) L°mNaBr = L°mNaCl + L°mKBr
(b) L°mNaBr = L°mNaCl + L°mKBr – L°mKCl
(c) L°mNaBr = L°mNaOH + L°mNaBr – L°mNaCl
(d) L°mNaBr = L°mNaCl – L°mNaBr
20. Choose the option with correct words to fill
in the blanks.
According to preferential discharge theory, out
of number of ions the one which requires _____
energy will be liberated ____ at a given electrode.
(a) least, first
(b) least, last
(c) highest, first
(d) highest, last
21. For the cell reaction :
2Cu +(aq) → Cu (s) + Cu 2+
(aq) , the standard cell
potential is 0.36 V. The equilibrium constant
for the reaction is
(a) 1.2 × 106
(b) 7.4 × 1012
(c) 2.4 × 106
(d) 5.5 × 108
22. E° values of three metals are listed below.
–
Zn2+
(aq) + 2e → Zn(s) ; E° = –0.76 V
–
Fe2+
(aq) + 2e → 2Fe(s) ; E° = –0.44 V
–
Sn2+
(aq) + 2e → Sn(s) ; E° = – 0.14 V
Which of the following statements are correct on
the basis of the above information?
(i) Zinc will be corroded in preference to iron if
zinc coating is broken on the surface.
(ii) If iron is coated with tin and the coating
is broken on the surface then iron will be
corroded.
(iii) Zinc is more reactive than iron but tin is less
reactive than iron.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i), (ii) and (iii)
(d) (i) and (iii)
23. Which of the following is the correct order in
which metals displace each other from the salt
solution of their salts.
(a) Zn, Al, Mg, Fe, Cu (b) Cu, Fe, Mg, Al, Zn
(c) Mg, Al, Zn, Fe, Cu (d) Al, Mg, Fe, Cu, Zn
24. The reaction which is taking place in nickel
- cadmium battery can be represented by which
of the following equation?
(a)
(b)
(c)
(d)
Cd + NiO2 + 2H2O → Cd(OH)2 + Ni(OH)2
Cd + NiO2 + 2OH– → Ni + Cd(OH)2
Ni + Cd(OH)2 → Cd + Ni(OH)2
Ni(OH)2 + Cd(OH)2 → Ni + Cd + 2H2O
25. Molar conductivity of 0.15 M solution of KCl
at 298 K, if its conductivity is 0.0152 S cm–1 will
be
(a) 124 W–1 cm2 mol–1 (b) 204 W–1 cm2 mol–1
(c) 101 W–1 cm2 mol–1 (d) 300 W–1 cm2 mol–1
26. Fluorine is the best oxidising agent because
it has
(a) highest electron affinity
(b) highest reduction potential
(c) highest oxidation potential
(d) lowest electron affinity.
27. During the electrolysis of dilute sulphuric
acid, the following process is possible at anode.
(a) 2H2O(l) → O2(g) + 4H+(aq) + 4e–
2–
–
(b) 2SO2–
4(aq) → S2O 8(aq) + 2e
+
–
(c) H2O(l) → H (aq) + OH (aq)
(d) H2O(l) + e– →
1
H
+ OH–(aq)
2 2(g)
28. Mark the correct choice of electrolytes
represented in the graph.
Λm(S cm2 mol–1)
(a) Reduction takes place at magnesium
electrode and oxidation at SHE.
(b) Oxidation takes place at magnesium
electrode and reduction at SHE.
(c) Standard electrode potential for Mg2+ | Mg
will be –2.36 V.
(d) Electrons flow from magnesium electrode to
hydrogen electrode.
A
B
C1/2 (mol L–1)
(a)
(b)
(c)
(d)
A
A
A
A
→ NH4OH, B → NaCl
→ NH4OH, B → NH4Cl
→ CH3COOH, B → CH3COONa
→ KCl, B → NH4OH
29. Molar conductivity of 0.025 mol L –1
methanoic acid is 46.1 S cm2 mol–1, the degree
of dissociation and dissociation constant
will be (Given : l°H+ = 349.6 S cm 2 mol –1 and
°
l°HCOO– = 54.6 S cm2 mol–1)
(a) 11.4%, 3.67 × 10–4 mol L–1
(b) 22.8%, 1.83 × 10–4 mol L–1
(c) 52.2%, 4.25 × 10–4 mol L–1
(d) 1.14%, 3.67 × 10–6 mol L–1
30. Electrode potential for Mg electrode
varies according to the equation,
E
Mg 2+ |Mg
= E°
Mg 2+ |Mg
−
0.059
1
.
log
2
[Μg 2+ ]
The graph of EMg2+|Mg vs log [Mg2+] is
7
Electrochemistry
(a)
(c)
(b)
(d)
31. E° values for the half cell reactions are given
below:
Cu2+ + e– → Cu+ ; E° = 0.15 V
Cu2+ + 2e– → Cu ; E° = 0.34 V
What will be the E° of the half-cell : Cu+ + e– → Cu?
(a) +0.49 V
(b) +0.19 V
(c) +0.53 V
(d) +0.30 V
32. Given below are two figures of Daniell cell
(X) and (Y). Study the figures and mark the
incorrect statement from the following.
33. Which of the following is/are an application
of electrochemical series?
(a) To compare the relative oxidising and
reducing power of substances.
(b) To predict evolution of hydrogen gas on
reaction of metal with acid.
(c) To predict spontaneity of a redox reaction.
(d) All of these
34. Two solutions of X and Y electrolytes are
taken in two beakers and diluted by adding
500 mL of water. Lm of X increases by 1.5 times
while that of Y increases by 20 times, what
could be the electrolytes X and Y ?
(a) X → NaCl, Y → KCl
(b) X → NaCl, Y → CH3COOH
(c) X → KOH, Y → NaOH
(d) X → CH3COOH, Y → NaCl
35. What would be the equivalent conductivity
of a cell in which 0.5 N salt solution offers
a resistance of 40 ohm whose electrodes are
2 cm apart and 5 cm2 in area?
(a) 10 ohm–1 cm2 eq–1
(c) 30 ohm–1 cm2 eq–1
(b) 20 ohm–1 cm2 eq–1
(d) 25 ohm–1 cm2 eq–1
36. The half-cell reactions with their appropriate
standard reduction potentials are
(i) Pb2+ + 2e– → Pb ; E° = –0.13 V
(ii) Ag+ + e– → Ag ; E° = +0.80 V
Based on the above data, which of the following
reactions will take place?
(a) Pb2+ + 2Ag → 2Ag+ + Pb
(b) 2Ag + Pb → 2Ag+ + Pb2+
(c) 2Ag+ + Pb → Pb2+ + 2Ag
(d) Pb2+ + 2Ag+ → Pb + Ag
(a) In fig (X), electrons flow from Zn rod to Cu
rod hence current flows from Cu to Zn (Eext
< 1.1 V).
(b) In fig (Y), electrons flow from Cu to Zn and
current flows from Zn to Cu (Eext > 1.1 V).
(c) In fig (X), Zn dissolves at anode and Cu
deposits at cathode.
(d) In fig (Y), Zn is deposited at Cu and Cu is
deposited at Zn.
37. Units of the properties measured are given
below. Which of the properties has not been
matched correctly?
(a) Molar conductance = S m2 mol–1
(b) Cell constant = m–1
(c) Specific conductance = S m2
(d) Equivalent conductance = S m2 (g eq)–1
38. When water is added to an aqueous solution
of an electrolyte, what is the change in specific
conductivity of the electrolyte?
(a) Conductivity decreases
(b) Conductivity increases
(c) Conductivity remains same
(d) Conductivity does not depend on number of
ions.
CBSE Board Term-II Chemistry Class-12
8
39. The specific conductance of a saturated
solution of AgCl at 25°C is 1.821 × 10 –5 mho
cm–1. What is the solubility of AgCl in water
(in g L–1), if limiting molar conductivity of AgCl
is 130.26 mho cm2 mol–1?
(a) 1.89 × 10–3 g L–1
(b) 2.78 × 10–2 g L–1
–2
–1
(c) 2.004 × 10 g L
(d) 1.43 × 10–3 g L–1
40. The standard reduction potential for the
half-cell reaction, Cl2 + 2e– → 2Cl– will be
(Pt2+ + 2Cl– → Pt + Cl2 , E°cell = –0.15 V ;
Pt2+ + 2e– → Pt, E° = 1.20 V)
(a) –1.35 V
(b) +1.35 V
(c) –1.05 V
(d) +1.05 V
41. Zn gives hydrogen with H2SO4 and HCl but
not with HNO3 because
(a) Zn acts as oxidising agent when reacts with
HNO3
(b) HNO3 is weaker acid than H2SO4 and HCl
(c) Zn is above the hydrogen in electrochemical
series
(d) NO3– is reduced in preference to H+ ion.
42. Given below are few reactions with some
expressions. Mark the expression which is not
correctly matched.
(a) For concentration cell,
Ag | Ag+(C1) || Ag+(C2) | Ag ; Ecell = −
0.0591
C
log 1
1
C2
(b) For the cell, 2Ag+ + H2 (l atm) → 2Ag + 2H+ (1 M) ;
° −
Ecell = Ecell
0.0591 [ Ag + ]2
log
2
[ H + ]2
(c) For a n e le ct ro che m ica l reac t i o n , at
ne−
equilibrium aA + bB cC + dD ;
° =
Ecell
0.0591 [C ]c [ D ]d
log
n
[ A]a [ B ]b
44. In a cell reaction, Cu(s) + 2Ag+(aq) → Cu2+
(aq) + 2Ag(s)
E°cell = +0.46 V. If the concentration of Cu2+ ions
is doubled then E°cell will be
(a) doubled
(b) halved
(c) increased by four times
(d) unchanged.
45. Molar conductivity of NH 4 OH can be
calculated by the equation,
(a) L°NH4OH = L°Ba(OH)2 + L°NH4Cl – L°BaCl2
(b)L°NH4OH = L°BaCl2 + L°NH4Cl – L°Ba(OH)2
(c) L°NH4OH =
(d) L°NH4OH =
2
(a) 3.66%
(c) 2.12%
(b) 3.9%
(d) 0.008%
47. Mark the incorrect statement.
(a) The limiting equivalent conductance for
weak electrolytes can be computed with the
help of Kohlrausch’s law.
(b) EMF of a cell is the difference in the
reduction potentials of cathode and anode.
(c) For cell reaction to occur spontaneously, the
EMF of the cell should be negative.
(d) Fluorine is the strongest oxidising agent as
its reducing potential is very high.
48. The process of chemical decomposition of the
electrolyte by the passage of electricity through
its melt or aqueous solution is called electrolysis.
The following apparatus is used for the
electrolysis process:
DC source
+
–
e–
0.0591
1
log
n
[M n+ ]
43. Which of the following is the cell reaction
that occurs when the following half-cells are
combined?
I2 + 2e– → 2I– (1 M) ; E° = +0.54 V
Br2 + 2e– → 2Br– (1 M) ; E° = +1.09 V
(a) 2Br– + I2 → Br2 + 2I–
(b) I2 + Br2 → 2I– + 2Br–
(c) 2I– + Br2 → I2 + 2Br–
(d) 2I– + 2Br– → I2 + Br2
2
Λ° NH 4Cl + Λ°Ba ( OH )2
46. The equivalent conductivity of N/10 solution
of acetic acid at 25°C is 14.3 ohm–1 cm2 equiv–1.
What will be the degree of dissociation of acetic
acid (L∞CH3COOH = 390.71 ohm–1 cm2 equiv–1)?
(d) For the cell, M n+ (aq) + ne – → M (s) ;
E = E° −
Λ°Ba ( OH )2 + 2 Λ° NH 4Cl − Λ°BaCl 2
Ammeter
e–
Anode
Cathode
–
–
–
–
9
Electrochemistry
Nandini, a young scientist, tried different
electrolysis experiments using various
electrolytes.
The incorrect observation of her experiment is
(a) cations which get reduced at cathode
preferentially are hydronium ions in
electrolysis of aqueous NaCl
(b) cations reaching to cathode are Cu2+ ions
during electrolysis of CuSO4 solution
(c) during electrolysis of conc. H2SO4, S2O82– is
formed at anode
(d) S2O82– is formed at anode during electrolysis
of CuSO4 solution.
49. Jiya, a class-12 student recorded Λm of various
electrolytes like acetic acid, sodium chloride and
AlPO4, etc., at various concentrations. Then she
plotted Λm versus C . Graphs obtained by her
are shown below:
Λm
Λm
KCl
C
I
CH3COOH
Λm
AlPO4
C
C
II
III
Which of the given graph(s) is/are correct?
(a) I only
(b) I and II only
(c) I and III only
(d) I, II and III
50. Which of the given Nernst equation
representation(s) is/are not correct for the given
cell?
Mg | Mg2+(0.130 M) || Ag+ (0.0001 M) | Ag
I.

Ecell
Ecell =
(
Mg 2+ 
RT

−
ln 
2
2F
 Ag + 


)
2
 Ag + 
IV. Ecell =
(
 Ag + 
0.059

+
log 
2
Mg 2+ 


E +
Ag / Ag
−
(a) I only
(c) II and IV only
E  2+
Mg / Mg
)
[ M 2+ ][H+ ]
0.059
log
2
[ M 4+ ]
2
2
III. Ecell =
52. Arun, a class-12 student has a good habit of
practicing the topic at home whichever taught
in the class. After learning Nernst equation in
class, he tried writing few Nernst equations for
different cells. Next day when he shown the
work to his class teacher she said all are correct
except one.
The incorrect Nernst equation is
2+
(a) Pt(s) | H2(g), (1 bar) | H+(aq), 1 M || M4+
(aq), M (aq) | Pt(s)
 −
Ecell = Ecell
Mg 2+ 
0.059
II. Ecell = E  +
− E  2+
−
ln 
2
Ag / Ag
Mg / Mg

Ecell
are very important and we can extract a lot of
useful informations from them. If the standard
electrode potential of an electrode is greater
than zero then its reduced form is more stable
compared to hydrogen gas. Similarly, if the
standard electrode potential is negative then
hydrogen gas is more stable than the reduced
form of the species.
Based on the given data,
–
Fe(s); E° = –0.44 V
Fe2+
(aq) + 2e
2+
–
Sn(s) ; E° = –0.14 V
Sn (aq) + 2e
2+
–
Zn(s) ; E° = –0.76 V
Zn (aq) + 2e
3+
–
Cr(s) ; E° = –0.74 V
Cr (aq) + 3e
He made following conclusions:
I. SnSO4 solution can be stored in Fe vessel.
II. FeSO4 solution can be stored in Zn vessel.
III. Cr2(SO4)3 solution can be stored in Sn vessel.
IV. ZnSO 4 solution cannot be stored in iron
vessel.
The correct conclusion(s) is/are
(a) I and II
(b) III and IV
(c) III only
(d) all of these.
0.059
(0.0001)2
−
log
2
(0.130)
(b) I and III only
(d) II, III and IV only
51. Shubh learnt during his electrochemistry
class that the standard electrode potentials
(b) Pt | M|M3+ (0.001 mol L–1) || Ag+ (0.01 mol L–1) | Ag

Ecell = Ecell
−
[ M 3+ ]
0.059
log
3
3
 Ag + 
(c) Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)
 −
Ecell = Ecell
[Cu 2+ ]
2.303 RT
log
2F
[Zn2+ ]
+
(d) Ni(s) | Ni2+
(aq) || Ag (aq) | Ag(s)
Ecell = ( E
+
Ag /Ag
− E
Ni
2+
/Ni
)−
[Ni2+ ]
0.059
log
2
2
 Ag + 
CBSE Board Term-II Chemistry Class-12
10
Case Based MCQs
Case I : Read the passage given below and
answer the following questions from 53 to 57.
The study of the conductivity of electrolyte
solutions is important for the development of
electrochemical devices, for the characterisation
of the dissociation equilibrium of weak electrolytes
and for the fundamental understanding of charge
transport by ions. The conductivity of electrolyte
is measured for electrolyte solution with
concentrations in the range of 10–3 to 10–1 mol L–1,
as solution in this range of concentrations can
be easily prepared. The molar conductivity (Λm)
of strong electrolyte solutions can be nicely fit
by Kohlrausch equation.
Λm = L°m – K C …(i)
Where, Λ°m is the molar conductivity at infinite
dilution and C is the concentration of the
solution. K is an empirical proportionality
constant to be obtained from the experiment.
The molar conductivity of weak electrolytes,
on the other hand, is dependent on the degree
of dissociation of the electrolyte. At the limit of
very dilute solution, the Ostwald dilution law
is expected to be followed,
Λm CA
1
1
…(ii)
=
+

Λm Λ
( Λ  )2 K d
m
m
where, CA is the analytical concentration of the
electrolyte and Kd is dissociation constant. The
molar conductivity at infinite dilution can be
decomposed into the contributions of each ion.
Λ°m = n+l°+ + n– l°–
…(iii)
Where, l+ and l– are the ionic conductivities of
positive and negative ions, respectively and n+
and n– are their stoichiometric coefficients in
the salt molecular formula.
53. Which statement about the term infinite
dilution is correct?
(a) Infinite dilution refers to hypothetical
situation when the ions are infinitely far
apart.
(b) The molar conductivity at infinite dilution of
NaCl can be measured directly in solution.
(c) Infinite dilution is applicable only to strong
electrolytes.
(d) Infinite dilution refers to a real situation
when the ions are infinitely far apart.
54. Which of the following is a strong electrolyte
in aqueous solution?
(a) HNO2
(c) NH3
(b) HCN
(d) HCl
55. Which of the following is a weak electrolyte
in aqueous solution?
(a) K2SO4
(b) Na3PO4
(c) NaOH
(d) H2SO3
56. If the molar conductivities at infinite dilution
for NaI, CH 3 COONa and (CH 3 COO) 2 Mg are
12.69, 9.10 and 18.78 S cm2 mol–1 respectively
at 25°C, then the molar conductivity of MgI2 at
infinite dilution is
(a) 25.96 S cm2, mol–1 (b) 390.5 S cm2 mol–1
(c) 189.0 S cm2 mol–1
(d) 3.89 × 10–2 S cm2 mol–1
57. Which of the following is the correct order of
molar ionic conductivities of the following ions
in aqueous solutions?
(a) Li+ < Na+ < K+ < Rb+
(b) Li+ > Na+ > K+ > Rb+
(c) Rb+ < Na+ < Li+ < K+
(d) Li+ < Rb+< Na+ < K+
Case II : Read the passage given below and
answer the following questions from 58 to 62.
The electrochemical cell shown below is
concentration cell.
M | M2+ (saturated solution of a sparingly soluble
salt, MX2) || M2+ (0.001 mol dm–3) | M
The emf of the cell depends on the difference in
concentrations of M2+ ions at the two electrodes.
The emf of the cell at 298 K is 0.059 V.
58. The solubility product (K sp , mol 3 dm –9 )
of MX 2 at 298 K based on the information
available for the given concentration cell is
(take 2.303 × R × 298/F = 0.059)
(b) 4 × 10–15
(a) 2 × 10–15
(c) 3 × 10–12
(d) 1 × 10–12
59. The value of DG (in kJ mol–1) for the given
cell is (take 1 F = 96500 C mol–1)
(a) 3.7
(b) –3.7
(c) 10.5
(d) –11.4
60. The equilibrium constant for the following
reaction is
Fe2+ + Ce4+  Ce3+ + Fe3+
(Given: E°Ce4+/Ce3+ = 1.44 V and E°Fe3+/Fe2+ = 0.68 V)
11
Electrochemistry
(a) 7.6 × 1012
(c) 5.2 × 109
(b) 6.5 × 1010
(d) 3.4 × 1012
61. The solubility product of a saturated solution
of Ag2CrO4 in water at 298 K if the emf of the cell
Ag | Ag+ (satd. Ag2CrO4 soln) || Ag+ (0.1 M) | Ag is
0.164 V at 298 K, is
(a) 3.359 × 10–12 mol3 L–3
(b) 2.287 × 10–12 mol3 L–3
(c) 1.158 × 10–12 mol3 L–3
(d) 4.135 × 10–12 mol3 L–3
62. To calculate the standard emf of the cell,
which of the following options is correct if E° is
reduction potential values?
(a) emf = E°cathode – E°anode
(b) emf = E°anode – E°cathode
(c) emf = E°anode + E°cathode
(d) None of these
Case III : Read the passage given below and
answer the following questions.
Nernst equation relates the reduction potential
of an electrochemical reaction to the standard
potential and activities of the chemical species
undergoing oxidation and reduction.
nM(s)
Let us consider the reaction, Mn+(aq)
For this reaction, the electrode potential
measured with respect to standard hydrogen
electrode can be given as
E
( M n+ / M )
= E
( M n+ / M )
−
RT
[M ]
ln n+
nF [M ]
In the following questions (Q. No. 63-67), a
statement of assertion followed by a statement
of reason is given. Choose the correct answer
out of the following choices on the basis of the
above passage.
(a) Assertion and reason both are correct
statements and reason is correct explanation
for assertion.
(b) Assertion and reason both are correct
statements but reason is not correct
explanation for assertion.
(c) Assertion is correct statement but reason is
wrong statement.
(d) Assertion is wrong statement but reason is
correct statement.
63. Assertion : For concentration cell,
Zn(s) | Zn2+(aq) || Zn2+(aq) | Zn
C1
C2
For spontaneous cell reaction, C1 < C2
Reason : For concentration cell, Ecell =
C
RT
log 2
nF
C1
For spontaneous reaction, Ecell = +ve so, C2 > C1.
64. Assertion : For the cell reaction,
Zn(s) + Cu2+
Zn2+(aq) + Cu(s)
(aq)
voltmeter gives zero reading at equilibrium.
Reason : At the equilibrium, there is no change
in concentration of Cu2+ and Zn2+ ions.
65. Assertion : The Nernst equation gives the
concentration dependence of emf of the cell.
Reason : In a cell, current flows from cathode
to anode.
66. Assertion : Increase in the concentration of
copper half cell in a cell, increases the emf of the
cell.

Reason : Ecell = Ecell
+
0.059
[Cu2+ ]
log
2
[Zn2+ ]
67. Assertion : Electrode potential for the
electrode Mn+/Mn with concentration is given
by the expression under STP conditions.
0.059
E = E° +
log[Mn + ]
n
Reason : STP conditions require the temperature
to be 273 K.
Case IV : Read the passage given below and
answer the following questions from 68 to 72.
The concentration of potassium ions inside a
biological cell is at least twenty times higher than
the outside. The resulting potential difference
across the cell is important in several processes
such as transmission of nerve impulses and
maintaining the ion balance. A simple model for
such a concentration cell involving a metal M is
M(s) | M+(aq.; 0.05 molar) || M+(aq; 1 molar) | M(s)
68. For the above cell,
(a) Ecell = 0 ; DG > 0
(c) Ecell < 0 ; DG° > 0
(b) Ecell > 0 ; DG < 0
(d) Ecell > 0 ; DG° = 0
69. If the 0.05 molar solution of M+ is replaced by
a 0.0025 molar M+ solution, then the magnitude
of the cell potential would be
(a) 130 mV
(b) 185 mV
(c) 154 mV
(d) 600 mV
CBSE Board Term-II Chemistry Class-12
12
70. The value of equilibrium constant for a
feasible cell reaction is
(a) < 1
(b) = 1
(c) > 1
(d) zero
71. What is the emf of the cell when the cell
reaction attains equilibrium?
(a) 1
(b) 0
(c) > 1
(d) < 1
72. The potential of an electrode change with
change in
(a) concentration of ions in solution
(b) position of electrodes
(c) voltage of the cell
(d) all of these.
Case V : Read the passage given below and
answer the following questions from 73 to 75.
All chemical reactions involve interaction of
atoms and molecules. A large number of atoms/
molecules are present in a few gram of any
chemical compound varying with their atomic/
molecular masses. To handle such large number
conveniently, the mole concept was introduced.
All electrochemical cell reactions are also based
on mole concept. For example, a 4.0 molar
aqueous solution of NaCl is prepared and
500 mL of this solution is electrolysed. This
leads to the evolution of chlorine gas at one of
the electrode. The amount of products formed
can be calculated by using mole concept.
73. The total number of moles of chlorine gas
evolved is
(a) 0.5
(b) 1.0
(c) 1.5
(d) 1.9
74. If cathode is a Hg electrode, then the
maximum weight of amalgam formed from this
solution is
(Given : Atomic mass of Na = 23u and
Hg = 200.59 u)
(a) 300 g
(b) 446 g
(c) 396 g
(d) 296 g
75. In electrolysis of aqueous NaCl solution
when Pt electrode is taken, then which gas is
liberated at cathode?
(a) H2 gas
(b) Cl2 gas
(c) O2 gas
(d) None of these
Assertion & Reasoning Based MCQs
For question numbers 76-90, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
76. Assertion : The conductivity depends on
the charge and size of the ions in which they
dissociate, the concentration of ions or ease with
which the ions move under potential gradient.
chloride ions respectively, then the limiting
molar conductivity for sodium chloride is given
Reason : The conductivity of solutions of
different electrolytes in the same solvent and
at a given temperature is same.
Reason : This is according to Kohlrausch law
of independent migration of ions.
77. Assertion : If standard reduction potential
for the reaction,
Ag+ + e– → Ag is 0.80 volt, then for the reaction,
2Ag+ + 2e– → 2Ag, it will be 1.60 volt.
Reason : If concentration of Ag+ ions is doubled,
the standard electrode potential remains same.
78. Assertion : If λ o + and λ o − are molar
Na
Cl
limiting conductivities of the sodium and
by the equation, Λ °NaCl = λ°
Na +
+ λ°
Cl −
.
79. Assertion : The conductivity of solution is
greater than pure solvent.
Reason : Conductivity depends upon number of
the ions present in solution.
80. Assertion : At the end of electrolysis using
platinum electrodes, an aqueous solution of
copper sulphate turns colourless.
Reason : Copper in CuSO 4 is converted to
Cu(OH)2 during the electrolysis.
13
Electrochemistry
81. Assertion : The electrical resistance of any
object decreases with increase in its length.
electrolyte at infinite dilution is equal to the
sum of molar conductance of cations and anions.
Reason : The electrical resistance of any object
decreases with increase in its area of crosssection.
Reason : Kohlrausch’s law is applicable for
strong electrolytes.
82. Assertion : Substances like glass, ceramics,
etc. having very low conductivity are known as
insulators.
Reason : They do not allow the passage of
electric current through them.
83. Assertion : Molar conductivity of a weak
electrolyte at infinite dilution cannot be
determined experimentally.
Reason : Kohlrausch law helps to find the molar
conductivity of a weak electrolyte at infinite
dilution.
84. Assertion : The observed conductance
depends upon the nature of the electrolyte and
the concentration of the solution.
Reason : The cell constant of a cell depends
upon the nature of the material of the electrodes.
85. Assertion : The molar conductivity of
strong electrolyte decreases with increase in
concentration.
Reason : At high concentration, migration of
ions is slow.
86. Assertion : The molar conductance of weak
87. Assertion : Equivalent conductance of
all electrolytes decreases with increasing
concentration.
Reason : More number of ions are available per
gram equivalent at higher concentration.
88. Assertion : Specific conductance decreases
with dilution whereas equivalent conductance
increases.
Reason : On dilution, number of ions per
millilitre decreases but total number of ions
increases considerably.
89. Assertion : The ratio of specific conductivity
to the observed conductance does not depend
upon the concentration of the solution taken in
the conductivity cell.
Reason : Specific conductivity decreases with
dilution whereas observed conductance increases
with dilution.
90. Assertion : Kohlrausch law helps to find
the molar conductivity of weak electrolyte at
infinite dilution.
Reason : Molar conductivity of a weak electrolyte
at infinite dilution cannot be determined
experimentally.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Express the relation between conductivity
and molar conductivity of a solution held in a
cell?
4. Give reason :
Molar conductivity of CH3COOH increases on
dilution.
2. Limiting molar conductivity of an electrolyte
cannot be determined experimentally. Why?
5. Give reason :
On the basis of E° values, O2 gas should be liberated
at anode but it is Cl2 gas which is liberated in the
electrolysis of aqueous NaCl.
3. Following reactions occur at cathode during
the electrolysis of aqueous silver chloride
solution :
Ag+(aq) + e– → Ag(s), E° = +0.80 V
1
H+(aq) + e– → H2( g) , E° = 0.00 V
2
On the basis of their standard reduction
electrode potential (E°) values, which reaction
is feasible at the cathode and why?
6. What is the necessity to use a salt bridge in
a Galvanic cell?
7. What is the use of platinum foil in the
hydrogen electrode?
8. Out of HCl and NaCl, which do you expect
will have greater value for Lm and why?
CBSE Board Term-II Chemistry Class-12
14
9. State Kohlrausch’s law of independent
migration of ions. Write its one application.
10. Following reactions occur at cathode during
the electrolysis of aqueous copper (II) chloride
solution :
–
Cu2+
(aq) + 2e → Cu(s) ; E° = +0.34 V
H+(aq) + e– →
1
H2( g) ; E° = 0.00 V
2
On the basis of their standard reduction
electrode potential (E°) values, which reaction
is feasible at the cathode and why?
Short Answer Type Questions (SA-I)
11. Define the term degree of dissociation. Write
an expression that relates the molar conductivity
of a weak electrolyte to its degree of dissociation.
12. (i) Explain why fluorine is the strongest
oxidising agent?
(ii) Lithium metal is the strongest reducing
agent. Why?
13. The standard electrode potential (E°) for
Daniell cell is +1.1 V. Calculate the DrG° for the
reaction.
2+
Zn(s) + Cu2+
(aq) → Zn (aq) + Cu(s)
(1 F = 96500 C mol–1)
14. What is the difference between electronic
and electrolytic conductors?
15. Define electrochemical cell. What happens if
external potential applied becomes greater than
E°cell of electrochemical cell?
16. Why a galvanic cell stops working after
sometime?
17. In the plot of molar conductivity (Lm) vs square
root of concentration (c1/2),following curves are
obtained for two electrolytes A and B.
Answer the following :
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of L m
to concentration approaching zero for
electrolytes A and B?
18. Two half-reactions of an electrochemical cell
are given below :
MnO–4(aq) + 8H+(aq) + 5e– → Mn2+
(aq) + 4H2O(l),
E° = + 1.51V
2+
4+
–
Sn (aq) → Sn (aq) + 2e , E° = + 0.15 V
Construct the redox equation from the standard
potential of the cell and predict if the reaction is
reactant favoured or product favoured.
19. The conductivity of 0.001 M acetic acid is
4 × 10–5 S/cm. Calculate the dissociation constant
of acetic acid, if molar conductivity at infinite
dilution for acetic acid is 390 S cm2/mol.
20. Equilibrium constant (Kc) for the given cell
reaction is 10. Calculate E°cell.
A2+
A(s) + B2+
(aq)
(aq) + B(s)
21. Given that the standard electrode potential
(E°) of metals are :
K+/K = –2.93 V, Ag+/Ag = 0.80 V,
Cu2+/Cu = 0.34 V,
Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74 V,
Fe2+ /Fe = –0.44 V.
Arrange these metals in an increasing order of
their reducing power.
Short Answer Type Questions (SA-II)
22. Mention few applications of electrochemical
series.
23. A voltaic cell is set up at 25°C with the
following half cells :
Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M)
Write an equation for the reaction that occurs
when the cell generates an electric current and
determine the cell potential.
E°Ni2+/Ni = – 0.25 V and E°Al3+/Al = – 1.66 V.
(log 8 × 10–6 = – 5.09)
24. A cell is prepared by dipping copper rod in
1 M copper sulphate solution and zinc rod in 1 M
15
Electrochemistry
zinc sulphate solution. The standard reduction
potential of copper and zinc are 0.34 V and –0.76 V
respectively.
(i) What will be the cell reaction?
(ii) What will be the standard electromotive
force of the cell?
(iii) Which electrode will be positive?
25. Resistance of a conductivity cell filled with
0.1 mol L–1 KCl solution is 100 W. If the resistance
of the same cell when filled with 0.02 mol L–1 KCl
solution is 520 W, calculate the conductivity and
molar conductivity of 0.02 mol L–1 KCl solution.
The conductivity of 0.1 mol L–1 KCl solution is
1.29 × 10–2 W–1 cm–1.
26. Calculate the potential for half-cell
containing
0.10 M K2Cr2O7(aq) , 0.20 M Cr3+
(aq) and
1.0 × 10–4 M H+(aq). The half cell reaction is :
+
–
3+
Cr 2O 2–
7(aq) + 14H (aq) + 6e → 2Cr (aq) + 7H 2 O (l)
and the standard electrode potential is given as
E° = 1.33 V.
27. Estimate the minimum potential difference
needed to reduce Al 2O 3 at 500°C. The Gibbs
energy change for the decomposition reaction,
2
4
Al2O3 → Al + O2 is 960 kJ.
3
3
(F = 96500 C mol–1)
28. For the cell reaction,
+
Ni(s) | Ni2+
(aq) || Ag (aq) | Ag(s)
Calculate the equilibrium constant at 25°C.
How much maximum work would be obtained
by operation of this cell?
E
(Ni2+ /Ni)
= − 0.25 V and E
Ag + /Ag
= 0.80 V
29. Calculate DrG° and logKc for the following
reaction.
2+
Cd2+
(aq) + Zn(s) → Zn (aq) + Cd(s)
Given : E°Cd2+/Cd = –0.403 V ; E°Zn2+/Zn = –0.763 V
30. The equivalent conductivity of 0.05 N
solution of a monobasic acid is 15.8 mho cm2 eq–1.
If equivalent conductivity of the acid at infinite
dilution is 350 mho cm 2 eq –1 , calculate the
(a) degree of dissociation of acid (b) dissociation
constant of acid.
31. The electrical resistance of a column of
0.05 M NaOH solution of diameter 1 cm and
length 50 cm is 5.5 × 10 3 ohm. Calculate its
resistivity, conductivity and molar conductivity.
32. Depict the galvanic cell in which the reaction
Zn (s) + 2Ag +(aq) → Zn 2+
(aq) + 2Ag (s) takes place.
Further show :
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
33. What is the difference between a chemical
and a concentration cell?
34. A copper-silver cell is set up. The copper
ion concentration is 0.10 M. The concentration
of silver ion is not known. The cell potential
when measured was 0.422 V. Determine the
concentration of silver ions in the cell.
Given : E
Ag + /Ag
= + 0.80 V , E
Cu 2+ /Cu
= + 0.34 V
35. The resistance of 100 cm3 aqueous solution of
0.025 M CuSO4 is 520 ohm at 298 K. Calculate
the molar conductivity if the cell constant of the
conductivity cell is 153.7 m–1.
36. When a certain conductance cell was filled
with 0.1 M KCl, it has a resistance of 85 ohms
at 25°C. When the same cell was filled with an
aqueous solution of 0.052 M unknown electrolyte,
the resistance was 96 ohms. Calculate the
molar conductance of the electrolyte at this
concentration.
[Specific conductance of 0.1 M KCl
= 1.29 × 10–2 ohm–1 cm–1]
Long Answer Type Questions (LA)
37. E°cell for the given redox reaction is 2.71 V.
Mg(s) + Cu2+ (0.01 M) → Mg2+ (0.001 M) + Cu(s)
Calculate E cell for the reaction. Write the
direction of flow of current when an external
opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V
38. (a) Calculate standard emf of the cell in
which following reaction takes place at 25°C.
Cu2+ + 2Cl–
Cu(s) + Cl2(g)
E°Cl2/Cl– = +1.36 V, E°Cu2+/Cu = + 0.34 V
Also calculate standard free energy change and
equilibrium constant of the reaction.
(b) The emf of a galvanic cell composed of two
hydrogen electrode is 0.16 volt at 25°C. Calculate
pH of the anode solution if the cathode is in a
solution with pH = 1.
CBSE Board Term-II Chemistry Class-12
16
39. (a) Calculate the cell emf and DG° for the
cell reaction at 25°C for the cell :
Zn(s) | Zn2+ (0.0004 M) || Cd2+ (0.2 M) | Cd(s)
E° values at 25°C : Zn2+/ Zn = – 0.763 V;
Cd2+/Cd = – 0.403 V; F = 96500 C mol–1;
R = 8.314 J K–1 mol–1.
(b) If E° for copper electrode is 0.34 V, how will
you calculate its emf value when the solution in
contact with it is 0.1 M in copper ions? How does emf
for copper electrode change when concentration
of Cu2+ ions in the solution is decreased?
OBJECTIVE TYPE QUESTIONS
1. (b) : Higher the oxidation potential, more easily it is
oxidised and hence greater is the reducing power. Hence,
increasing order of reducing power is Ag < Cr < Mg < K.
2.
(b) : E°cell= E°Ag2O/Ag – E°Zn2+/Zn
= 0.344 – (–0.76) = 1.104 V
DG° = –nFE°cell = –2 × 96500 × 1.104
= –2.13 × 105 J mol–1
3.
(a) : E
(M n + / M )
= E°
(M n + / M )
−
RT
[M ]
ln n +
nF [M ]
Since concentration of solid is taken as unity,
RT
1
E n+
= E ° n+ −
ln
(M / M )
(M /M ) nF [M n + ]
4. (c) : Lm° CaCl2 = l°Ca2+ + 2l°Cl–
= 119.0 + 2 × 76.3 = 271.6 S cm2 mol–1
Lm° CH3COONa = l°CH3COO– + l°Na+
= 40.9 + 50.1 = 91 S cm2 mol–1
Lm° NaCl= l°Na+ + l°Cl–
= 50.1 + 76.3 = 126.4 S cm2 mol–1
5. (d) : The electronic conductance depends on all these
factors.
6. (a) : If an external potential of 1.1 V is applied to the
cell, the reaction stops and no current flows through the cell.
Any further increase in external potential again starts the
reaction but in opposite direction and the cell functions as
an electrolytic cell.
7. (a) : Daniell cell converts the chemical energy liberated
during the redox reaction to electrical energy and has an
electrode potential of 1.1 V.
8.
(c) : At anode : Fe → Fe2+(0.001 M) + 2e–
40. (a) Equivalent conductance of a 0.0128 N
solution of acetic acid is 1.4 mho cm 2 eq –1
and conductance at infinite dilution is
391 mho cm2 eq–1. Calculate degree of dissociation
and dissociation constant of acetic acid.
(b) The equivalent conductances of sodium
acetate, sodium chloride and hydrochloric
acid are 83, 127 and 426 mho cm 2 eq –1 at
250°C respectively. Calculate the equivalent
conductance of acetic acid solution.
At cathode : 2H+ (1 M) + 2e– → H2 (1 bar)
Net reaction : Fe + 2H+ → Fe2+ + H2
Nernst equation for the given cell,
° −
E cell = E cell
9.
0.0591 [Fe2+ ][H2 ]
log
2
[Fe][H+ ]2
(a) : At cathode : Ag+(aq) + e– → Ag(s)
1
O2(g) + H2O(l) + 2e–
2
10. (c) : According to convention, the standard hydrogen
electrode is assigned a zero potential at all temperatures.
–
At anode : 2OH (aq) →
11. (a) : E cell = E °cell − 0.0591log [Ion]RHS
n
[Ion]LHS
12. (a) : Lower the reduction potential, more is the reducing
power. Thus, the order is
Br– < Fe2+ < Al.
nFE °cell
2.303RT
0
.
0591
[a ]
° −
(b) E cell = E cell
log 1
2
[a2 ]
(d) Expression is valid at 298 K, not at 273 K.
13. (c) : (a) log K =
14. (b) : κ = G ×
l
A
1
l
= κ × = κ × R = 0.0212 × 55 = 1.166 cm−1
A
G
15. (a) : Zn + 2Ag+ → Zn2+ + 2Ag can be represented as
2+
+
Zn(s) | Zn (aq) | | Ag (aq) | Ag(s)
16. (a) : L°Al2(SO4)3 = 2l°Al3+ + 3l°SO42–
λ Al3+
=
Λ Al2 (SO4 )3 − 3λ SO2−
4
2
858 − (3 × 160)
=
= 189 S cm2 mol−1
2
17
Electrochemistry
28. (d) : For strong electrolytes, the plot between Lm and
C1/2 is a straight line.
For weak electrolytes, Lm increases steeply on dilution,
especially near low concentrations.
17. (a) : The cell reaction will be
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
E °cell = E °cathode – E °anode
= 0.80 – (–0.25) = +1.05 V
29. (a) : l°HCOOH = l°H+ + l°HCOO–
DG° = –nFE°cell
= 349.6 + 54.6 = 404.2 S cm2 mol–1
As E°cell = +ve,
Λm
DG° = –ve, hence reaction is feasible.
α=
18. (a) : Mg2+(aq) + 2e– → Mg(s); E° = –2.36 V
2H+ + 2e– → H2(g); E° = 0.00 V
Thus, oxidation takes place at magnesium electrode and
reduction at hydrogen electrode.
Ka =
19. (b) : L°m NaBr = L°m NaCl + L°m KBr – L°m KCl
20. (a) : The ion which requires less energy is liberated first.
°
21. (a) : log K c = nE cell
0.0591
For the given reaction, n = 1
1 × 0.36
log K c =
= 6.09
0.0591
Kc = antilog 6.09 = 1.2 × 106
22. (c) : Iron coated with zinc does not get rusted even if
cracks appear on the surface because Zn will take part in
redox reaction not Fe as Zn is more reactive than Fe. If iron is
coated with tin and cracks appear on the surface, Fe will take
part in redox reaction because Sn is less reactive than Fe.
23. (c) : In reactivity series,
Mg > Al > Zn > Fe > Cu
=
46.1
= 0.114 × 100 = 11.4%
404.2
C α2 0.025 × (0.114)2
=
1− α
1 − 0.114
0.025 × 0.114 × 0.114
= 3.67 × 10–4 mol L–1
0.886
0.059
log[Mg2+].
30. (b) : E = E° +
2
Hence, plot of E vs log [Mg2+] will be linear with positive
slope and intercept = E°.
=
31. (c) : Cu2+ + e– → Cu+ ; E°1 = 0.15 V, DG°1, n1 = 1
E°2 = 0.34 V, DG°2, n2 = 2
Cu2+ + 2e– → Cu ;
+
–
E°3 = ?, DG°3, n3 = 1
Cu + e → Cu ;
DG°3 = DG°2 – DG°1
–n3 FE°3 = –n2FE°2 + n1FE°1
–E°3 = –2 × 0.34 + 1 × 0.15
E°3 = 0.68 – 0.15 = +0.53 V
32. (d) : In fig. (Y), zinc is deposited at the zinc electrode
and copper dissolves at copper electrode.
33. (d)
Reactivity decreases
Hence, Mg can displace Al, Al can displace Zn and so on.
24. (a) : Nickel-Cadmium battery
Anode - Cd ; Cathode - NiO2 ; Electrolyte - KOH
–
At anode : Cd(s) + 2OH(aq) → Cd(OH)2(s) + 2e–
–
At cathode : NiO2(s) + 2H2O(l) + 2e– → Ni(OH)2(s) + 2OH(aq)
Cd(s) + NiO2(s) + 2H2O(l) → Cd(OH)2(s) + Ni(OH)2(s)
25. (c) : Λ m =
Λ °m
κ × 1000 1.52 × 10 −2 × 1000
=
M
0.15
= 101 W–1 cm2 mol–1
26. (b) : Higher the reduction potential, stronger is the
oxidising agent.
27. (a) : During the electrolysis of dilute sulphuric acid, the
following process is possible at anode:
+
2H2O(l) → O2(g) + 4H (aq) + 4e–
34. (b) : Electrolyte X is strong electrolyte as on dilution
the number of ions remain same, only interionic attraction
decreases and hence not much increase in Lm as seen. While
Lm for a weak electrolyte increases significantly.
1 l
1 2
35. (b) : κ = × = ×
R A 40 5
1000 1 2 1000
Λ eq = κ ×
= × ×
= 20 ohm−1 cm2 eq −1
N
40 5 0.5
36. (c) : At cathode : Ag+ + e– → Ag ; E° = +0.80 V
At anode : Pb → Pb2+ + 2e–; E° = +0.13 V
E°cell = E°cathode – E°anode = 0.80 – 0.13 = 0.67 V
Hence, the reaction will be
2Ag+ + Pb → Pb2+ + 2Ag
37. (c) : Specific conductance = S m–1
38. (a) : Conductivity decreases because number of ions per
unit volume decreases.
κ × 1000
39. (c) : Solubility =
Λ °m
CBSE Board Term-II Chemistry Class-12
18
1.821 × 10 −5 × 1000
=13.97 × 10–5 mol L–1
130.26
=13.97 × 10–5 × 143.5
(AgCl = 108 + 35.5 = 143.5)
–2
–1
= 2.004 × 10 g L
=
–
40. (b) : Pt + Cl2 → Pt2+ + 2Cl ; E°cell = 0.15 V
+ Pt2+ + 2e– → Pt ; E° = 1.20 V
Cl2 + 2e– → 2Cl– ; E° = 1.35 V
42. (b) : For 2Ag+ + H2 (1 atm) → 2Ag + 2H+(1 M)
43. (c) : 2I → I2 + 2e
–
Br2 + 2e → 2Br
+2
0.0591
[H ]
log + 2
2
[Ag ]
–
–
(Oxidation)
–
(Reduction)
2I– + Br2 → I2 + 2Br–
is net cell reaction
44. (d) : E°cell = E°cathode – E°anode
It will remain unchanged.
45. (c) : L°Ba(OH)2 = L°Ba2+ + 2L°OH–
L°BaCl2 = L°Ba2+ + 2L°Cl–
L°NH4Cl = L°NH4+ + L°Cl–
After substituting the above in
Λ°NH4OH =
Λ°Ba(OH)2 + 2Λ°NH4Cl − Λ°BaCl2
we get, L°NH4OH = L°NH4+
2
+ L°OH–
46. (a) : L∞m CH3COOH = 390.71 ohm–1 cm2 equiv–1
Lcm CH3COOH = 14.3 ohm–1 cm2 equiv–1
Degree of dissociation (a)
=
Λ cm
∞
Λm
=
14.3
= 0.0366 i .e. 3.66%
390.71
47. (c) : E°cell should be positive for a spontaneous reaction
as DG° = –nFE°cell.
48. (d) : Possible reactions at the anode during electrolysis
of CuSO4 solution are
2SO2− → S O2− + 2e − ; E  = +2.07 V
4 ( aq )
2 8( aq )
red
 = +1.23 V
+ 4e ; E red
Comparing reduction potentials values,
H 2O molecules will be oxidised at anode, given oxygen
gas.
2H2O(l ) → O2( g ) + 4H(+aq )
C
50. (c) : At cathode (Reduction) : 2Ag+(aq) + 2e– → 2Ag(s)
–
At anode (oxidation) : Mg(s) → Mg2+
(aq) + 2e
Mg2+ 
Mg2+ 
RT
0.059

ln 
E
log
=
−
cell
2
2
2F
2
 Ag+ 
 Ag+ 




= E cathode
− E anode
= E Ag
− E Mg
+
2+
/Ag
/Mg

−
E cell = E cell
41. (d) : Due to reduction of NO–3 in preference to H+ ion.
H+ ion is not reduced to give H2 gas.
E cell = E °cell −
Strong electrolyte
Intermediate electrolyte
Weak electrolyte
Λm
−
49. (b) : Salts that have polyvalent cations or anions are
intermediate electrolytes (AlPO4).

E cell

E cell = E cell
−
0.059
(0.130)
log
2
(0.0001)2
51. (c) : A more negative E° value means that the redox
couple is stronger reducing agent than the other one.
52. (c) : For the cell given in option (c),

E cell = E cell
−
[Zn2+ ]
2.303RT
log
2F
[Cu2+ ]
53. (a) 54. (d)
55. (d) : Weak electrolytes do not dissociate in aqueous
solution.
56. (a) : According to Kohlrausch’s law
Λ°(MgI2) = Λ°[(CH3COO)2Mg] + 2Λ°(NaI) – 2Λ°(CH3COONa)
= 18.78 + 2(12.69) – 2(9.10) = 25.96 S cm2 mol–1
57. (a)
+0.059 0.001
log 2+
58. (b) : 0.059 =
2
[M ]
0.001
log 2+ = 2 or [M 2+ ] = 10 −5
[M ]
Let solubility of salt be s mol/litre.
M2+ + 2X –
Thus, MX2
s
0
\
2s
3
Ksp = 4s = 4 × (10–5)3 = 4 × 10–15
59. (d) : DG = –nFE = –2 × 96500 × 0.059
= –11387 J mol–1 = –11.4 kJ mol–1
0.059

60. (a) : E cell
=
log K C
1
E°cell= E°Ce4+/Ce3+ – E °Fe3+/Fe2+
= 1.44 – 1.68 = 0.76 V
0.76
log10 K c =
= 12.88
0.059
Kc = 7.6 × 1012

61. (b) : E cell = E cell
−
[Ag+ ]
0.059
log +
1
[Ag ]Satd. Ag CrO
2
4
19
Electrochemistry
0.164 =
0.059
0.1
log +
1
[Ag ]Satd. Ag CrO
2
+
4
–4
[Ag ]Satd. Ag2CrO4 = 1.66 × 10 M
So,
1.66 × 10
[CrO24− ] =
−4
2
Ksp (Ag2CrO4) = [Ag+]2 [CrO42–]
 1.66 × 10 −4 
= (1.66 × 10 −4 )2 

2


= 2.287 × 10–12 mol3 L–3
62. (a)
C 
63. (a) : log  1  < 0 for spontaneity.
C
\ C1 < C2  2 
64. (a)
65. (b)
66. (a)
67. (d) : Nernst equation is measured at 298 K. At STP
conditions, temperature to be 273 K.
68. (b) : M
(s) → M +(aq) (0.05 M) + e–
M +(aq) (1.0 M) + e– → M(s)
M +(aq) (1.0 M) → M +(aq) (0.05 M)
0.059 0.05
For concentration cell, E cell = −
log
1
1
0.059
−2
E cell = −
log(5 × 10 )
1
0.059
E cell = −
[( −2) + log5] = –0.059(–2 + 0.698)
1
= –0.059(–1.302) = 0.0768
DG = –nFEcell
If Ecell is positive, DG is negative.
E
log 0.05
69. (c) : 1 =
E 2 log 0.0025
E1 log5 × 10 −2
=
E 2 log25 × 10 −4
E1 = 0.0768
0.0168 −1.3 1
=
or E 2 = 154 mV
=
−2.6 2
E2
 nE ° 
70. (c) : K = antilog 
 0.0591
For feasible cell, E° is positive, hence from the above equation,
K > 1 for a feasible cell reaction.
71. (b)
73. (b) : nNaCl =
\
nCl2 = 1 mol
72. (a)
4 × 500
= 2 mol
1000
74. (b) : nNa deposited = 2 mol
\ nNa – Hg formed = 2 mol
\ Mass of amalgam formed = 2 × 223 = 446 g
75. (a) : H2 gas at cathode.
76. (c) : The conductivity of solutions of different electrolytes
in the same solvent and at a given temperature is different.
Effect of concentration on electrode potential is found by
Nernst equation.
77. (d) : Standard reduction potential of an electrode has a
fixed value.
78. (a) : According to Kohlrausch law, “limiting molar
conductivity of an electrolyte can be represented as the sum
of the individual contributions of the anion and cation of the
electrolyte.”
79. (a) : When electrolytes are dissolved in solvent they
furnish their own ions in the solution hence, its conductivity
increases.
80. (c) : Cu2+ ions are deposited as Cu.
81. (d) : The electrical resistance of any object is directly
proportional to its length, l, and inversely proportional to
its area of cross-section, A. So, it increases with increase in
length of object and decreases with increase in area of crosssection of object.
82. (a) : The substances which do not allow the flow of
electric current through them are termed as insulators.
83. (b) : In the plot of molar conductivity versus
concentration, the extrapolation to zero concentration is not
possible. Since the graph is not linear.
84. (c) : The cell constant depends upon the distance
between the electrodes and their area of cross section.
85. (a)
∞
∞
86. (c) : Λ ∞
AB = λ A + + λ B −
Kohlrausch’s law is applicable for weak electrolytes.
87. (c) : At higher concentration, mobility of ions decreases.
Hence, conductance decreases.
88. (c) : Total number of ions will increase slightly on
dilution (not considerably).
89. (b)
90. (a)
SUBJECTIVE TYPE QUESTIONS
κ × 1000
in CGS units
M
κ × 10 −3
in SI units
Lm =
M
where, k is the conductivity, M is the molar concentration and
Lm is molar conductivity.
Λm =
1.
CBSE Board Term-II Chemistry Class-12
20
2.In weak electrolyte, the conductivity of the solution
increases very slowly with dilution of solution and goes on
increasing up to infinity. Therefore, it cannot be measured
experimentally.
Λm
1/2
C
3. The species that get reduced at cathode is the one
having higher value of standard reduction potential. Hence,
the reaction that will occur at cathode is
Ag+(aq) + e– → Ag(s).
4. Molar conductivity increases with decrease in
concentration. This is because the total volume, V, of solution
containing one mole of electrolyte also increases. It has been
found that decrease in K on dilution of a solution is more than
compensated by increase in its volume.
5. The reaction at anode with lower value of E° is preferred
i.e., O 2 gas should be liberated but on account of over
potential of oxygen reaction at anode, preferred reaction is
1
Cl(−aq ) → Cl2( g ) + e −
2
i.e., Cl 2 gas is liberated at anode in the electrolysis of
aq. NaCl.
6. The salt bridge allows the movement of ions from one
solution to the other without mixing of the two solutions.
Moreover, it helps to maintain the electrical neutrality of the
solutions in the two half cells.
7.
It is used for the inflow and outflow of electrons.
8. HCl will have greater value of Lm because H+ ions are
smaller than Na+ ions and hence H+ ions have greater ionic
mobility than Na+ ions.
9. Kohlrausch’s law of independent migration
of ions : It states that limiting molar conductivity of an
electrolyte can be represented as the sum of the individual
contributions of the anion and cation of the electrolyte.
Kohlrausch’s law helps in the calculation of degree of
dissociation of weak electrolytes like acetic acid.
10. The species that get reduced at cathode is the one which
have higher value of standard reduction potential. Hence, the
reaction that will occur at cathode is
–
Cu2+
(aq) + 2e → Cu(s)
11. The fraction of the total number of molecules present in
solution as ions is known as degree of dissociation.
Molar conductivity (lm) = al°m
where l°m is the molar conductivity at infinite dilution.
12. (i) Because fluorine has highest reduction potential.
(ii) Lithium metal is strongest reducing agent because Li
has lowest reduction potential i.e., E°Li+/Li = –3.05 V
13. Here n = 2, E°cell = 1.1 V, F = 96500 C mol–1
DrG° = –nFE°cell
DrG° =
– 2 × 1.1 × 96500 = – 212300 J mol–1
= – 212.3 kJ mol–1
14. The substance which conducts electricity by ions present
in solution is called electrolytic conductor e.g., NaCl solution.
Substances which conduct electricity in solid state are called
electronic conductors. These are made up of metals. e.g.,
Cu, Zn, Al. (Electrolytes are electrolytic conductors while
electrodes are electronic conductors).
15. The device which converts the chemical energy liberated
during the chemical reaction to electrical energy is called
electrochemical cell.
If external potential applied becomes greater than E°cell of
electrochemical cell then the cell behaves as an electrolytic
cell and the direction of flow of current is reversed.
16. With time, concentrations of the electrolytic solutions
change. Hence, their electrode potentials change when the
electrode potentials of the two half-cells become equal, the
cell stops working.
17. (i) Electrolyte A is a strong electrolyte while electrolyte
B is a weak electrolyte.
(ii) For electrolyte A, the plot becomes linear near high
dilution and thus can be extrapolated to zero concentration
to get the molar conductivity at infinite dilution.
For weak electrolyte B, Lm increases steeply on dilution and
extrapolation to zero concentration is not possible. Hence,
molar conductivity at infinite dilution cannot be determined.
4+
–
18. At anode : Sn2+
(aq) → Sn (aq) + 2e ] × 5
–
+
–
At cathode : MnO4(aq) + 8H (aq) + 5e →
Mn2+
(aq) + 4H2O(l)] × 2
Net cell reaction :
+
2+
4+
2MnO–4(aq) + 5Sn2+
(aq) + 16H (aq) → 2Mn (aq) + 5Sn (aq) + 8H2O(l)
E°cell = E°cathode – E°anode = 1.51 V – 0.15 V = 1.36 V
Since, cell potential is positive therefore the reaction is
product favoured.
19. C = 0.001 M, k = 4 × 10–5 S cm–1,
L∞m = 390 S cm2/mol
κ ×1000
Lmc =
C
Substituting the values,
Lmc =
4 × 10 −5 × 1000
= 40 S cm2/mol
0.001
21
Electrochemistry
a=
Λ cm
∞
Λm
=
40
= 0.10256 ≈ 0.103
390
CH3COO– + H+
CH3COOH
c
c (1 – a)
0
ca
0
ca
Ka =
[CH3COO− ] [H+ ] cα ⋅ cα cα2
=
=
c (1 − α ) 1 − α
[CH3COOH]
Ka =
0.001(0.103)2 1.061 × 10 −5
–5
=
(1 − 0.103)
0.897 = 1.18 × 10
B2+
(aq)
20. A(s) +
Here, n = 2
using formula,
0.059
log K c
E°cell =
n
A2+
(aq)
+ B(s)
21. The reducing power increases with decreasing
value of electrode potential. Hence, the order is
Ag < Cu < Fe < Cr < Mg < K.
22. (i) Ions with higher reduction potentials are strong
oxidising agents while lower reduction potentials are strong
reducing agents.
(ii) The electrode with higher electrode potential (E°) acts
as cathode while with lower electrode potential will act as
anode.
(iii) Predicting the feasibility of redox reaction.
(iv) Predicting the capability of metal to evolve H2 gas from
acid.
–
+ 3e ] × 2
23. At anode
At cathode : Ni + 2e → Ni(s)] × 3
3+
Cell reaction : 2Al(s) + 3Ni2+
(aq) → 2Al (aq) + 3Ni(s)
Applying Nernst equation to the above cell reaction,
0.0591 [Al3+ ]2
−
log 2+ 3
2×3
[Ni ]



Now, E cell
= E Ni
− E Al
2+
3+
/Ni
/Al
= – 0.25 – (–1.66) = 1.41 V
\
E cell = 1.41 −
0.0591 (10 −3 )2
log
6
(0.5)3
0.0591
log (8 × 10 −6 )
6
0.0591
= 1.41−
( −5.09)
6
= 1.41 + 0.050 = 1.46 V
= 1.41 −
−1
cell constant 129 m
= 0.248 S m–1
=
520 Ω
R
Concentration, C = 0.02 mol L–1
= 1000 × 0.02 mol m–3 = 20 mol m–3
0.059
log10
2
E°cell = 0.0295 V

E cell = E cell
25. Resistance of 0.1 M KCl solution R = 100 W
Conductivity k = 1.29 S m–1
Cell constant G* = k × R = 1.29 × 100 = 129 m–1
Resistance of 0.02 M KCl solution, R = 520 W
Conductivity, k =
E°cell =
: Al(s) → Al3+
(aq)
2+
–
24. (i) The cell reactions are :
–
Zn(s) → Zn2+
(aq) + 2e (Anode)
–
Cu2+
(aq) + 2e → Cu(s) (Cathode)
Net reaction :
2+
Zn(s) + Cu2+
(aq) → Zn (aq) + Cu(s)
(ii) E°cell = E°cathode – E°anode = 0.34 V – (– 0.76 V)
= 1.10 V
(iii) Copper electrode will be positive on which reduction
takes place.
−1
κ 0.248 S m
=
C 20 mol m−3
= 0.0124 S m2 mol–1
Molar conductivity, Lm =
26. For half cell reaction,
+
–
3+
Cr2O2–
7(aq) + 14H (aq) + 6e → 2Cr (aq) + 7H2O(l)

−
E cell = E cell
0.0591
[Cr3+ ]2
log
n
[Cr2O27− ][H+ ]14
Given, E°cell = 1.33 V, n = 6, [Cr3+] = 0.2 M
+
–4
[Cr2O2–
7 ] = 0.1 M, [H ] = 1 × 10 M
E cell = 1.33 −
= 1.33 −
0.0591
(0.20)2
log
6
(0.1) (10 −4 )14
0.0591
log (4 × 1055 )
6
= 1.33 −
0.0591
[log 4 + log 1055 ]
6
= 1.33 −
0.0591
[log 4 + 55 log 10]
6
0.0591
[0.602 + 55]
6
= 1.33 – 0.548 = 0.782 V
= 1.33 −
3
27.Al2O3 (2Al3++ 3O2–) → 2Al + O2, n = 6e–
2
2
4
2
\
Al2O3 → Al + O2, n = × 6e– = 4e–
3
3
3
DG = 960 × 1000 = 960000 J
Now, DG = –nFEcell
∆G −960000
E°cell = –
= –2.487 V
=
nF 4 × 96500
CBSE Board Term-II Chemistry Class-12
22
Minimum potential difference needed to reduce Al 2O 3 is
–2.487 V.
28.At anode : Ni → Ni2+ + 2e–
At cathode : [Ag+ + e– → Ag] × 2
Cell reaction : Ni + 2Ag+ → Ni2+ + 2Ag
E°cell = E°cathode – E°anode
= E°Ag+/Ag – E°Ni2+/Ni = 0.80 V – (– 0.25) V
E°cell = 1.05 V
0.0591

E cell
=
log K c
n
E  × n 1.05 × 2
log K c = cell
=
0.0591 0.0591
logKc = 35.53
Kc = antilog 35.53 = 3.38 × 1035
29. E°cell = E°cathode – E°anode = –0.403 – (–0.763) = 0.36 V
DrG° = –nFE°cell = –2 × 96500 × 0.36
= –69480 J = –69.48 kJ
Using formula, log K c =

nE cell
2 × 0.36
=
= 12.20
0.059
0.059
Kc = antilog 12.20 = 1.58 × 1012
Λ eq
30.(a) Degree of dissociation, a = ∞
Λ eq
15.8
\ a=
= 0.04514
350
(b) For monobasic acid, HA
H+ + A–
Cα 2
= Cα2
(1 − α )
As a < < < 1 hence (1 – a) ≈ 1
\ K = 0.05 × (0.04514)2 ⇒ K = 1.019 × 10–4
K=
31.Given : Diameter = 1 cm, length = 50 cm
R = 5.5 × 103 ohm, M = 0.05 M
r = ? k = ? Lm = ?
Area of the column,
32. The reaction is
Zn(s) + 2Ag+(aq) → Zn2+
(aq) + 2Ag(s)
Cell can be represented as
+
Zn | Zn2+
(aq) || Ag (aq) | Ag
(i) The zinc electrode is negatively charged (anode) as it
pushes the electrons into the external circuit.
(ii) Ions are the current carriers within the cell.
(iii) The reactions occurring at two electrodes are :
–
At zinc electrode (anode) : Zn(s) → Zn2+
(aq) + 2e
At silver electrode (cathode) : Ag+(aq) + e– → Ag(s)
33.A chemical cell is a galvanic cell in which electrical
energy produced is due to chemical changes occurring within
the cell and no transfer of matter takes place. It involves the
use of two different electrode dipped in solutions of different
electrolytes.
A concentration cell is a galvanic cell in which electrical
energy is produced due to physical change involving transfer
of matter from one part of the cell to the other. It involves
the use of the same electrodes dipped in solutions of the
same electrolyte with different concentrations (or electrodes
of different concentration dipped in the same solution of the
electrolyte).
34.The given cell may be represented as
Cu(s) |Cu2+ (0.10 M)|| Ag+ (C)| Ag(s)
E°cell = E°cathode – E°anode = 0.80 V – 0.34 V = 0.46 V
Ecell = E°cell –
2
1
3.14 2
a = πr 2 = 3.14 ×  cm =
cm
2 
4
Resistivity,
a
3.14 cm2
ρ = R ⋅ = 5.5 × 103 ohm ×
= 86.35 ohm cm
l
4 × 50 cm
1
Again, conductivity, κ =
ρ
1
−2
=
= 1.158 × 10 ohm−1 cm−1
86.35
103
and molar conductivity, Λ m = κ ⋅
M
= 1.158 × 10 −2 ohm−1 cm−1 ×
–1
2
= 231.6 ohm cm mol
–1
103
5 × 10 −2
or
0.0591 [Cu2+ ]
log + 2
2
[Ag ]
0.422 V = 0.46 V –
0.0591
0.1
log + 2
2
[Ag ]
– 0.038 V = – 0.0295 log
or
or
\
0.1
[Ag+ ]2
− 0.038
0.1
=
= 1.288
+ 2
− 0.0295
[Ag ]
0.1
= antilog 1.288 = 19.41
[Ag+ ]2
0.1
[Ag+]2 =
= 5.1519 × 10–3
19.41
log
[Ag+] = 7.1 × 10–2 M
35. Given : V = 100 cm3, M = 0.025 M, R = 520 ohm
G° = 153.7 m–1 = 1.537 cm–1, Lm = ?
23
Electrochemistry
0.16 = 0.0591 [log [H+]c – log [H+]a]
0.16 = 0.0591 [pHa – pHc]
0.16 = 0.0591 [pHa – 1]
0.16
= 2.70
or pHa − 1 =
0.0591
or pHa = 2.70 + 1 = 3.70
39. (a) E°cell = E°cathode – E°anode = – 0.403 – (– 0.763)
= 0.36 V
The net cell reaction is
2+
Zn(s) + Cd2+
(aq) → Zn (aq) + Cd(s)
Here, value of n = 2
or
or
1
1
= 1.537 cm−1 ×
R
520 ohm
= 2.95 × 10–3 ohm–1cm–1
κ =G×
Again, Λ m =
=
κ × 103
M
2.95 × 10 −3 ohm−1 cm−1 × 103
0.025 mol cm−3
Lm = 118.0 ohm–1 cm2 mol–1
36. k = 1.29 × 10–2 ohm–1 cm–1
1
k=
× Cell constant
R
⇒ Cell constant = k × R
= 1.29 S m–1 × 85 W = 109.65 m–1
For second solution,
1
k = 1 × Cell constant =
× 109.65 m–1
96
Ω
R
= 1.142 W–1m–1
1.142 Ω −1m−1 × 1000 cm3
Lm = κ × 1000 =
M
0.052
Λm =
1.142 Ω −1cm−1 × 10 −2 × 1000 cm3
0.052 mol
= 219.62 S cm2 mol–1
2+
0.0591 [Mg ]
log 2+
n
[Cu ]
0.0591 0.001
= 2.71 −
log
= 2.73955 V
2
0.01
(i) If external opposing potential is less than 2.71 V
then current will flow from Cu to Mg.
(ii) If external opposing potential is greater than 2.71 V
then current will flow in opposite direction i.e. from Mg
to Cu.
37. E cell = E °cell −
38. (a) The given cell may be represented as
Cu(s) | Cu2+ || Cl2 | Cl–
(i)

E cell
= E c
− E a
0.0591 [Zn2+ ]
log 2+
2
[Cd ]
0.0591 0.0004
= 0.36 −
log
2
0.2
0.0591
= 0.36 −
( −2.69) = 0.36 + 0.08 = 0.44 V
2
\ DG = – nFEcell = – 2 × 96500 × 0.44
= – 84920 J/mol
–
+
2e
→
Cu(s)
(b) Cu2+
(aq)
0.059
[Cu]

E Cu2+ /Cu = E Cu
−
log 2+
2+
/Cu
2
[Cu ]
0.059
0.059
1
= 0.34 −
log
= 0.34 −
log 10
2
0.1
2
0.059
= 0.34 −
× (1) = 0.34 – 0.0295 = 0.3105 V
2
When the concentration of Cu 2+ ions is decreased, the
electrode potential for copper decreases.

−
E cell = E cell
40.(a) Given : Leq = 1.4 mho cm2 eq–1,
L∞eq = 391 mho cm2 eq–1, a = ?, Ka = ?
Using formula, α =
Ka =
∆r G ° = − nFE ° = − 2 × 96500 C × 1.02 V = 196.86 kJ
0.0591
log K
(iii) E°Cell =
n
2 × 1.02
K = antilog
= antilog (34.51)
0.0591
K = 3.236 × 1034
(b) The given cell may be represented as
Pt, H2 (1 atm) | H+ (pH = ?) || H+ (pH = 1) | H2 (1 atm)
[H+ ]c
Using formula, E cell = 0.0591 log
1
[H+ ]a
=
1.4 mho cm2 eq−1
∞
Λ eq
391 mho cm2 eq−1
= 0.00358
= ( +1.36 V) − ( + 0.34 V) = 1.02 V
(ii)
Λ eq
(0.00358)2 × 0.0128
α2C
=
1− α
1 − 0.00358
1.64 × 10 −7
= 1.64 × 10–7
0.99642
(b) Given : Λ°eq (CH3COONa) = 83 mho cm2 eq–1
Λ°eq (NaCl) = 127 mho cm2 eq–1
Λ°eq (HCl) = 426 mho cm2 eq–1
Λ°eq (CH3COOH) = ?
Using Kohlrausch law of independent migration of ions
Λ°eq (CH3COOH) = Λ°eq (CH3COONa) + Λ°eq (HCl) – Λ°eq (NaCl)
83 + 426 – 127
or Λ°eq (CH3COOH) =
= 382 mho cm2 eq–1

=
CHAPTER
2
Chemical Kinetics
Recap Notes
Chemical kinetics : It is the branch of
chemistry which deals with the study of
reaction rates and their mechanisms.
Rate of a reaction : The rate of a reaction
can be defined as the change in concentration
of a reactant or a product in unit time. For
the reaction,
∆[ R ]
∆[ P ]
P, Rate = –
or +
R
∆t
∆t
Units of rate : Concentration time–1 i.e.,
mol L–1 s–1 or atm s–1 for gaseous reactions.
Average rate of reaction : It is the average
value during a large time interval.
rav =
− ∆[ R ] + ∆[ P ]
=
∆t
∆t
Instantaneous rate of reaction : It is the
rate of a reaction at a particular instant of
time i.e., when Dt approaches zero.
rinst =
− d[ R] + d[ P ]
=
dt
dt
Factors influencing rate of a reaction :
X Concentration
:
Greater
the
concentrations of the reactants, faster is
the rate of reaction.
X Physical state of reactants : Reactions
involving gaseous reactants are faster
than reactions containing solid and liquid
reactants.
X Temperature : The rate of reaction
increases with increase of temperature.
For most of the reactions, rate of reaction
becomes almost double with 10° C rise of
temperature.
X
X
X
X
Presence of catalyst : A catalyst
generally increases the speed of a reaction.
Surface area of reactants : For a
reaction involving a solid reactant or
catalyst, the greater is the surface area,
the faster is the reaction.
Presence of light : Photochemical
reactions take place in the presence of
light only.
Activation energy : Lower the activation
energy faster is the reaction.
Rate law and rate constant : The equation
that correlates the rate of reaction with
concentration of reactants is known as rate
law.
X For a simple reaction, A + B → C + D
Rate = k[A][B] where k is the rate constant
which is equal to the rate of reaction when
concentration of each of the reactant is
unity.
X For a simple reaction, aA + bB → cC + dD
Rate = k[A]x[B]y and order of reaction =
x+y
Difference between order and molecularity
Order and molecularity :
Molecularity of
reaction
It is the total
number of species
taking part in a
chemical reaction.
It is a theoretical
concept.
Order of
reaction
It is the sum of
the powers of the
concentration terms of
reacting species in the
rate law equation.
It is an experimental
quantity.
25
Chemical Kinetics
It is derived from
the mechanism of
reaction.
It can neither be
zero nor fractional.
It is always a whole
number.
It is derived from the
rate expression.
It is applicable
It is applicable to
only to elementary
elementary as well as
reactions. The
complex reactions.
overall molecularity
of a complex reaction
has no significance.
It may be zero,
fractional or an
integer (may range
from 0 to 3).
Half life of reaction : The time in which the concentration of a reactant is reduced to one
half of its initial concentration is called half life of the reaction.
1
t1 / 2 ∝
; where n is the order of the reaction.
n −1
a
Rate law, integrated rate law, half-life, units of rate constant and graph for the reactions
of different orders :
Order
Rate law
Integrated rate
law
Units of rate
constant
Half-life
Graph
mol L–1 s–1
[A] vs t; slope = –k
ln[A]t = –kt + ln [A]0 t1/2 = 0.693/k
s–1
ln[A] vs t; slope = –k
Rate = k[A]2
1/[A]t = kt + 1/[A]0
L mol–1 s–1
1/[A] vs t; slope = k
2
Rate = k[A] [B]
kt =
n
Rate = k[A]n
( n − 1) kt =
0
Rate = k[A]0
[A]t = –kt + [A]0
1
Rate = k[A]1
2
t1/2 = [A]0/2k
t1/2 = 1/k [A]0
[ B ]0 [ A ]
1
ln
[ A ]0 − [ B ]0
[ A ]0 [ B ]
1
n −1
[ A]
−
1
n −1
[ A0 ]
t1 / 2 =
Pseudo first order reactions : Those
reactions which are not truly of the first
order but under certain conditions become
reactions of the first order are called pseudo
first order reactions. e.g.,
X
Acid hydrolysis of ethyl acetate :
CH3COOC2H5 + H2O
H+
L mol–1 s–1
–
CH3COOH
+ C2H5OH
2n −1 − 1
n −1
k ( n − 1) [ A ]0
(mol L–1 )1–n s–1
–
1
[ A ]n−1
vs t; slope = k
Rate = k′[CH3COOC2H5][H2O]
X
= k[CH3COOC2H5] where, k = k′[H2O]
Acid catalysed inversion of cane
sugar :
H+
C12H22O11 + H2O
C6H12O6
Cane sugar Glucose
+ C6H12O6
Fructose
Rate = k[C12H22O11]
Practice Time
OBJECTIVE TYPE QUESTIONS
1. The decomposition of dimethyl ether is a
fractional order reaction. The rate of reaction is
given by rate = k(pCH3COCH3)3/2. If the pressure
is measured in bar and time in minutes, then
what are the units of rate and rate constant
respectively?
(a) bar min–1, bar2 min–1
(b) bar min–1, bar–1/2 min–1
(c) bar–1/2 min–1, bar2 min–1
(d) bar min–1, bar1/2 min–1
2. The rate of a gaseous reaction is given by the
expression k[A]2[B]3. The volume of the reaction
vessel is reduced to one half of the initial volume.
What will be the reaction rate as compared to
the original rate a?
1
1
a
(b)
(a)
a
2
8
(c) 2a
(d) 32a
3. For a reaction P + Q → 2R + S. Which of the
following statements is incorrect?
(a) Rate of disappearance of P = Rate of
appearance of S
(b) Rate of disappearance of Q = 2 × Rate of
appearance of R
(c) Rate of disappearance of P = Rate of
disappearance of Q
1
(d) Rate of disappearance of Q =
× Rate of
2
appearance of R
4. Which of the following statements for order
of reaction is not correct?
(a) Order can be determined experimentally.
(b) Order of reaction is equal to the sum of
powers of concentration terms in rate law
expression.
(c) Order cannot be fractional.
(d) Order is not affected by stoichiometric
coefficient of the reactants.
5. The half-life of the reaction X → Y, following
first order kinetics, when the initial concentration
of A is 0.01 mol L –1 and initial rate is
0.00352 mol L–1 min–1 will be
(a) 19.69 min
(b) 1.969 min
(c) 7.75 min
(d) 77.5 min
6. The reaction 2X → Y + Z would be zero order
reaction when
(a) rate remains unchanged at any concentration
of Y and Z
(b) rate of reaction doubles if concentration of
Y is doubled.
(c) rate of reaction remains same at any
concentration of X
(d) rate of reaction is directly proportional to
square of concentration of X.
7. The rate constant of a first order reaction is
15 × 10–3 s–1. How long will 5.0 g of this reactant
take to reduce to 3.0 g?
(a) 34.07 s
(b) 7.57 s
(c) 10.10 s
(d) 15 s
8. The decomposition of a substance follows
first order kinetics. If its concentration is
reduced to 1/8 of its initial value in 12 minutes,
the rate constant of the decomposition system
is
1
 2.303
 2.303

(a) 
log  min −1 (b) 
log 8 min −1
 12
 12

8
 0.693 
min −1
(c) 
 12 
 1

(d)  log 8 min −1
 12

9. Which of the following statements is not
correct?
(a) For a zero order reaction, t1/2 is proportional
to initial concentration.
1
(b) For a reaction t1 / 2 ∝
, where n is order
n −1
a
of the reaction.
27
Chemical Kinetics
(c) The unit of rate constant for a reaction is
mol1 – n Ln – 1 s–1 where n is order of the
reaction.
(d) The unit of rate of reaction changes with
order of reaction.
10. Consider the reaction, 2N2O5
4NO2 + O2.
In the reaction NO2 is being formed at the rate
of 0.0125 mol L–1 s–1. What is the rate of reaction
at this time?
(a) 0.0018 mol L–1 s–1
(b) 0.0031 mol L–1 s–1
(c) 0.0041 mol L–1 s–1
(d) 0.050 mol L–1 s–1
11. C o n s i d e r t h e r e a c t i o n P → Q . T h e
concentration of both the reactants and the
products varies exponentially with time. Which
of the following figures correctly describes
the change in concentration of reactants and
products with time?
(b)
[P]
Concentration
Concentration
(a)
[Q ]
[Q ]
Time
[Q ]
(d)
[P]
[P]
Concentration
(c)
Concentration
Time
[P]
Time
[Q ]
Time
12. The number of molecules of the reactants
taking part in a single step of the reaction is
indicative of
(a) order of a reaction
(b) molecularity of a reaction
(c) fast step of the mechanism of a reaction
(d) half-life of the reaction.
13. For a reversible reaction, A + B
C + D,
the graph for rate of reaction with time is given
below.
Rate
(p)
(r)
(q)
Time
Mark the terms (p), (q) and (r).
(a) (p) - rate of backward reaction, (q) - rate of
forward reaction, (r) - equilibrium
(b) (p) - rate of forward reaction, (q) - rate of
backward reaction, (r) - equilibrium
(c) (p) - concentration of products, (q) concentration of reactants, (r) - rate of
reaction
(d) (p) - instantaneous rate of reaction, (q) variation of rate, (r) - average rate of reaction
14. For the reaction, 2N 2O5
4NO2 + O2,
the rate of reaction can be expressed in terms
of time and concentration by the expression:
d[ N2 O5 ]
1 d[ NO2 ] 1 d[ O2 ]
=−
=
dt
4
dt
2 dt
1 d[ N2 O5 ] 1 d[ NO2 ] d[ O2 ]
=
=
(b) Rate = −
2
4
dt
dt
dt
(a) Rate = −
1 d[ N2 O5 ]
=
4
dt
1 d[ N2 O5 ]
(d) Rate = −
=
2
dt
(c) Rate = −
1 d[ NO2 ] d[ O2 ]
=
2
dt
dt
d
d
[
]
NO
1
1 [ O2 ]
2
=
2
dt
2 dt
15. In a reaction 2HI → H2 + I2, the concentration
of HI decreases from 0.5 mol L–1 to 0.4 mol L–1 in
10 minutes. What is the rate of reaction during
this interval?
(a) 5 × 10–3 M min–1
(b) 2.5 × 10–3 M min–1
–2
–1
(c) 5 × 10 M min
(d) 2.5 × 10–2 M min–1
16.
for
(a)
(b)
(c)
(d)
The unit of rate and rate constant are same
a
zero order reaction
first order reaction
second order reaction
third order reaction.
17. In pseudo unimolecular reactions,
(a) both the reactants are present in low
concentration
(b) both the reactants are present in same
concentration
(c) one of the reactants is present in excess
(d) one of the reactants is non-reactive
18. For a reaction R → P, the concentration of
a reactant changes from 0.05 M to 0.04 M in
30 minutes. What will be the average rate of
reaction in minutes?
(a) 4 × 10–4 M min–1
(b) 8 × 10–4 M min–1
(c) 3.3 × 10–4 M min–1 (d) 2.2 × 10–4 M min–1
CBSE Board Term-II Chemistry Class-12
28
Conc. of X
19. For a general reaction
Y, the plot of conc.
X
of X vs time is given in
the figure. What is the
order of the reaction and
what are the units of rate
Time
constant?
(b) First, mol L–1 s–1
(a) Zero, mol L–1 s–1
–1
(c) First, s
(d) Zero, L mol–1 s–1
20. Which of the following is an example of a
fractional order reaction?
(a) NH4NO2 → N2 + 2H2O
(b) NO + O3 → NO2 + O2
(c) 2NO + Br2 → 2NOBr
(d) CH3CHO → CH4 + CO
21. The value of rate of a pseudo first order
reaction depends upon
(a) the concentration of both the reactants
present in the reaction
(b) the concentration of the reactant present in
small amount
(c) the concentration of the reactant present in
excess
(d) the value of DH of the reaction.
22. The rate law for a reaction,
A + B → C + D is given by the expression k[A].
The rate of reaction will be
(a) doubled on doubling the concentration of B
(b) halved on reducing the concentration of A
to half
(c) decreased on increasing the temperature of
the reaction
(d) unaffected by any change in concentration
or temperature.
23. Nitrogen dioxide (NO 2 ) dissociates into
nitric oxide (NO) and oxygen (O2) as follows:
2NO2 → 2NO + O2
If the rate of decrease of concentration of NO2
is 6.0 × 10–12 mol L–1 s–1, what will be the rate
of increase of concentration of O2?
(a) 3 × 10–12 mol L–1 s–1
(b) 6 × 10–12 mol L–1 s–1
(c) 1 × 10–12 mol L–1 s–1
(d) 1.5 × 10–12 mol L–1 s–1
24. For the reaction 4NH3 + 5O2
4NO + 6H2O,
if the rate of disappearance of NH 3 is
3.6 × 10 –3 mol L –1 s –1 , what is the rate of
formation of H2O?
(a)
(b)
(c)
(d)
5.4
3.6
4×
0.6
× 10–3 mol L–1 s–1
× 10–3 mol L–1 s–1
10–4 mol L–1 s–1
× 10–4 mol L–1 s–1
25. The rate constant for the reaction,
2N2O5 → 4NO2 + O2 is 2 × 10–5 s–1. If rate of
reaction is 1.4 × 10–5 mol L–1 s–1, what will be
the concentration of N2O5 in mol L–1?
(a) 0.8
(b) 0.7
(c) 1.2
(d) 1
26. When a chemical reaction takes place,
during the course of the reaction the rate of
reaction
(a) keeps on increasing with time
(b) remains constant with time
(c) keeps on decreasing with time
(d) shows irregular trend with time.
27. For a unimolecular reaction,
(a) the order and molecularity of the slowest
step are equal to one
(b) molecularity of the reaction can be zero, one
or two
(c) more than one reacting species are involved
in one step
(d) molecularity of the reaction can be
determined only experimentally.
28. In a reaction 2X → Y, the concentration of X
decreases from 3.0 moles/litre to 2.0 moles/litre
in 5 minutes. The rate of reaction is
(a) 0.1 mol L–1 min–1 (b) 5 mol L–1 min–1
(c) 1 mol L–1 min–1
(d) 0.5 mol L–1 min–1
29. The chemical reaction, 2O3 → 3O2 proceeds
as
(fast)
O3  O2 + [O]
(slow)
[O] + O3 → 2O2
The rate law expression will be
(a) Rate = k[O][O3]
(b) Rate = k[O3]2 [O2]–1
2
(c) Rate = k[O3]
(d) Rate = k[O2][O]
30.
(a)
(b)
(c)
(d)
Radioactive disintegration is an example of
zero order reaction
first order reaction
second order reaction
third order reaction.
31. In a first order reaction the concentration
of reactant decreases from 400 mol L –1 to
25 mol L–1 in 200 seconds. The rate constant for
the reaction is
29
Chemical Kinetics
(a) 1.01386 s–1
(c) 1.386 × 10–2 s–1
(b) 2 × 10–4 s–1
(d) 3.4 × 10–4 s–1
(c) First order and zero order
(d) None of these
32. What will be the rate equation for the
reaction 2X + Y → Z, if the order of the reaction
is zero?
(a) Rate = k[X][Y]
(b) Rate = k
0
(d) Rate = k[X][Y]0
(c) Rate = k[X] [Y]
33. The following data were obtained during the
first order thermal decomposition of SO2Cl2 at
a constant volume.
SO2Cl2(g) → SO2(g) + Cl2(g)
Experiment Time/s–1 Total pressure/atm
1
0
0.5
2
100
0.6
What is the rate of reaction when total pressure
is 0.65 atm?
(a) 0.35 atm s–1
(b) 2.235 × 10–3 atm s–1
(c) 7.8 × 10–4 atm s–1
(d) 1.55 × 10–4 atm s–1
37. For the reaction N2 + 3H2 → 2NH3, how are
the rate of reaction expressions inter-related
d[ NH3 ]
d[ H2 ]
and
?
dt
dt
1 d[ H2 ]
1 d[ NH3 ]
=+
(a) −
3 dt
2
dt
1 d[ H2 ]
1 d[ NH3 ]
=+
2 dt
3
dt
d
[
NH
d
[
H
]
1
1
3]
2
(c) +
=−
2 dt
3
dt
1 d[ H2 ]
1 d[ NH3 ]
(d) +
=−
3 dt
2
dt
(b) −
38. The expression to calculate time required
for completion of zero order reaction is
[R ]
(a) t = 0
(b) t = [R] – [R0]
k
[ R ] − [ R]
k
(d) t = 0
(c) t =
[ R0 ]
[ R0 ]
34. A first order reaction is 20% complete in 10
minutes. What is the specific rate constant for
the reaction?
(b) 0.009 min–1
(a) 0.0970 min–1
–1
(c) 0.0223 min
(d) 2.223 min–1
39. For the reaction 2NH3 → N2 + 3H2, if
35. In a pseudo first order hydrolysis of ester in
water, the following results were obtained.
then the relation between k1, k2 and k3 is
(b) k1 = 3k2 = 2k3
(a) k1 = k2 = k3
t /s
Ester/mol L–1
0
30
60
90
0.55
0.31
0.17
0.085
What will be the average rate of reaction between
the time interval 30 to 60 seconds?
(a) 1.91 × 10–2 s–1
(c) 1.98 × 10–3 s–1
(d) 2.07 × 10–2 s–1
36. Two plots are shown below between
concentration and time t. Which of the given
orders are shown by the graphs respectively?
slope = –k
slope = –k
ln(a – x)
t
(a) Zero order and first order
(b) First order and first order
d[ NH3 ]
= k1 [ NH3 ],
dt
d[ H2 ]
= k3 [ NH3 ]
dt
(c) 1.5k1 = 3k2 = k3
d[ N2 ]
dt
= k2 [ NH3 ],
(d) 2k1 = k2 = 3k3.
40. The decomposition of dinitrogen pentoxide
(N2O5) follows first order rate law. What will be
the rate constant from the given data?
At t = 800 s, [N2O5] = 1.45 mol L–1
At t = 1600 s, [N2O5] = 0.88 mol L–1
(b) 4.67 × 10–3 mol L–1 s–1
At
−
t
(a) 3.12 × 10–4 s–1
(b) 6.24 × 10–4 s–1
(c) 2.84 × 10–4 s–1
(d) 8.14 × 10–4 s–1
41. The overall rate of a reaction is governed by
(a) the rate of fastest intermediate step
(b) the sum total of the rates of all intermediate
steps
(c) the average of the rates of all the intermediate
steps
(d) the rate of slowest intermediate step.
42. Rate of reaction is the change in concentration
of any one of the reactants or products per unit
time.
CBSE Board Term-II Chemistry Class-12
30
For a hypothetical reaction, A → B
∆[ A] ∆[ B]
=
Rate of reaction = −
dt
dt
In a reaction, A + 2B → 3C + 2D, the concentration
of A decreases from 0.5 mol/L to 0.35 mol/L in
15 seconds. Then a student Ajinkya calculated
following rates :
I. Rate of formation of C is 0.03 mol/L-s.
II. Rate of formation of D is 0.025 mol/L-s.
III. Rate of disappearance of B is 0.02 mol/L-s.
Which is/are incorrect statement(s)?
(a) I only
(b) II only
(c) II & III both
(d) I & III both
–kt
43. For a first order reaction, [A] = [A] 0 e
concentration of reactant decreases exponentially
with time.
0.693
and t1 / 2 =
k
This relation shows that half-life is independent
of concentration and t 1/2 decreases with the
increase of temperature
For first order reaction,
2N2O5(g)
4NO2(g) + O2(g)
The reaction proceeds to 99.6% completion in
(a) 2 half lives
(b) 6 half lives
(c) 8 half lives
(d) 5 half lives
44. A student of class-12, Jayesh did few
experiments for the reaction,
2N2O5(g)
2N2O4(g) + O2(g)
and he plotted time against total pressure
If this reaction follows first order kinetics, value
of rate constant k is
(b) 4.98 × 10–4 s–1
(a) 5.96 × 10–3 s–1
(c) 4.13 × 10–3 s–1
(d) 5.85 × 10–4 s–1
45. Priyanshi after learning chemical kinetics
chapter in class, she made notes for zero order
and first order reactions as given below:
Differential Integral StraRate
Rate law ight
Line
law
[II]
plot
[I]
[III]
Zero
order
d [ R]
= −k
dt
0.523
0.520
0.512
Pt 0.510
0.5
0
100
t
200
kt = [R]0 ln[R] [Ro]/k mol
L–1 s–1
– [R]
vs t
First d [ R]
[R] =
order dt = − k[ R] [R]0e–kt
[R]
vs t
ln 2k
s–1
But the made few mistakes. Identify the wrong
listed equations.
(a) Only I and III
(b) Only I and II
(c) Only IV
(d) III and IV
C
46. For the reaction, A + B
following data has been observed
From the following data for the reaction between A
and B.
Exp.
[A],
No. mol L–1
–4
initial rate
mole L–1 s–1 at
[B],
mol L–1
–5
300 K
320 K
–4
I
2.5 × 10
II
5.0 × 10–4 6.0 × 10–5 4.0 × 10–3
III 1.0 × 10
0.530
Half Units
Life
[V]
[IV]
–3
3.0 × 10
6.0 × 10
–5
5.0 × 10
1.6 × 10
–2
2.0 × 10–3
—
—
The incorrect option about this reaction is
(a) the order of reaction with respect to A is 2
(b) the order of reaction with respect to B is 1
(c) the rate constant for the given reaction is
2.67 × 105 mol–2 L2 s–1
(d) none of these.
Case I : Read the following and answer the
questions from 47 to 51 given below.
The half-life of a reaction is the time required for
the concentration of reactant to decrease by half,
i.e., [ A ]t = 1 [ A ]
2
0.693
For first order reaction, t1 / 2 =
k
this means t1/2 is independent
of initial concentration.
Figure shows that typical
variation of concentration
of reactant exhibiting first
order kinetics. It may be
noted that though the major
Concentration
Case Based MCQs
0
1 2 3 4
Number of half-life
31
Chemical Kinetics
portion of the first order kinetics may be over in
a finite time, but the reaction will never cease as
the concentration of reactant will be zero only at
infinite time.
47. A first order reaction has a rate constant
k = 3.01 × 10 –3 /s. How long it will take to
decompose half of the reactant?
(a) 2.303 s
(b) 23.03 s
(c) 230.3 s
(d) 2303 s
48. The rate constant for a first order reaction is
7.0 × 10–4 s–1. If initial concentration of reactant
is 0.080 M, what is the half life of reaction?
(a) 990 s
(b) 79.2 s
(c) 12375 s
(d) 10.10 × 10–4 s
49. For the half-life period of a first order
reaction, which one of the following statements
is generally false?
(a) It is independent of initial concentration.
(b) It is dependent on rate of the reaction.
(c) At t1/2, the concentration of the reactant is
reduced by half.
(d) None of these.
50. The rate of a first order reaction is
0.04 mol L–1 s–1 at 10 minutes and 0.03 mol L–1 s–1
at 20 minutes after initiation. The half-life of
the reaction is
(a) 4.408 min
(b) 44.086 min
(c) 24.086 min
(d) 2.408 min
51. The plot of t1/2 vs initial concentration [A]0
for a first order reaction is given by
(a) t1/2
(b) t1/2
[A]0
(c) t1/2
[A]0
(d) t1/2
[A]0
[A]0
Case II : Read the following and answer the
questions from 52 to 55 given below.
For a first order reaction, A → Products,
a
2.303
k =
log
, where a is the initial
t
a−x
concentration of A and (a–x) is the concentration
of A after time t. k is rate constant. Its value
is constant at constant temperature for a
reaction. The time in which half of the reactant
is consumed is called half-life period. Half-life
period of a first order reaction is constant. Its
value is independent of initial concentration or
any other external conditions.
In the following questions (Q. No. 52-55), a
statement of assertion followed by a statement
of reason is given. Choose the correct answer
out of the following choices on the basis of the
above passage.
(a) Assertion and reason both are correct
statements and reason is correct explanation
for assertion.
(b) Assertion and reason both are correct
statements but reason is not correct
explanation for assertion.
(c) Assertion is correct statement but reason is
wrong statement.
(d) Assertion is wrong statement but reason is
correct statement.
52. Assertion : Rate of reaction doubles when
concentration of reactant is doubled if it is a first
order reaction.
Reason : Rate constant also doubles.
53. Assertion : For the first order reaction,
2.303
half-life period is expressed as t1/2 =
log 2.
k
Reason : The half-life time of a first order
reaction is not always constant and it depends
upon the initial concentration of reactants.
54. Assertion : For a first order reaction,
the concentration of the reactant decreases
exponentially with time.
Reason : Rate of reaction at any time depends
upon the concentration of the reactant at that
time.
55. Assertion : Half-life period for a first order
reaction is independent of initial concentration
of the reactant.
0.693
Reason : For a first order reaction, t1/2 =
,
k
where k is rate constant.
Case III : Read the following and answer the
questions from 56 to 59 given below.
Number of molecules which must collide
simultaneously to give product is called
molecularity. It is equal to sum of coefficients
of reactants present in stoichiometric chemical
equation.
For reaction, m1A + m2B → Product
CBSE Board Term-II Chemistry Class-12
32
Molecularity = [m1 + m2]
In complex reaction each step has its own
molecularity which is equal to the sum of
coefficients of reactants present in a particular
step. Molecularity is a theoretical property.
Its value is any whole number. Number of
concentration terms on which rate of reaction
depends is called order of reaction or sum of
powers of concentration terms present in the
rate equation is called order of reaction.
Case IV : Read the following and answer the
questions from 60 to 64 given below.
For the reaction : 2NO(g) + Cl2(g) → 2NOCl(g),
the following data were collected. All the
measurements were taken at 263 K.
Experiment
No.
Initial
[NO]
(M)
1.
0.15
0.15
0.60
2.
0.15
0.30
1.20
Then order of reaction = m1 + m2.
3.
0.30
0.15
2.40
In simple reaction, order and molecularity are
same.
In complex reaction, order of slowest step is the
order of over all reaction. This step is known as
rate determining step. Order is an experimental
property. Its value may be zero, fractional or
negative.
4.
0.25
0.25
?
m
m
If rate equation of reaction is : Rate = k ⋅ C A1 ⋅ C B 2
60.
(a)
61.
(a)
Initial Initial rate of
[Cl2] disapp. of Cl2
(M)
(M/min)
The molecularity of the reaction is
1
(b) 2
(c) 3
(d) 4
The expression for rate law is
r = k[NO][Cl2]
(b) r = k[NO]2[Cl2]
(c) r = k[NO][Cl2]2
(d) r = k[NO]2[Cl2]2
56. The rate of reaction, A + 2B → products, is
d[ A ]
= k[ A ][ B ]2
given by the following equation: −
dt
62. The overall order of the reaction is
(a) 2
(b) 0
(c) 1
(d) 3
If B is present in large excess, the order of the
reaction is
(a) zero
(b) first
(c) second
(d) third.
63. The value of rate constant is
(a) 150.32 M–2 min–1 (b) 200.08 M–1 min–1
(c) 177.77 M–2 min–1 (d) 155.75 M–1 min–1
57. The molecularity of the reaction :
6FeSO4 + 3H2SO4 + KClO3 →
(a) 6
(c) 10
KCl + 3Fe2(SO4)3 + 3H2O is
(b) 3
(d) 7
58. Which of the following statements is false
in the following?
(a) Order of a reaction may be even zero.
(b) Molecularity of a reaction is always a whole
number.
(c) Molecularity and order always have same
values for a reaction.
(d) Order of a reaction depends upon the
mechanism of the reaction.
59. The rate of the reaction, A + B + C → products,
d[ A ]
is given by r = −
=k[ A ]1 / 2 [ B ]1 / 3 [ C ]1 / 4 . The
dt
order of the reaction is
13
1
1
1
(b)
(c)
(d)
(a)
12
3
2
4
64. The initial rate of disappearance of Cl2 in
experiment 4 is
(a) 1.75 M min–1
(b) 3.23 M min–1
(c) 2.25 M min–1
(d) 2.77 M min–1
Case V : Read the following and answer the
questions from 65 to 69 given below.
A reaction is said to be of the first order if the rate
of the reaction depends upon one concentration
term only. For a first order reaction of the type
A → Products, the rate of the reaction is given as :
rate = k[A]. The differential rate law is given as :
dA
= –k[A]. The integrated rate law is :
dt
[ A]
= –kt, where [A] is the concentration
ln
[ A ]0
of reactant left at time t and [A]0 is the initial
concentration of the reactant, k is the rate
constant.
65. The unit of rate constant for a first order
reaction is
(a) s–1
(b) mol L–1 s–1
–1 –1
(c) L mol s
(d) L2 mol–2 s–1
33
Chemical Kinetics
66. Half-life period of a first order reaction is
10 min. Starting with initial concentration
12 M, the rate after 20 min is
(a) 0.693 × 3 M min–1
(b) 0.0693 × 4 M min–1
(c) 0.0693 M min–1
(d) 0.0693 × 3 M min–1
67. 50% of a first order reaction is complete in
23 minutes. Calculate the time required to
complete 90% of the reaction.
(a) 70.4 minutes
(b) 76.4 minutes
(c) 38.7 minutes
(d) 35.2 minutes
68. For a first order reaction, (A) → products,
the concentration of A changes from 0.1 M to
0.025 M in 40 minutes. The rate of reaction
when the concentration of A is 0.01 M, is
(a) 3.47 × 10–4 M/min (b) 3.47 × 10–5 M/min
(c) 1.73 × 10–4 M/min (d) 1.73 × 10–5 M/min
69. The half-life period of a 1st order reaction is
60 minutes. What percentage will be left over
after 240 minutes?
(a) 6.25%
(b) 4.25%
(c) 5%
(d) 6%
Assertion & Reasoning Based MCQs
For question numbers 70-80, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
70. Assertion : The rate of the reaction is the
rate of change of concentration of a reactant or
a product.
Reason : Rate of reaction remains constant
during the complete reaction.
71. Assertion : The rate law equation can be
found only by experiment.
Reason : It can be written from stoichiometric
equation.
72. Assertion : The order of the reaction
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
is 1.
Reason : The molecularity of this reaction is 2.
73. Assertion : For the reaction,
2N2O5 → 4NO2 + O2, Rate = k [N2O5]
Reason : Rate of decomposition of N 2 O 5 is
determined by slow step.
Reason : Nature of reactants, concentration of
reactants, products and catalyst affect the rate
of reaction.
76. Assertion : Formation of HI is a bimolecular
reaction.
Reason : Two molecules of reactants are involved
in this reaction.
77. Assertion : Hydrolysis of cane sugar is a
pseudo first order reaction.
Reason : Water is present in large excess during
hydrolysis.
78. Assertion : Rate of reaction can be
expressed as rate of change in partial pressure
of the gaseous reactants or products.
Reason : Partial pressure of a gas is equal to
its concentration.
74. Assertion : Half-life period of a reaction of
first order is independent of initial concentration.
Reason : The time taken for completion of 75%
of a first order reaction is equivalent to two half
lives.
79. Assertion : The decomposition of NH3 on
finely divided platinum surface is first order
when the concentration is low, however at higher
concentration, the reaction becomes zero order.
Reason : In first order reaction, the rate of
reaction is proportional to the first power of the
concentration of the reactant.
75. Assertion : Chemical kinetics deals with
the rate of reaction, the factors affecting the rate
of the reaction and the mechanism by which the
reaction proceeds.
80. Assertion : Instantaneous rate of reaction
is equal to dx/dt.
Reason : It is the rate of reaction at any
particular instant of time.
CBSE Board Term-II Chemistry Class-12
34
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Distinguish between molecularity and order
of a reaction.
2.
Define the half-life period of reaction (t½).
3. For a reaction R
P, half-life (t 1/2) is
observed to be independent of the initial
concentration of reactants. What is the order of
reaction?
6. Express the rate of the following reaction in
terms of the formation of ammonia.
N2(g) + 3H2(g)
2NH3(g)
7. If the rate constant of reaction is k = 3 × 10–4s–1,
then identify the order of the reaction.
4. If half-life period of a first order reaction is
x and 3/4th life period of the same reaction is y,
how are x and y related to each other?
For the reaction 3H2(g) + N2(g) → 2NH3(g),
−d[H2 ]
how are the rate of reaction
expression
dt
d[NH3 ]
interrelated?
and
dt
9. Define elementary step in a reaction.
5. Draw a graph between concentration and
time for a zero order reaction.
10. Distinguish between ‘rate expression’ and
‘rate constant’ of a reaction.
8.
Short Answer Type Questions (SA-I)
11. For a reaction A + B → P, the rate law is given
by, r = k[A]1/2 [B]2
What is the order of this reaction?
12. For a chemical reaction R → P,
the variation in the concentration
[R] vs. time (t) plot is given as
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
13. For a reaction
I−
alkaline medium
2H2 O2 
→ 2H2 O + O2
the proposed mechanism is as given below :
(1) H2 O2 + I − → H2 O + IO − (slow)
(2) H2 O2 + IO − → H2 O + I − +O2 (fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii)Out of steps (1) and (2), which one is rate
determining step?
14. For a first order reaction, the time taken
to reduce initial concentration by a factor of
1/4 is 10 minutes. What will be the time required
to reduce initial concentration by a factor of
1/16?
15. For a first order reaction, show that time
required for 99% completion is twice the time
required for the completion of 90% of reaction.
16. What is meant by rate of reaction? Differentiate
between average rate and instantaneous rate
of reaction.
17. For a reaction A + B
P, the rate is given
by Rate = k[A][B]2
(i) How is the rate of reaction affected if the
concentration of B is doubled?
(ii) What is the overall order of reaction if A is
present in large excess?
18. The thermal decomposition of HCO2H is
a first order reaction with a rate constant of
2.4 × 10–3 s–1 at a certain temperature. Calculate
how long will it take for three-fourth of initial
quantity of HCO2H to decompose.
(log 0.25 = – 0.6021)
19. For a reaction : 2NH3(g)
N2(g) + 3H2(g) ;
Rate = k
(i) Write the order and molecularity of this
reaction.
(ii) Write the unit of k.
20. For the reaction, 2N2O5(g)
4NO2(g) + O2(g),
the rate of formation of NO2(g) is 2.8 × 10–3 M s–1.
Calculate the rate of disappearance of N2O5(g).
35
Chemical Kinetics
Short Answer Type Questions (SA-II)
21. What do you understand by the ‘order of a
reaction’? Identify the reaction order from each
of the following units of reaction rate constant :
(i) L–1 mol s–1
(ii) L mol–1 s–1
22. For a chemical reaction
R → P, the variation in the
ln [R]
ln
concentration, ln [R] vs. time (s)
plot is given as
t(s)
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
(iii) Write the unit of the rate constant for this
reaction.
23. For a reaction, the rate law is :
Rate = k [A][B] 1/2 . Can this reaction be an
elementary reaction?
24. A reaction is first order in A and second
order in B.
(i) Write differential rate equation.
(ii) How is rate affected when concentration of
B is tripled?
(iii) How is rate affected when concentration of
both A and B is doubled?
(iv) What is molecularity of a reaction?
25. In a pseudo first order hyrolysis of ester in
water, the following results are obtained :
t in seconds
[Ester]M
0
30
60
90
0.55
0.31
0.17
0.085
(i) Calculate the average rate of reaction
between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant
for the hydrolysis of ester.
26. The rate of decomposition of ammonia is
found to depend upon the concentration of NH3
d[ NH3 ]
k1 [ NH3 ]
according to the equation −
=
dt
1 + k2 [ NH3 ]
What will be the order of reaction when
(i) concentration of NH3 is very high?
(ii) concentration of NH3 is very low?
27. A certain reaction takes 5 minutes for initial
concentration 0.5 mol L–1 to become 0.25 mol L–1
and another 5 minutes to becomes 0.125 mol L–1.
What is the order and specific rate constant of
the reaction?
28. Following data are obtained for the reaction :
N2O 5
2NO2 +
1
O2
2
t/s
0
300
600
[N2O5]/
mol L–1
1.6 × 10–2
0.8 × 10–2
0.4 × 10–2
(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given : log 2 = 0.3010, log 4 = 0.6021)
29. A first order reaction takes 160 minutes time
for 20% completion. Calculate time required for
half completion of reaction.
30. Hydrogen peroxide, H2O2(aq) decomposes to
H2O(l) and O2(g) in a reaction that is first order in
H2O2 and has a rate constant k = 1.06 × 10–3 min–1.
(i) How long will it take for 15% of a sample of
H2O2 to decompose?
(ii) How long will it take for 85% of the sample
to decompose?
31. When inversion of sucrose is studied at pH = 5,
the half-life period is always found to be 500
minutes irrespective of any initial concentration
but when it is studied at pH = 6, the half-life
period is found to be 50 minutes. Derive the rate
law expression for the inversion of sucrose.
32. What will be the rate of decomposition
of N 2 O 5 and rate of formation of NO 2 and
O 2 when [N 2 O 5 ] = 0.40 M for the reaction
2N2O5 → 4NO2 + O2. The rate constant for this
reaction is 3.1 × 10–4 min–1.
33. The following data were obtained during the
first order thermal decomposition of SO2Cl2 at
a constant volume :
SO2Cl2(g)
SO2(g) + Cl2(g)
Experiment Time/s Total pressure/atm
1
0
0.4
2
100
0.7
Calculate the rate constant.
(Given : log 4 = 0.6021, log 2 = 0.3010)
CBSE Board Term-II Chemistry Class-12
36
34. From the data given below, calculate order
of reaction.
S.No.
[A](M)
[B](M)
Rate (M s–1)
1.
1.0
0.20
0.10
2.
2.0
0.20
0.20
3.
2.0
0.40
0.80
35. For the first order thermal decomposition
reaction, the following data were obtained :
C2H4(g) + HCl(g)
C2H5Cl(g)
Time/sec
0
300
Total pressure/atm
0.30
0.50
Calculate the rate constant.
(Given : log 2 = 0.301, log 3 = 0.4771, log 4 =
0.6021)
Long Answer Type Questions (LA)
36. Calculate the order of the reaction and the
rate constant for the decomposition of N2O5 at
30°C from the following rate data.
Rate of
reaction
(Mol L–1 hr–1)
Concentration of
N 2O 5
(Mol L–1)
1.
0.10
0.34
2.
0.20
0.68
3.
0.40
1.36
S.No.
reaction, as the concentration of water
remains constant.
(ii) Calculate the average rate of reaction
between the time interval 30 to 60 seconds.
(Given : log 2 = 0.3010, log 4 = 0.6021)
39. For a homogeneous gas phase reaction
A (g) → B (g) + C (g) + D (g) , the pressure of the
reaction mixture increases by 40% in 20 minute.
Calculate rate constant of a reaction.
37. The half-life period of a first order reaction is
30 minutes. Calculate the specific reaction rate
of the reaction. What fraction of the reactant
remains after 70 minutes?
40. The following results have been obtained
during the kinetic studies of the reaction :
Experiment
No.
[CH3COOCH3]/mol L
–1
0
0.60
30
0.30
60
0.15
(i) Show that it follows pseudo first order
OBJECTIVE TYPE QUESTIONS
1. (b) : In terms of pressure, Rate = k(pCH3COCH3)3/2
Units of rate = bar min–1
rate
Units of rate constant =
( pCH 3OCH 3 )3/2
=
bar min−1
bar3/2
= bar −1/2 min− 1
[A]
[B]
Initial rate of
formation of D
3.
0.1 M 0.1 M 6.0 × 10–3 M min–1
0.3 M 0.2 M 7.2 × 10–2 M min–1
0.3 M 0.4 M 2.88 × 10–1 M min–1
4.
0.4 M 0.1 M 2.40 × 10–2 M min–1
1.
38. For the hydrolysis of methyl acetate in
aqueous solution, the following results were
obtained :
t/s
C+D
2A + B
2.
Calculate the rate of formation of D when
[A] = 0.5 mol L–1 and [B] = 0.2 mol L–1.
2. (d) : Rate = k[A]2 [B]3 = a
When volume is reduced to one half then conc. of reactants
will be doubled.
Rate = k[2A]2 [2B]3 = 32 k[A]2 [B]3 = 32a
1
3. (b) : Rate of disappearance of Q =
× rate of
2
appearance of R
4.
(c) : Order of reaction can be zero, fractional or negative.
37
Chemical Kinetics
dx
= k [X ] (For a first order reaction)
dt
0.00352 = k × 0.01 ⇒ k = 0.352
0.693 0.693
t1/2 =
=
= 1.969 min
k
0.352
6. (c) : Rate of a zero order reaction is independent of the
concentration of reactants.
2.303  a 
2.303
5 
log 
or t =
log   = 34.07 s
7. (a) : t =
−3
3
 a − x 
k
15 × 10
5.
8.
(b) :
(b) : k =
2.303  a 
log 
 a − x 
t
(for first order)
20. (d) : CH3CHO → CH4 + CO
Rate = k[CH3CHO]3/2
21. (b) : If one of the reactant is present in excess, the
reaction becomes independent of the concentration of that
reactant hence it becomes pseudo first order reaction.
Thus, rate of pseudo first order reaction depends upon the
concentration of the reactant present in small amount.
22. (b) : The rate of reaction depends upon concentration of
only A.
23. (a) : For the reaction, 2NO2 → 2NO + O2
2.303
1  2.303

k=
=
log
log 8  min−1

12
1 / 8  12
−
1 d [NO2 ] 1 d [NO] d [O2 ]
=
=
2 dt
2 dt
dt
9. (d) : The unit of rate of reaction is mol L–1 s–1. It does
not change with order.
−
d [NO2 ]
= 6 × 10−12 mol L−1 s−1
dt
1 d [NO2 ] 1
10. (b) : Rate =
= × 0.0125 = 0.0031 mol L–1 s–1
4 dt
4
11. (b) : In a reaction P → Q, concentration of reactant
decreases as concentration of product increases during the
course of a reaction.
12. (b) : The number of molecules of the reactants taking
part in a single step of the reaction tells about molecularity
of the reaction.
13. (b) : Rate of forward reaction decreases and rate
of backward reaction increases with passage of time. At
equilibrium both the rates become equal.
14. (b) : For 2N2O5 → 4NO2 + O2, the rate of reaction can
be expressed as
1 d [N2O5 ] 1 d [NO2 ] d [O2 ]
=
=
2 dt
4 dt
dt
1 ∆[R ]
1 0.4 − 0.5
=− ×
15. (a) : Average rate = −
2 ∆t
2
10
1 0.1
−3
−1
= 5 × 10 M min
= ×
2 10
−
16. (a) : For a zero order reaction, rate = k[A]0 = k
Units = mol L–1 s–1
17. (c) : When one of the reactants is in excess the reaction
behaves as a first order reaction.
18. (c) : Average rate = −
=−
(0.04 − 0.05)
30
=
([R ]2 − [R ]1 )
∆[R ]
=−
∆t
t 2 − t1
0.01
= 3.3 × 10−4 M min−1
30
19. (a) : For a zero order reaction, rate = k =
Units of k = mol L–1 s–1
d [O2 ]
= 3 × 10−12 mol L−1 s−1
dt
1 d [NH3 ]
1 d [H2O]
24. (a) : −
=+
4 dt
6 dt
d [H2O] 6
= × 3.6 × 10−3 = 5.4 × 10−3 mol L−1 s−1
4
dt
25. (b) : Rate = k[N2O5] (first order as unit of rate constant
is s–1)
[N2O 5 ] =
Rate 1.4 × 10 −5 mol L−1 s −1
=
= 0.7 mol L−1
−5 −1
k
2 × 10 s
26. (c) : Rate of reaction ∝ conc. of reactants
As the reaction proceeds, concentration of the reactants
decreases hence the rate also keeps on decreasing with time.
27. (a) : For a unimolecular reaction, both order and
molecularity are one in rate determining step.
28. (a) : Rate = −
1 ∆[X ]
1 (2 − 3)
=−
= − 0.1 mol L−1 min−1
2 ∆t
2 5
k1
29. (b) : O3 O2 + [O]
k
−1
(fast)
k
2
→ 2O2(slow)
[O] + O3 
Rate of reaction is determined by slow step hence,
Rate = k2[O][O3]
[O] is unstable intermediate so substitute the value of [O] in
above equation.
Rate of forward reaction = k1[O3]
Rate of backward reaction = k–1[O2][O]
dx
dt
At equilibrium,
Rate of forward reaction = Rate of backward reaction
k1[O3] = k–1[O2][O]
CBSE Board Term-II Chemistry Class-12
38
[O ] =
k 1[ O 3 ]
k −1[O 2 ]
k [O 3 ]2
 k [O ] 
Rate = k 2  1 3  [O 3 ] ; Rate =
[O 2 ]
 k −1[O 2 ] 
30. (b) : Radioactive disintegration is an example of first
order reaction.
k =
2.303  N 0 
log  
 Nt 
t
=
37. (a) : For the reaction N2 + 3H2
−
2.303  a  2.303  400 
log 
log 
=
 a − x  200
 25 
t
38. (a) : [R] = [R]0 – kt
For completion of reaction [R] = 0
d [N2 ] 1 d [H2 ]
1 d [NH3 ]
=
=
2 dt
3 dt
dt
1
1
k 1 [NH3 ] = k 2 [NH3 ] = k 3 [NH3 ]
2
3
Pressure at time t
SO2Cl2 → SO2 + Cl2
p0
0
0
p0 – p
p
p
Let initial pressure p0 ∝ R0
Pressure at time t, Pt = p0 – p + p + p = p0 + p
Pressure of reactants at time t, p0 – p = 2p0 – Pt ∝ R
2.303  p0 
k=
log 
 2p0 − Pt 
t
=
2.303 
0.5
 2.303
log 
log1.25
=

100
2 × 0.5 − 0.6  100
= 2.2318 × 10–3 s–1
Pressure of SO2Cl2 at time t (pSO2Cl2)
= 2p0 – Pt = 2 × 0.50 – 0.65 atm = 0.35 atm
Rate at that time = k × pSO2Cl2
= (2.2318 × 10–3) × (0.35) = 7.8 × 10–4 atm s–1
34. (c) : a − x =
k=
80
× a = 0.8a , t = 10 min
100
2.303  a  2.303  a 
log 
=
log 
 a − x 
 0.8a 
10
t
10
= 0.2303(1 − 0.9030)
8
= 0.2303 × 0.0970 = 0.0223 min–1
k = 0.2303log
35. (b) : Average rate during the time interval 30-60 s.
C −C
(0.17 − 0.31) 0.14
=
Rate = − 2 1 = −
t 2 − t1
60 − 30
30
= 4.67 × 10–3 mol L–1 s–1
N2 + 3H2
Rate = −
32. (b) : Rate = k[X]0 [Y]0 or rate = k
Initial pressure
[R ]0
k
39. (c) : 2NH3
2.303
× 1.204 = 0.01386 s−1 = 1.386 × 10−2 s−1
200
33. (c) :
2NH3,
1 d [H2 ]
1 d [NH3 ]
=+
3 dt
2 dt
or t =
31. (c) : For first order reaction,
k=
36. (a) : Linear plots are obtained in the graph of
A t vs t for zero order reaction and ln a – x vs t for first order
reaction.
1.5k1 = 3k2 = k3
40. (b) : k =
k=
[A ]
2.303
log 1
(t 2 − t1 ) [A2 ]
2.303
1.45 2.303
log
=
× 0.2169
(1600 − 800) 0.88 800
= 6.24 × 10–4 s–1
41. (d) : The slowest step is rate determining step.
42. (b) :
=
1 [C ] 1 d [D ] − d [ A ]
1 d [B ]
=
=
=−
3 dt 2 dt
dt
2 dt
− d [A ]
(0.035 − 0.5) 0.15
=−
=
= 0.01 mol / L - s
dt
15
15
− d [B ]
= 0.02 mol / L - s
dt
d [C ]
= 3 × 0.01 = 0.03 mol / L - s
dt
d [D ]
= 2 × 0.01 = 0.02 mol / L - s
dt
43. (c) : For 99.6% completion,
Let a = 100
x = 99.6
a – x = 0.4
k=
2.303 100 2.303
=
log
log 250
t 99.6
0.4 t 99.6
2.303
log 250
k
0.693
k=
t1/ 2
t 99.6% =
From equation (i) & (ii)
...(i)
...(ii)
39
Chemical Kinetics
t 99.6% =
Dividing equation (ii) by equation (i),
2.303
× 2.4 × t1/ 2 = 8t1/ 2
0.693
(Rate)2
44. (b) : The following data were obtained during the
first order thermal decomposition of N2O5(g) at constant
volume :
2N2O4(g) + O2(g)
2N2O5(g)
S.No.
Time/s
Total Pressure/(atm)
1.
0
0.5
2.
100
0.512
Let the pressure of N2O5(g) decrease by 2x atm. As two moles
of N2O5 decompose to give two moles of N2O4(g) and one
mole of O2 (g), the pressure of N2O4(g) increases by 2x atm
and that of O2(g) increases by x atm.
2N2O4(g) + O2(g)
2N2O4(g)
Start t = 0
At time t
0.5 atm
(0.5 – 2x) atm
0 atm
2x atm
0 atm
x atm
(Rate)1
or k =
\
2.303
× 0.0216 = 4.98 × 10 −4 s −1
100
[R ]0
.
2k
For first order reaction ln [R] vs t gives straight line plot and
ln2
half life =
k
46. (c) : (i) Rate law can be written as follows:
Rate = k [A]p [B]q
From experiments I, II and III
(Rate)1 = k [2.5 × 10–4]p [3.0 × 10–5]q = 5.0 × 10–4 ...(i)
(Rate)2 = k [5.0 × 10–4]p [6.0 × 10–5]q = 4.0 × 10–3 ...(ii)
(Rate)3 = k [1.0 × 10–3]p [6.0 × 10–5]q = 1.6 × 10–2 ...(iii)
Dividing equation (iii) by equation (ii),
(Rate)2
or
(1.0 × 10 −3 ) p
1.6 × 10 −2
=
(5.0 × 10 −4 ) p 4.0 × 10 −3
2p = 4 or 2p = 22 i. e. p = 2
t1/2 =
0.693
3.01 × 10−3
= 230.3 s
49. (b) : For a first order reaction t1/2 =
45. (d) : For zero order reaction,
=
(2.5×10 −4 )2 (3.0×10 −5 )
48. (a) : Half life (t1/2) of a first order reaction is given as :
0.693
0.693
t 1/2 =
=
= 990 s
k
7.0 × 10−4
p
2.303
2.303
0.5 atm
log t =0 =
log
t
pt =100
100
0.476 atm
(Rate)3
5.0×10 −4 (mol L−1)3
47. (c) : For a first order reaction :
0.693
, k = 3.01 × 10–3 s–1
t1/2 =
k
We know that,
[R] vs t is a straight line plot and half life t1/2 =
4.0 × 10 −3
=
−4
(2.5 × 10 −4 ) p (3.0 × 10 −5 )q 5.0 × 10
= 2.67 × 108 mol−2 L 2 s −1
= (0.5 – 2x) + 2x + x = 0.5 + x
x = pt – 0.5
pN2O5 = 0.5 – 2x
= 0.5 – 2 (pt – 0.5) = 1.5 – 2pt
At t = 100 s; pt = 0.512 atm
pN2O5 = 1.5 – 2 × 0.512 = 0.476 atm
=
(5.0 × 10 −4 ) p (6.0 × 10 −5 )q
or 2p. 2q = 8 or 22. 2q = 8
or 2q = 8/22 or 2q = 21 or q = 1
Thus the rate equation is Rate = k[A]2 [B]
∴ Order of reaction with respect to A = 2
Order of reaction with respect to B = 1
(ii) Rate constant (k) at 300 K :
From experiment 1, we have
Rate = k (2.5 × 10–4)2 (3.0 × 10–5)
or 5.0 × 10–4 mol L–1
= k (2.5 × 10–4 mol L–1)2 (3.0 × 10–5 mol L–1)
p t = p N2O5 + p N2O4 + p O2
k=
=
0.693
k
therefore t1/2 depends solely on k.
50. (c) : Let the concentrations of the reactant after
10 min and 20 min be C1 and C2 respectively.
\
Rate after 10 min = 0.04 × 60 mol L–1min–1 = k.C1
and rate after 20 min =
0.03 × 60 mol L–1min–1 = k.C2
∴
C 1 0.04 × 60 4
=
=
C 2 0.03 × 60 3
Let the reaction starts after 10 minutes.
2.303 C1 2.303 4
k=
log =
log = 0.02878
10
10
3
C2
0.6932 0.6932
=
= 24.086 min
0.02878
k
51. (b) : For a first order reactions, t1/2 = k[A]00 = k. Thus t1/2
is independent of initial concentration. Hence plot of t1/2 vs
[A]0 will be a horizontal line.
∴ t1/2 =
CBSE Board Term-II Chemistry Class-12
40
52. (c) : For first order reaction, Rate1 = k[A1]
According to question, [A2] = [2A1]
\ Rate2 = k[2A1] ⇒ Rate2 = 2 Rate1
For a given reaction, rate constant is constant and independent
of the concentration of reactant.
2.303
a
53. (c) : For a first order reaction k =
log
t
a−x
2.303
a
2.303
a 2.303
k=
log
=
log
=
log 2
a /2
t1/2
a − a /2
t1/2
t1/2
62. (d) : As the order w.r.t. NO is 2 and order w.r.t. Cl2
is 1, hence the overall order is 3.
2.303
log 2.
k
Thus t1/2 is independent of initial concentration of reactant
for first order reaction.
65. (a) : Unit of rate constant for a reaction of nth order =
(conc.)1–n time–1
For a first order reaction, n = 1
Unit of rate constant = (mol L–1)1 – 1 s–1 = s–1
Therefore half-life period t1/2 =
54. (b) : For a first order reaction, [A] = [A]0e–kt…(i)
According to eq. (i), concentration of reactant decreases
exponentially.
55. (a) : For a first order reaction, t1/2 is inversely proportional
to k, it does not depend on the initial concentration of the
reactant.
56. (b) : From the expression
d [A ]
−
= k [A][B ]2
dt
when B is present in large excess, rate will be independent
upon the change in conc. of B, therefore order of reaction
will be one.
57. (c) : The total number of reactant molecules participating
in a chemical reaction is known as its molecularity, hence the
molecularity = 6 + 3 + 1 = 10.
58. (c) : Molecularity may or may not be equal to the order
of a reaction.
1 1 1 6 + 4 + 3 13
=
59. (d) : Order of reaction = + + =
2 3 4
12
12
60. (c) : 2NO(g) + Cl2(g) → 2NOCl(g)
Molecularity = 3
61. (b) : Let the rate of this reaction, r = k[NO]m[Cl2]n
then
r1 0.60 k (0.15)m (0.15)n
=
=
r2 1.20 k (0.15)m (0.30)n
n
1  1
or,
=   ⇒ n =1
2 2
r 1.20 k (0.15)m (0.30)n
=
Again from 2 =
r3 2.40 k (0.30)m (0.15)n
m
1  1 2
1  1
=  ⋅
= 
or
2 2 1
4 2
Hence, expression for rate law is
r = k[NO]2 [Cl2]1
or
m
⇒ m =2
63. (c) : Substituting the values of experiment 1 in rate law
expression
0.60 M min–1 = k(0.15 M)2 (0.15 M)1
or k =
0.60 M min−1
3
0.0225 × 0.15 M
= 177.77 M−2 min−1
64. (d) : r = 177.7 M –2 min –1 × (0.25 M) 2 (0.25 M)
= 2.77 M min–1
t /2
/2
66. (d) : 12 M t1
→ 6 M 1
→ 3M
Initial conc.
t1/2 = 10 min
0.693
k=
= 0.0693 min−1
10
As t1/2 is 10 min, after 20 minutes the concentration will
be 3 M.
Hence, Rate = 0.0693 × 3 M min–1
67. (b) : t1/2 = 23 minutes
0.693
0.693
0.693
t1/2 =
⇒ k=
⇒ k=
min−1
k
23
t1/2
For 90% completion,
2.303  a 
2.303 × 23  100 
;
t=
log 
 a − x  t = 0.693 log  100 − 90 
k
t = 76.4 minutes
68. (a) : For the first order reaction, k =
2.303  a 
log 
 a − x 
t
a = 0.1 M, a – x = 0.025 M, t = 40 min
2.303
0.1 2.303
k=
log
=
log 4 = 0.0347 min–1
40
0.025
40
[A] → product
Thus, rate = k[A]
rate = 0.0347 × 0.01 M min–1 = 3.47 × 10–4 M min–1
69. (a) : t1/2 =
0.693 0.693
0.693
⇒
=k ⇒
=k
k
t1/2
60
k = 0.01155 min–1
k=
2.303  a 
log 
 a − x 
t
Let the initial amount (a) be 100.
0.01155 min–1 =
2.303
 100 
log 

240 min  a − x 
41
Chemical Kinetics
0.01155 min−1 × 240 min
 100 
= log 
 a − x 
2.303
1.204 = log 100 – log (a – x)
1.204 = 2 – log (a – x)
log (a – x) = 2 – 1.204 = 0.796
(a – x) = 6.25%
70. (c) : Rate of reaction does not remain constant during
the complete reaction because rate depends upon the
concentration of reactants which decreases with time.
71. (c) : The rate law equation cannot be written from
stoichiometric equation.
72. (b) : During hydrolysis of ester, water is always present
in high concentration, thus there is very little change in its
concentration and it practically remains constant. Thus, the
order of reaction is 1 (pseudo first order reaction).
73. (a) : Rate of any reaction is equal to the rate of its
slowest step and here rate of given reaction = k [N 2O 5]
because the slowest step has only N2O5 molecule involved.
74. (b)
75. (b)
76. (a) : A bimolecular reaction may involve combination
of two molecules or exchange of atoms or groups of atoms
between the two reactant molecules.
77. (a) : Hydrolysis of cane sugar is pseudo first order
reaction. Since, water is always in excess, rate of reaction
does not depend appreciably on its concentration, thus it is
an example of pseudo unimolecular reactions.
78. (c) : For a gaseous reaction at constant T, concentration
is directly proportional to the partial pressure of the species.
Thus, the rate of reaction can be expressed in terms of partial
pressure for gaseous reactants or products.
79. (b) : In a heterogeneous system, the reactant is
absorbed on the surface of a solid catalyst. The fraction
of the surface of the catalyst covered by the reactant is
proportional to its concentration at low values and the rate
of reaction is first order. However at higher concentration,
the surface of catalyst is fully covered and the reaction rate
becomes independent of concentration and it becomes zero
order reaction.
80. (b) : Instantaneous rate of a reaction is equal to small
change in concentration (dx) during a small interval of time
(dt) at that particular instant of time divided by the time
interval.
SUBJECTIVE TYPE QUESTIONS
1.
Order of a reaction Molecularity of a reaction
1. It is the sum of powers 1. It is the number of reacting
of the concentration
species (atoms, ions or
of the reactants in the
molecules) taking part in an
rate law expression.
elementary reaction which
must collide simultaneously
in
order to bring about a
chemical reaction.
2. It can be zero or even 2. It is always a whole number.
a fraction.
Generally, in a complex reaction the order of reaction is equal
to the molecularity of the slowest step.
2. The time taken for half of the reaction to complete, i.e.,
the time in which the concentration of a reactant is reduced
to half of its original value is called half-life period of the
reaction.
[R ]
t = t1/2 when [R ] = 0
2
3. Half-life of first order reaction is independent of the
initial concentration of reactants.
0.693
t1/2 =
k
0.693 0.693
=
t1/2
x
4.
For a first order reaction, k =
For
3
3
1
th life period, a − x = a − a = a
4
4
4
\
k=
0.693 2.303
a
2.303
a
=
log
;
log
a /4
x
y
(a − x )
y
0.693 2.303
0.693 0.693 × 2
2log2 ;
=
=
x
y
x
y
y = 2x
5.
[R] = –k0t + [R]0
6.
N2(g) + 3H2(g)
2NH3(g)
−d [N2 ]
1 d [H2 ]
1 d [NH3 ]
=−
=+
dt
3 dt
2 dt
CBSE Board Term-II Chemistry Class-12
42
7. First order reaction has s –1 as the unit of the rate
constant.
1 d [H2 ]
1 d [NH3 ]
=+
3 dt
2 dt
9. Elementary step : Each step of a complex reaction is
called an elementary step.
8.
−
10. Rate expression is a way of expressing rate of reaction
in terms of concentration of reactants, e.g., for a general
reaction, aA + bB
cC + dD
x
Rate = k[A] [B]y
Rate constant (k) is equal to the rate of reaction when molar
concentration of reactant is unity. Its units depends upon
the order of reaction.
11. Rate law, r = k[A]1/2 [B]2
Order of reaction is sum of the powers of concentration terms,
1
5
\ Order of reaction = + 2 = = 2.5
2
2
12. (i) The reaction is of zero order.
d [R ]
(ii) Slope of the curve = − k =
dt
–
13. (i) Rate = k[H2O2][I ]
(ii) Overall order of reaction is 2.
(iii) Step (1) being the slow step is the rate determining step
of the reaction.
14. Let initial concentration (a) = 1
then, final concentration (a – x) = 1/4
2.303
1
∴ t1/ 4 =
log
...(i)
k
1/ 4
2.303
1
...(ii)
log
k
1 / 16
Dividing equation (ii) by (i)
t1/16 log 16
1.204
= 20 minutes
=
⇒ t1/16 = 10 ×
10
0.6021
log 4
Similarly, t1/16 =
15. For first order reaction, t =
[R ]
2.303
log 0
k
[Rt ]
For 99% completion of reaction
t = t0.99, [R]0 = 1, [R]t = (1 – 0.99) = 0.01 = 10–2
2.303
1
2.303
2.303
log −2 =
log 102 =
× 2 ...(i)
k
k
k
10
For 90% completion of reaction
t = t0.90, [R]0 = 1, [R]t = (1–0.9) = 0.1 = 10–1
2.303
1
2.303
2.303 ...(ii)
t 0.90 =
log −1 =
log 10 =
k
k
k
10
Comparing equations (i) and (ii),
t0.99 = 2 × t0.90
t 0.99 =
16. Change in concentration i.e., either (decrease in
concentration of reactant or increase in concentration of
product) per unit time is called rate of reaction.
C − C ∆C
Rate of reaction = 2 1 =
∆t
t 2 − t1
The ratio of change of concentration of reactants to the time
consumed in that change is called average rate of reaction.
∆x
C − C1
rav =
=− 2
∆t
t 2 − t1
The rate of reaction at a particular instant (time) is called
instantaneous rate of reaction.
rins =
dx
dt
dx = small change in concentration
dt = small time interval
17. (i) From the rate law equation, order of reaction w.r.t. B
is 2. Hence, if concentration of B is doubled, rate will become
four times.
(ii) If A is present in large excess, rate of reaction will
be independent of concentration of A and hence, order of
reaction will be 2.
18. For a first order reaction, t =
[R ]
2.303
log 0
k
[R ]
Given k = 2.4 × 10–3 s–1
[R ]
[R ] = 0 , t = ?
4
Substituting these values in the equation, we get
t=
t=
2.303
2.4 × 10
−3 −1
s
2.303
2.4 × 10 −3 s −1
log
[R ]0
[R ]
log 4 =
2.303
2.4 × 10 −3
× 0.6021 s
t = 577.7 s = 578 s
19. (i) The decomposition of gaseous ammonia on a hot
platinum surface is a zero order reaction at high pressure.
In this reaction, platinum metal acts as a catalyst. At high
pressure, the metal surface gets saturated with gas molecules.
So, a further change in reaction conditions is unable to alter
the amount of ammonia on the surface of the catalyst making
rate of the reaction independent of its concentration.
However, two molecules of ammonia react to give products
thus, the molecularity is two.
(ii) For a zero order reaction, unit of rate constant is
mol L–1 s–1.
d [NO2 ]
= 2.8 × 10 −3 M s −1
dt
According to rate law expression,
20. Given,
43
Chemical Kinetics
\
−
1 d [N2O5 ] 1 d [NO2 ] d [O2 ]
=
=
2
dt
4 dt
dt
−
1 d [N2O5 ] 1
= × 2.8 × 10 −3
2 dt
4
−d [N2O5 ] 1
= × 2.8 × 10 −3 = 1.4 × 10 −3 M s −1
dt
2
21. Order of a reaction : It is the sum of the power of
reactant in the rate law expression.
(i) L–1 mol s–1 – Zero order reaction
(ii) s–1
– First order reaction
22. (i) The reaction is of 1st order.
(ii) For first order reaction
ln[R] = –kt + ln [R]0
comparing eqn. y = m × x + c
we get a straight line with slope = –k and intercept equal
to ln[R]0.
(iii) Unit of rate constant for first order reaction
mol L−1
1
×
= s −1
−
1
1
s
(mol L )
23. For an elementary reaction, order should be equal to
molecularity and further molecularity should be integral. For
the given reaction, order of reaction = 1 + 1/2 = 3/2. Since
molecularity cannot be fractional, therefore, for the given
reaction, order is not equal to molecularity. Hence given
reaction cannot be an elementary reaction.
24. (i) Differential rate equation of reaction is
dx
= k [A ]1[B ]2 = k [A ][B ]2
dt
(ii) When conc. of B is tripled, it means conc. of B becomes
[3 × B]
dx ′
= k [A][3B ]2
\ New rate of reaction,
dt
 dx 
= 9k[A][B]2 = 9  
 dt 
i.e., the rate of reaction will become 9 times.
(iii) When conc. of A is doubled and that of B is also doubled,
then conc. of A becomes [2A] and that of B becomes [2B].
dx ′
= k [2A ][2B ]2 = 8k [A][B]2
\ Now rate of reaction,
dt
i.e., the rate of reaction will become 8 times.
(iv) Molecularity of a reaction is the number of reacting
particles which collide simultaneously to bring about the
chemical change. It is a theoretical concept.
25. (i) Average rate of reaction between the time interval
30 to 60 seconds is
rav =
−[0.17 − 0.31] 0.14
=
60 − 30
30
= 4.67 × 10–3 s–1 [Taking only difference]
2.303 [R ]0
log
t
[R ]
At t = 30 s,
2.303 0.55 2.303
k=
=
× 0.249 = 1.91 × 10 −2 s −1
log
30
0.31
30
At t = 60 s,
2.303 0.55 2.303
k=
log
=
× 0.5099 = 1.96 × 10 −2 s −1
60
0.17
60
(ii) k =
At t = 90 s
2.303
0.55
k=
log
90
0.085
2.303
× 0.8109 = 2.07 × 10 −2 s −1
90
\ Average value of k
k=
=
1.91 × 10 −2 + 1.96 × 10 −2 + 2.07 × 10 −2
= 1.98 × 10 −2s −1
3
26. The given rate law equation can be written as
d [NH3 ]
k1
−
=
1 / [NH3 ] + k 2
dt
If [NH3] is very high, 1/[NH3] becomes negligible
d [NH3 ] k1
∴ −
= =k
dt
k2
(i)
i.e., rate becomes independent of concentration. Hence, it is
of zero order.
(ii) If [NH 3] is is very small, 1/[NH 3] will be very large
(>>k 2 ), so that k 2 can be neglected in comparison to
d [NH3 ]
k1
=
= k1[NH3 ]
1/[NH3]. Hence −
dt
1 / [NH3 ]
Thus, reaction is of 1st order.
27. The given data is
5 min
5 min
0.5 molL−1 
→ 0.25 molL−1 
→ 0.125 molL−1
Half life period is independent of initial concentration of the
reactant, hence reaction is of first order.
For first order reaction
k=
0.693 0.693
=
= 0.138 min−1
t1/2
5 min
28. (a) The formula of rate constant for first order reaction
is
2.303 [A ]0
k=
log
[A ]t
t
CBSE Board Term-II Chemistry Class-12
44
2.303 (1.6 × 10 −2 ) mol L−1
k1 =
= 2.3 × 10 −3 s −1
log
300 s (0.8 × 10 −2 ) mol L−1
−2
−1
Similarly, k 2 = 2.303 log (1.6 × 10 ) mol L
600 s (0.4 × 10 −2 ) mol L−1
= 2.3 × 10–3 s–1
Unit and magnitude of rate constant shows the given reaction
is of first order.
(b) The formula for half-life for first order reaction is
0.693
0.693
t1/2 =
=
= 301.30 s
k
2.3 × 10 −3 s −1
29. Given : [R]0 = 1, [R] = 0.80, t = 160 min
t1/2 = ?
[R ]0
2.303
For first order reaction, k =
log10
[R ]
t
2.303
2.303
1
× log 1.25
log10
or, k =
160 min
160
0.8
2.303 × 0.0969
or, k =
= 1.39 × 10 −3 min−1
160 min
0.693
0.693
=
= 498 min
Again t1/2 =
k
1.39 × 10 −3
or,
k=
2.303 [A ]0
log
30. (i) t =
[A ]
k
[A ]0 100
=
Given k = 1.06 × 10 min ,
85
[A ]
2.303
100
log
t=
−3
−1
85
1.06 × 10 min
–3
–1
2303
[2 log10 − log 85]min
1.06
2303 × 0.0706
2303
[2 × 1 − 1.9294] =
t=
1.06
1.06
t = 153.39 min = 153.4 min.
t=
[A]0 100
=
[A] 15
2.303
100 2303
log
[2 log10 − log15]min
=
t=
−3
−1
15
1.06
1.06 × 10 min
(ii) Given k = 1.06 × 10–3 min–1,
2303
2303 × 0.8239
=
[2 ×1−1.1761] =
min = 1790 min
1.06
1.06
31. At pH = 5, as half-life period is found to be independent
of initial concentration of sucrose, this means with respect to
sucrose, it is a reaction of first order, i.e., Rate = k[Sucrose].
If n is the order with respect to H+ ion, t1/2 ∝[H+]1–n,
i.e., 500 ∝ (10–5)1 – n [pH = 5 means [H+] = 10–5 M] ...(i)
...(ii)
and 50 ∝ (10–6)1 – n [pH = 6 means [H+] = 10–6M]
Dividing (i) by (ii), 10 = (10)1 – n i.e. 1 – n = 1 or n = 0, i.e.,
order with respect to H+ ion = 0. Hence, overall rate law is
Rate = k [Sucrose] [H+]0.
32. The given reaction is 2N2O5 → 4NO2 + O2
Unit of rate constant suggests that rate of reaction is first
order.
Hence, rate of reaction = k[N2O5]
= 3.1 × 10–4 min–1 × 0.40 M
= 1.24 × 10–4 M min–1
Rate of reaction,
1 d [N2O5 ]
1 d [NO2 ]
d [O2 ]
=−
=+
=+
2 dt
4 dt
dt
d [N2O5 ]
= 1.24 × 10 −4 M min−1 × 2
dt
= 2.48 × 10–4 M min–1
−
∴
Similarly
+
d [NO2 ]
= 4 × 1.24 × 10 −4 M min−1
dt
= 4.96 × 10–4 M min–1
d [O2 ]
= 1.24 × 10 −4 M min−1
dt
33. The given reaction is
SO2Cl2(g)
At t = 0 At time t
SO2(g) + Cl2(g)
0.4 atm 0 0
(0.4 – x) atm x atm x atm
Total pressure at time t will be
Pt = (0.4 – x) + x + x = 0.4 + x
x = (Pt – 0.4)
Pressure of SO2Cl2 at time t will be
pSO2Cl2 = 0.4 – x = 0.4 – (Pt – 0.4) = 0.8 – Pt
At time t = 100 s, Pt = 0.7 atm
\ pSO2Cl2 = 0.8 – 0.7 = 0.1 atm
According to first order kinetic equation
p SO2Cl2 (initial)
2.303
k=
log
t
p SO2Cl2 (after reaction)
=
2.303  0.4 
log   = 1.3 × 10 −2 s −1
 0.1 
100
34. Let rate of reaction r = k[A]a[B]b
From the data
r1 = 0.10 M s–1 = k(1.0 M)a (0.20 M)b
r2 = 0.20 M s–1 = k(2.0 M)a (0.20 M)b
r3 = 0.80 M s–1 = k(2.0 M)a (0.40 M)b
Dividing eqn. (i) by (ii)
r1 0.10 Ms −1 k (1.0 M)a (0.20 M)b
=
=
r2 0.20 Ms −1 k (2.0 M)a (0.20 M)b
... (i)
... (ii)
... (iii)
45
Chemical Kinetics
a
1  1
=   or a = 1
2 2
Dividing eqn. (ii) by eqn. (iii)
\
or,
b
1  1
r2 0.20 Ms −1 k (2.0)a (0.20)b
or,
=   or, b = 2
=
=
−
1
a
b
4 2
r3 0.80 Ms
k (2.0) (0.40)
Hence, order of reaction = 1 + 2 = 3
35. The given reaction is
C2H5Cl(g)
At time t = 0
At time t = 300 sec
C2H4(g) + HCl(g)
0.30 atm
0.30 – x
0
0
x
x
Total pressure = 0.30 – x + x + x = 0.50
or 0.30 + x = 0.50
\ x = 0.50 – 0.30 = 0.20
\ Initial pressure, P0 = 0.30 atm
Pressure of C2H5Cl after 300 sec,
Pt = 0.30 – 0.20 = 0.10 atm
Using formula for first order reaction,
2.303  P0 
k=
log  
 Pt 
t
k=
2.303  0.30 
log 
 0.10 
300
k=
2.303
2.303 × 0.4771
log 3 =
= 3.66 × 10–3 sec–1
300
300
36. (i) Let rate of reaction
∴
d [N2O5 ]
= k [N2O5 ]n
dt
r1 = 0.10 mol L–1 hr–1 = k × (0.34 mol L–1)n
r2 = 0.20 mol L–1 hr–1 = k × (0.68 mol L–1)n
r3 = 0.40 mol L–1 hr–1 = k × (1.36 mol L–1)n
r1 0.10 mol L−1 hr −1 k (0.34 mol L−1)n
=
=
r2 0.20 mol L−1 hr −1 k (0.68 molL−1)n
or
1
=
2
 1
 
2
n
\n=1
∴
Rate = k⋅[N2O5]
0.10 mol L–1 hr–1 = k × 0.34 mol L–1
or
k=
(ii)
0.10 mol L−1 hr −1 or k = 2.9 × 10–1 hr–1
0.34 mol L−1
37. For a first order reaction
0.6932 0.6932
k=
=
= 0.0231min−1
t1/2
30
Suppose the reaction is
A
t = 0 a t = 70 min (a – x) Products
0
x
Fraction of the reactant remains unreacted =
Now, k =
(a − x )
a
2.303  a 
2.303  a 
log 
 a − x  or 0.0231 = 70 log  a − x 
t
 a  0.0231 × 70
=
log 
= 0.7021
 a − x 
2.303
or
\
a
= antilog 0.7021 = 5.036
a−x
a−x
1
=
= 0.1985 ≈ 0.2
a
5.036
38. (i) For a first order reaction,
2.303 [A ]0
k=
log
[A ]
t
When t = 30 s
2.303  0.60  2.303
=
× 0.3010 = 0.0231 s–1
log 
 0.30 
30
30
When t = 60 s
k=
2.303  0.60  2.303
=
× 0.602 = 0.0231 s–1
log 
 0.15 
60
60
As the value of k is constant at different time intervals, the
reaction is first order w.r.t. ester when [H2O] is constant.
Hence, it is pseudo first order reaction.
C − C −(0.15 − 0.30)
(ii) Average rate = − 2 1 =
t 2 − t1
60 − 30
k=
0.15
= 5 × 10 −3 mol L−1s −1
30
39. The given reaction is
A(g) →
B(g) +
C(g) +
=
Initial pressure
Po
After 20 min. Po – x
0
x
0
x
Total pressure = Po – x + x + x + x = Po + 2x
40 2x Po
Po + 2x = Po + Po ×
=
;
or Po = 5x
100
1 2.5
Po
∴ x =
5
Now,
[R]0 = Po
P
[R ] = Po − x = Po − o = 0.8 Po
5
[R ]
2.303
log10 0
Using formula, k =
[R ]
t
Po
2.303
or, k =
log10
20 min
0.8 Po
or,
k=
2.303
log 1.25
20 min
or, k =
k = 1.115 × 10–2 min–1
2.303 × 0.0969
20 min
D(g)
0
x
CBSE Board Term-II Chemistry Class-12
46
40. Suppose order with respect to A is m and with respect
to B is n. Then the rate law will be
Rate = k[A]m[B]n
Substituting the value of experiments 1 to 4, we get
Expt. 1 : Rate = 6.0 × 10–3 = k (0.1)m (0.1)n...(i)
Expt. 2 : Rate = 7.2 × 10–2 = k (0.3)m (0.2)n...(ii)
Expt. 3 : Rate = 2.88 × 10
–1
m
n
= k (0.3) (0.4) ...(iii)
–2
= k (0.4)m (0.1)n...(iv)
Expt. 4 : Rate = 2.4 × 10
Comparing equation (i) and equation (iv),
\
or,
\
(Rate)1 6.0 × 10 −3 k (0.1)m (0.1)n
=
=
(Rate) 4 2.4 × 10 −2 k (0.4)m (0.1)n
1 (0.1)m  1 
=
= 
4 (0.4)m  4 
m=1
m
Comparing equation (ii) and equation (iii)
(Rate)2 7.2 × 10 −2
k (0.3)m (0.2)n
=
=
(Rate)3 2.88 × 10 −1 k (0.3)m (0.4)n
or,
2
(0.2)n  1 
 1
=
= 
 
2
(0.4)n  2 
n
\ n=2
Rate law expression is : Rate = k[A][B]2
The rate constant can be calculated from the given data of
any experiment using expression :
Rate
k=
[A][B ]2
6.0 × 10 −3
From expt. 1, k =
= 6.0
0.1 × (0.1)2
\
Rate constant k = 6.0 mol–2 L2 min–1
dx
Unit of k,
= k [A]1[B ]2 = 6.0[0.5][0.2]2
dt
= 6 × 5 × 4 × 10–3 = 1.2 × 10–1 mol L–1 min–1
\

CHAPTER
3
Surface Chemistry
Recap Notes
Surface chemistry : It deals with
phenomena that occur at the surfaces or
interfaces.
Adsorption : It is the process of accumulation
of molecular species at the surface rather
than in the bulk of a solid or liquid.
Adsorbate : The molecular species or
substance, which concentrates or accumulates
at the surface.
Adsorbent : The material on the surface of
which the adsorption takes place.
Distinction between adsorption
and absorption
Adsorption
Absorption
It is a surface
phenomenon, i.e.,
it occurs only at
the surface of the
adsorbent.
It is a bulk
phenomenon, i.e.,
occurs throughout
the body of the
material.
In this phenomenon,
the concentration
on the surface of
adsorbent is different
from that in the
bulk.
In this
phenomenon, the
concentration is
same throughout
the material.
Its rate is high in
the beginning and
then decreases
till equilibrium is
attained.
Its rate remains
same throughout
the process.
Desorption : The process of removing an
adsorbed substance from the surface.
Sorption : The term used when both absorption
and adsorption occur simultaneously.
DG, DH and DS all are – ve for adsorption.
Types of adsorption : Depending on forces
which hold the adsorbate on the surface of
adsorbent, adsorption is divided into two
classes :
X
Physical adsorption : When the
particles are held to the surface by
the physical forces like van der Waals’
forces, the adsorption is called physical
adsorption or physisorption.
X
Chemical adsorption : When the
particles are held to the surface by the
chemical forces or by chemical bonds, the
adsorption is called chemical adsorption
or chemisorption.
Differences between physisorption and chemisorption
Property
Physisorption
Chemisorption
Enthalpy
Low enthalpy, is of the order of High enthalpy, is of the order of
80-240 kJ mol–1
20-40 kJ mol–1
Reversibility
Reversible process
Irreversible process
CBSE Board Term-II Chemistry Class-12
48
Effect of
temperature
With the increase in temperature,
extent of adsorption decreases because
adsorption is an exothermic process
and kinetic energy of gas molecules
increases with temperature.
Selectivity
Not selective in nature. Does not Highly selective in nature.
depend upon the chemical properties
of adsorbent.
Nature and
state of
adsorbate
The extent of adsorption depends The state of adsorbed molecules may
upon the ease of liquefaction of the be different from that in the bulk.
gas.
Activation
energy
No appreciable energy needed
Pressure
I n c r e a s e i n p r e s s u r e i n c r e a s e s Increase in pressure decreases
adsorption
adsorption
Layers
Multimolecular layer
Factors affecting adsorption of gases
on solids :
X Nature of adsorbent : Greater the
strained forces on the surface, more is the
ease with which adsorption takes place
on the surface. The activated adsorbents
have high adsorbing power.
X Surface area of adsorbent : Greater
X
X
X
the surface area, more is the adsorption.
Nature of gas being adsorbed : Easily
liquefiable gases like NH3, HCl, Cl2, SO2,
CO2, etc. (whose critical temperature is
high) are adsorbed to greater extent.
Pressure : At constant temperature,
adsorption increases with increase in
pressure. The effect of pressure is large at
low temperature.
Temperature : Since adsorption is
an exothermic process so according to
Le-Chatelier’s
principle
adsorption
decreases with increase in temperature.
Freundlich adsorption isotherm :
x
The plot of
vs pressure at constant
m
temperature is called Freundlich adsorption
isotherm,
where, m = mass of the adsorbent, x =
mass of the adsorbate
Chemisorption first increases with
temperature upto a certain extent and
then decreases. A gas adsorbed at low
temperature by physical adsorption
may change into chemisorption at
high temperature.
High activation energy needed
Mono-molecular layer
For low pressure,
For high pressure,
x
∝p
m
x
∝ p0
m
For intermediate pressures,
log
x
∝ p1/ n (n > 1)
m
x
1
= log k + log p
n
m
Similarly, for adsorption of solutes from
x
= k . C1 / n where, C is the
m
equilibrium concentration, i.e., when
solutions,
adsorption is complete.
Plot of log
x
vs log C is linear.
m
49
Surface Chemistry
Waals’
Properties of colloidal solutions :
X
Colligative properties : Colloids
show colligative properties like relative
lowering of vapour pressure, elevation
of boiling point, etc. and magnitude of
colligative properties of colloids is much
less than true solutions due to larger size
of colloidal particles.
X
Tyndall effect (Optical property) :
Scattering of light by colloidal particles due
to which the path of light beam becomes
visible.
X
Brownian movement (Mechanical
property):
Zig-zag
movement
of
colloidal particles due to the unbalanced
bombardment by the molecules of
dispersion medium.
X
Charge on colloidal particles : Colloidal
particles always carry an electric charge
and nature of charge (+ve or –ve) is same
on all the particles in a given colloidal
solution. The charge is due to preferential
adsorption of ions from solution.
Positively
charged sols
Hydrated metallic
oxides, e.g.,
Al2O3 ⋅ xH2O,
Fe2O3 ⋅ xH2O, metal
hydroxides, e.g.,
Fe(OH)3, Al(OH)3,
basic dye stuff
like Prussian blue,
haemoglobin (blood),
etc.
Negatively
charged sols
Metallic particles,
e.g., Cu, Ag, Au
Metal sulphides,
e.g., As2S3, CdS,
Acidic dyes like
eosin, congo red
etc, sols of gelatin,
gum, starch, etc.
Electrophoresis (Electrical property):
Movement of colloidal particles towards
one of the electrodes on passage of
electricity through colloidal solution. The
direction depends on the type of charge on
colloidal particles.
X Coagulation of colloids : Precipitation of
colloidal solution by induced aggregation
of colloidal particles.
– Lyophobic sols: They can be coagulated
by electrophoresis, boiling, persistent
dialysis, mixing of oppositely charged sols
and addition of electrolytes.
X
CBSE Board Term-II Chemistry Class-12
50
– Hardy–Schulze rules :
• In case of electrolytes, the ion carrying
charge opposite to that of colloidal particles
is effective in causing coagulation and
greater the valency of the ion causing
coagulation, greater is the coagulating power.
• The minimum concentration of an
electrolyte in millimoles per litre required
to cause precipitation of a sol in two hours
is called coagulating value. The smaller
the quantity needed, the higher will be
the coagulating power of an ion.
– Lyophilic sols : They can be coagulated
by addition of electrolytes or addition of a
suitable solvent.
Practice Time
Surface Chemistry
51
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. Which of the following statements is not
correct about physiosorption?
(a) It is a reversible process.
(b) It requires less heat of adsorption.
(c) It requires activation energy.
(d) It takes place at low temperature.
2. The term activation of adsorbent is used
when
(a) adsorbing power is increased by increasing
surface area by making the surface rough
(b) adsorbing power is increased by dipping the
surface in acid to make it smooth
(c) adsorbing power is increased by dissolving
it in water
(d) adsorbing power is decreased to reduce the
extent of adsorption.
3. When a colloidal solution is viewed from the
direction at right angles of light beam, the path
of the beam is illuminated due to scattering of
light. In the figure (A) and (B) are
Microscope
Light source
(a)
(b)
(c)
(d)
A
A
A
A
-
(A)
(B)
Colloidal solution
Tyndall cone, B - Scattered light
Scattered light, B - Tyndall cone
Tyndall cone, B - Blind spot
Tyndall effect, B - Tyndall cone
4. Which of the following statements does not
show correct difference between adsorption and
absorption?
(a) In adsorption the substance is concentrated
only at the surface while in absorption it is
uniformly distributed in the bulk.
(b) Adsorption is instantaneous while absorption
is a slow process.
(c) A substance can be adsorbed as well as
absorbed simultaneously and the process is
called sorption.
(d) Only gases are adsorbed while solids and
liquids are absorbed.
5. Which of the following is not a method for
coagulation of lyophobic sols?
(a) By electrophoresis
(b) By mixing oppositely charged sols
(c) By adding electrolyte
(d) By adding a protective colloid
6. Mixing of positively charged colloidal
solution with negatively charged colloidal
. The decreasing order
solution brings
of coagulating power of Na+, Ba2+ and Al3+ for
negatively charged colloidal solution is
.
(a) mutual coagulation, Na+ > Ba2+ > Al3+
(b) mutual coagulation, Al3+ > Ba2+ > Na+
(c) coagulation, Na+ > Ba2+ > Al3+
(d) peptization, Al3+ > Ba2+ > Na+
7. Movement of dispersion medium under the
influence of electric field is known as
(a) electrodialysis
(b) electrophoresis
(c) electroosmosis
(d) cataphoresis.
8. Why is alum added to water containing
suspended impurities?
(a) To make a colloidal solution.
(b) To coagulate the suspended impurities.
(c) To remove impurities of calcium and
magnesium.
(d) To protect the colloidal solution from getting
precipitated.
9. The cause of Brownian movement which is
not shown by true solutions or suspensions is
due to
(a) unbalanced bombardment of particles by
molecules of the dispersion medium
CBSE Board Term-II Chemistry Class-12
52
10. Which of the following is less than zero
during adsorption?
(a) DG
(b) DS
(c) DH
(d) All of these.
11. Substances which behave as normal
electrolytes in solution at low concentration
and exhibit colloidal properties at higher
concentration are called
(a) lyophilic colloids
(b) lyophobic colloids
(c) macromolecular colloids
(d) associated colloids.
12. Which of the plots is adsorption isobar for
chemisorption?
(a)
(b)
(c)
(d)
13. A colloidal system in which liquid is
dispersed phase and solid is dispersion medium
is classified as
(a) gel
(b) sol
(c) emulsion
(d) aerosol.
14. Which of the following gases is least adsorbed
on charcoal?
(a) HCl
(b) NH3
(c) O2
(d) CO2
15. What is the role of activated charcoal in gas
masks used in mines?
(a) It acts as an adsorbent for poisonous gases
present in coal mines.
(b) It acts as an adsorbent for coal particles
present in coal mines.
(c) It acts as a mask through which exhaled
gases are diffused out.
(d) It acts as a base for scattering the light.
16. Which of the following statements is not
correct for chemisorption and physisorption?
(a) Physisorption and chemisorption are both
exothermic processes.
(b) Magnitude of chemisorption is favourable
at low temperature while physisorption is
favourable at high temperature.
(c) C h e m i s o r p t i o n i s i r r e v e r s i b l e a n d
physisorption is reversible.
(d) In physisorption activation energy is low
while in chemisorption it is high.
17. The combination of two layers of opposite
charges around the colloidal particles is called
Helmholtz electrical double layer. The potential
difference between the fixed layer and the
diffused layer of opposite charges is called
(a) electrode potential (b) zeta potential
(c) adsorption potential (d) diffused potential.
18. A lyophobic colloid cannot be formed by
(a) mixing dispersed phase and dispersion
medium
(b) chemical reactions like hydrolysis
(c) exchange of solvent
(d) peptisation.
19. The substances which behave as normal
electrolytes at low concentration but undergo
association at higher concentration and behave
as colloidal solutions are called
(a) associated colloids
(b) multimolecular colloids
(c) macromolecular colloids
(d) protective colloids.
20. Which of the processes is being shown in the
figure?
Water + Electrolyte
+
–
+
–
Sol particles
Water
Membrane
(a) Electrodialysis
(c) Electroosmosis
(b) Dialysis
(d) Electrophoresis
21. A graph is plotted between log (x/m) and log
P according to the equation
log (x/m)
(b) attractive forces between dispersed phase
and dispersion medium
(c) larger size of the particles due to which they
keep colliding and settling down
(d) conversion currents formed in the sol.
0
}
x
= kP1/ n
m
b Slope = 1/n
a
log K (intercept)
logP
Which of the following statements about this
graph is not correct?
53
Surface Chemistry
(a) The figure shows Freundlich adsorption
isotherm.
(b) The figure shows Langmuir adsorption
isotherm.
(c) The adsorption varies directly with pressure.
(d) The factor 1/n can have values between 0
and 1.
22. Lyophilic sols are also called reversible
colloids because
(a) they can be reformed by mixing residue
(dispersed phase) in dispersion medium
even after drying
(b) they can be easily precipitated from the
colloidal system
(c) once formed, the dispersion medium and
dispersed phase cannot be separated
(d) special reversible reactions are used to
prepare them.
23. Which of the following gases present in a
polluted area will be adsorbed most easily on
the charcoal gas mask?
(b) O3
(a) H2
(c) N2
(d) SO2
24. Fe(OH)3 sol can be more easily coagulated
by Na3PO4 in comparison to KCl because
(a) mass of Na3PO4 is more than KCl hence it
is more effective than KCl
(b) phosphate ion (PO43–) has higher negative
charge than Cl– ion hence are more effective
for coagulation
(c) KCl is more soluble than Na3PO4 hence less
effective for coagulation
(d) Na+ ions are more effective than K+ ions for
coagulation.
(b) Adsorption of pine oil on sulphide ore
particles.
(c) Adsorption of pine oil on impurities.
(d) Production of heat in the process of
exothermic reaction.
27. When an excess of a very dilute aqueous
solution of KI is added to a very dilute aqueous
solution of silver nitrate, the colloidal particles
of silver iodide are associated with which of the
following Helmholtz double layer?
(b) AgI / K+ NO3–
(a) AgI / Ag+ I–
–
(c) AgI / NO3– Ag+
(d) AgI / I K+
28. Which of the following will not form a
colloidal system?
(a) Solid-gas
(b) Liquid-gas
(c) Gas-gas
(d) Gas-liquid
29. Which of the following factors contribute
towards higher stability of lyophilic colloid?
(a) Charge on their particles.
(b) Attractive forces between particles.
(c) Small size of their particles.
(d) High solvation due to a layer of dispersion
medium.
30. Which of the following examples is correctly
matched?
(a) Butter – gel
(b) Smoke – emulsion
(c) Paint – foam
(d) Milk – aerosol
31. In Bredig’s arc method an electric arc is
struck between the metal electrodes under the
surface of water containing some stabilizing
agent. The process involves
Metal rods
25. Colloidal solutions of metals like gold can
be prepared when their salt solutions react with
certain substances like SnCl 2, formaldehyde,
phenyl hydrazine, etc.
2AuCl3 + 3SnCl2 → 3SnCl4 + 2Au
sol
The above method is an example of
(a) reduction method
(b) oxidation method
(c) hydrolysis method
(d) double decomposition method.
(a)
(b)
(c)
(d)
26. What is the role of adsorption in froth
floatation process used especially for
concentration of sulphide ores?
(a) Shape selective catalysts.
33. In these colloids, a large number of small
atoms or smaller molecules of a substance
aggregate to form colloidal particles having size
in colloidal range. These colloids are known as
Water + KOH
Ice arc
mechanical dispersion
condensation
both dispersion and condensation
ultrasonic dispersion.
32. Fog is an example of colloidal system of
(a) liquid in gas
(b) gas in liquid
(c) solid in gas
(d) gas in solid.
CBSE Board Term-II Chemistry Class-12
54
(a) multimolecular colloids
(b) macromolecular colloids
(c) associated colloids (d) lyophilic colloids.
(b)
log x/m
(d)
log x/m
(c)
log x/m
(a)
log x/m
34. Which of the following curves is in accordance
with Freundlich adsorption isotherm?
35. Which of the following acts as the best
coagulating agent for ferric hydroxide sol?
(a) Potassium ferrocyanide
(b) Potassium chloride
(c) Potassium oxalate (d) Aluminium chloride
36. Which of the following is not characteristic
of chemisorption?
(a) Adsorption is specific.
(b) Heat of adsorption is of the order of
200 kJ mol–1.
(c) Adsorption is irreversible.
(d) Adsorption may be multimolecular layers.
37. Why is ferric hydroxide colloid positively
charged when prepared by adding ferric chloride
to hot water?
(a) Due to precipitation of ferric hydroxide there
is an excess of Fe3+ ions.
(b) Due to preferential adsorption of Fe3+ ions
by the sol of Fe(OH)3.
(c) Due to absence of any negatively charged ion.
(d) Due to adsorption of OH– and Cl– ions, the
remaining sol has only Fe3+ ions.
38. Which out of the following electrolyte
solutions having the same concentration will
be most effective in causing the coagulation of
arsenic sulphide sol?
(a) KCl
(b) MgCl2
(c) AlCl3
(d) Na3PO4
39. At CMC (critical micelle concentration) the
surface molecules
(a) dissociate
(b) associate
(c) become bigger in size due to adsorption
(d) become smaller in size due to decomposition.
40. Tyndall effect is observed only when which
of the following conditions are satisfied?
(i) The diameter of the dispersed particles is
not much smaller than the wavelength of
the light used.
(ii) The refractive indices of dispersed phase
and dispersion medium differ greatly in
magnitude.
(iii) The size of the particles is generally between
10–11 and 10–9 m in diameter.
(iv) The dispersed phase and dispersion phase
can be seen separately in the system.
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (i) and (ii)
41. Which of the following
removing impurities from a
(a) Electrodialysis
(b)
(c) Ultra centrifugation (d)
is not a method of
colloidal sol?
Ultrafiltration
Distillation
42. In the given figure label the parts.
(A)
(Stearate ion)
–
A
A
A
A
–
+ Na+
(Sodium ion)
– is COO
where
(a)
(b)
(c)
(d)
(B)
-
Hydrophilic tail, B - hydrophobic head
Hydrophobic tail, B - hydrophobic head
Hydrophobic tail, B - hydrophilic head
Hydrophilic tail, B - hydrophilic head
43. After reading adsorption thoroughly Shubh
plotted these 4 curves :
The given graphs/data I, II, III and IV
represent general trends observed for different
physisorption and chemisorption processes
under mild conditions of temperature and
pressure. Which of the following choice(s) about
I, II, III and IV is(are) correct?
55
Surface Chemistry
Next day he showed these plots to his friend and
asked about the plots and explained.
(i) Plot I represents physisorption because
physisorption decrease with increase in
temperature.
(ii) Plot II represents chemisorption as it
initially increases upto certain extent and
then decreases regularly.
(iii) Plot III represents physiosorption because
it decreases with increase in both P & T.
(iv) Plot IV represents chemisorption because
activation energy for it is of the order
80 – 240 kJ/mol.
The incorrect observations or explanations are
(a) (i) & (iii)
(b) (ii) & (iii)
(c) (i) & (iv)
(d) (iii) & (iv)
46. Colloidal solutions when prepared, generally
contain excessive amount of electrolytes and
some other soluble impurities. These soluble
electrolytes if present in larger amount can
coagulate the sol. So it is necessary to reduce
their concentration to minimum amount.
Method I is used for purification and it can be
converted to method II also.
Water +
crystalloid
Sol particle
Water
Crystalloid
Method I
44. A plot between log x/m (almost adsorbed)
and log P has been plotted and it is shown below:
Dialysing
membrane
Water
+
Electrolyte
Cathode
log x/m
0.4
Anode
0.3
Water
0.2
Sol particle
0.1
Crystalloid
Method II
0
Advantage of method II over method I is
(a) it uses less energy
(b) it is cost friendly
(c) it speeds up the purification
(d) none of these.
0.2 0.4 0.6 0.8
log P
Value of n is
(a) 2
(c) 1
(b) 0.5
(d) 1.5
45. Colloids can be classified as
Dispersed phase
47. Observe the given figure:
Eye
Liquid
Solid
Gas
Solid Liquid Gas Solid Liquid Gas
Solid
sol
Sol
P
Gel Emulsion Q
Solid
Liquid
Solid
R
sol
Choose the correct statement.
(a) Examples of P are cheese and jellies.
(b) Examples of Q are pumice stone and foam rubber.
(c) Examples of R are froth, whipped cream and
soap lather.
(d) P is aerosol, Q is gel and R is foam.
Light
source
Microscope
Scattered
light
Tyndall
cone
Solution
Sanjeev performed this experiment with
different solutions listed here :
Sugar solution (I), salt solution (II), ink (III),
milk (IV), glucose solution (V), muddy water (VI),
slaked lime (VII).
Tyndall cone can not be seen in
(a) Only I, II and VII (b) I, IV and VIII
(c) II, III and IV
(d) I, II and V
CBSE Board Term-II Chemistry Class-12
56
Case Based MCQs
Case I : Read the passage given below and
answer the following questions from 48 to 52.
A graph between the amount adsorbed (x/m)
by an adsorbent and the equilibrium pressure
of the adsorbate at a constant temperature is
called the adsorption isotherm.
A relationship between the amount adsorbed
(x/m) and the equilibrium pressure (P) can be
obtained as follows :
0
In the intermediate range of pressure, x/m = kP1/n
(was originally put forward by Freundlich and
is known as Freundlich adsorption isotherm).
48. According to Freundlich adsorption isotherm,
which of the following is correct?
x
(a)
∝ P0
m
x
∝ P1
(b)
m
(c) x ∝ P1/n
m
(d) All the above are correct for different range
of pressure.
49. In the Freundlich adsorption isotherm
equation
x
1
log
= log k + log p, the value of 1/n is
m
n
amount of solute adsorbed per gram of adsorbent
will be (log 5 = 0.6990)
(a) 1 g
(b) 2 g
(c) 3 g
(d) 5 g
x
vs log p for an adsorption,
m
a straight line inclined at an angle of q = 14.04°
to the x-axis was obtained. The ‘n’ value for this
adsorption process is (tan 14.04° = 0.25)
(a) 5
(b) 8
(c) 4
(d) 2
51. In the plot of log
52. In the adsorption of a gas on solid, Freundlich
isotherm is obeyed. The slope of the plot is zero.
Then the extent of adsorption is
(a) directly proportional to the pressure of the
gas
(b) inversely proportional to the pressure of the
gas
(c) directly proportional to the square root of
the pressure of the gas
(d) independent of the pressure of the gas.
Case II : Read the passage given below and
answer the following questions from 53 to 57.
Adsorption is a spontaneous process and
involves unequal distribution of the molecules
of the gaseous substance on the surface of solid
or liquid. Adsorption is an exothermic process.
The attractive forces between adsorbate and
adsorbent are either van der Waals’ forces or
chemical bonds. Adsorption of gases on solids
is generally controlled by the factors like
temperature, pressure and nature of adsorbate
and adsorbent.
any value from 0 to 1
a negative integer
a positive integer
a positive or a negative fractional number.
53. In physisorption process, the attractive
forces between adsorbate and adsorbent are
(a) covalent bonds
(b) ionic bonds
(c) van der Waals’ forces
(d) H-bonds.
50. Plot of log x/m against log p is a straight line
inclined at an angle of 45°. When the pressure is
0.5 atm and Freundlich parameter, k is 10, the
54. Which of the following graph represents
the variation of physical adsorption with
temperature?
(a)
(b)
(c)
(d)
57
Surface Chemistry
x
(b) m
(a) x
m
t
t
x
(d) m
x
(c) m
t
t
55.
not
(a)
(b)
(c)
(d)
Which one of the following processes does
use adsorption?
Froth floatation process
Chromatography
Decolourisation of sugar liquors
Dissolution of sugar in water
56.
(a)
(b)
(c)
(d)
Which of the following statements is true?
Chemisorption forms unimolecular layer.
Chemisorption is a reversible process.
Chemisorption is independent of pressure.
Chemisorption has low enthalpy change.
57. Methylene blue, from its aqueous solution,
is adsorbed on activated charcoal at 25°C. For
this process, the correct statement is
(a) the adsorption requires activation at 25°C
(b) the adsorption is accompanied by a decrease
in enthalpy
(c) the adsorption increases with increase of
temperature
(d) the adsorption is irreversible.
Case III : Read the passage given below and
answer the following questions from 58 to 62.
Hardy Schulze rule states that the precipitating
effect of an ion on dispersed phase of opposite
charge increases with the valency of the ion.
The higher the valency of the flocculating
ion, the greater is its precipitating power.
Thus, for the precipitation of As 2 S 3 sol
(–ve sol) the precipitating power of Al 3+, Ba2+
and Na+ ions is of the order, Al3+ > Ba2+ > Na+.
Similarly, for precipitating Fe(OH) 3 sol
(+ve sol) the precipitating power of [Fe(CN)6]3–,
SO 42– and Cl – is of the order, [Fe(CN) 6 ] 3– >
SO42– > Cl–. The minimum concentration of an
electrolyte in millimoles per litre required to
cause precipitation of a sol in 2 hours is called
flocculation value. The smaller the flocculation
value, the higher will be the coagulating power
of the ion. The minimum mass of the protective
colloid (lyophilic colloid) in milligrams that must
be added to 10 mL of a standard red gold sol so
that no coagulation occurs when 1 mL of 10%
NaCl solution is rapidly added to it is called the
gold number of the protective colloid.
58. The gold number of four protective colloids
A, B, C and D are 0.03, 0.003, 10 and 30
respectively. Protective power of these colloids
will be of the order :
(a) B > A > C > D
(b) A > B > C > D
(c) C > B > D > A
(d) D > A > C > B
59. Which of the following has least flocculating
value for positive sol?
(b) SO42–
(a) Cl–
3–
(c) PO4
(d) [Fe(CN)6]4–
60. Which of the following colloidal solutions is
positively charged?
(a) TiO2
(b) As2S3
(c) Starch sol
(d) Gold sol
61. The coagulation value in millimoles per litre
of electrolytes used for the coagulation of As2S3
are as below :
I. NaCl = 52
II. KCl = 50
III. BaCl2 = 0.69
IV. MgSO4 = 0.72
The correct order of their flocculating power is
(a) I > II > III > IV
(b) I > II > IV > III
(c) I < II < IV < III
(d) IV < I < II < III
62.
(a)
(b)
(c)
(d)
1 mol of AgI/Ag+ is coagulated by
1 mol of KI
200 mL of 1 M K2SO4
300 mL of 1 M Na3PO4
2 mol of AgI
Case IV : Read the passage given below and
answer the following questions from 63 to 67.
Adsorption depends on the nature of the
adsorbent. The rough solid surface has more
number of pores and adsorb more number of
gases than the smooth surface. Most common
adsorbents are silica gel, activated charcoal.
The extent of adsorption also depends on the
surface area of the solid. Specific surface area
of an adsorbent is the surface area available
for adsorption per gram of the adsorbent. The
greater the surface area of the solid, the greater
would be the adsorption. Charcoal is a more
CBSE Board Term-II Chemistry Class-12
58
effective adsorbent than solid wood. Desorption
is a process of removing an adsorbed substance
from a surface on which it is absorbed.
Physisorption is non-specific and any gas can
be adsorbed. But the gases which are easily
liquefiable (e.g., NH3, HCl, CO2) are adsorbed
at a faster rate and to a large extent than the
gases which are difficult to liquefy (e.g., H 2,
O2, N2). It depends on the critical temperature.
Higher the critical temperature of a gas, more
easily liquefiable the gas is and more is the
rate of adsorption. Chemisorption is specific
in nature. Therefore, only those gases can be
adsorbed which are capable of forming chemical
bonds with the adsorbent.
63. Select the correct statement regarding
desorption of gases on solid.
(a) It is done by cooling or by increasing the
pressure applied.
(b) It is done by cooling or by reducing the
pressure applied.
(c) It is done by heating or by reducing the
pressure applied.
(d) It is done by heating or by increasing the
pressure applied.
64. Which of the following statements regarding
the physical adsorption of a gas on surface of
solid is not correct?
(a) On increasing temperature, adsorption
increases continuously.
(b) Enthalpy changes are negative.
(c) It is non-specific in nature.
(d) It is reversible in nature.
65. At the same temperature and pressure,
select the correct order of adsorption of the
following gases on the same mass of charcoal.
(a) SO2 > CH4 > H2
(b) CH4 < SO2 < H2
(c) H2 > CH4 > SO2
(d) CH4 < H2 < SO2
66. Select the correct option among the following
when adsorption of a gas on solid metal surface
is spontaneous and exothermic.
(a) DS increases.
(b) DS decreases.
(c) DG increases.
(d) DH increases.
67. Select the incorrect statement among the
following.
(a) Physical adsorption is favourable at low
temperature and chemisorption is favourable
at high temperature.
(b) In physisorption heat of adsorption
lies between 20-40 kJ mol –1 while in
chemisorption it lies between 80-240 kJ mol–1.
(c) C h e m i s o r p t i o n i s i r r e v e r s i b l e a n d
physisorption is reversible.
(d) Magnitude of chemisorption decreases with
rise in temperature while physisorption
increases with rise in temperature.
Assertion & Reasoning Based MCQs
For question numbers 68-80, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
68. Assertion : A colloidal sol of As 2 S 3 is
coagulated faster by 0.1 M BaCl 2 than by
0.1 M NaCl.
Reason : BaCl2 gives double the number of Cl–
ions than NaCl.
69. Assertion : According to Freundlich,
x
= k ⋅ P1/ n
m
Reason : The isotherm shows variation of the
amount of gas adsorbed by the adsorbent with
temperature.
70. Assertion : There is no interface between
gases.
Reason : The shape and volume of gases are
not definite.
59
Surface Chemistry
71. Assertion : Soap and detergent are
macromolecular colloids.
Reason : Macromolecular colloids are formed
by molecules of large size.
72. Assertion : Lyophilic colloids are called
reversible sols.
Reason : Lyophilic sols are liquid loving.
73. Assertion : Colloidal particles show
Brownian movement.
Reason : Brownian movement arises because
of the impact of the molecules of the dispersion
medium with the colloidal particles.
74. Assertion : Porous or finely divided forms of
adsorbents adsorb larger quantities of adsorbate.
Reason : The greater the surface area of the
solid, the greater would be its adsorbing capacity.
75. Assertion : NH3 is adsorbed more readily
than O2 on charcoal.
Reason : More easily liquefiable gases are
adsorbed easily.
76. Assertion : The molecules on the surface
have lesser energy than the molecules inside.
Reason : During adsorption, the surface of solid
is in a state of strain.
77. Assertion : Muddy water is an example of
sol.
Reason : A colloidal system in which solid is
dispersed in a liquid is called sol.
78. Assertion : Gold number is a measure
of protective action by a lyophilic colloid on a
lyophobic colloid.
Reason : Zeta potential (or electrokinetic
potential) is the potential difference between
fixed charged layer and the diffused layer having
opposite charge.
79. Assertion : When FeCl 3 is added to an
excess of hot water, a positively charged sol of
hydrated ferric oxide is formed.
Reason : When ferric chloride is added to NaOH
solution a negatively charged sol is obtained due
to adsorption of OH– ions.
80. Assertion : A colloidal sol scatters light but
a true solution does not.
Reason : The particles in a colloidal sol move
slowly than in a true solution.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1.
Explain the following :
Artificial rain is caused by spraying salt over
clouds.
2.
Physisorption is multi-layered, while
chemisorption is mono-layered. Explain.
3.
Based on the type of dispersed phase, what
type of colloid is micelles?
4.
A colloidal sol is prepared
by the given method in figure.
What is the charge on hydrated
ferric oxide colloidal particles
formed in the test tube? How
is the sol represented?
5. Give reasons for the following observations:
A delta is formed at the meeting point of sea
water and river water.
6. Out of NH 3 and CO 2 , which gas will be
adsorbed more readily on the surface of activated
charcoal and why?
7. Name of the temperature above which the
formation of micelles takes place.
8. Out of BaCl2 and KCl, which one is more
effective in causing coagulation of a negatively
charged colloidal sol? Give reason.
9. Why are medicines more effective in colloidal
state?
10. How can a colloidal solution and true solution
of the same colour be distinguished from each
other.
CBSE Board Term-II Chemistry Class-12
60
Short Answer Type Questions (SA-I)
(i) NH3 gas absorbs more readily than N2 gas
on the surface of charcoal.
16. (i) Out of MgCl2 and AlCl3, which one is more
effective in causing coagulation of negatively
charged sol and why?
(ii) Powdered substances are more effective
adsorbents.
(ii) Out of sulphur sol and proteins, which one
forms multimolecular colloids?
12. What happens when
17. (i) Same substances can act both as colloids
and crystalloids. Explain.
(ii) What will be the charge on AgI colloidal sol
when it is prepared by adding small amount of
AgNO3 solution to KI solution in water? What is
responsible for the development of this charge?
18. Write the differences between physisorption
and chemisorption with respect to the following:
(i) Specificity
(ii) Temperature
dependence
(iii) Reversibility and
(iv) Enthalpy change
19. Write one difference in each of the following:
(a) Multimolecular colloid and associated colloid
(b) Coagulation and peptization
20. Pressure of a closed vessel filled with gas
decreases when powdered charcoal is added.
Explain.
11. Give reasons for the following observations :
(a) a freshly prepared precipitate of Fe(OH)3 is
shaken with a small amount of FeCl3 solution?
(b) persistent dialysis of a colloidal solution is
carried out?
13. What is meant by coagulation of a colloidal
solution? Name any method by which coagulation
of lyophobic sols can be carried out.
14. (i) What is the role of activated charcoal in
gas mask?
(ii) How does chemisorption vary with temperature?
15. Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic
sol.
Short Answer Type Questions (SA-II)
21. Define the following terms :
(i) Lyophilic colloid
(ii) Zeta potential
(iii) Associated colloids
22. (i) Of physisorption and chemisorption,
which has a higher enthalpy of adsorption?
(ii) P h y s i s o r p t i o n i s r e v e r s i b l e w h i l e
chemisorption is irreversible. Why ?
(iii) What type of forces are responsible for the
occurrence of physisorption?
23.
(a)
the
(b)
(c)
Give reasons for the following :
Brownian movement provides stability to
colloidal solution.
True solution does not show Tyndall effect.
Addition of alum purifies water.
24. Distinguish between multimolecular, macromolecular and associated colloids. Give one
example of each.
25. Giving appropriate examples, explain
how the two types of processes of adsorption
(physisorption and chemisorption) are influenced
by the prevailing temperature, the surface area
of adsorbent and the activation energy of the
process?
26. Explain what is observed when :
(i) A beam of light is passed through a colloidal
solution.
(ii) NaCl solution is added to hydrated ferric
oxide sol.
(iii) Electric current is passed through a colloidal
solution.
27. (i) How are the following colloidal solutions
prepared?
(a) Sulphur in water
(b) Gold sol
(ii) Why is adsorption always exothermic?
61
Surface Chemistry
28. Classify colloids where the dispersion medium
is water. State their characteristics and write an
example of each of these classes.
29. Ranju is using normal water for washing
clothes. She observed that her clothes were
not getting very clean although she is using
more amount of soaps or detergents. Her friend
Swarna advised Ranju washing clothes in warm
water. Ranju was surprised to see that washing
of clothes with soaps or detergents is easier in
luke warm water than cold water.
Now answer the following questions:
(i) What are the processes involved in washing
of clothes?
(ii) Why washing of clothes using soap or
detergent is easier in warm water?
30. (i) Explain the following terms giving one
example for each.
(a) Micelles
(b) Aerosol
(ii) Write one similarity between physisorption
and chemisorption.
31. What is meant by coagulation of a colloidal
solution? Describe briefly any three methods
by which coagulation of lyophobic sols can be
carried out.
32. (i) Define the following terms giving an
example: Hydrosol
(ii) Which complex ion is formed when undecomposed
AgBr is washed with hypo solution in photography?
33. (i) What are protective colloids? Which
type of colloids are used as protective colloids?
(ii) Why does sky look blue?
34. (i) Write the dispersed phase and dispersion
medium of the following colloidal systems :
(a) Smoke
(b) Milk
(ii) In reference to Freundlich adsorption
isotherm write the expression for adsorption of
gases on solids in the form of an equation.
35. (i) What is the principle of separation of
inert gases from its mixture?
(ii) Why silica and alumina gels are used for
removing moisture and controlling humidity?
(iii) How does adsorption of a gas on a solid
surface vary with temperature?
Long Answer Type Questions (LA)
36. (a) Give reason, why a finely divided
substance is more effective as an adsorbent?
(b) Physical and chemical adsorptions respond
differently to rise in temperature. What is this
difference and why is it so?
(c) The volume of nitrogen gas at 0°C and
1.013 bar required to cover a sample of silica
gel with unimolecular layer is 129 cm 3 g –1
of gel. Calculate the surface area per gram
of the gel if each nitrogen molecule occupies
16.2 × 10–20 m2.
37. What is an adsorption isotherm? Describe
Freundlich adsorption isotherm.
38. (a) Assuming adsorption to be a spontaneous
process, show thermodynamically that it is
always an exothermic process.
(b) Why are all adsorption processes exothermic?
(c) How is the adsorption of a gas related to its
critical temperature?
(d) A small amount of silica gel and a small
amount of anhydrous calcium chloride are placed
separately in two corners of a vessel containing
water vapour. What phenomenon will occur?
39. (i) Explain why excessive dialysis should
be avoided for purification of a colloid?
(ii) What is the difference between dialysis and
ultrafiltration?
40. (i) 1 g of charcoal adsorbs 100 mL of 0.5 M
CH3COOH to form a monolayer, and thereby the
molarity of CH3COOH reduces to 0.49 M. Calculate
the surface area of the charcoal adsorbed by each
molecule of acetic acid. Surface area of charcoal
= 3.01 × 102 m2.
(ii) A solution of palmitic acid (M = 256g) in
benzene contains 4.24 g acid per litre. When
this solution is dropped on the water surface,
benzene evaporates and palmitic acid forms
monomolecular film of the solid type. If we wish
to cover an area of 500 cm2 with a monolayer,
what volume of solution should be used? The
area occupied by one palmitic acid molecule may
be taken to be 21 × 10–20 m2.
CBSE Board Term-II Chemistry Class-12
62
OBJECTIVE TYPE QUESTIONS
1. (c) : Physisorption does not require activation energy
since it takes place at low temperature.
2. (a) : Activation of adsorbent means increasing the
adsorbing power by making adsorbent’s surface rough by
subdividing it into smaller pieces or removing gases already
adsorbed on it.
3. (a) : The bright cone of the light is known as Tyndall
cone. The scattering of light is seen by the microscope.
4. (d) : Gases, liquids or solids can be adsorbed on the solid
surfaces.
5. (d) : Addition of protective colloid is a method of
prevention of coagulation.
17. (b) : The first layer of ions is held firmly and is termed as
fixed layer while the second layer is mobile or diffused layer.
The potential difference between fixed and diffused layer is
called electrokinetic potential or zeta potential.
18. (a) : Lyophobic colloids cannot be prepared by simple
mixing of dispersed phase and dispersion medium.
19. (a) : Substances whose molecules aggregate
spontaneously in a given solvent to form particles of colloidal
dimensions are called associated colloids or micelles.
20. (a) : When an electric field is applied to purify an impure
colloidal solution, the process is known as electrodialysis.
The ions present in the colloidal solution migrate out to the
oppositely charged electrodes.
21. (b) : This is Freundlich isotherm and not Langmuir.
6. (b) : According to Hardy Schulze rule, the coagulating
power of an ion depends upon its valency. Higher the valency
of ion, greater is its coagulating power.
22. (a) : If the dispersed phase is removed completely
from the colloidal system, it can be formed again by mixing
dispersion medium with it.
7. (c) : When movement of particles (electrophoresis) is
prevented by some suitable means, it is observed that the
dispersion medium begins to move in an electric field. This
phenomenon is termed as electroosmosis.
23. (d) : Easily liquefiable gases like CO2, NH3, SO2, etc. are
more easily adsorbed than the elemental gases like H2, N2,
O2, etc.
8. (b) : The water obtained from natural sources often
contains suspended impurities. Alum is added to such water
to coagulate the suspended impurities and make water fit for
drinking purposes.
9. (a) : The colloidal particles are in a continuous zig-zag
motion due to unbalanced bombardment of the particles by
molecules of the dispersion medium.
10. (d) : Adsorption is a spontaneous, exothermic process
with decrease in entropy, hence DG = –ve, DH = –ve and
DS = –ve.
11. (d) : At higher concentration the aggregated particles
called micelles are formed by electrolytes like soap which act
as colloidal particles.
12. (b) : The extent of adsorption first increases and then
decreases with increase in temperature.
13. (a) : Gel is a colloidal system in which liquid is dispersed
in a solid.
14. (c) : It has been found that more readily liquefiable
gases are adsorbed more than permanent gases.
15. (a) : The poisonous gases present in coal mines are
adsorbed on the activated charcoal.
16. (b) : Chemisorption is favoured at high temperature and
physisorption is favoured at low temperature.
24. (b) : Ferric hydroxide is a positively charged sol hence
ions carrying negative charge can coagulate it. Since PO3–
4 has
higher negative charge than Cl– hence it is more effective for
coagulation.
25. (a) : Salt solution of gold is being reduced to gold by
using a reducing agent like SnCl2.
26. (b) : Pine oil is adsorbed on sulphide ore particles
resulting in formation of emulsion and froth.
27. (d) : As excess of KI has been added, I – ions are adsorbed
on AgI forming a fixed layer (and giving it a negative charge).
It then attracts the counter ions (K+) from the medium forming
a second layer (diffused layer).
28. (c) : Gas-gas is a true solution.
29. (d) : Lyophilic colloids are highly solvated hence more
stable.
30. (a) : In gel, liquid is dispersed in solid.
31. (c) : The method involves both dispersion and
condensation. The intense heat of arc vapourises some of
the metal which condenses under cold water.
32. (a) : Liquid (water droplets) are dispersed in gas (air) in
fog.
33. (a) : In multimolecular colloids the smaller particles
aggregate and are held together by van der Waals’ forces.
e.g. sols of gold atoms and sulphur molecules.
63
Surface Chemistry
34. (c)
35. (a) : The efficiency of coagulation of an electrolyte
depends upon its valency. Thus [Fe(CN) 6] 4– is the best
coagulating agent for Fe(OH)3 sol.
36. (d) : Chemisorption is unimolecular or in one layer.
37. (b) : The adsorption of positively charged Fe3+ ions by
the sol of hydrated ferric oxide results in positively charged
colloid.
38. (c) : As 2S 3 is a negatively charged sol. To cause its
coagulation, the ions must be positively charged. Greater the
magnitude of positive charge, greater will be its coagulating
power. Thus AlCl3 containing Al3+ ion will be most effective
in causing coagulation of As2S3.
39. (b) : At CMC, the particles of an electrolyte aggregate
and form associated colloids known as micelles.
40. (d) : Tyndall effect is observed only when these two
conditions are satisfied.
41. (d) : Distillation cannot be used to remove impurities
from colloidal sol.
42. (c) :
The RCOO– ion formed in the water contains two parts, a long
hydrocarbon chain R (non-polar or hydrophobic tail) and a
polar group COO– (polar or hydrophilic head).
43. (d) : In physisorption as temperature increases, the extent
of adsorption decreases thus, I and III are physisorptions.
Chemical adsorption first increases with increase in temperature
upto a certain extent and then decreases regularly thus II is
chemisorption. In chemisorption, attractive forces between
adsorbent and adsorbate molecules are strong chemical
bonds, thus enthalpy of adsorption is high and of the order
80-240 kJ/mol. As in IV DH adsorption = 150 kJ/mol, thus it also
represents chemical adsorption.
x
= kp1/n
44. (b) :
m
x
1
log = log k + log p
m
n
1
Here = slope
n
0.2 − 0.1 2
Slope =
=
0.4 − 0.2 4
2
n = = 0.5
4
45. (c) : Example of P (aerosol) are smoke, dust. Example of
Q (aerosol) are fog, mist, cloud, etc.
46. (c) : (a) Method I is dialysis and Method II is electrodialysis
which is faster than dialysis.
47. (d) : Tyndall effect can be observed in colloidal solution
and suspension.
x
48. (d) : At low pressure, ∝ P
m
x
At high pressure, ∝ P 0
m
In the intermediate range of pressure,
49. (a)
1
x
∝P n
m
50. (d) : According to Freundlich equation
1
x 
log   = log k + log P
m 
n
\ Plot of log x/m vs log P is linear with slope = 1/n and
intercept = log k.
1
Thus, = tan θ = tan 45° = 1 or n = 1
n
At P = 0.5 atm and k = 10
x
= kP 1/n
m
x
= 10 × (0.5)1 = 5 ∴ x = 5 g
m
x
1
51. (c) : Plot of log vs log P is a straight line with slope =
m
n
Also, slope = tan q = tan 14.04° = 0.25
1
= 0.25 or n = 4
⇒
n
x
52. (d) :
= kP 1/ n
m
x
1
1
ln = ln k + lnP ; slope = = 0
m
n
n
x
0
Thus, = kP
m
53. (c) : In physisorption process, the attractive forces
between adsorbate and adsorbent are van der Waals’ forces.
54. (a)
55. (d)
56. (a)
57. (b) : The adsorption of methylene blue on activated
charcoal is physical adsorption. It is accompanied by a
decrease in enthalpy.
58. (a) : Lesser is the gold number, greater is the protective
power.
59. (d) : Charge of [Fe(CN)6]4– is highest, hence, it will be
most effective for the coagulation of positive colloids. More
is the coagulating power lesser will be the flocculating value.
60. (a)
61. (c) : Coagulation value is inversely proportional to their
flocculating power.
CBSE Board Term-II Chemistry Class-12
64
62. (a) : 1 mol Ag+ will combine with 1 mol I– to form
precipitate of 1 mol AgI.
63. (c) : Desorption is done by heating or by reducing the
pressure applied.
64. (a) : Physisorption is exothermic in nature. Therefore,
according to Le Chatelier’s principle, it occurs readily at low
temperature and decreases with increase in temperature.
Bonds between surface and adsorbate are weak so when
temperature is increased the bonds break easily, so rate will
decrease on increasing temperature.
65. (a) : Higher the critical temperature of a gas, greater is
the amount of gas adsorbed. Critical temperature (in Kelvin)
of the gases :
H2 < CH4 < SO2
33.2 190.6 430.3
66. (b) : Since for spontaneous and exothermic process,
DG = –ve, DH = –ve at all temperatures, therefore from
DG = DH – TDS, DS should be –ve. Also adsorption of gas
on solid surface gives more orderly arrangement.
67. (d) : Chemisorption first increases with increase of
temperature. Physisorption decreases with rise in temperature.
68. (b) : As2S3 being negatively charged is coagulated faster
by Ba2+ ions than by Na+ ions, according to Hardy Schulze
rule, which states that greater the valency of the flocculating
ion added, greater is its power to cause precipitation.
76. (d) : The molecules on the surface have higher energy
than those inside. The surface of a solid or a liquid is in a state
of strain or tension on account of the unbalanced or residual
forces.
77. (a)
78. (b) : Gold number gives a comparative idea of protective
power of various lyophilic colloids on a lyophobic colloid. Zeta
potential is the potential difference between fixed charged
layer and the diffused layer having opposite charge.
79. (b) 80. (b)
SUBJECTIVE TYPE QUESTIONS
1. Clouds are aerosols and the water particles in air
carry some charge over them. Rainfall can occur when two
oppositely charged clouds meet. Spraying a sol carrying
charge opposite to the one on clouds causes artificial rain.
2. In physisorption, the gas can be adsorbed one over the
other by van der Waals’ forces, thus is multilayered while
chemisorpiton involves formation of a chemical bond, which
can be formed only with the layer that is in direct contact with
the adsorbent. Therefore, it is mono-layered.
3.
Associated colloids.
4.
FeCl3 + NaOH → Fe2O3·xH2O/OH–
Negatively charged sol
5. Sea water contains electrolytes. River contains colloids of
sand and clay. When they meet the electrolytes neutralise the
charge on colloidal particles which results in the precipitation
of sand, clay etc. thus, resulting in a delta formation.
69. (c) : Freundlich adsorption isotherm gives an empirical
relationship between the quantity of gas adsorbed by unit
x 
mass of solid adsorbent   and pressure at a particular
m 
temperature.
6. NH3 gas will be adsorbed more readily on the surface
because it has higher critical temperature than CO2 gas.
70. (b) : There is no interface between gases due to their
complete miscibility.
7. The formation of micelles takes place only above a
particular temperature called Kraft temperature (Tk).
71. (d) : Soap and detergent are associated colloids as
they are formed by the aggregation of a large number of
ions due to attraction towards the oppositely charged ions
in concentrated solution.
8. BaCl 2 is more effective in causing coagulation of
negatively charged colloidal sol.
Because greater the valency of the coagulating ion, greater
is its power to bring about coagulation.
72. (b) : If the dispersion medium is separated from the
dispersed phase, the lyophilic sol can be reconstituted by
simply remixing with the dispersion medium. That is why
these sols are also called reversible sols.
9. Medicines are more effective in colloidal form because
in this form, these are more easily assimilated due to large
surface area.
73. (a) : The impact of the molecules of the dispersion
medium on the colloidal particles are unequal leading to
zig-zag motion i.e., Brownian movement.
74. (a) : Porous or finely divided forms of adsorbent possess
greater specific area which is available for adsorption per
gram of the adsorbent.
75. (a)
10. When a powerful beam of light is passed through
colloidal solution it exhibits tyndall effect whereas true
solution does not.
11. (i) Higher the critical temperature of the gas, more
readily it can get adsorbed on the surface of an adsorbent
since van der Waals’ forces are stronger at this temperature.
NH3 (132°C) has a higher critical temperature than dinitrogen
(–147°C) thus, it gets adsorbed more readily than N2.
65
Surface Chemistry
(ii) A finely divided substance is more effective as adsorbent
because it has more surface area and more number of active
sites (active centres) which increases the extent of adsorption.
12.(a) On treating a precipitate of iron (III) hydroxide with
a small amount of FeCl3 solution, a reddish brown coloured
colloidal solution is formed. In this case, Fe3+ ions from ferric
chloride are adsorbed by Fe(OH)3 precipitate.
Fe(OH)3⋅Fe3+
Fe(OH)3 + Fe3+
ions in
Ppt. Electrolyte
Colloidal sol
(b) When dialysis is persistent and prolonged, traces of
electrolyte are also removed. These electrolytes stabilise the
colloid and when removed completely, the unstable colloid
gets coagulated.
13. Coagulation : The process of aggregating together of
colloidal particles into large sized particle which ultimately
settles down under the force of gravity as a precipitate is
called coagulation.
Coagulation of lyophobic sols can be carried out by adding
electrolyte.
14. (i) Activated charcoal in gas masks adsorb the poisonous
gases present in air and thus purify the air for breathing.
(ii) Effect of temperature : Chemisorption is an exothermic
process but is very slow at lower temperature. High temperature
is more favourable thus with increasing temperature, rate of
adsorption increases.
15. (i) Animal hides are colloidal in nature. When a hide,
which has positively charged particles is soaked in tannin,
containing negatively charged colloidal particles, mutual
coagulation takes place. This results in the hardening of
leather.
(ii) Lyophilic sol is more stable than lyophobic sol because
it is highly hydrated in the solution.
16. (i) According to Hardy-Schulze rule, for a negatively
charged sol, greater is the valency of the positive ion added,
greater is its coagulation power.
3+
2+
Thus, AlCl3 (Al ) is more effective than MgCl2(Mg ) in
causing coagulation of negatively charged sol.
(ii) Proteins are macromolecules which cannot form
multimolecular colloids while sulphur sol have smaller S8
molecules which can form multimolecular colloids.
17. (i) When the size of the particles lies between 1 to
1000 nm, it behaves as a colloid. If particle size is less than
1 nm, it exists as a true solution and behaves as a crystalloid.
(ii) When AgNO3 solution is added to aqueous KI solution, a
negatively charged sol of AgI is formed. This is due to selective
adsorption of I– ions from the dispersion medium on AgI.
[Agl]I–
Agl + I–
Negative
sol
18.
S.No. Criteria
(i) Specificity
Physisorption
It is not
specific in
nature.
(ii) Temperature It decreases
dependence with increase in
temperature. Thus,
low temperature
is favourable.
Chemisorption
It is highly
specific in
nature.
It increases with
increase in
temperature. Thus,
high
temperature is
favourable.
(iii) Reversibility Reversible in
Irreversible in
nature.
nature.
(iv) Enthalpy
Low enthalpy of High enthalpy of
adsorption.
adsorption.
change
19. (a)
Multimolecular colloid
Associated colloid
The particles of this type
of colloids are aggregates
of large number of atoms
or smaller molecules. e.g.,
sulphur sol consists of
colloidal particles which are
aggregate of S8 molecules.
They are substances which
at low concentration
behave as electrolytes but
at higher concentration
exhibit colloidal behaviour
due to formation of
aggregated particles.e.g.,
micelles are associated
colloids.
(b)
Coagulation
Peptization
It is the process of settling It is the process of
of colloidal particles.
converting a precipitate
into colloidal sol.
20. Powdered charcoal is a good adsorbent. It adsorbs the
gases on its surface which reduces the pressure of the gas in
the enclosed vessel.
21. (i) A colloidal sol in which dispersed phase and
dispersion medium attract each other is called lyophilic
colloid. e.g., gum.
(ii) The difference in potential between the fixed layer and
the diffused layer of opposite charges in a colloidal sol is
known as electrokinetic or zeta potential.
(iii) Associated colloids : Micelles are associated colloids.
They are substances which at low concentrations behave
as strong electrolytes but at higher concentrations exhibit
colloidal behaviour due to formation of aggregates.
22. (i) Chemisorption has higher enthalpy of adsorption.
(ii) Physisorption takes place on account of weak van der
Waals’ forces and no chemical bond is formed, thus the
CBSE Board Term-II Chemistry Class-12
66
process is reversible. Chemisorption, on the other hand,
involves compound formation, thus it is irreversible in nature.
(iii) The forces operating in physisorption are weak
van der Waals’ forces.
23. (a) The Brownian movement has a stirring effect
which does not permit the particles to settle and thus, it is
responsible for the stability of colloidal solutions.
(b) Tyndall effect is not observed in true solutions as the
diameter of dispersed particles is very small to disperse the
light incident on it.
(c) The water obtained from natural sources often contains
suspended impurities. Alum is added to coagulate the
suspended impurities and make the water fit for drinking
purposes.
24. Multimolecular colloids : When a large number of small
molecules or atoms of a substance combine together in a
dispersion medium to form aggregates, having size in the
colloidal range, the colloidal solutions thus, formed are known
as multimolecular colloids. e.g., sulphur sol.
Macromolecular colloids: When macromolecules which
have large molecular masses are dispersed in a suitable
dispersion medium, they form a solution in which the size
of the macromolecule may be in the colloidal range. Such
colloids are called macromolecular colloids. e.g., cellulose,
starch, etc.
Associated colloids : They are substances which at low
concentration behave as electrolytes but at higher
concentration exhibit colloidal behaviour due to formation
of aggregated particles. e.g., micelles are associated colloids.
25. Effect of temperature : Physisorption decreases with
increase of temperature and chemisorption first increases
then decreases with increase of temperature.
Surface area : Greater the surface area, greater is the extent
of physisorption and chemisorption.
Activation energy : In physisorption, no appreciable activation
energy is needed. In chemisorption, sometimes high activation
energy is needed.
26. (i) Scattering of light by the colloidal particles takes place
and the path of light becomes visible (Tyndall effect).
(ii) The positively charged colloidal particles of ferric hydroxide
sol get coagulated by the oppositely charged Cl– ions provided
by NaCl.
(iii) On passing electric current through a sol, colloidal particles
start moving towards oppositely charged electrodes where
they lose their charge and get coagulated (electrophoresis).
27. (i) (a) Sulphur sol is prepared by the oxidation of H2S
with SO2.
Oxidation
SO2 + 2H2S
3S + 2H2O
(Sol)
(b) Gold sol is prepared by Bredig’s arc process or by the
reduction of AuCl3 with HCHO.
Oxidation
2Au (Sol) + 3HCOOH 2AuCl3 + 3HCHO + 3H2O
+ 6HCl
(ii) In adsorption, there is always a decrease in residual
unbalanced forces on the surface. This results in decrease in
surface energy which appears as heat. Hence, adsorption is
unconditionally an exothermic process.
28. (i) Sol : When solid is dispersed in water, it is called sol,
e.g., gold sol.
(ii) Emulsion: When liquid is dispersed in water, it is called
emulsion, e.g., milk.
(iii) Foam : When gas is dispersed in water, it is called foam,
e.g., soap lather, whipped cream.
29. (i) Washing of clothes involves micelle formation and
emulsification.
(ii) Washing of clothes is due to micelle formation. Micelles
are formed at a certain minimum temperature known as
Kraft’s temperature. This temperature is more readily achieved
in warm water as compared to cold water.
30. (i) (a) Aggregated particles of associated colloids at
high colloidal concentration are called micelles. e.g., soaps.
(b) Colloid of a liquid in a gas is called aerosol e.g., fog, etc.
(ii) Physical adsorption and chemical adsorption both
increase with increase in surface area of the adsorbent.
31. The process of settling of colloidal particles is called
coagulation of the sol. It is also known as precipitation.
Following are the three methods by which coagulation of
lyophobic sols can be carried out :
(i) Electrophoresis : In this process, the colloidal particles
move towards oppositely charged electrodes and get
discharged resulting in coagulation.
(ii) Mixing of two oppositely charged sols : When equal
proportions of oppositely charged sols are mixed, they
neutralise each other resulting in coagulation.
(iii) Persistent dialysis : On prolonged dialysis, electrolytes
present in sol are removed completely and colloids become
unstable resulting in coagulation.
32. (i) Hydrosol : It is a colloidal solution in which water is
the dispersion medium. e.g., starch solution.
(ii) The developed film is immersed in sodium thiosulphate
(hypo) solution which removes uncharged silver bromide as
a complex ion. This is known as fixing.
AgBr + 2Na2S2O3 → Na3[Ag(S2O3)2] + NaBr
After fixing, the film is not sensitive to light.
33. (i) The colloids which protect coagulation of other
colloids from the electrolytes are called protective colloids.
Lyophilic colloids are used as a protective colloid for lyophobic
colloids.
67
Surface Chemistry
(ii) Dust particles along with water suspended in air have
size smaller than wavelength of visible light and are more
effective in scattering light of shorter wavelength, blue light
which has smallest wavelength reaches our eyes and the sky
looks blue to us.
34. (i) (a) Dispersed phase of smoke = Solid
Dispersion medium of smoke = Gas
(b) Dispersed phase of milk = Fat (liquid)
37. Adsorption isotherm : It is the variation in the amount
of gas adsorbed by the adsorbent with pressure at constant
temperature.
195 K
224 K
x
m
273 K
Dispersion medium of milk = Water (liquid)
x
= kp1/ n (n > 1)
m
x
1
log = logk + logp
n
m
where, x is the mass of gas adsorbed on mass m of the
adsorbent at pressure p.
(ii)
35. (i) The separation of inert gases from a mixture is based
on the difference in degree of adsorption of gases by the
coconut charcoal.
(ii) Alumina and silica are good adsorbents. They can adsorb
even small amount of moisture present in atmosphere.
(iii) Adsorption of gas on solid surface decreases with rising
temperature.
Physical
adsorption
isobar
Temperature
Extent of
adsorption
Extent of
adsorption
36. (a) Adsorption of an adsorbate increases with increasing
surface area of adsorbent. Since surface area of a finely
divided substance is larger than any other form of adsorbent,
hence it is more effective as adsorbent.
(b) Adsorption isobar for physical adsorption shows that
the extent of adsorption decreases with the increase in
temperature. The adsorption isobar of chemical adsorption
shows that the extent of adsorption first increases and
then decreases with the increase in temperature. The initial
increase in the extent of adsorption with temperature is due
to the fact that the heat supplied acts as activation energy
required for chemical adsorption which is not required for
physical adsorption.
p
Adsorption isotherm
These curves indicate that on increasing temperature, physical
adsorption decreases at a fixed pressure.
Freundlich adsorption isotherm : It is an empirical relationship
between the quantity of gas adsorbed by unit mass of solid
adsorbent and pressure at a particular temperature.
x
= kp1/n(n > 1)
…(i)
m
x
x
when, n = 1, ⇒
= kp or
∝p
m
m
where, x is the mass of gas adsorbed on mass m of the
adsorbent at pressure p. k and n are constants which depend
on the nature of the adsorbent and the gas at the particular
temperature.
Taking log in Eq. (i), gives
1
x
= log k + log p
log
n
m
The validity of Freundlich isotherm can be verified by plotting
x
on y-axis and log p on x-axis.
m
If it comes to be a straight line, the Freundlich isotherm is
valid.
Chemical adsorption
isobar
Temperature
(c) PV = nRT
1.013 × 0.129 = n × 0.0821 × 273
1.013 × 0.129
= 0.00583 mol
⇒ n=
0.0821 × 273
Area occupied = 0.00583 × 6.023 × 1023 × 16.2 × 10–20
= 568 m2 g–1
38. (a) For spontaneous process, free energy decreases i.e.,
DG = –ve.
During adsorption process, the gas is adsorbed on the surface
of adsorbent, hence it involves the loss of degree of freedom
of the gas, therefore, entropy should also decrease during this
process i.e., DS = –ve.
Now from DG = DH – TDS
for adsorption –DG = DH – [T(–DS)] = DH + TDS
68
That means DG can be negative if DH has sufficiently high
negative value, hence the process is exothermic (DH = –ve).
(b) In adsorption, there is always a decrease in residual
unbalanced forces on the surface. This results in decrease in
surface energy which appears as heat. Hence, adsorption is
always an exothermic process.
(c) The amount of a gas adsorbed by solid depends on the
nature of the gas. In general, higher the critical temperature
of a gas, greater is the ease of liquefaction of gas i.e., larger
are the van der Waals’ forces of attraction. Therefore, greater
is the adsorption.
(d) Silica gel and calcium chloride are good adsorbents.
Therefore, adsorption of water vapours will occur in both the
cases. They will desiccate the vessel completely.
39. (i) Dialysis is the process of removing the electrolyte
particles from the colloidal solution by means of diffusion
through semi-permeable membrane. The charged nature of
the ‘sol’ is due to ions of the electrolyte adsorbed, which
makes it stable. If the electrolyte is completely removed from
the sol by excessive dialysis, then the uncharged particles
will come nearer to each other and coagulate resulting in
precipitation of the ‘sol’. Therefore excessive dialysis should
be avoided.
(ii) The colloidal particles can’t pass through semi-permeable
membrane, therefore electrolytes or other molecules from a
‘sol’ can be separated by diffusing through the membrane.
This process of purification of a ‘sol’ is known as dialysis. One
CBSE Board Term-II Chemistry Class-12
of the application of dialysis is purification of blood using
artificial kidney machine. The separation of ‘sol’ particles from
the liquid medium as well as from electrolytes by filtration
through an ultrafilter paper such as cellophane is called as
ultrafiltration. After separation of ‘sol’ particles it can be
further mixed with the dispersion medium to get pure ‘sol’.
40. (i) Number of moles of acetic acid in 100 mL before
adding charcoal = 0.05
Number of moles of acetic acid in 100 mL after adding
charcoal = 0.049
Number of moles of acetic acid adsorbed on the surface of
charcoal = (0.05 – 0.049) = 0.001
Number of molecules of acetic acid adsorbed on the surface
of charcoal = 0.001 × 6.02 × 1023 = 6.02 × 1020
Surface area of charcoal = 3.01 × 102 m2
Area occupied by one molecule of acetic acid on the surface
3.01 × 102
= 5 × 10−19 m2
6.02 × 1020
(ii) Suppose V litre volume of solution is taken,
W = 4.24 × V
W
× Avogadro’s number
Number of atoms =
Atomic mass
4.24V
=
× 6.023 × 1023
256
4.24V
Area = 500 × 10–4 =
× 6.023 × 1023 × 21 × 10 −20
256
V = 0.0000239 L = 0.0239 cm3
of charcoal =
The d- and f-Block
Elements
CHAPTER
4
Recap Notes
TRANSITION ELEMENTS
(d-BLOCK ELEMENTS)
Elements in which the last electron enters
any one of the five d-orbitals of their
respective penultimate shell are known as
transition elements or d-block elements.
Their general electronic configuration is
(n – 1)d1 – 10ns0 – 2.
Transition series : d-block consists of four
transition series,
1st Transition series or 3d series 21Sc – 30Zn
2nd Transition series or 4d series 39Y – 48Cd
3rd Transition series or 5d series 57La, 72Hf
– 80Hg
th
4 Transition series or 6d series 89Ac, 104Rf
– 112Cn
General characteristics :
Melting and boiling points
Enthalpies of atomisation
Ionisation enthalpies
Oxidation states
Atomic radii
Complex formation
Coloured compounds
Magnetic properties
Catalytic behaviour
Interstitial compounds
Alloy formation
High due to strong metallic bonding
High due to strong interatomic interactions
Generally increases from left to right in a series
Variable due to participation of ns and (n – 1)d electrons
Decrease from left to right but become constant when pairing of
electrons takes place
Form complexes due to high nuclear charge and small size and
availability of empty d-orbitals to accept lone pair of electrons
donated by ligands.
Form coloured compounds due to d-d transitions
Transition metal ions and their compounds are paramagnetic
due to presence of unpaired electrons in the (n – 1)d-orbitals
and it is calculated by using the formula, m = n(n + 2) where, n
is the no. of unpaired electrons.
Due to variable oxidation states and ability to form complexes
Due to empty spaces in their lattices, small atoms can be easily
accommodated
Due to similar atomic sizes
INNER TRANSITION ELEMENTS
( f-BLOCK ELEMENTS)
X Lanthanoids : Last electron enters one
of the 4f-orbitals. Cerium (At. no. 58) to
lutetium (At. no. 71).
X
Actinoids : Last electron enters one of
the 5f-orbitals. Thorium (At. no. 90) to
lawrencium (At. no. 103).
General electronic configuration :
(n – 2)f 1 – 14 (n – 1)d0 – 1 ns2
CBSE Board Term-II Chemistry Class-12
70
General characteristics of lanthanoids :
Atomic and ionic Decrease on going from La to Lu.
radii
Oxidation states
Most common oxidation state of lanthanoids is +3. Some elements exhibit
+2 and +4 oxidation states due to extra stability of empty, half-filled or fullyfilled f-subshell, e.g., Ce4+ acts as an oxidising agent and gets reduced to Ce3+,
Eu2+, Yb2+ act as strong reducing agents and get oxidised to Eu3+ and Yb3+.
Action of air
All the lanthanoids are silvery white soft metals and tarnish readily in
moist air. They burn in oxygen of air and form oxides (Ln2O3 type).
Coloured ions
They form coloured trivalent metal ions due to f-f transitions of unpaired
electrons. La3+ and Lu3+ are colourless ions due to empty (4f 0 ) or fully
(4f 14) orbitals.
Magnetic
properties
La3+, Lu3+ are diamagnetic while trivalent ions of the rest of lanthanoids
are paramagnetic.
Reducing agents
They readily lose electrons and are good reducing agents.
Electropositive
character
Highly electropositive because of low ionisation energies.
Alloy formation
They form alloys easily with other metals especially iron.
Tendency to form
complexes
Lanthanoids do not have much tendency to form complexes due to low
charge density because of their large size. The tendency to form complexes
and their stability increases with increasing atomic number.
Lanthanoid contraction : In lanthanoid
series, with increasing atomic number, there
is progressive decrease in atomic/ionic radii
(M3+ ions) from La3+ to Lu3+.
X Reason : Due to addition of new electrons
into f-subshell and imperfect shielding of
one electron by another in the f-orbitals,
there is greater effect of increased nuclear
charge than screening
contraction in size occurs.
X
effect
hence
Consequences : Their separation is
difficult, they have small differences in
properties and 4d and 5d transition series
have almost same atomic radii (Zr and Hf
have similar properties due to same size).
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. The correct order of ionic radii of Ce, La, Pm
and Yb in +3 oxidation state is
(a) La3+ < Pm3+ < Ce3+ < Yb3+
(b) Yb3+ < Pm3+ < Ce3+ < La3+
(c) La3+ < Ce3+ < Pm3+ < Yb3+
(d) Yb3+ < Ce3+ < Pm3+ < La3+
7. The equation
2–
–
3MnO4 + 4H+ → 2MnO4 + MnO2 + 2H2O
represents
(a) reduction
(b) disproportionation
(c) oxidation in acidic medium
(d) reduction in acidic medium.
2. CuSO 4 is paramagnetic while ZnSO 4 is
diamagnetic because
(a) Cu2+ ion has 3d9 configuration while Zn2+
ion has 3d10 configuration
(b) Cu2+ ion has 3d5 configuration while Zn2+
ion has 3d6 configuration
(c) Cu2+ has half filled orbitals while Zn2+ has
fully filled orbitals
(d) CuSO 4 is blue in colour while ZnSO 4 is
white.
8. Which of the following is not correctly
matched with the example given?
(a) An element of first transition series which
has highest second ionisation enthalpy – Cu
(b) An element of first transition series with
highest third ionisation enthalpy – Zn
(c) An element of first transition series with
lowest enthalpy of atomisation – Zn
(d) Last element of third transition series – Cd
3. The correct order of number of unpaired
electrons is
(a) Cu2+ > Ni2+ > Cr3+ > Fe3+
(b) Ni2+ > Cu2+ > Fe3+ > Cr3+
(c) Fe3+ > Cr3+ > Ni2+ > Cu2+
(d) Cr3+ > Fe3+ > Ni2+ > Cu2+
4. Arrange the oxides of manganese according
to increasing acidic strength.
(a) MnO < Mn3O4 < Mn2O3 < MnO2 < Mn2O7
(b) Mn2O7 < MnO2 < Mn2O3 < Mn3O4 < MnO
(c) MnO2 < Mn2O7 < Mn3O4 < Mn2O3 < MnO
(d) Mn3O4 < Mn2O3 < Mn2O7 < MnO2 < MnO
5. Which of the following d-block element
has half-filled penultimate as well as valence
subshell?
(a) Cu
(b) Au
(c) Ag
(d) Cr
6. The correct order of E M 2+ /M values with
negative sign for the four successive elements
Cr, Mn, Fe and Co is
(a) Fe > Mn > Cr > Co (b) Cr > Mn > Fe > Co
(c) Mn > Cr > Fe > Co (d) Cr > Fe > Mn > Co
9. The magnetic moment of a divalent ion in
aqueous solution with atomic number 25 is
(a) 5.9 B.M
(b) 2.9 B.M
(c) 6.9 B.M
(d) 9.9 B.M.
10. Highest oxidation state of manganese in
fluoride is +4 (MnF4) but highest oxidation state
in oxides is +7 (Mn2O7) because
(a) fluorine is more electronegative than oxygen
(b) fluorine does not possess d-orbitals
(c) fluorine stabilises lower oxidation state
(d) in covalent compounds, fluorine can form
single bond only while oxygen forms double
bond.
11. Which of the following transition metal ions
is colourless?
(b) Cr3+
(a) V2+
2+
(c) Zn
(d) Ti3+
12. E Mn3+/Mn2+ is highly positive than that of
E°Cr3+/Cr2+ or E°Fe3+/Fe2+ because
(a) Mn2+ (d5) can be easily oxidised to Mn3+(d4)
due to low ionisation enthalpy
(b) third ionisation enthalpy of Mn is much
larger due to stable half filled d5 electronic
configuration of Mn2+
72
(c) Mn3+ is more stable than Mn2+ due to higher
oxidation state.
(d) second ionisation enthalpy of Mn is higher
than third ionisation enthalpy.
13. I n t e r s t i t i a l c o m p o u n d s a r e n o n stoichiometric compounds formed by trapping
small atoms like C, H or N in crystal lattices
of transition metals. Which of the following
properties is not shown by these compounds?
(a) They have high melting points, higher than
those of pure metals.
(b) They are very hard, some borides are
comparable to diamond in hardness.
(c) They are chemically very reactive.
(d) They retain metallic conductivity.
14. Reactivity of transition elements decreases
almost regularly from Sc to Cu because of
(a) lanthanoid contraction
(b) regular increase in ionisation enthalpy
(c) regular decrease in ionisation enthalpy
(d) increase in number of oxidation states.
15. For Zn2+, Ni2+, Cu2+ and Cr2+ which of the
following statements is correct?
(a) Only Zn2+ is colourless and Ni2+, Cu2+ and
Cr2+ are coloured.
(b) All the ions are coloured.
(c) All the ions are colourless.
(d) Zn2+ and Cu2+ are colourless while Ni2+ and
Cr2+ are coloured.
16. Which of the following statement concerning
lanthanide elements is false?
(a) All lanthanides are highly dense metals.
(b) More characteristic oxidation state of
lanthanide elements is +3.
(c) Lanthanides are separated from one another
by ion exchange method.
(d) Ionic radii of trivalent lanthanides steadily
increases with increase in the atomic
number.
17. Which of the following statements is not
correct about magnetic behaviour of substances?
(a) Diamagnetic substances are repelled by an
applied magnetic field.
(b) Paramagnetic substances are attracted by
an applied magnetic field.
(c) Magnetic moment of n unpaired electrons is
given by µ = n(n − 2) B.M.
(d) Magnetic moment increases as the number
of unpaired electrons increases.
CBSE Board Term-II Chemistry Class-12
18. The electronic configuration of Cu(II) is
3d9 whereas that of Cu(I) is 3d10. Which of the
following is correct?
(a) Cu(II) is more stable.
(b) Cu(II) is less stable.
(c) Cu(I) and Cu(II) are equally stable.
(d) Stability of Cu(I) and Cu(II) depends on
nature of copper salts.
19. Which of the following are basic oxides?
Mn2O7, V2O3, V2O5, CrO, Cr2O3
(a) Mn2O7 and V2O3
(b) V2O3 and CrO
(c) CrO and Cr2O3
(d) V2O5 and V2O3
20. Which of the following catalysts is not
correctly matched with the reaction?
(a) Vanadium(V) oxide in contact process for
oxidation of SO2 to SO3.
(b) Finely divided iron in Haber’s process in
conversion of N2 and H2 to NH3.
(c) PtCl2 catalyses the oxidation of ethyne to
ethanal in the Wacker process.
(d) Ni in presence of hydrogen for conversion of
vegetable oil to ghee.
21. Which is the non-lanthanide element?
(a) La
(b) Lu
(c) Pr
(d) Pm
22. Magnetic moment of Ce3+ ion on the basis
of ‘spin-only’ formula will be _____ B.M.
(a) 1.232
(b) 1.332
(c) 1.532
(d) 1.732
23. In which of the following pairs of ions, the
lower oxidation state in aqueous solution is more
stable than the other?
(a) Tl+, Tl3+
(b) Cu+, Cu2+
2+
3+
(c) Cr , Cr
(d) V2+, VO2+ (V4+)
24. Which of the following compounds is expected
to be coloured?
(a) Ag2SO4
(b) CuF2
(c) MgF2
(d) CuCl
25. Consider the following statements with
respect to lanthanides :
1. The basic strength of hydroxides of
lanthanides increases from La(OH)3 to Lu(OH)3.
2. The lanthanide ions Lu3+, Yb2+ and Ce4+ are
diamagnetic.
Which of the statement(s) given above is/are
correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
73
The d- and f-Block Elements
26. Cu+ ion is not stable in aqueous solution
because
(a) second ionisation enthalpy of copper is less
than the first ionisation enthalpy
(b) large value of second ionisation enthalpy
of copper is compensated by much more
negative hydration energy of Cu2+(aq)
(c) hydration energy of Cu +(aq) is much more
negative than that of Cu2+(aq)
(d) many copper (I) compounds are unstable
in aqueous solution and undergo
disproportionation reaction.
27. Select the correct option, among Sc(III),
Ti(IV), Pd(II) and Cu(II) ions,
(a) all are paramagnetic
(b) all are diamagnetic
(c) Sc(III), Ti(IV) are paramagnetic and Pd(II),
Cu(II) are diamagnetic
(d) Sc(III), Ti(IV) are diamagnetic and Pd(II),
Cu(II) are paramagnetic.
28. Identify the species in which the metal atom
is in +6 oxidation state.
–
3–
(a) MnO4
(b) Cr(CN)6
2–
(c) NiF6
(d) CrO2Cl2
29. Which of the following statements is correct
about stability of the complexes of lanthanoids?
(a) Stability of complexes increases as the size
of lanthanoid decreases.
(b) Stability of complexes decreases as the size
of lanthanoid decreases.
(c) Lanthanoids do not form complexes.
(d) All the complexes of lanthanoids have same
stability.
30. Fe3+ ion is more stable than Fe2+ ion because
(a) more the charge on the atom, more is its
stability
(b) configuration of Fe2+ is 3d6 while Fe3+ is 3d5
(c) Fe2+ has a larger size than Fe3+
(d) Fe3+ ions are coloured hence more stable.
31. Transition metals make the most efficient
catalysts because of their ability to
(a) adopt multiple oxidation states and to form
complexes
(b) form coloured ions
(c) show paramagnetism due to unpaired
electrons
(d) form a large number of oxides.
32. General electronic configuration of transition
metals is
(b) nd10ns2
(a) (n – 1)d1-10ns0-2
10 2
(c) (n – 1)d ns
(d) (n – 1)d1-5ns2
33. Consider the following statements
I. La(OH)3 is least basic among hydroxides of
lanthanides.
II. Zr4+ and Hf 4+ possess almost the same ionic
radii.
III. Ce4+ can act as an oxidizing agent.
Which of the above is/are true?
(a) I and III
(b) II and III
(c) II only
(d) I and II
34. Arrange the following in increasing value of
magnetic moments.
(ii) [Fe(CN)6]3–
(i) [Fe(CN)6]4–
(iii) [Cr(NH3)6]3+
(iv) [Ni(H2O)4]2+
(a) (i) < (ii) < (iii) < (iv)
(b) (i) < (ii) < (iv) < (iii)
(c) (ii) < (iii) < (i) < (iv)
(d) (iii) < (i) < (ii) < (iv)
35. Fe3+ compounds are more stable than Fe2+
compounds because
(a) Fe3+ has smaller size than Fe2+
(b) Fe3+ has 3d5 configuration (half-filled)
(c) Fe3+ has higher oxidation state
(d) Fe3+ is paramagnetic in nature.
36. Following order is observed in oxidising
power of certain ions:
+
2–
–
VO2 < Cr2O7 < MnO4
The reason for this increasing order of oxidising
power is
(a) increasing stability of the lower species to
which they are reduced
(b) increasing stability of the higher species to
which they are oxidised
(c) increasing stability of the higher species to
which they are reduced
(d) increasing stability of the lower species to
which they are oxidised.
37. Which of the following transition metal ions
has highest magnetic moment?
(b) Ni2+
(a) Cu2+
2+
(c) Co
(d) Fe2+
CBSE Board Term-II Chemistry Class-12
74
39. Which one of the following is a ‘d-block
element’?
(a) Gd
(b) Hs
(c) Es
(d) Cs
40. Which of the following lanthanide is
commonly used?
(a) Lanthanum
(b) Nobelium
(c) Thorium
(d) Cerium
41. Transition elements form binary compounds
with halogens. Which of the following elements
will form MF3 type compounds?
(a) Cr
(b) Cu
(c) Ni
(d) All of these
42. Although zirconium belongs to 4d and
hafnium to 5d-transition series even they show
similar physical and chemical properties because
both
(a) belong to d-block
(b) have same number of electrons
(c) have similar atomic radius
(d) belongs to the same group of the periodic
table.
43. Most of the transition metals exhibit
(i) paramagnetic behaviour
(ii) diamagnetic behaviour
(iii) variable oxidation states
(iv) formation of coloured ions
(a) (ii), (iii) and (iv)
(b) (i), (iii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (ii) and (iv)
44. Which of the following has no unpaired
electrons but is coloured?
(a) K2Cr2O7
(b) K2MnO4
(c) CuSO4⋅5H2O
(d) MnCl2
45. The second and third row elements of
transition metals resemble each other much
more than they resemble the first row because
of
(a) lanthanoid contraction which results in
almost same radii of second and third row
metals
(b) diagonal relationship between second and
third row elements
(c) similar ionisation enthalpy of second and
third row elements
(d) similar oxidation states of second and third
row metals.
46. Which of the following compounds is not
coloured?
(b) Na2[CdCl4]
(a) Na2[CuCl4]
(c) K4[Fe(CN)]6]
(d) K3[Fe(CN)6]
47. The salts of Cu in +1 oxidation state are
unstable because
(a) Cu+ has 3d10 configuration
(b) Cu+ disproportionates easily to Cu(0) and
Cu2+
(c) Cu + disproportionates easily to Cu2+ and
Cu3+
(d) Cu+ is easily reduced to Cu2+.
48. Colour of transition metal ions are due to
absorption of some wavelength. This results in
(a) d-s transition
(b) s-s transition
(c) s-d transition
(d) d-d transition.
49. The observed values and calculated values
of E° of various 3d-series elements are shown in
the figure.
Standard electrode potential / V
38. Which of the following are amphoteric
oxides?
(i) Mn2O7
(ii) CrO3
(iii) Cr2O3
(iv) CrO
(vi) V2O4
(v) V2O5
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (iii) and (v)
(d) (ii) and (vi)
0.5
0
–0.5
–1
–1.5
–2
Ti
V
Cr Mn
Observed
values
Fe Co
Ni Cu Zn
Calculated
values
Which of the following facts cannot be explained
on the basis of the given graph?
(a) Inability of Cu to liberate H2 from acids
(b) Extra stability of d5(Mn 2+) and d 10(Zn 2+)
configuration
(c) Extra stability of Ni2+ due to d10 configuration
(d) All of these.
50. For the given reactions :
–
X + Y + H2O
[Fe(H2O)6]2+ + NO3 + H+
2+
Z + H2O
[Fe(H2O)6] + X
The incorrect statement about X, Y and Z is
(a) Magnetic moment of Y is 5.9 B.M.
(b) Oxidation state of Fe in Z is +1.
(c) Complex Z is reddish-brown in colour.
(d) X is an acidic oxide of nitrogen.
51. W, X, Y and Z are four consecutive members
of 3d-series.
Trend in their melting point are shown in the
given figure.
75
The d- and f-Block Elements
On the basis of these values, Krish concluded
M.P. /103 K
3
W
Y
2
the following statements:
Z
X
1
Atomic number
Correct statement about W, X, Y and Z is
(a) magnetic moment of X in its +2 oxidation
state is 2.83 B.M.
(b) W 3+ ion is green in colour.
(c) Y3+ catalyses reaction between iodide and
persulphate ions.
(d) stable oxidation states of Z are +1, +2 and
+6.
I.
Cr2+ is a reducing agent
II. Mn3+ is an oxidizing agent
III. both Cr2+ and Mn3+ exhibit d4 electronic
configuration
IV. when Cr 2+ is used as a reducing agent,
the chromium ion attains d 5 electronic
configuration.
The incorrect conclusion made by him is
(a) I
(b) II
(c) III
(d) IV
52. Few electrode potential values are given
below:
53. Find the incorrect analogy for lanthanoids.
Cr3+/Cr2+ = – 0.41 V
(b) Paramagnetic lanthanide ion : Yb2+
Cr2+/Cr = – 0.90 V
(c) Ions that can exist in aqueous solution :
Mn3+/Mn2+ = +1.57 V
Mn2+/Mn = –1.18 V
(a) Good oxidising agent : Ce4+
Eu2+, Yb2+
(d) Colourless ions : Ce3+, Yb3+
Case Based MCQs
Case I : Read the passage given below and
answer the following questions.
The lanthanide series is a unique class of
15 elements with relatively similar chemical
properties. They have atomic number ranging
from 57 to 71, which corresponds to the filling
of the 4f orbitals with 14 electrons. This
configuration leads to phenomenon known
as lanthanide contraction. The lanthanides
are sometimes referred to as the ‘rare earth
elements’, leading to misconception that they
are rare. In fact many of the rare earth elements
are more common than gold, silver and in some
cases, lead. The lanthanides are commonly
found in nature as a mixture in a number of
monazite (LnPO4) and bastnaesite (LnCO3F) in
the +3 oxidation state.
The chemical and physical properties of
lanthanides provide the unique features that set
them apart from other elements. Lanthanides
are most stable in the +3 oxidation state. Yb
and Sm though stable in the +3 state, also
have accessible +2 oxidation states. The ease
of accessibility of both oxidation states is quite
important in chemical synthesis and these
elements act as Lewis acid in the +3 oxidation
state and single electron reductant in the +2
oxidation state.
In the following questions (Q. No. 54-58), a
statement of assertion followed by a statement
of reason is given. Choose the correct answer out
of the following choices on the basis of the above
passage.
(a) Assertion and reason both are correct
statements and reason is correct explanation
for assertion.
(b) Assertion and reason both are correct
statements but reason is not correct
explanation for assertion.
(c) Assertion is correct statement but reason is
wrong statement.
(d) Assertion is wrong statement but reason is
correct statement.
54. Assertion : The elements scandium and
yttrium are called “rare earths”.
Reason : Scandium and yttrium are rare on
earth’s crust.
CBSE Board Term-II Chemistry Class-12
76
55. Assertion : Separation of lanthanide
elements is difficult.
Reason : They have similar chemical properties.
56. Assertion : There is continuous increase in
size among lanthanides.
Reason : Lanthanides show lanthanide contraction.
57. Assertion : Yb2+ is more stable than Yb3+.
Reason : Electronic configuration of Yb 2+ is
[Xe]4f 7.
58. Assertion : All lanthanides have similar
chemical properties.
Reason : Because the lanthanoids differ only in
the number of 4f - electrons.
Case II : Read the passage given below and
answer the following questions from 59 to 63.
The transition elements have incompletely filled
d-subshells in their ground state or in any of
their oxidation states. The transition elements
occupy position in between s- and p-blocks in
groups 3-12 of the Periodic table. Starting from
fourth period, transition elements consists of
four complete series : Sc to Zn, Y to Cd and
La, Hf to Hg and Ac, Rf to Cn. In general, the
electronic configuration of outer orbitals of these
elements is (n – 1)d 1–10 ns 0–2. The electronic
configurations of outer orbitals of Zn, Cd, Hg
and Cn are represented by the general formula
(n – 1)d 10 ns 2 . All the transition elements
have typical metallic properties such as high
tensile strength, ductility, malleability. Except
mercury, which is liquid at room temperature,
other transition elements have typical metallic
structures. The transition metals and their
compounds also exhibit catalytic property and
paramagnetic behaviour. Transition metal
also forms alloys. An alloy is a blend of metals
prepared by mixing the components. Alloys may
be homogeneous solid solutions in which the
atoms of one metal are distributed randomly
among the atoms of the other.
59. Which of the following characteristics of
transition metals is associated with higher
catalytic activity?
(a) High enthalpy of atomisation
(b) Variable oxidation states
(c) Paramagnetic behaviour
(d) Colour of hydrated ions
60. Transition elements form alloys easily
because they have
(a)
(b)
(c)
(d)
same atomic number
same electronic configuration
nearly same atomic size
same oxidation states.
61. The electronic configuration of tantalum
(Ta) is
(b) [Xe]4f14 5d2 6s2
(a) [Xe]4f 0 5d 1 6s2
(c) [Xe]4f 14 5d 3 6s2
(d) [Xe]4f 14 5d 4 6s2
62. Which one of the following outer orbital
configurations may exhibit the largest number
of oxidation states?
(b) 3d 54s2
(a) 3d 54s1
(c) 3d 24s2
(d) 3d 34s2
63. The correct statement(s) among the following
is/are
(i) all d- and f-block elements are metals
(ii) all d- and f-block elements form coloured
ions
(iii) all d- and f-block elements are paramagnetic.
(a) (i) only
(b) (i) and (ii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Case III : Read the passage given below and
answer the following questions from 64 to 68.
Transition metal oxides are compounds formed
by the reaction of metals with oxygen at high
temperature. The highest oxidation number in
the oxides coincides with the group number. In
vanadium, there is a gradual change from the
basic V2O3 to less basic V2O4 and to amphoteric
V2O5⋅ V2O4 dissolves in acids to give VO2+ salts.
Transition metal oxides are commonly utilized
for their catalytic activity and semiconductive
properties. Transition metal oxides are also
frequently used as pigments in paints and
plastic. Most notably titanium dioxide. One
of the earliest application of transition metal
oxides to chemical industry involved the use of
vanadium oxide for catalytic oxidation of sulfur
dioxide to sulphuric acid. Since then, many
other applications have emerged, which include
benzene oxidation to maleic anhydride on
vandium oxides; cyclohexane oxidation to adipic
acid on cobalt oxides. An important property of
the catalyst material used in these processes is
the ability of transition metals to change their
oxidation state under a given chemical potential
of reductants and oxidants.
64. Which oxide of vanadium is most likely to
be basic and ionic ?
(a) VO
(b) V2O3
(c) VO2
(d) V2O5
77
The d- and f-Block Elements
65. Vanadyl ion is
(a) VO2+
(c) V2O
+
(b)
(d)
+
VO2
VO43–
Which of the following statements is false?
With fluorine vanadium can form VF5.
With chlorine vanadium can form VCl5.
Vanadium exhibits highest oxidation state
in oxohalides VOCl3, VOBr 3 and fluoride
VF5.
(d) With iodine vanadium cannot form VI 5
due to oxidising power of V5+ and reducing
nature of I–.
66.
(a)
(b)
(c)
67. The oxidation state of vanadium in V2O5 is
(a) +5/2
(b) +7
(c) +5
(d) +6
68. Identify the oxidising agent in the following
reaction.
V2O5 + 5Ca
2V + 5CaO
(b) Ca
(a) V2O5
(c) V
(d) None of these
Case IV : Read the passage given below and
answer the following questions from 69 to 73.
The unique behaviour of Cu, having a positive
E° accounts for its inability to liberate H2 from
acids. Only oxidising acids (nitric and hot
concentrated sulphuric acid) react with Cu, the
acids being reduced. The stability of the halffilled (d5) subshell in Mn2+ and the completely
filled (d 10) configuration in Zn 2+ are related
to their E°(M3+/M2+) values. The low value for Sc
reflects the stability of Sc3+ which has a noble
gas configuration. The comparatively high value
for Mn shows that Mn 2+ (d 5 ) is particularly
stable, whereas a comparatively low value for
Fe shows the extra stability of Fe3+(d 5). The
comparatively low value for V is related to the
stability of V2+ (half-filled t2g level).
69. Standard reduction electrode potential of
Zn2+/Zn is – 0.76 V. This means
(a) ZnO cannot be reduced to Zn by H2 under
standard conditions
(b) Zn cannot liberate H 2 with concentrated
acids
(c) Zn is generally the anode in an electrochemical
cell
(d) Z n i s g e n e r a l l y t h e c a t h o d e i n a n
electrochemical cell.
70. E° values for the couples Cr 3+ /Cr 2+ and
Mn3+/Mn2+ are –0.41 and +1.51 volts respectively.
These values suggest that
(a) Cr2+ acts as a reducing agent whereas Mn3+
acts as an oxidizing agent
(b) Cr2+ is more stable than Cr3+ state
(c) Mn3+ is more stable than Mn2+
(d) Cr 2+ acts as an oxidizing agent whereas
Mn3+ acts as a reducing agent.
71. The reduction potential values of M, N and
O are +2.46, –1.13 and –3.13 V respectively.
Which of the following order is correct regarding
their reducing property?
(a) O > N > M
(b) O > M > N
(c) M > N > O
(d) M > O > N
72. Which of the following statements are true?
(I) Mn2+ compounds are more stable than Fe2+
towards oxidation to +3 state.
(II)Titanium and copper both in the first series
of transition metals exhibits +1 oxidation
state most frequently.
(III) Cu+ ion is stable in aqueous solutions.
(IV) The E° value for the Mn3+/Mn2+ couple is
much more positive than that for Cr3+/Cr2+
or Fe3+/Fe2+.
(a) (II) and (III)
(b) (I) and (IV)
(c) (I) and (III)
(d) (II) and (IV)
73. The stability of Cu2+(aq) rather than Cu+(aq)
is due to
(a) more negative Dhyd H° of Cu2+(aq)
(b) less negative Dhyd H° of Cu2+(aq)
(c) more positive Dhyd H° of Cu2+(aq)
(d) less positive Dhyd H° of Cu2+(aq).
Case V : Read the passage given below and
answer the following questions from 74 to 78.
The f-block elements are those in which the
differentiating electron enters the (n –2)f
orbital. There are two series of f-block elements
corresponding to filling of 4f and 5f-orbitals.
The series of 4f-orbitals is called lanthanides.
Lanthanides show different oxidation states
depending upon stability of f 0 , f 7 and f 14
configurations, though the most common
oxidation states is +3. There is a regular decrease
in size of lanthanides ions with increase in
atomic number which is known as lanthanide
contraction.
CBSE Board Term-II Chemistry Class-12
78
74. The atomic numbers of three lanthanide
elements X, Y and Z are 65, 68 and 70 respectively,
their Ln3+ electronic configuration is
(b) 4f 11, 4f 8, 4f 13
(a) 4f 8, 4f 11, 4f13
0
2
11
(c) 4f , 4f , 4f
(d) 4f 3, 4f 7, 4f 9
75. Lanthanide contraction is observed in
(a) Gd
(b) At
(c) Xe
(d) Te
76. Which of the following is not the configuration
of lanthanoid?
(a) [Xe]4f106s2
(b) [Xe]4f15d16s2
14
10 2
(c) [Xe]4d 5d 6s
(d) [Xe]4f75d16s2
77. Name a member of the lanthanoid series
which is well known to exhibit +4 oxidation
state.
(a)
(b)
(c)
(d)
Cerium (Z = 58)
Europium (Z = 63)
Lanthanum (Z = 57)
Gadolinium (Z = 64)
78. Identify the incorrect statement among the
following.
(a) Lanthanoid contraction is the accumulation
of successive shrinkages.
(b)The different radii of Zr and Hf is due to
consequence of the lanthanoid contraction.
(c) Shielding power of 4f electrons is quite
weak.
(d) There is a decrease in the radii of the atoms
or ions as one proceeds from La to Lu.
Assertion & Reasoning Based MCQs
For question numbers 79-95, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
79. Assertion : Transition metals form
substitutional alloys.
Reason : Alloys are made to develop some useful
properties which are absent in the constituent
elements.
Reason : It is radioactive and has been prepared
by artificial means.
80. Assertion : Reduction potential of Mn (+3 to +2)
is more positive than Fe (+3 to +2).
Reason : Ionisation potential of Mn is more
than that of Fe.
of H2SO4 by Contact process.
81. Assertion : Mn2+ is more stable than Mn3+.
Reason : Mn2+ has half-filled configuration.
82. Assertion : [Ti(H2O)6]3+ is a coloured ion.
Reason : Ti shows +2, +3, +4 oxidation states.
83. Assertion : Chromium is hard but mercury
is soft.
Reason : Chromium is a 3d transition element.
87. Assertion : Transition metals are good
catalysts.
Reason : V2O5 or Pt is used in the preparation
88. Assertion : Europium (II) is more stable
than cerium (II).
Reason : Cerium salts are used as a catalyst in
petroleum cracking.
89. Assertion : When Zn is placed in a magnetic
field, it is feebly magnetised in a direction
opposite to that of the magnetising field.
Reason : Zn has completely filled atomic
orbitals.
90. Assertion : The correct order of oxidising
+
2+
84. Assertion : Cu+ is paramagnetic.
Reason : Cu+ is less stable than Cu2+.
power is : VO2 < VO < VO .
85. Assertion : Co (IV) is known but Ni (IV) is
not.
Reason : Ni (IV) has d6 electronic configuration.
is +7.
86. Assertion : Promethium is a man-made
element.
Reason : The oxidation state of Mn in MnO4
91. Assertion : Transition metals form a large
number of interstitial compounds.
Reason : They have high melting point and
boiling point.
79
The d- and f-Block Elements
92. Assertion : Members of 4d and 5d series
of transition elements have nearly same atomic
radii.
Reason : The effective nuclear charge felt by
(n – 1)d electrons is higher as compared to that
by ns electrons.
Reason : Atomic and ionic radii for transition
elements are smaller than their corresponding
s-block elements.
94. Assertion : The maximum oxidation state of
chromium in its compounds is +6.
Reason : Chromium has only six electrons in ns
and (n – 1)d orbitals.
93. Assertion : In transition elements, ns
orbital is filled up first and (n – 1)d afterwards,
during ionization ns electrons are lost prior to
(n – 1)d electrons.
95. Assertion : Transition metals are poor
reducing agents.
Reason : Transition metals form numerous
alloys with other metals.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Give reasons for the following :
Eu2+ is a strong reducing agent.
7.
2. Write the formula of an oxoanion of
manganese (Mn) in which it shows the oxidation
state equal to its group number.
3. Name a member of the lanthanoid series
which is well known to exhibit +4 oxidation state.
4. Assign reason for the following :
Copper (I) ion is not known in aqueous solution.
How would you account for the following :
Among lanthanoids, La(III) compounds are
predominant. However, occasionally in solutions
or in solid compounds, +2 and +4 ions are also
obtained.
8.
Account for the following :
Zr and Hf have almost similar atomic radii.
5. Name a member of the lanthanoid series
which is well known to exhibit +2 oxidation state.
9. Write the formula of an oxoanion of chromium
(Cr) in which it shows the oxidation state equal
to its group number.
6. Account for the following :
Zn is not considered as a transition element.
10. Zn2+ salts are white while Cu2+ salts are
coloured. Why?
Short Answer Type Questions (SA-I)
11. Why is europium (II) more stable than
cerium (II)?
(ii) Which is the most stable ion in +2 oxidation
state and why?
12. The magnetic moment of a transition metal
ion is found to be 3.87 BM. How many number of
unpaired electrons are present in it ?
14. How would you account for the following :
(i) The E°M 2+/M for copper is positive (+0.34 V).
Copper is the only metal in the first series of
transition elements showing this behaviour.
(ii) The metallic radii of the third (5d) series of
transition metals are nearly the same as those
of the corresponding members of the second (4d)
series.
13. Use the data to answer the following and
also justify giving reasons :
Cr
Mn
Fe
Co
E°M2+/M
–0.91
–1.18
–0.44
–0.28
E°M3+/M2+
–0.41
+1.57
+0.77
+1.97
(i) Which is a stronger reducing agent in
aqueous medium, Cr2+ or Fe2+ and why?
15. Write the electronic configuration of Ce3+
ion, and calculate the magnetic moment on
the basis of ‘spin-only’ formula. [Atomic no. of
Ce = 58]
CBSE Board Term-II Chemistry Class-12
80
16. (i) Which metal in the first transition series
(3d-series) exhibits +1 oxidation state most
frequently and why?
(ii) Which of following cations are coloured in
aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
17. How would you account for the following :
(i) Cr2+ is reducing in nature while with the
same d-orbital configuration (d4), Mn3+ is an
oxidising agent.
(ii) In a transition series of metals, the
metal which exhibits the greatest number of
oxidation states occurs in the middle of the
series.
18. What is lanthanoid contraction and what is
it due to? Write two consequences of lanthanoid
contraction.
19. What is meant by ‘disproportionation’? Give
an example of a disproportionation reaction in
aqueous solution.
20. (i) Write two characteristic of the transition
elements.
(ii) Which of the 3d-block elements may not be
regarded as the transition elements and why?
Short Answer Type Questions (SA-II)
21. (i) Explain the cause of paramagnetism in
lanthanoid ions.
(ii) Nb and Ta exhibit similar properties. Give
reason.
(iii) Among the ionic species, Sc3+, Ce4+ and Eu2+,
which one is a good oxidising agent.
22. Following are the transition metal ions of 3d
series :
Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers : Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following :
(i) Which ion is most stable in aqueous solution
and why?
(ii) Which ion is strong oxidising agent and why?
(iii) Which ion is colourless and why?
23. Compare qualitatively the first and second
ionisation potentials of copper and zinc. Explain
the observation.
24. (i) Ce (IV) is a good analytical reagent.
Why?
(ii) Account for the following : Copper(I)
compounds are white whereas copper(II)
compounds are coloured.
(ii) The E° value for the Mn3+/Mn2+ couple is
much more positive than that for Fe3+/Fe2+
couple.
(iii) The highest oxidation state of a metal is
exhibited in its oxide or fluoride.
27. (a) Assign reasons for the following :
(i) Cu(I) ion is not known to exist in aqueous
solutions.
(ii) Transition metals are much harder than the
alkali metals.
(b) Name the lanthanoids which show
abnormally low value of third ionisation enthalpy.
28. Account for the following :
(i) The transition metals and their compounds
act as good catalysts.
(ii) The lowest oxide of transition metal is basic,
the highest is amphoteric/acidic.
(iii) The magnetic moment (B.M.) of Fe2+ ion
is 24 .
25. How do the oxides of transition elements in
lower oxidation states differ from those in higher
oxidation states and why?
29. (a) Account the following :
(i) Transition metals form large number of
complex compounds.
(ii) E°° value for the Mn3+/Mn2+ couple is highly
positive (+1.57 V) as compared to Cr3+/Cr2+.
(iii) Which of following cations are coloured in
aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
26. How would you account for the following :
(i) The atomic radii of the metals of the third
(5d) series of transition elements are virtually
the same as those of the corresponding members
of the second (4d) series.
30. (i) Out of the ions Ag+, Co2+ and Ti4+ which
will be coloured in aqueous solution?
(ii) If each one of the above ionic species is placed
in a magnetic field, how will they respond and
why?
81
The d- and f-Block Elements
31. (a) Explain the following :
The enthalpies of atomization of transition
metals are quite high.
(b) Explain the following observations.
(i) With the same d-orbital configuration
(d4), Cr2+ is a reducing agent while Mn3+ is an
oxidising agent.
(ii) There is hardly any increase in atomic size
with increasing atomic numbers in a series of
transition metals.
32. (i) Transition metals have very high melting
and boiling points. Why?
(ii) In d-block element, ionic radii of ions of
the same charge decreases progressively with
increasing atomic number in a series. Why?
33. (a)
E°(M2+/M)
Cr
–0.91
Mn
–1.18
Fe
Co
Ni
Cu
–0.44 –0.28 –0.25 +0.34
From the given data of E° values, answer the
following questions :
(i) Why is E°(Cu2+/Cu) value exceptionally positive?
(ii) Why is E°(Mn2+/Mn) value highly negative as
compared to other elements?
(b) Give reason and select one atom/ion which
will exhibit asked property :
(i) Sc3+ or Cr3+ (exhibit diamagnetic behaviour)
(ii) Cr or Cu (high melting and boiling point)
34. Give reasons for the following :
(i) Mn3+ is a good oxidising agent.
(ii) EM2+/M values are not regular for first row
transition metals (3d-series).
(iii) d-block elements exhibit more oxidation
states than f-block elements.
35. How would you account for the following :
(i) The oxidising power of oxoanions are in the
order
VO+2 < Cr2O72– < MnO4–
(ii) The third ionization enthalpy of manganese
(Z = 25) is exceptionally high.
(iii) Cr2+ is a stronger reducing agent than Fe2+.
Long Answer Type Questions (LA)
(iii) Co (II) is easily oxidised in the presence of
strong ligands.
38. (a) The elements of 3d transition series are
given as
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following :
(i) Write the element which shows maximum
number of oxidation states. Given reason.
(ii) Which element has the highest melting
point?
(iii) Which element shows only +3 oxidation
state?
(iv) Which element is a strong oxidising agent in
+3 oxidation state and why?
(v) Why Mn2O3 is basic whereas Mn2O7 is
acidic?
37. Compare the general characteristics of
the first series of the transition metals with
those of the second and third series metals in
the respective vertical columns. Give special
emphasis on the following points :
(i) electronic configuration
(ii) oxidation states
(iii) ionisation enthalpies and
(iv) atomic sizes.
39. (a) Account for the following :
(i) Transition metals show variable oxidation
states.
(ii) Zn, Cd and Hg are soft metals.
(b) Give reason :
Iron has higher enthalpy of atomization than
that of copper.
(c) What are interstitial compounds? Write their
properties.
36. (a) Give reason :
(i) Sc (21) is a transition element but Ca (20) is
not.
(ii) The Fe2+ is much more easily oxidised to
Fe3+ than Mn2+ to Mn3+.
(b) How would you account for the following :
(i) Metal-metal bonding is more extensive in
the 4d and 5d series of transition elements than
the 3d series.
(ii) Mn (III) undergoes disproportionation reaction
easily.
CBSE Board Term-II Chemistry Class-12
82
40. Give reasons for the following :
(i) Silver bromide is used in photography.
(ii) Most transition metal compounds are
coloured.
(iii) Zinc and not copper is used for the recovery
of metallic silver from complex
[Ag(CN)2]–. Explain.
(iv) The colour of mercurous chloride, Hg2Cl2,
changes from white to black when treated with
ammonia.
(v) The species [CuCl4]2– exists while [CuI4]2–
does not.
OBJECTIVE TYPE QUESTIONS
11. (c) : Zn 2+ → 3d 10 has no unpaired electrons to be
excited. Hence, it is colourless.
1. (b) : The overall decrease in atomic and ionic radii from
La3+ to Lu3+ is called lanthanoid contraction. Hence, the
correct order is
Yb3+ < Pm3+ < Ce3+ < La3+
2. (a) : Cu 2+ has 3d 9 configuration i.e., one unpaired
electron, hence, it is paramagnetic while Zn2+ has 3d 10
configuration i.e., all orbitals are filled, hence it is diamagnetic
in nature.
3. (c) : Fe3+ – 3d 5
Cr3+ – 3d 3
Ni2+ – 3d 8
Cu2+ – 3d 9
No. of unpaired electrons = 5
No. of unpaired electrons = 3
No. of unpaired electrons = 2
No. of unpaired electrons = 1
4. (a) : Acidic strength of oxides of transition metals
increases with increase in oxidation number.
+2
+8 /3
+3
+4
+7
Hence acidic strength is of the order of
MnO < Mn3O4 < Mn2O3 < MnO2 < Mn2O7
5.
Acidic
Amphoteric
(d) :
24 Cr → 1s
2
1 5
2s 2 2p 6 3s 2 3p 6 4
s
3
d
half -filled
6. (c) : E° values for M 2+/M with negative signs are
Cr = – 0.91 V, Mn = – 1.18 V, Fe = – 0.44 V, Co = – 0.28 V
Thus, the order is Mn > Cr > Fe > Co.
+6
+7
+4
→ 2MnO4− + MnO2 + 2H2O shows
7. (b) : 3MnO24− + 4H+ 
disproportionation since the oxidation state of Mn changes
from +6 to +7 (MnO4– ) and +4 (MnO2).
8.
(d) : Hg is the last element of third transition series.
9. (a) : µ = n (n + 2)
Electronic configuration of ion (25) = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 5 4 s 2
No. of unpaired electrons (n) = 5
µ = 5(5 + 2) = 35 = 5.9 B.M.
10. (d)
13. (c) : They are not chemically reactive. They are chemically
inert.
14. (b) : Reactivity of transition elements decreases almost
regularly from Sc to Cu because of regular increase in
ionisation enthalpy.
15. (a) : Zn2+(3d10) has zero unpaired electron (colourless).
Ni2+(3d 8) has 2 unpaired electrons (coloured).
Cu2+(3d 9) has 1 unpaired electron (coloured).
Cr2+(3d 4) has 4 unpaired electrons (coloured).
16. (d) : Ionic radii of trivalent lanthanides decrease with
increase in atomic number.
17. (c) : µ = n (n + 2) BM
. .
MnO, Mn3 O4 ,Mn2 O3 ,MnO2 ,Mn2 O7
Basic
12. (b) : The third ionisation energy of Mn required to
change Mn2+(d 5) to Mn3+(d 4) is much larger due to stable
half-filled d 5 electronic configuration.
18. (a) : Though Cu(I) has 3d10 stable configuration while
Cu(II) has 3d 9 configuration, yet Cu(II) is more stable due
to greater effective nuclear charge of Cu(II) (i.e., to hold 17
electrons instead of 18 in Cu(I)).
19. (b) : In case of transition metal oxides, the oxides with
metals in lower oxidation state are basic in nature.
O.S. of Mn in Mn2O7 = +7; V in V2O3 = +3; V in V2O5 = +5;
Cr in CrO = +2; Cr in Cr2O3 = +3
Thus in V2O3, CrO and Cr2O3 transition metal ion is in lower
oxidation state but Cr2O3 is amphoteric in nature. Hence V2O3
and CrO are basic in nature.
20. (c) : In the Wacker process, the oxidation of ethyne to
ethanal is catalysed by PdCl2.
21. (a) : Lanthanum is a d-block element which resembles
lanthanides.
22. (d) : The electronic configuration of Ce3+ is 4f 1
Hence, µ = n (n + 2) = 11
( + 2) = 1.732 B.M.
23. (a) : Tl+ is more stable than Tl3+ due to inert pair effect.
Cu2+ is more stable than Cu+. Cr3+ is more stable than Cr2+.
V4+ in aqueous solution is more stable than V2+.
83
The d- and f-Block Elements
24. (b) : Ag2SO4 → Ag+ (4d10) – colourless
CuF2 → Cu2+ (3d 9) – coloured
MgF2 → Mg2+ (no d-electrons) – colourless
CuCl → Cu+ (3d10) – colourless
25. (b) : Basic strength decreases from La(OH)3 to Lu(OH)3.
Hence, (1) is incorrect.
0
Ce : [Xe] 4f 1 5d1 6s2; Ce4+ : [Xe] 4f
Yb : [Xe] 4f 14 6s2; Yb2+ : [Xe] 4f 14
Lu : [Xe] 4f 14 5d1 6s2; Lu3+ : [Xe] 4f 14
The given ions contain no unpaired electrons and therefore,
are diamagnetic.
26. (b)
27. (d) : Sc3+ (3d 0), Ti 4+ (3d 0) are diamagnetic while Pd2+
(4d 8) and Cu2+ (3d 9) are paramagnetic.
28. (d) : In CrO2Cl2, O.S. of Cr = +6
MnO4–, O.S. of Mn = +7
3–
Cr(CN)6 , O.S. of Cr = +3
2–
NiF6 , O.S. of Ni = +4
29. (a)
30. (b) : Fe2+ − 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6
Fe3+ − 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5
In Fe3+, orbital is half-filled hence provides extra stability.
31. (a) : The transition metals and their compounds are
known for their catalytic activity because of their ability to
adopt multiple oxidation states and to form complexes.
32. (a) : (n – 1)d1-10 ns0 –2
33. (b) : La(OH)3 is most basic. Hence, (I) is wrong. (II) is
correct due to lanthanoid contraction. (III) is correct because
Ce4+ tends to change to stable Ce3+.
34. (b) : [Fe(CN)6]4– ; No. of unpaired electrons = 0
[Fe(CN)6]3– ; No. of unpaired electrons = 1
[Ni(H2O)4]2+ ; No. of unpaired electrons = 2
[Cr(NH3)6]3+ ; No. of unpaired electrons = 3
35. (b) : 3d 5 configuration is more stable due to singly
occupied half-filled orbitals.
36. (a) : The increasing order of oxidising power is due to
increasing stability of the lower species to which they are
reduced.
37. (d) : More the number of unpaired d-electrons, more is
the magnetic moment.
Cu2+ – 3d 9 ; No. of unpaired electrons = 1
Ni2+ – 3d 8 ; No. of unpaired electrons = 2
Co2+ – 3d 7 ; No. of unpaired electrons = 3
Fe2+ – 3d 6 ; No. of unpaired electrons = 4
38. (c) : Cr2O3 and V2O5 are amphoteric oxides.
39. (b) : Hs (Z = 108) belongs to 6d series with electronic
configuration – [Rn]5f 146d 67s 2.
40. (d) : Ce is most commonly used lanthanoid, nobelium
(No) and Th (thorium) are actinoids.
41. (a) : Cr forms CrF3 whereas Cu and Ni do not form CuF3
and NiF3.
42. (c) : Due to lanthanoid contraction, Zr and Hf have
nearly equal size.
43. (b) : Due to presence of unpaired electrons in (n – 1)d
orbitals, the most of the transition metal ions and their
compounds are paramagnetic. They form coloured ions and
show variable oxidation states due to presence of vacant
d-orbitals.
44. (a) : The electronic configuration of Cr is
5
1
24Cr → [Ar] 3d 4s (six unpaired electrons)
In K2Cr2O7, the oxidation number of Cr = +6. So it has no
unpaired electron.
45. (a) : Due to lanthanoid contraction, the atomic radii of
second and third row transition elements is almost same.
Hence, they resemble each other much more as compared
to first row elements.
46. (b) : Na2[CuCl4] ; Cu = +2 or
Na2[CdCl4] ; Cd = +2 or
K4[Fe(CN)6] ; Fe = +2 or
K3[Fe(CN)6] ; Fe = +3 or
Since Cd 2+ has completely filled
colourless.
Cu2+ → 3d 9
Cd2+ → 4d 10
Fe2+ → 3d 6
Fe3+ → 3d 5
d-subshell hence it is
47. (b) : Cu+ ions undergo disproportionation,
2Cu+ → Cu2+ + Cu
48. (d) : The colour of transition metal ions is due to d-d
transitions.
49. (c) : Negative value of E° for Ni2+/Ni is related to the
highest negative DH°hyd .
50. (d) : 3[Fe(H2O)6]2+ + NO3– + 4H+
NO + 3[Fe(H2O)6]3+ + 2H2O
[Fe(H2O)6]
2+
(X ) + NO
[Fe(H2O)5NO]
Brown (Z)
2+
(Y )
+ H 2O
In Y, Fe3+ is present : [Ar]3d 5
µ = n (n + 2) = 5.9 B.M.
NO is a neutral oxide.
51. (c) : From the melting point trend it is clear that
W → Cr
X → Mn
Y → Fe
Z → Co
CBSE Board Term-II Chemistry Class-12
84
Mn2+ ⇒ [Ar]3d 5
µ = 5 (5 + 2) = 5.9 B.M.
Cr3+ → Violet
Fe3+ catalyses the given reactions :
I2 + 2SO42–
2I– + S2O82–
Co shows +2 and +3 oxidation states.
52. (d) : (a) Cr2+ is a reducing agent, it gets oxidised to
Cr3+ (3d 3 or t 32g, stable half-filled configuration).
(b) Mn3+ is an oxidizing agent, it gets reduced to Mn2+
(3d 5, most stable, half-filled configuration).
Mn (25) : 3d 54s2
(c) Cr (24) : 3d 44s2
2+
4
Cr : 3d
Mn3+ : 3d 4
Both Cr2+ and Mn3+ exhibit d 4 electronic configuration.
(d) When Cr2+ is used as a reducing agent, the chromium
ion attains d 3 electronic configuration.
53. (b) : Ce3+ and Yb3+ are colourless despite having one
unpaired electron.
The lanthanide ions, other than 4f 0 type (La3+ and Ce4+) and
the 4f 14 type (Yb2+ and Lu3+) are all paramagnetic.
54. (c) : The elements scandium and yttrium are called “rare
earths” because they were originally discovered together
with lanthanides in rare minerals and isolated as oxides or
“earths”. Collectively, these metals are also called rare earth
elements.
55. (a)
64. (a) : Oxide of V in lowest oxidation state, i.e., VO is basic
and ionic in character.
65. (a) : Vanadyl ion is VO2+ where V is in +4 oxidation
state.
66. (b)
67. (c)
68. (a)
69. (a)
70. (a) : Lesser and negative reduction potential indicates
that Cr2+ is a reducing agent. Higher and positive reduction
potential indicates that Mn3+ is a stronger oxidizing agent.
71. (a) : The electrode which has more reduction potential
is a good oxidizing agent and has least reducing power.
72. (b) : (I) It is because Mn 2+ has 3d 5 electronic
configuration which has extra stability.
(II) Not titanium but copper, because with +1 oxidation
state an extra stable configuration, 3d10 results.
(III) It is not stable as it undergoes disproportionation;
2Cu+(aq) → Cu2+(aq) + Cu(s). The E° value for this is favourable.
(IV) Much larger third ionisation energy of Mn (where the
required change is d 5 to d 4) is mainly responsible for this.
+
73. (a) : The stability of Cu2+
(aq) rather than Cu (aq) is due to
+
than Cu+, which
the much more negative DHydH° of Cu2(aq)
more than compensates for the second ionisation enthalpy
of Cu.
56. (d) : In lanthanide series, with increasing atomic number,
there is a progressive decrease in the atomic as well as on
radii of trivalent ions form La3+ to Lu3+.
57. (c) : Yb 2+ is more stable than Yb3+ because it will
acquire stable configuration of completely filled 4f subshell
after losing 2 electrons. Electronic configuration of Yb2+ is
[Xe]4f14.
74. (a) : Terbium (Tb3+) : 4f 8, Erbium (Er3+) : 4f 11, Ytterbium
(Yb) : 4f 13.
75. (a)
76. (c)
58. (a)
79. (b) : Transition metals form substitutional alloys since
they have nearly the same size, they can substitute one
another in the crystal lattice.
59. (b) : The transition metals and their compounds are
known for their catalytic activity. This activity is ascribed
to their ability to adopt multiple oxidation states to form
complexes.
60. (c) : Because of similar radii and other characteristics of
transition metals, alloys are readily formed by these metals.
61. (c)
62. (b) : Greater the number of valence electrons, more will
be the number of oxidation states exhibited by the element.
63. (a) : All the d-block elements are metals, they exhibit
most properties of metals like lustre, malleability, ductility,
high density, high melting and boiling point, hardness,
conduction of heat and electricity, etc. All the f-block elements
are also metals but they are not good conductors of heat and
electricity.
77. (a)
78. (b) : The almost identical radii of Zr (160 pm) and
Hf (159 pm), a consequence of lanthanoid contraction.
80. (c) : Mn2+ = [Ar]3d 5, Mn3+ = [Ar]3d 4
Fe2+ = [Ar]3d 6, Fe3+ = [Ar]3d 5
Thus, Mn2+ has more stable configuration than Mn3+ while
Fe3+ has more stable configuration than Fe2+.
Hence, reduction potential for Mn3+/Mn2+ couple is more
positive than Fe3+/Fe2+.
As we move across the period, ionisation potential increases,
thus, ionisation potential of Fe is more than that of Mn.
81. (a) : A half-filled or fully-filled orbital is more stable than
incompletely filled orbital.
Hence Mn2+ (3d 5) is more stable than Mn3+ (3d 4).
85
The d- and f-Block Elements
82. (b) : Ti3+ has [Ar]3d1 configuration. Thus, d–d transition
is possible and thereby it shows colour.
83. (b) : Chromium has maximum number of unpaired
d electrons while Hg do not have any unpaired
d electrons. Thus, Cr is hard but Hg is soft metal.
84. (d) : Even though Cu+ has completely filled 3d orbitals,
[Ar]3d10 the nuclear charge in Cu is not enough to hold a
core of 18 electrons in Cu+ and thus, Cu+ is unstable in
comparison to Cu2+. Cu+ is diamagnetic in nature.
85. (d) : Both Co and Ni have (IV)
has 3d 6 electronic configuration.
Metals Outer electronic
configuration
Co 3d 74s2
Ni 3d 84s2
SUBJECTIVE TYPE QUESTIONS
1. Eu2+ has a strong tendency to lose electrons to attain
the more stable +3 oxidation state of lanthanoids hence, it
is a strong reducing agent.
2. Formula of oxoanion of manganese is MnO–4.
Oxidation state of Mn in this oxoanion = + 7
Group number of Mn is 7.
3.
Lanthanoids showing +4 oxidation state are
58Ce, 59Pr, 65Tb.
oxidation state. Ni (IV)
Oxi. states
+2, +3, +4
+2, +3, +4
86. (a)
87. (b) : Due to large surface area and variable valencies,
transition metals form intermediate activated complexes
easily, hence they are used as good catalysts.
88. (b) : The electronic configurations of europium (II) and
cerium (II) are Eu2+ : [Xe] 4f 7 and Ce2+ : [Xe] 4f 1 5d1
In Eu2+, f-subshell is half-filled thus, it is more stable.
89. (a) : Zinc has all electrons paired [Ar]3d104s2. So, it is
diamagnetic in nature.
90. (d) : The oxidation states of the given compounds are
the following,
VO+2 : x + 2(–2) = +1
⇒ x = +5
VO : x –2 = 0
⇒ x = +2
VO2+ : x + 1(–2) = +2
⇒ x=+4
The correct order of oxidising power is :
+
VO < VO2+ < VO 2
–
In MnO4 : x + 4 (–2) = –1
⇒ x – 8 = –1 ⇒ x = – 1 + 8 = +7
91. (b) : Some non-metallic atoms (e.g., H, B, C, N, etc.) are
able to fit in the interstitial sites of transition metals lattices
to form interstitial compounds.
92. (b) : It is due to lanthanide contraction.
93. (a)
94. (a)
95. (b) : In actual practice transition metals react with acid
very slowly and act as poor reducing agents. This is due to
the protection of metal as a result of formation of thin oxide
protective film. Further, their poor tendency as reducing agent
is due to high ionisation energy, high heat of vapourization
and low heat of hydration.
4. In aqueous solutions, Cu+ undergoes disproportionation
to form a more stable Cu2+ ion.
2Cu+(aq) → Cu2+(aq) + Cu(s)
Cu in aqueous solutions is more stable than Cu+ ion
because hydration enthalpy of Cu2+ is higher than that of
Cu+. It compensates the second ionisation enthalpy of Cu
involved in the formation of Cu2+ ions.
2+
5. Europium (Eu) is well known to exhibit +2 oxidation
state due to its half-filled f orbital in +2 oxidation state.
6. In the electronic configuration of Zn the d-orbitals are
completely filled in the ground state as well as in its common
oxidation state. So, it is not regarded as transition metal.
7. Lanthanum and all the lanthanoids predomi-nantly
show +3 oxidation state. However, some of the lanthanoids
also show +2 and +4 oxidation states in solution or in solid
compounds. This irregularity arises mainly due to attainment
of stable empty (4f 0), half-filled (4f 7) and fully filled (4f14)
sub shell.
e.g., Ce4+ : 4f 0 , Eu2+ : 4f 7
Tb4+ : 4f 7 , Yb2+ : 4f 14
8. Due to lanthanoid contraction the elements of 4d and
5d-series have similar atomic radii e.g., Zr = 145 pm and
Hf = 144 pm.
9. Oxoanion of chromium in which it shows +6 oxidation
state equal to its group number is Cr2O72– (dichromate ion).
10. Zn2+ ion has completely filled d-subshell and no d-d
transition is possible. So zinc salts are white. Configuration
of Cu2+ is [Ar] 3d 9. It has partly filled d-subshell and hence
it is coloured due to d-d transition.
11. Europium (II) has electronic configuration [Xe]4f 75d 0
while cerium (II) has electronic configuration [Xe] 4f 1 5d1. In
Eu2+, 4f subshell is half filled and 5d-subshell is empty. Since,
half filled and completely filled electronic configurations are
more stable, so Eu2+ ions is more stable than Ce2+ in which
neither 4f subshell nor 5d subshell is half filled or completely
filled.
CBSE Board Term-II Chemistry Class-12
86
12. Magnetic moment, meff = 3.87 BM corresponds to the
number of unpaired electrons, n = 3 by applying the formula.
meff =
n(n + 2) BM
For n = 1, m = 1.73 B.M, for n = 2, m = 2.83 BM
For n = 3, m = 3.87 B.M and so on.
13. Cr2+ is a stronger reducing agent than Fe2+.
E° Cr3+/Cr2+ is negative (–0.41 V) whereas E° Fe3+/Fe2+ is
positive (+ 0.77 V). Thus Cr2+ is easily oxidized to Cr3+ but
Fe2+ cannot be easily oxidized to Fe3+. Hence, Cr2+ is stronger
reducing agent than Fe2+.
(ii) More positive is the value of E°, reaction will be more
feasible.
As E°Co3+/Co2+ is maximum, thus Co2+ ion is most stable.
14. (i) Electrode potential (E°) value is the sum of three
factors :
(a) Enthalpy of atomisation ; DaH for Cu(s) → Cu(g)
(b) Ionisation enthalpy ; DiH for Cu(g) → Cu2+
(g)
Cu2+
(g)
Cu2+
(aq)
(c) Hydration enthalpy ; DhydH for
→
In case of copper the sum of enthalpy of atomisation and
ionisation enthalpy is greater than enthalpy of hydration.
This is why E°M2+/M for Cu is positive.
(ii) This is due to lanthanoid contraction.
15. Ce (Z = 58) = [Xe] 4f1 5d1 6s2
\ Ce3+ = [Xe] 4f1 5d 0 6s0
Therefore, it has only one unpaired electron. i.e., n = 1
\
µ = n (n + 2) = 11
( + 2) = 3 = 1.73B.M.
16. (i) Copper exhibits +1 oxidation state in its compounds.
Electronic configuration of Cu in the ground state is 3d10 4s1.
So, Cu can easily lose 4s1 electron to attain a stable 3d10
configuration. Thus, it shows +1 oxidation state.
(ii) Only those ions will be coloured which have partially filled
d-orbitals facilitating d-d transition. Ions with d0 and d10 will
be colourless.
From electronic configuration of the ions, V3+(3d 2) and
Mn2+(3d 5), are all coloured. Ti4+(3d 0) and Sc3+(3d 0) are
colourless.
17. (i) E° values for the Cr3+/Cr2+ and Mn3+/Mn2+ couples
are
–
Cr3+
Cr2+
(aq) + e
(aq); E° = –0.41 V
–
Mn3+
(aq) + e
Mn2+
(aq) ; E° = +1.551 V
These E° values indicate that Cr2+ is strongly reducing while
Mn3+ is strongly oxidising agent.
(ii) Middle of the transition series contains greater number of
unpaired electrons in (n –1)d and ns orbitals.
18. Lanthanoid contraction : The steady decrease in the
atomic and ionic radii of lanthanoid elements with increase
in atomic number is called lanthanoid contraction.
It is caused due to imperfect shielding of nuclear charge by
4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of
lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements
and 5d series elements are similar. e.g., atomic radii of
zirconium(Zr) is same as that of hafnium (Hf).
19. Disproportionation reaction involves the oxidation
and reduction of the same substance. The examples of
disproportionation reaction are :
(i) Aqueous NH3 when treated with Hg2Cl2 (solid)forms
mercury aminochloride disproportionatively.
Hg + Hg(NH2)Cl + NH4Cl
Hg2Cl2 + 2NH3
+
Cu + Cu2+
(ii) 2Cu
20. (i) Elements which have incompletely filled d-orbitals in
their ground state or in any one of their oxidation states are
called transition elements.
Characteristics of transition elements :
(a) They show variable oxidation states.
(b) They exhibit catalytic properties.
(ii) Zn, Cd, Hg are considered as d-block elements but not
as transition elements because they do not have partly filled
d-orbitals in their atomic state or their common oxidation
states (i.e., Zn2+, Cd2+, Hg2+).
21. (i) Due to presence of unpaired electrons in f-orbital
lanthanoid ions are paramagnetic in nature.
(ii) Due to lanthanoid contraction, their sizes are same.
Hence, their properties are similar.
(iii) Ce4+. The stable oxidation state of lanthanoids is +3.
Ce4+ tends to accept an electron to change to +3 state.
Hence, it acts as a good oxidising agent.
22. (i) Ti4+ has highest oxidation state among the given ions.
Ti4+ has stable inert gas configuration and hence, most stable
in aqueous solution.
On the other hand, V2+, Mn3+, Cr3+ have unstable electronic
configuration and hence, are less stable.
(ii) Due to presence of highest oxidation state of Ti, it acts
as the strongest oxidising agent among the given ions.
(iii) Due to absence of unpaired electron in Ti4+, it is a
colourless ion.
E.C. of Ti4+ : [Ar]3d 04s0
23. The electronic configuration of Zn and Cu are :
Zn: 1s2 2s2 2p6 3s2 3p6 3d10 4s2
Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1
87
The d- and f-Block Elements
From the above configuration it is clear that first ionisation
energy of Zn is greater than that of Cu (because of 4s2 and
4s1 configuration of Zn and Cu respectively). More energy is
needed to remove an electron from 4s2 than from 4s1.
The second I.E. of Cu is higher than that of Zn because for
Cu+ the configuration is 1s2 2s2 2p6 3s2 3p6 3d10 and for
Zn+ the configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s1, it is
easier to remove 4s1 electron of Zn+ than a 3d-electron from
3d10 (stable configuration).
24. (i) The E° value for Ce4+/Ce3+ is 1.74 V which suggests
that it can oxidise water. However the reaction rate is very
slow and hence Ce (IV) is a good analytical reagent.
(ii) Cu(I) compounds have completely filled d-orbitals and
there are no vacant d-orbitals for promotion of electrons
whereas in Cu(II) compounds have one unpaired electron
which is responsible for colour formation.
25. In the lower oxidation state, the transition metal
oxides are basic and they are acidic if the metal is in higher
oxidation state. The oxides are amphoteric when the metal
is in intermediate oxidation state.
For example,
+3
Mn2O3
Basic
+4
MnO2
Amphoteric
+7
Mn2O7
Acidic
In case of lower oxide of a transition metal, the metal atom
has a low oxidation state. This means some of the valence
electrons of the metal atom are not involved in bonding,
hence, these can be used for donation. Thus, these are act
as bases.
26. (i) This is due to lanthanoid contraction.
(ii) Much larger third ionisation energy of Mn(where change
is d 5 to d 4) is mainly responsible for this. This also explains
that +3 state of Mn is of little importance.
From the relation, DG° = –nFE°
More positive is the value of E°, reaction will be feasible.
–
+ e–
Mn3+
Mn2+
;
Fe3+ + e Fe2+
3d 4
3d 5
3d 5
3d 6
more stable
more stable
(half filled)
(half filled)
3+
2+
Hence, E°value for Mn /Mn couple is much more positive
than that for Fe3+/Fe2+.
(iii) Manganese can form pp-dp bond with oxygen by utilising
2p-orbital of oxygen and 3d-orbital of manganese due to
which it can show highest oxidation state of +7. While with
fluorine it cannot form such pp - dp bond thus, it can show
a maximum oxidation state of +4.
27. (a) (i) In aqueous solutions, Cu + undergoes
disproportionation to form a more stable Cu2+ ion.
2Cu+(aq) → Cu2+(aq) + Cu(s)
Cu 2+ in aqueous solutions is more stable than Cu + ion
because hydration enthalpy of Cu2+ is higher than that of
Cu+. It compensates the second ionisation enthalpy of Cu
involved in the formation of Cu2+ ions.
(ii) This is attributed to the involvement of greater number
of electrons from (n –1)d in addition to the ns electrons in
the interatomic metallic bonding.
(b) Third ionization enthalpy of lanthanoid is low if it leads
to stable empty, half filled or completely filled configuration,
as indicated by the abnormally low third ionization enthalpies
of La, Gd, and Lu.
28. (i) The transition metals and their compounds, are known
for their catalytic activity. This activity is ascribed to their
ability to adopt multiple oxidation states, ability to adsorb
the reactant(s) and ability to form complexes. Vanadium (V)
oxide (in Contact Process), finely divided iron (in Haber’s
Process), and nickel (in catalytic hydrogenation) are some of
the examples.
(ii) Lowest oxidation compounds of transition metals are basic
due to their ability to get oxidised to higher oxidation states.
Whereas, the higher oxidation state of metal and compounds
gets reduced to lower ones and hence acts as acidic in nature.
e.g., MnO is basic whereas Mn2O7 is acidic.
(iii) :
Fe2+ ⇒ 3d 6, \ n = 4
Magnetic moment =
n(n + 2) B.M. = 24 B.M.
29. (a) (i) Transition metals form a large number of complex
compounds due to following reasons :
– Comparatively smaller size of metal ions.
– High ionic charges.
– Availability of d-orbitals for bond formation.
(ii) Because Mn2+ is stable due to half filled configuration.
Thus Mn3+ has high tendency to form Mn2+ while Cr3+ is
more stable than Cr2+.
(iii) Only those ions will be coloured which have partially
filled d-orbitals facilitating d-d transition. Ions with d 0 and
d10 will be colourless.
From electronic configuration of the ions, V3+(3d 2) and
Mn2+(3d 5), are all coloured. Ti4+(3d 0) and Sc3+(3d 0) are
colourless.
30. (i) The ionic species which possesses unpaired electron
or electrons in (n – 1)d-subshell will show colour. Out of
the ions Ag+(4d10), Co2+(3d7) and Ti4+(3d 0), Co2+ will be
coloured as it contains three unpaired electrons, Ag+ and
Ti4+ will be colourless.
(ii) When placed in magnetic field, Co2+ will be attracted
because it is paramagnetic due to unpaired electrons. Ag+
and Ti4+ ions will be repelled by the magnetic field as they
are diamagnetic.
CBSE Board Term-II Chemistry Class-12
88
31. (a) As transition metals have a large number of unpaired
electrons in the d-orbitals of their atoms they have strong
interatomic attraction or metallic bonds. Hence, they have
high enthalpy of atomization.
(b) (i) E° values for the Cr3+/Cr2+ and Mn3+/Mn2+ couples
are
–
Cr3+
Cr2+
(aq) + e
(aq); E° = –0.41 V
–
Mn3+
(aq) + e
Mn2+
(aq) ; E° = +1.551 V
These E° values indicate that Cr2+ is strongly reducing while
Mn3+ is strongly oxidising agent.
(ii) As one proceeds along a transition series, the nuclear
charge increases which tends to decrease the size but the
addition of electrons in the d-subshell increases the screening
effect which counterbalances the effect of increased nuclear
charge. As a result, the atomic radii remain practically same
after chromium.
32. (i) The high melting and boiling points of transition
metals are attributed to the involvement of greater number
of electrons from (n – 1) d-orbital in addition to the ns
electrons in the interatomic metallic bonding (d-d overlap).
(ii)As the atomic number increases the new electron enters
the d-orbital and expected to increase in atomic size, but
due to poor shielding effect of d-orbitals the electrostatic
attraction between nucleus and outermost orbital increases
and hence, the ionic radii decreases.
33. (a) (i) Copper has high energy of atomisation and low
energy of hydration.
(ii) Mn2+ ion has stable half-filled (3d5) electronic configuration.
Its ionisation enthalpy value is lower in comparison to hydration
enthalpy. Hence E °Mn2+/Mn is more negative.
(b) (i) Sc3+ has 3d 0 outer electronic configuration, therefore
it is diamagnetic in nature whereas Cr 3+ has 3d 3 outer
electronic configuration. So, it is paramagnetic due to
presence of unpaired electrons.
(ii) In a particular series, the metallic strength increases
upto middle with increasing number of unpaired electrons,
i.e., upto d 5 configuration. After Cr, the number of unpaired
electrons goes on decreasing. Accordingly, the m.pt and b.pt.
decrease after middle (Cr) because of increasing pairing of
electrons.
34. (i) Mn2+ is more stable due to half filled d5 configuration
and Mn3+ easily changes to Mn2+ hence, it is a good oxidising
agent.
(ii) The E° M 2+ / M values are not regular which can be
explained from the irregular variation of ionisation enthalpes
i.e., IE1 + IE2 and also the sublimation enthalpies which are
relatively much less for manganese and vanadium.
(iii) All transition elements except the first and the last
member in each series show a large number of variable
oxidation states. This is because difference of energy in the
(n – 1)d and ns orbitals is very little.
Hence, electrons from both the energy levels can be used
for bond formation.
35. (i) Change in Cr2O72– to Cr(III) is 3 and in MnO4– to
Mn (II) is 5.
Change in oxidation state is large and the stability of reduced
products in V(III) < Cr(III) < Mn(II). This is why oxidising
power of VO2+ < Cr2O72– < MnO4–.
(ii) Third ionization enthalpy of Mn is very high because the
third electron has to be removed from the stable half-filled
3d-orbitals [Mn2+ (Z = 25) = 3d 5].
(iii) Cr 2+ is a stronger reducing agent than Fe 2+. E°Cr3+/
Cr2+ is negative (–0.41 V) whereas E°Fe3+/Fe2+ is positive
(+ 0.77 V). Thus Cr 2+ is easily oxidized to Cr 3+ but Fe2+
cannot be easily oxidized to Fe3+. Hence, Cr2+ is stronger
reducing agent than Fe2+.
36. (a) (i) Sc(21) is a transition element but Ca(20) is not
because of incompletely filled 3d orbitals.
(ii) Electronic configuration of Mn2+ is 3d 5 which is half
filled and hence stable. Therefore, third ionization enthalpy
is very high, i.e., 3rd electron cannot be lost easily. In case of
Fe2+, electronic configuration is 3d 6. Hence, it can lose one
electron easily to give the stable configuration 3d 5.
(b) (i) The metals of 4d and 5d-series have more frequent
metal bonding in their compounds than the 3d-metals
because 4d and 5d-orbitals are more exposed in space than
the 3d-orbitals. So the valence electrons are less tightly held
and form metal-metal bonding more frequently.
(ii) Mn3+ is less stable and changes to Mn2+ which is more
stable due to half-filled d-orbital configuration. That is why,
Mn3+ undergoes disproportionation reaction.
(iii) The tendency to form complexes is high for Co(III) as
compared to Co(II). Co2+ ions are very stable and are difficult
to oxidise. Co3+ ions are less stable and are reduced by water.
In contrast many Co(II) complexes are readily oxidised to
Co(III) complexes and Co(III) complexes are very stable, e.g.,
[Co(NH3 )6 ]2+ 
→ [Co(NH3 )6 ]3+
Air
This happens because the crystal field stabilisation energy of
Co(III) with a d 6(t26g) configuration is higher than for Co(II)
with a d 7 (t 62g eg1) arrangement.
37. (i)E l e c t r o n i c c o n f i g u r a t i o n : E l e c t r o n i c
configuration of group 3 elements (Sc, Y, La) is [Noble gas]
(n – 1)d1 ns2.
89
The d- and f-Block Elements
Elements of group 6 (Cr, Mo and W) show exception in
electronic configuration. For Cr and Mo [Noble gas] (n – 1)
d 5 ns1 and for W it is [Noble gas] 4f 14 5d 4 6s2.
Group 11 elements (Cu, Ag and Au) also show exceptional
electronic configuration :
Cu : [Ar] 3d10 4s1, Ag : [Kr]4d10 5s1,
Au : [Xe]4f 14 5d 10 6s1.
Group - 10 (Ni, Pd and Pt) also show anomalous electronic
configuration:
Ni : [Ar] 3d 8 4s2
Pd : [Kr] 4d10 5s0
Pt : [Xe] 4f 14 5d 9 6s1
(ii) Oxidation states : Elements within the same group
show similar oxidation states. Highest number of oxidation
states are shown by the elements lying in the middle of the
transition series. Minimum oxidation states are shown by the
elements lying near to left and far right side of the series.
Stability of higher oxidation states increases from first to third
series.
(iii) Ionisation enthalpies : Ionisation enthalpies generally
decrease down a group. This trend is followed from 3d to
4d-elements but the ionisation enthalpies either remain same
or increase in going from 4d to 5d-series with the same group.
This reverse trend is due to the poor shielding of the nuclear
charge by the inner 4f-electrons. This increases the Zeff and in
turn increases the ionisation enthalpy.
(iv) Atomic size : Due to poor shielding of nuclear charge
by 4f-electrons, increase in Zeff decreases the size. So, the
atomic size increase from 3d to 4d but decrease or remain
almost the same from 4d to 5d.
38. (i) Mn shows maximum no. of oxidation states from
+2 to +7 because Mn has maximum number of unpaired
electrons in 3d sub-shell.
(ii) Cr has maximum melting point, because it has 6 unpaired
electrons in the valence shell, hence it has strong interatomic
interaction.
(iii) Sc shows only +3 oxidation state because after losing
3 electrons, it has noble gas electronic configuration.
(iv) Mn is strong oxidising agent in +3 oxidation state
because change of Mn3+ to Mn2+ give stable half filled (d 5)
electronic configuration,
E°(Mn3+/Mn2+) = 1.5 V.
(v) Basic nature of oxides decreases and acidic nature
increases with increase in oxidation state of the metal.
Oxidation state of Mn in Mn2O3 is +3 while in Mn2O7 is +7.
39. (a) (i) Transition elements can use their ns and
(n – 1)d orbital electrons for bond formation therefore, they
show variable oxidation states.
For example, Sc has ns2(n – 1) d1 electronic configuration.
It utilizes two electrons from its ns subshell then its oxidation
state = +2. When it utilizes both the electrons then its
oxidation state = +3.
(ii) In Zn, Cd and Hg, all the electrons in d-subshell are
paired. Hence, the metallic bonds are weak. That is why they
are soft metals with low melting and boiling points.
(b) Greater the number of unpaired electrons, stronger is
the metallic bond and therefore, higher is the enthalpy of
atomisation. Since, iron has greater number of unpaired
electrons than copper hence has higher enthalpy of
atomisation.
(c) When small atoms of non metals like H, C, B, N etc can
occupy vacant interstitial spaces in condition metals it give
size to interstitial compound like hydrides, carbides. Few
properties are as follow
(i)
They have high melting points then pure metals.
(ii) They are conductive.
(iii) They are chemically inert.
40. (i) Silver bromide is used in photography because of its
sensitivity to sunlight. In light, AgBr reduces to metallic silver.
(ii) The colour of transition metal compound is due to
the presence of incompletely filled d-orbitals in transition
metal ions/atoms, because of this d-d transition can occur
in them. The colour is due to d-d transition for which the
energy is absorbed from visible region. The visible colour of
a compound is the complementary colour of the absorbed
light.
(iii) Zinc is a cheaper and stronger reducing agent as
compared to copper.
(iv) Mercurous chloride (white) changes to black on
treatment with ammonia because of the formation of finely
divided mercury (grey).
(v) Cu2+ is reduced to Cu+ by I– and thus CuI2 gets converted
to Cu2I2. This change cannot be brought about by Cl–.

CHAPTER
5
Coordination Compounds
Recap Notes
Werner’s coordination theory :
It explains the nature of bonding in
complexes. Metals show two different
kinds of valencies:
– Primary valency : Non directional and
ionisable. It is equal to the oxidation state
of the central metal ion.
– Secondary valency : Directional
and non-ionisable. It is equal to the
coordination number of the metal. It
is commonly satisfied by neutral and
negatively charged or sometimes by
positively charged ligands.
X
X
The ionisation of the coordination
compound is written as :
[Co(NH3)6]Cl3
[Co(NH3)6]3+ + 3Cl–
Representation of CoCl3 . 6NH3 complex
according to Werner's theory
Addition compounds : The compounds
formed by combination of two or more simple
compounds are called addition compounds.
They are of two types :
X
Double salt : A compound formed by
combination of two or more simple
compounds, which is stable in solid state
only is called double salt. In solution it
breaks into component ions. e.g.,
X
K2SO4⋅Al2(SO4)3⋅24H2O;
Potash alum
FeSO4⋅(NH4)2SO4⋅6H2O;
Mohr’s salt
KCl⋅MgCl2⋅6H2O;
Carnallite
Complex compound : A compound
formed by combination of two or more
simple compounds which retain its
identity both in solid and solution states
is called complex compound.
e.g., K4[Fe(CN)6], Potassium ferrocyanide
[Cu(NH3)4]SO4, Cupramine sulphate
Some important terms pertaining
coordination compound :
X Coordination entity : The central metal
atom or ion and ligand taken together
is called coordination entity. It may be
positive, negative or neutral.
e.g., [Cu(NH3)4]2+, [Fe(CN)6]4–, [Ni(CO)4]
X Central atom : The atom or ion with
which definite number of ligands are
attached in a definite geometry is called
central atom/ion. Any atom/ion which has
high positive charge density or vacant
orbitals of suitable energy may be central
atom or ion, e.g., transition metals,
lanthanoids. It is Lewis acid (electron
acceptor).
X Ligands : Molecules or ions which are
bound to the central atom/ion in the
coordination entity are called ligands. A
molecule or ion which has high negative
charge or dipole or lone pair of electrons
may be ligands. It is Lewis base (electron
donor).
91
Coordination Compounds
X
Classification of ligands :
Ligands
On the basis
of charge
On the basis
of bonding
On the basis of
number of donor sites
Negative ligands
–
–
–
–
CN , F , Cl , NO2 ,
–
–
2–
NO3 , OH , O
Monodentate : Only one
donor site e.g., H2O, NH3
Positive ligands
+
+
+
NO2 , NO , N2H5
e.g., (COO–)2, CH2
NH2
CH2
NH2
Neutral ligands
H2O, NH3, CO,
NH2OH, CH3NH2
Bidentate : Two donor sites
(Oxalato)
(Ethylenediamine)
Polydentate : More than
two donor sites e.g.,
EDTA (Hexadentate)
Chelating ligands : A bidentate or polydentate
ligand which forms more than one coordinate
bonds in such a way that a ring is formed.
CH2 NH2
CH2 NH2
(Ethylenediamine)
Ambidentate ligands : Monodentate ligand which
contains more than one coordinating atom (or
donor atom).
M
M
X
X
Coordination number (C.N.) : The
total number of coordinate bonds
through which the central metal atom
or ion is attached with ligands is known
as coordination number. Examples :
[Ag(CN)2]– : C.N. = 2, [Cu(NH3)4]2+ : C.N.
= 4, [Cr(H2O)6]3+ : C.N. = 6
Coordination sphere : The central
atom and the ligands which are directly
attached are collectively known as
coordination sphere. It is non-ionisable
and written enclosed in square brackets.
The ionisable groups are written outside
the brackets.
Example :
[Cu(NH3)4] SO4
Coordination
sphere
X
Ionisable
group
[Cu(NH3)4]2+ + SO42–
Coordination polyhedron : The spatial
arrangement of the ligand atoms which
are directly attached to the central atom
defines a coordination polyhedron about
the central atom, e.g., [(Co(NH3)6)]2+
is octahedral [Ni(CO)4] is tetrahedral.
Octahedral is most common coordination
polyhedron.
or
O
N
–
N O, M
O
O
–
M
SCN, M
or
NCS M
CN
or
NC
Homoleptic and heteroleptic complexes:
X Homoleptic complexes : Complexes in
which a metal is bound to only one kind of
ligands are called homoleptic complexes.
e.g., [Co(NH3)6]3+, [Ti(H2O)6]3+, [Cu(CN)4]3–
X Heteroleptic complexes : Complexes
in which the central atom is bound
to different type of ligands are called
heteroleptic complexes.
e.g., [Co(NH3)4Cl2], K2[Fe(CN)5NO],
[Fe(H2O)5NO]SO4
Nomenclature of coordination
compounds :
X Rules for writing the formula of
coordination compounds :
– Formula of the cation whether simple or
complex must be written first, followed by
anion.
– The coordination sphere is written in
square brackets.
– Within the coordination sphere the
sequence of symbols is, first the metal atom
followed by anionic ligand then neutral
ligand finally cationic ligand. Ligands of
same type are arranged alphabetically.
– Polyatomic ligands are enclosed in
parentheses.
– The number of cations or anions to be
written in the formula is calculated on the
basis that total positive charge must be
equal to the total negative charge, as the
complex as a whole is electrically neutral.
CBSE Board Term-II Chemistry Class-12
92
X
–
–
–
–
X
–
–
–
–
–
Rules for naming coordination
compounds :
The cation is named first then the anion.
In naming coordination sphere, ligands
are named first in alphabetical order
followed by metal atom and then oxidation
state of metal by a roman numeral in
parentheses.
The complex part is written as one word.
When the coordination sphere is anionic,
name of central metal ends in – ate.
Naming of ligands :
Name of anionic ligands end in – o.
–
e.g., Cl : Chlorido
Neutral ligands (with a few exceptions)
retain their names e.g., NH3 : Ammine
Name of cationic ligands end in – ium.
+
e.g., NO2 : Nitronium
Certain ligands are represented by
abbreviations in parentheses instead of
their complex structural formulae.
e.g., ethylenediamine(en).
Ambidentate ligands are named by using
different names of ligands or by placing
the symbol of donor atom.
e.g., —SCN (Thiocyanato-S or Thiocya–
nato), —NCS (Thiocyanato-N or Isothio–
cyanato), —ONO (Nitrito-O or Nitrito),
–
—NO2 (Nitrito-N or Nitro)
– The prefixes di-, tri-, tetra-, pentaand hexa- are used to indicate the
number of each ligand. If the ligand
name includes such a prefix, the ligand
name should be placed in parentheses
and preceded by bis-(2), tris-(3),
tetrakis-(4), pentakis-(5) and hexakis-(6).
Bonding in coordination compounds :
X Valence bond theory : It was developed
by Pauling.
– A suitable number of vacant orbitals must
be present in the central metal atom or
ion for the formation of coordinate bonds
with the ligands.
– Central metal ion can use appropriate
number of s, p or d-orbitals for
hybridisation depending upon the total
number of ligands.
– The outer orbital (high spin) or inner
orbital (low spin) complexes are formed
depending upon whether outer d-orbitals
or inner d-orbitals are used.
C. No.
Type of
hybridisation
2
sp
Linear
[Ag(NH3)2]+, [Ag(CN)2]–
3
sp2
Trigonal planar
[HgI3]
sp3
Tetrahedral
Ni(CO)4, [NiX4]2–, [ZnCl4]2–, [CuX4]2–,
where, X = Cl–, Br–, I–
dsp2
Square planar
[Ni(CN)4]2–, [Cu(NH3)4]2+, [Ni(NH3)4]2+
dsp3
Trigonal bipyramidal
[Fe(CO)5], [CuCl5]3–
sp3d
Square pyramidal
[SbF5]2–
d2sp3
Octahedral (Inner orbital)
[Cr(NH3)6]3+, [Fe(CN)6]3–
sp3d2
Octahedral (Outer orbital)
[FeF6]3–, [Fe(H2O)6]2+, [Ni(NH3)6]2+
4
5
6
Geometry
Inner orbital complexes
Examples
–
Outer orbital complexes
Involves inner d-orbitals i.e., (n – 1)d-orbitals.
Involves outer d-orbitals i.e., nd-orbitals.
Low spin complexes
High spin complexes
Have less or no unpaired electrons.
e.g., [Co(NH3)6]3+, [Co(CN)6]4–
Have large number of unpaired electrons.
e.g., [MnF6]3–, [CoF6]3–
93
Coordination Compounds
– Low spin complexes are generally
diamagnetic and high spin complexes are
paramagnetic.
– Paramagnetism ∝ No. of unpaired electrons.
– Magnetic moment = n(n + 2) B.M. where
n = number of unpaired electrons.
X Crystal field theory : It assumes the
ligands to be point charges and there is
electrostatic force of attraction between
ligands and metal atom or ion. When
ligands approach the central metal ion,
then the five degenerate orbitals do
not possess equal energy any more and
results in splitting, which depends upon
nature of ligand field strength.
– Greater the ease with which the ligand
can approach the metal ion, the greater
will be the crystal field splitting caused by it.
– Crystal field splitting in octahedral
coordination complexes is shown as :
– If Do > P, then pairing of electrons takes
place and a low spin complex is formed.
– Crystal field splitting in tetrahedral
complexes is shown as :
– Difference in energy between e and t2
level is less in tetrahedral complexes.
4
∆t = ∆o
9
– Spectrochemical series : Arrangement
of ligands in the order of increasing field
strength.
Weak field
ligands
Increasing order of CFSE (∆ )
o

→ Strong field
ligands
–
–
I– < Br– < SCN– < Cl– < S2– < F < OH <
–
C2O42– < H2O < NCS < edta4– < NH3 < en <
NO–2 < CN– < CO
– If Do < P (where ‘P’ is energy required
for forced pairing of electrons) then the
electrons will remain unpaired and a high
spin complex is formed.
Colour of coordination compounds :
The magnitude of CFSE (Do) for most of the
complexes is of the same order as the energy
of a photon of visible light. Hence, whenever
d-d transition takes place, it imparts colour
to the complex. The colour of the complex is
the colour complementary to the wavelength
absorbed.
94
CBSE Board Term-II Chemistry Class-12
Practice Time
OBJECTIVE TYPE QUESTIONS
1. The correct IUPAC name of the coordination
compound K3[Fe(CN)5NO] is
(a) potassium pentacyanonitrosylferrate(II)
(b) potassium pentacyanonitroferrate(III)
(c) potassium nitritopentacyanoferrate(IV)
(d) potassium nitritepentacyanoiron(II).
6. Hexaamminenickel(II) hexanitrocobaltate(III)
can be written as
(a) [Ni(NH3)6][Co(NO2)6]
(b) [Ni(NH3)6]3[Co(NO2)6]2
(c) [Ni(NH3)6] [Co(NO2)6]
(d) [Ni(NH3)6(NO2)6]Co
2. Ammonia acts as a very good ligand but
ammonium ion does not form complexes because
(a) NH3 is a gas while NH4+ is in liquid form
(b) NH3 undergoes sp3 hybridisation while NH4+
undergoes sp3d hybridisation
(c) NH 4+ ion does not have any lone pair of
electrons
(d) NH4+ ion has one unpaired electron while
NH3 has two unpaired electrons.
7. Which of the following is correct?
(a) Valence bond theory explains the colour of
the coordination compounds.
(b) [NiCl4]2– is diamagnetic in nature.
(c) EDTA is a chelating ligand.
(d) A bidentate ligand can have four coordination
sites.
3. Correct formula of tetraamminechloridonitroplatinum(IV) sulphate can be written as
(a) [Pt(NH3)4(ONO)Cl]SO4
(b) [Pt(NH3)4Cl2NO2]2SO4
(c) [Pt(NH3)4(NO2)Cl]SO4
(d) [PtCl(ONO)NH3(SO4)]
4. The magnitude of magnetic moment (spin
only) of [NiCl4]2– will be
(a) 2.82 B.M.
(b) 0
(c) 1.23 B.M.
(d) 5.64 B.M.
5. Consider the following coordination compounds.
(i) [Pt(NH3)4Cl2]Br2
(ii) [Pt(NH3)4Br2]Cl2
(iii) [Co(NH3)4Cl2]NO2
Which of the following observations is correct?
(a) (i) will give a pale yellow and (ii) will give a
white precipitate with AgNO3 solution.
(b) (iii) will give a white precipitate with AgNO3
solution.
(c) (i), (ii) and (iii) will give white precipitate
with AgNO3 solution.
(d) None of the above coordination compounds
will give white precipitate with AgNO3
solution.
8. Electronic configuration of [Cu(NH3)6]2+ on
the basis of crystal field splitting theory is
(a) t42g e5g
6 3
(b) t2g
eg
(c) t 92g e0g
(d) t 52g e4g .
9. Which of the following primary and secondary
valencies are not correctly marked against the
compound?
(a) [Cr(NH3)6]Cl3 p = 3, s = 6
(b) K2[PtCl4] p = 2, s = 4
(c) [Pt(NH3)2Cl2] p = 2, s = 4
(d) [Cu(NH3)4]SO4 p = 4, s = 4
10. What will be the correct order of absorption
of wavelength of light in the visible region
for the complexes, [Co(NH3)6]3+, [Co(CN)6]3–,
[Co(H2O)6]3+ ?
(a) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+
(b) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3–
(c) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3–
(d) [Co(CN)6]3– > [Co(H2O)6]3+ > [Co(NH3)6]3+
11. Which of the following does not depict the
correct name of the compound?
(a) K2[Zn(OH)4] : Potassium
tetrahydroxozincate(II)
95
Coordination Compounds
(b) [Co(NH3)5CO3]Cl : Pentaammine
carbonatochlorocobaltate(III)
(c) Na3[Co(NO2)6] : Sodium hexanitrocobaltate(III)
(d) K3[Cr(CN)6] : Potassium hexacyanochromate(III)
19. Which of the following shall form an
octahedral complex?
(a) d 4(low spin)
(b) d 8(high spin)
(d) None of these
(c) d 6(low spin)
12. When excess of ammonia is added to copper
sulphate solution, the deep blue coloured complex
is formed. The complex is
(a) tetrahedral and paramagnetic
(b) tetrahedral and diamagnetic
(c) square planar and diamagnetic
(d) square planar and paramagnetic.
20. The increasing order of crystal field splitting
strength of the given ligands is
–
–
–
(a) NH3 < Cl < CN < F < CO < H2O
–
–
–
(b) F < Cl < NH3 < CN < H2O < CO
–
–
–
(c) Cl < F < H2O < NH3 < CN < CO
–
–
(d) CO < CN < NH3 < H2O < F < Cl–
13. Arrange the following complexes in increasing
order of conductivity of their solutions.
(ii) [Co(NH3)4Cl2]Cl
(i) [Co(NH3)3Cl3]
(iv) [Co(NH3)5Cl]Cl2
(iii) [Co(NH3)6]Cl3
(a) (i) < (ii) < (iv) < (iii)
(b) (ii) < (i) < (iii) < (iv)
(c) (i) < (iii) < (ii) < (iv)
(d) (iv) < (i) < (ii) < (iii)
14. Which of the following complexes will have
tetrahedral shape?
(b) [Pd(CN)4]2–
(a) [PdCl4]2–
2–
(c) [Ni(CN)4]
(d) [NiCl4]2–
15.
(a)
(b)
(c)
(d)
The name of [Co(NH3)5NO2]Cl2 will be
pentaamminonitrocobalt(II) chloride
pentaamminenitrochloridecobaltate(III)
pentaamminenitrito-N-cobalt(III) chloride
pentanitrosoamminechlorocobaltate(III).
16. Which of the following ligands form a chelate?
(a) Acetate
(b) Oxalate
(c) Cyanide
(d) Ammonia
17. Copper sulphate dissolves in ammonia due
to the formation of
(b) [Cu(NH3)4]SO4
(a) Cu2O
(c) [Cu(NH3)4]OH
(d) [Cu(H2O)4]SO4
18. When excess of aqueous KCN solution is
added to an aqueous solution of copper sulphate,
the complex [Cu(CN)4]2– is formed. On passing
H2S gas through this solution no precipitate of
CuS is formed because
(a) sulphide ions cannot replace CN– ions
(b) [Cu(CN)4]2– does not give Cu2+ ion in the
solution
(c) sulphide ions from H2S do not form complexes
(d) sulphide ions cannot replace sulphate ions
from copper sulphate solution.
21. The number of
[Ni(CO)4] is
(a) one
(c) three
unpaired electrons in
(b) two
(d) zero
22. [Fe(CN)6]4– and [Fe(H2O)6]2+ show different
colours in dilute solution because
(a) CN– is a strong field ligand and H2O is a
weak field ligand hence magnitude of CFSE
is different
(b) both CN– and H2O absorb same wavelength
of energy
(c) complexes of weak field ligands are generally
colourless
(d) the sizes of CN– and H2O are different hence
their colours are also different.
23. In which of the following compounds, the
transition metal is in oxidation state of zero?
(a) [Fe(H2O)3(OH)3]
(b) [Ni(CO)4]
(d) [Co(NH3)6]Cl3
(c) [Fe(H2O)6]SO4
24. A substance appears coloured because
(a) it absorbs light at specific wavelength in
the visible part and reflects rest of the
wavelengths
(b) ligands absorb different wavelengths of light
which give colour to the complex
(c) it absorbs white light and shows different
colours at different wavelength
(d) it is diamagnetic in nature.
25. Which of the following statements is correct
about [Co(H2O)6]2+ complex?
(a) Electronic configuration = 3d7 → t52g e2g, no. of
unpaired electrons = 3, m = 3.87 B.M.
(b) Electronic configuration = 3d6 → t42g e2g, no. of
unpaired electrons = 2, m = 2.87 B.M.
(c) Electronic configuration = 3d7 → t62g e1g , no. of
unpaired electrons = 1, m = 2.87 B.M.
CBSE Board Term-II Chemistry Class-12
96
(d) Electronic configuration = 3d7 = t32g e4g , no. of
unpaired electrons = 3, m = 3.87 B.M.
26. Hexacyano complexes of metals in their
+2 oxidation state are usually yellow while the
corresponding hexaaqua compounds are often
blue or green. This is so because
(a) hexacyano complexes absorb orange or red
light thus appear yellow while hexaaqua
complexes absorb indigo thus appear yellow
(b) hexacyano complexes absorb indigo thus
appearing yellow while hexaaqua complexes
absorb orange or red light thus appear blue
or green
(c) hexacyano complexes absorb yellow light
while hexaaqua complexes absorb blue light
(d) CN– ions are yellow in colour while aqua ions
are blue or green in colour.
27. Low spin tetrahedral complexes are not
formed because
(a) for tetrahedral complexes, the CFSE is lower
than pairing energy
(b) for tetrahedral complexes, the CFSE is
higher than pairing energy
(c) electrons do not go to eg in case of tetrahedral
complexes
(d) tetrahedral complexes are formed by weak
field ligands only.
28. Which of the following sets of examples and
geometry of the compounds is not correct?
(a) Octahedral – [Co(NH3)6]3+, [Fe(CN)6]3–
(b) Square planar – [Ni(CN)4]2–, [Cu(NH3)4]2+
(c) Tetrahedral – [Ni(CO)4], [ZnCl4]2–
(d) Trigonal
[CuCl4]2–
bipyramidal
–
[Fe(NH3)6]2+,
29. A coordination compound CrCl3⋅4H2O gives
white precipitate of AgCl with AgNO3. The molar
conductance of the compound corresponds to two
ions. The structural formula of the compound is
(a) [Cr(H2O)4Cl3]
(b) [Cr(H2O)3Cl3]H2O
(d) [Cr(H2O)4Cl]Cl2
(c) [Cr(H2O)4Cl2]Cl
30. Among the following, which are ambidentate
ligands?
(ii) NO–3
(i) SCN–
–
(iii) NO2
(iv) C2O42–
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
31. The lowest value of paramagnetism is shown
by
(b) [Fe(CN)6]3–
(a) [Co(CN)6]3–
(c) [Cr(CN)6]3–
(d) [Mn(CN)6]3–
32. Which of the following is a tridentate ligand?
(a) EDTA4–
(b) (COO)22–
–
(c) dien
(d) NO2
33. Which of the
paramagnetism?
(a) [Cr(H2O)6]3+
(c) [Cu(H2O)6]2+
following
has
largest
(b) [Fe(H2O)6]2+
(d) [Zn(H2O)2]2+
34. Identify the statement which is not correct?
(a) Coordinate compounds are mainly known for
transition metals.
(b) Coordination number and oxidation state of
a metal are same.
(c) Tetrahedral complexes form low spin
complex.
(d) A ligand donates at least one electron pair to
the metal atom to form a bond.
35. When aqueous solution of potassium fluoride
is added to the blue coloured aqueous CuSO4
solution, a green precipitate is formed. This
observation can be explained as follows.
(a) On adding KF, H2O being weak field ligand
–
is replaced by F ions forming [CuF4]2– which
is green in colour.
(b) Potassium is coordinated to [Cu(H2O)4]2+ ion
present in CuSO4 and gives green colour.
(c) On adding KF, Cu2+ are replaced by K+
forming a green complex.
(d) Blue colour of CuSO4 and yellow colour of KI
form green colour on mixing.
36. The formula of the complex diamminechlorido
(ethylenediamine)nitroplatinum(IV) chloride is
(a) [Pt(NH3)2Cl(en)NO2]Cl2
(b) Pt[Pt(NH3)2(en)Cl2NO2]
(c) Pt[(NH3)2(en)NO2]Cl2
(d) Pt[(NH3)2(en)NO2Cl2]
37. Using valence bond theory, the complex
[Cr(H2O)6]3+ can be described as
(a) sp3d2, outer orbital complex, paramagnetic
(b) dsp2, inner orbital complex, diamagnetic
(c) d2sp3, inner orbital complex, paramagnetic
(d) d2sp3, outer orbital complex, diamagnetic.
Coordination Compounds
38. The ligand N(CH2CH2NH2)3 is
(a) bidentate
(b) tridentate
(c) tetradentate
(d) pentadentate.
39. Which of the following is not correctly
matched?
(a) Coordination compound containing cationic
complex ion : [Fe(H2O)2(C2O4)2]2SO4
(b) Coordination compound containing anionic
complex ion : [Ag(NH3)2]Cl
(c) Non-ionic
coordination
compound
:
[Co(NO2)3(NH3)3]
(d) Coordination compound containing cationic
and anionic complex ion : [Pt(NH3)4] [CuCl4]
40. A coordination compound X gives pale
yellow colour with AgNO3 solution while its
isomer Y gives white precipitate with BaCl2. Two
compounds are isomers of CoBrSO4⋅5NH3. What
could be the possible formula of X and Y?
(a) X = [Co(NH3)5SO4]Br, Y = [Co(NH3)5Br]SO4
(b) X = [Co(NH3)5Br]SO4, Y = [Co(NH3)5SO4]Br
(c) X = [Co(NH3)5Br(SO4)], Y = [CoBr(SO4)(NH3)5]
(d) X = [Co(Br)5NH3]SO4, Y = [CoBr(SO4)]NH3
41. When one mole of each of the following
complexes is treated with excess of AgNO3 which
will give maximum amount of AgCl?
(b) [Co(NH3)5Cl]Cl2
(a) [Co(NH3)6]Cl3
(c) [Co(NH3)4Cl2]Cl
(d) [Co(NH3)3Cl3]
42. Which of the following descriptions about
[FeCl6]4– is correct about the complex ion?
(a) sp3d, inner orbital complex, diamagnetic
(b) sp3d2, outer orbital complex, paramagnetic
(c) d2sp3, inner orbital complex, paramagnetic
(d) d2sp3, outer orbital complex, diamagnetic
43. According to Werner’s theory of coordination
compounds,
(a) primary valency is ionisable
(b) secondary valency is ionisable
(c) primary and secondary valencies are
ionisable
(d) neither primary nor secondary valency is
ionisable.
44. CuSO4⋅5H2O is blue in colour while CuSO4 is
colourless due to
(a) presence of strong field ligand in CuSO4⋅5H2O
(b) due to absence of water (ligand), d-d
transition are not possible in CuSO4
(c) anhydrous CuSO4 undergoes d-d transitions
due to crystal field splitting
(d) colour is lost due to loss of unpaired electrons.
97
45. Which of the following complexes will show
maximum paramagnetism?
(b) 3d 5
(a) 3d 4
(c) 3d 6
(d) 3d 7
46. Among the following compounds which is
both paramagnetic and coloured?
(b) [Co(SO4)]
(a) K2Cr2O7
(c) (NH4)2[TiCl6]
(d) K3[Cu(CN)4]
47. Which of the following rules is not correct
regarding IUPAC nomenclature of complex ions?
(a) Cation is named first and then anion.
(b) In coordination sphere, the ligands are
named alphabetically.
(c) Positively charged ligands have suffix ‘ate’.
(d) More than one ligand of a particular type are
indicated by using di, tri, tetra, etc.
48. Mark the correct statements regarding the
geometry of complex ions.
(i) The geometry of the complex ion depends
upon the coordination number.
(ii) If coordination number is 6, the complex is
octahedral.
(iii) If coordination number is 4, the geometry of
the complex may be tetrahedral or square
planar.
(a) (i), (ii) and (iii)
(b) (i) and (ii) only
(c) (i) and (iii) only
(d) (ii) and (iii) only
49. Which of the following is not a neutral
ligand?
(b) NH3
(a) H2O
(c) ONO
(d) CO
50. In coordination compounds metals show
two type of linkages : primary and secondary.
Primary valency is ionisable and corresponds to
conductivity. Several coordination compounds
are formed by Co(III) with ligand NH3 and Cl–
both.
Conductivity of complex I corresponds to 1 : 3
electrolyte, conductivity of complex II corresponds
to 1 : 2 electrolyte while conductivity of complex
3 and 4 corresponds to 1 : 1 electrolyte. So correct
option about these complexes is
(a) complex I is [(Co(NH 3 ) 6 ]Cl 3 with purple
colour
(b) complex II is [(Co(NH 3) 6]Cl 3 with purple
colour
(c) complex III is [Co(NH3)4Cl2]Cl with violet
colour
(d) both (b) and (c)
CBSE Board Term-II Chemistry Class-12
98
51. Ms. Anjali class 12 th chemistry teacher
explained IUPAC nomenclature of coordination
compounds in her class. Then she asked students
to write the names of 5 coordination compounds.
Kavya written these five names :
[Cr(NH 3 ) 3 (H 2 O) 3 ]Cl 3 - Triamminetriaqua
chromium(III)chloride, [Ag(NH 3 ) 2 ][Ag(CN) 2 ]
Diamminesilver(I) dicyanidosilver(I)
[CoCl2(en)2]Cl-Dichloridobis(ethane-1,
2-diammine) cobalt (III) chloride
K3[Al(C2O4)3] - Potassium trioxalatoaluminium (III)
[Ni(CO)4]-Tetracarbonylnickel(0)
Few names given by her were not correct as
she didn’t follow one rule while naming these
compounds. That one rule is
(a) the ligands are name in an alphabetical
order before the name of central atom/ion.
(b) prefixes mono, di, tri etc. are used to indicate
the number of individual ligands in the
coordination entity.
(c) if the complex ion is cation, the metal is
named same as the element.
(d) if the complex ion is anion, the metal ends
with suffix – ate.
Octahedral complex
52. Fex+ + SCN–
y+
–
Fe + CN
Octahedral complex
(x and y may or may not be equal)
The difference between the spin-only magnetic
moments is 4.2 B.M. approximately. The reason
for this difference in magnetic moment is
(a) CN– is a strong ligand while SCN– is a weak
ligand
(b) Fe is present in O.S. I in complex with SCN–
while in O.S. III in complex with CN–
(c) SCN– is a strong ligand while CN– is a weak
ligand
(d) x is 3 while y is 1.
53. If a ligand is weak, the complex will be
high spin while if the ligand is strong then the
complex will be low spin. Here few complexes
are listed:
I. [Cr(H2O)6]2+
II. [CoCl4]2–
III. [Fe(H2O)6]2+
IV. [Mn(H2O)6]2+
V. [Ni(CO)4]
VI. [Ni(CN)4]2–
The complexes which have zero magnetic
moment are.
(a) I and V
(b) II and VI only
(c) III and IV
(d) V and VI only
54. Some details of few Nickel complexes are
given below:
Complex I : Diamagnetic and square planar
Complex II : Paramagnetic and tetrahedral
Complex III : Diamagnetic and tetrahedral
Complex IV : Paramagnetic and Octahedral
Which is not correct option for the given
complexes?
(a) The ligand in complex I is CN– and it has
dsp2 hybridisation.
(b) The ligand in complex II is Cl– and it has
sp3 hybridisation.
(c) The ligand in complex IV is H2O and it has
d2sp3 hybridisation.
(d) The ligand in complex III is CO and it has
sp3 hybridiation.
Case Based MCQs
rise to on absorption spectrum consisting of a
single peak that can be represented as shown :
Absorbance
Case I : Read the passage given below and
answer the following questions from 55 to 57.
The extent to which the set of d-orbitals is
split in the electrostatic field produced by the
ligands depends upon several factors. Two of
the most important factors are the nature of
the ligands and the nature of the metal ion. In
order to see this effect, consider the complex ion
[Ti(H2O)6]3+. The Ti3+ ion has a single electron in
the 3d-orbital, and we refer to it as d1 ion. In the
octahedral field generated by six H2O molecules,
the single electron will reside in one of the three
degerate t2g orbitals. Under spectral excitation,
the electron is promoted to an e.g., orbital giving
20,300
u (cm–1)
The maximum absorption in the spectrum for
[Ti(H2O)6]3+ occurs at 20,300 cm–1 which is
equal to 243 kJ mol–1.This gives the value of
99
Coordination Compounds
Do directly, but only in case of simple d1 ions.
Other complexes containing the Ti3+ ion (e.g.,
[Ti(NH3)6]3+, [TiF6]3–, etc.) could also be prepared
and spectra obtained for these complexes. If this
was done, it would be observe that the absorption
maximum occurs at a different energy for each
complex. Because the maximum corresponds
to the splitting of d-orbitals, the ligands could
be ranked in terms of their ability to cause the
splitting of orbital energies. Such a ranking
is known as the spectrochemical series and for
several common ligands the following order of
decreasing energy is observed, CO > CN– >NO2– >
en > NH3 > H2O > OH– > F–, Cl– > Br–. In general,
the splitting in tetrahedral fields is only about
half as large as that in octahedral fields.
55. Which of the following ligands has lowest Do
value?
(a) CN–
(b) CO
–
(d) NH3
(c) F
56. The visible spectra of salts of the following
complexes are measured in aqueous solution
for which complex would the spectrum contain
absorption with highest Emax values?
(a) [Co(H2O)6]2+
(c) [Co(NH3)6]3+
(b) [Co(H2O)6]3+
(d) [Co(CN)6]3–
57. Which of the following statements is incorrect
for complex [Ti(H2O)6]3+ ?
(a) [Ti(H2O)6]3+ is violet in colour.
(b) [Ti(H2O)6]3+ is an octahedral complex.
(c) Exitation of electron in [Ti(H2O)6]3+ occurs
as, t12g e0g → t02g e1g
(d) The colour of the complex [Ti(H2O)6]3+ arises
due to d-d and f-f transition of the electron.
Case II : Read the passage given below and
answer the following questions.
Werner, a Swiss chemist in 1892 prepared and
characterised a large number of coordination
compounds and studied their physical and
chemical behaviour. He proposed that, in
coordination compounds, metals possess two
types of valencies, viz. primary valencies, which
are normally ionisable and secondary valencies
which are non-ionisable. In a series of compounds
of cobalt (III) chloride with ammonia, it was found
that some of the chloride ions could be precipitated
as AgCl on adding excess of AgNO3 solution in
cold, but some remained in solution. The number
of ions furnished by a complex in a solution can
be determined by precipitation reactions. The
measurement of molar conductance of solutions
of coordination compounds helps to estimate the
number of ions furnished by the compound in
solution.
In the following questions (Q. No. 58-62), a
statement of assertion followed by a statement
of reason is given. Choose the correct answer
out of the following choices on the basis of the
above passage.
(a) Assertion and reason both are correct
statements and reason is correct explanation
for assertion.
(b) Assertion and reason both are correct
statements but reason is not correct
explanation for assertion.
(c) Assertion is correct statement but reason is
wrong statement.
(d) Assertion is wrong statement but reason is
correct statement.
The following questions are multiple choice
questions. Choose the most appropriate answer :
58. Assertion : The complex [Co(NH3)3Cl3] does
not give precipitate with silver nitrate solution.
Reason : The given complex is non-ionisable.
59. Assertion : The complex [Co(NH3)4Cl2]Cl
gives precipitate corresponding to 2 mol of AgCl
with AgNO3 solution.
Reason : It ionises as [Co(NH3)4Cl2]+ + Cl–.
60. Assertion : CoCl3⋅4NH3 gives 1 mol of AgCl
on reacting with AgNO3, its secondary valency
is 6.
Reason : Secondary valency corresponds to
coordination number.
61. Assertion : 1 mol of [CrCl2(H2O)4]Cl ⋅ 2H2O
will give 1 mol of AgCl on treating with AgNO3.
Reason : Cl– ions satisfying secondary valanceis
will not be precipitated.
62. Assertion : CoCl3⋅3NH3 is not conducting
while CoCl3⋅5NH3 is conducting.
Reason : The complex of CoCl3⋅3NH3 is
[CoCl3(NH3)3] while that of CoCl3⋅5NH3 is
[CoCl(NH3)5]Cl2.
CBSE Board Term-II Chemistry Class-12
100
Case III : Read the passage given below and
answer the following questions from 63 to 66.
Valence bond theory considers the bonding
between the metal ion and the ligands as purely
covalent. On the other hand, crystal field theory
considers the metal-ligand bond to be ionic arising
from electrostatic interaction between the metal
ion and the ligands. In coordination compounds,
the interaction between the ligand and the metal
ion causes the five d-orbitals to split-up. This
is called crystal field splitting and the energy
difference between the two sets of energy level is
called crystal field splitting energy. The crystal
field splitting energy (Do) depends upon the
nature of the ligand. The actual configuration of
complexes is divided by the relative values of Do
and P (pairing energy).
If Do < P, then complex will be high spin.
If Do > P, then complex will be low spin.
63. The crystal field splitting energy for
octahedral (Do) and tetrahedral (Dt) complex is
related as
1
4
(a) ∆t = ∆ o
(b) ∆t = ∆ o
2
9
2
3
(c) ∆t = ∆ o
(d) ∆t = ∆ o
5
5
64. On the basis of crystal field theory, the
electronic configuration of d 4 in two situations :
(a) Do > P and (b) Do < P are
(a)
(b)
(a) t2g4 eg0
3 1
t2g
eg
3 1
(b) t2g
eg
4 0
t2g
eg
3 1
t2g
eg
4 0
t2g eg
t2g3 eg1
4 0
t2g
eg
(c)
(d)
65. Using crystal field theory, calculate magnetic
moment of central metal ion of [FeF6]4–.
(a) 1.79 B.M.
(b) 2.83 B.M.
(c) 3.85 B.M.
(d) 4.9 B.M.
66. Electronic configuration of d-orbitals in
[Ti(H2O)6]3+ ion in an octahedral crystal field is
1 0
(a) t2g
eg
2 0
(b) t2g
eg
(c) t02g e1g
(d) t12g e1g
Few rules for naming coordination compounds
are :
(I) In ionic complex, the cation is named first
and then the anion.
(II) In the coordination entity, the ligands are
named first and then the central metal ion.
(III)When more than one type of ligands are
present, they are named in alphabetical order
of preference without any consideration of
charge.
67. The IUPAC name of the complex
[Pt(NH3)3Br(NO2)Cl]Cl is
(a) triamminechlorobromonitroplatinum(IV)
chloride
(b) triamminebromonitrochloroplatinum(IV)
chloride
(c) triamminebromidochloridonitroplatinum
(IV) chloride
(d) triamminenitrochlorobromoplatinum(IV)
chloride.
68.
(a)
(b)
(c)
(d)
The IUPAC name of [Ni(CO)4] is
tetracarbonylnickel(II)
tetracarbonylnickel(0)
tetracarbonylnickelate(II)
tetracarbonylnickelate(0).
69. As per IUPAC nomenclature, the name of the
complex [Co(H2O)4(NH3)2]Cl3 is
(a) tetraaquadiamminecobalt(II) chloride
(b) tetraaquadiamminecobalt(III) chloride
(c) diamminetetraaquacobalt(II) chloride
(d) diamminetetraaquacobalt(III) chloride.
70. Which of the following represents correct
formula of dichloridobis(ethane-1, 2-diamine)
cobalt(III) ion?
(a) [CoCl2(en)]2+
(b) [CoCl2(en)2]2+
(c) [CoCl2(en)]+
(d) [CoCl2(en)2]+
71. Correct formula of pentaamminenitrito-Ocobalt(III) sulphate is
(a) [Co(NO2)(NH3)5]SO4
(b) [Co(ONO)(NH3)5]SO4
(c) [Co(NO2)(NH3)4](SO4)2
(d) [Co(ONO)(NH3)4](SO4)2
Case IV : Read the passage given below and
answer the following questions from 67 to 71.
Case V : Read the passage given below and
answer the following questions from 72 to 76.
Coordination compounds are formulated and
named according to the IUPAC system.
To explain bonding in coordination compounds
various theories were proposed. One of the
101
Coordination Compounds
important theory was valence bond theory.
According to that, the central metal ion in the
complex makes available a number of empty
orbitals for the formation of coordination bonds
with suitable ligands. The appropriate atomic
orbitals of the metal hybridise to give a set of
equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may
be either inner d-orbitals i.e., (n – 1)d or outer
d-orbitals i.e., nd.
For example, Co3+ forms both inner orbital and
outer orbital complexes, with ammonia it forms
[Co(NH3)6]3+ and with fluorine it forms [CoF6]3–
complex ion.
72.
(a)
(b)
(c)
(d)
Which of the following is not true for [CoF6]3–?
It is paramagnetic.
It has coordination number of 6.
It is outer orbital complex.
It involves d2sp3 hybridisation.
73. [Cr(H2O)6]Cl3 (at. no. of Cr = 24) has a
magnetic moment of 3.83 B.M. The correct
distribution of 3d-electrons in the central metal
of the complex is
(a) 3d1xy , 3d1 2 2 , 3d1yz
x −y
1
1
1
(b) 3d xy , 3d yz , 3dzx
(c) 3d1xy , 3d1zy , 3d12
z
(d) 3d1 2
x − y2
, 3d12 , 3d1xz
z
74. Which of the following is true for
[Co(NH3)6]3+ ?
(a) It is an octahedral, dimagnetic and outer
orbital complex.
(b) It is an octahedral, paramagnetic and outer
orbital complex.
(c) It is an octahedral, paramagnetic and inner
orbital complex.
(d) It is an octahedral, dimagnetic and inner
orbital complex.
75. The paramagnetism of [CoF6]3– is due to
(a) 3 electrons
(b) 4 electrons
(c) 2 electrons
(d) 1 electron.
76. Which of the following is an inner orbital or
low spin complex?
(a) [Ni(H2O)6]3+
(b) [FeF6]3–
(c) [Co(CN)6]3–
(d) [NiCl4]2–
Assertion & Reasoning Based MCQs
For question numbers 77-90, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
77. Assertion : The [Ni(en)3]Cl2 (en =
ethylenediamine) has lower stability than
[Ni(NH3)6]Cl2.
Reason : In [Ni(en)3]Cl2 the geometry of Ni is
octahedral.
78. Assertion : Ethylenediaminetetraacetate
ion forms an octahedral complex with the metal
ion.
Reason : It has six donor atoms which coordinate
simultaneously to the metal ion.
79. Assertion : All the octahedral complexes of
Ni2+ must be outer orbital complexes.
Reason : Outer orbital octahedral complexes are
given by weak ligands.
80. Assertion : The second and third transition
series elements have lesser tendency to form low
spin complex as compared to the first transition
series.
Reason : The CFSE (Do) is more for 4d and 5d.
81. Assertion : [Fe(CN)6]3– has d2sp3 type
hybridisation.
Reason : [Fe(CN)6]3– ion shows magnetic
moment corresponding to two unpaired electrons.
82. Assertion : [Ti(H2O)6]3+ is coloured while
[Sc(H2O)6]3+ is colourless.
Reason : d – d transition is not possible in
[Sc(H2O)6]3+.
83. Assertion : Thiocarbonyl is a neutral ligand.
Reason : Thiocarbonyl has three donor atoms
but behaves as a bidentate ligand.
CBSE Board Term-II Chemistry Class-12
102
84. Assertion : The ligand N–3 is named as
nitride.
Reason : N3– is derived from HN3.
85. Assertion : [CrCl2(H2O)4]NO3 is
dichlorotetraaquachromium(III) nitrate.
Reason : In writing the name of the complex
cation is written first followed by the anion.
86. Assertion
:
[Fe(CN)6]3–
4–
paramagnetic while [Fe(CN)6]
is
weakly
is diamagnetic.
Reason : [Fe(CN)6]3– has +3 oxidation state
while [Fe(CN)6]4– has +2 oxidation state.
87. Assertion : [Al(NH3)6]3+ does not exist in
aqueous solution.
Reason : NH3 is a neutral ligand.
88. Assertion : Low spin complexes have less
number of unpaired electrons.
Reason : [FeF6]3– is a low spin complex.
89. Assertion : [Pt(NH3)2Cl2] is square planar.
Reason : The oxidation state of platinum is + 2.
90. Assertion : Cu(OH)2 is soluble in NH4OH
but not in NaOH.
Reason : Cu(OH)2 forms a soluble complex with
NH3.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Write the hybridisation and number of
unpaired electrons in the complex [CoF 6 ] 3– .
(Atomic no. of Co = 27)
2. What do you understand by ‘denticity of a
ligand’?
3. Explain the term crystal field splitting in an
octahedral field.
4. Wr i t e t h e f o r m u l a o f t h e f o l l o w i n g
coordination compound :
Iron (III) hexacyanoferrate(II)
5. When a coordination compound CrCl3⋅6H2O
is mixed with AgNO 3 , 2 moles of AgCl are
precipitated per mole of the compound. Write
structural formula of the complex.
6. Why a solution of [Ni(H 2 O) 6 ] 2+ is green
while a solution of [Ni(CN)4]2– is colourless?
(At. no. of Ni = 28)
7.
Write the IUPAC name of
[Cr(NH3)6][Co(CN)6].
8. What is the difference between a complex
and a double salt?
9. Chelates are generally more stable than the
complexes of unidentate ligands. Explain.
10. Write the coordination number and oxidation
state of platinum in the complex [Pt(en)2Cl2].
Short Answer Type Questions (SA-I)
11. Why is [NiCl4]2– paramagnetic but [Ni(CO)4]
is diamagnetic? (At. no. : Cr = 24, Co = 27, Ni = 28)
12. (i) On the basis of crystal field theory, write
the electronic configuration of d4 ion if Do < P.
(ii) Write the hybridization and magnetic
behaviour of the complex [Ni(CO)4].
(At. no. of Ni =28)
13. Out of [CoF6]3– and [Co(C2O4)3]3–, which one
complex is
(i) diamagnetic (ii) more stable
(iii) outer orbital complex and
(iv) low spin complex ?
(Atomic no. of Co = 27)
14.
the
(i)
(ii)
Using IUPAC norms write the formulae for
following :
Pentaamminenitrito–O–cobalt(III) chloride
Potassium tetracyanonickelate(II)
15. Explain the following term giving a suitable
example : Ambidentate ligand
16. U s i n g v a l e n c e b o n d t h e o r y, e x p l a i n
the geometry and magnetic behaviour of
[Co(NH3)6]3+.
(At. no. of Co = 27)
17. Why Co 2+ is easily oxidised to Co 3+ in
presence of a strong ligand?
103
Coordination Compounds
18. Write down the IUPAC name of the following
complex : [Cr(NH3)2Cl2(en)]Cl
19. Ravi prepared a complex compound of cobalt
with NH3 and NO2 as donor ligands. He got a
red precipitate. Sohan also prepared the same
complex using same metal salt solution and
same ligands. He obtained yellow crystals.
Sohan complained his teacher that his chemicals
were different so he got different product. But
their teacher is satisfied with both the results.
Now answer the following questions :
(i) What type of ligand is present in the given
compounds which is responsible for changing
colour?
(ii) Write IUPAC name of both the compounds.
20. Give the formula of each of the following
coordination entities :
(i) Co 3+ ion is bound to one Cl – , one NH 3
molecule and two bidentate ethylene diamine
(en) molecules.
(ii) Ni2+ ion is bound to two water molecules and
two oxalate ions.
Write the name and magnetic behaviour of each
of the above coordination entities.
(At. nos. Co = 27, Ni = 28)
Short Answer Type Questions (SA-II)
21. Write the IUPAC names of the following
coordination compounds :
(ii) K3[Fe(CN)6]
(i) [Cr(NH3)3Cl3]
+
(iii) [CoBr2(en)2]
22. The splitting pattern of d-orbitals in
octahedral and tetrahedral geometry are reverse
of each other. Why?
23. (a) What is d-d transition?
(b) Tetrahedral complexes are always of high
spin. Explain.
24. For the complex [NiCl4]2–, write
(i) the IUPAC name
(ii) the hybridization type
(iii) the shape of the complex.
(Atomic no. of Ni = 28)
25. Write the name, the structure and the
magnetic behaviour of each one of the following
complexes :
(i) [Pt(NH3)2Cl(NO2)] (ii) [Co(NH3)4Cl2]Cl
(iii) Ni(CO)4
(At. nos. Co = 27, Ni = 28, Pt = 78)
26. [Mn(CN) 6 ] 3– has two unpaired electrons
whereas [MnCl6]3– has four unpaired electrons.
Why?
(ii) Hybrid orbitals and shape of the complex
(iii) Magnetic behaviour of the complex
(iv) Name of the complex.
29. Name the following coordination entities
and describe their structures.
(i) [Fe(CN)6]4–
(ii) [Cr(NH3)4Cl2]+
2–
(iii) [Ni(CN)4]
30. Explain the following :
(i) Anhydrous CuSO4 is white while hydrated
CuSO4 is blue in colour.
(ii) [Ti(H2O)6]Cl3 is violet in colour but becomes
colourless on heating.
31. For the complex [Fe(CN) 6 ] 3– , write the
hybridization type, magnetic character and spin
nature of the complex. (At. number : Fe = 26)
32. Give reason : [CoF6]3– is outer orbital but
[Co(NH3)6]3+ is inner orbital complex.
33. Write the state of hybridization, the shape
and the magnetic behaviour of the following
complex entities :
(i) [Cr(NH3)4Cl2]Cl
(ii) [Co(en)3]Cl3
(iii) K2[Ni(CN)4]
27. Write the IUPAC name, deduce the geometry
and magnetic behaviour of the complex
K4[Mn(CN)6].
[Atomic no. of Mn = 25]
34. (a) What is meant by crystal field splitting
energy? On the basis of crystal field theory,
write the electronic configuration of d 4 in
terms of t2g and eg in an octahedral field when
(ii) Do < P
(i) Do > P
(b) Write two limitations of crystal field theory.
28. For the complex [Fe(en)2Cl2]Cl, identify the
following :
(i) Oxidation number of iron
35. [Ni(H 2O) 6] 2+ is green and becomes violet
when ethane-1, 2-diamine is added to it. Identify
the observation.
CBSE Board Term-II Chemistry Class-12
104
Long Answer Type Questions (LA)
36. (i) Using crystal field theory, draw energy
level diagram, write electronic configuration of
the central metal atom/ion and determine the
magnetic moment value in the following :
(a) [CoF6]3–, (b) [FeF6]3–, (c) [Fe(CN)6]4–
(ii) FeSO 4 solution mixed with (NH 4 ) 2 SO 4
solution in 1:1 molar ratio gives the test of Fe2+
ion but CuSO 4 solution mixed with aqueous
ammonia in 1 : 4 molar ratio does not give the
test of Cu2+ ion. Explain why?
37. Write down the IUPAC name for each of the
following complexes and indicate the oxidation
state, electronic configuration and coordination
number. Also give stereochemistry and magnetic
moment of the complex:
(i) K[Cr(H2O)2(C2O4)2].3H2O
(ii) [Co(NH3)5Cl]Cl2
(iii) [CrCl3(py)3]
(iv) Cs[FeCl4]
(v) K4[Mn(CN)6]
OBJECTIVE TYPE QUESTIONS
1.
(a)
2. (c) : NH+4 ion does not have any lone pair of electrons
which it can donate to central metal ion hence it does not
form complexes.
3. (c) : Tetraamminechloridonitroplatinum(IV) sulphate can
be written as [Pt(NH3)4(NO2)Cl]SO4.
4.
2–
(a) : In [NiCl4] , Ni is in +2 oxidation state.
38. (a) What are bidentate ligands? Explain
with examples.
(b) Explain the coordination sites of polydentate
ligands taking an example of EDTA.
(c) Calculate charge on the central metal in the
following complexes:
(i) [Cu(NH3)6]2+ (ii) [Ag(CN)2]–
39. Give the oxidation state, d-orbital occupation
and coordination number of the central metal
ion in the following complexes :
(i) K3[Co(C2O4)3]
(ii) [Cr(en)2Cl2]Cl
(iii) (NH4)2[CoF4]
(iv) [Mn(H2O)6]SO4
40. (a) Explain hybridisation in the complex
which contains hexacyanidoferrate(III) ion.
(b) Based on the valence bond theory describe
the formation and nature of hexaaminecobalt(III)
chloride.
(c) How will you show that hexafluorocobaltate(III) ion is paramagnetic in nature?
6. (b) : [Ni(NH3)6]3[Co(NO2)6]2
Hexaamminenickel(II) hexanitrocobaltate(III)
7.
(c)
8. (b) : In [Cu(NH 3 ) 6 ] 2+ , oxidation state of Cu = +2,
Cu2+ = 3d 9
6 3
3d 9 = t 2g
eg
9. (d) : In [Cu(NH3)4]SO4 primary valency is 2 and secondary
valency is 4.
= 2(2 + 2) = 2.82 B.M.
10. (c) : The CFSE of the ligands is in the order :
H2O < NH3 < CN–
Hence, excitation energies is in the order:
[Co(H2O)6]3+ < [Co(NH3)6]3+ < [Co(CN)6]3–
hc
1
⇒E ∝
From the relation E =
λ
λ
5. (a) : [Pt(NH3)4Cl2]Br2
[Pt(NH3)4Cl2]2+ + 2Br–
Br– + AgNO3
AgBr + NO–3
The order of absorption of wavelength of light in the visible
region : [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3–
In [NiCl4]2– :
Magnetic moment, µ = n (n + 2)
[ n = number of unpaired electrons]
Pale yellow ppt.
[Pt(NH3)4Br2]Cl2
Cl– + AgNO3
[Pt(NH3)4Br2]
AgCl + NO–3
White ppt.
2+
+ 2Cl
–
11. (b) : [Co(NH3)5CO3]Cl
Pentaamminecarbonatocobalt(III) chloride
12. (d) : CuSO4 + 4NH3
[Cu(NH3)4]2+ + SO42–
deep blue
105
Coordination Compounds
In [Cu(NH3)4]2+, oxidation state of Cu = +2
3d
Cu2+ : 3d9 :
××
4s
4p
××
×× ××
1 electron from 3d is shifted to 4p
2
Hybridisation is dsp hence geometry is square planar and
paramagnetic due to presence of one unpaired electron.
23. (b) : In metal carbonyls, metal is in zero oxidation state.
24. (a) : A substance absorbs light at specific wavelength
in the visible part of the spectrum and reflects the rest of the
wavelengths. Each wavelength represents a different colour
hence corresponding colour is observed.
25. (a) : In [Co(H 2 O) 6 ] 2+ , oxidation state of Co = +2,
Co2+ = 3d 7
13. (a) : Higher the number of ions in the solution, higher
is the conductivity.
No. of ions : [Co(NH3)3Cl3] = 0; [Co(NH3)4Cl2]Cl = 2
[Co(NH3)5Cl]Cl2 = 3; [Co(NH3)6]Cl3 = 4
14. (d) : In [NiCl4]2–, oxidation state of Ni = + 2
[NiCl4]
2–
:
sp3 hybridisation and tetrahedral shape.
15. (c) : The name of [Co(NH3)5NO2]Cl2 is
pentaamminenitrito-N-cobalt(III) chloride.
16. (b) : Oxalate is a bidentate ligand hence forms a chelate.
17. (b) : CuSO4 + 4NH3
[Cu(NH3)4]SO4
18. (b) : When excess of aqueous KCN solution is added to
aqueous CuSO4 solution, highly stable [Cu(CN)4]2– is formed
which does not give Cu2+ ion in the solution, hence copper
sulphide, is not formed.
[Cu(CN)4]2– + 4H2O
[Cu(H2O)4]2+ + 4CN–
6
19. (c) : In d (low spin), electrons get paired up to make
two empty d-orbitals. Hybridisation is d 2sp3 (octahedral) and
the complex is low spin.
20. (c) : In general, the ligands can be arranged in a series
in the order of increasing field strength as
I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O2–
4 < H 2O <
NCS– < edta4– < NH3 < en < NO–2 < CN– < CO
8
2
21. (d) : Ni (Z = 28) : 3d 4s
Oxidation state of Ni in [Ni(CO)4] = 0
5 2
t 2g
eg
µ = n (n + 2) = 3(3 + 2) = 15 = 3.87 B.M.
[ n = 3]
26. (b) : The ligand CN– absorbs indigo, the high energy
radiation and thus appears yellow. The aqua complexes have
much smaller CFSE, they absorb orange or red light and thus
appear blue or green.
27. (a) : Crystal field stabilisation energy for tetrahedral
complexes is less than pairing energy hence they do not pair
up to form low spin complexes.
28. (d) : [CuCl4]2– – Tetrahedral
[Fe(NH3)6]2+ – Octahedral
29. (c) : It gives precipitate with AgNO3 it means it gives
Cl– ions in the solution. Since conductivity corresponds to two
ions, it shows one Cl– is outside the coordination sphere. The
structure will be
[Cr(H2O)4Cl2]+ + Cl–
[Cr(H2O)4Cl2]Cl
AgCl + NO–3
AgNO3 + Cl–
white ppt.
30. (a) : SCN and NO–2 are ambidentate ligands since they
have more than one donor atoms to attach to the central
metal atom.
–
31. (a) : Electronic configuration of Co3+ = 3d 6
[Co(CN)6]3– :
No. of unpaired electrons = 0, hence, shows no paramagnetism.
No. of unpaired electrons = 0
32. (c) : Dien (Diethylenetriamine) has the following
structure
⋅⋅
⋅⋅
⋅⋅
H2N — CH2 — CH2 — NH — CH2 — CH2 — NH2
22. (a) : CN – is a strong field ligand hence pairing of
electrons takes place while in case of H2O pairing does not
take place. Both ligands show different magnitude of crystal
field splitting energy hence absorb different wavelengths and
show different colours.
33. (b) : More the number of unpaired electrons, higher is
its paramagnetism.
Cr3+ : 3d 3, Fe2+ : 3d 6, Cu2+ : 3d 9, Zn2+ : 3d10
Fe2+ has four unpaired electrons hence it shows highest
paramagnetism.
In [Ni(CO)4] :
CBSE Board Term-II Chemistry Class-12
106
34. (b) : Coordination number is the number of ligands that
are directly bound to the central metal atom by coordinate
bonds. Oxidation state is the residual charge on the central
metal atom left after removing all ions.
35. (a) : Aqueous CuSO4 solution contains [Cu(H2O)4]2+
ions which are blue in colour. When aqueous solution of KF
is added, H2O being weak field ligand can be replaced by
F– ions forming [CuF4]2– which is green in colour.
[Cu(H2O)4]2+ + 4F–
(from KF)
[CuF4]2– + 4H2O
green
36. (a) : [Pt(NH3)2Cl(en)NO2]Cl2 :
Diamminechlorido(ethylenediamine)nitroplatinum(IV) chloride
37. (c) : In [Cr(H2O)6]3+, oxidation state of Cr = + 3,
Cr3+ = 3d 3
[Cr(H2O)6]3+ :
due to absence of ligand, crystal field splitting is not possible
hence no colour is observed.
45. (b) : 3d 5 has maximum number of unpaired electrons.
46. (b) : In [Co(SO 4)], the oxidation state of Co is +2.
Configuration of Co2+ = 3d7, it has unpaired electrons in
3d-orbitals so it is paramagnetic. Because of incompletely
filled d-orbitals it is coloured.
47. (c) : Positively charged ligands have suffix ‘ium’.
48. (a) : All the statements are correct with respect to the
geometry of complex ions.
49. (c) : ONO is an anionic ligand (ONO–).
50. (c) :
Colour
Yellow
Purple
It is an inner orbital (3d – 4p) complex. Due to presence of
three unpaired electrons, it is paramagnetic.
38. (c) : Number of donor atoms in N(CH2CH2NH2)3 is four
hence it is a tetradentate ligand.
39. (b) : [Ag(NH3)2]Cl is a coordination compound containing
cationic complex ion.
40. (a) : [Co(NH3)5SO4]Br
[Co(NH3)5SO4]+ + Br–
X
–
AgNO3 + Br
AgBr + NO–3
Pale yellow ppt.
[Co(NH3)5Br]SO4
Y
BaCl2 + SO42–
[Co(NH3)5Br]2+ + SO2–
4
Green
Violet
Formation
[Co(NH3)6]3+3Cl–
2+
[CoCl(NH3)5] 2Cl
41. (a) : [Co(NH3)6]Cl3 gives 3 moles of AgCl.
[Co(NH3)5Cl]Cl2 gives 2 moles of AgCl.
[Co(NH3)4Cl2]Cl gives 1 mole of AgCl.
1 : 3 electrolyte
–
1 : 2 electrolyte
+
–
1 : 1 electrolyte
+
–
1 : 1 electrolyte
[CoCl2(NH3)4] Cl
[CoCl2(NH3)4] Cl
51. (d) : [Ag(NH3)2] [Ag(CN)2] : Diamminesilver(I)
dicyanidoargentite(I)
K3[Al(C2O4)3] Potassium trioxalatoaluminate(III)
52. (a) : Fe (26) : 3d 64s2
Fe3+ : 3d 5
In [Fe(CN)6]3–, CN– is a strong field ligand which causes
pairing of electrons.
3d
4s
4p
d 2sp3 hybridisation
BaSO4 + 2Cl–
white ppt.
Solution conductivity
corresponds to
µ = n (n + 2) = 11
( + 2) = 3 = 1.732 BM
In [Fe(SCN)6]3–, SCN– being a weak field ligand does not
cause pairing of electrons.
3d
4s
4p
4d
[Co(NH3)3Cl3] will not give AgCl.
42. (b) : In [FeCl6]4–, oxidation state of Fe = +2, Fe2+ = 3d 6
[FeCl6]4– :
sp3d 2 hybridisation
µ = n (n + 2) = 5(5 + 2) = 35 = 5.916 BM
Difference = 5.916 – 1.732 = 4.184 ≈ 4.2 BM
53. (d) : [Cr(H2O)6]2+ : Cr2+ → [Ar]3d 44s0
Paramagnetic due to presence of four unpaired electrons.
43. (a) : Primary valency is ionisable according to Werner’s
theory of coordination compounds.
44. (b) : In CuSO4⋅5H2O, water acts as ligand and as a result
it causes crystal field splitting making d-d transition possible
in CuSO4⋅5H2O. Hence, it is coloured. In anhydrous CuSO4
3d
4 unpaired electrons
[CoCl4]2– : Co2+ →
3d
3 unpaired electrons
4s
[Ar]3d74s0
4s
107
Coordination Compounds
[Fe(H2O)6]2+ : Fe2+ → [Ar]3d 64s0
3d
4 unpaired electrons
[Mn(H2O)6]2+ : Mn2+
3d
4s
→ [Ar]3d 54s0
4s
67. (c) : Ligands are named in alphabetical order irrespective
of their charge.
71. (b) : Ligand NO–2 is ambidentate ligand as it can donate
electrons through either nitrogen (NO2) or oxygen (ONO).
[Ni(CO)4] : Ni0 → [Ar]3d 84s2
4s
72. (d) : It involves sp3d 2 hybridisation and not d 2sp 3.
Zero unpaired electrons
[Ni(CN)4]2– : Ni2+ → [Ar]3d 84s0
3d
66. (a) : In [Ti(H2O)6]3+, Ti is in +3 oxidation state and there
is only one electron in d-orbital.
68. (b) 69. (d) 70. (d)
5 unpaired electrons
3d
Magnetic moment (m) = n (n + 2) = 4(4 + 2) = 4.9 B.M.
4s
Zero unpaired electrons
Hence, [Ni(CO)4] and [Ni(CN)4]2– have zero unpaired electrons
i.e., zero magnetic moment.
54. (c) : [Ni(CN) 4]2– : dsp2 hybridisation (square planar
complex) and have zero unpaired electrons (diamagnetic)
[NiCl 4] 2– : sp 3 hybridisation (tetrahedral) and have two
unpaired electrons (paramagnetic)
[Ni(H2O)6]2+ : sp3d2 hybridisation (octahedral) and have two
unpaired electrons (paramagnetic)
[Ni(CO)4] : sp 3 hybridisation (tetrahedral) and have zero
unpaired electrons (diamagnetic)
73. (b) : Magnetic moment of 3.83 B.M. suggests that it
has 3 unpaired electrons,
\ n = 3 i.e., Cr3+ : 3d 3
It involves d 2sp 3 hybridisation so correct distribution of
electrons is 3dxy1 , 3dyz1 , 3dzx1 .
74. (d) : [Co(NH3)6]3+ is d2sp3 hybridised with all electrons
paired hence, it is diamagnetic and inner orbital complex.
75. (b) :
Co
:
[CoF6]3– :
3d
55. (c) : Because F– is a weak field ligand.
56. (d) : [Co(CN)6]3– complex would contain absorption
with highest Emax value because according to spectrochemical
series, the crystal field splitting energy of CN– ion is very high.
57. (d)
58. (a)
59. (d) : [Co(NH3)4Cl2]Cl + AgNO3
[Co(NH3)4Cl2]+ + AgCl↓
Thus it gives precipitate of 1 mol of AgCl.
60. (b) : CoCl 3 ·4NH 3 gives 1 mol AgCl on reaction
with AgNO 3, hence the complex can be represented as
[CoCl2(NH3)4]Cl.
61. (a) : The Cl– ions outside the coordination sphere can
only be precipitated.
62. (a)
4p
4s
3d
3+
4p
4s
F–
sp3d 2 hybridisation
F– F– F–
F–
4d
F–
76. (c) : Inner orbital complexes are formed with strong
ligands as they force electrons to pair up and hence the
complex will be either diamagnetic or will have less number
of unpaired electrons.
77. (d) : [Ni(en)3]Cl2 is a chelating compound and chelated
complexes are more stable than similar complexes with
unidentate ligands as dissociation of the complex involves
breaking of two bonds rather than one.
In [Ni(en)3]Cl2, Ni with d 8 configuration shows octahedral
geometry.
78. (a)
79. (b) : Ni2+ configuration
63. (b)
64. (a) : When Do > P, the electrons paired up in the t2g
level rather than going to the eg level, so
4 0
when Do > P : t2g
eg
and Do < P : t2g3 eg1
4
65. (d) : Fe2+ : 3d 6 ⇒ t 2g
eg2
(Since, F– is a weak field ligand)
Hence four unpaired electrons are present.
During rearrangement only one 3d-orbital may be made
available by pairing the electrons. Thus, inner d 2 sp 3
hybridisation is not possible. So, only sp 3 d 2 (outer)
hybridisation can occur.
CBSE Board Term-II Chemistry Class-12
108
82. (a) : [Sc(H 2 O) 6 ] 3+ has no unpaired electron in its
d subshell and thus d – d transition is not possible whereas
[Ti(H2O)6]3+ has one unpaired electron in its d-subshell which
gives rise to d – d transition to impart colour.
83. (c) : Thiocarbonyl (CS) has one donor atom.
84. (d) : N3– is named as azido. It is derived from HN3.
H2N CH2 CH2 NH2
:
81. (c) : [Fe(CN) 6 ] 3– ion shows magnetic moment
corresponding to one unpaired electron.
2. Denticity : The number of coordinating groups present
in a ligand is called the denticity of ligand.
For example, bidentate ligand ethane-1, 2-diamine has two
donor nitrogen atoms which can link to central metal atom.
:
80. (d) : 4d and 5d elements have greater tendency to
form low spin complex (allows better pairing of electrons)
in comparison to 3d because the difference in energy of t2g
and eg (CFSE, Do) increases in 4d and 5d.
Ethane-1, 2-diamine
3. The splitting of the degenerate d-orbitals into three
orbitals of lower energy, t2g set and two orbitals of higher
energy eg set due to the interaction of ligand in an octahedral
crystal field is known as crystal field splitting in an octahedral
field.
85. (d) : Correct IUPAC name is tetraaquadichloridochromi
um(III) nitrate.
86. (b) : [Fe(CN)6]3– has one unpaired electron, hence it
shows paramagnetic nature while [Fe(CN)6]4– possesses no
unpaired electron and thus shows diamagnetic nature.
87. (b) : The complex ion [Al(NH3)6]3+ undergoes the change
into new complex ion [Al(H2O)6]3+ in aqueous medium due
to higher heat of hydration of aluminium ion on account of
its small size.
[Al(NH3)6]3+ + 6H2O → [Al(H2O)6]3+ + 6NH3
88. (c) : [FeF6]3– is a high spin complex since F– is a weak
field ligand.
89. (b) : The outer electronic configuration of platinum in
ground state is 5d 96s1. The Pt2+ ion formed by the loss of
two electrons has outer electronic configuration of 5d 8. In the
presence of strong ligands (NH3) two unpaired electrons in
the 5d-subshell pair up. This is followed by dsp2 hybridisation
resulting in the formation of four hybridised vacant orbitals
which accommodate four pairs of electrons from four ligands
(two from ammonia and two from Cl–). As such the resulting
complex is square planar.
90. (a)
SUBJECTIVE TYPE QUESTIONS
Oxidation state of Co ion in [CoF6]3– is +3.
1.
Co
3+
:
Co3+ in [CoF6]3– :
No. of unpaired electrons = 4
4.
Fe4[Fe(CN)6]3
5. For one mole of the compound, two moles of AgCl
are precipitated which indicates that two ionisable chloride
ions in the complex. Hence, its structural formula is
[CrCl(H2O)5]Cl2.H2O
6. [Ni(H 2 O) 6 ] 2+ is a high spin complex (D o small)
while [Ni(CN)4]2– is a low spin square planar complex.
In [Ni(H2O)6]2+ complex, d-d transitions are taking place on
absorbing low energy radiation (red component of spectrum)
from visible region showing green as the complementary
colour.
In [Ni(CN)4]2– complex, d-d transitions do not take place in
the visible region of spectrum, d-d transitions take place in
the UV region and hence, complex is colourless.
7.
Hexaamminechromium(III) hexacyanocobaltate(III).
8. Double salts dissociate into ions completely when
dissolved in water. On the other hand, in complexes, the
complex ion does not dissociate.
9. Chelates are cyclic compounds so they are more stable
than normal complexes. In chelates ligands are held by two
or more bonds with the transition metals. e.g.,
10. Coordination number and oxidation state of Pt in the
complex [Pt(en)2Cl2] are 6 and +2 because en is a bidentate
and neutral ligand.
109
Coordination Compounds
11. [NiCl4]2– contains Ni2+ ion with 3d 8 configuration.
[CoF6]3– :
Ni2+
Ground state
Cl – is a weak field ligand. Hence, outer 4s and
4p-orbitals are used in hybridisation.
[NiCl4]
6
[Co(C2O4)3]3– :
2–
with 4Cl– ligands
3
It has two unpaired electrons hence, it is paramagnetic.
[Ni(CO)4] contains Ni(0) – 3d 84s2 configuration.
Ni(0)
3d
4s
4p
Ground state
CO is a strong field ligand hence, 4s-electrons will shift to
3d-orbital making 4s-orbital vacant.
(i) [Co(C2O4)3]3– is diamagnetic as all electrons are paired.
(ii) [Co(C2O4)3]3– is more stable as C2O42– is a chelating
ligand and forms chelate rings.
(iii) [CoF6]3– is an outer orbital complex as it undergoes
sp 3 d 2 hybridization using the outer 4d-orbital.
(iv) [Co(C2O4)3]3– is low spin complex due to absence of
any unpaired electron.
14. (i)[Co(NH3)5(ONO)]Cl2
[Ni(CO)4]
(ii)
K2[Ni(CN)4]
15. Ambidentate ligand : A unidentate ligand which can
coordinate to central metal atom through two different atoms
is called ambidentate ligand.
For example, NO2– ion can coordinate either through nitrogen
or through oxygen to the central metal atom/ion.
The complex has all paired electrons hence, it is diamagnetic.
12. (i) For d 4 ion, if Do < P, the fourth electron enters one
3
of the eg orbitals giving the configuration t2 g e g1. Ligands
for which Do < P are known as weak field ligands and form
high spin complexes.
16. Oxidation state of cobalt in [Co(NH3)6]3+ is +3.
Co
3+
3d6
:
4s0
4p0
Ground State
[Co(NH3)6]3+ ion :
(ii) [Ni(CO)4] contains Ni(0) – 3d 84s2 configuration.
Ni(0)
3d
4s
4p
CO is a strong field ligand hence, 4s-electrons will shift to
3d-orbital making 4s-orbital vacant.
Hybridisation – d 2sp3
Structure – Octahedral
(low spin)
Nature – Diamagnetic
[Ni(CO)4]
IUPAC name : Hexaamminecobalt(III) ion
Ground state
17. In presence of strong field ligand, Co(II) has electronic
6 1
eg .
configuration t2g
eg
The complex has all paired electrons hence, it is diamagnetic.
13. Formation of [CoF 6 ]
represented as :
3–
and [Co(C 2 O 4 ) 3 ]
3–
can be
o > P
t2g
It can easily lose one electron present in eg orbital to give stable
t 62g configuration. This is why Co2+ is easily oxidised to Co3+ in
the presence of strong field ligand.
CBSE Board Term-II Chemistry Class-12
110
18. Diamminedichlorido(ethane-1,2-diamine)chromium(III)
chloride.
19. (i) Ambindent ligands (NO2, ONO) present in the given
compounds.
(ii) [Co(NH3)5(NO2)]Cl2 (Yellow)
Pentaamminenitrito-N-cobalt(III) chloride
[Co(NH3)5(ONO)]Cl2 (Red)
Pentaamminenitrito-O-cobalt(III) chloride
20. (i) [Co(en)2Cl(NH3)]2+
Amminechloridobis(ethane-1,2-diamine)cobalt(III) ion
In presence of strong NH3 and en ligand, Co3+ (3d 6) forms
low spin complex. Hence, complex is diamagnetic.
(ii) [Ni(ox)2(H2O)2]2– : Diaquadioxalatonickelate(II) ion
In the presence of weak ox and H2O ligand, Ni(II) forms high
spin complex (sp3d 2 hybridisation). It is paramagnetic.
21. (i) Triamminetrichloridochromium(III)
(ii) Potassium hexacyanoferrate(III)
(iii) Dibromidobis(ethane-1,2-diamine)cobalt(III) ion
22. In octahedral complex ligands approach along the axes.
So axial orbitals (dx2 – y2, dz2) lie directly in the path of ligand
and experience greater repulsion than the non-axial orbitals.
Whereas in tetrahedral complex ligands are closer to nonaxial orbitals (i.e. dxy, dxz and dyz). So, non-axial orbitals
experience greater force of repulsion than the axial orbitals.
i.e. approach of ligands in octahedral and tetrahedral fields
is opposite of each other. This is why splitting pattern of
d-orbitals in octahedral and tetrahedral geometry is reverse
of each other.
Ground state :
Ni2+ ion :
[NiCl4]2– :
The complex ion has tetrahedral geometry and is paramagnetic
due to the presence of unpaired electrons.
25. (i) [Pt(NH3)2Cl(NO2)] :
Diamminechloridonitrito-N-platinum(II)
It is square planar and diamagnetic.
(ii) [Co(NH3)4Cl2]Cl :
Tetraamminedichloridocobalt(III) chloride
It is octahedral and diamagnetic.
(iii) Ni(CO)4 : Tetracarbonylnickel(0)
It is tetrahedral and diamagnetic.
26. In [Mn(CN)6]3–, Mn is in +3 state so, it has configuration
of 3d 4.
Since CN– is a strong field ligand hence pairing of electrons
in 3d-orbital takes place.
eg
Free metal ion
4
(3d )
t2g
In strong octahedral
ligand field
So, [Mn(CN)6]3– has two unpaired electrons. But in [MnCl6]3–,
Cl– is a weak field ligand, so no pairing takes place and it
has 4 unpaired electrons.
27. Mn (Z = 25)
23. (a) When ligands approach the central metal, atom or
ion of complex its d-orbital breaks into two parts t2g and eg
levels. When light falls on the complex the complex absorbs
light of suitable frequency for transfer of electron from lower
level to higher level. This jump of electron from one d-level
to another is called d-d transition.
(b) For tetrahedral complexes crystal field splitting energy ∆t
is always less than pairing energy. So, tetrahedral complexes
are always high spin.
24. Tetrachloridonickelate(II) ion
Ni atom (Z = 28)
Ground state :
Mn2+ ion :
[Mn(CN)6]4– :
IUPAC name : Potassium hexacyanomanganate(II)
Geometry : Octahedral
No. of unpaired electrons, n = 1
Magnetic behaviour : paramagnetic
28. (i) [Fe(en)2Cl2]Cl
x + 0 × 2 + (–1) × 2 + (–1) × 1 = 0
Oxidation number of iron = +3
\ x = +3
111
Coordination Compounds
(ii) d2sp3 hybridisation and octahedral shape.
(iii) Paramagnetic due to presence of one unpaired electron.
(iv) dichloridobis(ethane-1,2-diamine)iron(III) chloride
29. (i) [Fe(CN)6]4– : Hexacyanidoferrate(II) ion
Hybridisation - d 2sp3
Structure : Inner orbital octahedral complex
CN
CN
CN
Fe
CN
CN
CN
H3N
Cr
Cl
NH3
Since F– is a weak field ligand. So, outer d-orbitals will be
used.
Since, outer d-orbital is used for hybridisation. So, it is outer
orbital complex.
In [Co(NH3)6]3+, Co is in +3 state.
(ii) [Cr(NH3)4Cl2]+ :
Tetraamminedichloridochromium(III) ion
Hybridisation - d 2sp3
Structure : Inner orbital octahedral complex
H3N
:
4–
Octahedral
NH3
32. In [CoF6]3–, Co is in +3 state and has 3d 6 configuration.
:
Since NH3 is a strong field ligand pairing of electrons in
3d-orbital takes place to make two 3d orbitals vacant.
4s
4p
+
Cl
Octahedral
2–
(iii) [Ni(CN)4] : Tetracyanidonickelate(II) ion
Hybridisation - dsp2
Structure - Square planar
Square planar
30. (i) Anhydrous CuSO4 has no ligand. So, crystal field
splitting does not occur so, it does not show any colour but
in hydrated form it is linked with H2O ligand so, it shows
colour due to d-d transition.
(ii) [Ti(H2O)6]Cl3 is a complex compound. In presence of 6
H2O molecules the d-orbitals of Ti3+ undergo splitting. The
compound is coloured (violet) due to d-d transition. On heating
water molecules escape, d-orbitals become degenerate. There
is no d-d transition. Hence compound becomes colourless.
31. Fe atom (Z = 26)
Ground state :
0
Fe
3+
[Fe(CN)6]
33.
Complex Central Hybridi- Geometry
metal sation
of
ion/ of metal complex
atom
ion
involved
Magnetic
behaviour
[Cr(NH3)4
Cl2]Cl
Cr3+
d 2sp3
Octahedral Paramagnetic
[Co(en)3]Cl3
Co3+
d 2sp3
Octahedral Diamagnetic
K2[Ni(CN)4]
Ni
2+
dsp
2
Square
planar
Diamagnetic
34. (a) The difference of energy between two splitted
levels of d-orbitals is called crystal field splitting energy. It
is denoted by D or 10 Dq.
For octahedral Do, for tetrahedral it is Dt and for square
planar Dsp.
0
ion :
3–
Since it uses inner d-orbital for its hybridisation so, it is inner
orbital complex.
Dq
ion
The complex ion has inner orbital octahedral geometry
(low spin) and is paramagnetic due to the presence of one
unpaired electron.
4 0
(i) When Do > P, t 2g
eg
(ii) When Do < P,t 32g e1g
(b) (i) It assumes ligand to be point charges.
(ii) It does not take into account the covalent character of
bonding between the ligand and the central atom.
CBSE Board Term-II Chemistry Class-12
112
35. Ethane-1,2-diamine is stronger ligand than H2O. When
H2O molecule is replaced by ethane-1,2-diamine (en) the
crystal field splitting energy (∆) increases. Complex absorbs
light of higher frequency for d-d transition. This is why colour
of complex changes from green to violet.
[Ni(H2O)6 ]2+ − Green, [Ni(en )3 ]2+ − violet
∆ o small
36. (i) (a) [CoF6]
∆ o large
3–
:
No. of unpaired electrons = 4, m =
4(4 + 2) = 4.9 B.M.
(b) [FeF6]3– :
Oxidation state of Co = +3, coordination number = 6
Trichloridotripyridinechromium(III)
Oxidation state of Cr = +3, coordination number = 6
Stereochemistry : octahedral
E.C. of Cr3+, 3d3 = (t2g)3, (eg)0
Magnetic moment = 3.87 B.M.
eg
(iv) Cs[FeCl4]
5
t2g
No. of unpaired electrons = 5, m =
5(5 + 2) = 5.92 B.M.
(c) [Fe(CN)6]4– :
Caesium tetrachloridoferrate(III)
Oxidation state of Fe = +3; coordination number = 4
Stereochemistry = Tetrahedral
E.C. of Fe3+, 3d 5 = (e)2, (t2)3
Magnetic moment = 5(5 + 2) B.M. = 35 B.M. = 5.92B.M.
(v) K4[Mn(CN)6]
eg
Fe2+ : 3d 6
t2g
No. of unpaired electrons = 0, m = 0
(ii) When FeSO 4 and (NH 4) 2SO 4 solutions are mixed in
1 : 1 molar ratio, Mohr’s salt (a double salt) is formed.
FeSO4(aq) + (NH4)2SO4(aq) → FeSO4·(NH4)2SO4·6H2O
FeSO4·(NH4)2SO4·6H2O
+
2–
Fe2+
(aq) + 2NH 4(aq) + 2SO4(aq) + 6H2O
Because Fe2+ ions are formed on dissolution of Mohr’s salt,
its aqueous solution gives the test of Fe2+ ions.
When CuSO4 is mixed with ammonia, following reaction
occurs :
CuSO4(aq) + 4NH3(aq) → [Cu(NH3)4]SO4
This complex does not produce Cu2+ ion, so the solution of
CuSO4 and NH3 does not give the test of Cu2+ ion.
[Cu(NH3)4]SO4
Pentaamminechloridocobalt(III) chloride
(iii) [CrCl3(py)3]
t2g
: 3d
= 3(3 + 2) = 15 B.M. = 3.87 B.M.
(ii) [Co(NH3)5Cl]Cl2
Magnetic moment (µ) = 0
Co3+ : 3d 6
Fe
Magnetic moment (µ ) = n (n + 2) B.M.
Stereochemistry = octahedral
E.C. of Co3+, 3d6 = (t2g)6, (eg)0
eg
3+
E.C. of Cr3+, 3d 3 = (t2g)3, (eg)0
2–
[Cu(NH3)4]2+
(aq) + SO4(aq)
37. (i) K[Cr(H2O)2(C2O4)2].3H2O
Potassium diaquadioxalatochromate(III) trihydrate
Oxidation state of Cr = +3; coordination number = 6
Stereochemistry : octahedral
Potassium hexacyanomanganate(II)
Oxidation state of Mn = +2, coordination number = 6
Stereochemistry = octahedral
E.C. of Mn2+, 3d5 = (t2g)5 (eg)0
Magnetic moment
(µ) = 11
( + 2) B.M. = 3 B.M. = 1.73B.M.
38. (a) Ligands which can coordinate with the central metal
atom or ion through two donor atoms are known as bidentate
ligands. Examples of bidentate ligands are:
H2C
CH2
O
C
O
H2N
NH2
O
C
O
Ethylene diamine
(en)
Oxalate anion
2–
O
C
O–
O–
Carbonate ion
(b) Ligands which coordinate with the central ion through
more than two donor atoms present in the molecule are
called polydentate ligands. These are called tridentate (three),
tetradentate (four), pentadentate (five) and hexadentate (six)
ligands depending upon the number of coordinating donor
atoms present in their molecules. A common example of
hexadentate ligand is ethylenediamminetetraacetate ion as
113
Coordination Compounds
shown below:
O
(5)
–
O
(6)
–
O
C
O
H2C
(1)
N
C
(c) The charge carried by a complex ion is the algebraic
sum of charges carried by the central ion and the ligands
coordinated to it. Thus [Cu(NH3)6]2+ carries a charge of +2
and because ammonia molecule is neutral therefore, Cu2+
carries a charge of +2. [Ag(CN)2]–, ion carries a charge of
–1 and two cyanide ions coordinated to it carry a charge of
–1 each. So, Ag+ carries a charge of +1.
H2C
(2)
CH2
CH2
CH2
C
CH2
C
N
(3)
–
O
(4)
O
O–
O
Structure of ethylenediamminetetraacetate ion
39.
S.No.
Complex
Oxidation state of
metal atom
Coordination
number of central
metal atom
d-orbital occupation
(i)
K3[Co(C2O4)3]
+3
6
Co3+ = 3d 6; (t2g)6, (eg)0
(ii)
[Cr(en)2Cl2]Cl
+3
6
Cr3+ = 3d 3; (t2g)3
(iii)
(NH4)2[CoF4]
+2
4
Co2+ = 3d 7; (e)4, (t2)3
(iv)
[Mn(H2O)6]SO4
+2
6
Mn2+ = 3d 5; (t2g)3, (eg)2
40. (a) Formation of hexacyanidoferrate(III) ion; [Fe(CN)6]3– :
Electronic configuration of iron in the ground state is 3d 64s2.
The oxidation state of iron is +3 in this complex. Iron (III) has
outer electronic configuration 3d54s0. It has been experimentally observed that this complex has one unpaired electron.
To account for this, two unpaired electrons in 3d subshell
pair up, thus leaving two 3d-orbitals empty. These two vacant
3d-orbitals, alongwith one 4s-orbital and three 4p-orbitals
hybridise to give six equivalent d2sp3 hybridised orbitals. Six
pairs of electrons, one from each cyanide ion, occupy the six
vacant hybrid orbitals so produced. The complex has octahedral geometry and is paramagnetic due to the presence of
one unpaired electron.
2
3
The complex evidently involves (n – 1)d nsnp hybridisation
and is, therefore, called inner orbital or low spin complex.
(b) The outer electronic configuration of cobalt (III) ion
is 3d 6. According to Hund’s rule, four of the 3d-orbitals
are singly filled and one 3d-orbital has a pair of electrons.
Octahedral complexes are formed through d2sp3 hybridisation
for which the metal atom must have two of its 3d-orbitals
empty. This is achieved by the pairing of the two 3d-electrons
as a result of the energy released due to the approach of
the ligands. This results in the formation of an octahedral
complex. As is evident from the figure, the complex does not
contain any unpaired electron and is, therefore, diamagnetic.
(c) In [CoF6]3– complex, the 3d-orbitals remain undisturbed
while the outer 4d-orbitals are used for hybridisation, as
Co3+ ion
[CoF6]3– ion
The complex is paramagnetic since it contains four unpaired
electrons.

Aldehydes, Ketones and
Carboxylic Acids
CHAPTER
6
Recap Notes
ALDEHYDES AND KETONES
General formula : CnH2nO having C O
group.
R
C O ; where R = H, alkyl
X Aldehydes :
H
or aryl.
R
X Ketones :
C O ; where R = alkyl or
R
aryl.
Nomenclature : The common names of
most aldehydes are derived from the common
names of the corresponding carboxylic acids
by replacing the ending –‘ic’ of acid with
aldehyde.
– The IUPAC names of open chain aliphatic
aldehydes and ketones are derived from
the names of the corresponding alkanes
by replacing the ending –‘e’ with –‘al’ and
–‘one’ respectively.
Structure : The C-atom of carbonyl group
is sp2 hybridised and forms three s-bonds
and one p-bond with O atom. Carbonyl
carbon with three atoms attached to it lie in
a same plane with bond angle 120° (trigonal
coplanar structure) and p-electron cloud lies
above and below of this plane.
R2CHOH
X
X
Zn/ 2
2RCHO + ZnO
X
Rosenmund reduction :
RCOCl + H2
X
X
Cu
573 K
RCHO + HCl
Reduction of nitriles :
RC
N
RC
N
RC
N
(i) AlH(i-Bu)2
(ii) H2O
RCHO
(i) RMgX/dry ether
(ii) H3O+
(i) SnCl2 + HCl
Dry ether
(ii) H3O+
RCOR
RCHO + NH4Cl
(Stephen reduction)
From esters :
RCOOR
X
Pd-BaSO4, S
boiling xylene
(i) DIBAL-H, 195 K
(ii) H2O
RCHO
Gatterman-Koch reaction :
CHO
CO, HCl
Anhyd. AlCl3, CuCl
RCHO + H2O
R
C
X
RCHO + H2↑
Friedel-Crafts acylation :
COR
RCOCl
Anhyd. AlCl3
R + H2O
Catalytic dehydrogenation of alcohols :
RCH2OH
O
O
O
R2CHOH + [O]
R2CO + H2↑
Reductive ozonolysis of alkenes :
O
RCH CHR + O3
RCH CHR
Preparation :
X Oxidation of alcohols :
RCH2OH + [O]
Cu
573 K
X
From alkynes :
C
C
dil. H2SO4
HgSO4, 333 K
RCHO or RCOR
115
Aldehydes, Ketones and Carboxylic Acids
C
X
C
B2H6, THF
H2O2/OH–
RCHO or RCOR
Oxidation of 1,2-glycols :
R
CH CH R’ + Pb(OOCCH3)4
OH OH
RCHO + R’CHO
OH OH
R
R
C C
R
R
X
[O]
RCOR + RCOR
Etard reaction :
CH3
– Aromatic aldehydes and ketones are
much less soluble than corresponding
aliphatic aldehydes and ketones due to
larger benzene ring.
– All carbonyl compounds are fairly soluble
in organic solvents.
Chemical properties :
X Nucleophilic addition reactions :
CN
HCN
C O
C
OH
Cyanohydrin
C O
CHO
NaHSO3
Bisulphite
(i) CrO2Cl2/CS2
+
(ii) H3O
X
C O
Side chain chlorination :
CH3
CHO
(i) RMgX
C O
(ii) H2O, 373 K
C O
CH2OH
dry HCl
ROH
dry HCl
R
OH
C
(ii) H3O+
CH2OH
(i) Cl2/h
Physical properties :
X Physical state : Lower members of aldehydes
and ketones (upto C10) are colourless, volatile
liquids except formaldehyde which is gas at
ordinary temperature.
– Higher members of aldehydes and ketones
are solids with fruity odour.
– Lower aldehydes have unpleasant odour
but ketones possess pleasant smell.
X Boiling points : The boiling points of
aldehydes and ketones are higher than
hydrocarbons and ethers of comparable
molecular masses due to weak dipoledipole interactions.
– Their boiling points are lower than those
of alcohols of similar molecular masses
due to absence of intermolecular hydrogen
bonding.
– Among isomeric aldehydes and ketones,
ketones have slightly higher boiling
points due to the presence of two electron
releasing alkyl groups which make
carbonyl group more polar.
X Solubility : Lower members of aldehydes
and ketones (upto C4) are soluble in water
due to H-bonding between polar carbonyl
group and water. However, solubility
decreases with increase in molecular
weight.
SO3Na
OH
C
O CH2
O CH2
C
C
OR
OH
ROH
dry HCl
OR
OR
Acetal
Hemiacetal
X
C
Nucleophilic addition-elimination
reactions :
C O
C O
(i) NH3
(ii) C
NH,
Imine
(i) NH2 – Z
C
(ii) N
Z
where, Z Alkyl, Aryl, OH, NH2, NHC6H5,
NO2
NH
X
NO2 , NHCONH2
Oxidation :
K2Cr2O7/H+
R
C O
RCOOH
H
R
H
H
C O
2Cu2+, 5OH–
+ Rochelle salt
RCOO– + Cu2O
red ppt.
Fehling’s solution test
(Only aldehydes)
CBSE Board Term-II Chemistry Class-12
116
X
Reduction :
R
H2/Ni or
R
CHOH
C O Pt or Pd
R
R
R
LiAlH4 or NaBH4
R
CHOH
C O
R
R
R
R
R
R
R
R′
R
Zn-Hg/HCl
C O Clemmensen reduction
C O
R
HI/Red P, 423 K
R
R
CH2
CH2
R
NH2 NH2/KOH
C O Wolff-Kishner reduction
R′
CH2
Haloform reaction :
NaI + NaOI + H2O
2NaOH + I2
RCOONa +
RCOCH3 + 3NaOI
CHI3↓ + 2NaOH
X
X
Iodoform (yellow ppt.)
(Given by compounds having CH3CO—
group or CH3CH(OH)— group).
Aldol condensation :
O
dil. NaOH
2R CH2 C H
OH
–H2O R
R
R
CH2 CH
Aldol
O
CH2 CH C C
,-Unsaturated aldehyde
R
O
CH C H
H
(aldehydes and ketones having at least
one a-hydrogen)
– Intramolecular aldol condensation :
It takes place in diketones and give rise to
cyclic products.
– Cross aldol condensation : Aldol
condensation is carried out between two
different aldehydes and/or ketones and if
both of them contain a-hydrogen atoms, it
gives a mixture of four products.
X Cannizzaro reaction :
HCHO + HCHO
conc. KOH
Formaldehyde
X
CH3OH + HCOOK
Potassium
formate
Methanol
(aldehydes which do not have an
a-hydrogen atom)
Cross Cannizzaro reaction :
O
O
C6H5 C H + H C H
OH–
C6H5CH2OH + HCOONa
Benzyl alcohol
Sod. formate
– Intramolecular Cannizzaro reaction :
It is given by dialdehydes having no
a-hydrogen atoms.
X Electrophilic substitution reactions :
Aromatic aldehydes and ketones undergo
electrophilic substitution at the ring
in which the carbonyl group acts as a
deactivating and meta directing group.
X Distinction between aldehydes and
ketones :
Tests with
Schiff ’s
reagent
Fehling’s
solution
Tollens’
reagent
2,4dinitrophenylhydrazine
Aldehydes
Ketones
Pink colour
No colour
Red
precipitate
Silver mirror
No precipitate
No silver
mirror
OrangeOrangeyellow or red yellow or red
well defined
well defined
crystals with crystals with
melting points melting points
characteristic characteristic
of individual
of individual
aldehydes.
ketones.
CARBOXYLIC ACIDS
General Formula : C n H 2n O 2 having
—COOH group.
RCOOH where, R=H or alkyl or aryl.
Nomenclature : The common names end
with the suffix –‘ic acid’ and have been
derived from Latin or Greek names of their
natural sources.
– In the IUPAC system, aliphatic carboxylic
acids are named by replacing the ending
–‘e’ in the name of the corresponding
alkane with –‘oic acid’. In numbering the
carbon chain, the carboxylic carbon is
numbered one.
Structure : In carboxylic acids, the bonds to
the carboxyl carbon lie in one plane and are
separated by about 120°. The carboxylic carbon
is less electrophilic than carbonyl carbon
because of the possible resonance structure.
Classification : They are classified as mono,
di, tri and polycarboxylic acids depending
upon the number of carboxyl groups present
in a molecule.
117
Aldehydes, Ketones and Carboxylic Acids
HO
C
C OH
COOH
OH
C OH
O
C
OH
Benzoic acid
(Aromatic
CH2 C OH monocarboxylic acid)
Citric acid
(Tricarboxylic acid)
Aliphatic monocarboxylic acids and aliphatic
esters are known as functional isomers.
Some higher aliphatic monocarboxylic
acids (C12—C18) are known as fatty acids
because they occur in natural fats as esters
of glycerol, e.g., palmitic acid and stearic
acid are obtained on hydrolysis of fats.
Oxidation, alk. KMnO4, H3O+
or CrO3–H2SO4
RCH2OH
K2Cr2O7 + dil. H2SO4
Primary
alcohol
Oxidation
RCHO
K2Cr2O7 + dil. H2SO4
Aldehyde
Hydrolysis
RCN
RCOOH
Mineral acid
Alkyl
Cyanide HCl, H2SO4, etc.
RCN
Alkyl
Cyanide
Hydrolysis
Alkali
NaOH or KOH
RMgX
RCOONa
HCl
RCOOH
CO2 (dry ice)
Grignard
reagent
H3O+
Amide
or, (i) OH–/H2O
Acyl halide
(ii) H3O+
(RCO)2O
H2O
Acid anhydride
RCOOR
H3O+, Ester
RCOOR
Ester
NaOH, H3O+
RCOOH
RCOOH
, H3O+
RCOOH
COOH
KMnO4 / OH /
H3O+
Alkyl benzene
Physical Properties :
X
Physical state : The lower fatty acids
upto C9 are colourless liquids. The higher
ones are colourless waxy solids.
X
Odour : The first three members have a
sharp pungent odour. The middle ones, C4
to C9, have an odour of rancid butter. The
higher members do not possess any smell.
Solubility : Simple aliphatic carboxylic
acids having upto four carbon atoms are
miscible in water due to the formation of
hydrogen bonds with water.
– The solubility decreases with increasing
number of carbon atoms. Higher carboxylic
acids are practically insoluble in water due
to the increased hydrophobic interaction of
hydrocarbon part.
– Benzoic acid, the simplest aromatic carboxylic
acid is nearly insoluble in cold water.
– Carboxylic acids are also soluble in less
polar organic solvents like benzene, ether,
alcohol, chloroform, etc.
X
X
RCOOH
RCOOH
R OOH
, H3O+
R
RCOOH
H2O
RCOCl
KMnO4/OH–
KMnO4/OH–
CR
Alkyne
RCOOH
dry ether
RCONH2
RCOOH
RCOOH
or Fehling’s solution
or Tollens’ reagent
RC
O
Oxalic acid
(Dicarboxylic
acid)
Preparation :
CHR
Alkene
O
CH3 COOH
Acetic acid
(monocarboxylic
acid)
O
CH2 C
O
RCH
Boiling points : Carboxylic acids are
higher boiling liquids than aldehydes,
ketones and even alcohols of comparable
molecular masses due to more extensive
association of their molecules through
intermolecular hydrogen bonding. The
H-bonds are not broken completely even
in the vapour phase.
Chemical reactions :
X
Reactions involving cleavage of O—H
bond :
O
R C OH
CBSE Board Term-II Chemistry Class-12
118
X
Reactions involving cleavage of C—OH bond:
RCOOH/H+, or P2O5, –H2O
O
R C OH
ROH/H+
–H2O
O
X
Ring substitution in aromatic acids :
Aromatic
R C O C R
Anhydride
carboxylic
electrophilic
acids
substitution
undergo
reactions
in which the carboxyl group acts as a
RCOOR
Ester
PCl3 or PCl5
or SOCl2 in pyridine
NH3, –H2O
X
O
deactivating and meta directing group.
RCOCl
Acid chloride
RCONH2
Amide
Reactions involving —COOH group :
X
Distinction test between phenol and
carboxylic acid :
X
Hell—Volhard—Zelinsky reaction :
Test
Phenol
Carboxylic acid
NaHCO3
test
No reaction
Brisk
effervescence of
CO2 gas.
FeCl3 test Violet colour Buff coloured ppt.
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1.
Ketones
can be obtained in one
step by (where R and R′ are alkyl groups)
(a) hydrolysis of esters
(b) oxidation of primary alcohols
(c) oxidation of secondary alcohols
(d) reaction of alkyl halides with alcohols.
2. Aldehydes other than formaldehyde react
with Grignard’s reagent to give addition products
which on hydrolysis give
(a) tertiary alcohols
(b) secondary alcohols
(c) primary alcohols
(d) carboxylic acids.
3. Which of the following compounds will
undergo Cannizzaro reaction?
(b) CH3COCH3
(a) CH3CHO
(c) C6H5CHO
(d) C6H5CH2CHO
4. Propanal on treatment with dilute sodium
hydroxide gives
(a) CH3CH2CH2CH2CH2CHO
(b) CH3CH2CH(OH)CH2CH2CHO
(c) CH3CH2CH(OH)CH(CH3)CHO
(d) CH3CH2COOH
5. Various products formed on oxidation of
2,5-dimethylhexan-3-one are
(i)
(ii)
(iii) CH3COOH
(iv) HCOOH
(a) (i) and (iii)
(b) (i), (ii) and (iii)
(c) (i), (ii), (iii) and (iv) (d) (iii) and (iv)
6. Alkene (X) (C 5H 10) on ozonolysis gives a
mixture of two compounds (Y) and (Z). Compound
(Y) gives positive Fehling’s test and iodoform
test. Compound (Z) does not give Fehling’s test
but give iodoform test. Compounds (X), (Y) and
(Z) are
XY
Z
(a) C6H5COCH3
CH3CHO
CH3COCH3
(b)
CH3CHO
CH3COCH3
(c) CH3CH2CH == CH2
CH3CH2CHO HCHO
CH3CHO
(d) CH3 —CH == CH — CH3 CH3CHO
7. A compound (X) with a molecular formula
C 5 H 10 O gives a positive 2,4-DNP test but a
negative Tollen’s test. On oxidation it gives a
carboxylic acid (Y) with a molecular formula
C3H6O2. Potassium salt of (Y) undergoes Kolbe’s
reaction and gives a hydrocarbon (Z). (X), (Y)
and (Z) respectively are
(a) pentan-3-one, propanoic acid, butane
(b) pentanal, pentanoic acid, octane
(c) 2-methylbutanone, butanoic acid, hexane
(d) 2, 2-dimethylpropanone, propanoic acid,
hexane
8. Which of the following statements is
incorrect?
(a) FeCl3 is used in the detection of phenols.
(b) Fehling solution is used in the detection of
glucose.
(c) Tollens’ reagent is used in the detection of
unsaturation.
(d) NaHSO3 is used in the detection of carbonyl
compounds.
9. Which of the following compounds will
give a coloured crystalline compound with
?
(a) CH3COCl
(c) CH3COCH3
(b) CH3COOC2H5
(d) CH3CONH2
CBSE Board Term-II Chemistry Class-12
120
10. Which of the following reagents are not
correctly matched with the reaction?
(a) CH3CH CHCHO
CH3CH CHCOOH
: Ammonical AgNO3
(b) CH3CH CHCHO
CH3CH CHCH2OH
: H2/Pt
—
—
(c) R COOH
R CH2OH : NaBH4
(d) CH3CH2COCl
CH3CH2CHO
: H2, Pd/BaSO4
11. In the following reaction, product (P) is
O
R—C—Cl
H2
P
Pd/BaSO4
(a) RCHO
(c) RCOOH
(b) RCH3
(d) RCH2OH
12. Which of the following will not give aldol
condensation?
(a) Phenyl acetaldehyde
(b) 2-Methylpentanal
(c) Benzaldehyde
(d) 1-Phenylpropanone
13. Which of the following statements is correct
regarding formic acid?
(a) It is a reducing agent.
(b) It is a weaker acid than acetic acid.
(c) It is an oxidising agent.
(d) When its calcium salt is heated, it forms
acetone.
14. The condensation product of benzaldehyde
and acetone is
(a)
O
(b) C6H5CH2 C CH CH2
(c)
(d)
15. Which among the following is most reactive
to give nucleophilic addition?
(a) FCH2CHO
(b) ClCH2CHO
(c) BrCH2CHO
(d) ICH2CHO
16. Which of the following IUPAC names is not
correctly matched?
(a)
(b)
(c) PhCH2CH2COOH : 3-Phenylpropanoic acid
(d)
17. Which of the following is the most reactive
isomer?
(a)
(b)
(c)
(d)
18. The correct order of increasing acidic
strength is
(a) Phenol < Ethanol < Chloroacetic acid <
Acetic acid
(b) Ethanol < Phenol < Chloroacetic acid <
Acetic acid
(c) Ethanol < Phenol < Acetic acid < Chloroacetic
acid
(d) Chloroacetic acid < Acetic acid < Phenol <
Ethanol
19. To differentiate between pentan-2-one and
pentan-3-one a test is carried out. Which of the
following is the correct answer?
(a) Pentan-2-one will give silver mirror test
(b) Pentan-2-one will give iodoform test.
(c) Pentan-3-one will give iodoform test
(d) None of these.
20. What happens when a carboxylic acid is
treated with lithium aluminium hydride?
(a) Aldehyde is formed.
(b) Primary alcohol is formed.
(c) Ketone is formed.
(d) Grignard reagent is formed.
21. Which of the following will not undergo HVZ
reaction?
(a) Propanoic acid
(b) Ethanoic acid
(c) 2-Methylpropanic acid
(d) 2,2-Dimethylpropanoic acid
121
Aldehydes, Ketones and Carboxylic Acids
22. When propanal reacts with 2-methylpropanal
in presence of NaOH, four different products are
formed. The reaction is known as
(a) aldol condensation
(b) cross aldol condensation
(c) Cannizzaro reaction
(d) HVZ condensation.
23. Propanone can be prepared from ethyne by
(a) passing a mixture of ethyne and steam over
a catalyst, magnesium at 420°C
(b) passing a mixture of ethyne and ethanol
over a catalyst zinc chromite
(c) boiling ethyne with water and H2SO4
(d) treating ethyne with iodine and NaOH.
24. Match the column I with column II and
mark the appropriate choice.
Column I
(A)
RCOCH3
Zn-Hg
HCl
RCH2CH3
(B)
NaOH
2C6H5CHO
C6H5COONa +
C6H5CH2OH
Column II
(i) Kolbe’s
reaction
(ii) Clemmensen
reduction
(C) C H + CH COCl Anh. (iii) Friedel–Crafts
6 6
3
AlCl3
reaction
C6H5COCH3
(D) C6H5OH + CO2 + NaOH (iv) Cannizzaro
reaction
HOC6H4COONa
(a)
(b)
(c)
(d)
(A)
(A)
(A)
(A)
→
→
→
→
(ii), (B) → (iv), (C) → (iii), (D) → (i)
(i), (B) → (iii), (C) → (ii), (D) → (iv)
(iii), (B) → (ii), (C) → (i), (D) → (iv)
(iv), (B) → (i), (C) → (ii), (D) → (iii)
25. Arrange the following compounds in
increasing order of their reactivity in nucleophilic
addition reactions.
Ethanal, Propanal, Propanone, Butanone
(a) Butanone < Propanone < Propanal < Ethanal
(b) Propanone < Butanone < Ethanal > Propanal
(c) Propanal < Ethanal < Propanone < Butanone
(d) Ethanal < Propanal < Propanone < Butanone
26. Which of the following reactions will give
benzophenone?
(i) Benzoyl chloride + Benzene + AlCl3
(ii) Benzoyl chloride + Phenylmagnesium
bromide
(iii) Benzoyl chloride + Diphenyl cadmium
(a) (i) and (ii)
(c) (i) and (iii)
(b) (ii) and (iii)
(d) (i), (ii) and (iii)
27. What are the correct steps to convert
acetaldehyde to acetone?
(a) CH3MgBr, H2O, Oxidation
(b) Oxidation, Ca(OH)2, Heat
(c) Reduction, KCN, Hydrolysis
(d) Oxidation, C2H5ONa, Heat
28. Hydrocarbons are formed when aldehydes
and ketones are reacted with amalgamated zinc
and conc. HCl. The reaction is called
(a) Cannizzaro reaction
(b) Clemmensen reduction
(c) Rosenmund reduction
(d) Wolff-Kishner reduction.
29. The addition of HCN to carbonyl compounds
is an example of
(a) nucleophilic addition
(b) electrophilic addition
(c) free radical addition
(d) electromeric addition.
30. Identify reactant (X) in the given reaction
sequence.
CH3COCH3 + X
(a) CH3MgCl
(c) MgCl2
(CH3)3C — OMg — Cl
H 2O
(CH3)3C — OH + Mg
(b) CH3COCl + Mg
(d) CH3CH2MgCl
OH
Cl
31. Which of the following is the correct order of
relative strength of acids?
(a) ClCH2COOH > BrCH2COOH > FCH2COOH
(b) BrCH2COOH > ClCH2COOH > FCH2COOH
(c) FCH2COOH > ClCH2COOH > BrCH2COOH
(d) ClCH2COOH > FCH2COOH > BrCH2COOH
32. An organic compound of molecular formula
C3H6O did not give a silver mirror with Tollen’s
reagent but gave an oxime with hydroxylamine.
It may be
(a) CH2 == CH — CH2 — OH
(b) CH3COCH3
(c) CH3CH2CHO
(d) CH2==CH — OCH3
33. What is the test to differentiate between
pentan-2-one and pentan-3-one?
(a) Iodoform test
(b) Benedict’s test
(c) Fehling’s test
(d) Aldol condensation test
CBSE Board Term-II Chemistry Class-12
122
34. Which of the following carbonyl compounds
is most polar?
(a)
(c)
(b)
(d)
35. In nucleophilic addition reactions, the
reactivity of carbonyl compounds follows the
order
(a) HCHO > RCHO > ArCHO > R2CO > Ar2CO
(b) HCHO > R2CO > Ar2CO > RCHO > ArCHO
(c) Ar2CO > R2CO > ArCHO > RCHO > HCHO
(d) ArCHO > Ar2CO > RCHO > R2CO > HCHO
36. Which of the following statements is not
correct?
(a) Aldehydes and ketones are functional
isomers.
(b) Formaldehyde reacts with ammonia to form
hexamethylenetetramine.
(c) LiAlH4 converts ketones into sec-alcohols.
(d) Butanal and propanal can be distinguished
by iodoform test.
37. Study the given reaction and identify the
process which is carried out.
(a) It is used for purification of aldehydes and
ketones.
(b) It is used to distinguish aldehydes from
ketones.
(c) It is used to prepare cyclic aldehydes and
ketones.
(d) It is used to study polar nature of aldehydes
and ketones.
38. Which of the following aldehydes will show
Cannizzaro reaction?
(a) HCHO
(b) C6H5CHO
(d) All of these
(c) (CH3)3CCHO
39. a-Hydroxypropanoic acid can be prepared
from ethanal by following the steps given in the
sequence.
(a) Treat with HCN followed by acidic hydrolysis.
(b) Treat with NaHSO 3 followed by reaction
with Na2CO3.
(c) Treat with H2SO4 followed by hydrolysis.
(d) Treat with K2Cr2O7 in presence of sulphuric
acid.
40. Which is the correct method of synthesising
acetamide from acetone?
Pd/ BaSO4
(a) CH3COCH3 
→ CH3CHO
NH3
H2O

→ CH3CH2NH2 
→ CH3CONH2
I
NaOH
2
(b) CH3COCH3 
→ CH3COONa
H+
NH3
∆

→ CH3COONH4 
→ CH3CONH2
CrO3
(c) CH3COCH3 
→ CH3COOH
NH
3 → CH CONH

3
2
I2
(d) CH3COCH3 
→ CH3COOH
HCl
NaOH
NH
3 → CH CONH

→ CH3COCl 
3
2
41. There is a large difference in the boiling
points of butanal and butan-1-ol due to
(a) intermolecular hydrogen bonding in
butan-1-ol
(b) intramolecular hydrogen bonding in butanal
(c) higher molecular mass of butan-1-ol
(d) resonance shown by butanal.
42. A compound (X) having molecular formula
C4H8O2 is hydrolysed by water in presence of an
acid to give a carboxylic acid (Y) and an alcohol
(Z). (Z) on oxidation with chromic acid gives (Y).
(X), (Y) and (Z) are
X Y
Z
(a) CH3COOCH3
CH3COOH
CH3OH
CH3COOH
C2H5OH
(b) CH3COOC2H5
C2H5COOH
C2H5OH
(c) C2H5COOCH3
C2H5COOH
CH3OH
(d) CH3COOC2H5
43. –OH group present in alcohols is neutral
while it is acidic in carboxylic acid because
(a) in carboxylic acid –OH group is attached to
electron withdrawing carbonyl group
(b) in alcohols –OH group is attached to alkyl
group which is electron withdrawing
(c) carboxylic group is an electron releasing
group
(d) alcoholic group is an electron withdrawing
group.
44. Which of the following orders is not correct
for the decreasing order of acidic character?
(a) CH3CH2CH(Cl)COOH >
CH3CH(Cl) CH2COOH >
CH2(Cl)CH2CH2COOH > CH3CH2CH2COOH
(b) ICH2COOH > BrCH2COOH > ClCH2COOH
> FCH2COOH
123
Aldehydes, Ketones and Carboxylic Acids
(c) CCl3COOH > CHCl2COOH > CH2ClCOOH
> CH3COOH
(d) HCOOH > CH 3 COOH > C 2 H 5 COOH >
(CH3)2CHCOOH
49. Identify the products (X) and (Y) in the given
reaction :
45. The correct structure representation of
carboxylate ion is
(a) X = Acetophenone, Y = m-Nitroacetophenone
(b) X = Toluene, Y = m-Nitroacetotoluene
(c) X = Acetophenone, Y = o and
p-Dinitroacetophenone
(d) X = Benzaldehyde, Y = m-Nitrobenzaldehyde
(a)
(b)
(c)
(d)
46. Which of the following reactions does not
occur?
(a)
(b)
50. Benzaldehyde can be prepared from benzene
by passing vapours of ......... and ........... in its
solution in presence of catalyst mixture of
aluminium chloride and cuprous chloride. The
reaction is known as ........... .
(a) HCl, SnCl4, Rosenmund reduction
(b) CO, HCl, Gattermann–Koch reaction
(c) CO2, H2SO4, Clemmensen reduction
(d) O3, alcohol, Wolff–Kishner reduction
51. The best oxidising agent for oxidation of
is
(a) Baeyer’s reagent
(b) Tollen’s reagent
(c) Schiff’s reagent
(d) Acidified dichromate.
(c)
(d)
47. Which of the following names of the organic
compounds is not correctly written?
52. The product of hydrolysis of ozonide of
1-butene are
(a) ethanal only
(b) ethanal and methanal
(c) propanal and methanal
(d) methanal only.
53. CH3—C CH
(a)
40% H2SO4
1% HgSO4
Isomerisation
CH3—C—CH3
O
Structure of A and type of isomerism in the above
reaction are
(a) Prop-1-en-2-ol, metamerism
(b) Prop-1-en-1-ol, tautomerism
(c) Prop-2-en-2-ol, geometrical isomerism
(d) Prop-1-en-2-ol, tautomerism.
(b)
(c)
(d)
48. In the following sequence of reaction, the
final product (Z) is
54. A diene, buta-1,3-diene was subjected to
ozonolysis to prepare aldehydes. Which of the
following aldehydes will be obtained during the
reaction?
(a) CHO
(a) ethanal
(c) propanone
A
(b) propan-2-ol
(d) propan-1-ol
+ 2HCHO
CHO
(b) CH3CHO + 2HCHO
CBSE Board Term-II Chemistry Class-12
124
(c) CH3CH2CHO + CH3CHO
(d) 2CH3CH2CHO
(A) + C2H5OH
55. Addition of water to alkynes occurs in acidic
medium and in the presence of Hg2+ ions as a
catalyst. Which of the following products will be
formed on addition of water to but-1-yne under
these conditions?
(C) + HOH
(D)
[O]
H
(B) + (C)
+
(B) + (D)
(B)
(B) + Ca(OH)2
Calcium salt + H2O
dry distillation
(A)
(a)
(a)
(b)
(c)
(d)
(b)
(c)
(d)
56. An organic compound (X) with molecular
formula C 9 H 10 O gives positive 2,4-DNP and
Tollen’s tests. It undergoes Cannizzaro
reaction and on vigorous oxidation it gives
1,4-benzenedicarboxylic acid. Compound (X) is
(a) benzaldehyde
(b) o-methylbenzaldehyde
(c) p-ethylbenzaldehyde
(d) 2, 2-dimethylhexanal
O
57. R CH CH CHO + NH2 C NHNH2
(X) in the above reaction is
H+
X
(a)
(b)
(c)
(d)
58. Which of the following will not yield acetic
acid on strong oxidation?
(a) Butanone
(b) Propanone
(c) Ethyl ethanoate
(d) Ethanol
59. Which of the following compounds does not
react with NaHSO3?
(a) HCHO
(b) C6H5COCH3
(c) CH3COCH3
(d) CH3CHO
60. Study the following reactions and mark the
appropriate choice.
(B)
CH3COCH3
(Acetone)
(C)
(D)
(CH3CO)2O CH3COOH CH3COOC2H5
C2H5OH
CH3COCl
HCOOH
CH3COOCH3
CH3OH
CH3COOH
CH3OH
CH3COOCH3
CH3OH
CH3NH2
CH3COOH CH3COOCH3
C2H5OH
61. Compound (X) with molecular formula
C 3 H 8 O is treated with acidified potassium
dichromate to form a product (Y) with molecular
formula C3H6O. (Y) does not form a shining silver
mirror on warming with ammoniacal AgNO3.
(Y) when treated with an aqueous solution of
NH2CONHNH2. HCl and sodium acetate to give
a product (Z). The structure of (Z) is
(a)
(b)
(c)
(d)
62. Aldehydes and ketones are isomers as
they have same general formula but different
functional groups. Both these functional groups
can be distinguished by various tests.
A compound with molecular formula C9H10 has
two isomers P and Q which undergo ozonolysis
to give two functional isomers (R and S) with
formula, C8H8O.
(i) O3/CH2Cl2
P (ii) Zn/H O
2
(i) O /CH Cl
R + HCHO,
3
2 2
Q (ii) Zn/H O S + HCHO
2
Which of the given options can not be correct for
P, Q, R and S?
I. If P is 4-vinyl toluene then R gives Cannizzaro
reaction but not haloform reaction.
II. If Q is 4-vinyl toluene then S gives haloform
reaction but not Cannizzaro.
III. If Q is 2-phenylpropene then S gives haloform
reaction but not Cannizzaro.
IV. If P is 2-phenylpropene then R gives both
Cannizzaro and haloform reaction.
(a) I and II only
(b) I and III only
(c) II and III only
(d) II and IV only
125
Aldehydes, Ketones and Carboxylic Acids
63. Study the given reactions chart carefully :
Compound [A]
(C5H12O)
COOH
Oxidation
X2/NaOH
No
haloform
reaction
[B]
2,4-DNP
(C5H10O)
+
2,4- Dinitrophenylhydrazone
–
Ag /OH
No silver mirror is formed
Which is correct for compounds A and B?
(a) B is an aldehyde.
(b) B is a ketone but not methyl ketone.
(c) A is a primary alcohol.
(d) B convert to A using Zn–Hg/HCl.
64. Acidic nature of carboxylic groups depends
on various factors like presence of electron
withdrawing groups (–I, –R effect), presence of
electron donating groups (+I, +R effect), distance
of attached groups. Stability of carboxylate ion
plays an important role in the acidic nature of
carboxylic acid.
Vinita, a class 12 student has written the following
orders of acidity for various carboxylic acids :
COOH
COOH
COOH
COOH
>
I:
>
NO2
>
CH3
OCH3
II : CH3CH2CH COOH >
CH3
CH
CH3
F
CH
F
CH2COOH >
CH2COOH > CH2 CH2CH2COOH
Cl
Cl
>
III :
COOH
COOH
>
OH >
COOH
CH3
NO2
Which of the following orders is not correct and
what is the reason behind it?
(a) Order I, +R effect of –OCH 3 group is not
interpreted correctly.
(b) Order II, position of electronegative group
is not interpreted correctly.
(c) Order III, ortho effect is not considered.
(d) All the orders I, II and III are correct.
65. Carboxylic acids do not undergo Friedel Craft’s
reaction because
(a) —COOH group is meta directing
(b) —COOH group is resonance stabilised
(c) carboxyl group is deactivating and gets
bonded to Friedel Craft’s catalyst
(d) all of above.
66. A ketone ‘A’ (C 4H 8O), which undergoes a
haloform reaction, gives a compound ‘B’ on
reduction. ‘B’ on heating with sulphuric acid
gives a compound ‘C’ which forms mono-ozonide
‘D’. ‘D’ on hydrolysis with zinc dust gives only,
‘E’.
Identify the correct statement.
(a) A is butan-2-one; B is butan-2-ol.
(b) B is but-2-ene; C is acetaldehyde.
(c) D is acetaldehyde; E is butan-2-ol.
(d) B is butan-1-one; D is but-2-ene.
Case Based MCQs
Case I : Read the passage given below and
answer the following questions :
Carboxylic acids having an a-hydrogen atom
when treated with chlorine or bromine in the
presence of small amount of red phosphorus gives
a-halocarboxylic acids. The reaction is known as
Hell-Volhard-Zelinsky reaction.
R — CH2 — COOH + X2
red P
When sodium salt of carboxylic acid is heated
with soda lime it loses carbon dioxide and gives
hydrocarbon with less number of C-atoms.
R — COOH
Carboxylic
acid
R—CH—COOH
X
(X = Cl, Br)
NaOH
R— COONa
Sod.
carboxylate
NaOH + CaO
D
R— H + Na2CO3
Alkane
CBSE Board Term-II Chemistry Class-12
126
In the following questions (Q. No. 67-71), a
statement of assertion followed by a statement
of reason is given. Choose the correct answer out
of the following choices on the basis of the above
passage.
(a) Assertion and reason both are correct
statements and reason is correct explanation
for assertion.
(b) Assertion and reason both are correct
statements but reason is not correct
explanation for assertion.
(c) Assertion is correct statement but reason is
wrong statement.
(d) Assertion is wrong statement but reason is
correct statement.
67. Assertion : (CH 3) 3CCOOH does not give
H.V.Z reaction.
Reason : (CH3)3CCOOH does not have
a-hydrogen atom.
68. Assertion : H.V.Z. reaction involves the
treatment of carboxylic acids having a-hydrogens
with Cl2 or Br2 in presence of small amount of
red phosphorus.
Reason : Phosphorus reacts with halogens to
form phosphorus trihalides.
69. Assertion : Propionic acid with Br2/P yields
CH2Br— CHBr —COOH.
Reason : Propionic acid has two a-hydrogen
atoms.
70. Assertion : C6H5COCH2COOH undergoes
decarboxylation easily than C6H5COCOOH.
Reason : C6H5COCH2COOH is a b-keto acid.
71. Assertion : On heating 3-methylbutanoic
acid with soda lime, isobutane is obtained.
Reason : Soda lime is a mixture of NaOH + CaO
in the ratio 3 : 1.
Case II : Read the passage given below and
answer the following questions from 72 to 76.
Aldehydes and ketones having acetyl group
O
CH3—C— are oxidised by sodium hypohalate
(NaOX) or halogen and alkali (X2 + OH–) to
corresponding sodium salt having one carbon
atoms less than the carbonyl compound and give
a haloform.
O
NaOX
R — C — CH3 or X + NaOH
2
O
R— C— ONa + CHX3 (X = Cl, Br, I)
Sodium hypoiodite (NaOI) when treated with
compounds containing CH3CO — group gives
yellow precipitate of iodoform. Haloform reaction
does not affect a carbon-carbon double bond
present in the compound.
72. Which of the following compounds will give
positive iodoform test?
(a) Isopropyl alcohol
(b) Propionaldehyde
(c) Ethylphenyl ketone (d) Benzyl alcohol
73. Which of the following compounds is not
formed in iodoform reaction of acetone?
(a) CH3COCH2I
(b) ICH2COCH2I
(c) CH3COCHI2
(d) CH3COCI3
74. For the given set of reactions,
starting compound A corresponds to
O
O
(b)
(a)
CH2COOH
CH2COOH
O
O
(c)
(d)
COCH3
COCH3
75. In the following reaction sequence, the
correct structures of E, F and G are
*
(* implies 13C labelled carbon)
(a)
(b)
(c)
(d)
76. An organic compound ‘A’ has the molecular
formula C 3 H 6 O. It undergoes iodoform test.
127
Aldehydes, Ketones and Carboxylic Acids
When saturated with HCl it gives ‘B’ of molecular
formula C9H14O. ‘A’ and ‘B’ respectively are
(a) propanal and mesityl oxide
(b) propanone and mesityl oxide
(c) propanone and 2,6-dimethyl-2,5-heptadien‑4‑one
(d) propanone and propionaldehyde.
78. Which of the following compounds would
be the main product of an aldol condensation of
acetaldehyde and acetone?
(a) CH3CH CHCHO
(b) CH3CH CHCOCH3
(c) (CH3)2C CHCHO
(d) (CH3)2C CHCOCH3
Case III : Read the passage given below and
answer the following questions from 77 to 81.
The addition reaction of enol or enolate to the
carbonyl functional group of aldehyde or ketone is
known as aldol addition. The b-hydroxyaldehyde or
b-hydroxyketone so obtained undergo dehydration
in second step to produce a conjugated enone. The
first part of reaction is an addition reaction and the
second part is an elimination reaction. Carbonyl
compound having a-hydrogen undergoes aldol
condensation reaction.
O
O
79. Which combination of carbonyl compounds
gives phenyl vinyl ketone by an aldol
condensation?
O
2CH3CH2
C
–
OH
H D
CH3CH2CH
C
O
O
CH
H
CH3
Mechanism :
HO + H
C
C
H –H O H3C
2
CH
C
H
CH3
CH3CH2
O
O
C +CH
C
H
H CH3
O
CH3CH2
OH
H2O
CH3CH2 CH
O
CH
C
H
H
–H2O
D
CH3
CH3CH2
C
CH3
O
CH
CH
O
CH
C
C
H
CH3
77. Condensation reaction is the reverse of
which of the following reaction?
(a) Lock and key hypothesis
(b) Oxidation
(c) Hydrolysis
(d) Glycogen formation
(a)
(b)
(c)
(d)
Ph
Acetophenone and
Acetophenone and
Benzaldehyde and
Benzaldehyde and
Formaldehyde
acetaldehyde
acetaldehyde
acetone
80. Which of the following will undergo aldol
condensation?
(a) HCHO
(b) CH3CH2OH
(c) C6H5CHO
(d) CH3CH2CHO
81. Which of the following does not undergo
aldol condensation ?
(a) CH3CHO
(b) CH3CH2CHO
(c) CH3COCH3
(d) C6H5CHO
Case IV : Read the passage given below and
answer the following questions :
Aldehydes and ketones undergo nucleophilic
addition reactions.
Nu
Nu
R1 +d –d Nu
E+
C O
C
C
R2
R1 R O
R1 R OE
2
2
Carbonyl carbon is electron deficient hence acts
as an electrophile. Nucleophile attacks on the
electrophilic carbon atom of the carbonyl group
from a direction perpendicular to the plane of the
molecule.
R1 +d
C
R2
Nu
d
O
Nu
Slow
C
R1 R O
2
Nu
Fast
H+
C
R1 R OH
2
In this process, hybridisation of carbon atom
changes from sp2 to sp3 and a tetrahedral
alkoxide ion is formed as intermediate. This
intermediate captures proton from the reaction
medium to give the neutral product. Aldehydes
are generally more reactive than ketones in
nucleophilic addition reactions.
CBSE Board Term-II Chemistry Class-12
128
In the following questions (Q. No. 82-86), a
statement of assertion followed by a statement
of reason is given. Choose the correct answer out
of the following choices on the basis of the above
passage.
(a) Assertion and reason both are correct
statements and reason is correct explanation
for assertion.
(b) Assertion and reason both are correct
statements but reason is not correct
explanation for assertion.
(c) Assertion is correct statement but reason is
wrong statement.
(d) Assertion is wrong statement but reason is
correct statement.
82. Assertion : Benzaldehyde is more reactive
than ethanal towards nucleophilic attack.
Reason : The overall effect of –I and +R effect of
phenyl group decreases the electron density on
the carbon atom of C O group in benzaldehyde.
83. Assertion : (CH3)3CCOC(CH3)3 and acetone
can be distinguished by the reaction with
NaHSO3.
Reason : HSO3– is the nucleophile in bisulphite
addition.
Reason : Reactivity of carbonyl group is due to
electrophilic nature of carbonyl carbon.
Case V : Read the passage given below and
answer the following questions from 87 to 91.
When an aldehyde with no a-hydrogen reacts with
concentrated aqueous NaOH, half the aldehyde
is converted to carboxylic acid salt and other half
is converted to an alcohol. In other words, half of
the reactant is oxidized and other half is reduced.
This reaction is known as Cannizzaro reaction.
O
2
85. Assertion : The formation of cyanohydrin
from an aldehyde or ketone occurs very slowly
with pure HCN. This reaction is catalysed by a
base.
Reason : Base generates CN– ion which is a
stronger nucleophile.
CHO
86. Assertion :
is more reactive towards
NO2
CHO
.
nucleophilic addition reaction than
CH3
Conc. NaOH
C
Mechanism :
O
Ph
H
C
OH
Ph
ONa +
O
C
H+ C
C
O
H
+
H
C
H
Ph
H
Ph
Ph
O
O
Ph
CH2OH
O
OH
(I), CH3CHO(II) and
CH3COCH3(III) is I > II > III.
Reason : Aldehydes and ketones undergo
nucleophilic addition reactions.
H
O
84. Assertion : Ease of nucleophilic addition
COCH3
of the compounds
C
OH
O
C
O
+ H
C
Ph
H
87. A mixture of benzaldehyde and formaldehyde
on heating with aqueous NaOH solution gives
(a) benzyl alcohol and sodium formate
(b) sodium benzoate and methyl alcohol
(c) sodium benzoate and sodium formate
(d) benzyl alcohol and methyl alcohol.
88. Which of the following compounds will
undergo Cannizzaro reaction?
(b) CH3COCH3
(a) CH3CHO
(c) C6H5CHO
(d) C6H5CH2CHO
89. Trichloroacetaldehyde is subjected to
Cannizzaro’s reaction by using NaOH. The
mixture of the products contains sodium
trichloroacetate ion and another compound. The
other compounds is
(a) 2, 2, 2-trichloroethanol
(b) trichloromethanol
129
Aldehydes, Ketones and Carboxylic Acids
(c) 2, 2, 2-trichloropropanol
(d) chloroform.
90. In Cannizzaro reaction given below :
2PhCHO
step is
OH
–
PhCH2OH +
PhCO2–
the slowest
–
(a) the attack OH at the carboxyl group
(b) the transfer of hydride to the carbonyl group
(c) the abstraction of proton from the carboxylic
group
(d) the deprotonation of PhCH2OH.
91. Which of the following reaction will not
result in the formation of carbon-carbon bonds?
(a) Cannizzaro reaction
(b) Wurtz reaction
(c) Reimer-Tiemann reaction
(d) Friedel-Crafts’ acylation
For question numbers 92-105, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
92. Assertion : Aromatic aldehydes and
formaldehyde undergo Cannizzaro reaction.
Reason : Aromatic aldehydes are almost as
reactive as formaldehyde.
93. Assertion : a-Hydrogen atoms in aldehydes
and ketones are acidic.
Reason : The anion left after the removal of
a-hydrogen is stabilised by inductive effect.
94. Assertion : Hydrogen bonding in carboxylic
acids is stronger than alcohols.
Reason : Highly branched carboxylic acids are
more acidic than unbranched acids.
Reason : Carboxyl group increases the electron
density at the meta-position.
99. Assertion : Boiling point of aldehydes lie
in between parent alkanes and corresponding
alcohols.
Reason : Aldehydes cannot form intermolecular
hydrogen bonds like alcohols.
100. Assertion : Carboxylic acids are stabilised
by resonance.
Reason : Chloroacetic acid is weaker than acetic
acid.
101. Assertion : Benzaldehyde undergoes aldol
condensation.
Reason : Aldehydes having a-hydrogen atom
undergo aldol condensation.
95. Assertion : m-Chlorobenzoic acid is a
stronger acid than p-chlorobenzoic acid.
Reason : In m-chlorobenzoic acid both – I-effect
and +R-effect of Cl operate but in p-chlorobenzoic
acid only +R-effect of Cl operates.
102. Assertion : Formic acid is a stronger acid
than benzoic acid.
Reason : pKa of formic acid is lower than that of
benzoic acid.
96. Assertion : Ketones can be converted into
acids by haloform reaction.
Reason : Addition of Grignard reagents to dry
ice followed by hydrolysis gives ketones.
103. Assertion : NaHSO 3 is used for the
purification of carbonyl compounds.
Reason : They are used in the blending of
perfumes and flavouring agents.
97. Assertion : Acetic acid in vapour state
shows a molecular mass of 120.
Reason : It undergoes intermolecular hydrogen
bonding.
104. Assertion : Carboxylic acids have higher
boiling points than alkanes.
Reason : Carboxylic acids are resonance hybrids.
98. Assertion : Nitration of benzoic acid gives
m-nitrobenzoic acid.
105. Assertion : o-Substituted benzoic acids are
generally stronger acids than benzoic acids.
Reason : Increased strength is due to ortho-effect.
CBSE Board Term-II Chemistry Class-12
130
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Arrange the following in the increasing order
of their boiling points.
CH3CHO, CH3COOH, CH3CH2OH
2. Write chemical equations for the following
reactions :
Benzoyl chloride is hydrogenated in presence of
Pd/BaSO4.
3. Write structures of compounds A and B in
each of the following reactions.
4. Write structures of compounds A and B in
each of the following reactions :
5.
Give reasons :
Chloroacetic acid is stronger than acetic acid.
6. Write the IUPAC name of the following
compound :
7. Aldehydes and ketones have lower boiling
points than corresponding alcohols. Why?
8.
Complete the following reactions :
9.
Write the IUPAC name of the following :
10. Write the equation involved in Etard
reaction.
Short Answer Type Questions (SA-I)
11. A compound ‘A’ of molecular formula
C 2 H 3 OCl undergoes a series of reactions as
shown below. Write the structure of A, B, C
and D in the following reactions :
(C2H3OCl)A
B
C
D
12. Write chemical equations for the following
reactions :
(i) Propanone is treated with dilute Ba(OH)2.
(ii) Acetophenone is treated with Zn(Hg)/Conc.
HCl
13. Give reasons :
(i) Electrophilic substitution in benzoic acid
takes place at meta-position.
(ii) Carboxylic acids do not give the characteristic
reactions of carbonyl group.
14. Describe how the following conversions can
be brought about :
(i) Ethylbenzene to benzoic acid
(ii) Bromobenzene to benzoic acid
15. Which acid of each pair shown here would
you expect to be stronger?
(i) F—CH2—COOH or Cl—CH2—COOH
(ii)
or CH3COOH
16. The reaction of carbonyl compound with
pure HCN is very slow and becomes fast in
presence of a base.
17. A compound having the molecular formula
C 3 H 6 O forms a crystalline white ppt. with
sodium bisulphite and reduces Fehling’s
solution. Suggest the structural formula and
IUPAC name of this compound. Name an isomer
for it from a group other than its own.
131
Aldehydes, Ketones and Carboxylic Acids
18. Account for the following :
(a) Aromatic carboxylic acids do not undergo
Friedel–Crafts reaction.
(b) pKa value of 4-nitrobenzoic acid is lower than
that of benzoic acid.
19. (i) What is the advantage of using DIBAL-H
as reducing agent?
(ii) Which of the following can be nitrated more
easily and why? Benzoic acid or phenol.
20. Account for the following :
(i) CH3CHO is more reactive than CH3COCH3
towards reaction with HCN.
(ii) There are two –NH2 groups in semicarbazide
(H2NNHCONH2).
However, only one is involved in the formation of
semicarbazone.
Short Answer Type Questions (SA-II)
21. Write the equation of the reactions of ethanal
with
(i) Fehling’s solution (ii) Phenylhydrazine
(iii) Hydroxylamine.
22. Illustrate the following name reactions
giving a chemical equations in each case :
(i) Clemmensen reaction
(ii) Cannizzaro reaction
23. (i) Write the structures of compounds A, B
and C in each of the following reactions :
(b) Etard reaction
(ii) Distinguish between CH3COOH and HCOOH.
26. Two moles of organic compound ‘A’ on
treatment with a strong base gives two compound
‘B’ and ‘C’. Compound ‘B’ on dehydrogenation
with Cu gives ‘A’ while acidification of ‘C’ yields
carboxylic acid ‘D’ with molecular formula of
CH2O2. Identify the compounds A, B, C and D
and write all chemical reactions involved.
(b)
27. (a) Write the chemical reaction involved in
Wolff-Kishner reduction.
(b) Arrange the following in the increasing order
of their reactivity towards nucleophilic addition
reaction.
C6H5COCH3, CH3 CHO, CH3COCH3
(ii) Do the following conversion in not more than
two steps :
Benzoic acid to benzaldehyde
(c) A and B are two functional isomers of
compound C3H6O. On heating with NaOH and
I2, isomer B forms yellow precipitate of iodoform
whereas isomer A does not form any precipitate.
Write the formulae of A and B.
(a)
24. During practical exams, lab assistant
provided two test tubes containing 5 mL benzoic
acid and 5 mL acetaldehyde to every student. A
student, Rahul found that test tubes given to
him were unlabelled. He informed the teacher
before performing any experiment with the
given chemicals.
How can the chemicals be distinguished for
correct labelling?
25. (i) Write the equations involved in the
following reactions :
(a) Stephen reaction
28. (a) Write the main product in the following
equations :
(i) CH3 C CH3 LiAlH4 ?
O
(ii)
(b) Write the product in the following reaction :
CH3
CH
CH
CH2CN
(i) DIBAL-H
(ii) H2O
29. Write the products formed when ethanal
reacts with the following reagents :
CBSE Board Term-II Chemistry Class-12
132
(i) CH3MgBr and then H3O+
(ii) Zn-Hg/conc. HCl
(iii) C6H5CHO in the presence of dilute NaOH
30. (a) Draw the structures of the following :
(i) p-Methylbenzaldehyde
(ii) 4-Methylpent-3-en-2-one
(b) Describe how the following conversions can
be brought about :
Cyclohexanol to cyclohexan 1-one
31. (A), (B) and (C) are three non-cyclic functional
isomers of a carbonyl compound with molecular
formula C 4 H 8 O. Isomers (A) and (C) give
positive Tollens’ test whereas isomer (B) does
not give Tollens’ test but gives positive iodoform
test. Isomers (A) and (B) on reduction with
Zn(Hg)/conc. HCl give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is
least reactive towards addition of HCN?
32. In an industry aldehydes are being prepared
by controlled oxidation of primary alcohol using
acidified K2Cr2O7 or aqueous or alkaline KMnO4
as oxidant. Mohan suggested the owner of
factory to use Collin’s reagent instead of acidic
potassium dichromate. The yield of factory
increased sharply.
Now answer the following questions :
(i) What is Collin’s reagent?
(ii) What are the advantages of using Collin’s
reagent over conventional oxidising agent?
33. Identify A and E in the following series of
reactions :
NaOOC
34. (i)How will you bring about the following
conversions?
(a) Ethanal to but-2-enal
(b) Propanone to propene
(ii) Write the IUPAC name of the compound :
35. Write the structures of the main products of
the following reactions :
(i)
+ C6H5COCl
(ii) H3C
C
C
H
anhydrous AlCl3
CS2
Hg2+, H2SO4
CH3
1. CrO2Cl2
(iii)
2. H2O
NO2
Long Answer Type Questions (LA)
36. Write the structures of A, B, C, D and E in
the following reactions :
C6H6
CH3COCl
Anhyd. AlCl3
A
Zn–Hg/conc.HCl
NaOI
D+E
B
(i) KMnO4 – (ii) H3O+
KOH, C
37. Identify A to E in the following reactions :
38. (a)Give a plausible explanation for each
one of the following :
(i) There are two –NH2 groups in semicarbazide.
However, only one such group is involved in the
formation of semicarbazones.
(ii) Cyclohexanone forms cyanohydrin in good
yield but 2,4,6-trimethylcyclo-hexanone does not.
(b) An organic compound with molecular
formula C9H10O forms 2,4-DNP derivative,
reduces Tollens’ reagent and undergoes
Cannizzaro reaction. On vigorous oxidation it
gives 1,2-benzene-dicarboxylic acid. Identify the
compound.
39. (a) Identify A, B and C in the following
sequence of reactions :
CH3CHO
(i) C2H5MgCl
(ii) H2O
A
B
C
133
Aldehydes, Ketones and Carboxylic Acids
(b) Predict the structures of the products formed
when benzaldehyde is treated with
(i) conc. NaOH
(ii) HNO3/H2SO4 (at 273– 383 K)
40. An organic compound (A) on treatment with
ethyl alcohol gives a carboxylic acid (B) and
compound (C). Hydrolysis of (C) under acidified
OBJECTIVE TYPE QUESTIONS
1.
(c) : Ketones are formed by oxidation of secondary
conditions gives (B) and (D). Oxidation of (D)
with KMnO4 also gives (B). (B) on heating with
Ca(OH) 2 gives (E) having molecular formula
C 3 H 6 O. (E) does not give Tollen’s test and
does not reduce Fehling’s solution but forms a
2, 4-dinitrophenylhydrazone. Identify (A), (B),
(C), (D) and (E).
6.
(b) :
CH3 CH
alcohols.
(X)
(i) O3
C
CH3 (ii) Zn/H O
2
O
CH3
CH3CHO + CH3 C
Positive Fehling's
and iodoform test
(Y)
2.
(b) : Formaldehyde forms primary alcohol while all
other aldehydes form secondary alcohols on reaction with
CH3
Iodoform test
(Z)
7. (a) : Since the compound gives positive 2, 4-DNP test
and negative Tollen’s test, it is a ketone.
Grignard’s reagent followed by hydrolysis.
3.
(c) : Aldehydes with no a-H atom undergo Cannizzaro
reaction on heating with conc. alkali solution. Hence, only
C6H5CHO will undergo Cannizzaro reaction.
4.
(c) :
CH3CH2CHO + CH2CHO
dil. NaOH
CH3
8.
(c) : Tollens’ reagent is used to detect aldehyde group.
9.
(c) : CH3 C O + H2NHN
OH
CH3
CH2 CH CHCHO
CH3
CH3
5.
(c) :
CH3 O
CH3 CH
CH3 CH
CH
CH3
COOH + CH3
CH3
CH3
NO2
C
NO2
NHN
NO2
CH3
10. (b) :
CH3
C CH2
NO2
[O]
CH
CH2 COOH
CH3
11. (a) : Acid chlorides are reduced to aldehydes on reaction
with BaSO 4 and Pd.The reaction is called Rosenmund
reduction.
H
2
R COCl Pd/BaSO

→ R CHO + HCl
4
+ CH3
C CH3
O
HCOOH + CH3COOH
12. (c) : Benzaldehyde will not give aldol condensation due
to absence of a-H atom.
CBSE Board Term-II Chemistry Class-12
134
13. (a) : Formic acid acts as a reducing agent it reduces
Fehling’s and Tollen’s reagent, etc.
14. (d) :
C6H5
CH
CH2
C
CH3
CH
C
CO
+ HCl
O
C
+ (C6H5)2Cd
CH3
15. (a) : FCH2CHO is most reactive towards nucleophilic
addition since presence of most electronegative F withdraws
electrons from carbon of carbonyl group making it more polar.
16. (a) :
(iii)
AlCl3
27. (b) :
O
CH
+
COCl
–H2O
C6H5
26. (c) : (i)
O
OH
C6H5CHO + CH3COCH3
COCl
is 2-methylcyclopentanecarboxylic
acid.
17. (a) : Aldehydes are more reactive than ketones.
18. (c) : Due to –I effect of Cl, chloroacetic acid is a stronger
acid than acetic acid. Due to stabilization of phenoxide ion by
resonance, phenol is a stronger acid than ethanol.
28. (b) :
is called Clemmensen
reduction.
29. (a) :
Nucleophile attacks at the positive C centre of carbonyl group
hence the addition is nucleophilic addition.
30. (a) :
19. (b) : Pentan-2-one will give positive iodoform test while
pentan-3-one will not give this test.
20. (b) :
21. (d) : 2,2-Dimethylpropanoic acid will not undergo HVZ
reaction due to absence of an a-H atom.
22. (b) :
23. (a) : 2CH CH + 3H2O
420°C
Mg or
Zinc vanadate
CH3COCH3 + CO2 + 2H2
24. (a)
25. (a) : Ketones are less reactive than aldehydes.
31. (c) : The electron withdrawing strength of halogen
groups is in the order of F > Cl > Br. Hence, the strength of
acids is
FCH2COOH > ClCH2COOH > BrCH2COOH.
32. (b) : Aldehydes give silver mirror test with Tollen’s
reagent while ketones form oximes with hydroxylamine.
Hence the compound is a ketone. Alcohol and ethers do not
give this test.
33. (a) : Pentan-2-one and pentan-3-one can be
differentiated by iodoform test.
34. (d) : HCHO will be most polar due to lowest electron
density on carbon of carbonyl group.
35. (a) : Aldehydes are more reactive than ketones towards
nucleophilic addition reactions. Aromatic aldehydes and
ketones are less reactive than corresponding aliphatic
aldehydes and ketones.
36. (d) : Both butanal and propanal does not give iodoform
test, hence cannot be distinguished from each other.
135
Aldehydes, Ketones and Carboxylic Acids
37. (a) : Aldehydes and ketones form insoluble crystalline
compounds with NaHSO3 which can be filtered. These on
distillation with saturated solution of Na2CO3 again give the
aldehydes and ketones.
49. (a) :
38. (d) : All the given aldehydes will give Cannizzaro
reaction.
39. (a) :
40. (b) :
CH3COCH3
Acetone
I2
CH3COONa
NaOH
H+
CH3COOH
NH3
CH3CONH2
50. (b) : Benzaldehyde can be prepared from benzene by
passing vapours of CO and hydrochloric acid in its solution
in presence of catalyst mixture of AlCl3/CuCl. The reaction
is known as Gattermann–Koch reaction.
CH3COONH4
Acetamide
41. (a) : Butan-1-ol has higher boiling point due to
intermolecular hydrogen bonding.
42. (b) :
51. (b) : Tollen’s reagent oxidises only –CHO to –COOH
group.
52. (c) :
43. (a) : In alcohols –OH group is attached to an electron
releasing group while in carboxylic acids –OH group is
attached to an electron withdrawing group making it more
acidic.
44. (b) : ICH 2 COOH < BrCH 2 COOH < ClCH 2 COOH <
FCH2COOH
53. (d) : CH3
45. (b) :
C
CH
40% H2SO4
1% HgSO4
OH
CH 3
C
CH 2
Prop-1-en-2-ol (A)
Tautomerization
is a resonance hybrid of structures
O
CH3
C
Acetone
CH3
46. (d)
Prop-1-en-2-ol (A) and acetone show tautomerism.
47. (b) :
54. (a) :
3-Methylcyclohexanecarbaldehyde.
48. (c) :
CH CH
[O]
Hg2+
H2SO4
CH3CHO
(X)
CH3 CH
CH3COCH3
OH
(Y)
CH3MgX
H2O
H
CH3
CH3 C
OMgX
CH3
55. (b) :
CBSE Board Term-II Chemistry Class-12
136
56. (c) :
63. (b) :
O
57. (b) : R
CH
CH
CHO + H2N
C
NHNH2
H+
R CH
CH CH
N
NH
O
C
NH2
58. (c) : CH3COOC2H5 will not give acetic acid on oxidation.
59. (b) : Aromatic ketones are less reactive than aliphatic
ketones which in turn are less reactive than aldehyde. Hence,
acetophenone does not react with NaHSO3.
60. (a) :
Since (B) on reaction with 2,4-DNP forms a derivative, it
implies that (B) has
group.
(B) gives –ve Tollens’ test, hence it is not an aldehyde, but it
is a ketone.
(B) gives –ve haloform test, thus it is not a methyl ketone.
(B) is formed from the oxidation of (A), thus (A) is a 2°alcohol,
and among the given options,
(A) is
and (B) is
C == O
aldehyde/
ketones
Zn-Hg/
HCl
— CH2 —
alkanes
64. (c) : Ortho-effect says that all the o-substituted benzoic
acids are stronger acids than benzoic acid, so the correct order
is
61. (b) :
COOH
COOH
>
COOH
CH3 >
OH >
(ortho-effect
pronounced due
(-R effect of to H-bonding in
—NO2 group) carboxylate ion)
NO2
COOH
(ortho-effect)
Thus, Order III given is incorrect as in this ortho-effect is not
considered.
/C
62. (d) : H3C
/H
(P)
4-Vinyltoluene
Order I and II are correct.
65. (c) : Carboxylic acids do not undergo Friedel Craft’s
reaction because carboxyl group is deactivating and gets
bonded to the catalyst in Friedel Craft’s reaction.
66. (a) : A undergoes Iodoform reaction hence contains a
methyl ketone. So the structure of A is
/C
(Q)
CH3
/H
(Butan-2-one)
A (butan-2-one) on reduction gives butan-2-ol (B).
137
Aldehydes, Ketones and Carboxylic Acids
CH3CH2CHO : Propionaldehyde
B on heating with H2SO4 gives an alkene named but-2-ene
(C). CH3CH == CHCH3
C forms an ozonide D which on hydrolysis in presence of
Zn dust to form acetaldehyde (E) CH3CHO (2 moles)
The reaction sequence is as follows :
: Ethylphenyl ketone
C6H5 — CH2— OH : Benzyl alcohol
Therefore, isopropyl alcohol will give positive iodoform test.
73. (b) : Iodoform reaction of acetone occurs in following
steps :
67. (a)
68. (c) : Phosphorus converts a little of the acid into acid
chloride which is more reactive than the parent carboxylic
acid. Thus, it is the acid chloride, not the acid itself, that
undergoes chlorination at the a-carbon.
69. (d) : Bromination occurs at a-positions.
a
CH3—CH2—COOH
Br2/P
–HBr
CH3CHBr — COOH
74. (c) : Given reagents indicate the presence of —COCH3
group in the starting compound A. Further, since the —COOH
group introduced in B due to iodoform reaction is absent in
the final product, B should be a b-keto acid. Hence, A should
have structure given in option (c).
O
Br2/P –HBr
CH3—CBr2—COOH
(A)
70. (a) : b-ketoacids are unstable acids. They readily undergo
decarboxylation through a cyclic transition state.
O
(i) NaOI
(ii) H+
COCH3
COOH
(B)
O O
75. (d) : Ph * OH
(–CO2)
O
I2
*
Ph CH3 NaOH
(E)
O
Heat
β-keto Acid
Ph
71. (b) : CH3—CH—CH2COOH
CH3
NaOH/CaO
D
CH3—CH—CH3 + Na2CO3
CH3
72. (a) : Iodoform test is given by the organic compounds
or
: Isopropyl alcohol
group.
–+
*
ONa + CHI3
(F)
(G)
76. (c) : Since compound A(C3H6O) undergoes iodoform test,
it must be CH3COCH3 (propanone). Further, the compound
‘B’ obtained from ‘A’ has three times more the number of
carbon atoms as in ‘A’ (propanone), ‘B’ must be phorone,
i.e., 2, 6-dimethyl-2, 5-heptadien‑4-one.
(CH3)2C
O + H3CCOCH3 + O
C(CH3)2
A, propanone (3 molecules)
having
O
Heat
HCl
(CH3)2C
CHCOCH
C(CH3)2
2,6-dimethyl-2,5-heptadien-4-one
77. (c) : Condensation reaction is the reverse of hydrolysis,
which splits a chemical entity into two parts through the
action of the polar water molecule.
CBSE Board Term-II Chemistry Class-12
138
78. (b) : CH3CHO + CH3COCH3
Cl
CH3CH(OH)CH2COCH3
2Cl
CH3CH
C
NaOH
C
C
Cl
CHCOCH3
Cl
H
Cl
–H2O
Cl
O
O
C
O–
trichloroacetate ion
79. (a)
+ Cl
80. (d)
81. (d) : Benzaldehyde(C 6H 5CHO) with no a-hydrogen
cannot undergo aldol condensation.
82. (a)
83. (b) : HSO3– is a bulky nucleophile, hence, cannot attack
on sterically hindered ketones.
84. (d) : Aromatic aldehydes and ketones are less reactive
than the corresponding aliphatic analogues towards
nucleophilic addition reactions due to the +R effect of
benzene ring. Further, aldehydes are more reactive than
ketones due to +I effect and steric effect of alkyl group.
Therefore, the ease of nucleophilic addition will follow the
order :
Cl
OH
C
CH2
Cl
2, 2, 2-trichloroethanol
90. (b) : Hydride transfer is the slowest step.
O
Ph
C
–
O
OH + Ph
C
slow
H step
H
91. (a) : C
O
Ph
C
OH + Ph
CH2O–
C bond is not formed in Cannizzaro reaction
while other reactions result in the formation of C
C bond.
92. (c) : Aromatic aldehydes and formaldehyde do not
contain a-hydrogen and thus undergo Cannizzaro reaction.
Formaldehyde is more reactive than aromatic aldehydes.
85. (a) : Formation of cyanohydrin from an aldehyde or
ketone occurs very slowly with pure HCN because it is feebly
ionised. This reaction is catalysed by a base. Base generates
CN– ion which is a stronger nuclephile and readily adds to
carbonyl compound.
OH + HCN
CN + H2O
d+
O–
d
O + •• CN
C
C
CN
H+
OH
C
CN
86. (b) : Electron withdrawing group (–NO2) increases the
reactivity towards nucleophilic addition reactions, whereas
electron donating group (–CH3) decreases the reactivity
towards nucleophilic addition reactions.
87. (a) : It is an example of cross Cannizzaro reaction where
aromatic aldehyde gets reduced to alcohol and aliphatic
aldehyde gets oxidised to its sodium salt (both aldehydes
must not contain any a-hydrogen).
CH2OH
CHO
+ NaOH + HCHO
∆
93. (c) : The anion left after the removal of a-hydrogen is
stabilized by resonance effect.
94. (c) : Highly branched carboxylic acids are less acidic
than unbranched acids. The +I effect of alkyl groups in
branched acid increases the magnitude of negative charge.
Thus, –COOH group is shielded from solvent molecules
and cannot be stabilized by solvation as effectively as in
unbranched carboxylic acids.
95. (c) : In p-chlorobenzoic acid, both +R and –I effect
operate together but in m-chlorobenzoic acid only –I effect
operates. Therefore, m-chlorobenzoic acid is a stronger acid
than p-chlorobenzoic acid.
96. (c) : Addition of Grignard reagents to dry ice followed
by hydrolysis gives carboxylic acid not ketone.
97. (a)
+ HCOONa
88. (c)
89. (a) : The Cannizzaro product of given reaction yields 2,
2, 2-trichloroethanol.
98. (c) : Carboxyl group only marginally decreases the
electron density at m-position relative to o- and p-positions.
99. (b) : Aldehydes have higher molecular weight than
parent alkanes as well as polarity in aldehydes shows higher
boiling point than parent alkanes. Aldehydes do not have
139
Aldehydes, Ketones and Carboxylic Acids
any hydrogen atom attached directly to the oxygen so they
cannot form hydrogen bond with each other.
100. (c) : Chlorine atom has –I effect which increases
the ionisation of chloroacetic acid and stabilizes the
chloroacetate ion by dispersal of negative charge. In
acetic acid, methyl group due to +I effect destabilizes the
acetate ion by intensification of negative charge. Hence,
chloroacetic acid dissociates to a greater extent than acetic
acid.
101. (d) : Aldehydes having a methyl or methylene group
in the a-position or more correctly having atleast one
hydrogen atom in the a-position undergo dimerisation in
presence of a base at low temperature to form b-hydroxy
aldehydes called aldols. As benzaldehyde does not have any
a-hydrogen hence it does not undergoes aldol condensation.
102. (b) : Due to overall electron-donating effect of
the phenyl group, benzoate ion is less stable than formate
ion.
103. (b) : Carbonyl compounds form solid additive products
with NaHSO3 which are separated out. The solid bisulphites of
carbonyl compounds on hydrolysis with dilute acid regenerate
original carbonyl compounds and thus, this property is used
for the purification of carbonyl compounds as well as for
their separation.
104. (b) : Boiling points of carboxylic acids are higher due
to their tendency to associate and form dimers to a greater
extent by hydrogen bonding.
105. (a) : o-Substituted benzoic acids are generally
stronger acids than benzoic acid. This is regardless of the
nature (+I or –I) of the substituent. This is called orthoeffect and is probably due to a combination of steric and
electronic factors.
SUBJECTIVE TYPE QUESTIONS
1.
Increasing order of boiling point :
5. Chloroacetic acid has lower pKa value than acetic acid;
‘Cl’ in chloroacetic acid shows –I effect, it creates less electron
density on oxygen of carboxylic acid. Thus, release of proton
becomes easier. In case of acetic acid, the state of affair is
just opposite. Hence, chloroacetic acid is stronger than acetic
acid.
OH
6.
CHO
2-Hydroxybenzaldehyde
7. The boiling points of aldehydes and ketones are lower
than that of corresponding alcohols and acids due to absence
of intermolecular H–bonding in aldehydes and ketones.
8.
9. Hex-2-en-4-yn-oic acid
10. Etard reaction :
CH3
CS2
+ CrO2Cl2 →
CH(OCrOHCl2)2
Chromium
complex
Toluene
→
CH3 CHO < C2H5OH < CH3 COOH
4.
H 3O +
2.
CHO
Benzaldehyde
3.
11.
OH
O
CH3 CH CH2 C H
CH3 CH CH C H
(D)
O
(C)
CBSE Board Term-II Chemistry Class-12
140
16. With pure HCN reaction occurs very slowly because it
is a weak nucleophile. With base it produces CN– ion which
is a strong nucleophile and readily adds to the carbonyl
compound.
12. (i)
17. Since the compound forms crystalline white precipitate
with sodium bisulphite, it contains a carbonyl group. The
compound reduces Fehling’s solution so, the carbonyl group
is an aldehyde.
Structure
: CH3 CH2 CHO
IUPAC name : Propanal
O
(ii)
13. (i) Electrophilic substitution in benzoic acid takes place
at meta-position. Due to resonance in benzoic acid, there is
high electron density at meta-position. Therefore, electrophilic
substitution in benzoic acid takes place at meta-position.
O
C
OH
–O
+
C
OH
–O
C
OH
+
–O
C
O–
OH
C
OH
O
C
OH
+
+
COOH
COOH
+ (NO2+)
e.g.,
+ H2O
NO2
(ii) The carbonyl group in —COOH is inert and does not
show nucleophilic addition reaction like carbonyl compound.
It is due to resonance stabilisation of carboxylate ion :
R C O
R C O–
O–
CH2CH3
O
COOH
KMnO4/OH–
Vigorous oxidation
14. (i)
Benzoic acid
Ethylbenzene
MgBr
Br
(ii)
Mg/dry ether
Grignard reaction
Bromobenzene
Isomer
: CH3 C CH3 (Propanone)
18. (a) Due to presence of electron withdrawing group
(— COOH) in aromatic carboxylic acids, they do not undergo
Friedel-Crafts reaction.
(b) Due to presence of strong electron withdrawing group
(—NO2), 4-nitrobenzoic acid is more acidic than benzoic acid
and therefore, pKa value is lower.
19 (i) DIBAL–H reduces alkynes to alkenes but does not
reduce ethylenic double bonds and hence this reagent can
be used to reduce unsaturated nitriles to the corresponding
unsaturated aldehydes.
(ii) Phenol gets easily nitrated than benzoic acid. Because
carboxyl group is ring deactivating group whereas hydroxyl
group is ring activating group.
20. (i) It is a nucleophilic addition reaction, in which
CN – acts as a nucleophile. CH3CHO undergoes nucleophilic
addition reactions faster than CH3COCH3 as in CH3COCH3
there are two electron releasing methyl groups attached to
the carbonyl carbon that hinders the approach of nucleophile
to carbonyl carbon and reduce the electrophilicity of the
carbonyl group while in CH3CHO, there is only one methyl
group attached to carbonyl carbon.
(ii) Semicarbazide has the following resonance structures
arising due to the electron withdrawing nature of the O atom.
21. (i) CH3CHO + 2Cu2+ + 5OH– → CH3COO– + Cu2O
(Red ppt.)
+ 3H2O
CH
(ii)
3
C O + H NN H
H
2
Phenyl hydrozine
Phenylmagnesium
bromide
COOH
(i) Dry ice, (ii) H3O+
Benzoic acid
15. (i) F—CH2COOH > Cl—CH2COOH
OH
(ii) CH3COOH is stronger than
(iii) CH3
C O + NH2OH
H
Hydroxylamine
CH3
C N NH
H
CH3
C N OH
H Ethanal oxime
22. (a)(i) Clemmensen reduction : The carbonyl group of
aldehydes and ketones is reduced to CH2 group on treatment
with zinc amalgam and concentrated hydrochloric acid.
CH3
Zn – Hg CH3
C O HCl
CH2 + H2O
CH3
CH3
Propanone
Propane
141
Aldehydes, Ketones and Carboxylic Acids
(ii) Cannizzaro reaction : Aldehydes which do not contain
a-H atom undergo disproportionation when heated with
concentrated (50%) NaOH.
HCHO + HCHO
50% NaOH
Methanal
HCOONa + CH3OH
HCOOH + 2[Ag(NH3)2]+ + 2OH–
Formic acid
Warm
2Ag + CO2 + 2NH3 + 2NH4OH
Silver mirror
Acetic acid does not give this test.
Sodium formate Methanol
26. Since the molecular formula of D is CH2O2, thus, D is
HCOOH (formic acid). D is obtained by the acidification of C,
so, C is sodium formate (HCOONa).
Thus, A must be formaldehyde (as it undergoes Cannizzaro
reaction with a strong base).
23. (i) (a)
(b)
Thus, A = Formaldehyde (HCHO)
B = Methanol (CH3OH)
C = Sodium formate (HCOONa)
D = Formic acid (HCOOH)
(ii)
CHO
27. (a) Wolff-Kishner reduction : The carbonyl group of
aldehydes and ketones is reduced to CH2 group on treatment
with hydrazine followed by heating with potassium hydroxide
in a high boiling solvent such as ethylene glycol.
Benzaldehyde
24. Chemicals can be distinguished by sodium bicarbonate
test and iodoform test.
Benzoic acid will give brisk effervescence due to evolution of
carbon dioxide gas with sodium bicarbonate solution while
acetaldehyde does not.
Acetaldehyde will give yellow precipitate of iodoform with
iodine and sodium hydroxide solution while benzoic acid
does not.
25. (i) (a) Stephen reduction :
R—CN + SnCl2 + HCl → R —CH NH
H O+
3
→
R —CHO
(b) Etard reaction :
CH3
Toluene
CH(OCrOHCl2)2
CS2
+ CrO2Cl2 →
(b) Increasing order of reactivity towards nucleophilic
addition reaction :
C6H5COCH3 < CH3COCH3 < CH3CHO
(c) Formula of compounds A and B is C3H6O. B forms yellow
precipitate of iodoform. Hence, B must contain —COCH3
group. Therefore, compound ‘B’ must be
.
A does not give iodoform test and it is functional isomer of B
thus, it may be CH3CH2CHO.
28. (a) (i) CH3 C CH3
CHO
Chromium
complex
→
H 3O +
CHO
Benzaldehyde
(ii) Add Tollens’ reagent to formic acid and warm. Silver
mirror is formed.
LiAlH4
O
CH3 CH CH3
CHO
OH
HNO3/H2SO4
(ii)
273 – 283 K
Benzaldehyde
(b) CH3 — CH
CH3 — CH
NO2
m-Nitrobenzaldehyde
CH — CH2CN
(i) DIBAL-H
CH — CH2CHO
(ii) H2O
CBSE Board Term-II Chemistry Class-12
142
CH3
33.
29. (i)
(ii) CH3CHO
Zn-Hg
Conc. HCl
CH3
CH(OCOCH3)2
CrO3 + (CH3CO)2O
273 – 283 K
(A)
KMnO4, KOH
CH3
H3O+
COOK
(iii)
CHO
2
(D)
(B)
H3O+
conc. NaOH
COOH
+
30. (a) (i)
(E)
(C)
O
34. (a) (i) 2CH3 C H
Ethanal
(ii) 4-Methylpent-3-en-2-one :
CH3
CH3
O
OH–
Aldol
condensation
OH
CH3 CH CH2 CHO
H+
CH3
CH3 CH CH CHO
But-2-enal
OH
O
(b)
(ii) CH3
31. (a) As (A) and (C) give positive Tollens’ test thus these
two should be aldehyde while (B) should be a ketone (does
not give Tollens’ test) with
C CH3 group (as it gives
positve iodoform test).
35. (i)
Three isomers are
O
(A)
CH3
Propanone
(Acetone)
C
CH2
+ C6H5COCl
Zn(Hg)/conc. HCl
CH3
CH3
CH3
CH
Propene
Anyhd. AlCl3
CS2
CH3
CH3
CH
CHO
(C)
CH3CH2CH2CH3
(A)
CH2
CH
Propan-2-ol
Conc. H2SO4
443 K
(ii) CH3 C CH
(iii) O2N
Hg2+, H2SO4
CH3
CH3 C
CH3
1. CrO2Cl2
2. H3O+
O2N
(D)
Zn(Hg)/conc. HCl
O
Propanone
O
C
CH3
O
CH3
(B)
LiAlH4
C
(B)
CH3CH2CH2CHO
C
O
CH3CH2CH2CHO, CH3
CH3
COONa
CH2OH
CHO
p-Nitrobenzaldehyde
CH3CH2CH2CH3
(D)
(b) out of (A), (B) and (C) isomers, (B) is least reactive
towards addition of HCN.
32. (i) Collin’s reagent is a mixture of pyridine (C5H5N) and
CrO3 in dichloromethane (CH2Cl2).
(ii) Collin’s reagent is a mild oxidant. It oxidises 1°-alcohols
to aldehydes and 2°-alcohols are oxidised to ketones.
In case of using acidic K2Cr2O7 as oxidant, the aldehydes
and ketones formed by the oxidation of alcohols undergo
oxidation to give carboxylic acids.
36.
A
NaOI
(E)
COONa
+ CHI
3
(D)
CH2
143
Aldehydes, Ketones and Carboxylic Acids
37.
(b) (i)
38. (a) (i) Semicarbazide has the following resonance
structures arising due to the electron withdrawing nature of
the O atom.
1
O
2
H2N C NH
1
O
H2N C
2
3
+
H2N
NH2
O–
O
O
3
NH NH2
CHO
C NH NH2
H2N C
–
H2N C
CHO
HNO3/H2SO4
(ii)
NH
NH2
+
NH NH2
273 – 283 K
NO2
m-Nitrobenzaldehyde
Benzaldehyde
40.
O
H2N C
NH NH2
Lone pairs of N-1 and N-2 are involved in conjugation with
C O group while that of N-3 is not involved in resonance
thus, it is involved in the formation of semicarbazone.
(ii) Formation of cyanohydrin involves the nucleophilic attack
of cyanide ions (CN–) at the carbonyl carbon. In cyclohexanone,
reaction proceeds but in 2,4,6-trimethylcyclohexanone, the
methyl groups cause steric hindrance and yields are poor.
O
O
CH3
H3C
Cyclohexanone
CH3
2,4,6-Trimethylcyclohexanone
(b) The compound forms 2,4-DNP derivative. It shows that
it is a carbonyl compound. Further it reduces Tollens’ reagent
which shows that it contains aldehydic group. It undergoes
Cannizzaro reaction indicating that aldehyde group is
without any a-hydrogen. On vigorous oxidation, it gives
1,2-benzenedicarboxylic acid which shows that there are two
carbon residues on benzene ring. Since the molecular formula
is C9H10O, it fits into the structure, 2-ethylbenzaldehyde.
COOH
CHO
CH3COOC2H5 + H2O
2-Ethylbenzaldehyde
39. (a)
COOH
1,2-Benzenedicarboxylic acid
CH3COOH + CH3CH2OH
B
CH3CH2OH
KMnO4
D
D
CH3COOH
B
Ca(OH)2 Dry distillation
CH3COCH3 + CaCO3
E
E does not give Tollens’ test and does not reduce Fehling’s
solution as it is ketone.
NO2
CH3
CH3
C O+H2N NH
CH3
CH3
NO2
NO2
C N NH
NO2
2,4-dinitrophenyl hydrazone
Oxidation
CH2 CH3
H+
A=
CH3CO
CH3CO
O,
B = CH3COOH,
C = CH3COOC2H5,
D = CH3CH2OH,
E = CH3COCH3

CHAPTER
7
Amines
Recap Notes
AMINES
In secondary and tertiary amines, when two
or more groups are the same, the prefix di
or tri is appended before the name of alkyl
group.
Amines : These are alkyl or aryl derivatives
of ammonia and are obtained by replacing
one, two or three hydrogen atoms by alkyl/
aryl groups.
Structure : Nitrogen orbitals in amines are
sp3-hybridised and the geometry of amines is
pyramidal. Due to the presence of unshared
pair of electrons, the angle C N E, (where
E is C or H) is less than 109.5°.
Nomenclature : In common system, an
aliphatic amine is named by prefixing alkyl
group to amine, i.e., alkylamine. In IUPAC
system, amines are named as alkanamines.
Classification :
Amines
Aliphatic
Aromatic
Primary (1°) Amines
–NH2 group attached to
Secondary (2°) Amines
NH group attached to
one alkyl group (R—NH2)
two alkyl groups (R2NH)
Simple
Both alkyl groups
are same.
(R2NH)
Mixed
Both alkyl groups
are different.
(R′NHR)
Tertiary (3°) Amines
N group attached to three
alkyl groups (R3N)
Mixed
Simple
Either all three are different or
All three alkyl
groups are same. two are same, one is different
(R3N)
(R′RR′′N or RRR′N)
Preparation :
:
:
:
:
:
:
145
Amines
Limitations of Gabriel phthalimide
synthesis :
– It is used for the preparation of only
1°° amines. tert-Butylamine is a 1°° amine,
but cannot be prepared by this method. In
this case, elimination takes place.
– Aromatic amines cannot be prepared by
this method because aryl halides do not
undergo nucleophilic substitution reaction
with potassium phthalimide under mild
conditions.
X
Physical properties :
X
Lower amines are gases and liquids but
higher amines are solids.
X
Primary and secondary amines have
higher boiling points than other organic
compounds due to hydrogen bonding.
X
Primary and secondary amines are
soluble in water due to hydrogen bonding
between NH2 and H2O molecules.
X
Chemical properties :
X Basic character of amines :
– Amines are basic in nature due to the
presence of lone pair of electrons on
nitrogen atom.
– Aliphatic amines are stronger bases than
ammonia due to +I effect of alkyl groups
present in amines.
– Aromatic amines are weaker bases than
ammonia due to –I effect of aryl group.
– Beside inductive effect, there are other
effects like steric effect, solvation effect,
resonance effect which affect the basic
strength of amines.
– In gaseous phase, the order of basicity
of amines is 3°° amine > 2°° amine > 1°°
amine > NH3.
– In aqueous phase, despite of inductive
effect, solvation effect and steric hindrance
also play an important role. Thus, the
order of basicity of amines is
(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
and (CH3)2NH > CH3NH2 > (CH3)3N > NH3
Chemical reactions :
(Unpleasent smell)
X
Identification of primary, secondary and tertiary amines :
Test
Primary amine
Secondary amine
Tertiary amine
1.
Reaction with
nitrous acid.
Gives alcohol with
effervescence of N2
gas.
Gives oily
nitrosoamine which
gives Liebermann’s
nitrosoamine test.
Forms nitrite in cold
which is soluble in
water and on heating
gives nitrosoamine.
2.
Reaction with
benzene sulphonyl
chloride
(Hinsberg’s reagent)
Gives N-alkylbenzene
sulphonamide which
is soluble in alkali.
Gives N,
N-dialkylbenzene
sulphonamide which
is insoluble in alkali.
No reaction
3.
Carbylamine test :
Reaction with
chloroform and
alcoholic KOH
Forms carbylamine
or isocyanide (RNC)
with characteristic
unpleasant odour.
No reaction
No reaction
CBSE Board Term-II Chemistry Class-12
146
X
Electrophilic substitution reactions
of arylamines : Aniline undergoes
electrophilic
substitution
reactions.
NH2 group is ortho- and para-directing
and a powerful activating group.
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1.
Identify X, Y and Z in the given reaction :
LiAlH 4
Br2
NaCN
CH 2== CH 2 CCl

→ X → Y 
→Z
4
X
(a)
(b)
(c)
(d)
Y
Z
CH2Br–CH2Br CH3CH2CH2CN CH3CH2CH2CH2NH2
CH2Br–CH2Br CH3CH2CN
CH3CH2Br
CH3CH2CN
CH3CH2CH2NH2
CH3CH2CH2NH2
CH2Br–CH2Br NCCH2CH2CN H2NCH2CH2CH2CH2NH2
2. Benzoic acid is treated with SOCl2 and the
product (X) formed is reacted with ammonia to
give (Y). (Y) on reaction with Br2 and KOH gives
(Z). (Z) in the reaction is
(a) aniline
(b) chlorobenzene
(c) benzamide
(d) benzoyl chloride.
3. Amongst the given set of reactants, the most
appropriate for preparing 2° amine is
(a) 2° R—Br + NH3
(b) 2° R—Br + NaCN followed by H2/Pt
(c) 1° R—NH2 + RCHO followed by H2/Pt
(d) 1° R—Br (2 moles) + potassium phthalimide
followed by H3O+/heat.
4.
(a)
(b)
(c)
(d)
Which of the following can exist as zwitter ion?
p-Aminoacetophenone
Sulphanilic acid
p-Nitroaminobenzene
p-Methoxyphenol
5. Which of the following amines will give
carbylamine reaction?
(a) (C2H5)3N
(b) (C2H5)2NH
(c) C2H5NH2
(d) C3H7NHC2H5
6. Among the compounds: C3H7NH2, CH3NH2,
C2H5NH2 and C6H5NH2. Which is the least basic
compound?
(a) CH3NH2
(b) C2H5NH2
(c) C3H7NH2
(d) C6H5NH2
7. The correct order of boiling points of the
following isomeric amines is
C4H9NH2, (C2H5)2NH, C2H5N(CH3)2
(a) C2H5N(CH3)2 > (C2H5)2NH > C4H9NH2
(b) (C2H5)2NH > C2H5N(CH3)2 > C4H9NH2
(c) C4H9NH2 > (C2H5)2NH > C2H5N(CH3)2
(d) (C2H5)2NH > C4H9NH2 > C2H5N(CH3)2
8. Tertiary amines have lowest boiling points
amongst isomeric amines because
(a) they have highest molecular mass
(b) they do not form hydrogen bonds
(c) they are more polar in nature
(d) they are most basic in nature.
9.
In the following reaction,
.
The organic product X has the structure
(a)
(b)
(c)
(d)
10. Amides may be converted into amines by a
reaction named after
(a) Hoffmann
(b) Claisen
(c) Perkin
(d) Kekule.
11. The reaction of benzenesulphonyl chloride
with ethylamine yields
(a) N-ethylbenzenesulphonamide, insoluble in
alkali
(b) N, N-diethylbenzenesulphonamide, soluble
in alkali
(c) N, N-diethylbenzenesulphonamide, insoluble
in alkali
(d) N-ethylbenzenesulphonamide, soluble in alkali.
12. Basic strength of different alkyl amines
depends upon
(a) +I effect
(b) steric effect
(c) solvation effect
(d) all of these.
13. Which of the following amides will give
ethylamine on reaction with sodium hypobromite?
(a) Butanamide
(b) Propanamide
(c) Acetamide
(d) Benzamide
CBSE Board Term-II Chemistry Class-12
148
14. o-Chloroaniline is treated with a mixture
of NaNO2 and HCl and the product is reacted
with cuprous bromide. The final product in the
reaction will be
(a)
(b)
(c)
(d)
15. When p-toluidine reacts with chloroform
and alcoholic KOH, then the product is
(a)
(b)
(c)
(d)
16. Acetylation of a secondary amine in alkaline
medium yields
(a) N, N-dialkyl acetamide
(b) N, N-dialkyl amine
(c) N, N-dialkyl amide
(d) acetyl dialkyl amine.
17. Amino group is o, p-directing for electrophilic
substitution reaction. But, on nitration the
major product is m-nitroaniline because
(a) aniline gets protonated with strong acids to
give anilinium ion which is m-directing
(b) nitration requires nitric acid which oxidises
–NH2 to –NO2 group
(c) electrophile NO+2 is a m-directing group
(d) benzene ring exerts + I effect and deactivates
the ring.
18. Which of the following is amphoteric in nature?
(a) CH3NH2
(b) CH3NHCH3
(c) CH3CONH2
(d)
19. The shape of (CH3)3N is pyramidal because
(a) nitrogen forms three sp3 hybridised sigma
bonds with carbon atoms of methyl groups
and there is one non-bonding electron pair
(b) nitrogen forms three sp2 hybridised sigma
bonds with carbon atoms of methyl groups
and fourth orbital forms pi bond
(c) nitrogen has five valencies which are arranged
in pyramidal shape.
(d) the unpaired electron present on nitrogen is
delocalised.
20. Which of the following compounds will show
Hoffmann bromamide degradation reaction?
(a) C6H5NO2
(b) C6H5CH2NH2
(c) C6H5CONH2
(d) C6H5NHCH3
21. Which of the following statements is not
true?
(a) In aqueous solution (CH3)2NH is a stronger
base than (CH3)3N.
(b) Secondary amines show carbylamine
reaction.
(c) Nitrogen gas is evolved when ethylamine is
treated with nitrous acid.
(d) Secondary amines can show metamerism.
22. Reduction of aromatic nitro compounds
using Sn and HCl gives
(a) aromatic primary amines
(b) aromatic secondary amines
(c) aromatic tertiary amines
(d) aromatic amides.
23. Arrange the following compounds in
increasing order of basicity : CH3NH2, (CH3)2NH,
NH3, C6H5NH2
(a) C6H5NH2 < NH3 < (CH3)2NH < CH3NH2
(b) CH3NH2 < (CH3)2NH < NH3 < C6H5NH2
(c) C6H5NH2 < NH3 < CH3NH2 < (CH3)2NH
(d) (CH3)2NH < CH3NH2 < NH3 < C6H5NH2
24. Primary amines react with benzoyl chloride
to give
(a) benzamides
(b) ethanamides
(c) imides
(d) imines.
Ni / H 2
NaCN
Acetic
→ Y anhydride
→ Z
25. CH3CH 2Cl → X 
Z in the above reaction is
(a) CH3CH2CH2NHCOCH3
(b) CH3CH2CH2NH2
(c) CH3CH2CH2CONHCH3
(d) CH3CH2CH2CONHCOCH3
26. Anilinium hydrogensulphate on heating
with sulphuric acid at 453-473 K produces
(a) sulphanilic acid
(b) benzenesulphonic acid
(c) aniline
(d) anthranilic acid.
27. A compound (X) with molecular formula
C3H9N reacts with C6H5SO2Cl to give a solid
which is insoluble in alkali. (X) is
(a) CH3CH2CH2NH2
(b)
(c) CH3 – NH – CH2CH3 (d)
149
Amines
28. Electrophilic substitution of aniline with
bromine- water at room temperature gives
(a) 2-bromoaniline
(b) 3-bromoaniline
(c) 2, 4, 6-tribromoaniline
(d) 3, 5, 6-tribromoaniline.
29. The decreasing order of boiling points
of ethyldimethylamine, n-butylamine and
diethylamine is
n-Butylamine > Diethylamine > Ethyldimethylamine.
This trend of boiling point can be explained as
(a) boiling point increases with increase in
molecular mass
(b) tertiary amines have highest boiling point
due to highest basicity
(c) intermolecular hydrogen bonding is maximum
in primary amines and absent in tertiary
amines
(d) intramolecular hydrogen bonding is present
in tertiary amines.
30. Which of the following is used as Hinsberg’s
reagent?
(a) C6H5SO2Cl
(b) C6H5SO3H
(c) C6H5NHCH3
(d) C6H5COCH3
31. Canonical structures of anilinium ion
obtained by accepting a proton are given below.
Choose the correct statements.
(a) Anilinium ion has two stable canonical
structures I and III.
(b) II is not an acceptable structure because
carbonium ion is less stable.
(c) Only I and III are acceptable aromatic
canonical structures since II is non-aromatic.
(d) Anilinium ion has three stable canonical
structures I, II and III.
32. Which of the following gas evolved when
methylamine reacts with nitrous acid?
(b) Cl2
(c) N2
(d) H2
(a) CH4
33. What is obtained
reacts with aniline in
hydroxide?
(a) Benzoic acid
(c) Acetanilide
when benzoyl chloride
the presence of sodium
(b) Benzanilide
(d) Azobenzene
34. Aniline can be converted into benzylamine
by which of the following processes in sequence?
(a)
(b)
(c)
(d)
NaNO2 + HCl, CuCN, H2/Ni
Br2/CCl4, KCN, LiAlH4
HNO2, K2Cr2O7/H+, Sn + HCl
CH3OH, KMnO4, OH–, H3+O
35. For a nitration of aniline, which of the
following steps is followed?
(a) Direct nitration using nitrating mixture
(conc. HNO 3 + conc. H 2 SO 4 ) followed by
oxidation.
(b) Using fuming HNO3 carrying out reaction
at 273 K followed by hydrolysis.
(c) Using NaNO2 and HCl followed by reaction
with conc. HNO3 followed by hydrolysis.
(d) Acetylation followed by nitration and
hydrolysis.
36. What is the end product in the following
sequence of reactions?
(a) Aniline
(b) Phenol
(c) Benzene
(d) Benzenediazonium chloride
37. Which of the following has highest pK b
value?
(a) (CH3)3CNH2
(b) NH3
(c) (CH3)2NH
(d) CH3NH2
38. Which of the following reactions is not
correctly matched?
(a) Reaction used to convert amide into primary
amine with one carbon atom less – Hoffmann
bromamide reaction
(b) Reaction used to convert primary amines
into isocyanides – Carbylamine reaction
(c) Reaction used to distinguish primary,
secondary and tertiary amines – Hinsberg’s
reaction
(d) Preparation of primary amines using
phthalimide – Victor Meyer’s synthesis
39. Which of the following compounds cannot be
identified by carbylamine test?
(a) CH3CH2NH2
(b) (CH3)2CHNH2
(c) C6H5NH2
(d) C6H5NHC6H5
40. When excess of ethyl iodide is treated with
ammonia, the product is
(a) ethylamine
(b) diethylamine
(c) triethylamine
(d) tetraethylammonium iodide.
CBSE Board Term-II Chemistry Class-12
150
41. The amines are basic in nature, hence they
form salts with hydrochloric acid. Which of the
following will be insoluble in dil. HCl?
(b) (C6H5)3N
(a) C6H5NH2
(c) C2H5NH2
(d) CH3NHCH3
42. The most basic amine among the following is
(a)
(b)
(c)
(a)
(b)
(c)
(d)
49. A basicity order is given below :
(d)
PCl
44.
(a)
(b)
(c)
(d)
NO2
45. Which of the following amines does not react
with Hinsberg reagent?
(a) CH3CH2—NH2
(b) CH3—NH—CH3
(c) (CH3CH2)3N
(d) All of these
46. The reagent required to convert
NH
(c)
CH3
is
(a) KOH/Br2, LiAlH4 (b) KOH/Br2, CH3COCl
(c) HNO2, (CH3CO)2O
(d) KOH/Br2, CH3OH/Na
47. Primary and secondary amines react with
acid chloride or acid anhydride to form
(a) tertiaryamonium salts
(b) substituted amides
(c) diazonium salts
(d) nitro compounds
48. Read the given map carefully :
LiAlH4
R
C
N
H2/Na
A
>
CH3
(3)
OCH3
(2)
(1)
KCN
AgCN
[H]
(b)
H2/Ni
[H]
(d)
H2/Ni
KCN
H3O+
AgCN
H3O+
51. In chemistry class various reactions of
nitrogen containing compounds were taught, like
preparation of amines and chemical reactions of
amines.
Few reactions are shown here :
CH3CONH2
NaOH + Br2
A
CHX3/KOH
C
H2/Ni
B
[H]
A and C can be differentiated by
(a) reaction with C6H5SO2Cl
(b) treatment with diethyl oxalate
(c) both (a) and (b)
(d) A and C cannot be differenced as both are same.
52. The correct order of basicity of the following
amine in aqueous solution is
B
Na(Hg)/C2H5OH
Partial hydolysis
(4)
>
NH2
50. Science teacher Sunaina ask class – 12 students
to write a sequence of two reactions for converting
n-propylbromide to methyl propyl amine.
Four students written 4 different paths shown in
all options, then which one is correct ?
(a)
O
NH2
NO2
(5)
(b) triethylamine
(d) propylamine.
C
>
NH2
Which of the following cannot be predicted by the
given order?
(a) +R effect is more powerful than +I effect.
(b) At m-position only –I effect is applicable not
R effect
(c) Electron donating groups increase the
basicity of aniline.
(d) In (4), –NO2 is showing –R effect.
KCN
Secondary amines can be prepared by
reduction of nitro compounds
reduction of amides
reduction of isonitriles
reduction of nitriles.
O
NH2
>
5 → Y 
2 → X 
Ethylamine 
→Z
(a) propanenitrile
(c) diethylamine
NH2
NH2
43. The end product Z of the reaction is
HNO
A and B are chain isomers
B and D are position isomers
B and C are functional isomers
D and E are members of same homologous
series.
LiAlH4/H2O
C
Br2/KOH
D
E
The correct statement about these products are
(a)
(b)
(c)
(d)
NH3
NH3
NH3
NH3
<
<
<
<
CH3NH2 < (CH3)2NH < (CH3)3N
(CH3)2NH < CH3NH2 < (CH3)3N
(CH3)3N < CH3NH2 < (CH3)2NH
CH3NH2 < (CH3)3N < (CH3)2N
151
Amines
Case Based MCQs
Case I : Read the passage given below and
answer the following questions from 53 to 56.
RCONH2 is converted into RNH2 by means of
Hoffmann bromamide degradation. During the
reaction amide is treated with Br2 and alkali to get
amine. This reaction is used to descend the series
in which carbon atom is removed as carbonate ion
(CO32–). Hoffmann bromide degradation reaction
can be written as :
O
O
OH–
Cl
Cl
Br2
NH2
NH – Br
(ii)
(i)
–
O
OH
C
Cl
OH
O
H
N
(iii)
(iv)
H2O
N
O
Cl
Cl
(v)
–CO2
N – Br
••
NH2
Cl
(vi)
53. Hoffmann bromamide degradation is used
for the preparation of
(a) primary amines
(b) secondary amines
(c) tertiary amines
(d) secondary aromatic amines.
54. Which is the rate determining step in
Hoffmann bromamide degradation?
(a) Formation of (i)
(b) Formation of (ii)
(c) Formation of (iii)
(d) Formation of (iv)
55. Which of the following is used for the
conversion of (i) to (ii)?
(a) KBr
(b) KBr + CH3ONa
(c) KBr + KOH
(d) Br2 + KOH
56. What are the constituent amines formed when
the mixture of (i) and (ii) undergoes Hoffmann
bromamide degradation?
15
CONH2
D
(a)
(ii)
(i)
15
NH2,
D
CONH2
NH2
D
(b)
NH2,
D
(c)
(d)
15
NH2,
15
NH2
15
NH2
15
NHD,
Case II : Read the passage given below and
answer the following questions from 57 to 60.
The amines are basic in nature due to the presence
of a lone pair of electron on N-atom of the –NH2
group, which it can donate to electron deficient
compounds. Aliphatic amines are stronger bases
than NH3 because of the +I effect of the alkyl
groups. Greater the number of alkyl groups
attached to N-atom, higher is the electron density
on it and more will be the basicity. Thus, the
order of basic nature of amines is expected to be
3° > 2° > 1°, however the observed order is 2° > 1°
> 3°. This is explained on the basis of crowding
on N-atom of the amine by alkyl groups which
hinders the approach and bonding by a proton,
consequently, the electron pair which is present
on N is unavailable for donation and hence
3° amines are the weakest bases.
Aromatic amines are weaker bases than ammonia
and aliphatic amines. Electron-donating groups
such as –CH3, –OCH3, etc. increase the basicity
while electron-withdrawing substitutes such as –
NO2, –CN, halogens, etc. decrease the basicity of
amines. The effect of these substituents is more
at p than at m-positions.
57. Which one of the following is the strongest
base in aqueous solution?
(a) Methyl amine
(b) Trimethyl amine
(c) Aniline
(d) Dimethyl amine
58. Which of the following order of basicity is
correct?
(a) Aniline > m-toluidine > o-toluidine
(b) Aniline > o-toluidine > m-toluidine
(c) o-Toluidine > aniline > m-toluidine
(d) o-Toluidine < aniline < m-toluidine
CBSE Board Term-II Chemistry Class-12
152
59. What is the decreasing order of basicity of
primary, secondary and tertiary ethylamines
and NH3?
(c) N, N-dialkyl benzene sulphonamide soluble
in KOH solution
(d) N, N-dialkyl benzene sulphonamide insoluble
in KOH solution.
(b) (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3
63. To separate amines in a mixture, Hoffmann’s
method is used. The Hoffmann’s reagent is
(a) benzenesulphonyl chloride
(b) diethyl oxalate
(c) benzeneisocyanide
(d) p-toulenesulphonic acid.
(a) NH3 > C2H5NH2 > (C2H5)2NH > (C2H5)3N
(c) (C2H5)2NH > C2H5NH2 > (C2H5)3N > NH3
(d) (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
60. The order of basic strength among the
following amines in benzene solution is
(a) CH3NH2 > (CH3)3N > (CH3)2NH
(b) (CH3)3N > (CH3)2NH > CH3NH2
(c) CH3NH2 > (CH3)2NH > (CH3)3N
(d) (CH3)3N > CH3NH2 > (CH3)2NH
Case III : Read the passage given below and
answer the following questions from 61 to 61.
When the mixture contains the three amine salts
(1°, 2° and 3°) along with quaternary salt, it is
distilled with KOH solution. The three amines
distill, leaving the quaternary salt unchanged
in the solution. Then the mixture of amines is
separated by fractional distillation, Hinsberg’s
method and Hoffmann’s method.
(1°, 2° and 3° amines in mixture)
(i) C6H5SO2Cl
(ii) KOH
(iii)Distillation
Distillate
(Amine A)
Mother liquor
(Amine B and amine C)
Filtered
Filtrate
70% H2SO4
(Amine B)
61.
(a)
(b)
(c)
(d)
Hinsberg reagent is
aliphatic sulphonyl chloride
phthalamide
aromatic sulphonyl chloride
anhydrous ZnCl2 + conc. HCl.
Residue
70% H2SO4
(Amine C)
62. Primary amine with Hinsberg’s reagent forms
(a) N-alkyl benzene sulphonamide soluble in
KOH solution
(b) N-alkyl benzene sulphonamide insoluble in
KOH solution
64.
(a)
(b)
(c)
(d)
3° amines with Hinsberg’s reagent give
no reaction
product which is same as that of 1° amine
product which is same as that of 2° amine
products which is a quaternary salt.
Case IV : Read the passage given below and
answer the following questions from 65 to 68.
Amines are basic in nature. The basic strength
of amines can be expressed by their dissociation
constant, Kb or pKb.
RNH2 + H2O  RNH+3 + OH–
Kb =
[RNH3+ ][OH − ]
and pK b = − log K b
[RNH2 ]
Greater the Kb value or smaller the pKb value, more
is the basic strength of amine. Aryl amines such as
aniline are less basic than aliphatic amines due to
the involvement of lone pair of electrons on N-atom
with the resonance in benzene. In derivatives of
aniline, the electron releasing groups increase the
basic strength while electron withdrawing groups
decrease the basic strength. The base weakening
effect of electron withdrawing group and base
strengthening effect of electron releasing group
is more marked at p-position than at m-position.
o-Substituted aniline is less basic than aniline due
to ortho effect and is probable due to combination
of electronic and steric effect.
65. Which of the following has lowest pKb value?
(a)
NH2
NH2
(b)
NO2
(c)
N(CH3)2
(d)
NHCH3
153
Amines
66. The strongest base among the following is
(b) p-NO2 – C6H4NH2
(a) C6H5NH2
(c) m-NO2 – C6H4NH2 (d) C6H5CH2NH2
68. Which of the following statements is not
correct?
67. Which among the following shoes maximum
pKb value ?
(b) Primary amines form hydrogen bonds.
(a)
NH
(c) (CH3CH2)2NH
(b)
NHCH3
(d) (CH3)2NH
(a) Methylamine is more basic than NH3.
(c) Ethylamine has higher boiling point than
propane.
(d) Dimethylamine is less basic than
methylamine.
Assertion & Reasoning Based MCQs
For question numbers 69-80, a statement of assertion followed by a statement of reason is given. Choose
the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
69. Assertion : Ortho substituted anilines are
usually weaker bases than anilines.
Reason : This is due to ortho effect.
70. Assertion : In Hoffmann bromamide
reaction, the amine formed has one carbon atom
less than the parent 1° amide.
Reason : N-methyl acetamide undergoes Hoffmann
bromamide reaction.
71. Assertion : In strongly acidic solutions,
aniline becomes more reactive towards electrophilic
reagents.
Reason : The amino group being completely
protonated in strongly acidic solution, the lone
pair of electrons on the nitrogen is no longer
available for resonance.
72. Assertion : Ammonia is more basic than water.
Reason : Nitrogen is less electronegative than
oxygen.
73. Assertion : Me3N reacts with BF3 whereas
Ph3N does not.
Reason : The electron pair on nitrogen atom in
Ph3N is delocalised in the benzene ring and is not
available for boron in BF3.
74. Assertion : Aniline is a weaker base than
cyclohexylamine.
Reason : Aniline undergoes halogenation even
in the absence of a catalyst.
75. Assertion : Controlled nitration of aniline
at low temperature mainly gives m-nitroaniline.
Reason : In acidic medium, –NH2 group gets
converted into m-directing group.
76. Assertion : Ammonolysis of alkyl halides
involves the reaction between alkyl halides and
alcoholic ammonia.
Reason : Reaction can be used to prepare 1°, 2°, 3°
amines and finally quaternary ammonium salts.
77. Assertion : Aniline does not undergo
Friedel-Crafts reaction.
Reason : –NH2 group of aniline reacts with AlCl3.
78. Assertion : Boiling point of amines are
lower than those of alcohols and carboxylic acids.
Reason : Amines are much more soluble in water
than less polar solvents like alcohol, ether, etc.
79. Assertion : Nitration of aniline can be done
conveniently by protecting the amino group by
acetylation.
Reason : Acetylation increases the electron
density in the benzene ring.
80. Assertion : Aniline hydrogen sulphate, on
heating, forms p-aminosulphonic acid.
Reason : The sulphonic acid group is
electron-withdrawing.
CBSE Board Term-II Chemistry Class-12
154
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Write chemical equations for the following
conversion :
Benzyl chloride to 2-phenylethanamine.
2.
What carbylamine reaction ?
3. Rearrange the following in an increasing
order of their basic strengths :
C6H5NH2, C6H5N(CH3)2, (C6H5)2NH and CH3NH2
4. Write IUPAC name of the following
compound: (CH3CH2)2NCH3
5. Propanamine and N, N-dimethylmethanamine
contain the same number of carbon atoms, even
though propanamine has higher boiling point
than N, N-dimethylmethanamine. Why?
6. Arrange the following compounds in
increasing order of solubility in water :
C6H5NH2, (C2H5)2NH, C2H5NH2
7.
Write the structure for N-ethylmethylamine.
8. How will you convert the following :
Aniline into N-phenylethanamide
(Write the chemical equations involved.)
9. Why primary aromatic amines cannot be
prepared by Gabriel phthalimide synthesis?
10. Draw the structure of N, N-diethylethanamine.
Short Answer Type Questions (SA-I)
11. How are the following conversions carried out?
(i) CH3CH2Cl to CH3CH2CH2NH2
16. Give the structures of A, B and C in the
following reactions :
(ii) Benzene to aniline
(i) C6H5NO2
12. Account for the following :
(ii) CH3CN
(i) Tertiary amines do not undergo acylation
reaction.
(ii) Amines are more basic than comparable
alcohols.
13. Complete the following reactions :
(i) CH3CH2NH2 + CHCl3 + alc. KOH Heat
NH2
(ii)
+ HCl(aq)
14. Acetamide is less basic than ethanamine.
Why?
15. Write the structures of the main products of
the following reactions :
NH2
(i)
(ii)
(CH3CO)2O
A
B
C
C
(i) Methylamine and dimethylamine
(ii) Aniline and N-methylaniline
18. How would you account for the following :
(i) Aniline is a weaker base than cyclohexylamine.
(ii) Methylamine in aqueous medium gives
reddish-brown precipitate with FeCl3.
19. How will you convert the following :
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
20. Give the structure of of products A and B in
the given sequence of reactions.
(i) R
C
O
(CH3)2NH
B
17. Give the chemical tests to distinguish between
the following pairs of compounds :
Pyridine
SO2Cl
A
(ii) R
C
N
H2/Ni
Na(Hg)/C2H5OH
NH2
(i) LiAlH4
(ii) H2O
B
A
155
Amines
Short Answer Type Questions (SA-II)
21. How are the following reactions carried out?
Write the equations and conditions.
(i) Acetic acid to ethylamine
28. Write the structure of N-methylethanamine.
29. Write the structure of 2-aminotoluene.
(i)
30. Amit wants to manufacture aniline for the
synthesis of dye stuff. For this he has selected
tin and hydrochloric acid as reducing agent for
the reduction of nitrobenzene. But his friend
suggested to use iron scrap and hydrochloric
acid as the reducing agent.
(ii) CH3CH2CH2NH2 and CH3NHCH2CH3
(i) Write the chemical equation for the reduction
of nitrobenzene to aniline.
(ii) Bromocyclohexane to cyclohexanamine.
(iii) Methylamine to dimethylamine.
22. Which amine in each of the following pairs
is a stronger base? Give reason.
Now answer the following questions :
23. Write the structures of main products when
aniline reacts with the following reagents :
(i) Br2 water
31. (a) Give a simple chemical test to distinguish
between aniline and N, N-dimethylaniline.
(ii) HCl
(iii) (CH3CO)2O/pyridine
24. A compound X (C7H7Br) reacts with KCN
to give Y (C8H7N). Reduction of Y with LiAlH4
yields Z (C8H11N). Z gives carbylamine reaction,
reacts with Hinsberg’s reagent in the presence
of aq. KOH to give a clear solution. With NaNO2
and HCl at 0°C (Z) gives a neutral compound
which gives red colour with ammonium cerric
nitrate. What are X, Y and Z ?
25. (a) Give one chemical test to distinguish
between the compounds of the following pairs :
(i) CH3NH2 and (CH3)2NH
(ii) (C2H5)2NH and (C2H5)3N
(b) Why aniline does not undergo Friedel–Crafts
reaction?
26. Write the structures of A, B and C in the
following sequence of reactions :
(i) C6H5 CONH2
C
(ii) CH3 Cl
KCN
Br2/aq. KOH
KI
B
LiAlH4
A
C
A
NaNO2 +HCl
0-5°C
B
CHCl3 + alc. KOH
27. Write the structure of
(ii) Why Amit’s friends has suggested to use
scrap iron and HCl in place of tin and HCl?
2, 4-dinitrochlorobenzene.
(b) Arrange the following in the increasing
order of their boiling point :
C2H5NH2, C2H5OH, (CH3)3N
32. Write the chemical equations for the
following conversions :
(i) Ethyl isocyanide to ethylamine.
(ii) Aniline to benzonitrile
(iii) Aniline to p-nitroaniline.
33. An aromatic compound ‘A’ on treatment with
aqueous ammonia and heating forms compound
‘B’ which on heating with Br2 and KOH forms a
compound ‘C’ of molecular formula C6H7N. Write
the structures and IUPAC names of compounds
A, B and C.
34. Account for the fact that although N,
N-dimethyl aniline is only slightly more basic
than aniline, 2, 6-dimethyl N, N-dimethyl aniline
is much more basic than 2, 6-dimethyl aniline.
35. Give reasons :
(i) Acetylation of aniline reduces its activation
effect.
(ii) CH3NH2 is more basic than C6H5NH2.
(iii) Although —NH2 is o/p directing group, yet
aniline on nitration gives a significant amount
of m-nitroaniline.
CBSE Board Term-II Chemistry Class-12
156
Long Answer Type Questions (LA)
36. (a) Write structures of different isomers
corresponding to the molecular formula, C3H9N.
Write IUPAC name of the isomers which will
liberate nitrogen gas on treatment with nitrous
acid.
(b) Classify the following amines as primary,
secondary and tertiary :
NH2
N(CH3)2
(i)
(ii)
(iii) (C2H5)2CHNH2
(iv) (C2H5)2NH
37. (a)
38. (a) Write the reactions of (i) aromatic and
(ii) aliphatic primary amines with nitrous acid.
(b) Write the chemical equations for the following
conversions
(i) Chlorobenzene to p-chloroaniline
(ii) Aniline to p-bromoaniline
39. Write the structure of the reagents/organic
compounds A to F in the following sequence of
reactions :
X
Find X and Y. Is Y optically active? Write the
intermediate steps.
(b) Which of the following is more acidic and why?
OBJECTIVE TYPE QUESTIONS
1.
Br2 /
NaCN
(d) : CH 2 = CH 2 
→ CH 2Br − CH 2Br →
CCl
4
(X)
LiAlH 4
NCCH 2CH 2CN 
→ H 2 NCH 2CH 2CH 2CH 2 NH 2
(Y )
(1, 4- Diaminobutane)
(Z )
COOH
2.
SOCl2
(a) :
Benzoic
acid
3.
COCl
CONH2
NH3
Benzoyl
chloride
(X)
NH2
Br2
KOH
Benzamide
(Y)
Aniline
(Z)
40. Give plausible explanation for each of the
following :
(i) Why are amines less acidic than alcohols of
comparable molecular masses?
(ii) Why do primary amines have higher boiling
point than tertiary amines?
(iii) Why are aliphatic amines stronger bases
than aromatic amines?
5. (c) : Only primary amines will give carbylamine reaction.
6. (d) : Aryl amines are less basic than alkyl amines because
phenyl group exerts –I effect and one pair of electrons on
nitrogen is in conjugation resulting in decrease in electron
density on nitrogen atom for protonation.
7. (c) : The boiling points of amines vary in the order of
primary > secondary > tertiary.
8. (b) : Primary and secondary amines can form hydrogen
bonds whereas tertiary amines fail to do so. Hence, their
boiling points are lowest.
9. (c) :
(c) :
10. (a)
11. (d) :
4.
(b) :
157
Amines
12. (d)
13. (b) :
25. (a) : CH3CH2Cl
CH3CH 2 NH 2 + NaBr + Na 2CO3 + H 2O
Cl
14. (c) :
NH2
15. (b) :
Cl
NaNO2
HCl
N2Cl
NH2
(Y )
(Z )
Br
26. (a) :
NC
+ CHCl3 + 3KOH (alc.)
+ 3KCl + 3H2O
CH3
CH3
p-Toluidine
(4-Methylaniline)
17. (a) : Anilinium ion formed by protonation of aniline
deactivates o- and p-positions hence, substitution takes place
at m-position.
+
+
NH2
NH3
H
27. (c) : Since the compound reacts with benzenesulphonyl
chloride to give a product which is insoluble in alkali, it
shows there is no H attached to N in the product. Hence, the
compound X is a secondary amine.
CH3 − NH − C2 H5 + C6 H5SO 2Cl 
→
CH3 N
16. (a) :
+
NH3
HNO3
H2SO4
Anilinium
ion
–H
NO2
m-Nitroanilinium
cation
C2H5
N-Ethyl-N-methylbenzene sulphonamide
28. (c) :
+
NO2
m-Nitroaniline
19. (a) : Nitrogen forms three sp3 hybridised sigma bonds
with carbon atoms of methyl groups and has a non-bonding
electron pair in fourth sp3 orbital. Thus (CH3)3N has pyramidal
shape.
20. (c) : Amides on reaction with Br2 and KOH will give
amine containing one carbon atom less. This reaction is called
Hoffmann bromamide degradation.
∆
C6H5CONH2 + Br2 + 4KOH 
→
C6H5NH2 + 2KBr + K 2CO3 + 2H2O
21. (b) : Only primary amines show carbylamine reaction.
29. (c) : The trend in boiling point can be explained on the
basis of intermolecular hydrogen bonding which is maximum
in primary amines.
30. (a) : Benzenesulphonyl chloride (C6H 5SO 2Cl) acts as
Hinsberg’s reagent.
31. (a) : II is not an acceptable canonical structure because
nitrogen has 10 valence electrons in the structure. Anilinium
ion exists in two canonical structures only which are I and III.
32. (c) :
33. (b) : C6H5NH2 + ClCOC6H5 + NaOH →
Aniline Benzoyl chloride
C6H5NH – COC6H5 + NaCl + H2O
23. (c) : C6H5NH2 < NH3 < CH3NH2 < (CH3)2NH
24. (a) : Primary amines react with benzoyl chloride to give
benzamides and the reaction is known as benzoylation.
+ C6H5COCl 
→ CH3 NHCOC6H5 + HCl
N - Methylbenzamide
Benzanilide
N+2 Cl
NH2
34. (a) :
22. (a) :
Benzoyl chloride
SO2C6H5
NH2
18. (c) : Amines are basic in nature while amides are
amphoteric in nature.
CH3 NH 2
Methanamine
CH3CH2CH2NHCOCH3
Cl
CuBr
– HBr
+
(X)
CH3CH2CH2NH2
Pr opanamide
CH3CH2CN
Br2 / NaOH
CH3CH 2CONH 2 
→
( NaOBr )
NaNO2 + HCl
(0-4°C)
–
CN
CuCN
or KCN
CH2NH2
H2/Ni
Benzylamine
35. (d) : –NH2 group is oxidised on direct nitration hence –NH2
group is blocked by acetylation and then nitration is carried out.
CBSE Board Term-II Chemistry Class-12
158
NH2
NHCOCH3
NHCOCH3
RNH 2 + ( RCO) 2 O 
→ RNHCOR + RCOOH
Substituted
amides
HNO3
CH3COCl
H2SO4
48. (d) :
R
LiAlH
4
N H /Na
2
C
NO2
NH2
H2O/H
R CH2NH2
(A or B)
+
NO2
(C) R
C
Na(Hg)/C2H5OH
Partial Hydrolysis
LiAlH4/H2O
NH2
O
36. (a) :
Br2/KOH
R
CH2NH2 (D)
R
NH2 (E)
49. (d) : At meta position NO2 shows –I effect while at
o – and p – positions it shows –R effect.
50. (c) :
37. (b) : Higher the basicity lower is the pKb value. Since
NH3 is the weakest base, hence it has highest pKb value.
38. (d) : The synthesis of primary amines from phthalimide
is known as Gabriel phthalimide synthesis.
CH3CH2CH2Br
KCN
[H]
CH3CH2CH2CN H /Ni CH3CH2CH2CH2NH2
2
n-Butylamine
H 3O +
CH3CH2CH2COOH
CH3CH2CH2Br
AgCN
CH3CH2CH2NC
39. (d) : Secondary amines do not give carbylamine test.
40. (d) : NH3 + C2H5I 
→ [(C2H5 ) 4 N + ]I −
( Excess )
41. (b) : There is no free hydrogen in tertiary amines hence
they do not form salts and are not soluble in acids.
51. (c) : CH3CONH2
(C)
Secondary Amine
PCl
KCN
(Y )
C2H5CN
Propanenitrile
(Z )
H
44. (c) : RN ≡≡ C 4
→ RNHCH3
Other compounds give primary amines.
4H
RCONH 2 → RCH 2 NH 2 + H 2O
4H
RCN → RCH 2 NH 2
45. (c) : 3° amines do not react with Hinsberg reagent.
O
46. (b) :
NH2
NH2
KOH/Br2
CH3NH2 (A)
(Primary Amine)
H2/Ni
[H]
CH3N
(B)
C
52. (c) : The value of pKb describes the relative strength of
the bases.
NH3 CH3NH2 (CH3)2NH (CH3)3N
3.38
3.27
4.22
pKb 4.7
So, the correct order of basicity in aqueous medium is :
NH3 < (CH3)3N < CH3NH2 < (CH3)2NH
53. (a)
6H
RNO 2 → RNH 2 + 2H 2O
O
NaOH + Br2
CH3NHCH3
43. (a) :
5
C2H5 NH 2 2
→ C2H5OH 
→ C2 H5Cl
Methyl propyl amine
CHX3/KOH
42. (c) : Only –CH3 group is electron donating group hence it
increases the electron density on nitrogen making it most basic.
(X )
CH3CH2CH2NH CH3
CH3CH2CH2NH2 + HCOOH
Quaternary salt
HNO
[H] H2/Ni
H 3O +
NH
CH3COCl
47. (b) : RNH 2 + RCOCl 
→ RNHCOR + HCl
54. (d) : The rate determining step is probably loss of Br– to
form isocyanate as this is the slowest step.
O
KOH
55. (d) : Cl
+ Br2
NH2
CH3
Cl
O
NHBr
+ KBr + H2O
56. (b) : Since, the overall reaction is intermolecular, hence
there will be no effect on product formation.
159
Amines
NH2
Br2/NaOH
62. (a) : A primary amine forms N-alkylbenzene sulphonamide
with Hinsberg’s reasent because of the presence of an acidic
hydrogen on the N-atom, dissolves in aqueous KOH.
15
+
NH2
63. (b)
D
64. (a) : Tertiary amine does not contain a replaceable
hydrogen on the nitrogen atom. So, 3° amine does not react
with Hinsberg’s reagent.
cross product (not formed)
15
CONH2 +
CONH2
65. (c)
D
Br2/NaOH
NH2 +
15
NH2
D
57. (d) : The increasing order of basicity of the given
compounds is (CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2.
Due to the +I effect of alkyl groups, the electron density on
nitrogen increases and thus, the availability of the lone pair
of electrons to proton increases and hence, the basicity of
amines also increases. So, aliphatic amines are more basic
than aniline. In case of tertiary amine (CH3)3N, the covering
of alkyl groups over nitrogen atom from all sides makes the
approach and bonding by a proton relatively difficult, hence
the basicity decreases. Electron withdrawing groups decrease
electron density on nitrogen atom and thereby decreasing
basicity.
58. (d) : In general, electron donating (+R) group which
when present on benzene ring (–NH2, –OR, –R, etc.) at the
para position increases the basicity of aniline.
Ortho substituted anilines are weaker bases than aniline due
to ortho effect.
T
T
59. (d) : In case of ethylamines, the combined effect of
inductive effect, steric effect and solvation effect gives the
order of basic strength as
(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
(2°)
(3°)
(1°)
60. (b) : In non-aqueous solvents the basic strength of alkyl
amines follows the order :
tertiary amines > secondary amines > primary amines.
61. (c)
66. (d)
67. (a)
68. (d) : Dimethylamine is more basic than methylamine.
69. (a) : Ortho effect is a consequence of steric and
electronic factors.
70. (c) : Only primary amines can be prepared from amides
(RCONH2) by treating with Br 2 and KOH. Thus, N-methyl
acetamide i.e., CH3CONHCH3 does not undergo Hoffmann
bromamide reaction.
71. (d) : In strongly acidic medium, aniline gets protonated
and so the lone pair of electrons is not available to produce
+E or +M effects. On the other hand, the –NH3 group exerts
strong –I effect and thus it causes the deactivation of the
ring
.
72. (a) : Ammonia is more basic than water. It is because
nitrogen being less electronegative than oxygen, has a greater
tendency to donate electrons.
73. (a)
74. (b) : Aniline exists as resonance hybrid. As a result
of resonance, the lone pair of electrons on nitrogen gets
delocalized over the benzene ring and thus, is less easily
available for protonation than in case of cyclohexylamine
where no such resonance takes place.
75. (a) : Under acidic condition, aniline gets protonated
to anilinium ion (–NH 3+ group). This is deactivating and
m-directing group. Thus, controlled nitration of aniline mainly
gives m-nitroaniline.
76. (b)
77. (a)
78. (b) : It is because hydrogen bonding is less pronounced
in primary and secondary amines than that in alcohols or
carboxylic acids and nitrogen is less electronegative than
oxygen.
79. (c) : Acetylation decreases the electron density in
the benzene ring and deactivate the ring hence control the
reaction.
80. (b) : –NH2 being o, p-directing group directs –SO3H
group to less hindered p-position.
CBSE Board Term-II Chemistry Class-12
160
SUBJECTIVE TYPE QUESTIONS
CH2 Cl
1.
CH2 CH2 NH2
CH2 CN
LiAlH4
KCN
Benzyl
chloride
2-Phenylethanamine
12.
(i)In tertiary amines there are no acidic hydrogen due
to which they do not undergo acylation reaction.
(ii) N being less electronegative than O gives lone pair of
electron more easily than O atom. Therefore amines are more
basic than alcohols.
13.
(i)CH3
2.
Carbylamine reaction is the reaction in which 1° amines
produce a bad smelling compound when treated with
chloroform in the presence of alkali.
C + 3KCl + 3H2O
RNH2 + CHCl3 + 3KOH Heat R — N
It is the test for primary amines.
(ii)
3.
(C6H5)2NH < C6H5NH2 < C6H5N(CH3)2 < CH3NH2
14.
4.
H3CH2C
N — CH3
H3CH2C
CH3
CH2
NC + 3KCl + 3H2O
+
H2O
NH2 + HCl
NH3Cl–
;
In acetamide, lone pair on nitrogen atom is involved in resonance
and hence is not free for donation as in the case of ethanamine.
IUPAC name : N-Ethyl-N-methylethanamine
5.
Primary amines (R – NH2) have two hydrogen atoms
on nitrogen which can undergo intermolecular hydrogen
bonding whereas no such hydrogen bonding is present in
tertiary amines (R3N). So, primary amines boil at a higher
temperature than tertiary amines.
6.
C6H5NH2 < (C2H5)2NH < C2H5NH2
1° amines are more soluble in water than 2° amines. Aniline
due to large hydrophobic benzene ring is least soluble.
7.
NH2 + CHCl3 + alc 3KOH Heat
CH2
15. (i)
O
(ii)
S
CH3N C2H5 (N-ethylmethylamine)
Cl + H
CH3
N
–HCl
CH3
O
Benzene sulphonyl
chloride
H
O
S
CH3
N
CH3
O
N, N-dimethylbenzene
sulphonamide
8.
16.
(i)
9.
Aromatic amines cannot be prepared by Gabriel phthalimide
synthesis because aryl halides do not undergo nucleophilic
substitution with the anion formed by phthalimide.
10. C2H5
N
C 2H 5
(ii) CH3CN
A
H2O
C
H2O/H+
C6H5NH2
C6H5N+2 Cl–
NaNO2 + HCl
273 K
B
CH3COOH
NH3
CH3CONH2
B
Br2+KOH
CH3NH2
C
11.
(i)
17. (i)
Methylamine gives carbylamine test, i.e., on treatment
with alc. KOH and chloroform, followed by heating it gives
offensive odour of methyl isocyanide. Dimethylamine does
not give this test.
NO2
Nitration
Benzene
C6H5OH
Sn + HCl
A
C 2H 5
(ii)
C6H5NO2
NH2
Fe/HCl
Nitrobenzene
Aniline
(ii) Aniline gives carbylamine test, i.e., on treatment
with alc. KOH and chloroform followed by heating it gives
offensive odour of phenylisocyanide but N-methylaniline
being secondary amine, does not show this test.
161
Amines
18. (i)
Aniline is weaker base than cyclohexylamine because of
resonance. Due to electromeric effect, the lone pair on nitrogen
is attracted by benzene ring. Hence, donor tendency of —NH2
group decreases. There is no resonance in cyclohexylamine.
Electron repelling nature of cyclohexyl group further increases
the donor property of NH2 group. So, cyclohexylamine is a
stronger base.
;
(ii) Methylamine forms hydroxide ions when dissolved in water
due to the following acid - base equilibrium.
+
CH3 NH3 + OH–
CH3 NH2 + H2O
These OH– ions react with Fe3+ ions to form ferric hydroxide.
2Fe(OH)3
2Fe + 6OH–
NH2
CH3 CH CH3 i s m o r e b a s i c t h a n
CH3 CH COOCH3 because – COOCH 3 is an electron
22. ( i )
withdrawing group which decrease the electron density on
nitrogen atom.
(ii) 2° amines are more basic than 1° amines, because in
2° amine there are two electron releasing groups and in
1° amine only one electron releasing group is present, so,
CH3NHCH2CH3 is more basic than CH3CH2CH2NH2.
Br
23. (i)
C6H5NH2 + Br2(aq)
NH2
Br
+ 3HBr
Br
(ii)
NH2 + HCl
+
H2O
NH3Cl–
19. (i)
(iii)
(ii) CH3COOH
PCl5
Ethanoic acid
(Acetic acid)
CH3 NH2
O
CH3 C
Cl
Br2/KOH
Methanamine
20.
(i) R
C
N
O
(ii) R
C
NH2
O
CH3 C
H2/Ni
Na(Hg)/C2H5OH
(i) LiAlH4
(ii) H2O
NH3
R
R
CH2
24.
NH2
CH2
NH2
NH2
25.
(a) (i) Methylamine gives carbylamine test, i.e., on
treatment with alc. KOH and chloroform, followed by heating
it gives offensive odour of methyl isocyanide. Dimethylamine
does not give this test.
(ii) (C2H5)2NH and (C2H5)3N can be distinguish by Hinsberg’s
reagent.
(b) (i) In Friedel Crafts reaction, AlCl3 is added as a catalyst
which is a Lewis acid. It forms a salt with aniline due to which
the nitrogen of aniline acquires positive charge. This positively
charged nitrogen acts as a strong deactivating group, hence
aniline does not undergo Friedel Crafts reaction.
21.
(i)
(ii)
(iii) CH3NH2 + CHCl3 + KOH
D
CH3NC
H2
Na/C 2H5 OH
NH2
CH3 NHCH3
+ AlCl3
+
NH2
AlCl–3
CBSE Board Term-II Chemistry Class-12
162
Br2
26. (i)
0–5C
NaNO2+ HCl
aq. KOH
that of amines because oxygen is more electronegative than
nitrogen.
+
H
32.
(i) C 2H5 − N →
→
= C + 2 HOH 
Ethyl isocyanide
(A)
Hydrolysis
C 2 H5 NH2 + HCOOH
Ethylamine
KI
(ii) CH3—Cl
KCN
CH3CN
(C)
LiAlH4
(ii)
(B)
CH3CH2NH2
(A)
(B)
(iii)
27.
2, 4-Dinitrochlorobenzene
28.
CH3CH2NHCH3
CH3
29.
NH2
2-Aminotoluene
30. (i)
[Fe + 2HCl → FeCl2 + 2H] × 3
(ii) Amit’s friend has suggested to use scrap iron and
hydrochloric acid because in this reaction FeCl2 formed gets
hydrolysed to release HCl during the reaction. Thus only small
amount of HCl is required to initiate the reaction and scrap
iron is also cheaper.
31.
(a)A niline undergoes isocyanide test (carbylamine
reaction) whereas, N, N – dimethylaniline does not.
NH2
+ CHCl3 + 3KOH
NC
Aniline
+ 3KCl + 3H2O
Phenyl isocyanide
(foul smell)
(b)Increasing order of boiling points :
(CH3)3N < C2H5NH2 < C2H5OH
Tertiary amine does not have hydrogen to form hydrogen
bonding and hydrogen bonding in alcohol is stronger than
33.
Formula of the compound ‘C’ indicates it to be an amine.
Since it is obtained by the reaction of Br2 and KOH with the
compound ‘B’ so compound ‘B’ can be an amide. It is also
indicated because ‘B’ is obtained from compound ‘A’ by reaction
with ammonia following by heating. So compound ‘A’ could
be an aromatic acid. Formula of compound ‘C’ shows that it
is aniline, then ‘B’ is benzamide and compound ‘A’ is benzoic
acid. The sequence of reactions can be written as follows :
34.
Extended p bonding between the amino nitrogen and
the ring requires the s bonds on N atom to become coplanar
with the ring and its ortho bonds.
Presence of bulky substituents in the ortho positions
(2, 6-positions) sterically hinder the attainment of this
geometry (coplanarity) and thus interferes with the base—
weakening extended p bonding.
Because of this effect, (called steric inhibition of resonance)
2, 6-dimethyl N, N-dimethyl aniline is much more basic than
2, 6-dimethyl aniline.
163
Amines
35. (i) After acetylation of aniline, acetanilide is formed
in which due to the presence of
group having
CH3
CH3 CH CH CH3
–I effect, electron density on N-atom decreases and hence,
activation effect of aniline gets reduced.
(ii)
NH2 + HCl
(X)
NH2
NaNO2/HCl
–N2
CH3
+
CH3 C CH CH3
H
+
H2O
NH3Cl–
(iii) Nitration is carried out with conc. HNO3 in the presence
of conc. H2SO4. In the presence of these acids, the –NH2
+
group of aniline gets protonated and is converted into –NH3
group. This positively charged group acts as a strong electron
withdrawing and meta-directing group. Hence, the incoming
electrophile goes to m-position.
36.
(a)In all, four structural isomers are possible. These are
as follows :
(b)
is more acidic due to the strong electron
withdrawing power (–I effect) of fluorine.
38.
(a) (i) Aromatic primary amines react with nitrous acid
to form diazonium salts.
Primary amines :
(ii) Aliphatic primary amines also form diazonium salts
on reaction with nitrous acid but they are unstable and
decompose to give the corresponding alcohols as the major
product with the evolution of nitrogen.
Secondary amines :
HCl
R − NH2 + HNO2 
→  R − N2 + Cl − 
273 − 278 K


Tertiary amines :
H O
Only primary amines react with HNO2 to liberate N2 gas.
(i)
(ii)
(b) (i)
(ii)
(b) (i)
2

→ R − OH + N2 + HCl
(ii)
(iii)
(iv)
CH3
*
37.
(a) CH3 – CH – CH – CH3
(X)
NaNO2/HCl
NH2
CH3
CH3 – C – CH2 – CH3 + N2 + NaCl + H 2O
OH
Y (rearranged product)
39.
Structure of reagents/organic compounds :
A = Benzene
CBSE Board Term-II Chemistry Class-12
164
R − NH2 → R
B = Nitrobenzene
NH
–
H
+
R–O–H→R–O +H
(ii) Primary amines (R – NH2) have two hydrogen atoms
on nitrogen which can undergo intermolecular hydrogen
bonding whereas no such hydrogen bonding is present in
tertiary amines (R3N). So primary amines boil at a higher
temperature than tertiary amines.
(iii) In aromatic amines, the lone-pair of electrons on nitrogen
atom is involved in resonance with the benzene ring as shown
below for aniline.
C = Aniline
D = Phenyl isocyanide
E = Methyl phenyl amine
F = Acetic anhydride
40.
(i)In alcohols, the hydrogen atom is attached to more
electronegative oxygen atom whereas nitrogen of amines is
less electronegative. After the loss of H+ ion, the negative
charge is more easily accommodated on oxygen than in case
of nitrogen in amines. Hence, amines have lesser tendency to
lose H+ ions, so they are less acidic than alcohols.
It shows that this pair of electrons is less available
for protonation. In case of aliphatic amines electron
releasing alkyl groups increase electron density on nitrogen
atom. So, aliphatic amines are stronger bases than aromatic
amines.

PRACTICE PAPER 1
Time allowed : 2 hours
Maximum marks : 35
General Instructions : Read the following instructions carefully.
(a)
(b)
There are 16 questions in this question paper. All questions are compulsory.
Section A : Q. No. 1 to 8 are objective type questions. Q. No. 1 is passage based question carrying 4 marks while
Q. No. 2 to 8 carry 1 mark each.
Section B : Q. No. 9 to 12 are short answer questions and carry 2 marks each.
Section C : Q. No. 13 and 14 are short answer questions and carry 3 marks each.
Section D : Q. No. 15 and 16 are long answer questions carrying 5 marks each.
There is no overall choice. However, internal choices have been provided.
Use of calculators and log tables is not permitted.
(c)
(d)
(e)
(f)
(g)
SECTION - A (OBJECTIVE TYPE)
1.
Read the passage given below and answer the following questions :
Cyclohexanone is an important intermediate in the manufacture of polyamides in chemical industry but direct
selective hydrogenation of phenol to cyclohexanone under mild conditions is a challenge. Hydrogenation
of phenol to cyclohexanone has been investigated in the presence of the composite catalytic system of Pd/C
heteropolyacid. 100% conversion of phenol and 93.6% selectivity of cyclohexanone were achieved under 80°C
and 1.0 mPa hydrogen pressure.
O
OH
Pd/C + HPA
Phenol
Cyclohexanone
It has been found that a synergetic effect of Pd/C and heteropolyacid enhanced the catalytic performance of the
composite catalytic system which suppressed the hydrogenation of cyclohexanone to cyclohexanol.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i)
The palladium-based catalyst mention in the study above can be used to convert selectively only
(a) cyclohexanone to cyclohexanol
(b) phenol to cyclohexanone
(c) phenol to cyclohexane
(d) phenol to cyclohexanol.
(ii) The product formed during hydrogenation of phenol by using this Pd–C + HPA catalyst can also be obtained by
(a)
(c)
oxidation of cyclohexane in air
oxidation of cyclohexanol in air
What will be the product of following reaction.
O
2
dil NaOH
OR
(b) reduction of cyclohexane in Sn/HCl
(d) reduction of hexanone in air.
?
*The paper is for practice purpose. CBSE has yet not released the official sample paper.
So, the pattern is suggestive only. For latest information visit www.cbse.gov.in.
CBSE Board Term-II Chemistry Class-12
166
OH
(a)
O
OH
(b)
OH
OH
OH
(c)
O
(d)
(iii) An organic compound A, C6H10O on reaction with CH3MgBr followed by acid treatment gives compound B.
The compound, B on reaction with HBr gives compound, C. Identify compound A, B and C.
CH3
CH3
CH3 OMgBr
O
OH
OMgBr
(a)
A=
, B=
(c)
A=
H3C
CH3
O
, B=
(b) A =
C=
,
Br
, B=
H3C
O
(d) A =
, C=
,
Br
CH3
C=
CH3
, C=
, B=
(iv) The product formed during hydrogenation of phenol by using the catalyst given in the case study is
(a)
(c)
miscible with water
immiscible with organic solvents
(b) miscible with organic solvents
(d) none of these.
Following questions (Q. No. 2-6) are multiple choice questions carrying 1 mark each :
2.
The variation of concentration of the product (X) with time in the reaction A→X is shown in graph. Hence, the
d[ A]
graph between −
and time will be of the type
dt
(a)
(b)
(c)
(d)
OR
In a reaction between A and B, the initial rate of reaction r0 was measured for different initial concentrations of
A and B as given below :
A/mol L–1
B/mol L–1
r0/mol L–1s–1
0.20
0.30
0.20
0.10
0.40
0.05
5.07 × 10–5
5.07 × 10–5
1.43 × 10–4
The order of the reaction with respect to A is
(a) 1.5
3.
(b) 0.5
(c) 1
(d) 2
Electrode potential data for a few elements is given :
Fe3+(aq) + e– → Fe2+(aq); Eº = + 0.77 V
Al3+(aq) + 3e– → Al(s); Eº = – 1.66 V
Br2(aq) + 2e– → 2Br–(aq);Eº = + 1.08 V
Based on the data, the reducing power of Fe2+, Al and Br– will increases in the order
(a) Br – < Fe2+ < Al
(b) Fe2+ < Al < Br –
(c) Al < Br– < Fe2+
(d) Al < Fe2+ < Br –
167
Practice Paper - 1
4.
Identify A in the following sequence of reactions :
Reduction
(a) Ethyl halide
5.
(b) iso-Propylamine
(c) n-Propyl halide
(d) iso-Propyl halide
(c)
(d) 2 2
The magnetic moment (B.M.) of Fe2+ ion is
(a) 0
(b)
35
24
OR
Amongst TiF62–, CoF63–, Cu2Cl2 and NiCl42– (at. nos. Ti = 22, Co = 27, Cu = 29, Ni = 28), the colourless species are
(a) CoF63– and NiCl42– (b) TiF62– and CoF63–
6.
(c) Cu2Cl2 and NiCl42–
(d) TiF62– and Cu2Cl2
Which of the following statements is true?
(a) If Do > P, strong field ligands and low spin complexes.
(b) If Do < P, strong field ligands and high spin complexes.
(c) If Do > P, weak field ligands and low spin complexes.
(d) If Do < P, weak field ligands and low spin complexes.
In the following questions (Q. No. 7 and 8), a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
7.
Assertion : Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason : Aromatic aldehydes are almost as reactive as formaldehyde.
OR
Assertion : CH3COOH but not HCOOH can be halogenated in presence of red P and Cl2.
Reason : Both formic acid and CH3COOH are highly soluble in water.
8.
Assertion : Fe(OH)3 and As2S3 colloidal sols on mixing precipitate.
Reason : Fe(OH)3 and As2S3 combine and form precipitate of new composition.
SECTION - B
The following questions Q. No. 9-12 are short answer type and carry 2 marks each.
9.
A colloidal solution of ferric oxide is prepared by two different methods as shown below :
FeCl3
FeCl3
NaOH
Hot
water
(A)
(B)
(a) What is the charge on colloidal particles in two test tubes (A) and (B)?
(b) Give reasons for the origin of charge.
CBSE Board Term-II Chemistry Class-12
168
OR
Explain the following observations :
(i) Lyophilic colloid is more stable than lyophobic colloid.
(ii) Sky appears blue in colour.
10.
Account for the following :
(i) The pKb of aniline is more than that of methylamine.
(ii) Gabriel phthalimide synthesis is the preferred method for synthesizing primary amines.
11. (a)Give reason why aldehydes are more reactive than ketones towards nucleophilic reagents.
(b)Why pH of reaction should be carefully controlled while preparing ammonia derivatives of carbonyl
compounds ?
OR
Two moles of organic compound ‘A’ on treatment with a strong base gives two compounds ‘B’ and ‘C’.
Compound ‘B’ on dehydrogenation with Cu gives ‘A’ while acidification of ‘C’ yields carboxylic acid ‘D’
with molecular formula of CH2O2. Identify the compounds A, B, C and D and write all chemical reactions
involved.
12. Assign a reason for each of the following observations :
(i)The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling
points.
(ii)The ionisation enthalpies (first and second) in the first series of the transition elements are found to
vary irregularly.
SECTION - C
Q. No. 13 and 14 are short answer type II carrying 3 marks each.
13. (a)Propanamine and N, N-dimethylmethanamine contain the same number of carbon atoms, even though
propanamine has higher boiling point than N, N-dimethylmethanamine. Why?
(b) Illustrate the following reactions giving suitable example in each case :
(i) Ammonolysis
(ii) Acetylation of amines
OR
Write the chemical equations for the following conversions :
(i) Aniline to N-phenylethanamide.
(ii) Aniline to p-nitroaniline.
14.
(i)For a reaction, A + B → Product, the rate law is given by, Rate = k[A]1[B]2. What is the order of the reaction?
(ii) Write the unit of rate constant ‘k’ for the first order reaction.
(iii)For the reaction A → B, the rate of reaction becomes twenty seven times when the concentration of A is
increased three times. What is the order of reaction?
SECTION - D
Q. No. 15 and 16 are long answer type carrying 5 marks each.
15. (a) (i)Express the relation between the conductivity (k) and the molar conductivity (Lm) of a solution.
(ii)Electrolytic conductivity of 0.30 M solution of KCl at 295 K is 3.72 × 10–2 S cm–1. Calculate the
molar conductivity.
169
Practice Paper - 1
(b) (i)Solutions of two electrolytes ‘A’ and ‘B’ are diluted. It is found that Lm value of ‘B’ increases
2 times while that of ‘A’ increases 20 times. Which of the two is a strong electrolyte?
(ii)A galvanic cell has E°cell = 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will
happen to the cell reaction and current flowing through the cell?
(iii) How will the pH of brine solution be affected on electrolysis?
OR
(a) Calculate the cell emf and DG for the cell reaction at 25°C for the cell :
Zn(s) | Zn2+ (0.0004 M) || Cd2+ (0.2 M) | Cd(s) E° values at 25°C : Zn2+/ Zn = – 0.763 V; Cd2+/Cd = – 0.403
V; F = 96500 C mol–1; R = 8.314 J K–1 mol–1.
(b) If E° for copper electrode is 0.34 V, how will you calculate its emf value when the solution in contact
with it is 0.1 M in copper ions? How does emf for copper electrode change when concentration of Cu2+
ions in the solution is decreased?
16.
(a) Give the IUPAC name of [PtCl(NH2CH3)(NH3)2]Cl.
(b)Compare the magnetic behaviour of the complex entities [Fe(CN)6]4– and [FeF6]3–.
[Atomic number of Fe = 26].
(c) Tetrahedral complexes are always of high spin. Explain.
OR
(a)What is spectrochemical series? Explain the difference between a weak field ligand and a strong field
ligand.
(b)FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4
solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why?
1. (i) (b)
O
(ii) (a) :
+ O2
+ H2O
Cyclohexane
Cyclohexanone (also formed
by hydrogenation of phenol)
5.07 × 10–5 = k[0.20]x [0.30]y...(i)
OH
dil. NaOH
(a) : 2
O
5.07 × 10–5 = k[0.20]x [0.10]y...(ii)
O
H3C
1.43 × 10–4 = k[0.40]x [0.05]y...(iii)
On dividing eq. (i) by (ii), we get
1=
OMgBr
H+
CH3MgBr
(iii) (c) :
CH3
Br
H3C
HBr
(B)
[0.30]y
i.e., y = 0
[0.10]y
On dividing eq. (iii) by eq. (ii), we get
D
(A)
(C6H10O)
OR
(a) : Let the rate law, rate = k[A]x [B]y
From the data given
OR
O
2. (d) : Concentration of the product increases linearly
d[ A]
= constant.
with time t, hence order = 0, and −
dt
1.43 × 10 −4
[0.40]x [0.05]y
=
5.07 × 10 −5
[0.20]x [0.10]y
as y = 0
 0.40 
x
Thus, 2.82 = 
 0.20 
(C)
(iv) (b) : Cyclohexanone is formed from hydrogenation
of phenol which is slightly soluble in water and miscible
with organic solvents.
⇒ (2)x = 2.82 ⇒ x = 1.5
3. (a) : Lower the reduction potential more is the
reducing power.
CBSE Board Term-II Chemistry Class-12
170
4. (d) :
5. (c) : Fe2+ ⇒ 3d6, \ n = 4
Magnetic moment =
n(n + 2) B.M. = 24 B.M.
OR
(d) : Oxidation state of
Ti in TiF62– = +4 i.e., Ti4+
3d 0
Co in CoF63– = +3 i.e., Co3+
3d 6
Ni in NiCl42– = +2 i.e., Ni2+
3d 8
+
Cu in Cu2Cl2 = +1 i.e., Cu
3d10
Colour of salts is due to partially filled d-orbitals. Since
TiF62– has completely empty d-subshell and Cu2Cl2
involves completely filled d-subshell, these are colourless
salts.
6. (a) : Strong field ligands cause greater splitting
which leads to pairing of electrons.
7. (c) : Aromatic aldehydes and formaldehyde do
not contain a-hydrogen and thus undergo Cannizzaro
reaction. Formaldehyde is more reactive than aromatic
aldehydes.
OR
(b) : T he Hell—Volhard—Zelinsky (HVZ) reaction
shows halogen substitution at a-carbon atom. HCOOH
has no a-carbon atom and thus does not undergo HVZ
reaction.
8. (c) : On mixing Fe(OH) 3 (+ve sol) and As 2 S 3
(–ve sol), mutual coagulation occurs which causes
precipitation. No new compound is formed.
9. (i) (a) Colloidal particles of test tube (A) are
positively charged whereas colloidal particles of test
tube (B) are negatively charged.
(b) In test tube (A), Fe3+ ions are adsorbed on the ppt.
Fe2O3·xH2O [or Fe2O3⋅xH2O/Fe3+ is formed].
In test tube (B), OH– ions are adsorbed on the ppt.
Fe2O3⋅xH2O [or Fe2O3⋅xH2O/OH–is formed].
OR
(i) A lyophilic colloid is stable due to the charge as
well as solvation of the sol particles. Such a solution
can only be coagulated by adding an electrolyte and
by adding a suitable solvent which can dehydrate the
dispersed phase.
On the other hand, a lyophobic sol is stable due to
charge only and hence can be easily coagulated by
adding small amount of an electrolyte.
(ii) The atmospheric particles of colloidal range scatter
blue component of the white sunlight preferentially.
That is why sky appears blue.
10. (i) In case of aniline, the lone pair of electrons on
the N-atom is delocalized with the p-electrons of the
benzene ring, making the lone pair of electrons on
nitrogen less available for protonation.
NH2
NH2
NH2
NH2
NH2
On the other hand, in methylamine the electronreleasing methyl group increases the electron density
around nitrogen, thereby increasing the availability of
the lone pair of electrons.
CH3 → NH2
Therefore, aniline is a weaker base than methylamine and
hence, its pKb value is higher than that of methylamine.
(ii) Gabriel phthalimide synthesis is the only method
used to make primary amines because it is not
possible to alkylate potassium phthalimide with either
(CH3)2CH X or (CH3)3C X.
11. (a) Ketones are less reactive than aldehydes towards
nucleophilic addition reactions because :
The two electron releasing alkyl groups decrease the
magnitude of positive charge on carbonyl carbon and
make it less susceptible to nucleophilic attack.
R
R
C O
C O
R
H
Ketone
Aldehyde
The two bulkier alkyl groups hinder the approach of the
nucleophile to the carbonyl carbon. This is called steric
factor.
(b) In strongly acidic medium ammonia derivatives
being basic will react with acids and will not react with
carbonyl compound. In basic medium, OH– will attack
carbonyl group. Therefore, pH of a reaction should be
carefully controlled.
OR
Since the molecular formula of D is CH2O2, thus, D is
HCOOH (formic acid). D is obtained by the acidification
of C, so, C is sodium formate (HCOONa).
Thus, A must be formaldehyde (as it undergoes
Cannizzaro reaction with a strong base).
171
Practice Paper - 1
HCHO
NaOH
A
Formaldehyde
CH3OH + HCOONa
B
Methanol
C
Sodium formate
Cu (Dehy drogenation)
HCHO
Formaldehyde
A
Acidification
HCOOH
Formic acid
D
Thus, A = Formaldehyde (HCHO)
B = Methanol (CH3OH)
C = Sodium formate (HCOONa)
D = Formic acid (HCOOH)
12. (i) Hardness and high melting and boiling points
of these metals are attributed to the involvement of
greater number of electrons from (n – 1)d in addition
to the ns electrons in the interatomic metallic bonding.
(ii) Irregular variation of ionisation enthalpies is mainly
attributed to varying degree of stability of different
3d-configurations (e.g., d 0, d 5, d 10 are exceptionally
stable).
13.(a) Primary amines (R–NH2) have two hydrogen
atoms on nitrogen which can undergo intermolecular
hydrogen bonding whereas no such hydrogen bonding
is present in tertiary amines (R3N). So, primary amines
boil at a higher temperature than tertiary amines.
(b) (i) Ammonolysis : Alkyl halides when treated
with ethanolic solution of ammonia give a mixture of
primary, secondary, tertiary amines and quaternary
ammonium salt.
(ii) Acetylation of amines : The process of introducing
O
an acetyl group (CH3 C ) into a molecule is called
acetylation.
OR
(i)
(ii)
14. (i) Order of reaction is sum of powers of
concentration terms.
\ order of reaction = 1 + 2 = 3
(ii) Unit of rate constant for first order reaction is s–1.
...(i)
(iii) Let r = k[A]n
...(ii)
Then, 27r = k[3A]n
If eqn. (ii) is divided by eqn. (i), we get
27r k[3 A]n
or 33 = 3n
=
r
k[ A]n
n=3
\ Thus, order = 3
κ × 103
15. (a) (i) Λ m =
M
where M is the concentration of solution in molarity.
(ii) Electrolytic conductivity, k = 3.72 × 10–2 S cm–1
Molar conductivity,
κ (S cm −1 ) × 1000 (cm3 L−1 )
Λm =
M (mol L−1 )
3.72 × 10 −2 × 1000
= 124 S cm2 mol–1
=
0.30
(b) (i) For strong electrolytes, Lm increases slowly
with dilution. Since there is no wide effect on L m
value of electrolyte ‘B’ on dilution. Thus, ‘B’ is a strong
electrolyte.
(ii) If an external opposite potential is applied in the
galvanic cell and increased slowly, reaction continues
to take place till the opposing voltage reaches the value
1.1 V. After that reaction stops and no further chemical
reaction takes place. Hence, no current flows through
the cell.
(iii) The overall reaction for electrolysis of brine
solution can be written as
1
1
 NaOH(aq ) + H2( g ) + Cl 2( g )
NaCl(aq) + H2O(l) →
2
2
Hence, the pH of brine solution which is neutral will
increase due to formation of NaOH.
CBSE Board Term-II Chemistry Class-12
172
OR
(a) E°cell = E°cathode – E°anode = – 0.403 – (– 0.763)
= 0.36 V
The net cell reaction is
Zn(s) + Cd2+(aq) → Zn2+(aq) + Cd(s)
Here, value of n = 2
0.0591
[Zn2+ ]

Ecell = Ecell
−
log
2
[Cd 2 + ]
= 0.36 −
0.0591
0.0004
log
2
0. 2
0.0591
(−2.69) = 0.36 + 0.08 = 0.44 V
2
\ DG = – nFEcell = – 2 × 96500 × 0.44
= – 84920 J/mol
(b) Cu2+(aq) + 2e– → Cu(s)
0.059
[Cu]
°
ECu2+ /Cu = ECu
−
log
2+
/Cu
2
[Cu 2 + ]
0.059
0.059
1
log
= 0.34 −
log 10
2
0. 1
2
0.059
× (1) = 0.34 – 0.0295 = 0.3105 V
2
When the concentration of Cu2+ ions is decreased, the
potential for copper electrode decreases.
= 0.34 −
16. (a)Diamminechlorido(methylamine)
platinum(II) chloride
4–
(b) (i) [Fe(CN)6] ion
:
Fe2+ ion is hybridised under the influence of strong
field ligand.
[Fe(CN)6]4– ion formation :
4s
4p
OR
(a) The crystal field splitting, ∆ o, depends upon the
field produced by the ligand and charge on the metal
ion. Some ligands are able to produce strong fields in
which, the splitting will be large whereas others produce
weak fields and consequently result in small splitting of
d-orbitals. In general, ligands can be arranged in a series
in the order of increasing field strength as given below :
I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O
< NCS– < EDTA4– < NH3 < en < CN– < CO.
Such a series is termed as spectrochemical series.
Ligands for which ∆ o < P are known as weak field
ligands and form high spin complexes. In this case
∆ o, the fourth electron enters one of the e g orbitals
3 1
giving the configuration t2g
e g. Ligands for which Do >
P are known as strong field ligands and form low spin
complexes. In this case it becomes more energetically
favourable for the fourth electron to occupy a t2g orbital
with configuration t42g eg0.
(b) When FeSO4 and (NH4)2SO4 solutions are mixed
in 1 : 1 molar ratio, Mohr’s salt (a double salt) is formed.
FeSO4(aq) + (NH4)2SO4(aq) → FeSO4·(NH4)2SO4·6H2O
FeSO4·(NH4)2SO4·6H2O
hybridisation
Since the complex ion does not contain any unpaired
electron, so it is diamagnetic.
(ii) [FeF6]3– ion
:
As the complex ion contains five unpaired electrons, it
is highly paramagnetic in nature.
µ s = 5(5 + 2) = 35 = 5.9 B.M.
(c) For tetrahedral complexes, crystal field splitting
energy ∆t is always less than pairing energy. So,
tetrahedral complexes are always high spin.
= 0.36 −
= 0.34 −
Fe3+ ion is hybridised under the influence of weak field
ligand.
[FeF6]3– ion formation :
4s
4p
4d
Fe(2aq+ ) + 2NH +4 (aq ) +
2SO24(−aq ) + 6H2 O
Because Fe2+ ions are formed on dissolution of Mohr’s
salt, its aqueous solution gives the test of Fe2+ ions.
When CuSO 4 is mixed with ammonia, following
reaction occurs :
CuSO4(aq) + 4NH3(aq) → [Cu(NH3)4]SO4
This complex does not produce Cu2+ ion, so the solution
of CuSO4 and NH3 does not give the test of Cu2+ ion.

PRACTICE PAPER 2
Time allowed : 2 hours
Maximum marks : 35
General Instructions : Read the following instructions carefully.
(a)
(b)
There are 16 questions in this question paper. All questions are compulsory.
Section A : Q. No. 1 to 8 are objective type questions. Q. No. 1 is passage based question carrying 4 marks while
Q. No. 2 to 8 carry 1 mark each.
Section B : Q. No. 9 to 12 are short answer questions and carry 2 marks each.
Section C : Q. No. 13 and 14 are short answer questions and carry 3 marks each.
Section D : Q. No. 15 and 16 are long answer questions carrying 5 marks each.
There is no overall choice. However, internal choices have been provided.
Use of calculators and log tables is not permitted.
(c)
(d)
(e)
(f)
(g)
SECTION - A (OBJECTIVE TYPE)
1.
Read the passage given below and answer the following questions :
The molecular compounds which are formed from the combination of two or more simple stable compounds
and retain their identity in the solid as well as in the dissolved state are called coordination compounds. Their
properties are completely different from the constituents. In coordination compounds, the central metal atom
or ion is linked to a number of ions or neutral molecules, called ligands, by coordinate bonds. For example,
Dimethyl glyoxime (dmg) is a bidendate ligand chelating large amounts of metals.
When dimethyl glyoxime is added to alcoholic solution of NiCl2 and ammonium hydroxide is slowly added to it,
a rosy red precipitate of a complex is formed.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i) The structure of the complex is
(a)
CH3
C
CH3
C
Ni
(b)
2
CH3
C
NOH
CH3
C
NO
Ni
2
O
(c)
CH3
C
CH3
C
Ni
(d)
CH3
C
O
CH3
C
O
Ni
O
(ii) Oxidation number of Ni in the given complex is
(a) +3
(c) +2
(b) +1
(d) zero.
(iii) Hybridisation and structure of the complex is
(a) sp3, tetrahedral
(c) sp3, square planar
(b) dsp2, square planar
(d) sp3d, trigonal bipyramidal.
*The paper is for practice purpose. CBSE has yet not released the official sample paper.
So, the pattern is suggestive only. For latest information visit www.cbse.gov.in.
CBSE Board Term-II Chemistry Class-12
174
OR
Which of the following is true about this complex?
(a) It is paramagnetic, containing 2 unpaired electrons.
(b) It is paramagnetic, containing 1 unpaired electron.
(c) It is paramagnetic, containing 4 unpaired electrons.
(d) It is diamagnetic with no unpaired electron.
(iv) Which one will give test for Fe3+ ions in the solution?
(b) [Fe(CN)6]2–
(a) [Fe(CN)6]3–
(c) (NH4)2SO4 · FeSO4 · 6H2O
(d) Fe2(SO4)3
The following questions are multiple choice questions. Choose the most appropriate answer :
2.
The carboxylic acid which does not undergo Hell –Volhard –Zelinsky reaction is
(b) (CH3)2CHCOOH
(a) CH3COOH
(d) (CH3)3CCOOH
(c) CH3CH2CH2CH2COOH
3.
The strongest base among the following is
(a)
N
(b)
NH2
(c)
N
H
(d)
N
H
OR
The order of basic strength among the following amines in benzene solution is
(b) (CH3)3N > (CH3)2NH > CH3NH2
(a) CH3NH2 > (CH3)3N > (CH3)2NH
(c) CH3NH2 > (CH3)2NH > (CH3)3N
(d) (CH3)3N > CH3NH2 > (CH3)2NH
4.
How many ‘d’ electrons are present in Cr2+ ion?
(a) 4
(b) 5
5.
The E° for the cell reaction,
Cu2+(aq) + 2Ag(s) is 0.46 V, what is its equilibrium constant?
Cu(s) + 2Ag+(aq)
(c) 4 × 1015
(d) 1.56 × 1015
(a) 15.6
(b) 4 × 1016
6.
The time taken for 90% of a first order reaction to complete is approximately
(a) 1.1 times that of half-life
(b) 2.2 times that of half-life
(c) 3.3 times that of half-life
(d) 4.4 times that of half-life.
(c) 6
OR
Compound that is both paramagnetic and coloured is
(b) (NH4)2[TiCl6]
(a) K2Cr2O7
(d) K3[Cu(CN)4]
(c) VOSO4
(d) 3
In the following questions (Q. No. 7 and 8 ), a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
7.
Assertion : In complex [Cr(NH3)4BrCl]Cl, the ‘spin only’ magnetic moment is close to 2.83 B.M.
Reason: Mononuclear complexes of chromium(III) in strong field ligand have three unpaired electrons.
8.
Assertion : Hofmann degradation of benzamide gives aniline.
Reason : Hofmann bromamide degradation reaction can be used for descending amine series.
OR
Assertion : Ammonolysis of alkyl halides involves reaction between alkyl halides and alcoholic ammonia.
Reason : Ammonolysis of alkyl halides mainly produces 2° amines.
175
Practice Paper - 2
SECTION - B
The following questions, Q. No. (9 to 12) are short answer type and carry 2 marks each.
9.
Write the IUPAC names of the products (A) and (B) in the following reactions :
O
(a)
CH3COOH
NH3
D
SeO
2
A
(b)
B
10. Calculate the cell emf at 25°C for the following cell : Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.10 M) | Sn(s)
[Given : E°(Mg2+ / Mg) = – 2.34 V, E°(Sn2+/Sn) = – 0.136 V, 1 F = 96500 C mol–1].
11. Calculate the order of the reaction for the decomposition of N2O5 at 30°C from the following rate data.
S. No.
Rate of reaction (mol L–1 hr–1)
Concentration of N2O5 (mol L–1)
1.
2.
0.10
0.20
0.34
0.68
3.
0.40
1.36
OR
Hydrogen peroxide, H2O2(aq) decomposes to H2O(l) and O2(g) in a reaction that is first order in H2O2 and
has a rate constant k = 1.06 × 10–3 min–1. How long will it take for 15% of a sample of H2O2 to decompose?
12. Write two differences between physisorption and chemisorption.
OR
Write one difference between each of the following :
(i) Multimolecular colloid and macromolecular colloid
(ii) Sol and gel
SECTION - C
Q. No. (13 and 14) are short answer type II carrying 3 marks each.
13. Account for the following :
(i) Zirconium and hafnium exhibit almost similar properties.
(ii) Zinc salts are white while copper II salts are coloured [At. nos. Zn = 30, Cu = 29].
(iii) Europium (II) is more stable than cerium (II).
14. How will you convert :
(i) Ethanoic acid into methanamine
(ii) Ethanamine into methanamine
OR
Give reasons :
(i) Acetylation of aniline reduces its activation effect.
(ii) CH3NH2 is more basic than C6H5NH2.
(iii) Although —NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.
SECTION - D
Q. No. (15 and 16) are long answer type carrying 5 marks each.
15. (a)
(b)
(c)
Write the complete reaction for each of the following conversions stating the conditions necessary :
(i) Toluene to benzaldehyde
(ii) Aldehyde to acetal
Describe the preparation of acetic acid from acetylene.
How can the following be obtained from acetic acid?
(i) Acetaldehyde
(ii) Acetone
CBSE Board Term-II Chemistry Class-12
176
(a)
OR
Carry out the following transformations.
COOH
Br
(i)
(ii) HC
OH
CH
O
(b)
(c)
(iii) C6H5CH = CH2
C6H5CH2COOH
Out of nitrobenzoic acid and chlorobenzoic acid which one is a stronger acid and why?
The C – O bond in carboxylic acid is shorter than that in alcohol. Explain.
16. (a)The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant
volume :
Experiment
Time/s
Total pressure/atm
SO2(g) + Cl2(g)
SO2Cl2(g)
1
0
0.4
Calculate the rate constant.
2
100
0.7
(Given : log 4 = 0.6021, log 2 = 0.3010)
(b) The decomposition of NH3 on platinum surface
Pt
2 NH3( g ) → N2( g ) + 3 H2( g )
is a zero order reaction with k = 2.5 × 10–4 mol L–1 s–1. What are the rates of production of N2 and H2?
OR
(a)
(b)
1.
For a certain chemical reaction variation of ln[R] vs. time (s) plot is given below :
(i)
Predict the order of the given reaction.
(ii) What does the slope of the line and intercept indicate?
(iii) What is the unit of rate constant k?
A first order reaction takes 160 minutes time for 20% completion. Calculate time required for half
completion of reaction.
(i) (b) : NiCl2 + 2NH3 + 2
(ii) (c)
CH3
C
NOH
CH3
C
NOH
CH3
C
NOH
CH3
C
NO
to give ions in solution, but only Fe2(SO4)3 contains
Fe3+ ions. (NH4)2SO4·FeSO4·6H2O contains Fe2+ ions
not Fe3+ ions.
(d) : The acid does not contain a-hydrogen atom.
3. (c) : 2° amines are more basic than 1° and 3° amines.
Among the 2° amines, (b) and (c), (b) is less basic
since the lone pair of electrons on the nitrogen atom
is contributed towards the aromatic sextet formation.
Hence, piperidine, i.e., option (c) is correct.
2.
Ni
2
(iii) (b)
OR
(d) : It has no unpaired electrons hence, it is diamagnetic.
(iv) (d): (a) and (b) are coordination compounds
hence cannot give free Fe2+ or Fe3+ ions in solution.
(c) and (d) represent simple compounds hence are free
OR
(b) : In non-aqueous solvents and in vapour phase basic
strength of alkyl amines follows the order :
Tertiary amines > secondary amines > primary amines
177
Practice Paper - 2
4. (a) : Cr (Z = 24) → 3d 54s1 thus Cr2+ → 3d 4
i.e., No. of d-electrons = 4
OR
(c) : K2Cr2O7 contains Cr6+(3d0) which is diamagnetic
but coloured due to charge transfer spectra.
(NH4)2[TiCl6] contains Ti4+(3d0), which is diamagnetic
and colourless.
VOSO4 contains V4+(3d 1), which is paramagnetic and
coloured.
K3[Cu(CN)4] contains Cu+(3d 10), which is diamagnetic
and colourless.
0.059
log K c
5. (c) : E °cell =
2
0.46 × 2
or log K c =
= 15.6
0.059
\
Kc = 4 × 1015
6.
2.303
a
2.303
2.303
=
log
log 10 =
(c): t90% =
k
a − 0.9a
k
k
2.303
a
2.303
2.303
t1/2 =
=
× 0.3010
log
log 2 =
k
a −a/2
k
k
t90%
1
=
= 3.3 i.e., t90% = 3.3 times t1/2
t1/2 0.3010
7.
(d) : Cr3+ having 3d3 configuration always have
3 unpaired electrons with strong field as well as weak
field ligands with three unpaired electrons thus, the
magnetic moment is 3.83 B.M.
8. (a) : In this reaction the amine so formed contains
one carbon less than that present in the amide.
OR
(c)
9.
(a)
O
(b)
O
SeO2
O
(B)
Cyclohexane-1, 2-dione
10.
Mg(s)
2+
Sn (aq) + 2e–
Mg(s) + Sn2+(aq)
Mg2+(aq) + 2e– (Anodic half reaction)
Sn(s) (Cathodic half reaction)
Mg2+(aq) + Sn(s)
E°cell = E°Sn2+/Sn – E°Mg2+/Mg
= – 0.136 + 2.34 = 2.204 V
0.01 M
0.0591
log
2
0.10 M
°
Ecell = Ecell
−
= 2.24 – (0.0295 × –1) = 2.2335 V
11.
From the given data,
r1 = 0.10 mol L–1 hr–1 = k × (0.34 mol L–1)n
r2 = 0.20 mol L–1 hr–1 = k × (0.68 mol L–1)n
r3 = 0.40 mol L–1 hr–1 = k × (1.36 mol L–1)n
L−1 hr −1
r1 0.10 mol
=
r2 0.20 mol
or
t=
1  1
= 
2  2
L−1 hr −1
n
=
k(0.34 mol L−1 )n
k(0.68 mol L−1 )n
⇒ n =1
OR
[ A]0
2.303
log
[ A]
k
Given k = 1.06 × 10–3 min–1,
t=
2.303
1.06 × 10
−3
min
−1
log
100
85
[ A]0 100
=
[ A] 85
2303
[2 log 10 − log 85]min
1.06
2303
2303 × 0.0706
t=
[2 × 1 − 1.9294] =
1.06
1.06
t=
t = 153.39 min  153.4 min
12.
S.
Criteria
No.
(i) Specificity
Physisorption
Chemisorption
It is not
specific in
nature.
(ii) TempeIt decreases
rature
with increase
dependence in temperature.
Thus, low
temperature is
favourable for
physisorption.
(iii) Rever-sibility Reversible in
nature.
(iv) Enthalpy
Low enthalpy of
change
adsorption.
It is highly
specific in
nature.
It increases with
increase in
temperature.
Thus, high
temperature is
favourable for
chemisorption.
Irreversible in
nature.
High enthalpy of
adsorption.
(Any two)
CBSE Board Term-II Chemistry Class-12
178
OR
(i) The difference between multimolecular and
macromolecular colloids is :
Multimolecular
Colloids
When a large number of small
molecules or atoms (diameter
< 1 nm) of a substance
c ombi n e t o g e t h e r i n a
dispersion medium to form
aggregates, having size in the
colloidal range, the colloidal
solutions thus, formed are
known as multimolecular
colloids, e.g., gold sol,
sulphur sol, etc.
Macromolecular
Colloids
When substances which
possess very high
molecular
masses are dispersed
in suitable dispersion
medium, the colloidal
solutions thus,
formed are called
macromolecular
colloids, e.g., cellulose,
starch, etc.
(ii) The difference between sol and gel is :
Sol
Gel
Dispersed phase is solid
whereas dispersion
medium is liquid.
13. (i) D ue to lanthanoid contraction the elements
of 4d and 5d-series have similar atomic radii e.g.,
Zr = 160 pm and Hf = 159 pm. Thus Zr and Hf have
almost identical properties.
(ii) Zn 2+ ion has completely filled d-subshell and
no d-d transition is possible. So zinc salts are white.
Configuration of Cu2+ is [Ar] 3d 9. It has partially filled
d-subshell and hence it is coloured due to d-d transition.
(iii) Europium (II) has electronic configuration
[Xe]4f 7 5d 0 while cerium (II) has electronic
configuration [Xe] 4f 1 5d1. In Eu2+, 4f subshell is half
filled and 5d-subshell is empty. Since, half filled and
completely filled electronic configurations are more
stable, therefore Eu2+ ions is more stable than Ce2+ .
O
Ethanoic acid
(Acetic acid)
CH3 C Cl
O
CH3 C NH2
(ii) CH3 CH2
Ethanamine
K2Cr2 O7 / H+
CH3 CONH2
group having –I
which due to the presence of
effect, electron density on N-atom decreases and hence,
activation effect of aniline gets reduced.
(ii) CH 3NH 2 is more basic than C 6H 5NH 2 because
in aniline the lone pair of electrons on nitrogen are
involved in resonance.
(iii) Nitration is carried out with conc. HNO3 in the
presence of conc. H 2 SO 4 . In the presence of these
acids, the –NH2 group of aniline gets protonated and
is converted into –N+H3 group. This positively charged
group acts as a strong electron withdrawing and metadirecting group. Hence, the incoming electrophile goes
to m-position.
NH3
(ii)
(b)
(c) (i)
(ii)
Br 2 / KOH
NaNO2 / HCl
NH2
273 - 278 K
CH3 COOH
(i) After acetylation of aniline, acetanilide is formed in
15. (a) (i)
Dispersed phase is liquid
whereas dispersion
medium is solid.
14. (i) CH3COOH PCl5
OR
PCl5
OR
CH3 NH 2
Methanamine
Mg
ether
CH3 CH2 OH (a) (i)
CH3 COCl
Br2 / KOH
NH3
CH3 NH2
Methanamine
MgBr
Br
(ii) HC
COOH
1. CO2
2. H3O+
HBr
CH2
CH
H2C
CHMgBr
CHBr
Mg
ether
OH
1. CO2
2. H3O+
O
179
Practice Paper - 2
(iii) C6H5CH
CH2
1. B2H6/THF
2. H2O2/OH¯
C6H5CH2CH2OH
KMnO4/H+
C6H5CH2COOH
(b) Nitrobenzoic acid is more acidic than chlorobenzoic
acid due to greater –I effect of nitro group than chloro
group.
(c) Due to resonance structure of carboxylate ion the
C – O bond acquires some double bond character, due
to which its length is less than that in alcohol.
16. (a)The given reaction is
SO2Cl2(g)
SO2(g) + Cl2(g)
At t = 0
At time t
0.4 atm
(0.4 – x) atm
0
0
x atm
x atm
Total pressure at time t will be
PT = (0.4 – x) + x + x = 0.4 + x
x = (PT – 0.4)
Pressure of SO2Cl2 at time t will be
pSO2Cl2 = 0.4 – x = 0.4 – (PT – 0.4) = 0.8 – PT
At time t = 100 s, PT = 0.7 atm
\ pSO2Cl2 = 0.8 – 0.7 = 0.1 atm
According to first order kinetic equation
pSO Cl (initial)
2.303
2 2
k=
log
t
pSO Cl (after reaction)
2 2
=
2.303
 0.4 
log   = 1.3 × 10 −2 s −1
 0.1 
100
Pt
(b) 2 NH3( g ) N2( g ) + 3 H2( g )
k = 2.5 × 10–4 mol L–1 s–1
The order of reaction is zero i.e.,
Rate = k [Reactant]0
Rate = 2.5 × 10–4 × 1 = 2.5 × 10–4 mol L–1 s–1
d[N2 ] 1 d[H2 ]
\ Rate of reaction =
=
dt
3 dt
The rate of formation of N2 = 2.5 × 10–4 mol L–1 s–1
Again, 2.5 × 10 −4 =
1 d[H2 ]
3 dt
d[H2 ]
= 7.5 × 10 −4 mol L−1 s −1
dt
Therefore, rate of formation of H2
= 7.5 × 10–4 mol L–1 s–1
OR
(a) (i) First order
(ii) ln[R] = –kt + ln[R]0
C omparing this equation with y = mx + c, if we plot
ln[R] vs t, we get a straight line with slope = –k and
intercept = ln[R]0
(iii) For first order reaction, unit of k = s–1
(b)Given : [R]0 = 1, [R] = 0.80, t = 160 min
t1/2 = ?
[R]0
2.303
For first order reaction k =
log10
[R]
t
2.303
1
or, k =
log10
0.8
160 min
2
303
.
or, k =
× log 1.25
160 min
2.303 × 0.0969
or, k =
= 1.39 × 10 −3 min −1
160 min
∴
Again t1/2 =

0.693
0.693
=
min = 497 min
k
1.39 × 10 −3
PRACTICE PAPER 3
Time allowed : 2 hours
Maximum marks : 35
General Instructions : Read the following instructions carefully.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
There are 16 questions in this question paper. All questions are compulsory.
Section A : Q. No. 1 to 8 are objective type questions. Q. No. 1 is passage based question carrying 4 marks while
Q. No. 2 to 8 carry 1 mark each.
Section B : Q. No. 9 to 12 are short answer questions and carry 2 marks each.
Section C : Q. No. 13 and 14 are short answer questions and carry 3 marks each.
Section D : Q. No. 15 and 16 are long answer questions carrying 5 marks each.
There is no overall choice. However, internal choices have been provided.
Use of calculators and log tables is not permitted.
SECTION - A (OBJECTIVE TYPE)
1.
Read the passage given below and answer the following questions :
The half life of a reaction is the time required for the concentration of a given reactant to reach a value that is
arithmetic mean of its initial and final, or equilibrium, values. The half-life of a reaction has an exact quantitative
meaning only in the following cases:
(i)
For a first order reaction, t1/2 =
0.693
k
(ii)For a reaction involving more than one reactant, with their concentrations in their stoichiometric ratios. In
this case half-life of each reactant is same.
If the concentrations of reactants are not in their stoichiometric ratios, the half-life for the different reactants are
not the same.
–1
For second order reaction, t1/2 = k–1[A]0
The following questions are multiple choice questions. Choose the most appropriate answer :
(i)
In a first-order reaction A → B, if k is rate constant and initial concentration of the reactant A is 0.5 M, then the
half-life is
(a)
log 2
k
(b)
log 2
k 0. 5
(c)
ln 2
k
(d)
0.693
0.5k
(ii) 87.5% of the substance disintegrated in 45 minutes (first order reaction) what is its half life?
(a)
15 min
(b) 30 min
(c)
45 min
(d) 60 min
*The paper is for practice purpose. CBSE has yet not released the official sample paper.
So, the pattern is suggestive only. For latest information visit www.cbse.gov.in.
181
Practice Paper - 3
OR
Half-life period of a first order reaction is 1386 seconds. The specific rate constant of the reaction is
(a)
0.5 × 10–2 s–1
(b) 0.5 × 10–3 s–1
(c)
5.0 × 10–2 s–1
(d) 5.0 × 10–5 s–1
(iii) For a first order reaction, the ratio between the time taken to complete 3/4th of the reaction and time taken to
complete half of the reaction is
(a) t3/4 = 2t1/2
(b) t3/4 = t1/2
(c) t3/4 = 3t1/2
(d) t3/4 = 5t1/2
(iv) The half-life of a reaction is halved as the initial concentration of the reactant is doubled. The order of
reaction is
(a) 0.5
(b) 1
(c) 2
(d) 0
Following questions (Q. No. 2 to 6) are multiple choice questions carrying 1 mark each :
2.
Find the value of l°eq for potash alum.
Given : λ°
m(K + )
3.
= 73.5 Ω−1 cm2 mol −1 , λ°
m(Al 3+ )
= 189 Ω−1 cm2 mol −1 , λ°
m(SO24− )
(a)
145.6 W–1 cm2 eq–1
(b) 1165 W–1 cm2 eq–1
(c)
532 W–1 cm2 eq–1
(d) 195.5 W–1 cm2 eq–1
= 160 Ω−1 cm2 mol −1
The correct order of reactivity of aldehydes and ketones towards hydrogen cyanide is
(a)
CH3COCH3 > CH3CHO > HCHO
(b) CH3COCH3 > HCHO > CH3CHO
(c)
CH3CHO > CH3COCH3 > HCHO
(d) HCHO > CH3CHO > CH3COCH3
OR
In a set of reactions m-bromobenzoic acid gave a product D. Identify the product D.
COOH
SOCl2
B
NH3
C
Br
NaOH
Br2
D
A
SO2NH2
(a)
COOH
(b)
NH2
Br
NH2
(c)
Br
4.
CONH2
(d)
Br
Identify ‘X’.
C6H5COCl + (CH3)2NH
(a)
(c)
Pyridine
N, N-Dimethylbenzamide
N-Methyl-N-phenylamine
‘X’
(b) N, N-Dimethylbenzene
(d) N, N-Diphenylmethanamine
CBSE Board Term-II Chemistry Class-12
182
5.
A compound of vanadium has a magnetic moment of 1.73 BM. Choose the correct electronic configuration of
the vanadium ion in the compound.
(a)
1s2, 2s2 2p6, 3s2 3p6 3d2
(b) 1s2, 2s2 2p6, 3s2 3p6 3d3
(c)
1s2, 2s2 2p6, 3s2 3p6 3d1
(d) 1s2, 2s2 2p6, 3s2 3p6 3d0
OR
Which of the following pairs of ions have same paramagnetic moment ?
(a)
6.
Mn2+, Cu2+
(b) Cu2+, Ti3+
(c)
Ti4+, Cu2+
(d) Ti3+, Ni2+
Primary and secondary valency of platinum in the complex [Pt(en)2Cl2] are
(a)
4, 6
(b) 2, 6
(c)
4, 4
(d) 6, 4
In the following questions (Q. No. 7 and 8), a statement of assertion following by a statement of reason is given
choose the correct answer out of the following choices :
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
7.
Assertion : K
ohlrausch law helps to find the molar conductivity of weak electrolyte at infinite dilution.
Reason : Molar conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally.
OR
Assertion : When a copper wire is dipped in silver nitrate solution, there is no change in the colour of the
solution.
Reason : Copper can displace silver from its salt solution.
8.
Assertion : Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason : Those aldehydes which have a-H atom undergo Cannizzaro reaction.
SECTION - B
The following questions (Q. No. 9-12), are short answer type and carry 2 marks each.
9.
The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 is a first order reaction
(i) After 55 seconds at 400 K the pressure of Cl2O7 falls from 0.062 to 0.044 atm. Calculate the rate constant.
(ii) Calculate the pressure of Cl2O7 after 100 seconds of decomposition at this temperature.
10. Write the structures of the products formed in the following reactions :
O
CH2 — C — OCH3
(i)
NaBH4
O
(ii)
11. Explain the following :
(a) Low spin octahedral complexes of nickel are not known.
(b)
∆t =
4
∆
9 o
183
Practice Paper - 3
OR
Write the IUPAC nomenclature of the following complex along with hybridisation and structure.
K2[Cr(NO)(NH3)(CN)4], m = 1.73 BM
z is r = k[x]3/2 [y]–1/2
12. The rate law for the reaction, x + y
Find the order and molecularity of reaction.
OR
A substance A decomposes in solution following the first order kinetics. Flask I contains 1 litre of 1 M solution
of A and flask II contains 100 mL of 0.6 M solution of A. After 8 hours the concentration of A in flask I becomes
0.25 M. What will be the time for concentration of A in flask II to become 0.3 M?
SECTION - C
Q. No. 13 and 14 are short answer type II carrying 3 marks each.
13. (i)
Calculate the standard electrode potential of Ni2+/Ni electrode if emf of the cell
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s) is 0.059 V. [Given : E°
Cu2 + /Cu
= + 0.34 V]
(ii) Define the terms specific conductance and equivalent conductance.
OR
For M2+/M and M3+/M2+ systems, the E° values for some metals are as follows :
Cr2+/Cr : –0.9 V ; Cr3+/Cr2+ : –0.4 V
Mn2+/Mn : –1.2 V ; Mn3+/Mn2+ : +1.5 V
Fe2+/Fe : –0.4 V ;
Fe3+/Fe2+ : +0.8 V
Use this data to comment upon :
(i) The stability of Fe3+ in acid solution as compared to Cr3+ or Mn3+.
(ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or
manganese metal.
14. (a)Two moles of organic compound ‘A’ on treatment with a strong base give two compounds ‘B’ and ‘C’.
Compound ‘B’ on dehydrogenation with Cu gives ‘A’ while acidification of ‘C’ yields carboxylic acid ‘D’
having molecular formula of CH2O2. Identify the compounds A, B, C and D.
(b)Explain Hell–Volhard–Zelinsky reaction.
SECTION - D
Q. No. 15 and 16 are long answer type carrying 5 marks each.
15. (a)An organic compound A having molecular formula C2H7N on treatment with HNO2 gave an oily yellow
substance. Identify ‘A’. Give equation.
(b) Which amine in each of the following pairs is a stronger base? Give reason.
(i)
(c)
(ii) CH3CH2CH2NH2 and CH3NHCH2CH3
Account for the following :
(i) Tertiary amines do not undergo acylation reaction.
(ii) Amines are more basic than comparable alcohols.
CBSE Board Term-II Chemistry Class-12
184
OR
(a) Suggest a convenient scheme for separating aniline, N-methylaniline, toluene and phenol present together
in mixture. Distillation is not to be used.
(b)Identify A and B.
(c)
Account for the following :
(i)Ammonolysis of alkyl halides does not give a corresponding amine in pure state.
(ii)If – NO2 or – COOH group is attached to a carbon of benzene ring, electrophilic substitution becomes
difficult.
16. (a) Assign reason for the following :
(i) The enthalpies of atomisation of transition elements are high.
(ii)Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as transition element.
(b)What may be the possible oxidation states of the transition metals with the following d electronic
configurations in the ground state of their atoms :
3d34s2, 3d54s2 and 3d64s2. Indicate relative stability of oxidation states in each case.
OR
(a) How would you account for the following :
(i)The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those
of the corresponding members of the second (4d) series.
(ii)The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple or
Fe3+/Fe2+ couple.
(iii) The highest oxidation state of a Mn metal is exhibited in its oxide or fluoride.
(b) Which of following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+ (At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
1. (i) (c) : For a 1st order kinetics,
2.303
a
log
t
a−x
2.303
a
2.303
ln 2
⇒ t1/2 =
log
log 2 =
At t1/2, k =
a
k
k
t1/2
a−
2
(ii) (a) : a = 100 ; a – x = 100 – 87.5 = 12.5
0.693
⇒ k = 5 × 10–4 s–1 = 0.5 × 10–3 s–1
k
⇒ 1386 =
k=
2.303
100
2.303
a
log
log
=
= 0.046
45
12.5
t
a−x
0.693
0.693
=
t1/2 =
= 15 min
k
0.046
OR
k=
(b) : Given, t1/2 = 1386 s
For a first order reaction,
0.693
t1/2 =
(k = rate constant)
k
(iii) (a) : t1/2 =
t3/4 =
=
0 ⋅ 693
k
2 ⋅ 303
 a  2 ⋅ 303
log
=
log 4

3a 
k
k
 a − 
4
2 ⋅ 303
0 ⋅ 693 × 2
× 2 × 0.310 =
k
k
t3/4 0 ⋅ 693 × 2
k
=
×
⇒ t3/4 = 2t1/2
t1/2
k
0.693
(iv) (c) : For nth order reaction,
t1/2 ∝
1
an−1
;
(t1/2 )1 [a2 ]n−1 [a2 ] 
=
=
(t1/2 )2 [a1 ]n−1  [a1 ] 
n−1
185
Practice Paper - 3
On solving this equation we get, n = 1
So, vanadium atom must have one unpaired electron and
thus its configuration is
4+
2
2
6
2
6
1
23V : 1s , 2s 2p , 3s 3p 3d
n−1
t1/2
 2a 
=   ⇒ 2 = (2)n−1
1 / 2t1/2  a 
⇒ (n – 1) = 1 ⇒ n = 2
2. (a) : K2SO4⋅Al2(SO4)3⋅24H2O → 2K+(aq) + 2Al3+(aq)
+ 4SO2–4(aq)
Λ°m(Potash alum) = 2 λ° + + 2 λ° 3+ + 4 λ°
m(K )
m(Al )
m(SO24− )
= 2 × 73.5 + 2 × 189 + 4 × 160 = 1165 W–1 cm2 mol–1
Valency factor for potash alum = 8 (total positive
charge)
°
Λeq
(Potash alum) =
°
Λm
(Potash alum)
8
=
1165
8
–1
= 145.6 W cm2 eq–1
3. (d) : Alkyl group attached to carbonyl carbon
increases the electron density on the carbonyl carbon
and lowers its reactivity towards nucleophilic addition
reactions.
Also, as the number and size of alkyl group increases,
the attack of nucleophile on the carbonyl group becomes
more and more difficult due to steric hindrance.
Hence, the reactivity order will be :
HCHO > CH3CHO > CH3COCH3
OR
2+
(b) : Cu
OR
and Ti , both have one unpaired electron.
3+
6. (b) : Primary valency corresponds to oxidation
number while secondary valency corresponds to
coordination number.
7. (a)
OR
(d) : When a copper wire is dipped in silver nitrate
solution it turns blue. Cu is more reactive than Ag hence
displaces silver from its salt solution.
8. (c) : The aldehydes which have a-H atoms, undergo
aldol condensation.
9. (i) 2Cl2O7(g) → 2Cl2(g) + 7O2(g)
For 1st order reaction,
2.303
 P  2.303
 0.062 
= 6.2 × 10–3 s–1
log  0  =
log 
k=
 0.044 
P
t
55
(ii) k =
2.303
P 
log  0 
P
t
Here k = 6.2 × 10–3 s–1, t = 100 s, P0 = 0.062 atm
2.303
 0.062 
log 
∴ 6.2 × 10–3 =
 P 
100
0.062 6.2 × 10 −3 × 100
⇒ P = 0.033 atm
=
P
2.303
O
CH2 C OCH3
NaBH4
10. (i)
O
OH
or log
(c) :
NH3
CH3
O
C
CH2
O
4. (a) :
(ii)
5. (c) : Magnetic moment (m) =
n(n + 2) BM
(n = number of unpaired electrons)
Given that, m = 1.73 BM.
\ 1.73 =
n(n + 2)
11. (a) Nickel forms octahedral complexes mainly in +2
oxidation state which has 3d8 configuration. In presence
of strong field ligand also it has two unpaired electrons in
eg orbital.
CBSE Board Term-II Chemistry Class-12
186
Hence, it does not form low spin octahedral complexes.
(b) Number of ligands in tetrahedral geometry is 4
whereas in octahedral geometry it is 6.
In tetrahedral geometry no orbital lies directly in the
path of ligand whereas in octahedral geometry axial
orbitals interact directly with the ligand.
4
That is why, ∆t = ∆ o .
9
OR
µ = n (n + 2) = 1.73 which gives n = 1
This means that chromium ion has one unpaired
electron, i.e., it is present as Cr+ or Cr (I). This implies
that NO is present as nitrosonium ion. Hence, the name
will be potassium amminetetracyanidonitrosonium
chromate(I).
Cr+ :
3d
4s
In the complex, as there is only one unpaired electron
and coordination number is 6,
Cr+ :
0.059
(– log 10)
2
0.059
0.059
0.059 = E°cell +
⇒ E°cell = 0.059 –
2
2
0.059
= 0.0295 ≈ 0.03
\ E°cell =
2
0.059 = E°cell –
Now, E°cell = E°cathode – E°anode
0.03 = 0.34 – E°anode
E°anode = 0.34 – 0.03 = 0.31 V
Hence, E°Ni2+/Ni = + 0.31 V
(ii) Specific conductance : It is the conductance due to
ions present in 1 cm3 of electrolytic solution.
Equivalent conductance (Leq) : It is the conductance of
an electrolytic solution containing 1 gram-equivalent of
the electrolyte. The solution is contained in between two
electrodes which are 1 cm apart.
OR
is negative (–0.4 V), this means
(i) As E°
Cr3+ ions in solution cannot be reduced to Cr2+
easily, i.e., Cr3+ ions are very stable. As E°Mn3+/Mn2+
is more positive (+1.5 V) as compared to E°Fe3+/Fe2+
(+0.8 V), Mn3+ ions can easily be reduced to Mn2+ ions
in comparison to Fe3+ ions. Thus, the relative stability of
these ions is : Mn3+ < Fe3+ < Cr3+
(ii) Oxidation potentials for Cr, Mn and Fe will be
+0.9 V, +1.2 V and +0.4 V. Thus, the ease of getting
oxidised will be in the order, Mn > Cr > Fe.
Cr3+/Cr2+
d2sp3 hybridisation
it will undergo d2sp3 hybridisation to give octahedral
geometry.
2
NO
CN
NC
Cr
CN
NC
NH3
12.As the above reaction involves two species to form
product, thus the molecularity of the reaction is 2.
3  1
The order of reaction is = +  −  = 1
2  2
H
For flask I
2.303
2.303
1
2.303
k=
log
=
× 0.6021
log 4 =
8
0.25
8
8
For flask II
2.303 × 8
2.303
0. 6
log
× 0.3010 = 4 hours
=
0.3 2.303 × 0.6021
k
 Ni2+ 
0.059
log 
n
Cu2+ 
0.059
 0.01 
0.059 = E°cell –
log 
[Here, n = 2]
 0.1 
2
H C OH + H
H
(B)
H
H
Cu
C
OH 573 K HCHO
(A)
H
(B)
O
H C
–
+
H
+
HCOOH
O Na
(C )
13. (i) Ni(s) + Cu2+(aq) → Cu(s) + Ni2+(aq)
° −
Ecell = Ecell
NaOH
14. (a) 2HCHO
(A)
OR
t=
0.059
1
log  
 10 
2
0.059 = E°cell –
(D )
H
A : HCHO B : H C OH
H
C:H
O
C
ONa
D : HCOOH
O
C
–
+
O Na
(C )
187
Practice Paper - 3
(b) Carboxylic acids having α-hydrogen are halogenated
at the α-position on treatment with chlorine or bromine
in the presence of small amount of red phosphorus.
15. (a) The compound A gives a yellow oily substance on
treatment with HNO2. So, it must be 2°-amine. So, A is
(CH3)2NH.
(CH3)2 NH + HNO2 → (CH3)2N—N == O + H2O
(b) (i)
NH2
CH3 CH CH3
is
(A)
more
basic
than
CH3 CH COOCH3 because – COOCH3 is an electron
withdrawing group which decrease the electron density
on nitrogen atom.
(ii) 2° amines are more basic than 1° amines, because in
2° amine there are two electron releasing groups and in
1° amine only one electron releasing group is present, so,
CH3NHCH2CH3 is more basic than CH3CH2CH2NH2.
(c) (i) In tertiary amines there are no acidic hydrogen
due to which they do not undergo acylation reaction.
(ii) N being less electronegative than O gives lone pair of
electron more easily than O atom. Therefore, amines are
more basic than alcohols.
OR
(a) First the mixture is washed with HCl when aniline
and N-methylaniline are dissolved in it. Then the
remaining part is washed with NaOH, when phenol
dissolves leaving behind pure toluene. Phenol dissolved
in alkali is precipitated by adding HCl. The hydrochloric
acid solution of aniline and N-methyl aniline is treated
first with benzene sulphonyl chloride and then sodium
hydroxide. Benzene sulphonyl derivative of aniline
dissolves in NaOH. It is filtered off, the filtrate as well
as the residue are treated separately first with HCl and
then with KOH to obtain pure and N-methylaniline
separately.
(b)
(A)
(c) (i) If alkyl halide is in excess, the hydrogen atoms of
ammonia are successively replaced by alkyl group to form
primary, secondary and tertiary amines which further
react with alkyl halide to form quaternary ammonium
salt as follows :
(ii) Both –NO2 and –COOH groups are electron
withdrawing groups. They decrease the electron density
at the benzene ring and hence deactivate it towards
electrophilic substitution reactions.
16. (a) (i) As transition metals have a large number of
unpaired electrons in the d-orbitals of their atoms they
have strong interatomic attractions or metallic bonds.
Hence they have high enthalpy of atomization.
(ii) Scandium (Z = 21) has incompletely filled 3d-orbitals
in the ground state (3d1). Hence it is considered as a
transition element.
(b) The possible oxidation states for 3d34s2 = +5, +4,
+3, +2.
The possible oxidation states for 3d54s2 = +7, +6, +5, +4,
+3, +2
The possible oxidation states for 3d64s2 = +6, +4, +3, +2.
In a transition series the oxidation states which lead to
exactly half filled or completely filled d-orbitals are more
stable.
OR
(a) (i) This is due to lanthanoid contraction.
(ii) Much larger third ionisation energy of Mn(where
change is d5 to d4) is mainly responsible for this. This also
explains that +3 state of Mn is of little importance
and from the relation, DG° = –nFE°.
More positive
is the value of E°, reaction
will be feasible.
+ e–
+ e–
Mn2+ ;
Fe3+
Fe2+
Mn3+
3d4 3d5 3d5 3d6
more stable
(half filled)
more stable
(half filled)
Hence, E°value for Mn3+/Mn+2 couple is much more
positive than that for Fe3+/Fe2+.
(iii) Manganese can form pp-dp bond with oxygen
by utilising 2p-orbital of oxygen and 3d-orbital of
manganese due to which it can show highest oxidation
state of +7. While with fluorine it cannot form such
pp-dp bond thus, it can show a maximum of +4 oxidation
state.
(b) Only those ions will be coloured which have partially
filled d-orbitals facilitating d-d transition. Ions with d0
and d10 will be colourless.
From electronic configuration of the ions, V3+(3d2) and
Mn2+(3d5) are coloured while Ti4+(3d0) and Sc3+(3d0)
are colourless.

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