केंद्रीय विद्यालय संगठन, एरणाकुलम संभाग
KENDRIYA VIDYALAYA SANGATHAN
ERNAKULAM REGION
STUDENT SUPPORT MATERIAL
CHEMISTRY (043)
CLASS XII
SESSION 2021-22 (TERM-2)
STUDENT SUPPORT MATERIAL
CHEMISTRY
OUR PATRON
Shri. R Senthil Kumar
Deputy Commissioner
KVS RO, Ernakulam Region
OUR MENTORS
Smt. Deepti Nair
Assistant Commissioner
KVS RO, Ernakulam
Shri. Santhosh Kumar N
Assistant Commissioner
KVS RO, Ernakulam
CO-ORDINATOR
Shri. G Sasikumar
Principal, Kendriya Vidyalaya Kollam
CONTENT DEVELOPMENT TEAM
S
No
1.
2.
3.
4.
5.
6.
7.
Name of the
Chapter
Electrochemistry
Name of the Contributors
Mrs. SWAPNA SASIKALA
Mrs. RAJANI P
Mr. PURUSHOTHAMAN V
Mrs. REMADEVI EDWIN
Mr. N K LAL
Chemical Kinetics
Ms. P S ANITHA
Mrs. MARIETTE P SEBASTIAN
Ms. MANJULA P
Mr. SUMESH M S
Surface Chemistry
Mrs SUDHA PILLAI
Mrs. SUCHITRA SURENDRAN
Mrs. BIJI KURIAKOSE
d & f- Block Elements Mrs. MAYA REGHURAJAN
Mrs. O VIMA
Mrs. GRACE GEORGE K
Mr. SREENIVASAN S V
Mr. SIBU JOHN
Coordination
Mr. SHADANANAN P A
Compounds
Mr. MEENA D
Aldehydes, Ketones Mr. PRAMOD N T
and Carboxylic Acids Mr. SUBIN THOMAS
Mrs. SULEKHA RANI R
Mr. PARTHIBAN G
Mrs. SHYLA P
Amines
Mrs. DIVYA S
Mr. PRAVEEN KUMAR V
Mr. AJITH KUMAR K
Name of the KV
KV No.1 CPCRI, Kasargod
KV No.1 Calicut
KV No.1, Calicut
KV No.2, Naval Base
KV CRPF Palllipuram
KV Pallipuram
KV RB Kottayam
KV Kannur
KV Port Trust
KV Pangode
KV Adoor (Shift-2)
KV No.2 Naval Base
KV Pattom (Shift-1)
KV No.2, Calicut
KV No.1, Naval base
KV Ernakulam
KV Kollam
KV NAD Aluva
KV Pangode
KV Payyannur
KV Chenneerkara
KV INS Dronacharya
KV Kavaratti
KV Port Trust
KV Adoor (Shift-2)
KV Kanhangad
KV Adoor (Shift-2)
CONTENT REVIEW TEAM
S No
1
Name of the teacher
Ms Shyla P
Name of the KV
KV Port Trust
2
Mr Sumesh M S
KV Port Trust
3
Ms Raji K J
KV Port Trust
4
Mr Sibu John
KV Kollam
INDEX
CONTENT
PAGE
No.
ELECTROCHEMISTRY
Gist of the Lesson
1
Multiple choice questions
6
Assertion-Reason type of questions
8
Case based questions
10
Descriptive questions
11
CHEMICAL KINETICS
Gist of the Lesson
24
Multiple choice questions
30
Assertion-Reason type of questions
38
Case based questions
41
Descriptive questions
43
SURFACE CHEMISTRY
Gist of the Lesson
64
Multiple choice questions
72
Assertion-Reason type of questions
81
Case based questions
84
Descriptive questions
87
d AND f BLOCK ELEMENTS
Gist of the Lesson
95
Multiple choice questions
107
Assertion-Reason type of questions
113
Case based questions
116
Descriptive questions
119
COORDINATION COMPOUNDS
Gist of the Lesson
123
Multiple choice questions
128
Assertion-Reason type of questions
131
Case based questions
134
Descriptive questions
136
ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
Gist of the Lesson
148
Multiple choice questions
159
Assertion-Reason type of questions
166
Case based questions
169
Descriptive questions
171
AMINES
Gist of the Lesson
195
Multiple choice questions
200
Assertion-Reason type of questions
207
Case based questions
211
Descriptive questions
213
SYLLABUS FOR SESSION 2021-22 CLASS XII Term-II
S.No
UNIT
1 Electrochemistry
2
3
4
5
Chemical Kinetics
Surface Chemistry
d-and f-Block Elements
Coordination Compounds
6 Aldehydes, Ketones and Carboxylic Acids
7 Amines
TOTAL
No. of
Periods
7
5
5
7
8
10
7
49
MARKS
13
9
13
35
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 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
Term II: 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.
TERM-II Evaluation Scheme
S. No
1.
2.
3
4
Practical
Volumetric Analysis
Salt Analysis
Content Based Experiment
Project Work and Viva(Internal and External Both)
TOTAL
Marks
4
4
2
5
15
1) Volumetric analysis (4 marks)
Determination of concentration/ molarity of KMnO 4 solution by titrating it against a
standard solution of:
i. Oxalic acid,
ii. Ferrous Ammonium Sulphate
(Students will be required to prepare standard solutions by weighing themselves).
2) Salt analysis (Qualitative analysis) (4 marks)
Determination of one cation and one anion in a given salt.
Cations- Pb2+, Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2+, Zn2+, Co2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4+
Anions – (CO3)2- , S2- , NO2 - , SO3 2- , SO42- , NO3-, Cl- , Br- , I- , PO4 3- , C2O4 2- ,CH3COO- (Note:
Insoluble salts excluded)
3) Content based experiment
A. Preparation of Inorganic Compounds
Preparation of double salt of Ferrous Ammonium Sulphate or Potash Alum.
Preparation of Potassium Ferric Oxalate.
B.
Tests for the functional groups present in organic compounds:
Unsaturation, alcoholic, phenolic, aldehydic, ketonic, carboxylic and amino (Primary)
groups.
1
CHAPTER 3
ELECTROCHEMISTRY
GIST OF THE LESSON

Electrochemistry is the branch of chemistry which deals with the relationship between electrical
energy and chemical energy and inter-conversion of one form into another.

Oxidation is defined as a loss of electrons while reduction is defined as a gain of electrons.

In a redox reaction, both oxidation and reduction reaction take place simultaneously.

Direct redox reaction: In a direct redox reaction, both oxidation and reduction reactions take
place in the same vessel. Chemical energy is converted to heat energy in a direct redox reaction.

Indirect redox reaction: In indirect redox reactions, oxidation and reduction take place in
different vessels.
In an indirect redox reaction, chemical energy is converted into electrical energy. The device
which converts chemical energy into electrical energy is known as an electrochemical cell.

An electrochemical cell consists of two metallic electrodes dipped in electrolytic solutions. The
cells are of two types:
(a) Electrolytic cells (b) Galvanic cells

In a Galvanic cell:
1. The half-cell in which oxidation takes place is known as oxidation half-cell
2. The half-cell in which reduction takes place is known as reduction half-cell.
3. Oxidation takes place at anode which is negatively charged and reduction
takes place at cathode which is positively charged.
4. Transfer of electrons takes place from anode to cathode while electric current
flows in the opposite direction.
5. An electrode is made by dipping the metal plate into the electrolytic solution
of its soluble salt.
6. A salt bridge is a U-shaped tube containing an inert electrolyte in agar-agar
and gelatin.

Salt bridge: A salt bridge maintains electrical neutrality and allows the flow of electric current
by completing the electrical circuit.

Representation of an electrochemical cell:
Anode(s)/Anodic electrolyte(a)//Cathodic electrolyte(aq)/Cathode(s)
1. Anode is written on the left while the cathode is written on the right.
2
2. Anode represents the oxidation half-cell and is written as: Metal/Metal ion
(Concentration).
3. Cathode represents the reduction half-cell and is written as: Metal ion
(Concentration)/Metal
4. Salt bridge is indicated by placing double vertical lines between the anode and
the cathode

Electrode potential E is the potential difference that develops between the electrode and its
electrolyte. The separation of charges at the equilibrium state results in the potential difference
between the metal and the solution of its ions. It is the measure of tendency of an electrode in the
half cell to lose or gain electrons.

Standard electrode potential (E0): When the concentration of all the species involved in a half
cell is unity, then the electrode potential is known as standard electrode potential.

Types of electrode potential: There are 2 types of electrode potentials namely,
 Oxidation potential: It is the tendency of an electrode to lose electrons or get oxidized.
 Reduction potential: It is the tendency of an electrode to gain electrons or get reduced.
Oxidation potential is the reverse of reduction potential.

According to the present convention, standard reduction potential is called standard electrode
potential.

The electrode having a higher reduction potential have higher tendency to gain electrons
and so it acts as a cathode whereas the electrode having a lower reduction potential act as
an anode.

The standard electrode potential of an electrode cannot be measured in isolation.

According to convention, the Standard Hydrogen Electrode is taken as a reference electrode and
it is assigned a zero potential at all temperatures.

SHE: Standard hydrogen electrode consists of a platinum wire sealed in a glass tube and
carrying a platinum foil at one end. The electrode is placed in a beaker containing an
aqueous solution of an acid having 1 Molar concentration of hydrogen ions. Hydrogen
gas at 1 bar pressure is continuously bubbled through the solution at 298 K. The
oxidation or reduction takes place at the Platinum foil.
The standard hydrogen electrode can act as both anode and cathode.
If the standard hydrogen electrode acts as an anode:

If the standard hydrogen electrode acts as a cathode:

In the electrochemical series, various elements are arranged as per their standard reduction
potential values.

A substance with higher reduction potential value means that it has a higher tendency to get
reduced. So, it acts as a good oxidising agent.
3





A substance with lower reduction potential value means that it has a higher tendency to get
oxidised. So, it acts as a good reducing agent.
The electrode with higher reduction potential acts as a cathode while the electrode with a lower
reduction potential act as an anode.
The potential difference between the 2 electrodes of a galvanic cell is called cell potential and is
measured in Volts.
Standard cell potential is the difference between the standard reduction potential of cathode and
anode.
E0cell = E0Cathode – E0Anode
The cell potential is the difference between the reduction potential of cathode and anode.
E cell = E cathode – E anode

Cell potential is called the electromotive force of the cell (EMF) when no current is drawn from the
cell.

Nernst equation: Nernst formulated a relationship between standard electrode potential EΘ and
electrode potential E.

Electrode potential increases with increase in the concentration of the electrolyte and decrease in
temperature.

Nernst equation when applied to a cell, it helps in calculating the cell potential.

At equilibrium, cell potential Ecell becomes zero.

4

Relationship between equilibrium constant Kc and standard cell potential EΘcell:

Work done by an electrochemical cell is equal to the decrease in Gibb’s energy
If E0cell is positive, ΔG° would be negative and reaction would be spontaneous.
If E0cell is negative, ΔG° would be positive and the reaction would be non-spontaneous.

The substances which allow the passage of electricity through them are known as conductors.
Every conducting material offers some obstruction to the flow of electricity which is called
resistance. It is denoted by R and is measured in ohm.
 The resistance of any object is directly proportional to its length l and inversely proportional to its
area of cross section A.
where ρ is called specific resistance or resistivity.
 The SI unit of specific resistivity is ohm metre.
 The inverse of resistance is known as conductance, C
 Unit of conductance is ohm-1 or mho. It is also expressed in Siemens denoted by S.
 The inverse of resistivity is known as conductivity. It is represented by the symbol
.
 The SI unit of conductivity is Sm-1. But it is also expressed in Scm-1. 
 Conductivity = Conductance × Cell constant 
 Conductivity cell: The problem of measuring the resistance of an ionic solution can be resolved
by using a source of alternating current and the second problem is resolved by using a specially
designed vessel called conductivity cell. 
 A conductivity cell consists of 2 Pt electrodes coated with Pt black. They have area of cross
section A and are separated by a distance ‘l’. Resistance of such a column of solution is given
by the equation:
Where l/A is called cell constant and is denoted by the symbol
 Molar conductivity of a solution: It is defined as the conducting power of all the ions produced
by dissolving 1 mole of an electrolyte in solution
.
Molar conductivity
5
Where = Conductivity and M is the molarity Unit of Molar conductivity is Scm2 mol-1


For measuring the resistance of an ionic solution, there are 2 problems:
1. Firstly, passing direct current changes the composition of the solution
2. Secondly, a solution cannot be connected to the bridge like a metallic wire or a solid
conductor.
 Kohlrausch’s Law of independent migration of ions: According to this law, molar
conductivity of an electrolyte, at infinite dilution, can be expressed as the sum of individual
contributions from its individual ions. 
 If the limiting molar conductivity of the cations is denoted by
then the limiting molar conductivity of electrolyte is:
and that of the anions by
,
Molar conductivity,
Where v+ and v- are the number of cations and anions per formula of electrolyte
 Degree of dissociation: It is ratio of molar conductivity at a specific concentration ‘c’ to the
molar conductivity at infinite dilution. It is denoted by. 
 Dissociation constant:
Where Ka is acid dissociation constant, ‘c’ is concentration of electrolyte, α is degree of ionization.
 Faraday constant: It is equal to charge on 1 mol of electrons. It is equal to 96487 C mol-1 or
approximately equal to 96500 C mol-1.
 Faraday’s first law of electrolysis: The amount of substance deposited during electrolysis is
directly proportional to quantity of electricity passed.
 Faraday’s second law of electrolysis: If same charge is passed through different electrolytes,
the mass of substance deposited will be proportional to their equivalent weights.
 Products of electrolysis: The products of electrolysis depend upon the nature of electrolyte
being electrolyzed and the nature of electrodes.
If electrode is inert like platinum or gold, they do not take part in chemical reaction i.e., they
neither lose nor gain electrons.
If the electrodes are reactive then they will take part in chemical reaction and products will be
different as compared to inert electrodes.
 Some of the electrochemical processes although feasible but slow in their rates at lower voltage,
these require extra voltage, i.e., over voltage at which these processes will take place.
 The products of electrolysis also differ in molten state and aqueous solution of electrolyte.
6
MULTIPLE CHOICE QUESTIONS
1. Point out the incorrect statement
a) An oxidizing agent is that which can release electrons
b) on undergoing oxidation, the oxidation state of a substance increases
c) reduction is the process of gain of electron by a substance
d) reducing agent releases electrons
2. An electrochemical cell behaves like an electrolytic cell when
b) Ecell= 0
c) Eexternal⦢Ecell
d) Eexternal ≤Ecell
a) Ecell= Eexternal
3. In an electrochemical process, salt bridge is used as
a) reducing agent
b) oxidizing agent
c) complete the circuit to complete the reaction d) none
4. The redox reaction to proceed spontaneously in a given direction, the emf should have
a) positive value
b) negative value
c) zero
d) a and b
5 The oxidation potential of hydrogen electrode is taken as
a) Zero
b) half
c) one
d) negative
6. The reduction potential of 4 elements A, B, C and D are 1.4V, -0.44V, -0.68V and 0.48V, the
reactivity is maximum for
a) A
b) B
c) C
d) D
7 The reduction potential of Cu/Cu2+ and Ni/Ni2+ are 0.34 and -0.25 volts respectively. The emf of the
cell is
a) 0.58v
b) 0.59 V
c) 0.09
d) none
8. An electrochemical cell stops working after some because
a) Electrode potential of both the electrodes decreases
b) one electrode is eaten up
c) reduction potential of one electrode becomes equal to that of other
d) all the above
9. In an electrochemical cell, electrical energy
a) has to be supplied
b) is released
b) may be released or absorbed
d) neither supplied or released
10. The EMF of the for the following, E0 (Mg2+/Mg) -2.37 V, Ag/Ag+ 0.80 V is
a) -1.57V
b) -3.17V
c)3.17V
d) none
11. The mf obtained by coupling electrode with SHE was 1.37 V. If SHE was the positive terminal,
the potential of the electrode was
a) +1.37V
b) -1.37V
c) 0
d) none
12. In SHE, the hydrogen gas is at a pressure of
a) 2atm
b) 0.5atm
c) 1atm
13. The cell constant is
d) 1.5 atm
7
a) l/a
b) a/l
c) k/R
d) none
14. In an electron transfer reaction, the substance losing electrons is said to be
a) oxidized
b) reduced
c) neither a nor b
15. The unit of equivalent conductance is
a) ohm-1cm-1
b) ohm-1cm2equ-1
c) ohm-1cm-2
d) dissociated
d) none
16. Effect of dilution on conduction is as follows
a) specific conductance decreases, molar conductance decrease
b) specific conductance decreases, molar conductance increase
c) both increases with dilution
d) no change
17. Which of the following constitute Daniel cell
a) Zn-Ag cell
b) Cu-Ag cell
c) Zn-Cu cell
d) none
18. The negative terminal of the electrochemical cell is
a) Cathode
b) anode
c) neither anode nor cathode
d) unpredictable
19. In a galvanic cell
a) electrical energy is used to produce chemical change
b) chemical energy is used to produce electrical energy
c) chemical energy is used to produce light
d) no chemical change occurs
20 The chemical change in galvanic cell is
a) Oxidation
b) oxidation and reduction
21. Ohms is the unit of
a) Resistance
b) potential difference
c) reduction
c) quantity of electricity
d) dissociation
d) current
22) In the reaction Cu + H2SO4 --------⦢ CuSO4 +H2
a) Copper is oxidized
b) copper acts as oxidizing agent
c) copper gains electrons and becomes Cu2+
d) H2SO4 acts as reducing agent
23) If limiting molar conductivity of Ca2+ and Cl– are 119.0 and 76.3 S cm2 mol-1, then the value of
limiting molar conductivity of CaCl2 will be
a) 195.3 S cm2 mol-1
b) 271.6 S cm2 mol-1
c) 43.3 S cm2 mol-1
d) 314.3 S cm2 mol-1
24) The standard emf of a galvanic cell involving cell reaction with n = 2 is formed to be 0.295 V at
25° C. The equilibrium constant of the reaction would be
a) 1.0 × 1010
b) 2.0 × 1011
c) 4.0 × 1012
8
d) 1.0 × 102
[Given F = 96500 (mol-1); R = 8.314 JK-1 mol-1]
25) A 0.5M NaOH solution offers a resistance of 31.6 ohm in a conducting cell at room temperature.
What shall be the approximate molar conductance of this NaOH solution if the cell constant of the
cell is0.367 cm-1
a) 234 Scm2 mole-1
b) b)23.2 Scm2 mole-1
c) 4645 Scm2 mole-1
d) 5464 Scm2 mole-1
ANSWERS
1 a 2c 3c 4a 5 a 6c 7 b 8c 9 b 10c 11b 12c 13 a 14 a 15 a 16b 17c 18 b 19b 20b 21a 22a 23b
24 d 25d
ASSERTION-REASON TYPE OF QUESTIONS
The question given below consist of an assertion and a reason use the following key to choose
appropriate answer
a) Both assertion and reason are correct and reason is the correct explanation of the assertion
b) Both assertion and reason are correct and reason is not the correct explanation of the assertion
c) Assertion is correct but reason is incorrect
d) Assertion is wrong Reason is correct.
1) Assertion: The electrode potential of SHE is zero
Reason: In SHE HCl 1M and H2 gas at one bar pressure is taken
2) Assertion: H+ ion cannot oxidize copper
Reason: Reduction potential of Cu2+/ Cu is greater than H+/H
3) Assertion: The reduction potential of F-/F is highest among all electrodes
Reason: Fluorine is the strongest oxidising agent
4) Assertion: Electronic conduction decreases with temperature
Reason: The flow of electrons hindered on increasing the temperature
5) Assertion: conductivity decreases with increasing dilution
Reason: No of ions increases per unit volume increases with dilution.
6) Assertion: Electrolytic conduction increases with temperature
Reason: On increasing the temperature mobility of ion increases
9
7) Assertion: Molar conductivity of electrolytes decreases with dilution
Reason: For weak electrolytes degree of dissociation increases with dilution.
8) Assertion It is difficult to measure the conductivity of ionic solutions
Reason: Electrolytes conduct electricity and undergoes chemical change.
9) Assertion Molar conductivity of strong electrolytes increases with dilution
Reason: On dilution inter ionic interaction increases.
10) Assertion: Molar conductivity of acetic acid increases sharply with dilution
Reason: Degree of dissociation of acetic acid decreases with dilution
11) Assertion: An external potential of 1.1V is passed through Daniel cell, no current flow through it.
Reason: Standard emf of galvanic cell is 1.1 volt.
12) Assertion: Out of Li and K Potassium is the strongest reducing agent
Reason: Reduction potential of K is greater than that of Li
13) Assertion: Electrolytic conduction of electrolytes depends on size of ions
Reason: Larger the size of ion lesser will be the mobility of ions.
14) Assertion: Limiting molar conductivity of weak electrolytes can be obtained graphically
Reason: Limiting molar conductivity of weak electrolytes increases with dilution
15) Assertion: Electrolysis of aqueous NaCl produces Oxygen at the anode
Reason: Oxidation potential of Oxygen is lower than chlorine.
16) Assertion: e m f of Daniel cell increases with increase in concentration of Cu2+ ions and decrease
in concentration of Zn2+ ion
Reason: Emf of cells depends on the conc of ions in anode and cathode.
17) Assertion: limiting molar conductivity of an electrolyte is the sum of contribution of cation and
anion
Reason: Electrolytic conduction is due to the movement of ions
18) Assertion: calcium can be used to reduce Na+ ions
Reason: Electrode potential of Na is higher than Calcium
19) Assertion: The cell constant of a conductivity cell depends upon the nature of the material of the
electrodes
Reason: The electrodes of the cell are coated with platinum black to avoid polarization effects
20) Assertion: Daniel cell become dead after some time
Reason: oxidation potential of zinc anode decreases and that of copper increases
21) Assertion: Copper liberates H2 from a dilute solution of HCl
Reason: Hydrogen is above copper in the electrochemical series
10
22) Assertion: For the Daniel cell Zn/Zn2+ // Cu2+/Cu with E0 cell = 1.1 Volt the application of opposite
potential greater than 1.1 V results in to flow of electrons from cathode to anode
Reason: Zn is deposited at the cathode and Cu is dissolved at cathode.
23) Assertion: 0.1M NH4OH AT 250 C has more conductance than at 500C
Reason: Molar conductance of weak electrolytes increases with Temperature.
24) Assertion:Limiting molar conductivity of weak electrolytes can be calculated using Kohlrausch law
Reason: At low concentrations conductivity of the solution is so low that it cannot be measured.
25) Assertion: Limiting molar conductivity of KCl increases slightly with dilution and reaches
maximum value.
Reason: On dilution inter ionic interaction decreases and ions are free to move.
CASE BASED QUESTIONS
26) Both conductivity and molar conductivity changes with the concentration. Conductivity always
decreases with concentration for both weak and strong electrolytes. This is because the no of ions per
unit volume that carry the current in solution decreases on dilution. Molar conductivity decreases with
increase in concentration. This is because the total volume V of solution containing one mole of
electrolyte also increases. Molar conductivity can be defined as the conductance of the electrolyte
solution kept between the electrodes of a conductivity cell at unit distance but having large area of cross
section large enough to accommodate sufficient volume of solution that contains one mole of electrolyte.
For strong electrolytes molar conductivity increases slowly and reaches maximum value. For weak
electrolytes. it increases sharply with dilution and reaches maximum value.
Read the passage given above. A statement and reason are given below. Choose the correct answer
from the option given below.
a) Both assertion and reason are correct and reason is the correct explanation of the assertion
b) Both assertion and reason are correct and reason is not the correct explanation of the assertion
c) Assertion is correct but reason is incorrect
d) Assertion is wrong Reason is correct.
1) Assertion: conductivity of strong electrolytes decreases with increase in concentrations
Reasons: No of ions per unit volume decreases with dilution
2) Assertion: Molar conductance is the conductance offered by one molar solution
Reason: decrease in conducted on dilution of a solution is more than compensated by the increase in
volume.
3) Assertion: Molar conductivity increases with decrease in concentration of the electrolyte.
Reason: on decreasing the conc the ions move faster as inter ionic interaction decreases.
11
4) Assertion: For Weak electrolytes Molar conductivity increases sharply and reaches a maximum
value.
Reason.: Degree of dissociation increases and more ions are produced.
5) Assertion: The molar conductivity of strong electrolyte is high when compared to weak
electrolytes. At a particular concentration.
Reason: For strong electrolytes almost, complete dissociation takes place in aqueous medium.
ANSWER KEY
1-b 2) a 3) b 4) a 5) c 6) a 7) d 8) a 9) c 10) c 11) b 12) d 13) a 14) b 15) d 16) a 17) b
18) a 19) d 20) a 21) d 22) b 23) d 24) a 25) a 26) 1) a 2) d 3) b 4) a 5) a
SHORT ANSWER QUESTIONS
1. When can an electrochemical cell behave like an electrolytic cell?
Ans: Eext > Ecell
2. Find quantity of charge in faraday is required to obtain one mole of aluminium from Al3+.
Ans: 3F
3. What will happen during the electrolysis of aqueous solution of CuSO4 by using platinum electrodes?
Ans: The reaction with lower value of E° will be preferred at anode, hence o2 is released at anode
4. What will happen during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?
Ans: Copper will dissolve at anode
Copper will deposit at cathode
5. On which factors the molar conductivity of ionic solution depends?
Ans: Molar conductivity of ionic solution depends on temperature and concentration of electrolytes in
solution.
6. What is limiting molar conductivity?
Ans: It is the molar conductivity when the concentration approaches zero.
7. Define the term molar conductivity
Ans: Molar conductivity is defined as the conducting power of all the ions produced by one gram mole
of an electrolyte in a solution.
8. What are the units of specific conductance?
Ans: ohm-1 cm-1 or Scm-1.
12
9. How is cell constant calculated from conductance values?
Ans: Cell constant = specific conductance/observed conductance.
10. What is the effect of decreasing concentration on the conductivity of an electrolyte.?
Ans: Conductivity of an electrolyte decreases with decrease in concentration.
11. What is meant by Faraday’s constant?
Ans: Faraday’s constant is the quantity of electricity carried by one mole of electrons.
1 F = 96500 C/mol
12. What does negative value of Ecell indicate?
Ans: ΔG will be positive and the reaction is non spontaneous
13. Write 1 use of Kohlrausch’slaw?
Ans: To evaluate the molar conductivity of any electrolyte at infinite dilution or zero concentration
which cannot be experimentally measured
14. What is cell constant?
Ans: Cell constant is defined as the ratio of the distance between conductance-titration electrodes to the
area of electrodes
Cell constant = l/a= length of the conductor/area of cross section
15. Describe the characteristics of variation of molar conductivity with dilution for (a) weak and (b) strong
electrolytes.
Ans: For strong electrolytes, molar conductivity slowly increases with dilution.
For weak electrolytes, molar conductivity increases steeply on dilution
16. Write the reaction taking place at cathode and anode during the operation of Daniel cell.
Ans: Zn - Cu cell is known as Daniel cell. The reactions taking place are described as under:
Zn (s)→Zn2+ + 2e- (Anode)
Cu2+ + 2e- →Cu (s) (Cathode)
Zn (s) + Cu2+→Zn2+ + Cu (s) (Overall reaction)
17. Two half-cell reactions of an electrochemical cell are given below:
MnO4–(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e–, E° = + 0.15 V
Construct the redox equation from the two half-cell reactions and predict if this reaction favours
formation of reactants or product shown in the equation.
Ans: The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
[Sn2+ → Sn4+ (aq) + 2e– ] × 5 E° = + 0.15 V
At cathode (reduction) :
[MnO4-(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4-(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
13
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.
18. Express the relation among the cell constant, the resistance of the solution in the cell and the
conductivity of the solution. How is the conductivity of a solution related to its molar conductivity?
Ans:(1/R) ×(l/a) = Conductance (G) × Cell constant(G*) =K (Specific Conductance)
Molar conductance: (Λm) = (K×1000)/c =(Kx1000)/M(molarity)
19. The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1. Calculate
the conductivity of this solution.
Ans: C = 1.5 M, Λm = 138.9 S cm2 mol-1
Λm = (K×1000)/c
∴K = (Λm×C)/1000= (138.9×1.5)/1000 = 0.20835 S cm-1
20. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity.
Ans: Molar conductivity Λm = (1000×κ)/M
Given: K = 0.025 S cm-1, M = 0.20 M
Hence, Λm = (0.025×1000)/0.20
∴ Λm = 125 S cm2 mol-1
21. The standard electrode potential (E°) for Daniel cell is +1.1 V. Calculate the ΔG° for the reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(1 F = 96500 C mol-1).
Ans: We know, ΔG° = -nF E°cell
Given: E°cell = 1.1 volt
∴ ΔG° = -2 × 96500 C mol-1 × 1.1 volt
= -212300 CV mol-1
= -212300 J mol-1 = -212.3 KJ mol-1
22. i) 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.
14
22 ii) State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution
decrease with dilution?
Ans: Kohlrausch law of independent migration of ions states that the limiting molar conductivity can be
represented as the sum of the individual contribution of the anion and cation of the electrolyte.
Λ°m for AxBy = xλ°+ + yλ°–
For acetic acid Λ° (CH3COOH) = λ°CH3COO– + λ°H+
Λ°(CH3COOH) = Λ° (CH3COOK) + Λ° (HCl) – Λ° (KCl)
23. Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the
solution of CuSO4.
(Molar mass of Cu = 63.5 g mol-1,1 F = 96500 C mol-1)
Ans: CuSO4 → Cu+ + SO42Cu2+ + 2e– → Cu
63.5 gram of copper is deposited = 2 × 96500 C
1.27 gram of Cu is deposited = [(2×96500)/63.5 ]× 1.27
= I × t (Q = I × t)
t = (2×96500×1.27)/63.5×2 = 1930 seconds.
24. Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Λ m) is 39.05 S cm2 mol-1.
Given: λ°(H+) = 349.6 S cm2 mol-1 and λ°(CH3COO–) = 40.9 S cm2 mol-1
15
Ans:Λ°m(HAC) = λ°H+ + λ°AC–
= λ°CH3COOH = λ°H+ + λ°CH3COO–
= 349.6 S cm2 mol-1 + 40.9 S cm2 mol-1
= 390.5 S cm2 mol-1
25.
Calculate the emf of the following cell at 25°C : Ag(s) | Ag+ (10-3 M) || Cu2+ (10-1 M) | Cu(s) Given
E0cell = +0.46 V and log 10n = n.
Ans: As Fe + 2H+ → Fe2+ + H2 (n = 2)
According to Nernst equation
26. Conductivity of 2.5 × 10-4 M methanoic acid is 5.25 × 10-5 S cm-1. Calculate its molar conductivity
and degree of dissociation.
Ans: Given: λ0(H+) = 349.5 Scm2 mol-1
and λ0(HCOO–) = 50.5 Scm2 mol-1. (All India 2015)
Answer:
Concentration is 2.5 x 10-4 M
K = 5.25 × 10-5 Scm-1.
Λcm= (K×1000)/Concentration
27. Three electrolytic cells A, B, C containing solution of ZnSO 4, AgNO3 and CuSO4 respectively all
connected in series. A steady current of 1.5 amperes was passed through then until 1.45 g of silver
16
deposited at the cathode of cell B. How long did the current flow? What mass of copper and of
zinc were deposited?
Ans:
28. Calculate the emf of the following cell at 298 K: Fe(s)Fe2+ (0.001 M) | | H+ (1 M) | H2(g) (1bar),Pt(s)
(Given E°cell = + 0.44 V)
Ans: (Given E°cell = + 0.44 V)
17
= 0.44 V + 0.0885 V = 0.5285 v
29. Write an expression for the molar conductivity of acetic acid at infinite dilution according to
Kohlrausch law. (b) Calculate ⋀°m for acetic acid. Given that ⋀° m (HCl) = 426 S cm2 mol–1 ⋀°m
(NaCl) = 126 S cm2 mol–1 ⋀°m (CH3COONa) = 91 S cm2 mol–1
Ans:
18
30. 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 Eo Ag+/Ag=+0.80V, Eo Cu2+/Cu=0.34V)
Ans:
31. (a)How many moles of mercury will be produced by electrolysing 1.0 M. Hg(NO 3)2 solution with a
current of 2.00 A for 3 hours? (b) A voltaic cell is set up at 25° C with the following half-cells Al3+
(0.001 M) and Ni2+ (0.50 M). Write an equation for the reaction that occurs when the cell
generates an electric current and determine the cell potential.
Ans: (a)Mass of mercury produced at the cathode,
19
Applying Nernst equation to the above cell solution
32. (a) Calculate the standard free energy and equilibrium constant of the reaction:
2Al(s) + 3Cu2+(aq) →2Al3+(aq) + 3Cu(s)
E0cell=2.02V
(b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of
unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of
Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the
concentration of AgNO3 used.
20
(b) Half-cell reactions:
E0Zn2+/Zn=−0.76V, E0Ag+/Ag = + 0.80 V
33. Predict the products of electrolysis in each of the following.
A dilute solution of H2SO4 with platinum electrodes.
Ans: The dissociation of sulphuric acid gives protons and sulphate ions.
At cathode, either hydrogen ions or water molecules can be reduced. Since protons have higher reduction
potential than water, hydrogen ions are reduced to hydrogen gas.
At anode, either sulphate ions or water molecule can get oxidized. Since, during oxidation of sulphate ions, more
bonds are broken than oxidation of water molecules, sulphate ions have lower oxidation potential than water.
Hence, water is oxidized at the anode to liberate oxygen molecules.
H2SO4→2H++SO4−
H2O→H++OH−
Reaction at cathode;
2H++2e−→H2
Reaction at anode;
Due to platinum electrode, self ionization of water will take place.
H2O→2H++1/2O2+2e−
21
Hence, hydrogen gas will generate at cathode and oxygen gas will generate at anode.
34. Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution: Ag+(aq)
+ e- → Ag(s) E° = +0.80 V
H+(aq) + e- → 1/2 H2(g) E° = 0.00 V On the basis of their standard
reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
Ans:Ag+(aq) + e- → Ag(s) ; E° = +0.80 V
H+(aq) + e- → 1/2 H2(g) ;
E° = 0.00 V On the basis of their standard reduction potential (E°) values,
cathode reaction is given by the one with higher E° values. –
→ Ag(s) reaction will be more feasible at cathode.
Thus Ag+(aq) + e- → Ag(s) reaction will be more feasible at cathode.
35. Conductivity of 0.00241 M acetic acid solution is 7.896 x10-5 S cm-1. Calculate its molar conductivity
in this Solution if λ0m acetic acid is 390.5 Scm2 mol-1 What would be its dissociation constant.?
22
36. The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905 x 10-5 S cm-1. Calculate its molar
conductivity and degree of dissociation (α). Given λ° (H+) = 349.6 S cm2 mol-1 and i0 (CH3COO-) = 40.9
S cm2 mol-1
Ans:
37. An (a) Define molar conductivity of a solution and explain how molar conductivity changes with change
in concentration of solution for a weak and a strong electrolyte.
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω . What is
the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10–3 S cm–1 ?
Ans:(a) Molar conductivity of a solution at a given concentration is the conductance of the volume V of
the solution containing one mole of electrolyte kept between two electrodes with area of cross section A
and distance of unit length.
where k is the conductivity and V is the volume of solution containing one mole of the electrolyte and C
is the molar concentration. Molar conductivity increases with decrease in concentration or increase in
dilution as the number of ions as well as the mobility of ions increases with increase in dilution. For
23
strong electrolytes, the number of ions does not increase appreciably on dilution and only mobility of
ions increases due to decrease in interionic attractions. Therefore, Am increases a little as shown in
figure by a straight line. For weak electrolytes, number of ions as well as mobility of ions increases on
dilution which results in a very large increase in molar conductivity, especially at infinite dilution
, Cell constant = (b) R = 1500 Ω ; k = 0.146 x 10–3 s cm–1 Substituting the values in the expression,
Cell constant = k × R
Cell constant = 0.146 x 10 –3 S cm–1 x 1500Ω = 0.219 cm–1
38. Calculate the equilibrium constant for the equilibrium reaction
Fe(s) + Cd2+(aq) ⇌ Fe2+(aq) + Cd(s)
(Given : E0Cd2+/Cd = – 0.40 V, E0Fe2+/Fe = -0-44V)
E0Fe2+/Fe = -0-44V)
24
CHAPTER 4
CHEMICAL KINETICS
GIST OF THE LESSON
Rate of a Chemical Reaction: It is the change in concentration of a reactant or product in unit time (or) it
is the rate of change of concentration of reacting species.
Consider a reaction,
R →P
Rate of a Chemical Reaction = −[𝑅] =  [𝑃]
− 𝑡
𝑡
Since, Δ[R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to
make the rate of the reaction a positive quantity.
Units of rate of a reaction: mol L-1s–1.
NOTE: Rate of reaction always expressed for every one mole.
Consider a general reaction: a A + b B  c C + d D, then
Rate of reaction
=
- 1 [ Rate of disappearance of A]
= - 1 [ Rate of disappearance of B]
𝑎
=
1
𝑐
𝑏
[ Rate of disappearance of C]
=
1
𝑑
[ Rate of disappearance of D]
Factors Influencing Rate of a reaction: (i) effect of concentration of reactants (pressure in case of gases),
(ii) effect of temperature and (iii) effect of catalyst.
Average Rate
Instantaneous Rate
It is the change in concentration of reactants in a It is the change in concentration of reactants at
given interval of time
instantaneous time
Rate Law Expression: It is a mathematical expression in which rate of reaction is expressed in terms of
molar concentration of reactants with each term raised to power, which may or may not be equal to the
stoichiometric coefficient of the reacting species in a balanced chemical equation.
Consider a general reaction: a A + b B  c C + d D, then
Rate Law = k [A]x [B]y,
where,
‘x’ may/ may not be equal to ‘a’ and ‘y’ may/ may not be equal to ‘b’
Order of a Reaction: It is the sum of powers of concentrations of reactants expressed in rate law.
Consider a general reaction: a A + b B  c C + d D, then
Rate Law = k [A]x [B]y, overall order of reaction = (x+y) and x and y represent the order with respect to
the reactants A and B respectively.
Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero order reaction means that the rate of
reaction is independent of the concentration of reactants.
Units of rate constant (k)
REACTION
ORDER
UNITS OF RATE CONST ANT
Zero order reaction
0
mol L-1S-1
first order reaction
1
S-1
Second order reaction
2
mol-1 L S-1
nth order of reaction
nth
mol(1-n)L(n-1)S-1
25
NOTE: To know the order of reaction when unit of rate constant is given, then just add 1 to the number
given in power of litre factor.
Molecularity of a Reaction: The number of reacting species (atoms, ions or molecules) taking part in an
elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called
molecularity of a reaction.
Difference between order of reaction and molecularity of reaction:
Order of reaction
1 It is the sum of powers of
concentrations
of
reactants
expressed in rate law.
Molecularity of reaction
The number of reacting species (atoms, ions or
molecules) taking part in an elementary reaction, which
must collide simultaneously in order to bring about a
chemical reaction
an Molecularity of a reaction is a theoretical quantity.
2 Order of a reaction is
experimental quantity.
3 It can be zero and even a fraction
Molecularity cannot be zero or a non-integer.
4 Order is applicable to elementary Molecularity is applicable only for elementary
as well as complex reactions
reactions.
Integrated Rate Equations
(i) Zero Order Reactions: Consider zero order reaction
R  products
Let [R]o be the initial concentration of reactant and [R] be the final concentration at time‘t’.
For zero order reaction, the rate of the reaction is proportional to zero power of the concentration of
reactants.
Rate = − 𝑑[𝑅] = k [R ]o

Rate = − 𝑑[𝑅]=k

d[R] = – k dt
𝑑𝑡
𝑑𝑡
Integrating on both sides
[R] = – k t + I, where, I is the constant of integration.
when t = 0, the concentration of the reactant R becomes [R]0, where [R]0 is initial concentration of the
reactant.
[R]0 = –k × 0 + I
 [R]0 = I
 [R] = –kt + [R]0
Comparing with equation of a straight line, y = mx + c, a graph is drawn between [R] against t, it gives a
straight line with slope = –k and intercept equal to [R]0.
Some enzyme catalysed reactions and reactions which occur on metal surfaces (metal catalyst) are a few
examples of zero order reactions.
26
(ii) First Order Reactions
Consider first order reaction
R P
Let [R]o be the initial concentration of reactant and [R] be the final concentration at time‘t’.
For zero order reaction, the rate of the reaction is proportional to the first power of the concentration of the
reactant R.
Rate = −
𝑑[𝑅]
𝑑𝑡
𝑑[𝑅]
=kR 
𝑅
= −kdt
Integrating on both sides,
ln [R] = – kt + I, I is the constant of integration.
When t = 0, R becomes [R]0, where [R]0 is the initial concentration of the reactant.
Therefore, ln [R]0 = –k × 0 + I  ln [R]0 = I
Substituting the value of I, ln[R] = – kt + ln[R]0
Rearranging this equation,
Or, k = 1 ln [𝑅]𝑜
𝑅
𝑡
ln
 k=
[𝑅]
[𝑅]𝑜
2.303
𝑡
log
= kt
[𝑅]𝑜
𝑅
A graph is drawn between log [R] against ‘t’ gives a straight line with slope = –k and intercept equal to log
[R]0
Half-Life of a Reaction: It is the time required to reduce the concentration of reactant to half of its initial
concentration.
Case 1: For a zero order reaction, rate constant is given by equation
K=
[R0 − R]
𝑡
when t = t ½ , then [R] = ½ [R0]
The rate constant at
t½
becomes
k =
[R0 −1 R0]
2
𝑡1
2

t½ =
[𝑅𝑜]
2𝑘
t½ for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely
proportional to the rate constant.
Case 2: For the first order reaction,
k=
2.303
𝑡
log
[𝑅𝑜]
[𝑅]
At t = t ½ , then [R] = ½ [R0]  k =
t½ =
2.303
𝑡
log 2 
t½ =
0.693
2.303
𝑡
log
[𝑅𝑜]
[𝑅𝑜]/2
𝑘
For a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the
reacting species.
For zero order reaction t ½ 𝖺 [R]0.
For first order reaction t ½ is independent of [R]0.
In general for n th order, t ½ = 1/ [R] o(n-1)
27
Pseudo First Order Reaction: A reaction which appear to follow higher order but follows lower order
kinetics.
Effect of temperature on rate of a reaction
Rate of a reaction always increases with increase of temperature
(i) According to Kinetic theory of Molecular gases, the average kinetic energy of reacting gaseous
molecules is directly proportional to absolute temperature.
K.E  T  ½ mv2  T
As temperature increases, speed of reacting molecules increases. Thereby, the frequency of collisions also
increases and hence rate of reaction increases.
(ii) Experimental evidence shows that for every 10 degree rise in
temperature, the rate of a reaction will be doubled.
(iii) If ‘E’ is the energy barrier, then the number of molecules that crosses
over the energy barrier is more at higher temperature than at low
temperature.
Dependence of temperature on rate of a reaction – Arrhenius Equation (Deleted for CBSE 22)
k = A e-Ea / RT
-Ea /RT
e
represents the fraction of molecules with energies equal to or greater than Ea.
where A is the Arrhenius factor or the frequency factor (or pre-exponential factor, a constant specific to a
particular reaction. The number of collisions per second per unit volume of the reaction mixture is known
as collision frequency). R is gas constant and Ea is activation energy measured in J mol –1 .
Taking natural logarithm of both sides of equation
ln k = –Ea/RT + ln A
2.303log k = –Ea/RT + 2.303 log A
log k = –Ea
+ log A
2.303 𝑅 𝑇
Let k1 & k2 be the rate constant for a reaction at temperature T1 & T2, then
–Ea
At temperature T1,  log k1 =
+ log A
At temperature T2,  log k2 =
2.303 𝑅 𝑇1
–Ea
2.303 𝑅 𝑇2
+ log A
(Since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and
T2 respectively.
–Ea
–Ea
 log k2 – log k1 =
 log 𝑘1 =
[ 1 - 1 ]
–
𝐸𝑎
2.303 𝑅 𝑇1
2.303 𝑅 𝑇2
𝑘2
2.303 𝑅
𝑇1
𝑇2
Activated complex (Transition State theory): It is the reaction intermediate possessing energy that
correspond to top of energy barrier.
28
Activated complex state is highly energy state and hence it is highly unstable state. It is always reversible
state.
Consider a reaction:
H2 (g) + I2 ( g ) → 2HI ( g )
If the total energy of reacting species are equal or higher than threshold energy then the reaction proceed in
forward direction and hence gives the products otherwise it retain as reactants only.
Effect of Catalyst: A catalyst is a substance which alters the rate of a reaction without itself undergoing
any permanent chemical change.
Catalyst help to increase the rate of chemical reaction. It carry the reaction through the path of lower
activation energy.
Collision Theory:
1. A reaction is possible when all reacting molecules come closer to each other and then collide.
2. All collisions may not result into products. Only those collisions which result into products are called
‘effective collisions’ or ‘fruitful Collisions’.
3. In a reaction, all colliding molecules must have minimum energy called ‘Threshold Energy’ (It is the
minimum energy required by the reacting molecules to get the products).
4. Most of the reacting molecules have energy less than Threshold energy. Therefore, additional energy
required by the reacting molecules to attain Threshold energy is called Activation energy.
Activation Energy = Threshold Energy – average energy of reacting molecules
5. For effective collisions, all reacting molecules must be properly oriented.
29
To get the products, all reacting molecules have to cross over the energy barrier. Only those reacting
molecules whose total energy is equal or greater than Threshold energy can only cross over the energy
barrier and hence gives the product.
Important Concepts
Changes taking place in a
Chemical Reaction RàP
Rate of change of
concentration of reactants
Rate of change of
concentration of reactants
Rate of reaction
Instantaneous Rate
Average Rate
Rate law / Rate equation /
Rate expression
For aA + bB → cC + dD
Order of a Reaction
Elementary reactions
Complex reactions
Unit of rate constant
Molecularity of a Reaction
Rate Determining Step
1) Reactants decreases
2) Products increases
Decrease in concentration of reactant/time taken
Increase in concentration of Product/time taken
Rate of the slowest change
Change in concentration of a reactant or product in unit time. Unit =>
mol L-1s–1
Rate of a reaction in a very small time period
Rate of the reaction when the time period is large.
Average of all the instantaneous rate during the time period
Rate law is the expression in which reaction rate is given in terms of
molar concentration of reactants with each term raised to some power,
which may or may not be same as the stoichiometric coefficient of the
reacting species in a balanced chemical equation.
The rate expression for this reaction is :
Rate α [A]x [B]y
=> Rate = k [A]x [B]y k is the Rate const. (x ,
y from expts)
The sum of powers of the concentration of the reactants in the rate law
expression is called the order of that chemical reaction.
If Rate = k [A]x [B]y Order = x + y
The reactions taking place in one step are called Elementary Reactions
The reactions taking place in more than one step are called Elementary
Reactions
For n th order reaction: mol1-n Ln-1 s-1
The number of reacting species taking part in an elementary reaction,
which must collide simultaneously in order to bring about a chemical
reaction is called molecularity of a reaction (values are limited from 1
to 3)
The overall rate of the reaction is controlled by the slowest step in a
reaction called the rate determining step
30
Half-life (t1/2)
Pseudo First Order
Reaction
Activation Energy
The half-life of a reaction is the time in which the concentration of a
reactant is reduced to one half of its initial concentration. It is
represented as t1/2.
When one of two reactants are present in large excess, a second order
reaction will follow first order kinetics
Eg-During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of
water the concentration of water does not get altered much during the
course of the reaction and the reaction behaves as first order reaction.
Such reactions are called pseudo first order reactions.
Activation energy is given by the energy difference between activated
complex and the reactant molecules
MULTIPLE CHOICE QUESTIONS
1. 1. If the rate of a reaction is expressed by, rate = k [A]² [B], the order of reaction will be
(a) 2
(b) 3
(c) 1
(d) 0
2.I n a reaction, 2X → Y, the concentration of X decreases from 0.50 M to 0.38 M in 10 min. What
is the rate of reaction in Ms-1 during this interval ?
(a) 2 × 10-4
(b) 4 × 10-2
(c) 2 × 10-2
(d) 1 × 10-2
3.The decomposition of dimethyl ether is a fractional order reaction. The rate is given by rate
=k(PCH3OCH3)3/2. If the pressure is measured in bar and time in minutes, then what are the units of
rate and rate constant?
(a) bar min-1,
bar2 min-1
(b) ba r min-1,
ba r -1/2 min-1
(c) bar1/2 min-1, bar2 min-1
(d) bar min-1,
bar1/2 min-1
4A reaction in which reactants (R) are converted into products (P) follows second order kinetics. If
concentration of R is increased by four times, what will be the increase in the rate of formation of
P?
a) 9 times
(b) 4 times
(c) 16 times
(d) 8 times
5. The unit of rate constant for the reaction 2H2 + 2NO → 2H2O + N2
which has rate = K|H2||NO|², is
31
(a) mol L-1 s-1
(b) s-1
(c) mol-2 L² s-1
(d)mol L-1
6. The unit of rate and rate constant are same for
(a) zero order reaction
(b) first order reaction
(c) second order reaction
(d) third order reaction
7. 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
8. The order of reaction is decided by
a) temperature
(b) mechanism of reaction as well as relative concentration of reactants
(c) molecularity
(d) pressure
9. Rate constant in case of first order reaction is
a) inversely proportional to the concentration units
(b) independent of concentration units
(c) directly proportional to concentration units
(d) inversely proportional to the square of concentration units
10. Half-life period of a first order reaction is 10 min. What percentage of the reaction will be
completed in 100 min?
a) 25%
(b) 50%
(c) 99.9%
(d) 75%
11. 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
12. A first order reaction takes 40 min for 30% decomposition. What will be t12?
32
(a) 77.7 min
(b) 52.5 min
(c) 46.2 min
(d) 22.7 min
13. A first order reaction has a rate constant 1.15 × 10-3s-1. How long time will 5 g of this reactant
take to reduce to 3 g?
a) 444 s
(b) 400 s
(c) 528 s
(d) 669 s
14. A plot of log (a – x) against time t is a straight line. This indicates that the reaction is of
a) zero order
(b) first order
(c) second order
(d) third order
15. For a chemical reaction, X + 2Y → Z, if the rate of appearance of Z is 0.50 moles per litre per
hour, then the rate of disappearance of Y is
a) 0.5 mol L-1 hr-1
(b) 1.0 mol L-1 hr-1
(c) 0.25 mol L-1 hr-1
(d) cannot be predicted
16. The rate of a reaction is primarily determined by the slowest step. This step is called
(a) rate determining step
(b) activation step
(c) reaction rate step
(d) none of these.
17. For a chemical reaction A → B, it is found that the rate of the reaction quardruples when the
concentration of A is doubled. The rate expression for the reaction is, rate =k[A] n where the value
of n is
(a) 1
(b) 2
(c) 0
(d) 3
18The half-life period of any first order reaction
(a) is half the specific rate constant
(b) is always the same irrespective of the reaction
(c) is independent of initial concentration
(d) in directly proportional to initial concentration of reactants.
33
19. A hypothetical reaction 2p + q → s + r has rate constant as 2.0 × 10 -3 mol L-1 s-1. The order of
the reaction is
a) unpredictable
(b) zero
(c) one
(d) two
20Consider the reaction A —> B. The 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?
21 Which of the following statement is not correct about order of a reaction?
(a) The order of a reaction can be a fractional number.
(b) Order of a reaction is experimentally determined quantity.
(c) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the
balanced chemical equation for a reaction.
d) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law
expression.
22. For a first order gas phase reaction: A(g)→2B(g)+C(g). P0 be initial pressure of A and Pt, the total
pressure at time 't′. Integrated rate equation is:
34
23.
24. Compounds ′A′′and ′B′′react according to the following chemical equation. A(g)+2B(g)→2C(g)
Concentration of either ‘A’ or ‘B ’ were changed keeping the concentrations of one of the reactants
constant and rates were measured as a function of initial concentration. Following results were
obtained. Choose the correct option for the rate equations for this reaction.
25. What are the units of k for the rate law: Rate= k[A][B]2, when the concentration unit is mol/L-1 ?
(a) s-1
(b) s
(c) L mol-1 s-1
(d) L2 mol-2 s-1
(e) L2 s2 mol-2
26. For the reaction, 2H2S(g) + O2(g)
is absolutely true?
2S(s) + 2H2O(l), which one of the following statements
(a) The reaction is first order with respect to H2S and second order with respect to O2.
(b) The reaction is fourth order overall.
(c) The rate law is: rate = k[H2S]2[O2].
d) The rate law cannot be determined from the information given.
27. The half-life for a first-order reaction is 32 s. What was the original concentration if, after 2.0
minutes, the reactant concentration is 0.062 M?
(a) 0.84 M
(b) 0.069 M
(c) 0.091 M
(d) 0.075 M
(e) 0.13 M
35
28. Suppose the reaction: A + 2B
Step -1
A+B  AB
AB2 occurs by the following mechanism:
SLOW
STEP 2 AB + B  AB2 FAST
OVERALL REACTION
A+2B  AB2
The rate law expression must be Rate =
.
(a) k[A]
(b) k[B]
(c) k[A][B]
(d) k[B]2
(e) k[A][B]2
29. The decomposition of carbon disulfide, CS2, to carbon monosulfide, CS, and sulfur is first order with
k = 2.8 x 10-7 s-1 at 1000oC
CS2
CS + S What is the half-life of this reaction at 1000oC?
a) 5.0 x 107 s
(b) 4.7 x 10-6 s
(c) 3.8 x 105 s
(d) 2.5 x 106 s
30 A zero order reaction X →Product , with an initial concentration 0.02M has a half life of 10 min.
if one starts with concentration 0.04M, then the half life is
a) 10 s
b) 5 min
c) 20 min
d) cannot be predicted using the given information
31. For a first order reaction A → product with initial concentration x mol L−1 , has a half life
period of 2.5 hours . For the same reaction with initial concentration (x/2) mol L−1 the half life is
a)
b)
c)
d)
1.25 hrs
2.5 hrs
5 hrs
Cannot be predicted
32. For a first order reaction, the rate constant is 6.909 min-1.the time taken for 75% conversion in
minutes is
36
33. In a first order reaction x → y ; if k is the rate constant and the initial concentration of the
reactant x is 0.1M, then, the half life is
34. The rate constant of a reaction is 5.8 × 10−2 s−1 . The order of the reaction is
a) First order
b) zero order
c) Second order
d) Third order
35 For the reaction N2O5(g) → 2NO2 (g) + ½ O2(g), the value of rate of disappearance of N2O5 is
given as 6.5 × 10-3mol L-1 s-1. The rate of formation of NO2 and O2 is given respectively as
a) (3.25 × 10−2 mol L−1s−1 ) and (1.3 × 10−2 mol L−1s−1 )
b) (1.3 × 10−2 mol L−1s−1 ) and ( 3.25 × 10−2mol L−1s−1 )
c) (1.3 × 10−2 mol L−1s−1 ) and ( 3.25 × 10−3 mol L−1s−1 )
d) None of these
36. During the decomposition of H2O2 to give dioxygen, 48 g O2 is formed per minute at certain
point of time. The rate of formation of water at this point i
a) 0.75 mol min−1
b) 1.5 mol min−1
c) 2.25 mol min−1
d) 3.0 mol min−1
37 Which of the following observations is incorrect about the order of a reaction?
a. Order of a reaction is always a whole number
b. The stoichiometric coefficient of the reactants doesn’t affect the order
c. Order of reaction is the sum of power to express the rate of reaction to the concentration terms of the
reactants.
d. Order can only be assessed experimentally
38 The half life period of 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 × 10-3 s-1
39. The rate constant of a reaction A → B is 0.6 × 103 mole per second. If the concentration of [A] is
5 M, then what will be concentration of [B] after 20 minutes
(a) 0.36 M
(b) 0.72 M
(c) 1.08 M
(d) 3.60 M
37
40 A first order reaction has specific reaction rate 10-2s-1. How much time it will take for 20g of
reactant to reduce to 5g?
a) 138.6 s
b) 346.5 s
c) 693.0 s
d) 238.6 s
41. If the initial concentration of reactant is doubled, t1/2 is also doubled, the order of reaction is
a) Zero
b) 1
c) 2
d) 3
42. If conc, of reactant ‘A’ is increased 10 times and rate of reaction becomes 100 times. What is
order with respect to ‘A’?
a) zero
(b) 1
(c) 2
(d) 3
43. The half life of the first order reaction having rate constant K = 1.7 x 10-5s-1 is
a) 12.1 h
b) 9.7 h
c) 11.3 h
d) 1.8 h
44 RCOOR’ +H2O(EXCESS )  RCOOH + R-OH . What type of reaction is this?
a) Second order
b) Unimolecular
c) Pseudo-unimolecular
d) Third order
45. Which out of the following does not represents pseudo-order reaction.
a) Acid hydrolysis of ester
b) Inversion of cane sugar
c) Decomposition of PCI5
d) Conversion of methyl acetate into methanol and acetic acid in presence
46The rate constant of a reaction has same units the rate of the reaction. The reaction is of.
a)
b)
c)
d)
first order
zero order
second order
third order
47. If the concentration of a reactant A is doubled and the rate of its reaction increased by a factor
of 2, the order of reaction with respect to A is...
a) first
38
b) zero
c) third
d) second
48 For a reaction nx+ y → z the rate of reaction becomes twenty seven times when the
concentration of X is increased three times. What is the order of the reaction?
a)
b)
c)
d)
2
1
3
0
49 Radioactive disintegration is an example of
a) zero order reaction
b) first order reaction
c) second order reaction
d) third order reaction
50. The average rate and instantaneous rate of a reaction are equal
(a) at the start
(b) at the end
(c) in the middle
(d) when two rates have a time interval equal to zero
ANSWER KEY
1.b
2.a
3. b
4.c
5.c
6.a
7.b
8.b
9.b
10.c
11.c
12.a
13.a
14.b
15.b
16.a
17.b
18.c
19.b
20.b
21.c
23.c or
iii
33.c
24.a
25.d
26.d
27.a
28.c
29.d
30.c
31.b
22.b or
ii
32.b
34.a
35.c
36.d
37.a
38.b
39.b
40.a
41.a
42.c
43.c
44.c
45.c
46.b
47.a
48.c
49.b
50.d
ASSERTION-REASON TYPE AND CASE BASED QUESTIONS
A statement of assertion is followed by a statement of a reason. Mark the correct
choice from the options given below:
A) Both assertion and reason are true and reason is the correct explanation of
assertion.
B) Both assertion and reason are true but reason is not the correct explanation of
assertion.
C) Assertion is true but reason is false.
D) Assertion is false and reason is true
39
1
Assertion: Hydrolysis of methyl ethanoate is a pseudo first order reaction.
Reason: Water is present in large excess and therefore its concentration remained
constant throughout the reaction.
2
Assertion: Order of a heterogenous catalytic reaction can be zero.
Reason: In the case of heterogeneous catalysis, the reaction becomes independent of
concentration at high concentration of the reactants.
3
Assertion: The slowest elementary step in a complex reaction decides the rate of the
reaction.
Reason: The slowest elementary step always has the smallest molecularity.
4
Assertion: A catalyst increases the rate of a reaction.
Reason: The catalyst increases the activation energy which in turn increases the rate
of the reaction.
Assertion: Activated complex for the forward reaction will have lower energy than
that for the backward reaction in an exothermic reaction.
Reason: Reactants have greater energy than products for an exothermic reaction
5
6
Assertion: Increase in temperature increases rate of reaction.
Reason: More colliding molecules will have energy greater than threshold energy.
7
Assertion: Increase in concentration of reactant will not change the rate for a zeroorder reaction.
Reason: Rate constant for a zero-order reaction is a constant for a particular initial
concentration
8
Assertion: The half-life of a reaction is independent of initial concentration for a first
order reaction.
Reason: The integrated rate law for a first order reaction is independent of initial
concentration.
9
Assertion: Half-life of a reaction can be used to predict order of a reaction.
Reason: The relationship between half-life and initial concentration of the reactant is
dependent on order.
10
Assertion: Order of a reaction with respect to any reactant can be zero, positive,
negative or fractional.
Reason: Rate of a reaction cannot decrease with increase in concentration of a reactant.
Assertion: Rate constant of a zero-order reaction has same units as the rate of reaction.
Reason: Rate constant of a zero-order reaction does not depend upon the concentration
of reactant.
11
12
13
14
Assertion: In a zero-order reaction, if concentration of the reactant is doubled, halflife period is also doubled.
Reason: The total time taken for a zero-order reaction to complete is double of the
half-life period.
Assertion: 50% of a reaction is completed in 50 sec, 75% of the reaction will be
completed in 75 sec.
Reason: The rate constant of a zero-order reaction depends upon time.
Assertion: With increase in temperature the rate of reaction decreases.
Reason: For every 10 0C rise in temperature, the rate of the reaction doubles for most
of the reaction.
40
15
Assertion: Order of the reaction can be determined experimentally
Reason: We can determine order from balanced chemical equation.
16
Assertion: Order and molecularity are always the same.
Reason: Order is determined experimentally and molecularity is the sum of the
stoichiometric coefficient of rate determining elementary step.
Assertion: The enthalpy of reaction remains constant in the presence of a catalyst.
Reason: A catalyst participating in the reaction, forms different activated complex and
lowers down the activation energy but the difference in energy of reactant and product
remains the same.
Assertion: All collision of reactant molecules lead to product formation.
Reason: Only those collisions in which molecules have correct orientation and
sufficient kinetic energy lead to compound formation.
17
18
19
20
21
22
23
24
25
26
Assertion: Diamond shall convert to graphite.
Reason: The rate is so slow that the change is not perceptible at all.
Assertion: The rate of a reaction A→B becomes 4 times when concentration of A is
doubled
Reason: it is a first order reaction.
Assertion: In gaseous reactions rate cannot be expressed as rate of change of partial
pressure.
Reason: For gaseous reactions at const. temperature concentration is directly
proportional to the partial pressure of the species.
Assertion: Oxidation of ethane to CO2 is a complex reaction.
Reason: Reactions taking place in a sequence of elementary steps are called complex
reactions.
Assertion: The order for a reaction with rate constant k=3x104s-1 is 1.
Reason: Order of a reaction is the sum of the powers to which concentration terms are
raised in the rate law.
Assertion: The reaction X→Y follows second order kinetics. If concentration of X
increased 3 times rate become 3 times.
Reason: The rate of a reaction is directly proportional to the concentration of
reactants.
Assertion: The thermal decomposition of HI on gold is a zero-order reaction.
Reason: The thermal decomposition of HI on gold depends on the initial
concentration.
Assertion: All natural and artificial radioactive decay of unstable nuclei take place by
first order kinetics.
Reason: In a first order reaction half-life is independent of initial concentration.
27
Assertion: A catalyst does not alter the free energy change of a reaction.
Reason: A catalyst lowers the activation energy of a process.
28
Assertion: According to collision theory the rate of reaction does not depend on
collision frequency.
Reason: The collisions in which molecules collide with sufficient kinetic energy and
proper orientation are called effective collisions.
29
Assertion: The rate of a reaction is independent of temperature.
Reason: The rate at a particular moment of time is called instantaneous rate.
30
Assertion: The rate of a reaction quadrupled when concentration is doubled.
Reason: It is second order reaction.
41
CASE BASED QUESTIONS
31
32
Observe the following graphs and answer the questions based on these graphs.
(b) What is slope in graph II?
(c) How does t 1/2 varies with initial concentration in zero order reaction.
(d) If t 1/2 of first order reaction is 40 minutes, what will be t 99.9% for first order
reaction?
(e) What is t 1/2 of zero-order reaction in terms of ‘k’?
Observe the table given showing volume of CO2 obtained by reaction of CaCO3 and
dilute HCl after every minute. Answer the questions that follow:
Table showing volume of CO2 at one minute interval by reaction of CaCO3 with
dilute HCl.
Time/mm
0
1
2
3
4
5
6
Volume of CO2/cm
0
24 cm3
34 cm3
38 cm
40 cm
40 cm
40 cm
(a) What happens to rate of reaction with time?
(b) Why does CaCO3 powder react faster than marble chips?
(c) What happens to rate of reaction if concentrated HCl is used?
(d) In manufacture of NH3, N2(g) + 3H2(g)→ 2NH3+ heat, what is effect of pressure on
rate of reaction?
(e) Why does rate of reaction becomes almost double for energy 10° rise in
temperature?
42
33
Read the passage given below and answer the following questions:
The rate of a reaction, which may also be called its velocity or speed, can be defined
with relation to the concentration of any of the reacting substances, or to that of any
product of the reaction. If the species chosen is a reactant which has a concentration c
at time t the rate is
-dc/dt, while the rate with reference to a product having a concentration x at time t is
dx/dt. Any concentration units may be used for expressing the rate; thus, if moles per
litre are employed for concentration and seconds for the time, the units for the rate are
moles litre- 1 sec-1. For gas reactions pressure units are sometimes used in place of
concentrations, so that legitimate units for the rate would be (mm. Hg) sec-1 and atm.
sec-1 The order of a reaction concerns the dependence of the rate upon the
concentrations of reacting substances; thus, if the rate is found experimentally to be
proportional to the
α th power of the concentration of one of the reactants A, to the β th power of the
concentration of a second reactant B, and so forth, via., rate = k [A] α [B] β (1)the overall order of the reaction is simn = α + β + (2) Such a reaction is said to be of the
αth order with respect to the substance A, the β th order with respect to B and so on...
(Laidler, K. J., & Glasstone, S. (1948). Rate, order and molecularity in chemical
kinetics. Journal of Chemical Education, 25(7), 383.)
In the following questions, 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.
1. Assertion: Rate of reaction is a measure of decrease in concentration of reactant
with respect to time.
Reason: Rate of reaction is a measure of increase in concentration of product with
respect to time.
2. Assertion: For a reaction: P +2Q → Products,
Rate = k [P]1/2[Q]1 so the order of reaction is 1.5
Reason: Order of reaction is the sum of stoichiometric coefficients of the reactants.
3. Assertion: The unit of k is independent of order of reaction.
Reason: The unit of k is moles L-1 s -1 for a zero order reaction
4. Assertion: Reactions can occur at different speeds.
Reason: Rate of reaction is also called speed of reaction.
ANSWERS
1
A
16
D
2
A
17
A
3
C
18
D
4
C
19
B
5
D
20
C
6
A
21
D
7
B
22
A
8
C
23
B
9
A
24
D
10
C
25
C
11
A
26
B
12
B
27
B
13
C
28
D
14
D
29
D
15
C
30
A
43
31 a) 0
b) k /2.303 where k is the rate constant
c) t 1/2 is directly proportional to initial concentration.
d)
t 99.9% = 10t 1/2 = 10 × 40 = 400 minutes
e) t 1/2 = R 0 /2k for zero order reaction
32 a) The rate of reaction first decreases with time then become constant
b) CaCO3 powder has more surface area than marble chips therefore, more rate of
c)
reaction
The rate of reaction will increase because rate of reaction increases with the increase in
concentration
d) It is because number of molecules undergoing effective collisions become almost double,
rate of reaction almost doubled.
33) 1--B
hence
2--C
3--D
4--B
SHORT ANSWER QUESTIONS CARRYING TWO OR THREE MARKS
1. Define rate of a reaction? : What is the SI Unit of rate of reaction?
Ans: Change in molar concentration of reactant or product in per unit time is
called rate of reaction. Mol /L /s
2
1+1
2. For the reaction R--P, the conc. of reactant changes from 0.03M to 0.02M in 25 min.
Calculate average rate of the reaction using the unit of time in second
2
Rav = - ∆[R]/ ∆t
= - (0.02-0.03 ) /25x60
1/2
1
=-[-0.01] 1500
= 6.66x10-6 M/s
3. Mention any two factors which influence the rate of reaction.
Ans 1) Pressure or conc. of reactants
2) Temperature
3) Catalyst.
4. (i) State rate law?
(ii) Define rate constant of a reaction
1/2
2
(any 2)
2
44
Ans. (i) Expression in which reaction rate is given in terms of molar conc. of reactants
with
each term rose to some power which may or may not be same as the
stoichiometric coefficient of
the reacting species in a balanced chemical equation.
(ii) Rate constant is equal to rate of reaction when the product of the molar conc. of
reactants is unity
5. i) Define order of a reaction.
2
ii) Calculate the overall order of a reaction which has the rate expression.
Rate= K [A]1/2 [B]3/2
ans.
i) Sum of the powers of the concentration of the reactants in the rate equation is
order of reaction.
1+1
called
ii)Order of reaction = 1/2 + 3/2 = 2
6. i) What are complex reactions?
2
ii) What is the order of reaction whose unit of rate constant and rate of reaction are same?
Ans. i) Reactions taking place in more than one step are called complex reaction.
ii) Zero order
7. i) Identify the reaction order from the rate constant K=2.3x10-5 mol-1 .L.S-1
ii) Define molecularity of a reaction.
ans i) 2
1+1
2
1+1
ii) The number of reacting species taking part in an elementary reaction which must colloid
simultaneously in order to bring about a chemical reaction is called
molecularity of reaction.
8) a) In a complex reaction which step controls the overall rate of reaction and
2
what is it called?
b) From the rate expression for the following reactions determine their order of reaction and the
dimensions of the rate constants:
Ans: a) slowest step, which is called rate determining step.
1+1
Ans.
9. Define pseudo first order reaction. Give an example.
2
45
Ans: Chemical reactions which are not first order but behave as fist order reaction under suitable
conditions are called pseudo first order Reactions. 1+1
Ex: Inversion of cane sugar. C12 H22O11+H2O------- C6H12O6 + C6H12O6
10) Analyze the given graph, drawn between concentrations of reactant vs. time.
2
(a) Predict the order of reaction.
(b) Theoretically, can the concentration of the reactant reduce to zero after infinite time? Explain.
Ans. (a) 1 st order (b) No, due to exponential relation / the curve never touches
1+1/2+1/2
the x-axis.
11a). Calculate the overall order of the reaction whose rate law expression was predicted as:
Rate = k [ NO]3/2[O2]1/2
2
b) Write the slope value obtained in the plot of ln[R] vs. time for a first order reaction.
Ans. a) 2
1+1
b) Slope = -k
12. A reaction is first order w.r.t. reactant A as well as w.r.t. reactant B. Give the rate law.
Also give one point of difference between average rate and instantaneous rate.
2
Ans. Rate = k[A][B]
1+1
Average rate - Rate of a reaction for a particular period or interval of time.
Instantaneous rate –Rate of a reaction at a particular instant of time. (or any other suitable difference)
13. A first order reaction is 40% complete in 80 minutes. Calculate the value of rate Constant (k).
In what time will the reaction be 90% completed?
[Given: log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021, log 5 = 0·6771, log 6 = 0·7782]
Ans.
3
46
1+1/2+1+1/2
14. Define order of reaction?
2
Write the condition under which a bimolecular reaction follows first order kinetics.
Ans. Sum of powers of the concentration of the reactants in the rate law expression.
When one of the reactant is present in large excess
1+1
15. For the reaction A + B →products, the following initial rates were obtained at various given
initial concentrations.
Determine the half-life period.
3
Ans.
1+1+1
47
16. The rate constant for a first order reaction is 60 s-1 . How much time will it take to reduce 1g of
the reactant to 0.0625 g?
3
Ans.
1+1+1
17. The following data were obtained during the first order thermal decomposition of N2O5(g) at a
constant volume:
3
ANS.
18. Observe the graph in diagram and answer the following questions.
1+1+1
2
48
(ii) If slope is equal to -2.0x10-6 sec-1 ,what will be the value of rate constant?
(ii) How does the half-life of zero order reaction relate to its rate constant?
Ans: i) SLOPE = -K
2.0x10-6 sec-1=-K
1+1
Hence k = 2.0x10-6 sec-1
ii) t ½= R0/2k
19. Derive integrated rate equation of zero order reaction?
20. Show that half life of a zero order reaction is directly proportional to initial concentration of
reactant from integrated rate equation.
2
OR
Derive the relation between half life and rate constant of zero order reaction.
-
49
21. Show that half life of a first order reaction is independent of initial Conc. of reactant from
integrated rate equation.
2
Or
Derive the relation between half life of a first order reaction and its rate constant.
22. Show that the time required for 99/. Completion of a first order reaction is twice the time
required for the completion of 90% of reaction.
3
Ans.
1+1+1
23. 23.
50
Ans .
3
24. 24.
Ans:
25. The decomposition of NH3 on platinum surface is zero order reaction.
What are the rates of production of N2 and H2 if R=2.5 x 10-4 mol/1 /s
Ans.
3
3
1+1+1
51
26. Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by
iodide ions.
2
This reaction takes place in two steps as given below:
Step -1 H2O2 + I– → H2O + IO– (slow)
Step – II H2O2 + IO– → H2O + I– (fast)
(i) Write the rate law expression and determine the order of reaction w.r.t. H2O2.
(ii) What is the molecularity of each individual step?
Ans
(i) Rate = k[H 2O2] [I–]
Order w.r.t. H2O2 = 1
(ii) Molecularity: Step I = 2, Step II = 2
1+1
LONG ANSWER TYPE OF QUESTIONS
1 a) Vijay plotted a graph between concentration of R and time for a reaction R  P. On the basis
of this graph, answer the following questions:
3+2
(i)
(ii)
(iii)
Predict the order of reaction.
What does the slope of the line indicate?
What are the units of rate constant?
(b) A first order reaction takes 25 minutes for 25% decomposition. Calculate t1/2. [Given : log 2
= 0·3010, log 3 = 0·4771, log 4 = 0·6021]
Ans:
(a) (i) zero order
(b) (ii) slope = - k
(c) (iii) molL-1 s
1+1+1
(b) k = 2.303/t log [A0]/[A]
1+1
= 2.303/25 log 100/75
= 2.303/25 log4/3
= 2.303/25(0.6021 – 0.4771)
= 0.0115 min-1
52
t1/2 = 0.693/k
= 0.693/0.0115
= 60.26 min or 60.2 min
2. (a) The rate constant for a first order reaction is 60 s–1 . How much time will it take to reduce
the initial concentration of the reactant to its 1/16th value?
3
(b) Write two factors that affect the rate of a chemical reaction.
Ans. (a) t = 2.303/k log [A0]/[A]
2
1+1+1
= 2.303/ 60 log 1/1/16
= 0.046 s
(or by any other suitable method)
(b) Concentration of reactants, Temperature, Pressure, surface area and catalyst
(any two factors )
1+1
3) a)The conversion of molecules X to Y follows second order kinetics. If concentration of X is
increased to three times how will it affect the rate of formation of Y ?
b) A first order reaction has a rate constant 1.15 x 10-3 s-1. How long will 5 g of this reactant take to
reduce to 3 g?
2+3
Ans. a) The reaction is : X—>Y
According to rate law,
rate = k[X]2
If [X] is increased to 3 times, then the new rate is
rate = k[3X]2
rate= 9 k [X]2 = 9 rate
Thus, rate of reaction becomes 9 times and hence rate of formation of Y increases 9b)
times.
4) a) The decomposition of dimethyl ether leads to the formation of CH 4, H2 and CO and the
reaction rate is given by Rate=k [CH3OCH3]3/2 . The rate of reaction is followed by increase in
pressure in a closed vessel, so the rate can also, be expressed in terms of the partial pressure of
dimethyl ether, i.e., Rate= k (PCH3OCH3)3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate
constants?
53
b) A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B is doubled?
Ans a)
2+3
1+1
b)
5) a) In a reaction, 2A —-> Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 molL-1 in
10 minutes. Calculate the rate during this interval?
b) A first order reaction has a rate constant 1.15 x 10 -3 s-1. How long will 5 g of this reactant take to
reduce to 3 g?
2+3
Ans. a)
2
b) 3
6) (a) 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?
2+3
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required
for 90% completion of this reaction. (log 2 = 0.3010)
54
Ans. a) i) Rate will increase 4 times of the actual rate of reaction
ii) 2nd order
b) t1/2=0.693/k
30min =0.693/k
K=0.0231min-1
2+3
7) For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained
(i)
(ii)
Show that it follows pseudo first order reaction, as the concentration of water remains
constant.
Calculate the average rate of reaction between the time intervals 30 to 60 seconds. (Given
log 2 = 0.3010, log 4 = 0.6021)
Ans
As k is same for both the readings .hence it is pseudo first order reaction.
55
ii)
8) a) The following data were obtained during the first order thermal decomposition of
at a constant volume:
SO2Cl2 (g) → SO2(g) + Cl2 (g)
b) For a reaction: 2NH3 (g)––→ N2 (g) + 3H2 (g)
Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the units of K
Ans. a)
b. (i) zero order, bimolecular/unimolecular
(ii) mol L-1 s -1
3+2
SO2Cl2
56
9) a) Consider a certain reaction A → Products with k = 2.0 × 10-2 s-1. Calculate the concentration
of a remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
b) The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing
wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
3+2
Ans a)
b)
10) (i)Express the rate of the following reaction in terms of the formation of ammonia: 5
N2(g) + 3H2(g) → 2NH3(g)
(ii) If the rate constant of a reaction is k = 3 × 10-4 s-1, then identify the order of the
reaction.
(iii) For a reaction R → P, half-life (t1/2) is observed to be independent of the initial
concentration of reactants. What is the order of reaction?
(iv) Define the following:
(a) Elementary step in a reaction
(b) Rate of a reaction
Ans (i)
1+1+1+1+2
(ii) S-1 is the unit for rate constant of first order reaction.
(iii) The t1/2 of a first order reaction is independent of initial concentration of reactants.
(iv) (a) Elementary step in a reaction: Those reactions which take place in one step are
elementary reactions.
Example: Reaction between H2, and I2 to form 2HI
H2 + I2 → 2HI
(b) Rate of a reaction: The change in the concentration of any one of the reactants or
per unit time is called rate of reaction.
called
products
57
11) a) A reaction is of second order with respect to a reactant. How will the rate of reaction be
affected if the concentration of this reactant is
(i) doubled, (ii) reduced to half?
b) The rate constant for a reaction of zero order in A is 0.0030 mol L-1 s-1. How long will it take
for the initial concentration of A to fall from 0.10 M to 0.075 M?
Ans
(a) Since Rate = K [A]2
For second order reaction Let [A] = a then Rate = Ka2
(i) If [A] = 2a then Rate = K (2a)2 = 4 Ka2
∴Rate of reaction becomes 4 times
ii) If [A] = a/2 then Rate = K (a/2)2=K(a/4)
Then rate of reaction becomes ¼ times.
b)
12) a) Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction.
2+3
b) Rate constant k for a first order reaction has been found to be 2.54 × 10-3 sec-1.
Calculate its 3/4th life, (log 4 = 0.6020)
Ans
a) Rate expression: The expression which expresses the rate of reaction in terms of
molar
concentrations of the reactants with each term raised to their power, which
may or may not be
same as the stoichiometric coefficient of that reactant in the
balanced chemical equation.
Rate constant: The rate of reaction when the molar concentration of each reactant is
taken as unity.
b)
13) a) Write two differences between ‘order of reaction’ and ‘molecularity of reaction 3+2
b) If the half-life period of a first order reaction in A is 2 minutes, how long will it take [A] to
reach 25% of its initial concentration?
Ans a)
Order of reaction
Molecularity of reaction
(i) It is the sum of tire concentration terms on
which die rate of reaction actually depends.
It is the number of atoms, ions or molecules
that must collide with one another
simultaneously so as to result into a
chemical reaction.
(ii) It can be fractional as well as zero.
it is always a whole number.
58
(b)
14) (a) For a reaction A + B —> P, the rate is given by Rate = k[A]2 [B]
2+3
(i) How is the rate of reaction affected if the concentration of A is doubled?
(ii) What is the overall order of reaction if B is present in large excess?
(b) A first order reaction takes 23.1 minutes for 50% completion. Calculate the time
required for 75% completion of this reaction.
(Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021).
Ans:
15) (a)Plot a graph showing variation in the concentration of reactants against 1+1+3
time for a zero-order reaction.
(b) What do you mean by zero-order reaction?
(c) The initial concentration of the first-order reaction, N2O5( g)→2NO2( g)+1/2O2(g) was 1 24
-2
x 10 mol L-1 at 300 K. The concentration of N2O5 after ‘1’ hour was 0.20 x 10-2 mol L-1.
Calculate the rate constant of the reaction at 300 K.
Ans: a)
59
b) It is a reaction for which the rate of the reaction is proportional to zero power of the
concentration of reactants, i.e, and order is zero. Or, this is a reaction for which the rate of the
reaction is independent of the concentration of the reactants.
16) Rate of a reaction is the change in concentration of any one of the
2+1+1+1
Reactants or any one of the products in unit time.
i) Express the rate of the following reaction in terms of reactants and
Products:
2H I → H2 + l2
ii) If rate expression for the above reaction is, rate = k[H I] 2, What is the
Order of the reaction?
iii) Define order of a reaction.
iv) Whether the molecularity and the order of the above reaction are the same? Give reason.
(iii) Order is the sum of the powers of the concentration terms in the rate law.
rate = K [A]x [B]y
∴ Order = x + y
iv) Yes.
2H I = H2 + l2
rate = K [H I]2 ∴ Order = 2
Molecularity is the number of reacting species taking part in an elementary reaction which must
collide simultaneously in order to bring about a chemical reaction.
∴ Here molecularity = 2
[N2O5](M)
Time(min)
0.400
0.00
0.289
20.0
0.209
40.0
0.151
60.0
0.109
80.0
60
17) Nitrogen pentoxide decomposes according to equation:
3+1+1
2N2O5(g) → 4 NO2(g) + O2(g).
This first order reaction was allowed to proceed at 40°C and the data below were collected:
(a) Calculate the rate constant. Include units with your answer.
(b) What will be the concentration of N2O5 after 100 minutes?
(c) Calculate the initial rate of reaction.
Ans a) K = 2.303/t log [ A0]/[A]
Substituting the values, we get
18) a)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, what is the order of the reaction with respect to A
and B.
3+1+1
b) For a reaction R → P, half-life (t1/2) is observed to be independent of the initial concentration of
reactants. What is the order of reaction?
c. Identify the order of a reaction if the units of its rate constant are L-1 mol s-1
a)
61
b) The t1/2 of a first order reaction is independent of initial concentration of reactants.
c) Zero order reaction.
19) a)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-fourths of the initial
quantity of HCO2H to decompose, (log 0.25 = -0.6021)
3+2
(b) The decomposition of phosphine, PH3, proceeds according to the following equation:
It is found that the reaction follows the following rate equations:
4PH3(g)→P4(g)+ 6H2(g)
It is found that the reaction follows the following rate equations:
Rate =k[PH3]
The half-life of PH3 is 37.9 s at 120°C
(i) How much time is required for 3/4th of PH3 to decompose ?
(ii) What fraction of the original sample of PH3 remains behind after 1 minute?
Ans
3+2
(b) (i) Hence t1/2 = 37.9 s
62
20) (i) For the reaction
5
2NO(g) + Cl2(g) → 2NOCl(g) the following data were collected. All the measurements were
taken at 263 K:
(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the initial rate of disappearance of Cl2 in exp. 4?
Experiment
No.
Initial
[NO](M)
Initial
[Cl2](M)
Initial rate of disappearance of
Cl2 (M/min)
1
0.15
0.15
0.60
2
0.15
0.30
1.20
3
0.30
0.15
2.40
4
0.25
0.25
?
63
If half life period of first order reaction is x and 3/4 th life period of same reaction is ’y’,
how are V and ’y’ related to each other?
(iii)
Ans:
(i)
1+2+3
2
(a) Rate law = K[NO] [Cl2]
(b) (ii) 0.60 M min-1 = K[0.15]2 [0.15] M3
∴ K = 177.7 M-2 min-1
(iii) Initial rate of disappearance of Cl2 in exp. 4
Formula : Rate = K[NO]2 [Cl2]
∴ Initial rate = 177.7M-2 min-1 × (0.25)2 × (0.25)M3= 2.8 M min-1
(ii)
64
CHAPTER 5
SURFACE CHEMISTRY
GIST OF THE LESSON
Surface chemistry: Study of phenomena that occur at the surfaces or interfaces between two
phases.
Adsorption: The accumulation of molecular species at the surface rather than in the bulk of a
solid or liquid.
Components of adsorption
Adsorbent: Material on the surface of which the adsorption takes place.
Adsorbate: Substance which concentrates or accumulates at the surface of another substance
(adsorbent)
Example: Adsorption of water vapour (adsorbate) by silica gel (adsorbent).
Desorption: The process of removing an adsorbed substance from a surface on which it is
adsorbed.
Difference between Adsorption and Absorption
Adsorption
The substance is concentrated only at the
surface and does not penetrate through the
surface to the bulk of the adsorbent
Absorption
The substance is uniformly distributed
throughout the bulk of another
substance
Sorption: The process in which both adsorption and absorption take place simultaneously.
Mechanism of adsorption (Reason for adsorption): The unbalanced or residual attractive
forces of the surface particles (of adsorbent). Hence the extent of adsorption increases with
increase in surface area of the adsorbent.
Thermodynamics of adsorption: Adsorption is invariably an exothermic process (ΔH<0),
because it leads to a decrease in surface energy. As the entropy decreases after adsorption, ΔS
is also < 0. During a spontaneous adsorption process, ΔH becomes less negative and finally
becomes equal to TΔS so that ΔG=0.
Types of adsorptions: Physisorption and chemisorption; based on the type of force of
attraction between adsorbent and adsorbate.
Difference between physisorption and chemisorption
Physisorption
1. Arises because of van der Waals’ forces
2. Not specific in nature.
3. Reversible in nature.
Chemisorption
1. It is caused by chemical bond formation
2. It is highly specific in nature.
3. It is irreversible.
65
4. Enthalpy of adsorption is low
(20-40 kJ mol–1 )
5. Low temperature is favourable for
adsorption. It decreases with increase of
temperature
6. No appreciable activation energy is
needed
7. Results into multimolecular layers on
adsorbent surface under high pressure
4. Enthalpy of adsorption is high
(80-240 kJ mol–1).
5. High temperature is favourable for
adsorption. It increases with the increase
of temperature
6. High activation energy is needed
sometimes
7. It results into unimolecular layer
Similarities between physisorption and chemisorption
 Both depends on the surface area. Both increases with an increase of surface area of
adsorbent.
 Both depends on the nature of adsorbent (gas). During physisorption, more easily
liquefiable gases are adsorbed readily; In chemisorption those gases which can react
with the adsorbent get adsorbed readily.
Adsorption isotherms: Graph showing the variation in the amount of gas adsorbed by the
adsorbent with pressure at constant temperature.
Freundlich adsorption isotherm equation : x/m= kp 1/n (n > 1)
‘x’ is mass of the gas adsorbed; ‘m’, mass of adsorbent, ‘p’, pressure, k and n are
constants which depend on the nature of the adsorbent and the gas at a particular temperature.
Freundlich adsorption isotherm curve representation
at different temperatures (T1 < T2)
Adsorption on solid from different phases; Comparison
Adsorption of gas
x/m = kp1/n (p=pressure of adsorbent)
Absorption of solute from solution
x = kC1/n (C=concentration of solution)
m
log[x/m] = logk + 1logp
n
 There is decrease in physical
adsorption with increase in temperature
 The adsorption varies directly with
pressure when 1/n = 1 and independent of
pressure when 1/n = 0.
Graphical determination of k and n:
log[x/m] = logk + 1logC
n
 The extent of adsorption decreases with an
increase in temperature
 The extent of adsorption depends on the
concentration of the solute in solution
Graphical determination of k and n:
66
During heterogeneous
catalysis, the catalyst
adsorbs the reactants
and increase the rate of
reaction
Traces of air can be
adsorbed by charcoal
from a vessel to give
a very high vacuum
Silica and
aluminium gels
adsorb moisture
and control
Drugs kill germs by
getting adsorbed
on them
During Froth floatation
process in metallurgy,
the sulphide ores are
concentrated by
adsorption using pine oil
and frothing agent
Surfaces of certain
precipitates have the
property of adsorbing
some dyes, producing a
characteristic colour at
the end point in titrations
Activated charcoal
on gas mask
adsorbs poisonous
gases
Applications of adsorption
In chromatographic
method, the
movement of mobile
phase depends on its
adsorption on
stationary phase
Animal charcoal
removes colours of
solutions by
adsorbing coloured
impurities
A mixture of noble gases
can be separated by
adsorption on coconut
charcoal at different
temperatures due to their
difference in degree of
67
COLLOIDS
A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as
very fine particles in another substance called dispersion medium. Size range of particles
between 1 and 1000 nm (10–9 to 10–6 m)
Classification of colloids:
I. Based on physical state of dispersed phase and dispersion medium
Dispersed
phase
Solid
Solid
Solid
Liquid
Liquid
Liquid
Gas
Gas
Dispersion
medium
Solid
Liquid
Gas
Solid
Liquid
Gas
Solid
Liquid
Type of
colloid
Solid sol
Sol
Aerosol
Gel
Emulsion
Aerosol
Solid sol
Foam
Examples
Some coloured glasses and gem stones
Paints, cell fluids
Smoke, dust
Cheese, jellies
Milk, hair cream, butter
Fog, mist, cloud, insecticide sprays
Pumice stone, foam rubber
Froth, whipped cream, soap lather
II. Based on nature of interaction between dispersed phase and dispersion medium
Lyophilic colloids
1. Stronger force of attraction between
dispersed phase and dispersion medium
2. Reversible sols
3. They are quite stable do not easily
coagulated
4. Do not need stabilising agents for their
preservation.
Lyophobic colloids
1. Weaker force of attraction between
dispersed phase and dispersion medium
2. Irreversible sols
3. Not stable and easily coagulated
4. Need stabilising agents for their
preservation.
III. Based on the type of the particles of the dispersed phase
Multimolecular colloids
On dissolution, a large
number of atoms or
smaller molecules of a
substance aggregate
together to form species
having size in the colloidal
range.
Eg. gold sol, sulphur sol
Macromolecular colloids
Macromolecules in
suitable solvents form
solutions in which the
size of the
macromolecules may be
in the colloidal range.
Eg. Starch, nylon etc
Associated colloids (micelles)
There are some substances
which at low concentrations
behave as normal strong
electrolytes, but at higher
concentrations (CMC), exhibit
colloidal behaviour due to the
formation of aggregates
(micelles). Eg soap
Mechanism of micelle formation and cleansing action of soap
Soap is sodium salt of long chain fatty acids like stearic acid. Its formula is CH3 (CH2 )16 COO –
Na+, which is represented as R-COO-Na+. The ionic part COO–Na+ is polar and non-ionic
68
(organic) part R is non-polar. When it is dissolved in water and the concentration is above
CMC, micelle formation takes place. The soap molecules are arranged spherically in such a
way that the polar ends (heads) are oriented towards water and non-polar ends (tail) towards
the centre.
During the cleansing action, soap molecules form micelle around the oil droplet (dirt) in such
a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part
projects out of the grease droplet. Thus the oil (or dirt) is pulled out into water and get
washed off.
Preparation of colloids
a. Chemical method:
b. Electrical method (Bredig’s arc method)
In this process, an electric arc is struck between electrodes of the metal immersed in
the dispersion medium. The intense heat produced vapourises the metal, which then
condenses to form particles of colloidal size.
c. Peptization
It involves the conversion of a precipitate into colloidal sol by shaking it with
dispersion medium in the presence of a small amount of electrolyte known as
peptizing agent. The precipitate adsorbs one of the ions of the electrolyte on its
surface. This causes the development of charge on precipitates, which ultimately
break up into smaller particles of the size of a colloid.
Purification of colloid: The process used for reducing the amount of impurities to a requisite
minimum.
69
a. Dialysis
Process of removing dissolved substances from a colloid by means of
diffusion through a suitable semipermeable membrane.
b. Electrodialysis
Process of removing dissolved substances from a colloid by means of
diffusion through a suitable semipermeable membrane and the compartment is fitted
with two electrodes so as to increase the rate of flow.
c. Ultrafiltration
The colloidal particles are separated from rest of the materials by filtration through
ultra-filter paper.The colloidal particles left on the ultra-filter paper are then stirred
with fresh dispersion medium (solvent) to get a pure colloidal solution.
[Ultra-filter paper preparation: Soaking the ordinary filter paper in a collodion
solution, hardening by formaldehyde and then finally drying].
[Collodion is 4% solution of nitro-cellulose in a mixture of alcohol and ether].
Properties of colloids
i.
Colligative properties
The values of colligative properties are of small order as compared to values
shown by true solutions at same concentrations due, to bigger size of particles.
ii.
Optical property (Tyndall effect)
When a beam of light is passed through colloid, the path of light becomes visible
due to the scattering of light by the colloidal particles.
Conditions to observe Tyndall effect are;
a) The diameter of the dispersed particles is not much smaller than the
wavelength of the light used
b) The refractive indices of the dispersed phase and the dispersion medium differ
greatly in magnitude.
70
iii)
Colour of colloid:
The colour of colloidal solution depends on;
a) The wavelength of light scattered by the dispersed particles, which in turn
depends on particle size.
b) The colour changes with the manner in which the observer receives the light
(By reflected/transmitted)
iv)
Mechanical property (Brownian movement)
The continuous zig-zag motion of colloidal particles due to the unbalanced
bombardment of the particles by the molecules of the dispersion medium.
Significance: It has a stirring effect which does not permit the particles to settle and
is responsible for the stability of sols.
vi)
Charge on colloidal particles
Colloidal particles always carry similar electric charge which prevent them from
aggregating when they come closer to one another and thus providing stability.
Reasons for charge development:
i) Electron capture by sol particles during electrodispersion of metals
ii) Preferential adsorption of ions common to the colloidal particle and dispersion
medium/ formulation of electrical double layer
Helmholtz electrical double layer: The combination of the layer of charge acquired
by selective adsorption on the surface of colloidal particles together with layer of
counter ions of opposite charge.
Electrokinetic potential or zeta potential: The potential difference between the fixed
layer of charge acquired by selective adsorption and the diffused layer of counter
ions, around the colloidal particles. (OR Potential difference between the Helmholtz
electrical double layers).
Electrophoresis: The movement of colloidal particles under an applied electric
potential towards oppositely charged electrode.
Electroosmosis: Movement of dispersion medium in an electric field when the
movement of colloidal particles is prevented.
Coagulation or precipitation: The process of settling of colloidal particles.
71
Methods for coagulation





By electrophoresis
By mixing two oppositely charged sols
By boiling
By persistent dialysis
By addition of electrolytes
Hardy-Schulze rule: The greater the valence of the flocculating ion added, the greater is
its power to cause precipitation.
Coagulating value: The minimum concentration of an electrolyte in millimoles per litre
required to cause precipitation of a sol in two hours.
Reason for the stability of lyophilic sols: Charge and solvation.
Coagulation of lyophilic sol:(i) By adding an electrolyte, (ii) By adding a suitable solvent.
Protective colloids: The lyophilic colloidal particles when mixed with lyophobic sol, form a
layer around lyophobic particles and protect the latter from electrolytes. Lyophilic colloids
used for this purpose are called protective colloids.
Sky appears as blue
due to the scattering
of blue light by
water and dust in
Fertile soils are
colloidal in nature in
which humus acts as
a protective colloid
Colloids around us
River water is a
colloidal
solution of clay
Milk, butter,
halwa, ice
creams, fruit
Blood is a
colloidal solution
of an albuminoid
substance
Dust in air when cooled
below its dew point,
moisture gets condensed
on its surface forming
colloids (mist/fog)
Applications of colloids
 The smoke, before it comes out from the chimney, is led through Cottrell precipitator
chamber containing plates having a charge opposite to that carried by smoke particles.
The particles on coming in contact with these plates lose their charge and get
precipitated.
 Suspended impurities in water obtained from natural sources, is coagulated by adding
alum.
72
 Colloidal medicines are more effective because they have large surface area and are
therefore easily assimilated.
 Hardening of leather involves the coagulation of positively charged colloidal particles
of animal hide by the negatively charged particles in tannin.
 Cleansing action of soaps and detergents.
 Photographic plates or films are prepared by coating an emulsion of the light sensitive
silver bromide in gelatine over glass plates or celluloid films.
 Rubber is obtained by coagulation of latex by using formic acid.
 Paints, inks, synthetic plastics, rubber, graphite lubricants, cement, etc., are all
colloidal solutions.
MULTIPLE CHOICE QUESTIONS
1. Ammonia is adsorbed on activated charcoal to a larger extent as compared to CO
because
(a) Ammonia reacts with oxygen
(b) Ammonia exist in liquid state
(c ) CO reacts with oxygen
(d) Ammonia is more easily liquefiable
2. Freundlich equation for adsorption of gases (in amount of x g) on a solid (in amount
of m g) at constant temperature can be expressed as
(a) log x/m = log P + 1/n log K
(b) log x/m = log K + 1/n log P
(c ) x/m = log K + 1/n log P
(d) x/m = log P + 1/n log K
3. Which one of the following is not an example of adsorption?
(a) SO 2 incontact with Charcoal
(b) anhydrous CaCl2 with water
(c ) silica gel in contact with water vapours
(d). Aqueous solution of raw sugar with animal charcoal
4. Which of the following is related to adsorption?
(i) ∆H = – ve
(ii) ∆S = – ve
(iii) –T∆S = – ve
(iv) ∆G = – ve
(a) (i), (ii) and (iv)
(b) (ii) and (iii)
(c) (iii) only
(d) (i), (iii) and (iv)
73
5. What is the name of a phenomenon in which both adsorption and absorption takes
place:
(a) Chemisorption
(b) Physisorption
(c) Desorption
(d) Sorption
6. Tanning of leather is the
(a) Coagulative hardening of leather by chemicals
(b)Drying process to make the leather hard
(c )Polishing leather to make it look attractive
(d)Colouring of leather by chemicals
7. In Freundlich adsorption isotherm, the value of 1/n is :
(a) between 0 and 1 in all cases
(b) between 2 and 4 in all cases
(c) 1 in case of physical adsorption
(d) 1 in case of chemisorption
8. A graph of extent of adsorption versus pressure at constant temperature is called
(a) adsorption isobar.
(b) adsorption isotherm.
(c) adsorption isostere.
(d) none of these.
9. A plot of log x/m versus log P for the adsorption of a gas on a solid gives a straight
line with slope equal to
(a) log K
(b)1/ n
(c ) -log K
(d) n
10. Which of the following is most favourable for Vander Waal’s adsorption?
(a) High pressure and high temperature.
(b) Low pressure and low temperature.
(c) Low pressure and high temperature.
(d) High pressure and low temperature.
11. Which of the following statement about Adsorption is correct
(a) an exothermic process hence increase in temperature decreases adsorption in cases
where Vander Waals forces exits between adsorbate and adsorbent .
(b) an endothermic process hence increases in temperature increases adsorption
(c) an exothermic process hence increase in temperature increases adsorption.
(d) None of these.
12. Which one of the following is correctly matched?
74
(a) Emulsion-smoke
(b) Gel-butter
(c) Aerosol-hair cream
(d) Sol-whipped cream
13. On the basis of data given below predict which of the following gases shows least
adsorption on a definite amount of charcoal?
Gas
CO2
SO2
CH4
H2
Critical temp./K
304
630
190
33
(a) CO2
(b) SO2
(c) CH4
(d) H2
14. Among the electrolytes Na 2 SO4 , CaCl2 , Al2 (SO4 )3 and NH4Cl, the most effective
coagulating agent for Sb2 S3 sol is
(a) Na2 SO4
(b) CaCl2
(c) Al2 (SO4 )3
(d) NH4 Cl
15. Colloidion is 4% solution of which one of the following in alcohol-ether mixture.
(a) Nitro-glycerine
(b) Cellulose acetate
(c) Glycol dinitrate
(d) Nitrocellulose
16. Which one is an example of multimolecular colloid system
(a) Soap dispersed in water
(b) Protein dispersed in water
(c) Gold dispersed in water
(d) Gum dispersed in water
17. The formation of micelles takes place only above
(a) inversion temperature
(b) Boyle temperature
(c) critical temperature
(d) Kraft temperature
18. Hardy-Schulze rule explains the effect of electrolytes on the coagulation of colloidal
solution. According to this rule, coagulation power of cations follow the order
(a) Ba+2 > Na+ > Al+3
(b) Al+3 > Na+> Ba+2
75
(c) Al+3 > Ba+2 > Na+
(d) Ba+2 > Al+3 > Na +
19. Brownian movement depends on
(a) Viscosity of the solution
(b) Size of the particles
(c) Bombardment of particles by the molecules of dispersion medium.
(d) All the above
20. Flocculation value of BaCl2 is much less than that of KCl for sol A and flocculation
value of Na2 SO4 is much less than that of NaBr for sol B. The correct statement among
the following is :
(a) Both the sols A and B are negatively charged.
(b) Sol A is positively charged arid Sol B is negatively charged.
(c). Sol A is negatively charged and sol B is positively charged.
(d)Both the sols A and B are positively charged
21. How does a delta form at the meeting place of sea and river water?
(a) The electrolyte present in sea water coagulate the clay
(b) the electrolyte present in sea water has no role
(c) the electrolyte present in river water coagulate the clay
(d) Both (a) and (c) are correct
22. Which of the following will show Tyndall effect?
(a) Aqueous solution of soap below critical micelle concentration.
(b) Aqueous solution of soap above critical micelle concentration.
(c) Aqueous solution of sodium chloride
(d) Aqueous solution of sugar
23. On adding few drops of dil. HCl to freshly precipitated ferric hydroxide, a red
coloured
colloidal solution is obtained. This phenomenon is known as
(a) peptisation
(b) dialysis
(c) protective action
(d) dissolution
24. At high concentration of soap in water, soap behaves as_______ _.
(a) Molecular colloid
(b) Associated colloid
(c) Macromolecular colloid
(d) Lyophilic colloid
25. Lyophilic sols are more stable than lyophobic sols because
(a) They are solvent loving
76
(b) The interphase between colloidal particle and dispersion medium is stable
(c) They are extensively solvated
(d) All of the above
26. The gold numbers of four protective colloids O, P, Q and R are 0.005, 0.01, 0.1 and
0.5 respectively. The decreasing order of their protective power is
(a) R, Q, P, O
(b) O, P, Q, R
(c) P, Q, R, O
(d) Q, R, O, P
27. The blue colour of the sky is due to
(a) Reflection of blue light by colloidal particles
(b) Refraction of blue light by colloidal particles
(c) Dispersion of blue light by colloidal particles
(d)Scattering of blue light by colloidal particles
28. Which of the following statement is correct
(a) Mixing two oppositively charged sols neutralises their charges and stabilises the
colloid
(b) Presence of equal and similar charges on the colloidal particles provide stability to
the colloids
(c ) Brownian movement de stabilises sols
(d) Lyophobic colloids are called protective colloids
29. When a sulphur sol is evaporated sulphur is obtained whichon mixing with water
sulphur sol is not formed. The sol is
(a) Reversible
(b) Hydrophobic
(c ) Hydrophilic
(d) Lyophilic
30. Potential difference of the electrical double layer formed in a colloidal sol is called:
(a) EMF
(b) Zeta potential
(c) Brownian potential
(d) Nernst potential
31. Separation of colloidal particles from those of molecular dimension with electricity
isknown as
(a) electrolysis
(b) electrophoresis
(c) electrodialysis
(d) none of the above
32. Movement of dispersion medium under the influence of electric field is known as
77
(a) Electrodialysis
(b) Electrophoresis
(c) Electro osmosis
(d) cataphoresis
33. 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
34. Name the phenomenon observed when a beam of light is passed through a starch sol
(a) Electrophoresis
(b) Tyndall effect
(c ) Brownian Effect
(d)Coagulation
35. Coagulating value is the minimum concentration of an electrolyte in
(a) moles per litre of electrolyte required to coagulate a sol
(b) milli moles per litre of electrolyte required to coagulate a sol
(c) moles of electrolyte required to coagulate a sol
(d) grams per litre of electrolyte required to coagulate a sol
36. A protective colloid is
(a) Lyophilic sol which protects a lyophobic sol from coagulation
(b) Lyophobic sol which protects a lyophilic sol from coagulation
(c) Lyophilic sol which protects a lyophilic sol from coagulation
(d) Lyophobic sol which protects a lyophobic sol from coagulation
37. The coagulation of lyophobic sols can be carried out by
(a) Boiling
(b)Electrophoresis
(c )Persistent dialysis
(d) All of the above
38. Cottrell precipitator acts on which of the following principle?
(a) HardySchulze rule
(b) Distribution law
(c) Le Chatelier’s principle
(d) Neutralization of charge on the colloidal particles
39. Critical micelle concentration is
(a) Concentration above which a colloid becomes a solution
(b) Concentration above which a solution becomes colloid
(c) Concentration at which colloid becomes a suspension
(d)None of these
78
40. On adding AgNO3 Solution into KI solution, a negatively charged colloidal sol is
obtained when they are in:
(a) 100 mL of 0.1 M AgNO3 + 100 mL of 0.1 M KI
(b) 100 mL of 0.1 M AgNO3 + 100 mL of 0.2 M KI
(c) 100 mL of 0.2 M AgNO3 + 100 mL of 0.1 M KI
(d) 100 mL of 0.15 AgNO3 + 100 mL of 0.15 M KI
41. Dialysis is based on the fact that
(a) colloidal particles are retained by animal membrane.
(b) animal membrane allows the passage of colloidal particles from low to high
concentration.
(c) animal membrane allows the passage of colloidal particles from high to low
concentration.
(d) colloidal particles have affinity for dispersion medium.
42. Following are some of the examples of solutions
I: Milk
IV:Air
II:Smoke
V: Sea water
III :Gasoline
VI : Blood
Select colloidal solutions:
(a) I, II, III, IV
(b) II, III, IV, V
(c) I, II, VI
(d) II, IV, VI
43. During electrophoresis of a colloidal solution, colloidal particles move towards
(a) anode
(b) cathode
(c) both cathode as well as anode
(d) either cathode or anode
44. Which of the following colloids are formed when Hydrogen sulphide gas is passed
through cold solutionof Arsenious oxide
(a) As2 S3
(b) As2 O3
(c ) As2 S
(d) As2 H2
45. Hydrophilic sols are
(a) Reversible
(c )Unstable
(b)Irreversible
(d)None of these
46. In adsorption of oxalic acid on activated charcoal, the activated charcoal is known as
(a) Adsorbent
(b)Adsorbate
79
(c )Adsorber
(d)Absorber
47. Bredig’s arc method involves
(a) Dispersion
(c )Peptization
(b)Condensation
(d)Both Dispersion and Condensation
48. Which of the following reaction is not used for the preparation of colloidal solution
(a) 2H2 S + SO2 → 3S +2H 2 O
(b) 2Mg + CO2 → 2MgO + C
(c ) FeCl3 + 3H2 O → Fe(OH)3 + 3HCl
(d) 2AuCl3 + 3SnCl2 → 2Au + 3 SnCl4
49. Colligative properties of colloidal sols of same compound are very low in
comparison to their true solutions of same concentration because
(a) The number of particles in a colloidal solution is less.
(b) The number of particles in a colloidal solution is high
(c) colloidal sols do not show colligative properties.
(d) true solutions do not show colligative properties
50. The Enthalpy of chemisorption lies in the range of
(a) 20-40 KJ mol-1
(b)40-80 KJ mol-1
(c )80-160 KJ mol-1
(d )80-240 KJ mol-1
ANSWERS
1(d) Ammonia is more easily liquefiable
2 (b) log x/m = log K + 1/n log P
3 (b) Anhydrous CaCl2 with water
.4 (a) (i), (ii) and (iv)
5 (d) Sorption
6 (a)Coagulative hardening of leather by chemicals
7 (a) between 0 and 1 in all cases
8 (b) adsorption isotherm.
9 (b)1/ n
10 (d) High pressure and low temperature.
11 (a) an exothermic process hence increase in temperature decreases adsorption in cases
where Vander Waals forces exits between adsorbate and adsorbent .
80
12 (b) Gel-butter
13 (d) H2
14(c) Al2 (SO 4)3
15 (d) Nitrocellulose
16.(c) Gold dispersed in water
17. (d) Kraft temperature
18.(c) Al+3 > Ba +2 > Na+
19. (d) All the above
20.(c). Sol A is negatively charged and sol B is positively charged.
21.(a)The electrolyte present in sea water coagulate the clay
22. (b) Aqueous solution of soap above critical micelle concentration
23. (a) peptisation
24. (b) Associated colloid
25. (d) All of the above
26. (b) O, P, Q, R
27.(d)Scattering of light by colloidal particles
28. (b) Presence of equal and similar charges on the colloidal particles provide stability to the
Colloids
29. (b) Hydrophobic
30. (b) Zeta potential
31. (c) electrodialysis
32. (c) Electro osmosis
33. (b) To coagulate the suspended impurities
34. (b) Tyndall effect
35. (b) milli moles per litre of electrolyte required to coagulate a sol
36. (a) Lyophilic sol which protects a lyophobic sol from coagulation
37. (d) All of the above
38. (d) Neutralization of charge on the colloidal particles
39. (b) Concentration above which a solution becomes colloid
40. (b) 100 mL of 0.1 M AgNO3 + 100 mL of 0.2 M KI
81
41. (a) colloidal particles are retained by animal membrane.
42. (c) I, II, VI
43. (d) either cathode or anode
44.(a) As2 S3
45.(a)Reversible
46.(a)Adsorbent
47.(d)Both Dispersion and Condensation
48. (b) 2Mg + CO2 → 2MgO + C
49.(b) The number of particles in a colloidal solution is high
50.(d )80-240 KJ mol-1
ASSERTION AND REASON TYPE QUESTIONS
In these questions , 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
1. Assertion :
Reason:
2. Assertion :
Physisorption decrease with the increase of temperature.
Physisorption is an exothermic process
Powdered substances are more effective as adsorbents than their
Crystalline forms
Reason:
The powdered form of the adsorbent has lesser surface area as
compared to the crystalline form
3. Assertion:
Haber’s
Reason:
4. Assertion :
electrolysis of
It is necessary to remove CO when ammonia is obtained by
process
CO is a good reducing agent.
Platinum and palladium are not used in for carrying out
aqueous solutions .
82
Reason:
5. Assertion :
Reason:
Platinum and palladium are inert electrodes.
Physical adsorption of molecules takes place on surface only.
In this process, the bonds of the absorbed molecules are broken.
6. Assertion :
Colloidal solutions do not show Brownian motion.
Reason:
Brownian motion is responsible for stability of sols.
7. Assertion :
Reason:
8. Assertion :
Reason:
A colloidal sol scatters light but a true solution does not
The particles in a colloidal sol move slowly than in a true solution
Coagulation power of Al
3+
is more than Na+ .
Greater the valency of the flocculating ion added, greater is its
coagulating power
9. Assertion :
Reason:
The enthalpy of physisorption is greater than chemisorption
Molecules of adsorbate and adsorbent are held by van der Waals forces
in physisorption and by chemical bonds in chemisorption
10. Assertion :
Reason:
The coagulation of lyophobic sols can be carried out by boiling.
When a sol is boiled the adsorbed layer is disturbed due to increased
collisions.
11. Assertion :
Reason:
12. Assertion :
Reason:
13. Assertion :
Reason:
Adsorption always exothermic
Adsorption is accompanied by increase of randomness.
Lyophilic colloids are called as reversible sols.
Lyophilic sols are liquid loving
Colloidal particles show Brownian movement.
Brownian movement arises because of the impact of the molecules of
the dispersion medium with the colloidal particles
14. Assertion :
Reason:
15. Assertion :
When river water meets the sea water, delta formation happens.
Peptization occurs
For the coagulation of sols carrying positive charge, PO 4 -3 ions are
more efficient than SO4 -2 or Cl−
Reason:
16. Assertion :
Reason:
17. Assertion :
Reason:
18. Assertion :
This follows Hardy Schulze rule
Deep electric shock causes death of an animal
Electric shock coagulate the blood.
A reaction cannot become fast by itself unless a catalyst is added.
A catalyst alters the speed of a reaction
Increase in surface area, increase in rate of evaporation
83
Reason:
Stronger the intermolecular attractive forces, fast is the rate of
evaporation at a given temperature
19. Assertion :
Reason:
20. Assertion :
Aqueous gold colloidal solution is red in colour
The colour arises due to scattering of light by colloidal gold particles
Hydrated ferric oxide can be easily coagulated by sodium phosphate in
comparison to KCl.
Reason:
Phosphate ions have higher negative charge than chloride ions. Hence,
they are more effective for coagulation.
21. Assertion :
An ordinary filter paper impregnated with collodion solution stops the
flow of colloidal particles.
Reason:
Pore size of the filter paper becomes more than the size of colloidal
particle
22. Assertion :
The micelle formed by sodium stearate in water has −COO −groups at
the surface
Reason:
23. Assertion :
Reason:
24. Assertion :
Surface tension of water is reduced by the addition of stearate. .
Silica gel is used for drying air.
Silica gel adsorbs moisture from air
When a finely divided active carbon or clay is stirred into a dilute
solution of a dye, the intensity of colour in the solution is decreased
Reason:
25. Assertion :
Reason:
The dye is adsorbed on the solid surface
According to Freundlich: x/m=k.p1/n(n >1)
The isotherm shows variation of the amount of gas adsorbed by the
adsorbent with temperature
26. Assertion :
Lyophilic colloids are more stable than lyophobic colloids
Reason:
Due to (i) solvation (ii) charge on the colloidal particles.
27. Assertion :
Potassium chloride preferred over ferric chloride in the
case of a cut leading to bleeding.
Reason:
Blood is a +vely charged colloid. Greater the –ve charge, the faster the
coagulation.
28. Assertion :
When AgNO3 is added to excess of KI, the colloidal particles get
attracted towards anode in electrophoresis.
Reason:
Colloidal particles adsorb common ions (excess) and thus
become charged.
84
29. Assertion :
Sky appears blue colour.
Reason:
Colloidal particles of dust scatter blue light
30. Assertion :
1g of activated charcoal adsorbs more sulphur dioxide
Critical temperature 630K), than methane (critical temperature 190K)
Reason:
Easily liquefiable gases are less readily adsorbed
ANSWERS
1
2
3
4
5
6
7
8
9
10
a
c
d
d
c
d
b
a
d
a
11
12
13
14
15
16
17
18
19
20
c
b
a
c
a
a
d
c
a
a
21
22
23
24
25
26
27
28
29
30
c
b
a
A
c
a
d
a
a
c
CASE BASED QUESTIONS
I. Adsorption is the selective transfer of certain components of a fluid phase, called solutes to
the surface of an insoluble solid. The adsorbed solutes are referred to as adsorbates, and the
solid material as adsorbent. When an adsorbent is exposed to a fluid phase, molecules in the
fluid phase diffuse to its surface (including its pores if it is a porous adsorbent), where they
either chemically bond with the solid surface or are held there physically by weak van der
Waals intermolecular forces . When adsorption is caused by van der Waals forces, it is
referred to as physical adsorption or physisorption, whereas it is called chemical adsorption
or chemisorption if it is caused by chemical forces.
1. Physical adsorption is appreciable at
(a) Higher temperature (b) Lower temperature (c) At room temperature (d) 1000 C
2. Which of the following is not a characteristic of chemisorption
(a) Adsorption is irreversible (b) Enthalpy change is of the order of 40 kJ (c)
Adsorption is specific (d) Adsorption increases with increase of surface area
3. Which one of the following is not a correct statement ?
(a) Physical adsorption is reversible in nature
(b) Physical adsorption involves vanderwaals forces
85
(c) Rate of physical adsorption increases with increase of pressure on the adsorbate
(d) High activation energy is involved for physical adsorption
4. Which characteristic of adsorption is wrong :(a) Physical adsorption in general decreases with temperature
(b) Physical adsorption in general increases with temperature
(c) Physical Adsorption is a reversible process
(d) Adsorption is limited to the surface only
II. Adsorption process is usually studied through graphs known as adsorption isotherm. That is
the amount of adsorbate on the adsorbent as a function if its pressure or concentration at
constant temperature . Normally adsorption decreases with temperature because adsorption is
exothermic process. Physical adsorption shows regular decrease with temperature, but
chemical adsorption first increases then decreases with temperature because it is specific,
requires activation energy and in this process heat supplied may be used as its energy of
activation
1. What will be the effect of increase in temperature on physical adsorption
a) It will decrease
b) It will increase
c) First increase then decrease
d)None of these
2. The extent of adsorption of a gas on a solid depends on
a) Nature of the gas
b) Pressure of the gas
c) Temperature of the gas
d) All are correct
3. The Langmuir adsorption isotherm is deduced using the assumption.
86
(a) The adsorption sites are equivalent in their ability to adsorb the particles.
(b) The heat of adsorption varies with coverage.
(c) The adsorption molecules interact with each other.
(d) The adsorption takes place in multilayers
4. In Freundlich adsorption isotherm x/m = Kp1/n, the value of ‘n’ at low pressure is
(a) more than one.
(b) less than one.
(c) equal to one.
(d) from zero to one.
III. Dialysis is the separation of colloids from dissolved ions or molecules of small
dimensions, or crystalloid, in a solution. A colloid is any substance that is made of particles
that are of an extremely small size: larger than atoms but generally have the size of 10 -7 cm
ranging to 10 -3 cm. A crystalloid is a substance that has some or all of the properties of a
crystal or a substance that forms a true solution and diffuses through a membrane by dialysis.
Dialysis is a process that is like osmosis. Osmosis is the process in which there is a diffusion
of a solvent through a semipermeable membrane.
In 1861, chemist Thomas Graham used the process of dialysis, a process used to separate
colloidal particles from dissolved ions or molecules. Dialysis is possible because of
the unequal rates of diffusion through a semipermeable membrane. A semipermeable
membrane is a membrane that lets some molecules to pass through it while not letting others.
Examples of semipermeable membranes include parchment and cellophane.
With the help of a machine, blood can be filtered and purified, it is known as dialysis. It
assists in maintaining balance in electrolytes and fluids when the kidney fails to perform and
carry out its function. This process of dialysis has been in use since 1940 for the treatment of
kidney related disorders. Dialysis can be of two types – Peritoneal dialysis and
Haemodialysis.
1. The mineral that must be consumed in limited quantities by those undertaking
Dialysis is
(a) Mo
(b) Zn
(c) Fe
(d) K
2. This is the correct definition for Dialysis
(a) Stomach is implanted
(b) Waste materials are eliminated
(c) Substitution of liver enzymes
(d) Increase in the pumping of heart
3.This is the chief principle of Dialysis
87
(a) Capillary action
(b) Adhesion
(c) Reverse – osmosis
(d) Cohesion
4.The membrane used for dialysis is made of _____ _ __
a) cellulose
b) polyvinyl chloride
c) polyethylene
d) chitin
ANSWERS
I
III
1
2
3
4
b
b
c
b
1
2
3
4
d
b
c
a
II
1
2
3
4
b
d
a
c
VERY SHORT ANSWER TYPE
1. Define adsorption
Ans. The phenomenon of accumulation or higher concentration of molecular species (gases
or , liquids) at the surface rather than in the bulk of a solid -or liquid is called adsorption.
2. Write two applications of adsorption.
Ans:
i. In decolourisation of sugar.
ii.In gas masks, charcoal is used which adsorbs poisonous gases in mine.
3. Of physisorption or chemisorption, which has a higher enthalpy of adsorption.
Ans:
chemical
Chemisorption has higher enthalpy of adsorption than physisorption due to
bond formation.
4. What is the effect of temperature on chemisorption.
Ans :Chemisorption increases with increase of temperature.
5. Why is adsorption always exothermic.
Ans: Adsorption is accompanied by decrease of randomness, ΔS is -ve. For the process to be
spontaneous,ΔG must be negative.Hence, according to equation ΔG = ΔH – TΔS, ΔG can be
-ve only if ΔH is negative.
6. Write one similarity between physisorption and chemisorption.
Ans: Both physisorption and chemisorption increase with increase in surface area.
7. Out of SO2 and CH4 which will be adsorbed more readily on 1g activated charcoal.
88
Ans: SO2 will be more adsorbed as it's critical temperature is higher than that of CH4 .Greater
the critical temperature easier will be the ease of liquefaction of gas and more will be the
extent of it's adsorption. Vander waals force will be stronger.
8. What is an adsorption isotherm.
Ans: The variation of the amount of the gas adsorbed by the adsorbent with pressure at
constant temperature can be expressed by means of a curve. This curve is termed as
adsorption isotherm at a particular temperature.
9. What is the empirical relationship between the quantity of gas adsorbed by unit mass
of solid adsorbent and pressure at a particular temperature.
Ans: x/m= k.p 1/n where x is the mass of the 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 gas at a
particular temperature.
10. Out of anhydrous Calcium chloride and Silica gel which one adsorbs water vapour.
Ans: Silica gel.
11. What is an emulsion.
Ans: Emulsion is liquid-liquid colloidal system.
12. What are lyophobic colloids.
Ans: The colloidal solution in which the particles of the dispersed phase have little affinity,
rather have hatred for the dispersion medium are called lyophobic colloids.
13. How can a colloidal solution and true solution of the same colour be distinguished
from each other.
Ans: When a beam of light is passed through both the solution ,path of light is visible only in
colloidal solution due to scattering of light.
14. Give one example each of lyophobic sol and lyophilic sol.
Ans: Lyophobic sol — Metal sulphides
Lyophilic sol — Starch
15. What are associated colloids? Given an example.
Ans:Colloids which act as electrolyte at low concentration and show colloidal behaviour at
high concentration are called Associated colloids. Example : Soap solution, Detergents.
16. What type of colloid is formed when a liquid is dispersed in a solid? Give an
example.
Ans:Gel is formed when a liquid is dispersed in a solid, e.g., Cheese, butter, etc.
17. What type of colloid is formed when a solid is dispersed in a liquid? Give an
example.
Ans: Sol is formed when a solid is dispersed in a liquid, e.g., paints
18. What is Kraft temperature.
Ans: A particular temperature above which the formation of micelles takes place is called
Kraft temperature.
89
19. What is critical micelle concentration.
Ans: A particular concentration above which the formation of micelle takes place is
called critical micelle concentration.
20. Define the term ‘Tyndall effect’.
Ans: When a beam of light is passed through a colloidal solution and viewed perpendicular to
the path of the incident light, the path of light becomes visible as a bright streak due to
scattering of light.The illuminated path is called Tyndall cone and the phenomenon is called
Tyndall effect.
21. What is the ‘coagulation’ process.
Ans:The process of settling of colloidal particles is called coagulation or precipitation of the
solution.
22. Define ‘peptization’.
Ans: Peptization may be defined as the process of converting a precipitate into colloidal sol
by shaking it with dispersion medium in the presence of a small amount of electrolyte.
23. Define ‘electrophoresis’.
Ans:When electric current is passed through a colloidal solution, the positively charged
particles move towards cathode while negatively charged particles move towards anode
where they lose their charge and get coagulated. The phenomenon is known as
Electrophoresis.
24. What is electrophoresis due to?
Ans: Electrophoresis is due to electrical charge on the colloidal particles.
25. In reference to surface chemistry, define dialysis.
Ans :The process of removing the dissolved substances from a colloidal solution by means of
diffusion through a suitable membrane is called dialysis.
Short Answer Type [SA – I]
26. Describe a conspicuous change observed when
(i) a solution of NaCl is added to a sol. of hydrated ferric oxide.
(ii) a beam of light is passed through a solution of NaCl and then through a sol.
Ans: (i) Coagulation or precipitation of sol takes place.
(ii) Path of the light gets illuminated due to Scattering of light(Tyndall effect is
observed when light is passed through a sol.Path of light is not visible in a solution of NaCl.
27. Explain the following terms giving one example for each :
(i) Micelles
(ii) Aerosol
90
Ans: (i) Micelles : They are associated colloids showing colloidal behaviour at high
concentration and strong electrolytes at low concentration.
Example : soap, detergent.
(ii) Aerosol : Aerosol is a colloidal solution of solid or liquid (dispersed phase) in gas
(dispersion medium).
Example : smoke, dust, fog, mist, cloud.
28. What is the difference between multi-molecular and macromolecular colloids? Give
one example of each.
Ans: Multi-molecular colloid is aggregation of large number of atoms or smaller molecules
of a substance having size in the colloidal range. Whereas macromolecular colloid is the
solution containing macromolecules in the colloidal range.
Example: Multimolecular colloids: Gold sol, Sulphur sol
Macromolecular colloids: Proteins,Cellulose .
29. Explain the following :
(a) Same substance can act both as colloids and crystalloids.
(b) Artificial rain is caused by spraying salt over clouds.
Ans: (a) Sodium chloride behaves as a crystalloid when dissolved in water but behaves as a
colloid when dissolved in benzene.
(b) Artificial rain is caused by spraying common salt over the clouds, as it is an electrolyte
and brings about coagulation of water particles in clouds which is a colloid.
30. How are the following colloidal solutions prepared?
(a) Sulphur in water
(b) Gold in water
Ans: (a) Sulphur in water : An alcoholic solution (true solution) of sulphur in excess of water
is added, then colloidal solution is obtained. Or, Oxidation of H 2 S by passing SO 2 gas
through it.
SO2 +2 H2 S → 2H2 O + 3S(sol)
(b) Gold in water : Gold solution can be prepared by reduction of AuCl3 solution with
formaldehyde.
2AuCl3 + 3HCHO + 3H2 O → 2Au Gold Sol. + 3HCOOH + 6HCl
Short Answer Type – II [SA – II]
31. i. Which of the following is most effective in coagulating negatively charged
hydrated ferric oxide sol?
(i) NaN03 (ii) MgSO 4 (iii) AlCl3
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ii. Which of the following is most effective in coagulating positively charged hydrated ferric
oxide sol?
(i) NaNO 3 (ii) Na2 SO4 (Hi) (NH 4 )3 PO4.
iii. Which of the following is most effective in coagulating positively charged methylene blue
sol?
(i) Na3 PO4 (ii) K4 [Fe(CN)6 ] (iii) Na2 SO4
Ans: i.AlCl3 (Aluminium chloride) is most effective in coagulating negatively charged
hydrated ferric oxide sol.
ii. Ammonium phosphate (NH4)3PO4 is most effective in coagulating positively charged
hydrated ferric oxide sol
iii. Potassium ferrocyanide K 4 [Fe(CN) 6 ]
32. What is the difference between multimolecular and macromolecular colloids? Give
one example of each. How are associated colloids different from these two types of
colloids.
Answer:
Multimolecular colloids
Macromolecular colloids
Associated colloids
1. They are formed by the
aggregation of a large
number of atoms or
molecules which generally
have diameter less than 1
mm, e.g. sols of gold,
sulphur etc.
They are molecules of large
size but in colloidal particle
size. e.g. polymers like
rubber, nylon, starch, proteins
etc.
They are formed by
aggregation of a large
number of ions in
concentrated solution
e.g. soap sol.
2. Their molecular masses
are not very high.
They have high molecular
masses.
Their molecular masses
are generally high.
3. Their atoms or molecules
are held together by weak
Van der Waals forces.
Due to long chain, the Van
der Waals forces holding
them are comparatively
stronger.
Higher is the
concentration, greater
are the Van der Waals
forces.
33. Give reason for the following
(i) Cottrell precipitator is fitted in the chimney of factories.
(ii) Peptizing agent is added to convert a precipitate into a colloidal solution.
(iii) Colloidal gold is used for intramuscular injection.
Answer:
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(i). smoke is a colloid, due to electrophoresis carbon particles of smoke gets coagulated
and pure air will go out from the chimney
(ii).A precipitate adsorb certain ions (positive or negative) from the electrolyte depending
upon its nature and gets charged. These charged ions repel one another and change to
colloidal size particles.
(iii).Particles of colloidal gold get easily adsorbed or assimilated by the particles of blood
which are also of colloidal nature.
34. How are the following colloids different from each other in respect of their dispersion
medium and dispersed phase.
(i) Aerosol
(ii)Alcosol
(iii) Hydrosol
Answer:
(i) An aerosol: It is a colloidal solution in which dispersed phase is a solid and dispersion
medium is a gas. Example : Smoke.
(ii) Alcosol : : The colloidal solution where alcohol is used as the dispersion medium is called
alcosol.
(iii) A hydrosol : The colloidal solution where water is used as the dispersion medium is
called hydrosol or aquasol. Example : Starch sol.
35. Explain the following:
(i) Deltas are formed when river and sea water meet.
(ii) Finely divided substance is more effective as an adsorbent.
(iii) Physisorption is multi-layered, while chemisorption is mono-layered.
Ans: (i) River water is a colloidal solution of clay. Sea water contains a number of
electrolytes. When river water meets the sea water, the electrolytes present in sea water
coagulate the colloidal solution of clay resulting in its deposition with the formation of deltas.
(ii) Finely divided substances have more surface area so they are more effective as an
adsorbent.
(iii) In physical adsorption, layers of the gas can be adsorbed one over the other by van der
Waals forces. Multi-molecular layers are formed under high pressure. In chemical adsorption,
chemical bond can be formed only with the layer of molecules coming in direct contact with
the surface of the adsorbent, hence this type of adsorption is mono-layered.
36. Give reasons for the following observations :
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) Colloidal medicines are more effective
Answer:
(i) Due to mutual coagulation of leather by tanning. [Positively charged animal hide (leather)
with negatively charged colloidal particles of tannin].
(ii) Lyophilic sols are more stable because there is strong interaction between dispersed phase
and dispersion medium.
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(iii) Colloidal medicines are more effective because they have large surface area and are
therefore easily assimilated.
37. Write any three differences between Physisorption and Chemisorption.
Answer:
Basis
Physisorption
Chemisorption
(i) Specificity
It is not specific in nature i.e.
all gases are adsorbed on all
solids to some extent.
It is highly specific in nature and occurs only
when there is some possibility of compound
formation between the gas being adsorbed
and the solid being adsorbent.
(ii)
Temperature
dependence
It is independent of
temperature as it occurs at low
temperature and deceases with
increase in temperature.
It is temperature dependent and increase with
increase in temperature.
(iii)
Reversibility
It is reversible i.e. desorption
of gas takes place by
increasing the temperature or
decreasing the pressure.
It is irreversible in nature as it involves
formation of compound instead of release of
gas.
(iv) Enthalpy
change
It has low enthalpy of
adsorption i.e., 20-40 kj mol-1 .
It has high enthalpy of adsorption i.e., 40240 kj mol
38. i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCl3 which one is more effective in causing coagulation of negatively
charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids?
Ans:
Adsorption
(a) It is the phenomenon by which one
substance gets concentrated mainly on the
surface of the other substance rather than in
the bulk of a solid or liquid.
(b) It is a surface phenomenon.
(c) Its concentration is different at surface
from the bulk.
Absorption
(a) It is the phenomenon by which one
substance gets uniformly distributed
thoughout the body of the other substance.
(b) It is a bulk phenomenon.
(c) Its concentration is same throughout
the bulk.
94
(ii) AlCl3 is more effective than MgCl2 in causing coagulation of negatively charged sol
because coagulating power of an electrolyte is directly proportional to the valency of the
active ion i.e., Al3+> Mg2+.
(iii) Sulphur sol forms multimolecular colloids.
39.(a) Write the dispersed phase and dispersion medium of milk.
(b) What is observed when an electric current is passed through a colloidal solution.
(c) Write the chemical method by which Fe(OH) 3 sol is prepared from FeCl3 .
Ans: (a) In milk dispersed phase is liquid fat and dispersion medium is water.
(b). When electric current is passed through a colloidal solution, the positively charged
particles move towards cathode while negatively charged particles move towards anode
where they lose their charge and get coagulated. The phenomenon is known as
Electrophoresis
(c) Fe(OH)3 is prepared from FeCl3 by hydrolysis.
40. Define the following terms :
(i) Helmholtz electrical double layer
(ii) Hardy -Schulze rule.
(iii) Electrokinetic potential or Zeta potential
Ans: i.The combination of the two layers of opposite charges around the colloidal particle is
called Helmholtz electrical double layer.
ii. The greater the valence of the floculating ion added ,the greater is its power to cause
precipitation.
iii. The potential difference between the fixed layer and the diffused layer of opposite
charges in colloids is known as electrokinetic potential or Zeta potential.
41. What is Brownian movement. what is the cause of it. How is it responsible for the
stability of colloids.
Ans: The zig -zag or random motion of colloidal particles is called Brownian movement.
It arises due to the unbalanced bombardment of the colloidal particles by the molecules
of the dispersion medium.
The Brownian movement has a stirring effect which does not permit the particles to
settle and thus is responsible for the stability of sols.
95
CHAPTER 8
The d - and f- BLOCK ELEMENTS
GIST OF LESSON
 Transition elements
 Configurations of d block elements
 General characteristics
 Metallic Character
 Melting point
 Atomic size and ionic size
 Density
 Ionisation Enthalpies
 Oxidation states
 Standard electrode potentials and trends in oxidation states
 Stability of higher oxidation states
 Chemical reactivity and electrode potentials
 Magnetic properties
 Formation of coloured compounds
 Formation of complex compounds
 Catalytic properties
 Formation of interstitial compounds
 Alloy formation
 Inner transition elements
 General properties of f-block elements
 Oxidation state of lanthanoids
 Uses of d & f block elements
Deleted Topics :1. Preparation and properties of K2Cr2O7 and KMnO4 , (2). Chemical reactivity of Lanthanoids
And (3). Actinoids- Electronic configuration, oxidation states and comparison with lanthanoids.
CONCEPTS AND EXPLANATIONS
TRANSITION ELEMENTS:
Transition elements include elements from group III to XII. These elements have properties which are
intermediate between those of s and p block elements. A transition element is defined as the one which has
incompletely filled d orbitals in its ground state or in any one of its oxidation states. These are placed in the
middle part (between s and p) of the periodic table. There are four series of elements.
First series
Sc to Zn (At: No: 21 to 30)
Second series
Y to Cd (At: No: 39 to 48)
96
Third series
La, Hf to Hg (At: No: 57, 72 to 80)
Fourth series
Ac, Rf to Cn (At: No: 89, 104 to 112)
CONFIGURATIONS OF d BLOCK ELEMENTS:
General electronic configuration of d block elements is (n-1) d 1-10 ns 1-2. The inner d orbital is being filled
with increase in atomic number. There are four different transition series. Electronic configuration of first d
series is as follows
Sc
[Ar] 3d1 4s2
Ti
[Ar] 3d2 4s2
V
[Ar] 3d3 4s2
Cr
[Ar] 3d5 4s1
Mn
[Ar] 3d5 4s2
Fe
[Ar] 3d6 4s2
Co
[Ar] 3d7 4s2
Ni
[Ar] 3d8 4s2
Cu
[Ar] 3d10 4s1
Zn
[Ar] 3d10 4s2
Special configuration of Cr and Cu is due to the extra stability of half- filled and fully filled orbitals (d5 and
d10). Also the energy difference between the 3d and 4s is not very large.
GENERAL CHARACTERISTICS:


Almost all transition elements show typical metallic properties, high tensile strength, ductility,
malleability, high thermal and electrical conductivity and metallic lustre.
Zn, Cd and Hg is considered as non-transition elements as these elements have completely filled d
orbitals (d10 configuration)
METALLIC CHARACTER:
Except Hg which is a liquid, all transition elements have typical metallic structure. Metallic character is due
to their relatively low ionization enthalpy and one or two electrons in the outermost orbit. These elements are
electropositive.
MELTING POINT AND BOILING POINTS:
Melting and boiling points of transition elements depends on the number of unpaired electrons. Greater the
number of unpaired electrons stronger is the metallic bonding and hence high melting points.




In a particular series, melting point increases up to the middle and then decreases with decrease in
number of unpaired electrons. So Zn, Cd and Hg have low melting and boiling points.
There is a dip in the melting points of Mn and Tc in the figure. Mn has stable configuration and
electrons are more tightly held by the Mn atoms. This reduces the delocalization of electrons resulting
weaker metallic bonding.
Transition elements have high enthalpy of atomization due to the overlap of unpaired electrons
between the atoms resulting strong metallic bonding.
Zn, Cd and Hg have low enthalpies of atomization as there is no unpaired electrons.
97
Enthalpy of atomization of first row transition
elements.
ATOMIC SIZE AND IONIC SIZE:








Size of transition elements are
intermediate between s and p block
elements.
The atomic size first decreases with
increase in atomic number due to
increase in nuclear charge but low
shielding effect.
Atomic size become almost steady
towards the middle of the series
because of filling of d orbitals which
screens outer shell of electrons. So
increase in nuclear charge is partly
cancelled by increase in screening
effect.
Towards the end of the series, there is
slight increase in size due to the Variation of atomic size transition elements
increased repulsion between the
electrons in the same orbitals.
Atomic size of second transition series
are found to be larger than first series.
Ionic size of lanthanoids also follow the same order as that of atomic size. Ionic radii are different in
different oxidation states due to increase in effective nuclear charge. So M 3+cations are smaller than
M2+
Atomic sizes of third transition series are found to be almost same as that of second series. This is
explained on the basis of lanthanoid contraction.
Steady decrease in atomic size of lanthanoids due to poor shielding effect of 4f electrons is called
lanthanoid contraction. As shielding effect is less, nuclear charge increases along the series and hence
size decreases.
98
CONSEQUENCES:




Similarity in size of Zr and Hf (160 pm and 159pm respectively) is due to lanthanoid contraction.
Difficulty in separation of lanthanoids in pure state. Since the change in ionic size of lanthanoids is
small, their chemical properties are similar. This makes the separation of lanthanoids difficult in pure
state.
Basic strength of lanthanoid hydroxides decrease from La3+ to Lu3+ which is due to increase in
covalent character of hydroxides. So La (OH)3 is more basic than Lu (OH)3
Ionic size of lanthanoids also follow the same order as that of atomic size. Ionic radii are different in
different oxidation states due to increase in effective nuclear charge. So M 3+cations are smaller than
M2+
DENSITY:
Density of transition elements increase along the series due to decrease in atomic size and increase in nuclear
charge. (atomic volume decreases and atomic mass increases).
IONISATION ENTHALPIES:











The first ionization enthalpy of d – block elements are higher than s and p block elements. This is due
to increase in nuclear charge and poor shielding effect of ‘d’ electrons.
Increase in ionization enthalpy is not regular. This is because removal of one electron alters the relative
energies of 4s and 3d orbitals, so there is reorganization of energy accompanying ionization.
First ionization enthalpy of Zn is very high because the electron has to be removed from 4s orbital of
stable configuration (3d10 4s2).
Second ionization of Zn is found to be lower. This is because after the removal of one electron from
Zn+ it attains stable 3d10 configuration.
Cr and Cu have exceptionally high 1E2 (second 1E) than the neighboring elements due to the stability
of half- filled and fully filled configuration respectively. (3d5 and 3d10)
Third 1E of Zn2+ and Mn2+ is very high due to stability of 3d10 and 3d5 configuration.
Third 1E of Fe (Fe+2 to Fe+3) is low because loss of 3rd electron gives stable 3d5 configuration.
Very high values of third 1E of Cu, Zn, Ni indicate that it is difficult to show +3 oxidation state for
these elements.
The first 1E of third transition series in very high. This is due to the poor shielding effect of 4f electrons.
So outer electrons are attracted by the nucleus with a greater force.
Ni+2 compounds are well-known but Pt+2 compounds are rare. This is because the first two ionization
enthalpies of Ni are lower than Platinum. (1E1 + 1E2)
Pt+4 compounds are well-known but Ni+4 compounds are rare. This is because the first four1Es (1E1
+1E 2, 1E3 +1E4) of Pt are lower than that of Ni. That is, K2 [Pt Cl6] is available but K2 [Ni Cl6] is not
known.
In K2 [Pt Cl6]
Pt is +4
In K2 [Ni Cl6]
Ni is +4
OXIDATION STATES:
Transition elements exhibit variable oxidation state. This is due to the participation of electrons in the ns and
(n-1) level which are of almost same energy. Also incomplete d orbitals. Both ns and (n-1) d orbitals involve
in bonding.
99
Stability of a particular oxidation state depends on the nature of the element with which the element is
combined. The highest oxidation states are found in fluorides and oxides because of their high
electronegativity and small size.















Out of fluorides and oxides, stability of higher oxidation states of oxides are more than fluorides due
to multiple bond formation of oxygen with metals (oxo anions).
So highest fluorides of Mn is MnF4 but oxide is Mn2O7.
Maximum number of oxidation states are shown by elements which are present almost middle of the
series.
Transitions elements show oxidation states from +1 to +7. +8 is shown by Rh and Os (Ruthenium and
Osmium)
Except Sc all other elements from the first transition series show +2 oxidation state.
Sc is the element which shows only one oxidation state i.e. +3 and is the stable oxidation state. [Refer
table 8.3 textbook)
The lowest oxidation states correspond to the number of ns electrons. Eg; Cu. It shows +1oxidation
state. For others it is +2.
The highest oxidation state is equal to the sum of ns and (n-1) electrons.
In the first series highest oxidation state and maximum number of oxidation states are shown by Mn.
The highest oxidation state is +7 and others +2, +3, +4, +5, +6.
Ti+4 is more stable than Ti+2 and Ti+3 because Ti+4 is 3d0 4s0
For Zn, Zn+2 is the most stable oxidation state. Zn+2 is 3d10 4s0
Some elements show zero oxidation state in compounds eg Ni, Fe; [Ni (CO) 4 ] , [Fe(CO)5]
In p block elements, higher oxidation states are stable for lighter elements. Eg Tl+1 is more stable than
Tl +3 due to inert pair effect. Also, Pb+2, Sn+2 etc. are stable than Pb+4 and Sn+4
For transition elements heavier elements show more stable higher oxidation state. Eg: Mo +6, W+6 etc.
are more stable than Cr+6. So, Cr+6 act as oxidizing agent. Eg: K2Cr2O7 is a powerful oxidizing agent.
But WO3 and MoO3 are not oxidizing agents.
+2 and +3 oxidation states are shown generally by ionic compounds and in higher oxidation states the
compounds are covalent in nature.
STANDARD ELECTRODE POTENTIALS AND TRENDS IN OXIDATION STATES:
The stability of a particular oxidation state in aqueous solution depends on ionization enthalpy (IE),
sublimation enthalpy (∆H sub) and enthalpy of hydration (∆H hyd)
∆H


= ∆H sub
+ IE
+ ∆H hyd
Smaller the value of ∆H for a particular oxidation state, greater will be the stability of that oxidation
state.
E0 values of first row transition elements are irregular (M +2/M). This is explained with the irregular
variation in IE (IE1 + IE2) and sublimation enthalpies which are very less for Mn and V
TRENDS IN OXIDATION STATE (M+2/M).
The general trend towards less negative E0 values across the series is related to increase in sum of first and
second ionization enthalpies (IE1 + IE2).
100






E0 of Mn and Ni, Zn are more negative
than expected from the general trend.
For Mn and Zn it is related to the halffilled (d5) and fully filled (d10) state
respectively. The exceptional behavior
of Ni towards E0 is due to its high
negative ∆H hyd
Zn has low ∆Ha (atomization) and high
∆Hhyd. But it has low reduction
potential due high IE (IE1 + IE2).
Cu has positive value of E0 (+0.34v)
shows that this is the least reactive
metal among the elements of first
series. This unique behavior (+E0) is
responsible for the inability of Cu to
release hydrogen from acids. The very
high IE is not balanced by its high
hydration enthalpy.
E0 of Mn and Ni, Zn are more negative than expected from the general trend. For Mn and Zn it is
related to the half-filled (d5) and fully filled (d10) state respectively. The exceptional behavior of Ni
towards E0 is due to its high negative ∆H hyd
Zn has low ∆Ha (atomization) and high ∆H hyd. But it has low reduction potential due high IE (IE1 +
IE2).
Cu has positive value of E0 (+0.34v) shows that this is the least reactive metal among the elements of
first series. This unique behavior (+E0) is responsible for the inability of Cu to release hydrogen from
acids. The very high IE is not balanced by its high hydration enthalpy.
TRENDS IN OXIDATION STATE (M+3/M+2).






Except Cu and Zn all elements show +3 oxidation state.
Very low value for E0 Sc3+/Sc2+, shows that Sc3+ is highly stable.
Highest E0 value of Zn is due to its stable configuration in Zn+2 state. It is difficult to remove electron
from d10 configuration to change into +3 state.
High value of E0 Mn3+/Mn 2+ is due to half- filled stability of Mn2+ (d5)
Comparatively low value of E0 Fe 3+/ Fe2+ is due to the extra stability of d5 state. It has low value of
third ionization enthalpy.
Low value of E0 V+3/V+2 is related to half-filled t2g3 configuration
CHEMICAL REACTIVITY AND ELECTRODE POTENTIALS
Transition elements vary widely in their chemical reactivity. Some metals are highly electropositive and
dissolve in acids. But some are noble metals which are not affected by acids.


First row transition elements (except Cu) are more reactive than the other series. These metals oxidized
by acids. But Ti and V are passive to non-oxidizing acids.
E0 values show that Mn+3 and Co+3 are the strongest oxidizing agents in aqueous solution whereas Ti+2,
V+2, Cr+2 are the strongest reducing agents and liberate hydrogen from dilute acids
101
STABILITY OF HIGHER OXIDATION STATES:
METAL HALIDES:










The reactivity order of halogens with metal is F > Cl > Br >I.
Transition metals show highest oxidation state in fluorides as it has high lattice enthalpy,
electronegativity or due to high bond enthalpies.
CoF3 is stable due to its high lattice enthalpy. But VF5 and CrF6 are stable due to high bond enthalpies
The highest oxidation state in halides vary from +4 to +6. For Ti, it is Ti+4, for V, it is V+5 and for Cr,
it is Cr+6. Mn does not show simple oxidation state with +7 in halides. But in MnO3F is available with
+7
In first transition series, beyond Mn, no element shows oxidation state more than +3
Fe and Co form trihalides – FeCl3 and CoF3
V shows +5 only in VF5 but other halides undergo hydrolysis to give Oxo halides (VOX 3) in which V
is in +5 state.
In lower oxidation states fluorides are unstable. Cu+2 halides are known, but +1 are unstable as it
disproportionate to Cu+2 and Cu.
2 Cu +1 → Cu 2+ + Cu
Cu I2 is not known as Cu+2 oxidises iodide to iodine (I− to I2)
2Cu+2 + 4 I− → Cu2I2 + I2
Cu+2 is more stable than Cu +1. This due to much more negative ∆H hyd for Cu+2 than Cu+1 which
compensates the high second ionization enthalpy of Cu
METAL OXIDES:





The ability of oxygen to stabilize highest oxidation state is more than fluorine as it can form multiple
bonds with metals.
The highest oxidation state is same as that of group number up to group 7. Beyond group 7 maximum
oxidation state is +3.
Higher oxidation states are seen in oxo cations and oxo anions. Eg: VO2+, VO2+ , TiO2+, MnO4− , CrO4−
, Cr2O72−etc
Among halides, highest halides are with oxidation number +4.But in oxides +7 state is found in Mn.
Eg: Highest fluoride of Mn in MnF4 whereas highest oxide is Mn2O7.
The order of increasing oxidation power in the first series VO2+ < Cr2O72− < MnO4− are explained
with the increasing stability of the lower species to which it is reduced.
MAGNETIC PROPERTIES
Paramagnetic substances are substances which are attracted in the magnetic field. these are having one or more
unpaired electrons. Most of the transition elements are paramagnetic in nature and are attracted in the magnetic
field. This is due to the presence of unpaired electrons in their metal ions and in their compounds.
Magnetic properties are expressed in terms of magnetic moments in Bohr Magneton units (BM) Formula for
calculating spin only Magnetic moment μ = √n(n+2) BM where ‘n’ is the number of unpaired
electrons.
FORMATION OF COLOURED COMPOUNDS
Most of the transition metal compounds are coloured both in solid state as well as in aqueous state. Colour is
due to the presence of incomplete d orbitals with one or more unpaired electrons. Transition of electron from
the lower energy d orbital to high energy d orbital (d-d transition) occur and the energy of excitation
102
corresponds to the frequency of light. This frequency lies in the visible region. The colour observed
corresponds to the complementary colour of light absorbed.




Sc+3 is colourless Sc+3
[Ar] 3d0 4s0 No unpaired electrons
+2
+2
+2
Zn , Cd and Hg are also colourless. All are having d10 configuration
Ti+3 is purple coloured Ti+3
[ Ar] 3d1 4s0
Cu+1 salts are colourless but Cu+2 salts are coloured Cu+2 [Ar] 3d9 4s0 One unpaired electron in 3d
orbital.
FORMATION OF COMPLEX COMPOUNDS
Most of the transition elements form complex compounds due to small size and high ionic charges of the
metal ions and availability of d orbitals of suitable energy to accept electron pairs
Eg:
K4 [Fe (CN) 6]
CATALYTIC PROPERTIES
Most of the transition elements are used as catalysts. This is because the ability of transition elements to adopt
multiple oxidation states and to form complexes or intermediate compounds of variable oxidation state. Most
of these metals are in solid state and hence it can provide enough surface area for intermediate formation.
Eg: Fe+3 catalases the following reaction, 2 I− + S 2O 82− → I 2 + 2SO
This reaction proceeds as
2Fe3+ + 2 I− →
2Fe +2 + S2O82−
→
2−
4
2Fe+2 + I2
2Fe3+ + 2SO 42−
Ni, Pd, Pt etc. are used in hydrogenation and reduction reactions, V2O5 is used in Contact process, Fe and Mo
in Haber’s process and MnO2 in decomposition of H2O2
FORMATION OF INTERSTITIAL COMPOUNDS
Transition elements can trap some smaller atoms like carbon, hydrogen or nitrogen inside the crystal lattices
of metals. These compounds are non-stoichiometric in nature. These compounds are chemically inert, hard,
have high melting point, retain metallic conductivity.
ALLOY FORMATION:
Transition elements form homogeneous solid solutions or alloys as these metals have similarity in size and
characteristics. So substitution of one metal by the other in its crystal lattice become easier. Eg: Brass, Bronze,
alloy steels, stainless steels etc.
INNER TRANSITION ELEMENTS:
These are f –block elements which includes lanthanoids and actinoids. These are called inner transition
elements as the last electron enter into the (n-1) f-orbital. Fourteen elements comes after lanthanum (At No 58
to 71) are called lanthanoids and fourteen elements comes after Actinium (At No 90 to103) are called
Actinoids.


General configuration of these elements are (n-2) f1-14 (n-1) d 0-1 n s2
There is decrease in size for lanthanoids throughout the series which is called as lanthanoid
contraction. [Explanations and consequences of lanthanoid contraction are given along with d
block elements]
103
GENERAL PROPERTIES OF 4f – SERIES (LANTHANOIDS)








Lanthanoids are silvery in appearance and soft metals.
These have high melting points
All have typical metallic structure and good conductors of heat and electricity
These elements have high density
Most of the trivalent cations are coloured both in solid as well as in aqueous solution. This due to
partially filled f-orbitals which permit f-f transition. La +3 and Lu+3 are colourless
Most of the lanthanoids show para magnetism due to the presence of unpaired electrons in the 4f orbital.
Except Ce+4, Yb+2 and Lu+3, others are paramagnetic. Their spin only magnetic moments can be
calculated using the formula:
Magnetic moment
μ = √n (n+2) BM
The first and second ionization enthalpies are 600KJ/mol and 1200KJ/mol respectively which are
comparable with those of calcium. Variation in the third ionization enthalpies involves exchange
enthalpy considerations to explain the stability of empty, half-filled and completely filled 4f orbitals.
These elements are electropositive and readily lose their electrons. So, these are good reducing agents
OXIDATION STATE OF LANTHANOIDS:
Lanthanoids show +3 as very common oxidation state. +2 and +3 also shown by some elements. This
irregularity arises mainly from extra stability of empty, half-filled and completely filled f subshell.
Lanthanoids show limited number of oxidation states because the energy gap between 4f and 5d orbitals is
large. When the oxidation states are compared with transition metals which show many oxidation states the d
electrons are in (n-1) d orbital and can take part in bond formation. But in lanthanoids the f-electrons are
present in deeper (n-2) f level which cannot take part in bond formation.



Ce +4/Ce+3 has noble gas configuration. But it is a strong oxidant reverting to common oxidation state
of Ce+3. Tb +4 has half-filled f orbital, but it is an oxidant.
E0 value for Ce+4/Ce+3 is 1.74V suggests that it can oxidise water. But reaction rate is very slow. So,
Ce (IV) is used as a good analytical reagent
Eu+2 is a strong reducing agent though it has f 7 configuration. It changes to very common+3 oxidation
state. Yb+2 has f 14 configuration is a reluctant.
USES OF d- & f- BLOCK ELEMENTS:




Transition elements are used for making wide varieties of alloys, Mish metal is lanthanoid alloy used
for making bullets, shells and lighter flints.
Many metals and compound are used as catalysts, some complexes are used as catalysts and in medical
field. Eg Cis-Platin is used in cancer chemotherapy, 60Co is used in radiation therapy.
Some compounds of transition elements are used as powerful oxidizing agents.
Transition metals widely used in industry. Zn metal is used for galvanization and for making calamine
lotions.
104
MINIMUM LEVEL OF LEARNING


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.

General characteristics :
Transition series : d-block consists of fourtransition 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
4th Transition series or 6d series 89Ac, 104Rf – 112Cn
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 ofelectrons 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
paramagneticdue to presence of unpaired electrons in the
(n – 1)d-orbitals and it is calculated by using the formula,
magnetic moment
μ = √n(n+2)BM 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 easilyaccommodated
Due to similar atomic sizes
Inner Transition Elements( f-Block Elements)
▶ Lanthanoids : Last electron enters one of the 4f-orbitals. Cerium (At. no. 58) to
Lutetium(At.no.71).
▶ 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 – 1ns2
105
General characteristics of lanthanoids :
Atomic and ionic
radii
Oxidation states
Action of air
Coloured ions
Decrease on going from La to Lu.
Most common oxidation state of lanthanoids is +3. Some elements
exhibit
+2and+4oxidationstatesduetoextrastabilityofempty,half-filledorfully- filled fsubshell, 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+.
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).
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
lanthanoidsare 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+.
 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 effect hence contraction in size occurs.
 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 havesimilarpropertiesdue to samesize).
106
107
MULTIPLE CHOICE QUESTIONS
1.(n-1)d10ns2 is the electronic configuration of
(a)Fe ,Co ,Ni (b) Cu ,Ag .Au (c) Zn ,Cd ,Hg (d)Se ,Y ,La
2. The lowest melting point metals among d-block metals belongs to
(a)Group 3 b) Group 11 (c) Group 6 (d) Group 12
3. Cu+ ion in aqueous medium undergoes
(a)Oxidation only (b )Reduction only (c) Neither oxidation nor reduction
(d) Disproportionation
4. Which of the following oxide is least basic?
(a) CrO (b)Cr2O3
(c)CrO3
(d)Cr2O4
5. Electronic configuration of a transition element X in +3 oxidation state is [Ar] 3d 5 .
What is its atomic number?
(a) 25 (b) 26 (c) 27 (d) 24
6. Which of the following is not a transition element?
(a)Zn
(b) Ru ( c)Ag (d)Pb
7. In the first transition series which of the following has lowest enthalpy of atomisation?
(a)Sc (b) Cu (c) Ti (d)Zn
8. Which element does not show variable oxidation state ?
(a)Sc ( b )V (c)Fe (d) Hg
9. The maximum oxidation state of osmium is
(a) +6 (b)+7
(c)+8 ( d)+5
10. Which of the following ion is colourless in aqueous solution ?
(a) Fe ²+ (b) Mn 2+ (c)Ti 3+ (d)Sc 3+
11. Which of the following statement about transition elements is incorrect?
(a) They show variable oxidation states (b) All the ions are coloured
(c) They exhibit diamagnetic and paramagnetic properties (d) They exhibit catalytic property
12. Permanent magnets are generally made of alloys of
(a) Fe (b) Co (c) Ni (d) All of these
108
13. Name a member of lanthanide series which is well known to exhibit
+4 oxidation state
(a) Ce
(b) La (c) Lu (d) Pr
14. Zn and Hg do not from coloured ions because
(a)They are soft (b)Their inner d-orbitals are completely filled(c)They have only 2 -electron in
the outermostshell (d)None of these
15. Hardness of transition metals is due to
(a)Large atomic size (b) Metallic bonding (c) Ionic bond (d) High ionization enthalpy
16. Interstitial compounds are formed when small atoms are trapped inside the crystal
lattice of metals. Which of the following is not the characteristic property of interstitial
compounds?
(a) They have high melting points in comparison to pure metals
(b) They are very hard
(c) They retain metallic conductivity
(d) They are chemically very reactive
17. Highest oxidation state of manganese in fluoride is +4 (MnF4 ) but highest oxidation
state in oxides is +7 (Mn2 O7 ) 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.
18. Although zirconium belongs to 4d transition series and hafniun to 5d transition series
even then they show similar physical and chemical properties because ............. .
(a) both belong to d-block
(b) both have same number of electrons
(c) both have similar atomic radius
(d) both belong to the same group of the Periodic Table
19. Which of the following statements is not correct?
(a) Copper liberates hydrogen from acids
(b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine
109
(c) Mn 3+ and Co
3+
are oxidising agents in aqueous solution
(d) Ti 2+ and Cr 2+ are reducing agents in aqueous solution
20. The magnetic moment is associated with its spin angular momentum and orbital
angular momentum. Spin only magnetic moment value of Cr 3+ ion is
(a) 2.87 BM
(b) 3.87 BM
(c) 3.47 BM
(d) 3.57 BM
21. Which is the strongest base among
(a) La(OH)3 b) Lu(OH)3 c) Ce(OH)3 d) Yb(OH)3
22. In blue vitriol, number of water molecules joined by co-ordinate bonds with copper and
by hydrogen bond is respectively
(a) 3,2
(b) 2,3
(c) 4,1
(d) 5,0
23. Correct electronic configuration of lanthanum is
(a) [Xe]5d16s2
(b) [Xe]4f15s2 (c) [Xe]4f15s2
(d) [Xe]5f16s2
24. First synthetic element is
(a)Tc
(b) La (c) Po
(d) Y
25. Which is correct?
(a) Ce(OH)3, is weaker base than Lu(OH)3
(b) E⁰ Mn 3 +/Mn 2+is more positive than Fe 3+/Fe 2+
(c) Ti> Zr> Hf (Group 4 atomic radii)
(d) Lanthanides are not separated by ion exchange method
26. Lanthanides and Actinides are similar in
(a) Oxolon formation (b) Radioactive nature (c) Tendency towards complex formation (d)
Electronic configuration
27. The basic nature of the Lanthanide hydroxides decreases from Ce(OH)3 to Lu(OH)3.
This is due to
(a) Decrease in ionic radius (b) Due to lanthanoid contraction
(c) Increase in metallic character. (d) All of these
28. Which element among the lanthanides has the smallest atomic radius
(a) Cerium (b) Lutetium (c) Europium (d)Gadolinium
29. Which of the oxide of Manganese is acidic in nature?
(a) MnO
(b) Mn2O7 (c) Mn2O3 (d) MnO2
110
30. The trends in ionization enthalpy of an element is not regular because
(a) Removal of one electron alters the relative energies of 4s and 3d orbitals
(b) Due to different E.C (stability ) (c)Poor screening of 3p orbital
(d) Due to decrease in effective nuclear charge
31. Which of the following does not have abnormal electronic configuration?
(a) Cr (b) Pd (c) Pt (d) Hg
32. The element having lowest IE1
(a) Fe (b) Co (c) Ni (d) Cu
33. Choose the correct pair regarding IE 3
(a) Mn >Cr (b) Mn > Fe (c) Zn > Cu (d) All of these
34 With F highest stable oxidation state of Mn is
(a) +6 (b) +4 (c) +7 (d) +3
35. Oxygen stabilises higher oxidation state because
(a) It is electronegative (b) Of its tendency to form double bond (c) Of small size (d)Of large
size
36. The catalytic activity of transition metals and their compounds is ascribed mainly to
(a)Their magnetic behaviour (b)Their unfilled d orbital (c)Their ability to adopt variable
oxidation states (d) Their chemical reactivity
37. 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
38. Metallic radii of some transition elements are given below. Which of these elements will
have highest density?
Element
Fe
Co
Ni
Cu
Metallic
126
125
225
128
radii/pm
(a) Fe (b) Ni (c) Co (d) Cu
39. Which of the following statements is not correct?
(a) Copper liberates hydrogen from acids.
(b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.
(c) Mn 3+ and Co 3+ are oxidising agents in aqueous solution.
(d) Ti 2+ and Cr 2+ are reducing agents in aqueous solution.
40. The electronic configuration of Cu(II) is 3d 9 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
41. Which one of the following does not correctly represent the correct order of the
property indicated against it ?
(a) Ti < V < Cr < Mn, increasing number of oxidation states
111
(b) Ti < V < Mn < Cr, increasing second ionisation enthalpy
(c) Ti < V < Cr < Mn, increasing melting point
(d) Ti 3+ < V 3+ < Cr 3+ < Mn 3+, increasing magnetic moment
42. Identify the incorrect statement among the following:
(a) Lanthanoid contraction is the accumulation of successive shrinkages.
(b) There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.
(c) As a result of lanthanoid contraction, the properties of 4d series of the transition elements
have no similarities with the 5d series of elements.
(d) Shielding power of 4f electrons is quite weak.
43Which of the following oxidation state is common for all lanthanoids?
(a) +2 (b) +3 (c) +4 (d) +5.
44. Generally transition elements form coloured salts due to the presence of unpaired
electrons. Which of the following compounds will be coloured in solid state?
(a) Ag2SO4 (b) CuF2 (c) ZnF2 (d) Cu2Cl2
45.Which one alloy does not contain copper?
(a) Bronze (b) Brass (c) Germansilver (d) Mischmetal
46.The inner transition element that is radioactive
(a) Pm (b) Gd (c) Lu (d) Sm
47. In the following transition elements, the lowest m.p and b.p is exhibited by
(a)Cr (b) Hg (c) Cu (d) Au
48.The tendency of 3d-metal ions to form stable complexes is due to their
(a)Variable oxidation state (b) Strong electronegative nature (c) High charge/size ratio and
vacant d- orbitals (d) Very low ionization energies
49.How many unpaired electrons are present in chromium(Cr)?
(a) 0 (b) 3 (c) 2 (d) 6
50.The diamagnetic ion is
(a)V 2+(b) Cr 2+ (c) Ti 3+ (d) Sc 3+
ANSWER KEY
1. c) Zn,Cd,Hg
2. d)Group 12
3. (d) Disproportionation
112
4. (c)CrO3
5. (b) 26
6.(d)Pb
7. (d)Zn
8. (a)Sc
9. (c)+8
10. (d) Sc3+
11. (b) All the ions are coloured
12. (d) All of these
13. (a)Ce
14. (b)Their inner d-orbitals are completely filled
15.(b) Metallic bonding
16. (d) They are chemically very reactive
17. (d) in covalent compounds, fluorine can form single bond only while oxygen forms double
bond
18. (c) both have similar atomic radius
19. (a) Copper liberates hydrogen from acids
20. (b) 3.87 BM
21. (a) La(OH)3
22. (c) 4,1
23. (a) [Xe]5d16s2
24. (a)Tc
25. (b) E⁰ Mn 3 +/Mn 2+is more positive than Fe 3+/Fe 2+
26. (d) Electronic configuration
27. (a) Decrease in ionic radius
28. (b) Lutetium
29. (b) Mn2O7
30. (d) Hg
31. (a) Removal of one electron alters the relative energies of 4s and 3d orbitals
32. (c) Ni
33. (d) All of these
34. (b) +4
35. (b) Of its tendency to form double bond
36. (c)Their ability to adopt variable oxidation states
37.(d) d-d transition
38. (d) Cu
39. (a) Copper liberates hydrogen from acids.
40. (a) Cu(II) is more stable
41. (c) Ti < V < Cr < Mn, increasing melting point
42. (c) As a result of lanthanoid contraction, the properties of 4d series of the transition elements
have no similarities with the 5d series of elements.
43. (b) +3
113
44. (b) CuF2
45. (d) Mischmetal
46. (a) Pm
47. (b) Hg
48. (c) High charge/size ratio and vacant d- orbitals
49. (d) 6
50. (d) Sc 3+
ASSERTION- REASON TYPE OF QUESTIONS
In the following questions there are two statements labelled as Assertion and Reason. Select
the most appropriate answer from the options given below:
[A] Both assertion and reason both are correct and the reason is the correct explanation of
assertion.
[B] Both assertion and reason are true but the reason is not the correct explanation of assertion.
[C] Assertion is correct but Reason is incorrect
[D] Assertion is a wrong statement but the reason is a correct statement
1. Assertion : Both Cu2+ and V4+ salts are blue
Reason : Cu2+ and V4 contain the same number of unpaired electrons
Ans [A]
2. Assertion : Melting point of Mn is more than Fe.
Reason : Mn has higher number of unpaired electron than Fe.
Ans [D]
3.
Assertion : TiCl4 is colourless compound.
Reason : Ti4+ has no unpaired electron.
Ans [A]
4.
Assertion : Mn atom losses ns electrons first during ionisation as compared to (n – 1)d
electrons.
Reason : The effective nuclear charge experienced by (n – 1)d electrons is greater than
that by ns electrons.
Ans [A]
5.
Assertion : Ionisation of transition metals involve loss of ns electrons before (n – 1) d
electrons.
Reason : Filling of ns orbitals take place before the filling of (n – 1)d orbitals.
Ans [B]
6. Assertion : The spin only magnetic moment of Sc3+ is 1.73 B. M.
Reason : The spin only magnetic moment of an ion equal to ,√n(n+2) where n is the
number of unpaired electrons.
Ans [D]
114
7. Assertion : Cuprous ion (Cu+) has unpaired electrons while cupric ion (Cu++) does not
Reason : Cuprous ion (Cu+) is colourless whereas cupric ion (Cu++) is blue in the aqueous
solution
Ans [D]
8. Assertion: MnO is basic whereas Mn2O7 is acidic.
Reason: Higher the oxidation state of a transition metal in its oxide, greater is the acidic
character
Ans A
9. Assertion: Fe3+ is more stable than Fe2+
Reason: Fe3+ has 3d5 configuration while Fe2+ has 3d6 configuration.
Ans A
10. Assertion: Vanadium had the ability to exhibit a wide range of oxidation states.
Reason: The energy difference between ns & (n-1)d orbitals of Vanadium is very less.
Ans A
11. Assertion: Transition metals like Fe, Cr and Mn form oxoions
Reason: Oxygen is highly electronegative and has a tendency to form multiple bonds.
Ans B
12. Assertion: The highest oxidation states of the 3d metals depends only on electronic
configuration of the metal.
Reason: The number of electrons in the (n-1)d and ns subshells determine the oxidation
states exhibited by the metal.
Ans D
13. Assertion : Cu2+ iodide is not known.
Reason : Cu2+ oxidises I- to iodine.
Ans A
14. Assertion: Separation of Zr and Hf is difficult.
Reason: Zr and Hf lie in the same group of the periodic table.
Ans(B)
15. Assertion : Cu can not liberate hydrogen from acids .
Reason : Copper has positive electrode potential.
Ans [A]
16. Assertion : The highest oxidation state of osmium is +8.
Reason : Osmium is a 5d-block element.
Ans [B]
17. Assertion : Transition elements exhibit higher enthalpies of atomization
Reason : They have stronger interatomic interaction and stronger bonding between
atoms due to the presence of large number of unpaired electrons.
115
Ans [A]
18. Assertion: Transition metals generally form coloured compounds
Reason: The energy of excitation of an electron in d-orbital corresponds to the infrared
region.
Ans [C]
19. Assertion: The highest oxidation state of a metal exhibited in its oxide or fluoride only
Reason: Oxygen and fluorine are small in size and are highly electronegative.
Ans [A]
20. Assertion: Zinc is not considered as a transition element.
Reason: Zinc forms only one oxidation state, ie, of +2
Ans [B]
21. Assertion: 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.
Reason: The filling of 4f orbitals before 5d orbitals results in a regular decrease in atomic
radii .
Ans [A]
22. Assertion: Transition metals and many of their compounds show paramagnetic behaviour.
Reason: Transition metals can exhibit variable oxidation states.
Ans [B]
23. Assertion: Transition metals can form complex compounds.
Reason: They have large sizes of the ions, high ionic charges and availability of porbitals for bond formation.
Ans [C]
24. Assertion: Transition metals and many of their compounds act as good catalysts
Reason: Transition metals can adopt multiple oxidation states and can form complexes
Ans [A]
25. Assertion: Transition metals show variable valence
Reason: In transition metals there is large energy difference between the ns 2 and (n-1)d
electrons.
Ans [ C]
26. Assertion: Fe2+ is paramagnetic.
Reason: Fe2+ contains four unpaired electrons.
Ans [A]
27. Assertion: Scandium does not exhibit variable oxidation state and yet it is regarded as a
transition element.
Reason: Scandium does not have incomplete d-orbital.
Ans [C]
.
116
28. Assertion: Transition elements can not form interstitial compounds
Reason: Small atoms like H, C, or N are trapped inside the crystal lattices of Transition
metals
Ans [D]
29. Assertion: Mercury is not considered as a transition element
Reason: Mercury is a liquid.
Ans [B]
30. Assertion: Ce4+ is used as an oxidizing agent in volumetric analysis.
Reason: Ce4+ has the tendency of attaining a +3 oxidation state.
Ans [A]
CASE BASED QUESTIONS
Case Study Question 1:
Read the passage given below and answer the following questions:
The f-block elements are those in which the differentiating electrons enters the (n-2)f orbitals.
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 f0, f7 and f14 configurations, though the most common oxidation states is +3. There is a
regular decrease in the size of lanthanides ions with increase in atomic number which is known as
lanthanides contraction.
The following questions are multiple choice question. Choose the most appropriate answer:
(i) The atomic number of three lanthanides elements X, Y and Z are
respectively, their Ln3+ electronic configuration is
(a) 4f8, 4f11, 4f13
(b) 4f11, 4f8, 4f13
(c) 4f0, 4f2, 4f11
(d) 4f3, 4f7, 4f9
(ii) lanthanide contraction is observed in
(a) Gd
(b) At
(c) Xe
(d) Te
(iii) Which of the following is not the configuration of lanthanide?
(a) [Xe]4f106s2
(b) [Xe]4f14 5d16s2
65, 68 and 70
117
(c) [Xe]4d145d106s2
(d) [Xe]4f75d16s2
OR
Name a member of the lanthanide series which is well known to exhibit +4 oxidation state.
(a) Cerium (X=58)
(b) Europium (Z=63)
(c) Lanthanum (Z=57)
(d) Gadolinium (Z=64)
(iv) Identify the incorrect statement among the following.
(a) Lanthanide contraction is the accumulation of successive shrinkages.
b) the different radii of Zr and Hf due to consequences of the lanthanide contraction.
(c) Shielding power of 4f electrons is quite weak.
(d) There is a decrease in the radii of the atoms or ions proceeds from La to Lu
Answers
(i) a
(ii) a
(iii) c or a
(iv) b
Case Study Question 2:
1. Read the passage given below and answer the following questions:
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) d1-10 ns1-2. The electronic
configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n 1)d10 n2. 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.
The following questions are multiple choice questions. Choose the most appropriate answer
(i) Which of the following characteristics of transition metals is associated with higher
catalytic activity?
(a) High enthalpy of atomisation
(c) Paramagnetic behaviour
(b) Variable oxidation states
(d) Colour of hydrated ions
118
(ii) Transition elements form alloys easily because
they have
(b) same electronic configuration
(a) same atomic number
(d) same oxidation states.
(c) nearly same atomic size
(iii) The electronic configuration of tantalum (Ta) is
(a) [Xe]4f05d16s2
(b) [Xe)4f145d26s2
(d) [Xe]4f145d46s2
(c) [Xe]4f145d36s2
(iv) Which one of the following outer orbital configurations may exhibit the largest
number of oxidation states?
(a) 3d54s1
(b) 3d54s2
(c) 3d24s2
(d) 3d34s2
Answers
(i) (b)
(ii). (c)
(iii). (c)
(iv). (b)
Case Study Question 3:
Read the passage given below and answer the following questions:
Transition elements are elements that have partially filled d-orbitals. The configuration of these
elements corresponds to (n - 1)d1-10 ns1-2. It is important to note that the elements mercury,
cadmium and zinc (Ire not considered transition elements because of their electronic
configurations, which corresponds to (n - 1)d1-10 ns2.
Some general properties of transition elements are :
These elements can form coloured compounds and ions due to d-d transition;
These elements-.exhibit many oxidation states;
A large variety of ligands can bind themselves to these elements, due to this, a wide variety of
stable complexes formed by these ions. The boiling and melting point of these elements are high.
These elements have a large ratio of charge to the radius.
In these questions (i-iv), 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.
119
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
(i) Assertion: Tungsten has very high melting point.
Reason: Tungsten is a covalent compound.
(ii) Assertion: Zn, Cd and Hg are normally not considered transition metals.
Reason: d-Orbitals in Zn, Cd and Hg elements are completely filled, hence these metals do
not show the general characteristics properties of the transition elements.
(iii) Assertion: Copper metal gets readily corroded in acidic aqueous solution such as HCI and
dil. H2SO4
Reason: Free energy change for this process is positive.
(iv) Assertion: Tailing of mercury occurs on passing ozone through it.
Reason: Due to oxidation of mercury.
Answers
(i) c
(ii) a (iii) d
(iv) a
DESCRIPTIVE QUESTIONS
1. Though copper, silver and gold have completely filled sets of d-orbitals yet they are
considered as transition metals. Why?
Ans. These metals in their common oxidation states have incompletely filed d-orbitals, e.g. Cu2
+
has 3d9 and Au3+ has 5d 8 configuration.
2. Explain why Fe is a transition metal but Na is not ?
Ans. Fe contains incompletely filled 3d subshell (3d6 4s2). Hence, it is a d-block element. i.e., a
transition element. i.e. a transition element. Na has no d-subshell. Last electron in it enters
3s orbital (1s2 2s2 sp6 3s1). Hence, it belongs to s-block.
3. Why is there striking similarities (horizontal and vertical) in successive members of the
transition series ?
Ans. This is because of the fact that along a horizontal row, electrons enter an incomplete inner
shell in building up an atom, while the outer level remains almost unchanged. In vertical
columns, similarities are due to similar electronic configurations even in the d-subshells.
4. Why the properties of third transition series are very similar to second transition series?
OR
Why the second and third members in each group of the transition elements have
very similar atomic radii?
Ans. In the third transition series after lanthanum, there is lanthanoid contraction. Due to this
contraction, the size of any atom of the third transition series is almost the same as that of
120
the element lying just above in the second transition series. This leads to similarity in their
properties.
5. The melting and boiling points of Zn, Cd and Hg are low. Why ?
Ans. In Zn, Cd and Hg, all the electrons in d-subshell are paired. Hence, the metallic bonds
present in them are weak. That is why they have low melting and boiling points.
6. K2PtC16 is a well known compound whereas corresponding Ni compound is not known.
State a reason for it.
Ans. This is because Pt4+ is more stable than Ni4+ as the sum of four ionization enthalpies of Pt
is less than that of Ni.
7. Most of the transition metals do not displace hydrogen from dilute acids. Why?
Ans. This is because most of the transition metals have negative oxidation potentials.
8. Why do transition elements show variable oxidation states?
Ans. In the transition elements, the energies of (n-l) d-orbitals and ns orbitals are very close.
Hence, electrons from both can participate in bonding.
9. Why generally there is an increase in density of elements from titanium (Z= 22) to copper
(Z = 29) in the first series of transition elements.
Ans. Due to decrease in size and increase in mass.
10. Giving reasons indicate which one of the following would be coloured?
Cu+, VO2+, Sc3+, Ni2+ (At. Nos. Cu = 29, V = 23, Sc = 21, Ni = 28)
Ans. Ni2+ due to incompletely filled d-orbitals.
11. Scandium forms no coloured ions, yet it is regarded as a transition element. Explain why?
Ans. Scandium in the ground state has one electron in the 3d - subshell. Hence, it is regarded as
a transition element.
However, in its common oxidation state +3, it has no electron in 3d subshell (3d0). Hence,
it does not form coloured ions.
12. A substance is found to have a magnetic moment of 3.9 B.M. How many unpaired
electrons does it contain ?
Ans. Using the formula, = √𝑛 (𝑛 + 2) B.M., n = 3.
13. The 4d and 5d series of transition metals have move frequent metal-metal bonding in
their compounds than do the 3 d metals. Explain.
Ans. In the same group of d-block elements, the 4d and 5d transition elements have larger size
than that of 3d element. Thus, the valence electrons are less rightly held and hence can form
121
metal-metal bond more frequently. (That is why melting points of 4d and 5d series as well
as enthalpies of atomisation are higher than those of 3d series.
14. Give reasons for each of the following:
(i)
(ii)
Ans. (i)
(ii)
Transition metal fluorides are Ionic in nature whereas bromides and chlorides are
usually covalent in nature.
Chemistry of all the lanthanoids is quite similar.
As electrogrativity of halogens decreases in the order F > C1 > Br, the ionic character
of transition metal halides decreases in the order M – F > M – Cl > M – Br. Hence,
fluorides are ionic, whereas chlorides and bromides are covalent.
The change in the size of the lanthnoids due to lanthanoid contraction is very small
as we proceed from La (Z = 57) to Lu (Z – 71). Hence, their chemical properties are
similar. Moreover, their valence shell configuration remains the same because the
electrons are added into the inner 4 f-subshell. Hence, they show similar chemical
properties.
15. Assign reason for each of the following statements :
(i) The largest number of oxidation states are exhibited by the elements in the middle of
the first row transition elements.
(ii) The atomic radii decrease in size with the increasing atomic in the lanthanoid series.
Ans. (i) The maximum number of oxidation states are shown by that element in a transition
series which has maximum number of unpaired electrons. This is so in the middle
of the series.
(ii)
Imperfect shielding of e-s in the diffused 4f subshell.
16. The +3 oxidation states of lanthanum (Z = 57), gadolinium (Z = 64) and lutetium
(Z = 71) are especially stable. Why ?
Ans. This is because in the +3 oxidation state, they have empty, half-filled and completely filled
4f sub-shell respectively.
17. Why Zr and Hf or Nb and Ta exhibit similar properties?
Or
Zirconium (atomic number 40) and hafnium (atomic number 72) occur together in
minerals and they exhibit similar properties. Give reasons.
Ans. Due to lanthanoid contraction. Hf (Z = 72) has size similar to that of Zs (Z=40), Hence, their
properties are similar and therefore, occur together. For the same reason. Nb and Ta have
similar size and hence similar properties.
18. One among the lanthanides, Ce (III), can be easily oxidized to Ce (IV) (At. No. of Ce =
58). Explain. Why ?
OR
Of the lanthanides, cerium (At. No. 58) forms a tetrapositive ion, Ce 4+ in aqueous
solution. Why ?
122
Ans. Ce (III) having the configuration 4f 1 5d0 6s0 can easily lose electron to acquire the
configuration 4 f 0 and form Ce (IV). In fact, this is the only (+ IV) lanthanide which exists
in solution.
19. Which out of the two, La(OH)3+ and Lu (OH)3 is more basic and why ?
Ans. La(OH)3 is more basic than Lu(OH)3+. As the size of the lanthanoid ions decreases from
La3+ to Lu3+, the covalent character of the hydroxides increases (Fajan’s rules). Hence, the
basic strength decreases from La(OH), to Lu(OH).
20. Account for the following :
(i)
Oxidising power in the series:
VO2+ < Cr2 O7 2- + < Mn O4 Ans. (i)
21.
This is due to increasing stability of the lower species to which they are reduced.
Account for the following :
(a)
Europium (II) is more stable than cerium (II).
(b)
Transition metals have high enthalpies of atomisation.
Ans. (a)
(b)
Europium has stable electronic configuration. ie. Xe 4f 7 5d 0 6s 0.
Due to large number of unpaired electrons in their atoms, there is larger and
stronger metallic bonding.
22.
Name the element in the 3d series that
(i)
shows maximum oxidation state.
(ii)
is diamagnetic
Ans. (i)
Manganese (ii) Zinc.
23.
Why Zn2+ salts are white while Ni2+ salts or Cu2+ salts are blue.
Ans. Zn2+ has completely filled d-orbitals (3d10) while Ni2+ has incompletely filled d-orbitals
(3d8) or Cu2+ has 3d 9. Hence d-d transition can take place in Ni2+ salts or Cu2+ salts but not
in Zn2+ salts.
24.
Calculate the magnetic moment of Ni2+.
Ans. For Ni2+ = 3 d 8 =





Hence, n = 2.
= √𝑛 (𝑛 + 2) = B √2 (2 + 2) = √8 = 2.84 BM.
25.
How would you account for the following
Cobalt (III) is stable in aqueous solution but in presence of complexing agent it is easily
oxidised.
Ans. Cobalt (III) has greater tendency to form coordination complexes than cobalt (II).
123
CHAPTER 9
COORDINATION COMPOUNDS
GIST OF THE LESSON

Double salt:
When two salts in stoichiometric ratio are crystallised together from their saturated solution
they are called double salts
Example:
(Mohr’s salt)
They dissociate into simple ions when dissolved in water.

Co-ordination compounds:
A coordination compound contains a central metal atom or ion surrounded by number of
oppositely charged ions or neutral molecules. These ions or molecules re bonded to the metal
atom or ion by a coordinate bond.
Example:
They do not dissociate into simple ions when dissolved in water.

Ligands:
An ion or molecule capable of donating a pair of electrons to the central atom via a donor
atom.
 Unidentate ligands: Ligands with only one donor atom, e.g. NH3, Cl-, F- etc.
 Bidentate ligands: Ligands with two donor atoms, e.g. ethylenediamine, C2O42(oxalate ion) etc.
 Tridentate ligands: Ligands which have three donor atoms per ligand, e.g. (dien)
diethyl triamine.
 Hexadentate ligands: Ligands which have six donor atoms per ligand, e.g. EDTA.

Ambidentate ligand:
Ligands which can ligate (link) through two different atoms present in it are called
ambidentate ligand.
Example:
and
.
─
Here, NO2 can link through N as well as O while
can link through S as well as N
atom.

Coordination number: 
The coordination number (CN) of a metal ion in a complex can be defined as the number of
ligand donor atoms to which the metal is directly bonded.
Example: In the complex

, the coordination number of Fe is 6.
Chelating Ligands:
Multidentate ligand simultaneously coordinating to a metal ion through more than one site is
called chelating ligand.
124
Example: Ethylenediamine (NH2CH2CH2NH2)
These ligands produce a ring like structure called chelate.
Chelation increases the stability of complex.

Homoleptic complexes:
Those complexes in which metal or ion is coordinate bonded to only one kind of donor atoms.
For example: [Co(NH3)6]3+

Heteroleptic complexes:
Those complexes in which metal or ion is coordinate bonded to more than one kind of donor
atoms. For example: [Co(NH3)4CI2]+
 Werner’s Theory:




Metals possess two types of valencies i.e. primary (ionizable) valency and secondary(nonionizable) valency. 
Secondary valency of a metal is equal to the number of ligands attached to it i.e. coordination
number.
Primary valencies are satisfied by negative ions, while secondary valencies may be satisfied
by neutral, negative or positive ions. 
Secondary valencies have a fixed orientation around the metal in space. 
[Co(NH3)6]Cl3
Primary Valencies = 3 Cl─
Coordination Sphere = [Co(NH3)6]3─
Secondary Valencies = 6 NH3
Nomenclature of Complexes:

Positive ion is named first followed by negative ion.

Negative ligands are named by adding suffix - o.

Positive ligands are named by adding prefix – ium.

Neutral ligands are named as such without adding any suffix or prefix.

Ligands are named in alphabetical order.
125

Name of the ligands is written first followed by name of metal with its oxidation number
mentioned in roman numbers in simple parenthesis.

Number of the polysyllabic ligands i.e. ligands which have numbers in their name, is indicated
by prefixes bis, tris etc,

Number and name of solvent of crystallization if any, present in the complex is written in the
end of the name of complex.

When both cation and anion are complex ions, the metal in negative complex is named by
adding suffix-ate.
Types of Ligands
Negative
CH3COO
Positive
NO2─
─
Acetato
CN─
Cyano
NC─
Isocayano
SCN─
Br─
Bromo
NCS─
Cl
Chloro
2─
F─
Fluoro
OH─
NO+
Nitrosoniu
m
Hydraziniu
m
H2O
Aqua
NH3
Ammine
CO
Carbonyl
CH3NH
2
Methylamin
e
Sulfito
NO
Nitrosyl
O2─
Superoxo
C5H5N
Pyridine
Hydroxo
O22─
Peroxo
N3─
Nitrido
O2─
Oxo
C2O42-─
Oxalato
Thiosulfat
o
Mercapto
NH2─
Imido
SO42─
Sulphato
─
S2O32─
HS─
ONO
─
SO3
Nitrito-N-
Neutral
Notrito-O
NH2NH3
+
Thiocayanat
o
Isothiocyanat
o
Valence Bond Theory:
1. According to this theory, the metal atom or ion under the influence of ligands can use its (n1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of
definite geometry such as octahedral, tetrahedral, and square planar.
2. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron
pairs for bonding.
Hybridization:
Find out the hybridization of central metal ion using following steps:

Write down the electronic configuration of metal atom.

Find out oxidation state of metal atom.

Write down the electronic configuration of metal ion.

Write down the configuration of complex to find out hybridization.
126

Strong field ligands cause the pairing of electrons.
Strong Field Ligands: CO, CN -, NO2-, en, py, NH3.
Weak Filed Ligands: H2O, OH-, F-, Cl-, Br-,I -
When the d orbital taking part in hybridization is inside the s and p orbital taking part in hybridization
with respect to the nucleus, it is called an inner orbital complex. Example: d2sp3 hybridization of
[Co(NH3)6]3+ involves 3d, 4s and 4p orbital, hence it is an inner orbital complex.
When the d orbital taking part in hybridization outside the s and p orbital taking part in hybridization
with respect to the nucleus, it is called an outer orbital complex.
Example: sp3d2 hybridization of [CoF6]3- involves 4d, 4s and 4p orbital, hence it is an inner orbital
complex.
Geometry:
Coordination
Number

Type of hybridisation
Shape of hybrid
4
Tetrahedral
4
Square planar
5
Trigonal bipyramidal
6
(nd orbitals are involved – outer orbital complex
or high spin or spin free complex)
Octahedral
6
d orbitals are involved –inner orbital or
low spin or spin paired complex)
Octahedral
Magnetic Properties:
Diamagnetic: All the electrons paired.
Paramagnetic: Contains unpaired electrons.
A coordination compound is paramagnetic in nature if it has unpaired electrons and
diamagnetic if all the electrons in the coordination compound are paired.
Magnetic moment
where n is number of unpaired electrons.
Spin:
 Spin paired: All electrons paired.
 Spin free: Contains unpaired electrons.
Colour:
 Compound must contain free electrons in order to show colour.
Limitations of VBT:
This theory could not explain the quantisation of the magnetic data, existence of inner
orbital and outer orbital complex, change of magnetic moment with temperature and colour
of complexes.
127
Crystal Field Theory:
1. It assumes the ligands to be point charges and there is electrostatic force of attraction
between ligands and metal atom or ion.
2. It is theoretical assumption.
Crystal field splitting in octahedral coordination complexes:
Strong field ligand causes greater repulsion and thus results in the formation of low spin complexes
by pairing of electrons.

Crystal field splitting in tetrahedral coordination complexes:
For the same metal, the same ligands and metal-ligand distances, the difference in energy between
eg and t2g level is


Weak field ligands result in the formation of high spin complexes
Order of strength of ligands: CO > CN- > NO2- > en > py = NH3 > H2O > OH- > F- > Cl- >
Br- >I-
128

Octahedral Complexes: eg orbital are of higher energy than t2g orbital.

Tetrahedral Complexes: eg orbitals are of lower energy than t2g orbitals. 
Δt = (4/9) Δo
Crystal Field Stabilization Energy:
System
High Spin
Electronic
Configuration
Low Spin
CFSE
Electronic
Configuration
CFSE
Octahedral Complex
d4
t2g3 e g1
-(3/5)Δ0
t2g4 e g0
-(8/5)Δ0+P
d5
t2g3 e g2
0
t2g5 e g0
-(10/5)Δ0+2P
d6
t2g4 e g2
-(2/5)Δ0+P
t2g6 e g0
-(12/5)Δ0+3P
d7
t2g5 e g2
-(4/5)Δ0+2P
t2g6 e g1
-(9/5)Δ0+3P
Tetrahedral Complexes
d4
eg2 t 2g2
-(2/5)Δt
eg4 t 2g0
-(12/5)Δt +2P
d5
eg2 t 2g3
0
eg4 t 2g1
-2 Δt +2P
d6
eg3 t 2g3
-(3/5)Δt +P
eg4 t 2g2
-(8/5)Δt+2P
Drawbacks of Crystal Field Splitting:
(a) From the assumption that ligands are point charges, it follows that anionic ligands should exert
greatest splitting effect. But anionic ligands actually are found at the low end of the spectro
chemical series.
(b) It does not take into account for the covalent character of bonding between the ligand and the
central atom.
Importance and applications of coordination compounds:
(a) In many quantitative and qualitative chemical analysis.
(b) In extraction processes of metals, like silver and gold.
(c) Purification of metals like Ni can be achieved through formation and subsequent decomposition
of their coordination compounds.
(d) In biological systems the pigment responsible for photosynthesis is chlorophyll, is a
co-ordination compound of magnesium. Haemoglobin, coordination compound, of Fe, acts as an
oxygen carrier.
(e) Case of chelate therapy in medicinal chemistry.
MUTIPLE CHOICE QUESTIONS
1. Which of the following is a complex of metal other than transition metal?
(a) Haemoglobin
(b) Chlorophyll
(c) Ferrocene
(d) Vitamin B12
129
2. Which of the following is not a double salt but a coordinate compound?
(a) KCl.MgCl2.6H2O (b) FeSO4.(NH4)2SO4.6H2O (c) K2SO4.Al2(SO4)3.24H2O (d)
4KCN.Fe(CN)2
3. The donor atoms in ethylenediaminetetraacetate ion is
(a) two N and two O
(b) two N and four O (c) four N and two O (d) three N and three O
4. The correct I.U.P.A.C. name of the complex, Fe(C5H5)2 is
(a) cyclopentadienyl iron (II)
(b) bis(cyclopentadienyl) iron (II)
(c) dicyclopentadienyl iron (II)
(d) ferrocene (0)
5. Which of the following is not optically active?
(a) [Co(en)3]3+
(b) [Cr(ox)3]3(c) cis-[CoCl2(en)2]+
(d) trans-[CoCl2(en)2]+
+
6. The complex ion [Cu(NH3)4] is
(a) tetrahedral and paramagnetic
(b) tetrahedral and diamagnetic
(c) square planar and paramagnetic
(d) square planar and diamagnetic
7. The hybrid state of Co in high spin complex, K3[CoF6] is (a) sp3d2 (b) sp3 (c) d2sp3 (d) sp3d
8. In an octahedral crystal field, the t2g orbital are
(a) raised in energy by 0.4 ∆o
(b) lowered in energy by 0.4 ∆o
(c) raised in energy by 0.6 ∆o
(d) lowered in energy by 0.6 ∆o
∆ < P, then the correct electronic
configuration for d04 system
will be
9. If
(a) to 4 e 0
(b) t 3 e 1
(c) t e 4
(d) t 2 e 2
2g
g
2g
g
2g
g
2g
g
10. The tetrahedral complexes are generally high spin. This is because
(a) ∆t < P
(b) ∆t > P
(c) ∆t = P
(d) none of these
11. Wilkinson’s catalyst, [(Ph3P)3RhCl] is used for
(a) hydrogenation of carboxylic acids
(b) hydrogenation of alkynes
(c) hydrogenation of alkenes
(d) polymerization of alkenes
12. Zeigler Natta catalyst is used for
(a) synthesis of methanol
(b) polymerization of olefins
(c) cracking of hydrocarbons
(d) hydrogenation of alkenes
2—
213. Among the compounds, [Ni(CO)4] -1, [Ni(CN)4] - 2 and [NiCl4] - 3, the correct statement is
(a) 1, 3 are diamagnetic while 2 is paramagnetic (b) 2, 3 are diamagnetic while 1 is paramagnetic
(c) 1, 2 are diamagnetic while 3 is paramagnetic (d) 1 is diamagnetic while 2, 3 are paramagnetic
14. Which of the following is a complex salt?
(a) Fischer’s salt
(b) Mohr’s salt
(c) Glauber’s salt
(d) Microcosmic
salt
15. Which of the following will show maximum paramagnetic nature?
(a) [Cr(H2O)6]3+
(b) [Fe(CN)6]4(c) [Fe(CN)6]3(d) [Cu(H2O)6]2+
16. The correct formula of the complex formed in the brown ring test of nitrates is
(a) [Fe(H2O)5NO]+
(b) [Fe(H2O)5NO]2+
(c) Fe(H2O)5NO]3+
(d) [Fe(H2O)4(NO2)]
The
primary
valency
of
the
Fe
in
the
complex,
K
[Fe(CN)
]
is
17.
4
6
(a) 3
(b) 2
(c) 4
(d) 6
18. Which of the following compound will exhibit linkage isomerism?
(a) [Co(en)3]Cl3
(b) [Co(NH3)6][Cr(en)3] (c) [Co(en)2(NO2)Cl]Br (d) [Co(NH3)5Cl]Br2
19. Which of the following will form an octahedral complex?
(a) d4 ( low spin )
(b) d8 ( high spin )
(c) d6 ( high spin )
(d) none of these
20. The shape of cuprammonium ion is (a) tetrahedral (b) octahedral (c) trigonal (d) square planar
21. In the complex, [Co(NH3)5(CO3)]ClO4, the C.N., O.N., the number of d-electrons and the number of
unpaired electrons in d-orbital are respectively
(a) 6, 3, 6, 0
(b) 7, 2, 7, 1
(c) 7, 1, 6, 4
(d) 6, 2, 6, 3
22. Which of the following is paramagnetic in nature?
(a) [Cr(CO)6]
(b) [Fe(CO)5]
(c) [Fe(CN)6]4(d) [Cr(NH3)6]3+
130
23. When AgNO3 is added to a solution of complex CoCl3.(NH3)5, the precipitate of AgCl shows two
ionizable Cl-. It indicates that
(a) two Cl- satisfy primary valency while one Cl- satisfy both primary and secondary valency
(b) two Cl- satisfy primary valency while one Cl- satisfy only secondary valency
(c) all three Cl- satisfy only primary valency
(d) all three Cl- satisfy only secondary valency
24. Among the following, the most stable complex is
(a) K3[Al(C2O4)3]
(b) [Pt(en)2]Cl2
(c) [Ag(NH3)2]Cl
(d)
K2[Ni(edta)]
25. According to I.U.P.A.C. nomenclature, sodium nitro-prusside is named as
(a) sodium nitroferricyanide
(b) sodium nitroferrocyanide
(c) sodium pentacyanonitrosylferrate (II)(d) sodium pentacyanonitrosylferrate (III)
(a) 2
(b) 3
(c) 4
(d) 5
26. The number of unpaired electrons in [CoF6]3- is
27. The O.N. of central atom in Wilkinson’s catalyst is (a) + 1 (b) + 2 (c) + 3 (d) + 4
28. Which of the following acts as a ligand but does not possess any unshared pair of electrons?
(a) C2H4
(b) NH3
(c) en
(d) CN29. The hybridization of central atom and shape of Wilkinson’s catalyst is
(a) dsp2, square planar (b) sp3, tetrahedral (c) sp3d, trigonal bipyramidal (d) d2sp3, octahedral
30. Which of the following is a complex salt?
(a) KCl.MgCl2.6H2O (b) FeSO4.(NH4)2SO4.6H2O (c) K4[Fe(CN)6]
(d)
K2SO4.Al2SO4.24H2O
31. The number of moles ions produced on dissolving one mole Mohr’s salt is
(a) 3
(b) 4
(c) 5
(d) 6
What
is
coordination
of
metal
in
[Co(en)
Cl
]?
32.
2 2
(a) 3
(b) 4
(c) 5
(d) 6
33. Which of the following high spin complex has the largest CFSE?
(a) [Cr(H2O)6]2+
(b) [Cr(H2O)6]3+
(c) [Mn(H2O)6]2+
(d) [Mn(H2O)6]3+
34. One third of total chlorine of the complex, CrCl3.6H2O, is precipitated by adding AgNO3 to its
aqueous solution. The correct formula of the complex
(a) [Cr(H2O)3Cl3].H2O
(b) [Cr(H2O)6]Cl3
(c) [CrCl(H2O)5]Cl2.H2O (d)
[CrCl2(H2O)4]Cl.2H2O
35. The number of chloride ions produced by complex, tetraamminedichloridoplatinum(IV)chloride in
aqueous solution is
(a) 1
(b) 2
(c) 3
(d) 4
36. The blue color obtained in the Lassaigne’s test is due the formation of compound
(a) Fe4[Fe(CN)6]3
(b) Fe3[Fe(CN)6]
(c) Na4[Fe(CN)6]
(d) Fe3[Fe(CN)6]4
37. The coordination number and oxidation number of central atom in the compound [M(SO4)(NH3)5]
are
(a) 6 and 3
(b) 2 and 6
(c) 6 and 2
(d) 3 and 6
38. The number of unpaired electrons in the complex [CoF6]3- is
(a) 4
(b) 5
(c) 2
(d) 3The compound
39. The compound Na2[Fe(CN)5(NO)] is called
(a) sodium pentacyanonitrosoylferrate (III)
(b) sodium nitroprusside
(c) sodium pentacyanonitrosoniumferrate (III) (d) sodium nitrosoferrocyanide
(a) Fe2+
(b) Fe3+
(c) Cu2+
(d) Cd2+
40. K4[Fe(CN)6] is used for the identification of
41. Which of the following is correct about [Fe(H2O)5(NO)]SO4?
(a) it is brown ring in the test for NO3- ion (b) O.N. of Fe is + 1 (c) charge on NO is +1 (d) C.N. is 6
42. Which of the following acts as catalysts?
(a) [(Ph3P)3RhCl]
(b) TiCl4 + (C2H5)3Al (c) (C2H5)4Pb
(d) (CH3)2Cd
43. Ammonia acts as a very good ligand but ammonium ion does not form complexes because
(a) NH3 is a gas while NH+4 is in liquid form.
131
44.
45.
46.
47.
48.
49.
50.
(b) NH3 undergoes sp3 hybridization while NH+4 undergoes sp3 d hybridization
(c) NH4+ ion does not have any lone pair of electrons
(d) NH4 + ion has one unpaired electron while NH3 has two unpaired electrons
The ligand NH2-CH2-CH2-NH2 is
(a) bidentate
(b) tridentate
(c) tetradentate
(d) pentadentate
Among the following which arc ambidentate ligands?
(i) SCN–
(ii) NO−3
(iii) NO−2
(iv) C2O2−4
(a) (i) and (iv)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Which of the following ligands form a chelate?
(a) Acetate
(b) Oxalate
(c) Cyanide
(d) Ammonia
Which of the following is not a neutral ligand?
(a) H2O
(b) NH3
(c) ONO
CO2
Correct formulae of tetraaminechloridonitro-N-platinum (IV) sulphate can be written as
(a) [Pt(NH3)4 (ONO) Cl]SO4
(b) [Pt(NH3)4Cl2NO2]2
(c) [Pt(NH3)4 (NO2) Cl]SO4
(d) [PtCl(ONO)NH3(SO4)]
The hybridisation involved in [Co(C2O4)3]3- is
(a) sp3d2
(b) sp3d3
(c) dsp3
(d) d2sp3
The oxidation state ad coordination number of nickel in |(Ni(CO4)| are respectively
(a) 0,4
(b) 4,4
(c) 2,0
(d) 3,2
(d)
Answer key
1
b
11
c
21
a
31
c
41
a
2
d
12
b
22
d
32
d
42
a
3
b
13
c
23
a
33
a
43
c
4
b
14
a
24
d
34
a
44
a
5
d
15
a
25
d
35
c
45
b
6
c
16
b
26
c
36
a
46
b
7
a
17
b
27
a
37
c
47
c
8
b
18
c
28
a
38
a
48
c
ASSERTION-REASON TYPE QUESTIONS
Given below are two statements labeled as Assertion (A) and Reason (R)
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1
2
Assertion: K4 Fe(CN)6 do not dissociate into Fe2+ and CN- ions
Reason: K4Fe(CN)6 is a double salt.
Ans c
Assertion: [Co(NH3) 4Cl2]+ is a heteroleptic complex
Reason: it contains more than one kind of donor groups.
Ans a
9
b
19
a
29
a
39
b
49
d
10
a
20
d
30
c
40
b
50
a
132
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Assertion: in tetracarbonyl nickel, the oxidation state of Ni is +2
Reason: the coordination compounds are named as carbonyl when CO is the ligand
Ans d
Assertion: Co-ordination entities of the type [Co(NH3) 3(NO2)3] show mer and fac isomerism
Reason: Mer and fac isomers are geometrical isomers.
Ans a
Assertion: Number of moles of AgCl precipitated per mole of PdCl2.4NH3 with excess AgNO3 is 2
Reason: secondary valency of Pd is 4
Ans a
Assertion: (ethane-1,2-diamine) and (oxalate) are didentate ligands
Reason: these are Lewis bases which can bind with the central atom to form coordination
compounds
Ans b
Assertion: When a di- or polydentate ligand uses its two or more donor atoms simultaneously to
bind a single metal ion, it is said to be a chelate ligand.
Reason: chelates are acyclic compounds with greater stability than ordinary complexes
Ans c
Assertion: Mohr’s salt, FeSO4(NH4)2SO4.6H2O dissociate into simple ions completely when
dissolved in water.
Reason: Mohr’s salt is a double salt
Ans a
Assertion: in the complex ions, [Fe(C2O4)3]3– and [Co(en)3]3+, the coordination number of both, Fe
and Co, is 3
Reason: C2O42– and en (ethane-1,2-diamine) are didentate ligands.
Ans d
Assertion: in the complex K4[Fe(CN)6], the coordination sphere is [Fe(CN)6]4– and the counter ion
is K+.
Reason: The ionisable groups are known as secondary valencies
Ans c
Assertion: [Cr(NH3)3(H2O)3]Cl3 is named as triamminetriaquachromium(III) chloride
Reason: The complex ion inside the square bracket is a cation and the ammine ligands are named
before the aqua ligands according to alphabetical order.
Ans a
Assertion: [Co(NH3)6]3+ is called an outer orbital or high spin or spin complex.
Reason: NH3 molecule is a strong ligand the complex has octahedral geometry.
Ans d
Assertion: [Ni(CN)4]2– is a square planar complex.
Reason: Here nickel is in +2 oxidation state and its electronic configuration is 3d8.
Ans b
Assertion: [Ni(CO)4] has tetrahedral geometry but is diamagnetic
Reason: nickel is in zero oxidation state and contains no unpaired electron.
Ans a
Assertion: [Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes
Reason: the central atoms in them involve in sp3d2 hybridisation.
Ans c
Assertion: The spin only magnetic moment of [MnBr4]2– is 5.9 BM.
Reason: it is a tetrahedral complex of Mn2+ ion with coordination number 4 and having five
unpaired electrons in the d orbitals.
Ans a
Assertion: In an octahedral coordination entity with six ligands surrounding the metal atom/ion,
the dx2- y2 and dz2 orbitals experience more repulsion and will be raised in energy
133
18
19
20
21
22
23
24
25
26
27
28
29
30
Reason: splitting of the degenerate levels due to the presence of ligands in a definite geometry is
termed as crystal field splitting
Ans b
Assertion: 1 mol of [Co(NH3)3Cl3] forms three moles of AgCl with excess of silver nitrate solution
Reason: chlorine is present as a ligand and not as counter ion.
Ans d
Assertion: EDTA forms chelate with metal ions.
Reason: EDTA is a hexa dentate ligand capable of forming ring structured complexes
Ans a
Assertion: [Co(NH3)6][Cr(CN)6] shows coordination isomerism
Reason: coordination isomerism arises when the counter ion in a complex salt can act as a ligand
and can displace a ligand which can then become the counter ion.
Ans c
Assertion: [Ni(CO)4] has tetrahedral geometry and is diamagnetic
Reason: CO is a neutral ligand which forms synergic bonding with the metal.
Ans b
Assertion: Permanganate ion MnO4− is pink in colour
Reason: complexes are coloured due to d-d transition of electron within the metal ion
Ans b
Assertion: The ‘g’ subscript is used for the d-orbital splitting in octahedral and square planar
complexes.
Reason: octahedral and square planar complexes have centre of symmetry.
Ans a
Assertion: In tetrahedral coordination entity formation, low spin configurations are rarely
observed.
Reason: the orbital splitting energies are sufficiently large in tetrahedral field, forcing the electrons
to get paired.
Ans c
Assertion: the complex [Ti(H2O)6]3+, is violet in colour.
Reason: This is an octahedral complex where the single electron (Ti3+ is a 3d1 system) in the metal
d orbital is in the t2g level in the ground state of the complex.
Ans b
Assertion: removal of water from [Ti(H2O)6]Cl3 on heating renders it colourless.
Reason: in the absence of ligand, crystal field splitting does not occur and hence the substance is
colourless.
Ans a
Assertion: [CoF6]3– is an example of outer orbital complexes involving sp 3d2 hybridisation
Reason: the oxidation state of cobalt in it is +3
Ans b
Assertion: [Ni(CN)4]2– is a square planar complex
Reason: the hybridisation involved is sp3.
Ans c
Assertion: K2[Zn(OH)4] is named as Potassium tetrahydroxidozincate(II)
Reason: If the complex ion is an anion, the name of the metal ends with the suffix – ate.
Ans a
Assertion: Pi bonds formed between the ligand and the central atom/ion, are not counted for
calculating the coordination number.
Reason: coordination number of the central atom/ion is determined only by the number of sigma
bonds formed by the ligand with the central atom/ion.
Ans a
134
CASE BASED QUESTIONS
CASE STUDY - I
Transition metal complexes often have spectacular colors caused by electronic transitions by the
absorption of light. For this reason they are often applied as pigments. Most transitions that are
related to colored metal complexes are either d–d transitions or charge transfer bands. In a d–d
transition, an electron in a d orbital on the metal is excited by a photon to another d orbital of higher
energy, therefore d–d transitions occur only for partially-filled d-orbital complexes. For complexes
having d0 or d10 configuration, charge transfer is still possible even though d–d transitions are not.
1. Zinc salts are generally colourless. This is due to:
a) lack of d-orbital
b) d0 configuration
c) d10 configuration
d) lack of charge transfer
ans c
2. Potassium dichromate, K2Cr2O7, is a red-orange solid. The intense colour arises due to:
a) d-d transition of electron within the Cr +6 orbital
b)transition of electron from 4s level to the d-orbital of Cr+6
c)electron transfer from oxide ligand to d-orbital of Cr+6
d) charge transfer from metal ion to the ligand
ans c
3. anhydrous CuSO4 is white, because:
a) in the absence of ligand, crystal field splitting does not occur.
b)electron transition is absent as copper ion is having d10 configuration
c) bigger sulphate anions inhibit electron transitions
d) the E0 Cu2+/Cu is positive
ans a
CASE STUDY - II
An interesting characteristic of transition metals is their ability to form magnets. Metal complexes
that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this
magnetism must be due to having unpaired d electrons. Considering only monometallic complexes,
unpaired electrons arise because the complex has an odd number of electrons or because electron
pairing is destabilized.
1. regardless of the geometry or the nature of the ligands monomeric Ti(III) species are:
have one "d-electron" and must be ,
a) paramagnetic.
b) diamagnetic.
c) ferromagnetic.
135
d) antiferromagnetic.
Ans a
2. For d4 ions, for a weak field ligand in an octahedral field, the electron configuration is:
a) t2g 4
b) eg 2 t2g 2
c) t2g 3 eg1
d) t2g2 eg2
ans c
3. The spin only magnetic moment of [NiF4]2-.is
a) 5.9 BM
b) 2.8 BM
c) 3.9 BM
d) 4.9 BM
ans b
CASE STUDY - III
The atom within a ligand that is bonded to the central atom or ion is called the donor atom. A
typical complex is bound to several donor atoms, which can be same or different elements. The ions
or molecules surrounding the central atom are called ligands. These are generally bound to the
central atom by a coordinate covalent bond (donating electrons from a lone electron pair into an
empty metal orbital). There are also organic ligands such as alkenes whose pi (π) bonds can
coordinate to empty metal orbitals. An example is ethene in the complex known as Zeise’s salt,
K+[PtCl3(C2H4)]
1. The denticity of EDTA -4 is
a) 2
b) 4
c) 6
d) 0
ans c
2. the formula of mercury tetrathiocyanatocobaltate(III) is:
a) Hg[Co(SCN)3]
b) Hg[Co(SCN)4]
c) Hg[Co(CNS)3]
d) Hg[Co(CNS)4]
ans b
3. Chelates are:
a) Ambidentate ligands
b) Polydentate ligands
c) cyclic complexes
d) Heteroleptic complexes
ans c
136
SHORT ANSWER QUESTIONS
1. Give examples of two neutral ligands.
Ans. NH3 and H2O
2. A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms A & B.
The form A reacts with AgNO3 solution to give a white precipitate soluble in aq. Ammonia
whereas B gives a pale yellow precipitate soluble in concentrated ammonia. Write the
formulae of A&B.
Ans. A is [Cr(NH3)4ClBr]Cl
B is [Cr(NH3)4Cl2]Br
3. Calculate the oxidation number of metal atom or ion in the following.
[Fe (CN)6]3- , [Cu (NH3)4]2+ , [Ni(CO)4], [Co(NH3)4Cl2]+
Ans. Fe(+3) , Cu(+2) , Ni(0) , Co(+3)
4. What are ambidentate ligands ? Give two examples.
Ans. Ligand which has two different doner atoms and either of the two ligates in the complex is
called ambidentate ligands. Examples NO2- , SCN5. Calculate the magnetic moments of the following complexes :
a. (i) [Fe(CN)6]4 (ii) [FeF6] 3+2
6
−
Ans. (i) Fe → 3d . CN is a strong ligand so pairing of electrons takes place. Since there is no
unpaired e− , magnetic moment is zero(ii) In Fe 3+ there are 5 (d5) unpaired electrons
b. n = 5, So μ = √ n(n + 2)
i. μ =√ 5(5 + 2)
1. μ =√ 35 = 5.96 BM
6. Explain the following.
(i) NH3 act as a ligand but NH4+ does not
(ii) CN- is an ambidentate ligand
Ans. ( i. ) NH3 has one lone pair while NH4+ has no lone pair of electrons.
(ii) because it has two donor atoms
7. UsingIUPAC norms write the formula for the following
(i)Tetrahydroxidozincate(II) (ii) Hexaamminecobalt(III) sulphate
Ans. (i) [Zn(OH)4]2
(ii) [Co(NH3)6]2 (SO4)3
8. Using IUPAC norms write the systematic name of the following:
(i)[Pt(NH3)2Cl(NH2CH3)]Cl (ii) [Co(en)3]3+
Ans. (i)Diamminechlorido(methylamineplatinum(II) chloride
(ii) Tris(ethane 1,2-diamminecobalt(III) ion
9. What is a chelate ligand? Give one example.
Ans. When a di or polydentate ligand uses its two or more donor atoms simultaneously to bind
a single metal ion it is said to be a chelate ligand. Example : ethane1,2-diammine.
10. On the basis of the following observations made with aqueous solutions assign
secondary valences to metals in the following
Formula
Moles of AgCl precipitated per mole with excess of AgNO3
137
(i)PdCl2. 4NH3
2
(ii)CoCl3.4NH3
1
Ans:
(i) 4 (ii) 6
11. Give the formula of each of the following coordination entities:
(i) Co 3+ ion bound to one Cl- , one NH3 molecule and two ethylene diamine molecules.
(ii) Ni 2+ ion is bound to two water molecules and two oxalate ions.
Ans. (i) [Co (NH3)Cl (en)2]2+
(ii) Ni(H2O)2(ox)2] 212. Write the IUPAC names of the following coordination compounds.
(i) K3[Fe(C2O4)3] (ii) K2[PdCl4]
Ans. (i) Potasium trioxalatoferrate(III)
(ii) Potasium tetrachloridopalladate(II)
13. What are t2g and eg orbitals?
Ans. In a free transition metal ion ,the d orbitals are degenerate.When it form complex ,the
degeneracy is split and d orbitals split into t2g and eg orbitals
14. 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
i. Δ0 > P
ii Δ0 < P
Ans. The difference of energy between two sets of degenerate orbitals after crystal field splitting is
known as crystal field splitting energy.
i. t2g 4 eg0
ii. t2g 3 eg1
15. CuSO4 5H2O is blue in colour while CuSO4 is colourless. Why?
Ans. In CuSO4 5H2O , water acts as ligand and causes crystal field splitting. Hence d-d transition
is possible and shows colour.
In anhydrous CuSO4 crystal field splitting is not possible due to absence of ligand and is
colourlesss.
16. What is the coordination number of metal ion in the following complexes?
(i) [Ni(NH3)4]2+ (ii) [Fe(C2O4)3]3- (iii) [Pt(NH3)2Cl(NO2)]
Ans. (i) 4 (ii) 6
(iii)4
17 . Write the formula of the following coordination compounds:
(i) Dichloridobis(ethane-1,2-diammine) platinum(IV) nitrate
(ii)Iron(III)hexacyanidoferrate(II)
Ans. (i) [PtCl2(en)2](NO3)2
(ii) Fe4 [Fe(CN)6]3
18. What is the difference between double salt and a complex?
Ans. Double salts dissociate into simple ions completely when dissolved in water. Eg. Mohr’s salt
FeSO4.(NH4)2SO4 .6H2O dissociates to form Fe2+, NH4+ and SO42-
138
Complexes do not dissociate into ions. [Fe(CN)6]4- does not give Fe2+ and CN- ions
19. Give the denticity of (i) C2O4 2- (ii) EDTA4Ans. (i) 2 (ii) 6
20. Give one example each for square planar, tetrahedral and octahedral complexes.
Square planar [PtCl4]2-
Tetrahedral [Ni(CO)4]
Octahedral[Co(NH3)6]3+
21. Differentiate between weak field and strong field coordination entity.
Ans.
Weak field coordination entity
Formed when the crystal field stabilisation
energy in octahedral complexes is less than
the energy required for an electron pairing
in an orbital.
They are also called high spin complexes
They are mostly paramagnetic in nature
Strong field coordination entity
Formed when the crystal field stabilisation
energy in octahedral complexes is stronger
than the energy required for an electron
pairing in an orbital.
They are called low spin complexes
They are mostly diamagnetic or less
paramagnetic in nature than weak field.
22. What is the magnetic moment of [Fe(CN)6]4-
139
Fe2+ →3d6
CN causes pairing. The number of unpaired electron is 0. Therefore magnetic
moment is also 0.
23. [Ni(CO)4] possesses tetrahedral geometry while [Ni(CN)4]2- is square planar. Why?
Ans. In [Ni(CO)4] Ni is in 0 oxidation state and its electronic configuration is 3d8 4s2. CO is strong
ligand. Hybridization is sp3 and it is tetrahedral.
In [Ni(CN)4]2- ] Ni is in +2 oxidation state and its electronic configuration is 3d 8 . CN- is strong
ligand. Hybridization is dsp2 and it is square planar.
24. Explain the valence bond theory of bonding in coordination compounds.
Ans. According to this theory a metal atom or ion under the influence of ligands can use its (n1)d,ns,np or ns,np, nd orbitals for hybridization to yield a set of equivalent orbitals of definite
geometry such as tetrahedral, square planar ,octahedral . These hybrid orbitals are allowed to
overlap with ligand orbitals that can donate electron pairs for bonding.
25. What is spectrochemical series?
Ans. Spectrochemical series is the arrangement of common ligands in the increasing order of their
field strength. It is an experimentally determined series based on the absorption of light by
complexes with different ligands.
I- < Br-< SCN- < Cl- <S2- < F- <OH- <C2O42- <H2O <NCS- < NH3 <en < CO
140
LONG ANSWER QUESTIONS
26. 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?
Ans.
(NH4)2SO4 + FeSO4 + 6H2O → FeSO4.(NH4)2SO4.6H2O(Mohr’s salt)
CuSO4 + 4NH3 +5H2O
→ [Cu(NH3)4]SO4.5H2O(Teraaminecopper(II) sulphate)
Both the compounds fall under the category of addition compounds with only one major difference
i.e., the former is an example of a double salt, while the latter is a coordination compound.A double
salt is an addition compound that is stable in the solid state but that which breaks up into its
constituent ions in the dissolved state. These compounds exhibit
individual properties of their constituents. For e.g. FeSO 4.(NH4)2SO4.6H2O breaks into
Fe2+, NH4+, and SO42− ions. Hence, it gives a positive test for Fe2+ ions.
A coordination compound is an addition compound which retains its identity in the solid as
well as in the dissolved state. However, the individual properties of the constituents are lost.
[Cu(NH3)4]SO4.5H2O does not show thw test for Cu2+.The ions present in the solution of
[Cu(NH3)4]SO4.5H2O are [Cu(NH3)4]2+ and SO4 227. Explain with two examples each of the following: coordination entity, ligand, coordination
number, coordination polyhedron, homoleptic and heteroleptic complexes.
Answer
(i) Coordination entity:
A coordination entity is an electrically charged radical or species carrying a positive or
negative charge. In a coordination entity, the central atom or ion is surrounded by a
suitable number of neutral molecules or negative ions ( called ligands). For example:
[PtCl4]2- , [Co(NH3)4]3+
(ii) Ligands
The neutral molecules or negatively charged ions that surround the metal atom in a
coordination entity or a complex are known as ligands. For example,, Cl−, NH3. Ligands are
usually polar in nature and possess at least one unshared pair of valence electrons.
. (iii) Coordination number
The total number of ligand donor atoms to which the metal is directly bonded is called coordination
number of the central metal atom.
For example:
(a) In the complex, K2[PtCl6], there as six chloride ions attached to Pt in the coordinate
sphere. Therefore, the coordination number of Pt is 6.
(b) Similarly, in the complex [Ni(NH3)4]Cl2, the coordination number of the central
atom (Ni) is 4.
(iv) Coordination polyhedron:
Coordination polyhedrons about the central atom can be defined as the spatial
arrangement of the ligands that are directly attached to the central metal ion in the
coordination sphere. For example:
(a) Tetrahedral
(b) Square planar
(v) Homoleptic complexes:
These are those complexes in which the metal ion is bound to only one kind of a donor
141
group. For eg: K2[PtCl6 ] , [Ni(NH3)4] etc.
Heteroleptic complexes are those complexes where the central metal ion is bound to more
than one type of a donor group.
For e.g.: [Co(NH3)4 Cl2]+ , [Cr(NH3)3 (NO2)3]
28. What is meant by denticity of a ligand? Explain unidentate, didentate and polydentate
ligands with examples.
Ans. When a ligand uses its two or more donor atoms simultaneously to bind a single metal ion , the
number ligating groups is called denticity of the ligand.
When a ligand is bound to a metal ion through a single donor atom the ligand is said to be
unidentate. Eg. Cl- , NH3 , H2O etc
When a ligand can bind through two donor atoms the ligand is said to be didentate.
2Eg Ethane-1,2-diamine (NH2CH2CH2NH2) , Oxalate C O
2 4
When several donor atoms are present in a ligand it is said to be a polydentate ligand .
Eg. Ethylenediaminetetraacetate (EDTA)
29. Using IUPAC norms write the formula for the following.
(i) Potasiumtetrachloridoplatinum(II)
(ii) Potasiumhexacyanidoferrate(III)
(iii) Tetrabromidocuprate(II)
(iv) Tetraamineaquachloridocobalt(III) chloride
(v) Tetracarbonylnickel(0)
Ans
(i) K2[PtCl4]
(ii)K3[Fe(CN)6] (iii) [CuBr4]2(iv) [Co(NH3)4(H2O)Cl]Cl2
(v) [Ni(CO)4]
30. How many ions are produced from each of the following compounds in solution.
(i) [Ni(NH3)6]Cl (ii) K3[Fe(CN)6] (iii) [Pt(NH3)2Cl(NH2NH2)]Cl
(iv) [Mn(H2O)6]SO4 (v) [Co(NH3)5Cl]Cl2
Ans. (i) 3 (ii) 4 (iii) 2 (iv) 2 (v) 3
31. For the following types of hybridization predict the distribution of hybrid orbitals in
space. Sp3,dsp2,sp3d,sp3d2 and d2sp3.
Ans.
Sp3
Tetrahedral
dsp2
Square planar
sp3d
Trigonal bipyramidal
sp3d2
Octahedral
d2sp3
Octahedral
32. Draw the figure to show the splitting of d orbitals in an octahedral crystal field. How does
the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity.
142
Ans.
For d4 configuration ,if ∆o < P the 4th electron enters one of the eg orbital giving the configuration
t2g3 eg1
If ∆o > P , 4th electron will occupy t2g orbital giving the electronic configuration t2g4 eg0
33.(i) Draw the figure to show the splitting of d orbitals in an tetrahedral crystal field.
(ii) Low spin configurations are rarely observed in tetrahedral coordination entities. Why?
Ans. (i)
(ii) The orbital splitting energies are not sufficiently large for forcing pairing and therefore
low spin configurations are rarely observed in tetrahedral coordination entities
34. (i) Explain why [Ni(CN)4]2- with square planar structure is diamagnetic and [NiCl4]2- with
tetrahedral structure is paramagnetic.
143
(ii) Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer
orbital complex.
Ans. (i) In [Ni(CN)4]2- ] Ni is in +2 oxidation state and its electronic configuration is 3d8 . CNis strong ligand. Hybridization is dsp2 and it is square planar. Due to the absence of unpaired
electrons it is diamagnetic.
In [Ni(Cl)4]2- ] Ni is in +2 oxidation state and its electronic configuration is 3d8 . Cl- is a weak
ligand. Hybridization is sp3 and it is tetrahedral in structure. is paramagnetic. Due to the Presence
of unpaired electrons it is paramagnetic.
(ii)
[Co(NH3)6]3+
Coordination number of Co is 6 and
oxidation state is +3
Electronic configuration of Co 3+ is 3d6
NH3being strong ligand causes pairing.
Therefoe Co can undergo d2sp3
hybridization and is an inner orbital
complex.
[Ni(NH3)6]2+
Coordination number of Co is 6 and
oxidation state is +3
Electronic configuration of Ni 2+ is 3d8
If NH3 causes the pairing then only one 3d
orbital is empty. Thus it cannot undergo
d2sp3 hybridization and undergoes sp3d2
hybridization. Hence it forms outer orbital
complex.
35. Calculate the oxidation number of the central metal ion in the following coordination
compounds.
(i) [Sc(H2O)6]3+ (ii) [Co(NH3)4Cl2]+ (iii) [Cr(H2O)6]3+
(iv) K4[Fe(CN)6] (v) [Cu(CN)4]3Ans. (i) Sc3+ (ii) Co3+ (iii) Cr3+ (iv) Fe2+ (v) Cu+
36. Write IUPAC names of the following coordination compounds.
(i) [Cr(NH3)3(H2O)3]Cl3
(ii) [Co (H2NCH2CH2NH2)3]2(SO4)3
(iii) [Ag(NH3)2][Ag(CN)2]
(iv) K3[Fe(C2O4)3]
(v) [Cr(en)3]Cl3
Ans . (i) Triaminetriaquachromium(III) chloride
(ii) Tris(ethane-1,2-diaminecobalt(III) sulphate
(iii) Diaminesilver(I)dicyanidoargentate(I)
(iv) Potasiumtrioxalatoferrate(III)
(v) Tris(ethane-1,2-diamine)chromium(III) chloride.
37. Write IUPAC names of the following coordination compounds
(i) [Cr(en)3][Co(C2O4)3]
(ii) [FeCl4(H2O)2]-
144
(iii) [CuBr4]2(iv) [CoCl(NO2)(NH3)4]Cl
(v) [PtCl(NH2CH3)(NH3)2]Cl
Ans. (i) Tris(ethylenediamine)chromium(III)trioxalatocobaltate(III)
(ii) Diaquatetrachloridoferrate(III)
(iii) Trabromidocuprate (II)
(iv) Tetraaminechloridonitrito-N- cobalt(III) chloride
(v) Diaminechlorido(methylamine)platinum(II) chloride
38. Predict the number of unpaired electrons in the following complex ions.
(i) [Pt(CN)4]2(ii) [FeCl6]3(iii) [Ni(CO)4]
(iv) [Cr(NH3)6]3+
(v) [Co(ox)3]3Ans. (i) Pt2+ , electronic configuration is 5d8. CN- causes pairing. Therefore no unpaired electrons.
(ii) Fe3+ , electronic configuration is 3d5. Cl- does not cause pairing. Therefore there are 5
unpaired electron.
(iii) ) Ni is in 0 oxidation state , electronic configuration is 3d8 4s2. CO causes pairing of 4s
electrons also in the 3d orbitals. Therefore no unpaired electrons.
(iv) Cr3+ , electronic configuration is 3d3. So undergo d2sp3 hybridisation. There are 3 unpaired
electrons.
(v) Co3+ , 3d6. Oxalate causes pairing . Therefore no unpaired electron.
39. (i) What is spectrochemical series?
(ii) What is the coordination entity formed when excess of aqueous KCN is added to an
aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is
obtained when H2S (g) is passed through the solution?
Ans .(i) Ans. Spectrochemical series is the arrangement of common ligands in the increasing order
of their field strength. It is an experimentally determined series based on the absorption of light by
complexes with different ligands.
I- < Br-< SCN- < Cl- <S2- < F- <OH- <C2O42- <H2O <NCS- < NH3 <en < CO
(ii)[Cu(CN)4]2- is the coordination entity formed. Due non availability of Cu 2+ ions ,no CuS is
formed.
40.Predict the coordination number and geometry of the following compounds.
(i) [Co(NH3)6]3+ (ii) CoF6]3Ans. (i) 6, octahedral
(iii) [NiCl4]2- (iv) [Ni(CN)4]2- (v) [PtCl4]2-
145
(ii) 6 , octahedral
(iii) 4 tetrahedral
(iv) 4 , square planar
(v) 4, square planar
41. What is meant by crystal field splitting ? How t2g and eg orbitals are formed in an
octahedral complex?
Ans. In case of free metal ion all the five d orbitals have the same energy and are called degenerate
orbitals. The five degenerate orbitals of metal ion split into different sets of orbitals having different
energies in the presence of electrical field of ligands.This is called crystal field splitting.
In an octahedral complex ,the ligands are present at the corners of an octahedron. As lobes of d x2- y2
and dz2 lie along the axes,ie along the ligands the repulsions are more along the axis. The
degeneracy of d orbitals is lifted and these split into two set of orbitals t 2g and eg, slightly different
in energy.
42. What are the important features of crystal field theory?
Ans. (a) All types of ligands are considered as point charges.
(b) The interaction between the metal ion and the ligands is purely electrostatic.
© The ligands surrounding the metal ion produce electrical field and this electrical field
influences the energies of the orbitals of the central metal atom ,particularly d-orbitals.
(d) In case of free metal ion all the five d orbitals have the same energy and are called
degenerate orbitals. The five degenerate orbitals of metal ion split into different sets
of orbitals having different energies in the presence of electrical field of ligands. This
is called crystal field splitting.
( e) The magnetic properties ,spectra and preferences for particular geometry can be
explained in terms of splitting of d-orbitals in different crystal fields.
43. Explain the hybridization ,geometry and magnetic behaviour of [Co(NH 3)6]3+ using
valence bond theory.
Ans.
Hybridisation - d2sp3, geometry – octahedral , magnetic behaviour – diamagnetic
44. On the basis of the following observations made with aqueous solutions write the
correct formula of the compound in the following
146
Formula
(i)PdCl2. 4NH3
Moles of AgCl precipitated per mole with excess of AgNO3
2
(ii)CoCl3.4NH3
1
(iii)CoCl3.5NH3
2
(iv) CoCl3.6NH3
3
(v) PdCl2. 2NH3
0
Ans. (i) [Pd(NH3)4]Cl2
(iii) ) [Co(NH3)5Cl]Cl2
(ii) [Co(NH3)4Cl2]Cl
(iv) ) [Co(NH3)6]Cl3
(v) [Pd(NH3)2Cl2]
45. . For the complex [Fe(CN)6]4- , write the hybridisation type ,magnetic and spin nature of
the complex (at no.of Fe = 26 )
Ans.
Fe 2+ ( 3d6)
Diamagnetic, d2sp3 hybridization , low spin complex
46. . (i) Why is [NiCl4]2– paramagnetic while [Ni(CN)4]2– is diamagnetic? (Atomic number of
Ni = 28)
iii) Why are low spin tetrahedral complexes rarely observed?
Ans.
(i) In [NiCl4]2- : Ni 2+ (3d8). due to the presence of Cl-, a weak field ligand no pairing occurs and
due to two unpaired electrons it is paramagnetic.
Ni(CN)4]2-, CN- is a strong field ligand and pairing takes place and due to the absence of unpaired
electron it is diagrammatic
(ii) Because of very low CFSE which is not able to pair up the electrons.
47. [Ni(CN)4]2- is colourless where as [Ni(H2O)6]2+ is green. Why?
Ans. Atomic number of Ni is 28.
In [Ni(CN)4]2-, Ni is in +2 oxidation state with electronic configuration 3d 8. In the presence of
strong CN- ligand the two unpaired electron in 3d orbital pair up. As there is no unpaired electron it
is colourless.
In [Ni(H2O)6]2+ Ni is +2 oxidation state and electronic configuration 3d8. The two unpaired
electrons do not pair up in the presence of weak ligand H2O. The d-d transition absorbs red light
and complementary green light is emitted.
48. . [Fe(CN)6]3- is weakly paramagnetic while [Fe(CN)6]4- is diamagnetic. Why?
147
Ans. In . [Fe(CN)6]3- Fe is in +3 oxidation state. Electronic configuration is 3d5.
Weakly paramagnetic due to one unpaired electron.
In [Fe(CN)6]4- Fe is in the +2 oxidation state, electronic configuration is 3d6
Diamagnetic due to absence of unpaired electrons.
49.Classify the following into homoleptic and heteroleptic complexes.
(i) [ Co(en)3]3+ (ii) [Cr (NO2)Cl(NH3)4]+ (iii) [Cu(NH3)4]2+
(iv) [PtCl4]2- (v) [ Fe (H2O)5 Cl)]2+
Ans. Homoleptic (i),(iii) and (iv)
heteroleptic (ii) and (v)
50. A metal ion Mn+ having d4 valence electronic configuration combines with three didentate
ligands to form a complex compound. Assuming ∆o > P ,
(i) write the coordination number of the metal ion.
(ii) What type of hybridisation will Mn+ have?
(iii) Draw the diagram showing d-orbital splitting during this complex formation.
(iv) Write the electronic configuration of the metal.
Ans. (i) 6
(ii) d2sp3
(iii)
(iv) t2g 4 eg0
148
CHAPTER 12
ALDEHYDES KETONES AND CARBOXYLIC ACIDS
GIST OF THE LESSON
ALDEHYDES & KETONES




Aldehydes and ketones are carbonyl compounds with general formula R-CHO where R =
alkyl/aryl group or hydrogen atom and R-CO-R, where R- is alkyl /aryl group.
Structure of carbonyl group
In carbonyl group both the carbon and oxygen atoms are in sp2 hybridised state.
Comparison of C=O Bond with C=C Bond
(a) Both atoms in both the cases are in the sp2 hybridised state.
(b) Both the cases contain one σ-bond and one π-bond.
The difference between C=O and C=C
Carbonyl group is a polar group and due to this polarity of >C=O bond , carbonyl compounds
undergo nucleophilic addition reaction in the carbonyl compound on other hand in alkene (C
= C) there is electrophilic addition reaction
 General Methods of Preparation for both R–CHO and R–CO–R
1) From hydrocarbons
(i)
By ozonolysis of alkenes followed by hydrolysis give aldehydes and ketones
O
R
CH
CH
R'
+
Starting alkene
Pent -1-ene
2-Methylbut-2-ene
2,3-Dimethylbut-2-ene
(ii)
R
O3
CH
CH
O
O
R'
H2O/Zn
-ZnO
R
CHO
Product/s of ozonolysis followed by hydrolysis
Butanal + Methanal
Ethanal + Propanone
Propanone + Propanone
By hydration of alkynes
Acetylene/ethyne gives Acetaldehyde
+R'
CHO
149
Propyne gives Acetone.
2) From alcohols
(i)
Oxidation
(ii)
Primary alcohols give aldehydes and secondary alcohols give ketones. Stronger
oxidizing agents further oxidize aldehydes into acids.
 PCC (Pyridinium chlorochromate) or Corey’s reagent is a better selective
oxidizing agent for conversion of 10 alcohols to aldehyde. It will not attack
>C=C<
 Allylic and benzylic alcohols are oxidized with CrO3/H2SO4
Catalytic dehydrogenation (Cu/573K)
Primary and secondary alcohols give aldehydes and ketones, respectively
 Preparation of aldehydes only
(1) From acid chlorides
(i)
Rosenmund reduction.
Pd−BaSO4
RCOCl + H 2 →−−−−−−−−−−−−−→ RCHO + HCl
Boiling Xylene
Note: Formaldehyde cannot be prepared by this method as formyl chloride is unstable at room temperature
(ii)
From nitriles and esters by reduction
 Stephen reaction.
 Using reducing agent DiBAL-H
150
Nitriles are reduced by diisobutylaluminium hydride, (DIBAL-H) to imines
followed by hydrolysis to aldehydes:
 Esters are also reduced to aldehydes with DIBAL-H.
(iii)
From aromatic hydrocarbons
a) Oxidation of methyl benzene
(i)
Etard’s reaction (Using Chromyl chloride)
(ii)
Use of chromic oxide (CrO3):
Toluene or substituted toluene is converted to benzylidene diacetate
on treating with chromic oxide in acetic anhydride. The benzylidene
diacetate can be hydrolysed to corresponding benzaldehyde with
aqueous acid.
b) By side chain chlorination followed by hydrolysis
c) By Gatterman – Koch reaction
 Preparation of ketones only
(i)
From acyl chlorides using dialkyl cadmium
(ii)
Acetyl chloride & benzoyl chloride give butanone and acetophenone respectively.
From nitriles
Treating a nitrile with Grignard reagent followed by hydrolysis yields a ketone.
151
(iii)
From benzene or substituted benzenes
 Physical Properties of Aldehydes and Ketones
 The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of
comparable molecular masses. It is due to dipole-dipole interactions.
 Their boiling points are lower than those of alcohols of similar molecular masses due to
absence of intermolecular hydrogen bonding.
 Order of Increasing order of boiling points of the a few compounds
 Lower members form hydrogen bonds with water and are soluble. increasing the length of alkyl
chain decreases the solubility.
 Vanillin (from vanilla beans), salicylaldehyde (from meadowsweet) and cinnamaldehyde (from
cinnamon) have very pleasant fragrances.
 Chemical Reactions of Aldehydes and Ketones
(1) Nucleophilic Addition Reactions
 Mechanism
 Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to
steric and electronic reasons.
(a) Sterically, the presence of two alkyl groups in ketones hinders the
approach of nucleophile to carbonyl carbon than in aldehydes which have
only one alkyl group.
(b) Electronically, in ketones two alkyl groups (+I effect, electron releasing
effect) reduce the electrophilicity (positive charge) of the carbonyl carbon
more effectively than in aldehyde.
 The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of
the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in
benzaldehyde due to resonance hence it is less reactive than propanal.
HCHO>RCHO>PhCHO>RCOR>RCOPh>PhCOPh
 Summary of important reactions
152
(2) Nucleophilic addition-elimination reactions (Addition of Addition of ammonia and its
derivatives)
Ammonia derivative used
(NH2-Z)
Product Obtained
Note :
In semicarbazide , there are two -NH2 group, but only one is involved
in the formation of semicarbazone. This is due to the involvement of
lone pair of electron of one of the NH2 groups in resonance with >C=O
group.
153
Note :
 While preparing ammonia derivatives of carbonyl compounds, pH must be carefully
controlled.The reaction is usually carried out in weakly acidic medium.
(i) In strongly acidic medium, ammonia derivatives being basic will react
with acids and will not react with carbonyl compounds.
(ii) In basic medium, OH- will attack carbonyl group.
If
the
attacking
Nu
is
weak like R-OH or NH2OH etc. then, the nucleophilic attack gets

accelerated if the medium is weakly acidic. The carbonyl group gets protonated and the resulting
ion is resonance stabilized.
(3) Oxidation -Reduction reactions

Tollens’ test: (Silver mirror test)
∆
Aldehydes + Tollen’s reagent (Ammoniacal silver nitrate) →−−−→ Silver mirror

Fehling’s test (Test for aldehydes except aromatic aldehydes like benzaldehyde)
Fehling solution A : aqueous copper sulphate and
Fehling solution B : alkaline sodium potassium tartarate (Rochelle salt).
(4) Haloform reaction: It is given by aldehydes & ketones having CH3CO group or
alcohols with CH3CH(OH) group
2NaOH + I2 →−−−→ NaI + NaOI + H2O
RCOCH3 + 3NaOI →−−−→ RCOONa + CHI3 (iodoform -yellow ppt)+ 2NaOH
Iodoform test is a distinction test for aldehydes and ketones.
Reactions due to α-hydrogen :

α-hydrogen atoms of carbonyl compounds are acidic in nature
154

The acidity of a-hydrogen atoms of carbonyl compounds is due to the strong
electron withdrawing effect of the carbonyl group and resonance stabilisation of
the conjugate base.
(5) Aldol condensation
Aldehydes and ketones having at least one a-hydrogen undergo a reaction in the presence of
dilute alkali as catalyst to form b-hydroxy aldehydes (aldol) or b-hydroxy
ketones (ketol), respectively. This is known as Aldol reaction

When aldol condensation is carried out between two different aldehydes and / or
ketones, it is called cross aldol condensation.
 If both of them contain a-hydrogen atoms, it gives a mixture of four products.
Reactants
Products of cross -aldol condensation
But-2-enal + 2 -Methylpent-2-enal
Ethanal + Propanal
2 -Methylbut-2-enal + Pent-2-enal
Benzaldehyde Acetophenone
1,3-Diphenylprop-2-en-1-one
(6) Cannizzaro reaction:
 Aldehydes which do not have an a-hydrogen atom, undergo self oxidation and
reduction (disproportionation) reaction on heating with concentrated alkali.
 In this reaction, one molecule of the aldehyde is reduced to alcohol while another is
oxidised to carboxylic acid salt.
(7) Electrophilic substitution reaction:
Carbonyl group acts as a deactivating and meta-directing group.
155
Distinction between aldehydes and ketones
Test with
Aldehydes
Schiff’s reagent
Pink colour
Fehling’s solution
Red ppt
Tollen’s reagent
Silver mirror
2,4-DNP
Orange yellow or red
ppt/colour
Ketones
No colour
No ppt
No silver mirror
Orange yellow or red
ppt/colour
Note : Formic acid also gives Tollen’s test whereas acetic acid does not.
CARBOXYLIC ACIDS (R-COOH)
 Higher members of aliphatic carboxylic acids (C12 – C18) known as fatty acids, occur in natural fats as
esters of glycerol.
 Structure of Carboxyl Group
 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 resonance
structure. Therefore, carboxylic acids do not show the characteristic reactions of
carbonyl compounds (Nucleophilic addition reactions)
 Methods of Preparation of Carboxylic Acids
 Physical Properties of Carboxylic Acids
 Aliphatic carboxylic acids upto nine carbon atoms are colourless liquids at room temperature
with unpleasant odours.
 The higher acids are wax like solids and are practically odourless due to their low volatility.
 Boiling points of carboxylic acids are higher than aldehydes, ketones and even alcohols of
comparable molecular masses. It is due to strong intermolecular hydrogen bonding.
156

Lower members are soluble in water due to inter molecular hydrogen bonding. The solubility
decreases with increasing number of carbon atoms.
 Chemical Reactions
(1) Reactions Involving Cleavage of O–H Bond
(2) Reactions Involving Cleavage of C–OH Bond
(3) Reactions Involving –COOH Group
(4) Substitution Reactions in the Hydrocarbon Part
(1) Reactions Involving Cleavage of O–H Bond
(i)
Acidic Strength of Carboxylic Acids
Carboxylic acids dissociate in water to give resonance stabilized carboxylate anions and
hydronium ion.
The strength of an acid is generally indicated by its pKa . Smaller the pKa, the stronger the
acid.
 The carboxylate ion is more stabilised than phenoxide ion, so carboxylic acids are more
acidic than phenols even though phenol has more number of resonance structures.
Reasons:

The negative charge is delocalised over two electronegative oxygen atoms in
carboxylate ion whereas it is less effectively delocalised over one oxygen atom
and less electronegative carbon atoms in phenoxide ion.

Carboxylate ion, is stabilised by two equivalent resonance structures in which the
negative charge is at the more electronegative oxygen atom. The phenoxide ion,
has resonance structures in which the negative charge is at the less
electronegative carbon atom.
 Effect of substituents on the acidity of carboxylic acids:
Order of various groups based on their power as electron withdrawing tendency is given below:
CF3>NO2>CN>F>Cl>Br>I>Ph
Order of acidity of a few acids
HCOOH>CH3COOH>CH3CH2COOH
CCl3COOH>CHCl2COOH>ClCH2COOH>CH3COOH
ClCH2COOH>ClCH2CH2COOH
157
(ii)
Reaction with active metals and bases.
(2) Reactions Involving Cleavage of C–OH Bond
(3) Reactions Involving –COOH Group and
(4) Reactions Involving hydrocarbon Part
Halogenation
Carboxylic acids having an a-hydrogen are halogenated at the a-position on treatment with chlorine
or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids.
The reaction is known as Hell-Volhard-Zelinsky (HVZ) reaction.
Ring substitution
 Aromatic carboxylic acids undergo electrophilic substitution reactions.
 carboxyl group -deactivating and meta-directing group.
 They do not undergo Friedel-Crafts reaction (because the carboxyl group is deactivating and
the catalyst aluminium chloride (Lewis acid) gets bonded to the carboxyl group).
158
Nomenclature of aldehydes ketones and carboxylic acids
159
MULTIPLE CHOICE QUESTIONS
1. Carbonyl compounds undergo nucleophilic addition because of
(a) More stable anion with negative charge on oxygen and less stable carbocation
(b) Electromeric effect
(c) Electronegativity difference of carbon and oxygen atoms
(d) None of the above.
2. Wolf Kishner reduction is carried with
(a)
(b)
(c)
(d)
LiAlH4 in ether
Zn-Hg and HCl
H2 in the presence of Pd
NH2-NH2/ethylene glycol and KOH
3. The addition of HCN to carbonyl compounds is an example of
(a)
(b)
(c)
(d)
Electrophilic addition
Nucleophilic addition
Nucleophilic substitution
Electrophilic substitution
4. Oxidation of cyclohexene using acidified KMnO4 will give ………
(a) adipic acid
(b) hexane -1,6-dial
(c) cyclohexane carboxylic acid
(d) cyclopentane carboxylic acid
5. Which aldehyde will give Cannizzaro’s reaction?
(a) CH3 CH2 CH2 CHO
(b) CH3 CH2 CHCHO
160
(c) (CH3)3CCHO
(d) (CH3)2CH2CH2 CHO
6. Carboxylic acids are more acidic than phenol and alcohol because of
(a)
(b)
(c)
(d)
Formation of dimers
Highly acidic hydrogen
Resonance stabilization of their conjugate base
Intermolecular hydrogen bonding
7. Which of the following orders of relative strengths of acids is correct?
(a)
(b)
(c)
(d)
ClCH2 COOH > FCH2 COOH > BrCH2 COOH
ClCH2 COOH > BrCH2 COOH > FCH2 COOH
BrCH2 COOH > ClCH2 COOH > FCH2 COOH
FCH2 COOH > ClCH2 COOH > BrCH2 COOH
8. When ethanal is heated with Fehling’s solution it gives a precipitate of
(a)
(b)
(c)
(d)
CuO
Cu
Cu + Cu2 O + CuO
Cu2 O
9. Formic acid and acetic acid can be distinguished chemically by
(a)
(b)
(c)
(d)
Reaction with HCl
Iodoform test
Reaction with NH3
Tollens test
10. The major product in the given reaction is
𝐷𝐼𝐵𝐴𝐿−𝐻
CH3-CH=CH-CH2-CH2-CN →−−−−−→
𝐻2𝑂
(a)
(b)
(c)
(d)
CH3-CH2-CH2-CH2-CH2-CN
CH3-CH2-CH2-CH2-CH2-CHO
CH3-CH=CH-CH2-CH2-CH2-NH2
CH3-CH=CH-CH2-CH2-CHO
11. Which of the following conversions can be carried out by
Clemmensen Reduction?
(a) Benzaldehyde into benzyl alcohol
(b)Cyclohexanone into cyclohexane
(c)Benzoyl chloride into benzaldehyde
(d)Benzophenone into diphenyl methane
12. In Clemmensen Reduction carbonyl compound is treated with…………
(a) Zinc amalgam + HCI
(b)Sodium amalgam + HCI
(c)Zinc amalgam + nitric acid
161
(d)Sodium amalgam + HNO3
13. Which is the most suitable reagent for the following conversion?
CH3-CH=C(CH3)-CO-CH3 →
CH3-CH=C(CH3)-COO-
(a) Tollen’s reagent
(b) Benzoyl peroxide
(c) I 2 and NaOH solution
(d) Sn and NaOH solution
14. Cannizzaro's reaction is not given by………..
(a) C6H5CHO
(b)H CHO
(c)CCl3-CHO
(d)CH3-CHO
15. Which of the following compound is most reactive towards nucleophilic addition
reactions?
(a) C6H5CHO
(b) H CHO
(c) CH3-CO- CH3
(d) CH3-CHO
16. 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
17. The correct order of decreasing acidic strength is………….
(a) Dichloro acetic acid< Trichloroacetic acid < Chloroacetic acid < Acetic acid
b)Trichloro acetic acid< Dichloroacetic acid < Chloroacetic acid < Acetic acid
c)Trichloro acetic acid > Dichloroacetic acid > Chloroacetic acid > Acetic acid
d)Dichloro acetic acid< Trichloroacetic acid < Chloroacetic acid < Acetic acid
18. Tollens reagent is
(a)Ammoniacal silver bromide
(b)Ammoniacal silver nitrate
(c)Ammoniacal silver iodide
(d)Ammoniacal cuprous oxide
19. Which of the following does not react with NaHSO3
(a)Acetone
(b)Acetaldehyde
(c)Acetophenone
(d) benzophenone
20 . Aldol condensation will not occur in
a) HCHO
(b) CH3 CHO
(c) CH3 CO CH3
162
(d) CH3- CH2- CHO
21. Arrange the following molecules in the increasing order of Pka values
a)Dichloro acetic acid< Trichloroacetic acid < Chloroacetic acid < Acetic acid
b)Trichloro acetic acid< Dichloroacetic acid < Chloroacetic acid < Acetic acid
c)Trichloro acetic acid > Dichloroacetic acid > Chloroacetic acid > Acetic acid
d)Dichloro acetic acid< Trichloroacetic acid < Chloroacetic acid < Acetic acid
22.Stephen reduction is used to prepare aldehydes from
(a) Acid chloride
(b) Alkane nitriles
(c) Acid amides
(d) Grignard reagent
23. Ozonolysis of 2,3dimethyl but-2-ene gives
(a) one molecule of acetone and one molecule of butanone
(b) two molecule of acetone
(c) two molecules of propanal
(d) one molecule of propanal and one molecule of propanone
24. Arrange the following molecules in the increasing order BP
(a) propane< popanal <propanone < propanol < propanoic acid
(b) propane> popanal >propanone > propanol > propanoic acid
(c) propane< popanone <propanal < propanol < propanoic acid
(d) propane< popanal <propanone < propanoic acid < propanol
25.Acetophenone and benzophenone can be distinguished by
(a) Iodoform test
(b) Tollens Test
(c) Fehling’s test
(d) Benedicts test
26.Acetaldehyde and benzaldehyde can be distinguished by
(a) Iodoform test (b) Tollens Test
(c) Fehling’s test (d) neutral FeCl3
(a) a only
(b) both a and b
(c) both a and c
(d) d only
27. Arrange the following compounds in the decreasing order of their reactivity in
nucleophilic addition reactions.
(I) Benzaldehyde, (II) p-Tolualdehyde, (III) p-Nitro benzaldehyde,
(IV)Acetophenone.
(a) I >II >III > IV
(b) I >III >II > IV
(c) II >I >III > IV
(d) III >I >II > IV
163
𝐶𝑂 + 𝐻𝐶𝑙
→
∆
28. C6H6
(a)
(b)
(c)
(d)
X + HCl , X is
C6 H5CHO
CH3 CHO
CH3 CO CH3
C6 H5 CO CH3
29. The IUPAC name of CH3 COCH(CH3)2 is
(a) 3- methylbutan-2-one
(b) 3,3-dimethyl propanone
(c) 1,1-dimethyl propanone
(d) isopropyl methyl ketone
30. The IUPAC name of CH3 CH(OH)COOH is
(a) lactic acid
(b) 2-hydroxypropanoic acid
(c) 3- carboxypropanol
(d) 3-carboxy propan -2-ol
31.The purpose of S in Rosemonds reduction is
(a) to prevent the reduction of aldehyde to alcohol
(b) to prevent the oxidation of aldehyde to carboxylic acid.
(c) to increase the rate of reaction.
(d)None of the above
𝐻𝑔+2
32.CH3-CH≡CH →−→ X, X is
𝐻2𝑂
(a)
(b)
(c)
(d)
CH3-CO-CH3
CH3-CH2-CHO
CH3-CH2-CH2-OH
CH3-CH(OH)-CH3
𝑟𝑒𝑑 𝑃
→
33. R-CH2-COOH
𝑋2
R-C H2(X) -COOH + HX, the reaction is
(a) Cannizaro reaction
(b)Clemensons reduction
(c) HVZ reaction
(d) Kolbes reaction
34. The reagent that can be used to distinguish phenol and ethanoic acid is.
(a) Na metal
(b) Tollens reagent
(c) NaHCO3
(d) Fehling solution
𝐶𝑂2
35. C6H5-MgBr →−−→
𝐻3𝑂+
(a) C6H5-CH2-COOH
(b) C6H5-OH
X + Mg(OH)Br
164
(c) C6H5-COOH
(d) C6H5-O- C6H5
36.A substance X is heated with soda lime and gives ethane ,the substance X is
(a) ethanol
(b) ethanoic acid
(c) propanoic acid
(d) methanoic acid
37. The major product of nitration of benzoic acid is,
(a) 2-nitrobenzoic acid
(b) 4-nitrobenzoic acid
(c) 3-nitrobenzoic acid
(d) 2,4- dinitro benzoic acid
𝑑𝑖𝑙. 𝑁𝑎𝑂𝐻
CH3-CH = CH-CHO, X is
→
∆
(a) HCHO
(b) CH3-CHO
(c) CH3-CH = CH-CH 2OH
(d) CH3-CH (OH) CH-CHO
38. 𝑋
39. Benzoyl chloride is prepared from benzoic acid by
(a) Cl2/ FeCl3
(b) Cl2/ uv
(c) SOCl2
(d) HCl
40. Which of the following can be reduced with Zn/Hg and HCl to give the
corresponding hydrocarbon.
(a) Butanal
(b) Butan-2-one
(c) Butanoic acid
(d) Benzoyl chloride
(a)
(b)
(c)
(d)
a only
both band d
both a and c
both a and b
41. The product of the reaction,
CH3 –COOH
(a)
(b)
(c)
(d)
𝑃2𝑂5
→
……….. is
CO and H2O
(CH3 –CO) 2O
CH3 –COO-CH3
None of the above
42. Electrolysis of an aqueous solution of CH3 –COONa gives
165
(a)
(b)
(c)
(d)
methane
ethane
ethene
butane
43. Which of the following have the highest Ka value
(a)
(b)
(c)
(d)
formic acid
acetic acid
chloroacetic acid
fluroacetic acid
44. Toluene is oxidized to benzaldehyde by
(a) alkaline KMnO4
(b) PCC
(c) acidified K2Cr2O7
(d) CrO2Cl2
45. The oxidation of toluene to benzaldehyde using chromyl chloride is
(a) Stephan’s reaction
(b) Étards reaction
(c) Gatterman Koch reaction
(d) ozonolysis
46. Benedict solution provides
(a) Cu+2
(b) Cu+
(c) Ca+2
(d) Ba+2
47. Fehlings solution B is
(a) Copper sulphate solution
(b) alkane sodium potassium tartarate
(c) Alkaline sodium citrate
(d) ammoniacal silver nitrate
48. Carbonyl compounds can be converted to hydrocarbons by
(a) Clemensons reduction
(b) reduction using NaBH4 or LiAlH4
(c) Wolf Kishner reduction
(d) by both a and c
49. Oxidation of acetone using HNO3 will give
(a) two molecules of acetic acid
(b) one molecule of propionic acid
(c) CH3COOH + H2O + CO2
(d) two molecules of formic acid
50. C6H5-COCl +H2
H2OH
(a) C6H5C
166
𝑃𝑑/𝐵𝑎𝑆𝑂4
→
𝑆
X +
HCl,
X is
167
(b) C6H5-CHO
(c) C6H5-COOH
(d) C6H5-CO- C6H5
Answer key
1
2
3
4
5
6
7
8
9
10
A
D
B
A
C
C
D
D
D
D
11
12
13
14
15
16
17
18
19
20
B
A
C
D
B
C
C
B
D
A
21
22
23
24
25
26
27
28
29
30
B
B
B
A
A
C
D
A
A
B
31
32
33
34
35
36
37
38
39
40
A
A
C
C
C
C
C
B
C
D
41
42
43
44
45
46
47
48
49
50
B
B
D
D
B
A
B
D
C
B
ASSERTION - REASON TYPE QUESTIONS
Read the assertion and reason carefully to mark the correct option out of the options given below:
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the
Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the
Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
1. Assertion: The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers
of comparable molecular masses.
Reason: There is a weak molecular association in aldehydes and ketones arising out of the
dipole-dipole interactions.
2. Assertion: Compounds containing –CHO group are easily oxidised to corresponding carboxylic
acids.
Reason : Carboxylic acids can be reduced to alcohols by treatment with LiAlH4
3. Assertion: Acetic acid does not undergo a haloform reaction.
Reason: Acetic acid has no alpha hydrogens.
4. Assertion: Chloroacetic acid has a lower pKa value than Acetic acid.
Reason: The – I effect of Cl further stabilizes the carboxylate anion formed by the loss of the
proton.
5. Assertion: Lower aldehyde and ketones are soluble in water but the solubility decreases as
molecular mass increases.
Reason: Aldehydes and ketones can be distinguished by Tollen’s reagent.
6. Assertion: Aldehydes undergo aldol condensation only if it has α-hydrogen.
Reason: The 𝖺-hydrogen in aldehydes are acidic in nature because the anion formed by the loss
of the 𝖺-hydrogen is resonance stabilized.
7. Assertion: Acetophenone and benzophenone can be distinguished by the iodoform test.
Reason: Acetophenone and benzophenone both are carbonyl compounds.
168
8. Assertion: Benzaldehyde is more reactive than ethanol towards nucleophilic attack.
Reason: The overall effect of –I and +R effect of phenyl group decrease the electron density on
the carbon atom of > C = O group in benzaldehyde.
9. Assertion: Ketones are less reactive than aldehydes.
Reason: Ketones do not have alpha hydrogens
10. Assertion: Dry HCl is a must for the reaction of an aldehyde with alcohol to form hemiacetal.
Reason: HCl protonates the carbonyl oxygen increasing the electrophilicity of the carbonyl
carbon.
11. Assertion: Even though there are two NH2 groups in semicarbazide, only one reacts with
carbonyl compounds
Reason: Semicarbazide has two NH2 groups out of which one is in resonance with the carbonyl
group.
12. Assertion : Lower aldehydes and ketones are soluble in water but the solubility decreases as the
molecular masses increase
Reason: Distinction between aldehydes and ketones can be made by Tollens’ test.
13. Assertion: Aldehydes react with Tollen's reagent to form silver mirror.
Reason: Both, aldehydes and ketones contain a carbonyl group.
14. Assertion: Aldol condensation can be catalysed both by acids and bases.
Reason: ββ Hydroxy aldehydes or ketones readily undergo acid catalysed dehydration.
15. Assertion: Ethanal and Propanone both gives iodoform test
Reason: both having –CO- CH3 group
16. Assertion : Benzaldehyde gives a positive test with Tollens reagent but not with Fehling’s
solution
Reason: Under normal circumstances, aldehydes that lack alpha hydrogens and so cannot form
an enolate do not produce a positive test using Fehling's solution, which is a weaker oxidising
agent than Tollen's reagent.
17. Assertion: Propanal is more reactive than propanone.
Reason- Due to the presence of alkyl groups on both sides of the carbonyl carbon, propanone is
sterically more hindered than propanal, making it less reactive to nucleophilic attack.
18. Assertion : Carboxylic acid is a stronger acid than phenol.
Reason : Carboxylate ions are more polar and has less stearic hindrance
19. Assertion: CH3CHO is more reactive than CH3COCH3 towards reaction with HCN.
Reason : CH3CHO is more polar and has less stearic hindrance, therefore, more reactive with
HCN than CH3COCH3.
20. Assertion : Carboxylic acids do not give reactions of carbonyl group.
Reason: Due to resonance
21. Assertion : There are two -NH2 groups in semicarbazide (H2NNHCONH2).
any one NH2 group can involved in the formation of semicarbazone.
Reason : Its due to steric hindrance and electronic effect
22. Assertion : Benzoic acid does not undergo Freifel Craft reaction
Reason : Carbonyl group in benzoic acid donate electron into thr ring through resonance
23. Assertion : Although phenoxide ion has more number of resonating structures than Carboxylate
ion, Carboxylic acid is a stronger acid than phenol.
Reason : In carboxylate ions, negative charge is delocalised over two oxygen atoms which is
more stable. Phenoxide is less stable as negative charge is delocalised over one oxygen atom
and carbon atoms of benzene ring.
24. Assertion : Carboxylic acids have greater boiling point than alcohols
Reason : Carboxilic acid have the greater ability to form dimers in solution
25. Assertion : Chloro acetic acid has lower pKa value than acetic acid
169
38.
Reason : The –I effect of Cl further stabilises the carboxylate anion formed by the loss of the
proton
26. Assertion : Benzaldehyde resists nucleophilic addition in comparison with ethanal
Reason: The phenyl group in benzaldehyde is too bulky and therefore prevents attack of
Nucleophile
27. Assertion : Aldehydes undergo aldol condensation only if it has alpha hydrogen
Reason : The alpha hydrogen in aldehydes are acidic in nature because the anion formed by the
loss of the alpha hydrogen is resonance stabilised .
28. Assertion : Reactivity of ketones is more than aldehyde..
Reason : The carbonyl carbon of ketones is less electrophilic as compared to aldehyde.
29. Assertion : Cannizzaro reaction is a self oxidation reaction of aldehyde donot having alpha
hydrogens
Reason : Benzaldehyde shows cannizzaro reaction and gives benzyl alcohol and sodium salt of
benzoic acid
30. Assertion: Acetic acid does not undergo a haloform reaction.
Reason: Acetic acid has no alpha hydrogens.
31. Assertion: Carboxylic acids do not show the properties of a carbonyl group.
Reason: The -C=O in the carboxyl group is in resonance with the -OH group.
32. Assertion: Lower aldehyde and ketones are soluble in water but the solubility decreases as
molecular mass increases.
Reason: Aldehydes and ketones can be distinguished by Tollen’s reagent.
33. Assertion: Aldehydes undergo aldol condensation only if it has α-hydrogen.
Reason: The 𝖺-hydrogen in aldehydes are acidic in nature because the anion formed by the loss
of the 𝖺-hydrogen is resonance stabilized.
34. Assertion: Acetophenone and benzophenone can be distinguished by the iodoform test.
Reason: Acetophenone and benzophenone both are carbonyl compounds.
35. Assertion: Dry HCl is a must for the reaction of an aldehyde with alcohol to form hemiacetal.
Reason: HCl protonates the carbonyl oxygen increasing the electrophilicity of the carbonyl
carbon.
36. Even though there are two NH2 groups in semicarbazide, only one reacts with carbonyl
compounds
Reason: Semicarbazide has two NH2 groups out of which one is in resonance with the carbonyl
group.
37. Assertion: 2, 2–Dimethyl propanal undergoes Cannizzaro reaction with concentrated NaOH.
Reason: Cannizzaro is a disproportionation reaction.
Assertion: Acetaldehyde on treatment with dil NaOH it gives aldol.
Reason: Acetaldehyde molecules contain alpha hydrogen atom.
39. Assertion : The boiling points of aldehydes and ketones are higher than hydrocarbons and
ethers of comparable molecular masses.
Reason : There is a weak molecular association in aldehydes and ketones arising out of the
dipole-dipole interactions.
40. Assertion : Butan -2 – ol will give butanone on oxidation with alkaline KMnO4 solution.
Reason : Aldehydes are easily oxidised to carboxylic acids on treatment with common oxidising
agents like nitric acid, potassium permanganate, potassium dichromat, etc.
ANSWER KEY
1
a
11
a
21
d
31
a
2
b
12
b
22
c
32
b
170
3
4
c
a
13
14
b
b
23
24
a
a
33
34
a
b
5
b
15
a
25
a
35
a
6
a
16
a
26
c
36
a
7
b
17
a
27
a
37
b
8
a
18
c
28
d
38
a
9
10
b
a
19
20
a
a
29
30
b
c
39
40
a
a
CASE BASED QUESTIONS
1. An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid produces an
ester (B). (A) on mild oxidation gives (C). (C) with 50% KOH followed by acidification with dilute
HC1 generates (A) and (D). (D) with PC15 followed by reaction with ammonia gives (E). (E) on
dehydration produces hydrocyanic acid. Identify the compounds A, B, C, D and E.
CASE – 2
Read the passage given below and answer the following questions :
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
The following questions are multiple choice questions. Choose the most appropriate answer :
(i)
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
171
(ii) Which of the following compounds will undergo Cannizzaro reaction?
(a) CH3CHO (b) CH3COCH3 (c) C6H5CHO (d) C6H5CH2CHO
(iii) 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
(c) 2, 2, 2-trichloropropanol
(d) chloroform.
Ans ;
(i)
a
(ii) c (iii) a
Case – 3
Read the passage given below and answer the following questions :
`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.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i) Condensation reaction is the reverse of which of the following reaction?
(a) Lock and key hypothesis (b) Oxidation
(c) Hydrolysis (d) Glycogen formation
(ii) Which of the following compounds would be the main product of an aldol condensation of
acetaldehyde and acetone?
(iii) Which of the following will undergo aldol condensation?
Ans (i) C (ii) b (iii) d
Case – 4
The carboxylic acids like alcohols evolve hydrogen with electropositive metals and form
salts with alkalies similar to phenols. However, unlike phenols they react with weaker
172
bases such as carbonates and hydrogen carbonates to evolve carbon dioxide. This reaction
is used to detect the presence of carboxyl group in an organic compound
Carboxylic acids dissociate in water to give resonance stabilised carboxylate anions and
Hydronium ion.
Smaller the pKa, the stronger the acid (the better it is as a proton donor). Strong acids
have pKa values < 1, the acids with pKa values between 1 and 5 are considered to be
moderately strong acids, weak
Acids have pKa values between 5 and 15, and extremely weak acids have pKa values >15.
(i)
Which acid of each pair shown here would you expect to be stronger?
(a) CH2FCO2H or (b) CH2ClCO2H
(ii) Which one of the following is the correct order of acidic strength?
(a) CF3COOH > CHCl2COOH > HCOOH > C6H5CH2COOH > CH3COOH
(b) CH3COOH > HCOOH > CF3COOH > CHCl2COOH > C6H5CH2COOH
(c) HCOOH > C6H5CH2COOH > CF3COOH > CHCl2COOH > CH3COOH
(d) CF3COOH > CH3COOH > HCOOH > CHCl2COOH > C6H5CH2COOH
(iii) Which of the following acids has the smallest dissociation constant?
(a) CH3CHFCOOH
(b) FCH2CH2COOH
(c) BrCH2CH2COOH
(d) CH3CHBrCOOH
Ans (i)
a
(ii)
a
(iii) c
SHORT ANSWERS
1. Why is there a large difference in the boiling points of butanal and butan-1-ol?
Answer: Butan-1-ol has higher boiling point due to intermolecular hydrogen bonding.
2. Write a test to differentiate between pentan-2-one and pentan-3-one.
Answer: Iodoform test: Pentan-2-one gives yellow precipitate on reaction with NaOH and I2
3. What type of aldehydes undergo Cannizzaro reaction?
Answer: Aldehydes which do not contain α-hydrogen atom undergo Cannizzaro reaction e.g.
Formaldehyde (HCHO) and Benzaldehyde (C6H5CHO).
4. Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence
of anhydrous AlCl3. Name the reaction also.
173
Answer: benzoylium cation or Friedel Craft’s acylation reaction.
5. Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on
oxidation of 2, 5-dimethylhexan-3-one.
Answer: 2-methylpropanoic acid and 3-methylbutanoic acid. (poppot’s rule)
6. Arrange the following in decreasing order of their acidic strength.
CH3CH2OH, CH3COOH, ClCH2COOH, FCH2COOH, C6H5CH2COOH.
Answer: FCH2COOH > ClCH2COOH > C6H5CH2COOH > CH3COOH > CH3CH2OH
7. (a) How will you bring about the following conversions :
(i) Ethanol to 3-hydroxybutanal (ii) Benzaldehyde to Benzophenone
Answer:
(a) (i) Ethanol to 3-hydroxybutanal:
8. Compound ‘A’ was prepared by oxidation of compound ‘B’ with alkaline KMnO4. Compound
‘A’ on reduction with lithium aluminium hydride gets converted back to compound ‘B’. When
compound ‘A’ is heated with compound B in the presence of H 2SO4 it produces fruity smell of
compound C to which family the compounds ‘A’, ‘B’ and ‘C’ belong to?
Answer: ‘A’ is a carboxylic acid, ‘B’ is an alcohol and ‘C’ is an ester.
9. Arrange the following in decreasing order of their acidic strength. Give explanation for the
arrangement. C6H5COOH, FCH2COOH, NO2CH2COOH.
Answer: NO2CH2COOH > FCH2COOH > C6H5COOH (due to -I effect)
10. Alkenes (>C=C<) and carbonyl compounds ( >C=O), both contain a π bond but alkenes show
electrophilic addition reactions whereas carbonyl compounds show nucleophilic addition
reactions. Explain.
Answer: Carbon atom in carbonyl compounds acquires slight positive charge and is attacked by
nucleophile.
11. Carboxylic acids contain carbonyl group but do not show the nucleophilic addition reaction
like aldehydes or ketones. Why?
Answer: Due to resonance, the partial positive charge on carbonyl carbon atom is reduced.
12. Why are carboxylic acids more acidic than alcohols or phenols although all of them have
hydrogen atom attached to a oxygen atom (—O—H)?
Answer: Compare the stability of anion formed after the loss of H+ ion. More stable the anion
formed, more easy will be the dissociation of O—H bond, stronger will be the acids.
174
13. Ethylbenzene is generally prepared by acetylation of benzene followed by reduction and not
by direct alkylation. Think of a possible reason.
Answer: The direct alkylation cannot be performed because there is polysubstituion product is
formed. Due to disadvantage of polysubstituion that Friedel-Craft's alkylation reaction is not used
for preparation of alkylbenzenes. Instead of Friedel-Craft's acylation is used.
14. Write the equations involved in the following reactions:
(i) Wolff-Kishner reduction
(ii) Etard reaction
Answer:(i) Wolff-Kishner reduction
(ii) Etard reaction
15. Write the reactions involved in the following:
(i) Hell-Volhard Zelinsky reaction
(ii) Decarboxylation reaction
Answer:
16. Arrange the following carbonyl compounds in increasing order of their reactivity in
nucleophilic addition reactions :
(a) Ethanal, propanal, propanone, butanone
(b) Benzaldehyde, p-tolualdehyde, p-nitrobenzaldehyde, acetophenone.
Answer: (a) butanone< propanone < propanal < Ethanal.
(b) acetophenone < p-tolualdehyde < Benzaldehyde < p-nitrobenzaldehyde.
17. Describe how the following conversions can be brought about :
(i) Cyclohexanol to cyclohexan-1-one (ii) Ethylbenzene to benzoic acid
(iii) Bromobenzene to benzoic acid
(i)
Cyclohexanol to cyclohexan-1-one
175
18. (a) Give reasons :
(i) CH3—CHO is more reactive than CH3COCH3 towards HCN.
(ii) 4-nitrobenzoic acid is more acidic than benzoic acid.
Answers::
(a) (i) Because carbonyl carbon of CH3—CHO is more electrophilic than CH3COCH3 due to
only one electron donating CH3– group.
(ii) Because of electron withdrawing nature of -NO2 group.
19. (a) What is meant by the following terms? Give an example of the reaction in each case.
(i) Aldol (ii) Semicarbazone
Answer:
(a) (i) Two molecules of aldehyde and ketones containing a-hydrogen atom react in the presence
of aqueous alkali giving product known as Aldol.
20. Write the product(s) in the following reactions.
Answer:
176
21. (a) Account for the following :
(i) Propanal is more reactive than propanone towards nucleophilic reagents.
(ii) Electrophilic substitution in benzoic acid takes place at meta position.
(iii) Carboxylic acids do not give characteristic reactions of carbonyl group.
Answer:
(a) (i) Due to steric and + I effect of two methyl groups in propanone.
(ii) The benzene ring of benzoic acid undergoes electrophilic substitution reaction such as
nitration, sulphonation etc. Since the — COOH group in benzene is an electron withdrawing
group, therefore it is meta directing group.
(iii) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible
resonance structure.
22. Write structures of A, B, C and D in the following reaction sequence :
Answer:
23. (a) Give reasons for the following:
(i) Ethanal is more reactive than acetone towards nucleophilic addition reaction.
(ii) (CH3)3C—CHO does not undergo aldol condensation.
(iii) Carboxylic acids are higher boiling liquids than alcohols.
Answer:
(a) (i) It is because ethanal is more polar than acetone due to only one methyl group, whereas in
acetone, there are two methyl groups which are electron releasing and reduce positive charge on
carbonyl carbon.
(ii) It is because (CH3)3C—CHO does not have alpha (a) hydrogen.
(iii) It is due to greater intermolecular H-bonding in carboxylic acids than alcohols and they
exist as dimers.
24. Describe the mechanism of the addition of Grignard reagent to the carbonyl group of
compound to form an adduct which on hydrolysis yields an alcohol.
177
25. How will you prepare the following compounds starting with benzene:
(i) Benzaldehyde
(ii) Acetophenone
LONG ANSWERS:
1. (a) Give a plausible explanation for each one of the following :
(i) There are two – NH2 groups in semi carbazide. However, only one such group is involved
in the formation of semi carbazones.
(ii) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcyclohexanone does
not.
(b) An organic compound with molecular formula C 9H10O forms 2, 4, – DNP derivative,
reduces Tollens’ reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives
1, 2-benzene-di- carboxylic acid. Identify the compound.
ANSWER:
178
(a) (i) Because one of the – NH2 in semicarbazide is involved in the resonance with -CO-group.
(b) Compound C9H10O forms 2, 4-DNP derivative, so it contains a carbonyl group. Also it reduces
Tollens’ reagent therefore carbonyl group is an aldehyde group. Since it undergoes Cannizzaro’s
reaction, aldehyde has no a hydrogen atom, so compound is C8H9-CHO. On vigorous oxidation,
compound gives l, 2-benzene di-carboxylic acid hence the compound is
2. (a) Give chemical tests to distinguish between
(i)
Phenol
and
Benzoic
acid
(ii)
Benzophenone
(b) Write the structures of the main products of following reactions
and
Acetophenone
Answer:
(a) (i)Phenol and Benzoic acid: On addition of NaHCO3 to both solutions carbon dioxide gas is
evolved with benzoic acid while phenol does not form CO2
(iii)
Benzophenone and Acetophenone :
Iodoform test. Warm each organic compound with I2 and NaOH solution.
Acetophenone-yellow precipitate of iodoform is formed. Benzophenone does not respond to
179
this
test.
3. (a) Write a suitable chemical equation to complete each of the following transformations: 4Methylacetophenone
to
benzene-1,
4-dicarboxylic
acid
(b) An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid
produces an ester (B). (A) on mild oxidation gives (C). (C) with 50% KOH followed by
acidification with dilute HC1 generates (A) and (D). (D) with PC15 followed by reaction with
ammonia gives (E). (E) on dehydration produces hydrocyanic acid. Identify the compounds
A, B, C, D and E.
Answer:
(b)
4. (a) Give chemical tests to distinguish between :
(i)
Propanol
and
propanone
(ii)
Benzaldehyde
and
acetophenone
(b) Arrange the following compounds in an increasing order of their property as indicated :
(i) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
(iii) CH3CH2CH (Br) COOH, CH3CH (Br) CH2COOH, (CH3)2CHCOOH (acid strength)
(a) (i) Propanone and Propanol
It can be done by iodoform test where propanone gives yellow ppt. of iodoform while propanol
does
not
respond
to
test.
(ii) Benzaldehyde and acetophenone :
180
By Iodoform test : Acetophenone being a methyl ketone on treatment with I2/NaOH (NaOI)
undergoes iodoform test to give yellow ppt. of iodoform but benzaldehyde does not.
(b) (i) Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) 4-Methoxybenzoic acid < Benzoic acid < 3, 4-Dinitro benzoic acid
5. (a) An organic compound ‘A’ with molecular formula C8H8O forms an orange red precipitate
with 2,4-DNP reagent and gives yellow precipitate on heating with I 2 and NaOH. It neither
reduces Tollen’s reagent nor Fehling’s reagent nor does it decolourize bromine water or Baeyer’s
reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid ‘W having molecular
formula C7H6O2. Identify the compounds ‘A’ and ‘B’ and explain the reactions involved.
(b) Write the mechanism of esterification of carboxylic acids
(a) (i) Molecular formula : C8H8O
(ii) Since it gives orange red ppt with 2, 4-DNP reagent Therefore it must be either aldehyde or a
ketone.
(iii) Since, it does not reduce Tollen’s reagent nor Fehling’s reagent, solution ‘A’ must be a ketone,
(iv) ‘A’ on treatment with I2/NaOH gives yellow ppt
(v) It does not decolorise Br2 water or Baeyer’s reagent so the unsaturation of ‘A’ is due to
benzene
ring.
181
(b) Mechanism of esterification of carboxylic acids
:
6. (a) An organic compound ‘A’ which has characteristic odour, on treatment with NaOH forms
two compounds ‘B’ and ‘C’. Compound ‘B’ has the molecular formula C 7H8O which on
oxidation with CrO3 gives back compound ‘A’. Compound ‘C’ is the sodium salt of the acid.
‘C’ when heated with soda lime yields an aromatic hydrocarbon ‘D’. Deduce the structures of
‘A’, ‘B’, ‘C’ and ‘D’.
(b) Give reasons :
(i) Electrophilic substitution in Benzoic acid takes place at meta position.
(ii) Carboxylic acids do not give characteristic reactions of carbonyl group
Answer:
(a) (A) gives characteristic odour which on treatment with NaOH and forms two compounds B and C.
(b) (i) The benzene ring of benzoic acid undergoes electrophilic substitution reaction such as
nitration, sulphonation etc. Since the — COOH group in benzene is an electron withdrawing
group, therefore it is meta directing group.
(ii) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible
resonance
structure.
7. (a) Give chemical tests to distinguish between the following pairs of compounds :
(i) Benzene amide and 4-aminobenzoic acid (ii) Methyl acetate and Ethyl acetate
(b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative and reduces
Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1,2benzenedicarboxylic acid. Identify the compound and write chemical equations for the reactions.
182
(a) (i) NaHCO3 test : p-Aminobenzoic acid is an acid which decomposes NaHCO3 solution to
produce effervescence due to evolution of CO2 gas while benzene amide does not give this test.
(ii) Ethyl acetate when boiled with excess of NaOH gives ethyl alcohol and if it is heated with I2
and
NaOH
solution,
it
gives
yellow
precipitate
of
iodoform.
Methyl acetate on hydrolysis with NaOH gives methyl alcohol which does not respond to
iodoform test.
(b) (i) Molecular formula of compound : C9H10O
(ii) It forms a 2, 4-DNP derivative and reduces Tollen’s reagent. It must be an aldehyde.
(iii) Since it undergoes Cannizzaro’s reaction, therefore CHO group is directly attached to the
benzene
ring.
(iv) On oxidation it gives 1,2-benzenedicarboxylic acid, therefore it must be an ortho-substituted
benzaldehyde, so the structure should be
8. (a) Give chemical tests to distinguish between the following pairs of compounds :
(i) Benzoic acid and Phenol (ii) Benzaldehyde and Acetophenone
(b) An organic compound with molecular formula C5H10O does not reduce Tollen’s reagent but
forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test.
On vigorous oxidation, it gives ethanoic acid and propanoic acid. Identify the compound and
write all chemical equations for the reactions.
(a) (i) Benzoic acid is an acid which decomposes NaHCO3 solution to give effervescence of
CO2 whereas phenol does not respond to this test.
(ii) Benzaldehyde and acetophenone :
By Iodoform test : Acetophenone being a methyl ketone on treatment with I2/NaOH (NaOI)
undergoes iodoform test to give yellow ppt. of iodoform but benzaldehyde does not.
(b) (i) Molecular formula of compound : C5H10O.
(ii) It gives an addition product with NaHS03 so it must be either an aldehyde or methyl ketone.
183
(iii) It does not reduce Tollen’s reagent so it is not an aldehyde.
(iv) It gives +ve iodoform test, therefore it is methyl ketone.
(v) On oxidation it gives a mixture of ethanoic acid and propanoic acid
Therefore, possible structure is pentan-2-one.
Reactions :
9. (a) Although phenoxide ion has more number of resonating structures than carboxylate ion,
carboxylic acid is a stronger acid than phenol. Give two reasons.
(b) How will you bring about the following converstions?
(i) Propanone to propane (ii) Benzoyl chloride to benzaldehyde
(iii) Ethanal to but-2-enal.
Answer:
(a) Carboxylic acid is a stronger acid than phenol because:
(i) In the resonating structure of phenol and carboxylic acid, the negative charge on the
carboxylate ion is delocalised over two oxygen atoms while they are localized on oxygen atom.
(ii) The release of a proton from carboxylic acids is much easier than from phenols,
(b) (i) Propanone to Propane
184
(iii)
10. (a) Complete the following reactions:
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal (ii) Benzoic acid and Phenol
(b) (i) Ethanal and Propanal : Ethanal and propanal can be distinguished by iodoform test. Warm
each compound with iodine and sodium hydroxide solution in water. Ethanal gives yellow
crystal of iodoform while propanal does not respond to iodoform test.
(ii) Phenol and Benzoic acid: On addition of NaHCO3 to both solutions carbon dioxide gas is
evolved with benzoic acid while phenol does not form CO2
185
11. (a) How will you convert the following :
(i) Propanone to Propan-2-ol (ii) Ethanal to 2-hydroxy propanoic acid
(iii) Toluene to benzoic acid
(b) Give simple chemical test to distinguish between :
(i) Pentan-2-one and Pentan-3-one (ii) Ethanal and Propanal.
(a) (i) Propanone to Propan-2-ol
Add I2 and NaOH in both the solutions. Pentan-2-one gives yellow coloured precipitate, but
pentan- 3-one does not.
(ii) Ethanal and Propanal : Add I2 and NaOH in both the solutios. Ethanal gives yellow coloured
precipitate but propanal does not.
12. (a) Write the products of the following reactions :
(b) Which acid of each pair shown here would you expect to be stronger?
186
Answer:
(b) (i) Due to much stronger I-effect of F over Cl, the FCH2COO– ion is much more stable than
ClCH2COO– ion and hence FCH2COOH is a stronger acid than ClCH2COOH.
(ii) Ethanoic acid is more stronger acid than phenol due to less pKa than that of phenol and the
carboxylate ion is much more resonance stabilized than phenoxide ion.
13. 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.
∴ A = Formaldehyde, B = Methyl alcohol, C = Sodium formate, D = Formic acid
14. (a) How will you carry out the following conversions?
(i) Acetylene to Acetic acid (ii) Toluene to m-nitrobenzoic acid
(iii) Ethanol to Acetone
187
(b) Give reasons :
(i) Chloroacetic acid is stronger than acetic acid.
(ii) pH of reaction should be carefully controlled while preparing ammonia derivatives of
carbonyl compounds
(i) Acetylene to Acetic acid
(b) (i) Because – I effect of Cl decreases the electron density in the O-H bond thereby making
the release of a proton easier.
(ii) 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.
15. (a) How will you obtain the following :
(i) Benzaldehyde from Phenol (ii) Benzoic acid from Aniline
(b) Give reasons :
(i) Aldehydes are more reactive than ketones towards nucleophilic reagents.
(ii) Electrophilic substitution in benzoic acid takes place at meta position.
(iii) Carboxylic acids do not give the characteristic reactions of carbonyl group.
188
(a) (i) Benzaldehyde from Phenol
(b) (i) Aldehydes are more reactive than ketones due to the following two reasons :


Due to smaller +1 effect of one alkyl group in aldehydes as compared to larger +1 effect
of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in
aldehydes than in ketones. As a result nucleophilic addition reactions occur more readily
in aldehydes than in ketones.
Due to presence of a H-atom on the carbonyl group, aldehydes can be more easily oxidised
than ketones.
(ii) The benzene ring of benzoic acid undergoes electrophilic substitution reaction such as
nitration, sulphonation etc. Since the — COOH group in benzene is an electron withdrawing
group, therefore it is meta directing group.
(iii) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible
resonance structure.
16. (a) State and illustrate the following :
(i) Wolff-Kishner reduction (ii) Aldol condensation
(b) An organic compound (A) which has characteristic odour, on treatment with NaOH forms
two compounds (B) and (C). Compound (B) has the molecular formula C7H8O which on
oxidation with CrO3 gives back compound (A). Compound (C) is the sodium salt of the acid.
Compound (C) when heated with soda lime yields an aromatic hydrocarbon (D). Deduce the
structures of (A), (B), (C) and (D). Write chemical equations for all reactions taking place.
Answer:
189
(a) (i) Wolff-Kishner reduction reaction : The reduction of aldehydes and ketones to the
corresponding hydrocarbons by heating them with hydrazine and KOH or potassium tertbutoxide in a high boiling solvent like ethylene glycol is called Wolff-Kishner reduction.
(ii) Aldol condensation: In this two molecules of aldehydes or ketones containing a-hydrogen
atoms on treatment with dil. NaOH undergoes condensation to form β-hydroxyaldehydes or βhydroxy ketones.
17. (a) Write the products of the following reactions :
(b) Give simple chemical tests to distinguish between the following pairs of compounds :
(i) Benzaldehyde and Benzoic acid (ii) Propanal and Propanone.
Answer:
190
(ii) Propanal and Propanone : Propanal gives positive test with Fehling solution in which a red
ppt. of cuprous oxide is obtained while propanone does not respond to test.
18. (a) Account for the following :
(i) CH3CHO is more reactive than CH3COCH3 towards reaction with HCN.
(ii) Carboxylic acid is a stronger acid than phenol.
(b) A compound “A” of molecular formula C2H3OCl undergo a series of reactions as shown
below. Write the structures of A, B, C and D in the following reactions:
(a) (i) Aldehydes are more reactive than ketones due to the following two reasons :


Due to smaller +1 effect of one alkyl group in aldehydes as compared to larger +1 effect of
two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in
aldehydes than in ketones. As a result, nucleophilic addition reactions occur more readily
in aldehydes than in ketones.
Due to presence of a H-atom on the carbonyl group, aldehydes can be more easily oxidised
than ketones.
(ii) The release of a proton from carboxylic acid is much easier than phenols because carboxylate
ion is much more resonance stabilized than phenoxide ion.
(b)
191
19. (a) Write the products formed when CH3CHO reacts with the following reagents :
(i) HCN (ii) H2N – OH (iii) CH3CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds :
(i) Benzoic acid and Phenol (ii) Propanal and Propanone.
(b) (i) Benzoic acid and Phenol : On addition of NaHCO 3 to both solutions carbondioxide gas is
evolved with benzoic acid while phenol does not form CO2.
(ii) Propanal and Propanone : Propanal gives positive test with Fehling solution in which a red
ppt. of cuprous oxide is obtained while propanone does not respond to test.
20. (a) Account for the following :
(i) Cl – CH2COOH is a stronger acid than CH3COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Write the chemical equations to illustrate the following name reactions :
(i) Rosenmund reduction (ii) Cannizzaro’s reaction
(c) Out of CH3CH2 – CO – CH2 – CH3 and CH3CH2 – CH2 – CO – CH3, which gives iodoform
test?
(a) (i) Because – I effect of Cl decreases the electron density in the O-H bond thereby making
the release of a proton easier.
(ii) Carboxylic acids do not give the reactions of carbonyl group as there is no carbonyl group
192
present due to resonance.
(b) (i) Rosenmund reduction
(ii) Cannizzaro’s reaction : Aldehydes, which do not have an oe-hydrogen atom undergo self
oxidation and reduction on treatment with conc. alkali and produce alcohol and carboxylic acid
salt.
(c) CH3CH2 – CH2 – CO – CH3 gives Iodoform test.
21. (a) How will you convert:
(i) Benzene to acetophenone (ii) Propanone to 2-Methylpropan-2-ol
(b) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The
molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition
compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous
oxidation it gives ethanoic and propanoic acids. Derive the structure of the compound.
Answer:
(i) Benzene to acetophenone
193
22. (a) Draw the structures of the following compounds:
(i) 4-chloropentan-2-one (it) But-2-en-l-al
(b) Write the product(s) in the following:
Answer:
23. (a) Write the IUPAC names of the following compounds :
(i) CH3CO(CH2)4CH3 (ii) Ph — CH = CH — CHO
(b) Describe the following conversions in not more than two steps :
(i) Ethanol to 3-Hydroxybutanal (ii) Benzoic acid to m-Nitrobenzyl alcohol
(iii) Propanone to Propene.
Answer:
(a) (i) CH3CO(CH2)4CH3 :
IUPAC name : Heptan-2-one
(ii) Ph — CH = CH — CHO :
IUPAC name : 3-phenylprop-2-en-l-al
(b) (i) Ethanol to 3-Hydroxybutanal
(ii) Benzoic acid to m-nitrobenzyl alcohol :
(iii)
194
24. (a) Draw the structures of the following compounds :
(i) 4-Chloropentan-2-one (ii) p-Nitropropiophenone
(b) Give reasons :
(i) Electrophilic substitution in benzoic acid takes place at meta position.
(ii) Carboxylic acids are higher boiling liquids than aldehydes, ketones and alcohols of
comparable molecular masses.
(iii) Propanal is more reactive than propanone in nucleophilic addition reactions.
(b) (i) Because -COOH group is electron withdrawing group and deactivates the benzene ring.
As a result of this ortho and para position acquires positive charge but only meta does not, so
electrophile can attack on rneta position.
(ii) Because -COOH group of carboxylic acids is capable to do intermolecular hydrogen
bonding forming a dimer while alcohols, aldehydes and ketones cannot.
(iii) Because of smaller +1 effect of one alkyl group in propanal as compared to larger + I effect
ol 2 alkyl groups of propanone, the magnitude of positive charge on the carbonyl carbon is more
in propanal than propanone.
25. (a) Describe the following giving chemical equations :
(i) De-carboxylation reaction (ii) Friedel-Crafts reaction
(b) How will you bring about the following conversions?
(i) Benzoic acid to Benzaldehyde (ii) Benzene to m-Nitroacetophenone (iii) Ethanol to 3Hydroxybutanal.
(i) De-carboxylation reaction: When sodium salt of carboxylic acid is heated with soda lime
(NaOH + CaO), then corresponding alkane and CO2 will be evolved
(ii) Friedel-Crafts reaction: The introduction of alkyl or acetyl group in the presence of
anhydrous aluminium chloride (AlCl3) as catalyst to ortho and para positions of an aromatic
compound is called Friedel-Crafts reaction,
195
(a)
Complete the following equations : (Comptt. All India 2016)
Answer:
(b)
(i) On adding NaHCO3, CH3COOH produces brisk effervescence of CO2 gas whereas
phenol does not.
(ii) On heating with Tollen’s reagent, CH3CHO forms silver mirror whereas
CH3COCH3 does not.
196
CHAPTER 13
AMINES
GIST OF THE LESSON
Classification: - Amines are classified according to the number of carbon atoms bonded directly
to the nitrogen atom. A primary (1°) amine has one alkyl (or aryl) group on the nitrogen atom, a
secondary (2°) amine has two, and a tertiary (3°) amine has three.
Physical Properties of Amines
1. The lower aliphatic amines are gases with fishy smell. Primary amines with three or more
carbon atoms are liquid and higher members are all solids.
2. Lower aliphatic amines are water soluble because they can form hydrogen bonds with water
molecules, however the solubility decreases with increase in hydrophobic alkyl group.
3. Boiling points order: primary amine > secondary amine > tertiary amine
Preparation
HOFFMANN BROMAMIDE REACTION: -
197
GABREIL PHTHALIMIDE SYNTHESIS: -
Reduction of nitro compounds:
Ammonolysis of alkyl halides
Reduction of nitriles and amides:
THE ORDER OF BASICITY OF ETHYL AMINES AD METHYL AMINES
Amines act as Lewis bases due to the presence of lone pair of electrons on the nitrogen atom.
More the Kb, (dissociation constant of base), higher is the basicity of amines. Lesser the pKb
higher is the basicity of amines.
Aliphatic amines (R-NH2) are stronger bases than NH3 due to the electron releasing +I effect of
the alkyl group.
Among aliphatic methyl amines, the order of basic strength in aqueous solution is as follows:
198
(C2H5)2NH>(C2H5)3N > C2H5NH2> NH3
(CH3)2NH > CH3NH2> (CH3)3N > NH3
Aromatic amines are weaker bases than aliphatic amines and NH3, due to the fact that the
electron pair on the nitrogen atom is involved in resonance with the re-electron pairs of the ring.
BENZOYLATION:
CH3NH2 + C6H5COCl → CH3NHCOC6H5 + HCl
CARBYLAMINE REACTION:
HINSBERG TEST:
Tertiary amines do not react with benzene sulphonyl chloride.
ELECTROPHILIC SUBSTITUTION REACTIONS: BROMINATION: -
To prepare monosubstituted derivative, activating effect of –NH2 by acetylation with acetic
anhydride.
199
NITRATION: Direct nitration of aniline is not possible as it is susceptible to oxidation, thus
amino group is first protected by acetylation.
In strongly acidic medium, aniline is protonated to from anilinium ion which is meta directing so
it gives meta product also
SULPHONATION:
Aniline does not undergo Friedel Craft reactions due to salt formation with aluminium chloride,
the Lewis acid, which is used as catalyst.
DIAZOTISATION:
200
MIND MAP
201
MULTIPLE CHOICE QUESTIONS
1. What is the correct IUPAC name of H2N-(CH2)5-NH2?
A. Pentan-1,5-diamine
B. 1,5-Diaminopentane
C. Pentamethylenediamine
D. Pentane-1,5-diamine
2. Which of the following does not react with Hinsberg reagent?
A. C2H5NH2
B. (CH3)2NH
C. (CH3)3N
D. CH3 CH(NH2)CH3
3. Which of the following is incorrect for primary amines?
A. On reaction with nitrous acid alkylamines produce alcohol
B. On reaction with nitrous acid arylamines produce phenol
C. Alkylamines are more basic than ammonia
D. Alkylamines are more basic than arylamine
4. If the pKb value of ammonia is 4.75, predict the pKb value of methanamine?
A. 3.38
B. 4.70
C. 8.92
D. 9.38
5. Which of the following amines are insoluble in water?
A. Methanamine
B. Ethanamine
C. Propanamine
D. Benzenamine
6. Which of the following is formed when an alkyl primary amine reacts with nitrous acid?
A.
B.
C.
D.
Alkyl nitrite
Secondary amine
Nitroalkane
Alcohol
7. Arrange the following compounds in increasing order of basicity:
CH3NH2, (CH3)2 NH, NH3, C6H5NH2
A.
B.
C.
D.
C6H5NH2 < NH3 < (CH3)2NH < CH3NH2
CH3NH2 < (CH3)2NH < NH3 < C6H5NH2
C6H5NH2 <NH3 < CH3NH2<(CH3)2NH
(CH3)2NH < NH3 < C6H5NH2 < CH3NH2
8. Which of the following will have the highest pKb value?
202
A.
B.
C.
D.
C6H5NH2
p-C6H5(CH3)NH2
p-C6H5(OCH3)NH2
p-C6H5(NH2)NH2
9. What is the characteristic odour of relatively lower aliphatic amines?
A. Fruity odour
B. Fishy odour
C. Rotten egg smell
D. Odourless
10. The correct IUPAC name for CH2 = CHCH2NHCH3 is
A. Allylmethylamine
B. 2-amino-4-pentene
C. 4-aminopent-1-ene
D. N-methylprop-2-en-1-amine
11. Which of the following is not a final product of the reaction between propylamine and
nitrous acid?
A. CH3CH2CH2N2Cl
B. CH3CH2CH2OH
C. N2 gas
D. H2O
12. Which of the following test cannot be used in the distinction between ethylamine and
diethylamine?
A. Hinsberg’s test
B. p-toluenesulphonylchloride
C. Reaction with CH3CH2Br
D. Carbylamine test
13. The order of basic strength of amines in aqueous solution is
A. (CH3)3N > (CH3)2NH > CH3NH2 > NH3
B. CH3NH2 > (CH3)2 NH > (CH3)3N > NH3
C. NH3 > (CH3)3N > (CH3)2NH > CH3NH2
D. (CH3)2NH > CH3NH2 > (CH3)3N > NH3
14. In this reaction acetamide is converted to methanamine
A. Gabriel phthalimide synthesis
B. Carbylamine reaction
C. Stephen’s reaction
D. Hoffmann bromamide reaction
15. What are the products formed from the reaction between aniline and ethanoic anhydride?
A. N-Phenylethanamide and hydrochloric acid
B. N-Phenylethanamide and acetic acid
203
C. N-Methylbenzamide and acetic acid
D. N-Methylbenzamide and hydrochloric acid
16.
Hinsberg’s reagent is
A. Benzenesulphonic acid
B. Benzenesulphonyl chloride
C. p-toluenesulphonyl chloride
D. Chlorosulphuric acid
17. Identify the correct naming of H2NCH2CH2OH.
A. 2-Hydroxyethanamine
B. 2-Aminoethanol
C. Ethane-2-hydroxy-1-amine
D. 1-Aminoethan-2-ol
18. Amongst the following, the strongest base in aqueous medium is
A. CH3NH2
B. NCCH2NH2
C. (CH3)2 NH
D. C6H5NHCH3
.
19. Starting from propanoic acid, the following reactions were carried out, what is the
compound Z?
A.
B.
C.
D.
CH3−CH2−Br
CH3−CH2−NH2
CH3-CH2-COBr
CH3−CH2−CH2−NH2
20. Treatment of NH3 with excess of ethyl chloride gives:
Diethylamine
A. Ethane
B. Tetraethylammonium chloride
C. Methylamine
21. What is the correct order of basic strength of the following compounds?
A. Aniline > Diphenylaniline > Triphenylaniline
B. Aniline > Triphenylaniline > Diphenylaniline
C. Diphenylaniline > Triphenylaniline > Aniline
D. Triphenylaniline > Diphenylaniline > Aniline
22. An organic compound (X) was treated with sodium nitrite and HCl in ice cold
conditions. Bubbles of nitrogen gas were seen coming out. The compound (X) may be
A. a secondary aliphatic amine.
B. a primary aliphatic amine
C. a secondary aromatic amine
204
D. a tertiary amine
23. Which of the following amines, on reaction with benzenesulphonyl chloride, will give a
sulphonamide that is insoluble in alkali?
A. Ethylamine
B. Ethylmethylamine
C. Aniline
D. Trimethylamine
24. The best reagent for converting 2–phenylpropanamide into 2-phenylpropanamine is
A. excess H2
B. Br2 in aqueous NaOH
C. iodine in the presence of red phosphorus
D. LiAlH4 in ether
25. Which of the following amines can be prepared by Gabriel synthesis?
A. Isobutyl amine
B. Dimethylamine
C. N-methyl benzylamine
D. Aniline
26.
NH2 group in aniline is
A. Ortho directing
B. Meta directing
C. Ortho and para directing
D. Para directing
27. If the Kb values of ammonia, methylamine and ethylamine are x, y and z respectively,
identify the correct relation between x, y and z from the following.
A. x > y
B. y < z
C. x > z
D. x > y > z
28. Which of the following is formed from the reaction between RNHR’ and R” COCl,
where R, R’ and R” are three different alkyl groups?
A. RNR’COR”
B. R2NCOR’
C. R’2NCOR”
D. R”NR’COR
29. Reduction of aromatic nitro compounds using Fe and HCl gives
A. Aromatic oxime
B. Aromatic hydrocarbon
C. Aromatic primary amine
D. Aromatic amide
.
205
30. The product of the given reaction is
31. How many structural isomers are possible for C3H9N?
A. 4
B. 2
C. 5
D. 3
32. The hybrid state of N of R2NH
A. sp3
B. sp2
C. sp
D. dsp2
33. Which of the following methods of preparation of amines will not give the same number
of carbon atoms in the chain of amines as in the reactant?
A. The reaction of nitrile with LiAlH4
B. The reaction of the amide with LiAlH4 followed by treatment with water.
C. Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis.
D. Treatment of amide with bromine in the aqueous solution of sodium hydroxide.
34. Which of the following is associated with decrease in pKb value of amines?
A. Increase in acidic strength
B. Increase in basic strength
C. Better proton donation
D. Better electron acceptor
35. When aniline is heated with cone. H2SO4 at 455-475 K, it forms
A. Sulphurous acid
B. Sulphanilic acid
C. 2-Amino benzene sulphonic acid
D. Benzene sulphonic acid
206
36. Which of the following aromatic amines has lower pKb value than ammonia?
A. Benzylamine
B. Benzenamine
C. N-Methylbenzenamine
D. N, N-Dimethylbenzenamine
37. Which of the following should be most volatile?
I. CH3CH2CH2NH2
II. (CH3)3N
III CH3CH2NHCH3
IV. CH3CH2CH3
A. II
B. IV
C. I
D. III
38. Aniline in a set of reactions yielded a product D.
The structure of D would be
A. C6H5CH2OH
B. C6H5CH2NH2
C. C6H5NHOH
D. C6H5NHCH2CH3.
39. Acid anhydrides on reaction with primary amines give
A. Amide
B. Imide
C. Secondary amine
D. Imine
.
40. Which of the following is true for the basicity of amines?
A. Alkylamines are generally less basic than arylamines because N is sp hybridised
B. Arylamines are generally more basic than alkylamines due to aryl group
C. Arylamines are generally less basic than alkylamines due to delocalisation of lone
pair of electrons in the benzene ring
D. Alkylamines are generally less basic than arylamines because lone pair of
electrons on N in the arylamines are not delocalised in the benzene ring.
41. Among the compounds C3H7NH2, CH3NH2, C2H5NH2, and C6H5NH2. Which is the least
basic compound?
A. CH3NH2
B. C2H5NH2
C. C3H7NH2
D. C6H5NH2
207
42. Identify the incorrect name of the compound formed when two hydrogen atoms of
ammonia are replaced by phenyl groups.
A. Diphenylamine
B. N-Phenylaniline
C. N-Phenylbenzenamine
D. N-Diphenylaniline
43. What is the IUPAC name of CH3-NH-CH2-CH(NO2)-CH3?
A.
B.
C.
D.
N-Methyl-2-nitropropanamine
1-Aminomethyl-2-nitropropane
N-(2-Nitropropyl)methanamine
N-Methyl-2-methylnitropropanamine
44. Amine that cannot be prepared by Gabriel-Phthalmide synthesis is
A.
B.
C.
D.
Aniline
Benzyl amine
Methyl amine
Iso-butylamine
45. Which of the following can exist as zwitter ion?
A.
B.
C.
D.
p-Aminoacetophenone
Sulphanilic acid
p-Nitroaminobenzene
p-Methoxyphenol
46. Which of the following cannot be identified by carbylamine test?
1. C2H5NH2
2. C6H5NH2
3. C6H5 — NH — C6H5
4. (C2H5)3N
A. 1,2
B. 1,2,4
C. 3,4
D. 2,4
47. Which of the following groups when present at para position increases the basic strength
of aniline?
A. NO2
B. Br
C. NH2
D. COOH
208
48. If ‘a’ is the pKb value of aniline and ‘b’, ‘c’ and ‘d’ are the p Kb values of o-, m- and pisomers of methylaniline respectively, what is the correct order of the values a, b, c and
d?
A. a > b > c > d
B. d > c > a > b
C. d > c > b > a
D. b > a > c > d
49. 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
50. 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
ANSWERS
1
2
3
4
5
6
7
8
9
10
D
11
A
21
A
31
A
41
D
C
12
C
22
B
32
A
42
D
B
13
D
23
B
33
D
43
A
A
14
D
24
D
34
B
44
A
D
15
B
25
A
35
B
45
B
D
16
B
26
C
36
A
46
C
C
17
B
27
B
37
B
47
C
A
18
C
28
A
38
A
48
D
B
19
B
29
C
39
A
49
B
D
20
C
30
C
40
C
50
C
ASSERTION -REASON TYPE QUESTIONS
a
b
c
d
1
Choose the correct answer from the following choices
Both assertion and reason are correct statements and reason is correct explanation of
assertion
Both assertion and reason are correct statements but reason is not correct explanation of
assertion
Assertion is correct statement but reason is wrong statement
Assertion is wrong statement but reason is correct statement
Assertion: Alkylation of amines gives polysubstituted product where as acylation of
amines gives a monosubstituted product
209
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Reason: Steric hinderence of an acyl group prevents the approach of further acyl
groups.
Assertion: Anilinium chloride is more acidic than ammonium chloride
Reason: Anilinium ion is resonance stabilised.
Assertion: Gabriel phthalimide reaction can be used to prepare aryl and alkyl amines
Reason: Aryl halides have same reactivity as alkyl halides towards nucleophilic
substitution reactions.
Assertion: Aniline does not undergo Friedel -Crafts reaction
Reason: Friedel-Crafts reaction is electrophilic substitution reaction
Assertion:CuCl2 gives a deep blue coloured solution with ethyl amine
Reason: Ethylamine molecules coordinate with cupric ions forming a blue coloured
complex.
Assertion: Aniline reacts with bromine water to form 2,4,6-tribromoaniline
Reason: Aniline is resonance stabilised
Assertion: Acetanilide is less basic than aniline
Reason: Acetylation of aniline results in decrease of electron density on nitrogen
Assertion: N,N diethyl benzene sulphonamide is insoluble in alkali.
Reason: Sulponyl group attached to nitrogen is strong electron withdrawing group
Assertion: Amides on oxidation yields amines
Reason: Amides on treating with lithiumaluminiumhydride gives amines
Assertion: The order of boiling points of isomeric amines is
Primary>Secondary>Tertiary
Reason: Intermolecular association is more in primary ,then in secondary and least in
tertiary amines.
Assertion: Aliphatic amines are weaker base than ammonia
Reason:+I effect of alkyl group results in high electron density on nitrogen atom.
Assertion: Aniline is weaker base than ammonia
Reason: Aryl groups are electron withdrawing in nature.
Assertion: The order of basicity of amines in gaseous state is different than those in
aqueous solution
Reason: In aqueous solution solvation is also to be takes in to account.
Assertion: Amines react with carboxylic acids to form amides
Reason: They form salts with amines at room temperature
Assertion: Hinsberg’s reagent does not react with tertiary amines
Reason: No hydrogen atom is attached to nitrogen of amino group.
Assertion: On nitrating aniline in acidic medium, significant amount of meta product
also formed
Reason: In acidic medium aniline is protonated to form anilinium ion,which is meta
directive.
Assertion: Pkb value of aniline is low,
Reason: The unshared pair on nitrogen atom to be in conjugation with the benzene ring
making it less available
Assertion: Triethyl amine is less basic than diethyl amine
Reason: In Triethyl amine, steric hinderence outweigh solvation effect and inductive
effect.
Assertion: Solubility of amines increases with increase in molar mass
Reason: As molar mass increases the size of the hydrophobic alkyl groups also
increases. it makes with the solvent also increases.
Assertion: Hoffmann degradation of propanamide gives ethanoic acid
210
21
22
23
24
25
26
27
28
29
Reason: Hoffman degradation of amide gives an amine with one carbon less than the
parent amide.
Assertion: Ammonia is more soluble in water than methylamine
Reason: Ammonia is more soluble due to more hydrogen bonding
Assertion: Both anilinium hydrochloride and p-chloroaniline will give white ppt of
AgCl with aqueous AgNO3
Reason: Ammonolysis gives a mixture of primary , secondary and tertiary amines and
also a quarternary ammonium salt.
Assertion: CH3CH2NH2 is less basic than F-CH2CH2NH2
Reason: Presence of F atom will decrease the availability of lone pair on Nitrogen due
to its strong -I effect.
Assertion: A foul smelling compound is formed when primary amines are heated with
chloroform and ethanolic KOH
Reason: Benzene sulphonyl chloride can be used to distinguish between primary
,secondary and tertiary amines.
Assertion: All amines react with nitrous acid to form diazonium salts .
Reason: Aliphatic diazonium salts being unstable, liberate Nitrogen gas and form
alcohol.
Assertion: Ethanamine reacts with Benzoyl chloride to give Acetanilide.
Reason: The reaction of amines with benzoyl chloride is called benzoylation
Assertion: Nitration of aniline gives only ortho and paraproducts.
Reason: If nitration is carried out in strongly acidic medium ,Aniline is protonated to
form aniliniumion.
Assertion: Nitration of aniline can be conveniently done by protecting the nitro group by
acetylation
Reason: Acetylation increases the electron density in the benzene ring
Assertion: Nitrating mixture used for carrying out nitration consists of con.HNO 3 and
conH2SO4
Reason: In presence of H2SO4 ,HNO3 act as a base and produce NO 2+ ion
30 Assertion: Only a small amount of HCl is required to initiate the reduction of Nitro
compounds with iron scrap and HCl
Reason: FeCl2 formed gets hydrolysed to release HCl during the reaction.
ANSWERS
1
2
3
4
5
6
C
Due to delocalization of the lone pair of electrons of the nitrogen atom over the
carbonyl group in the acyl derivative, the electron density on the N atom decreases to
such an extent that it does not act as a nucleophile at all.
C
Ammonia is a stronger base than aniline and the conjugate acid of aniline is more
acidic than conjugate acid of ammonia
D
Gabriel phthalimide reaction is used to prepare only primary alkyl amines.
Aryl halides do not undergo nucleophilic substitution reactions.
B
AlCl3 forms a salt with aniline which deactivates the benzene ring
A
B
211
7
8
9
10
11
12
13
14
15
16
-NH2 group is ortho para directive
A
In amides the -CO group is stronger dipole than N-C dipole, hence the ability of NC
group to act as a base is restricted in presence of C=O dipole,
A
N,N-diethyl benzene does not have acidic hydrogen, hence insoluble.
D
Amides on reduction gives amines
A
Tertiary amines do not have intermolecular association due to the absence of hydrogen
atom available for hydrogen bond formation.
D
Aliphatic amines are stronger bases than ammonia
A
A
A
A
A
17 D
pKb value of aniline is high
18 A
19 D
Solubility decreases with increase in molar mass
20 D
Hoffmann degradation of propanamide gives ethanamine
21 A
22 D
Anilinium hydrochloride will give a white ppt of AgCl but p-chloroaniline will not
react with silver nitrate solution
23 A
24 B
Aliphatic and aromatic primary amines on heating with CHCl3 and ethanolic KOH
form isocyanides or carbyl amines which are foul smelling substances.
Hinsberg reagent gives alkali soluble product with primary amines, alkali insoluble
product with secondary amine and tertiary amines do not react with Hinsberg reagent.
25 D
Primary aliphatic amines gives aliphatic diazonium salts and aromatic amines give
stable diazonium salts where as secondary and tertiary amines react with nitrous acid
in a different manner.
26 D
Ethanamine undergoes benzoylation to give N-Ethyl benzamide
27 D
In strongly acidic medium nitration products of aniline includes both ortho.meta and
para products. Anilinium ion produced is meta directing in nature.
28 C
Acetylation decreases the electron density in the benzene ring
29 A
212
30 A
CASE BASED QUESTIONS
1
The basicity of amines of different classes do not follow a simple pattern because the
number of groups bonded to nitrogen affects the electron density at the nitrogen
atom. And, the stability of the conjugate acid in the solvent has a major effect on
basicity. Thus, the basicity of amines can be explained only for amines with similar
structures at the nitrogen atoms.
The basicity of an amine is increased by electron-donating groups and decreased by
electron-withdrawing groups. Aryl amines are less basic than alkyl-substituted
amines because some electron density provided by the nitrogen atom is distributed
throughout the aromatic ring. Basicity is expressed using Kb values measured from
the reaction of the amine with water. An alternate indicator of basicity is pKb, which
is −log Kb. A strong base has a large Kb and a small pKb. The basicity of amines is
also expressed by the acidity of their conjugate acids. A strong base has a weak
conjugate acid, as given by a small value of Ka and a large pKa. [Robert
J. Ouellette, J. David Rawn, in Organic Chemistry Study Guide, 2015]
a) pKb values for NH3 , CH3NH2,(CH3)2NH and (CH3)3N has 4.75,3.38,3.27 and
4.22 respectively. Write them in the decreasing order of basic strength.
b) Usually the order of basicity of amines will be different from the expected
order. Which are the three main factors affecting basicity of amines?
c) Write the decreasing order of basicity for CH3CH2NH2,(CH3CH2)2NH and
(CH3CH2)3N
d) Compare the basicity of m-toluidine and Aniline.
Answer
a)
b)
c)
d)
(CH3)2NH> CH3NH2>(CH3)3N>NH3
+I effect ,extent of hydrogen bonding with water molecules and steric effects of the
alkyl group
(CH3CH2)2NH>(CH3CH2)3N> CH3CH2NH2>NH3
m-Toluidine is more basic than aniline due to +I effect from meta position
Amines have higher boiling points than hydrocarbons of comparable molecular
weight because the C—N bond is more polar than a C—C bond. Also, primary
and secondary amines can form intermolecular hydrogen bonds because they can act
as both hydrogen bond donors and acceptors. Tertiary amines have no hydrogen
atoms bonded to the nitrogen atom and therefore are not hydrogen bond donors.
Thus, tertiary amines cannot form intermolecular hydrogen bonds. As a result, they
have lower boiling points than primary and secondary amines of comparable
molecular weight.
II
213
Amines have lower boiling points than alcohols because nitrogen is less
electronegative than oxygen. As a result the N—H bond is less polar than the O—H
bond, and the hydrogen bond in amines is weaker than the hydrogen bond in
alcohols.[ Robert J. Ouellette, J. David Rawn, in Organic Chemistry (Second
Edition), 2018]
a) The boiling point of Ethyl alcohol is more than that of ethyl amine.
Give reason
b) The order of boiling points for the following amines is
CH3CH2CH2NH2> (CH3CH2CH2) NH>(CH3CH2) N(CH3)
Explain
c) Arrange the following in the increasing order of molecular mass.
CH2CH2NH2,HCOOH,CH3CH2OH,CH3CH2CH3
ANSWERS
a) N is less electronegative than oxygen and the hydrogen bond in amines are
weaker than that in alcohols.
b) The boiling points of amines follows the order primary>secondary>tertiary
Primary amines have 2 ,secondary amines have 1 while tertiary amines have no
hydrogen linked to nitrogen.
c) Among these compounds of comparable molecular masses ,order of extend of
hydrogen bond is hydrocarbons <amines<alcohols<carboxylic acids
Hence the order of given compounds of comparable molecular masses is
CH3CH2CH3< CH2CH2NH2< CH3CH2OH< HCOOH
III. Basicity of aliphatic amines in aqueous solution depends on
+I effect, extent of hydrogen bonding with water molecules, steric effects of the alkyl
group. In aromatic amines it depends on effect of substituents on the ring. Basicity of
aralkyl amines, as the electrons are not conjugated with benzene ring and not delocalized. Hence electron pair is available for protonation. In aromatic amines ortho
effect refers mainly to the set of steric effects and some bonding interactions along
with polar effects caused by the various substituents which are in a given molecule
altering its chemical properties and physical properties. In a general sense the ortho
effect is associated with substituted benzene compounds.
Anything ortho to the amine, no matter whether it is electron donating or
withdrawing, will decrease the basicity of the aromatic amine. This is because of the
ortho effect, which is basically steric effect. The protonated amine will have a greater
steric interaction with the ortho group, so it will be less stable.
a) Compare the basicity of aniline, m-nitroaniline, p- nitroaniline and ortho
nitroaniline
b) Compare the basicity of benzylamine, aniline, methyl amine and ammonia
c) Aniline has pKb =9.38,p-aminophenol has pKb =8.50 and o-aminophenol
has pKb=9.28, but m-aminophenol has pKb=9.80
Explain.
ANSWER
a) o-nitroaniline<p-nitroaniline<m-nitroaniline<Aniline(ewg decreases the
basicity due to ortho effect o-nitro aniline is the weak one among
substituted anilines)
b) aniline<ammonia<benzylamine<methyl amine
214
(methyl group is more electron donating than benzyl group, hence methyl
amine is stronger base, in benzyl amine electrons are not delocalised
,hence better base than aniline but ammonia with pKb value4.75 is more
basic than benzylamine(pKb=4.70)
c) Here though -OH group is in ortho position o-aminophenol is stronger
base than aniline (pkb=9.38) and m-aminophenol(pKb=9.80) due to
stabilisation of anilinium ion by hydrogen bonding.
SHORT ANSWER QUESTIONS (2 marks)
1. Give one test to distinguish methanamine and N,N-dimethyl amine and give chemical
equation for the reaction involved.
An: Methanamine gives carbylamines reaction .When methanamine is heated with
chloroform and alcoholic KOH ,it gives a foul smelling methyl isocyanide.
CH3-NH2 + CHCl3 + 3KOH --------->CH3-NC + 3KCl + 3H2O
2. Write balanced chemical equation for the Hoffmann Bromamide degradation reaction.
Why is it called degradation reaction?
R CONH2 + Br2 + 4NaOH  RNH2 + Na2 CO3 + 2NaBr + 2H2 O
It involves decrease in number of carbon atoms .
3. Name the reagent and name of test used for distinguishing (CH 3 )2 NH and (CH3 )3 N ?
An:They can be distinguished using Hinsberg’s reagent, i.e., benzene sulphonyl chloride. When
treated with Hinsberg’s reagent ,secondary amine gives the product which is insoluble in alkali
but tertiary amine gives no reaction.
4. Arrange the following in the decreasing order of basicity and give reason
Aniline , p-Nitroaniline and p-methylaniline
An: p -Nitroaniline < aniline < p-methylaniline
CH3 group is electron releasing groups which increases electron density on nitrogen of amino
group . But presence of electron withdrawing group like nitro group decrease electron density
on nitrogen .
5. Give the IUPAC name of (a) Acetanilide (b) CH3 NHCOC6H5
An: (a) N-phenyl ethanamide (b) N-Methylbenzamide
6 (a) Which reagent can be used instead of Benzene sulphonyl chloride
(b) Which reaction is used in the quantitative estimation of aminoacids and proteins
An: (a) p- toluene sulphonyl chloride
(b) Primary aliphatic amines react with nitrous acid to form unstable diazonium salts which
liberate nitrogen gas quantitatively.
7. What will be the increasing order of basicity in gaseous state among the following compounds
and mention the reason involved .
ammonia, methyl amine, dimethyl amine and tri methyl amine
An: Ammonia< methyl amine <dimethyl amine<tri methyl amine
215
Greater the number of +I group ,greater will be the basicity
8. What will be the increasing order of basicity in aqueous state among the following
compounds? Give reason
ammonia, methyl amine dimethyl amine and tri methyl amine
An: Ammonia< trimethyl amine<methyl amine <dimethyl amine
It is due to interplay of 3 factors given below
(i)Steric hindrance
(ii)No of + I groups
(iii) Solvation effect
9. Convert ethyl amine to ethanol (b) Aniline to 4-Bromoaniline
(a) C2H5 NH2 + HNO2 ( NaNO2+ HCl at 278K)
(b)
 C2 H5 OH+N2+HCl
10. (a) Convert aniline to benzene diazonium chloride
(b)Out of alkyl amine and ammonia which is more basic why ?
(a)
(b) Alkyl amine is more basic due to electron releasing nature of alkyl groups.
11. How will you distiguish (a)ethylamine and diethylamine (b) aniline and methyl amine
An: (a)ethyl amine gives carbylamines reaction ie when ethyl amine is heated with chloroform
and alcoholic KOH ,it gives a foul smelling ethyl isocyanide gas.
Diethyl amine,being a secondary amine does not give this test
(b) Test with Bromine water is used to distinguish them. Aniline gives a white ppt of 2,4,6- tri
bromo aniline and methyl amine does not give.
216
.
12 How will you distiguish (a) Ethylamine and aniline (b) Aniline and N,N-di methyl aniline
An :(a) Test with Bromine water is used. Aniline gives a white ppt of 2,4,6- tri bromo aniline
and ethyl amine does not give.
(b) Aniline being a primary amine gives carbylamine test, i.e., when heated with an alcoholic
solution of KOH and CHCl3, it gives foul smell of phenyl carbylamine whereas N, Ndimethylaniline being a tertiary amine does not give this test.
13. Give equations for the Gabreil phthalimide synthesis to get primary amine
.
14. Convert (a)nitro benzene to Aniline
(b) Acetamide to methanamine
CH3 CO NH2 +4NaOH+Br2  CH3 NH2 + Na2 CO3 + 2NaBr + 2H2O
15. Account for the following
(3 marks)
(a) Basicity of aniline is less than that of methylamine
(b) Methylamine has higher boiling point than trimethylamine .
(c) Aniline does not undergo Friedel-Crafts reaction.
a. Due to resonance and activating effect of -NH2 group ,the lone pair on nitrogen atom of NH2 can conjugate with benzene ring, there is +ve charge on N in some resonating structure
therefore electron density on N is less and it is less basic. Where as in methyl amine, methyl
group is electron donating group and electron density on N increases that’s why it is more
basic.
(b) Inter molecular hydrogen bonding is more in CH3 NH2 but the same cannot be present in the
case of ( CH3)3 N Hence the boiling points of CH3 NH2 is higher.
217
(c ) iii) Aniline does not undergo Friedel-Crafts reaction as aniline ,being a Lewis base,reacts
with Lewis acid AlCl3 to form a salt.
16. Describe a method for the identification of primary, secondary and tertiary amines. Also write
chemical equation
An; Hinsberg Reagent, benzene sulphonyl chloride is used to distinguish primary, secondary and
tertiary amines
Primary amine gives the product which is soluble in alkali
secondary amine gives the product which is insoluble in alkali
Tertiary amines do not react with benzene sulphonyl chloride.
17. i) Amines are less acidic than alcohols of comparable molecular masses.
ii) Primary amines have higher boiling point than tertiary amines?
iii) Aliphatic amines are stronger bases than aromatic amines?
An: (i ) Amines are less acidic than alcohols because oxygen is more electronegative and
small in size than nitrogen, therefore O-H bond breaks easily than N – H bond.
ii) Primary amines have higher boiling point than tertiary amines. In primary amines
RNH2 , the molecules have intermolecular hydrogen bonding but the same cannot be present in
the case of tertiary amines (RN3 )Hence the boiling points of primary amines are higher than the
tertiary amines
iii) Aliphatic amines are stronger bases than aromatic amines because in aliphatic amines
electron releasing group is present, and in aniline, the lone pair of electron of nitrogen atom is
involved in resonance with benzene ring hence is not easily available for donation
18 i) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis?
ii) pKb of aniline is more than that of methylamine.
iii) Ethylamine is soluble in water whereas aniline is not.
Answer i)Aromatic primary amines cannot be prepared by Gabriel phthalimide
synthesis, because in preparation by Gabriel phthalimide synthesis, Ar-X is a reactant and aryl
halides do not undergo nuleophilic substitution easily due to presence of partial double character
of C-X bond .
218
ii) pKb of aniline is more than that of methylamine .Since there is +ve charge on N
in some resonating structure therefore it is less basic. Where as in methyl amine, methyl group
donate electron towards N. Electron density increases that’s why it is more basic.
iii) Ethylamine is soluble in water due to the formation of hydrogen bonds with
water whereas aniline due to large hydrocarbon part ,the extent of hydrogen bonding is low.
19. i) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
ii) Although amino group is o– and p– directing in aromatic electrophilic substitution
reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
iii) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
An : i)Methylamine in water reacts with ferric chloride to precipitate hydrated ferric
oxide because methylamine being more basic than water ,accepts a proton from water liberate
OH - ions ,these OH- ions combine with Fe 3+ to form a brown ppt of hydrated ferric oxide.
ii) Although amino group is o– and p– directing in aromatic electrophilic substitution
reactions, aniline on nitration gives a substantial amount of m-nitro aniline because nitration is
usually carried out with a mixture of conc HNO3 and conc H2SO4.In presence of these acids
aniline gets protonated to form the anilinium ion which is meta directing.
iii) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Because aryl diazonium ion is stabilized by dispersal of +ve charge and no such dispersal occurs
in alkyl diazonium ion . More over, alkyl diazonium ion immediately loses N 2 molecule to form
alkyl carbocation while aryl diazonum ion does not lose N 2 so easily because aryl carbocation is
not stable.
20 i)How do you account for the miscibility of ethoxyethane with water.
ii) Direct nitration of aniline is not carried out. Explain why?
iii) The presence of a base is needed in the ammonolysis of alkyl halides.
An : (i) Ethoxyethane is miscible with water, This is due to the fact that oxygen of ether
can form hydrogen bonds with water
ii) Direct nitration of aniline yields tarry oxidation products in addition to the nitro
derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium
ion which is meta directing. That is why besides the ortho and para derivatives, significant
amount of meta derivative is also formed. So we get only a mixture
iii) The presence of a base is needed in the ammonolysis to remove HX and make the
reaction forward.
21.(Word problems)
( i)An aromatic compound A on treatment with aqueous ammonia and heating forms
compound B which on heating with Br2 and KOH forms compound C of molecular formula
C6H7N .Write the structures and IUPAC names of compounds A,B,C.
Answer- Since C is formed by Hoffmann Bromamide reaction, it is a primary amine So
Structure is C6H5 NH2 , Aniline. B is C6H5 CO NH2 ,Benzamide .A is carboxylic acid
C6H5COOH, Benzoic acid
22. A compound A has the molecular formula C7H7NO.On treatment with Br2 and NaOH it
gives B. B gives a foul smelling substance C on treatment with CHCl3 and KOH.. Identify A to
C.
Answer-: A is C6H5 CO NH2, Benzamide.B is C6 H5 NH2, Aniline, C is phenyl isocyanide ,
C6H5NC
219
23) A compound X having molecular formula C3H7NO reacts with Br2 in presence of KOH to
give another compound Y .The Compound Y reacts with HNO2 to form ethanol and N2 gas
.Identify the compounds X and Y and write the reactions involved.
Answer :X=CH3CH2CONH2 Y= CH3CH2NH2
CH3CH2CONH2+4NaOH+Br2 CH3 CH2 NH2 + Na2 CO3 + 2NaBr + 2H2O
CH3CH2NH2+ HNO2  CH3CH2OH+N2
24) A compound A with molecular formula C2H5Cl on treatment with AgCN gives two isomeric
compounds of unequal amounts with the molecular formula C 3H5N.The minor of these two
products on complete reduction with H2 in the presence nickel gives a compound B with
molecular formula C3H9N. Deduce the structures of A,B,& C and write the reactions involved.
Answer: A=CH3CH2Cl ,On reaction with AgCN it gives CH3CH2CN (Minor) &
CH3CH2NC (Major) B= CH3CH2CH2NH2
25.) A compound A has the molecular formula C7H7NO. On heating with P2O5 gives B C7H5N.
This compound B on hydrolysis gives an acid C with formula C7H6O2. The compound B on
reduction with Na/C2H5OH gives D C7H9N a basic compound. D on treatment with NaNO2 and
HCl evolves N2 forming E C7H8O. This on oxidation gives C. C is also formed on heating A
with dilute acid. Identify A to E.
Answer : A-Benzamide C6H5 CO NH2, B is phenylcyanide C6H5CN.
C is Benzoic acid C6H5COOH, D is Benzylamine C6H5 CH2NH2 and E –Benzyl
alcohol C6H5-CH2-OH
Benzamide C6H5 CO NH2 undergoes dehydration to form B C6H5CN which on
hydrolysis gives C6H5COOH.
B , C6H5CN ,on reduction gives D, C6H5-CH2NH2
Aralkyl amine on reaction with NaNO2 and HCl gives unstable
aliphatic diazonium chloride which decomposes to alcohol,E (Benzyl alcohol C6H5-CH2-OH)
SAMPLE PAPER QUESTION (2021-22)
TERM – II
CHEMISTRY THEORY (043)
MM:35
Time: 2 Hours
GENERAL INSTRUCTIONS:
Read the following instructions carefully.
1. There are 12 questions in this question paper with internal choice.
2. SECTION A - Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
3. SECTION B - Q. No. 4 to 11 are short answer questions carrying 3 marks each.
4. SECTION C- Q. No. 12 is case based question carrying 5 marks.
5. All questions are compulsory.
6. Use of log tables and calculators is not allowed
SECTION A
1. Arrange the following in the increasing order of their property indicated (any 2):
a. Benzoic acid, Phenol, Picric acid, Salicylic acid (pka values).
b. Acetaldehyde, Acetone, Methyl tert butyl ketone (reactivity towards NH 2OH).
c. ethanol, ethanoic acid, benzoic acid (boiling point)
(1x2=2)
2. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times
while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your
answer. Graphically show the behavior of ‘A’ and ‘B’.
(2)
3. Give reasons to support the answer:
a. Presence of Alpha hydrogen in aldehydes and ketones is essential for aldol condensation.
b. 3 –Hydroxy pentan-2-one shows positive Tollen’s test.
(1x2=2)
SECTION B
4. Account for the following:
a. Aniline cannot be prepared by the ammonolysis of chlorobenzene under normal
conditions.
b. N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are
isomeric in nature.
c. Acylation of aniline is carried out in the presence of pyridine.
(1x3=3)
OR
4. Convert the following:
a. Phenol to N-phenylethanamide.
b. Chloroethane to methanamine.
c. Propanenitrile to ethanol.
(1x3=3)
5. Answer the following questions:
a. [Ni(H2O)6 ] 2+ (aq) is green in colour whereas [Ni(H2O)4 (en)]2+(aq)is blue in colour , give
reason in support of your answer .
b. Write the formula and hybridization of the following compound:
tris(ethane-1,2–diamine) cobalt(III) sulphate
(1+2)
OR
5. In a coordination entity, the electronic configuration of the central metal ion is t 2g3 eg1
a. Is the coordination compound a high spin or low spin complex?
b. Draw the crystal field splitting diagram for the above complex.
(1+2)
6.
Account for the following:
a. Ti(IV) is more stable than the Ti (II) or Ti(III).
b. In case of transition elements, ions of the same charge in a given series show progressive
decrease in radius with increasing atomic number.
c. Zinc is a comparatively a soft metal, iron and chromium are typically hard.
(1x3=3)
7. An alkene ‘A’ (Mol. formula C5H10) on ozonolysis gives a mixture of two compounds ‘B’
and ‘C’. Compound ‘B’ gives positive Fehling’s test and also forms iodoform on treatment
with I2 and NaOH. Compound ‘C’ does not give Fehling’s test but forms iodoform. Identify
the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform
from B and C.
(3)
8. Observe the figure given below and answer the questions that follow:
a. Which process is represented in the figure?
b. What is the application of this process?
c. Can the same process occur without applying electric field? Why is the electric field
applied?
9. What happens when reactions:
a. N-ethylethanamine reacts with benzenesulphonyl chloride.
b. Benzylchloride is treated with ammonia followed by the reaction with Chloromethane.
c. Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide. (1x3=3)
OR
9. a. Write the IUPAC name for the following organic compound:
CH3 – N—CH2CH3
b.Complete the following:
Sn/HCl
C6H5NO2
Br2/H2O
A
NaNO2/HCl
B
HBF4
C
D
273-278K
(1x3=3)
10. Represent the cell in which the following reaction takes place.The value of E˚ for the cell is
1.260 V. What is the value of Ecell ?
2Al(s) + 3Cd2+ (0.1M)  3Cd (s) + 2Al3+ (0.01M)
(3)
11. a. Why are fluorides of transition metals more stable in their higher oxidation state as
compared to the lower oxidation state?
b. Which one of the following would feel attraction when placed in magnetic field: Co 2+ ,
Ag+ ,Ti4+ , Zn2+
c. It has been observed that first ionization energy of 5 d series of transition elements are
higher than that of 3d and 4d series, explain why?
(1x3=3)
OR
11. On the basis of the figure given below, answer the following questions:
(source: NCERT)
a. Why Manganese has lower melting point
than Chromium?
b. Why do transition metals of 3d series have lower melting points as compared to 4d
series?
c. In the third transition series, identify and name the metal with the highest melting point.
(1x3=3)
SECTION C
12. Read the passage given below and answer the questions that follow.
Are there nuclear reactions going on in our bodies?
There are nuclear reactions constantly occurring in our bodies, but there are very few of them
compared to the chemical reactions, and they do not affect our bodies much. All of the physical
processes that take place to keep a human body running are chemical processes. Nuclear
reactions can lead to chemical damage, which the body may notice and try to fix.
The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less
stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable
nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons
to protons in a stable nucleus is thus around 1:1 for small nuclei (Z < 20). Nuclei with too many
neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to
a stable form through radioactive decay. Wherever there are atoms with unstable nuclei
(radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that
there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the
food you eat, and yes, in your body.
The most common natural radioactive isotopes in humans are carbon-14 and potassium-40.
Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the
body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building
them into the different parts of the cells, without knowing that they are radioactive. In time,
carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium
atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a
certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the
chemical. Normally, such changes are so rare, that the body can repair the damage or filter away
the damaged chemicals.
The natural occurrence of carbon-14 decay in the body is the core principle behind carbon
dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a
nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he
stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen
without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which
carbon-14 decays is constant and follows first order kinetics. It has a half - life of nearly 6000
years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate
when the person died. All living organisms consume carbon, so carbon dating can be used to date
any living organism, and any object made from a living organism. Bones, wood, leather, and
even paper can be accurately dated, as long as they first existed within the last 60,000 years. This
is all because of the fact that nuclear reactions naturally occur in living organisms.
(source: The textbook Chemistry: The Practical Science by Paul B. Kelter, Michael D. Mosher
and Andrew Scott states)
a. Why is Carbon -14 radioactive while Carbon -12 not? (Atomic number of Carbon: 6)
b. Researchers have uncovered the youngest known dinosaur bone, dating around 65 million
years ago. How was the age of this fossil estimated?
c. Which are the two most common radioactive decays happening in human body?
d. Suppose an organism has 20 g of Carbon -14 at its time of death. Approximately how much
Carbon -14 remains after 10,320 years? (Given antilog 0.517 = 3.289)
OR
d. Approximately how old is a fossil with 12 g of Carbon -14 if it initially possessed 32 g of
Carbon -14? (Given log 2.667 = 0.4260)
(1+1+1+2)
MARKING SCHEME (2021-22)
TERM – II
CHEMISTRY THEORY (043)
MM:35
1.
2.
Time: 2 Hours
(a)Picric acid < salicylic acid < benzoic acid <phenol
1
(b)Methyl tert – butyl ketone < acetone< Acetaldehyde
1
(c)ethanol <ethanoic acid < benzoic acid (boiling point of carboxylic acids is higher
than alcohols due to extensive hydrogen bonding , boiling point increases with
increase in molar mass)
1
B is a strong electrolyte. The molar conductivity increases slowly with dilution as
there is no increase in number of ions on dilution as strong electrolytes are
completely dissociated.
½+½
1
3.
4
(a) The alpha hydrogen atoms are acidic in nature due to presence of electron
withdrawing carbonyl group. These can be easily removed by a base and the
carbanion formed is resonance stabilized.
(b) Tollen’s reagent is a weak oxidizing agent not capable of breaking the C-C bond
in ketones . Thus ketones cannot be oxidized using Tollen’s reagent itself gets
reduced to Ag.
1
a) In case of chlorobenzene, the C—Cl bond is quite difficult to break as it acquires
a partial double bond character due to conjugation.
So Under the normal conditions, ammonolysis of chlorobenzene does not yield
aniline.
b) Primary and secondary amines are engaged in intermolecular association due to
hydrogen bonding between nitrogen of one and hydrogen of another molecule. Due
to the presence of three hydrogen atoms, the intermolecular association is more in
1
1
1
primary amines than in secondary amines as there are two hydrogen atoms available
for hydrogen bond formation in it.
c) During the acylation of aniline, stronger base pyridine is added. This done in order
to remove the HCl so formed during the reaction and to shift the equilibrium to the
right hand side.
1
OR
1
1
1
5
(a) The colour of coordination compound depends upon the type of ligand and dd transition taking place .
H2O is weak field ligand , which causes small splitting , leading to the d-d
transition corresponding green colour , however due to the presence of ( en )
which ia strong field ligand , the splitting is increased . Due to the change in
t2g -eg splitting the colouration of the compound changes from green to blue.
(b)Formula of the compound is [Co(H2NCH2CH2NH2 )3 ]2 (SO4 )3
The hybridisation of the compound is: d2sp3
1
1
OR
1
a)As the fourth electron enters one of the eg orbitals giving the configuration t2g3 eg1
,which indicates ∆o < P hence forms high spin complex.
b)
1
2
6
7
(a) Ti is having electronic configuration [Ar] 3d2 4s2. Ti (IV) is more stable as
Ti4+ acquires nearest noble gas configuration on loss of 4 e-.
(b)In case of transition elements, ions of the same charge in a given series show
progressive decrease in radius with increasing atomic number.
As the new electron enters a d orbital each time the nuclear charge increases by
unity. The shielding effect of a d electron is not that effective, hence the net
electrostatic attraction between the nuclear charge and the outermost electron
increases and the ionic radius decreases.
(c)Iron and Chromium are having high enthalpy of atomization due to the presence
of unpaired electrons, which accounts for their hardness. However, Zinc has low
enthalpy of atomization as it has no unpaired electron. Hence zinc is comparatively
a soft metal.
1
1
1
Compound A is an alkene, on ozonolysis it will give carbonyl compounds. As both
B and C have >C=O group,
B gives positive Fehling’s test so it is an aldehyde and it gives iodoform test so it is
so it has CH3C=O group. This means the aldehyde is acetaldehyde
C does not give Fehling’s test, so it is a ketone. It gives positive iodoform test so it is
a methyl ketone means it has CH3C=O group
Compound A (C5H10) on ozonlysis gives B (CH3CHO) + C (CH3COR)
So “C” is CH3COCH3
CH3CH=C(CH3)2 (i)O3 (ii) Zn/H3O+
CH3CHO + CH3COCH3
½
CH3CHO + 2Cu2+ + 5OHCH3COCH3 + 2Cu2+ + 5OHCH3CHO + 3I2 + 3 NaOH
CH3COCH3 + 3I2 + 3 NaOH
CH3COO- + Cu2O (red ppt) + 3H2O
No reaction
CHI3 (yellow ppt) + 3HI + HCOONa
CHI3 (yellow ppt) + 3HI + CH3COONa
½
½
A = CH3CH=C(CH3)2
B = CH3CHO
C = CH3COCH3
½
½
½
8
(a)electrodialysis
(b)purification of colloidal solution
(c)Yes. Dialysis is a very slow process to increase its speed electric field is applied
1
1
½+½
9
(a)When N-ethylethanamine reacts with benzenesulphonyl chloride ,
N,N-diethylbenzenesulphonamide is formed.
b)When benzylchloride is treated with ammonia , Benzylamine is formed which on
reation with Chloromethane yields a secondary amine ,
N-methylbenzylamine .
c)When aniline reacts with chloroform in the presence of alcoholic potassium
hydroxide ,
phenyl isocyanides or phenyl isonitrile is formed .
OR
(i)
N-Ethyl-N-methylbenzenamine or N-Ethyl-N-ethylaniline
1
(ii)
½
each
10
Al(s) /Cd2+ (0.1M) // Al3+ (0.01M) /Cd(s)
2Al(s) + 3Cd2+ (0.1M)  3Cd (s) + 2Al3+ (0.01M)
Ecell = Eocell -0.059 log [Al3+]2
n
[Cd2+]3
Ecell= 1.26 – 0.059 log (0.01)2
6
(0.1)3
½+½
1
1
½
½
½
= 1.26 – 0.059 (-1)
6
= 1.26+0.009
= 1.269 V
1
½
ans
unit
11
(a) The ability of fluorine to stabilize the highest oxidation state is attributed to
the higher lattice energy or high bond enthalpy.
(b) Co2+ has three unpaired electrons so it would be paramagnetic in nature, hence
Co2+ ion would be attracted to magnetic field.
(c) The transition elements of 5d series have intervening 4f orbitals. There is greater
effective nuclear charge acting on outer valence electrons due to the weak shielding
by 4f electrons. Hence first ionisation energy of 5 d series of transition elements are
higher than that of 3d and 4d series.
OR
a)Manganese is having lower melting point as compared to chromium , as it has
highest number of unpaired electrons , strong interatomic metal bonding , hence no
delocalisation of electrons .
b) There is much more frequent metal – metal bonding in compounds of the heavy
transition metals i.e 4d and 5d series , whixh accounts for lower melting point of 3d
series.
c) Tungsten
12
1
1
1
1
1
1
(a) Ratio of neutrons to protons is 2.3: 1 which is not the stable ratio of 1:1
(b)Age of fossils can be estimated by C-14 decay. All living organisms have C-14
which decays without being replaced back once the organism dies.
(c) carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to
stable calcium
(d)t = 2.303/ k log (Co/Ct)
Co = 20 g Ct = ?
t = 10320 years k = 0.693/6000 (half-life given in passage)
substituting in equation:
10320 = 2.303 / (0.693/6000) log 20/ Ct
0.517 = log 20 / Ct anlilog (0.517) = 20/Ct
3.289 = 20/Ct
Ct = 6.17 g
OR
t = 2.303/ k log (Co/Ct)
Co = 32 g Ct = 12
t = ? k = 0.693/6000 (half life given in passage)
substituting in equation:
t = 2.303 / (0.693/6000) log 32/ 12
1
1
t = 2.303 x 60000 /0.693 log 2.667
t = 2.303x6000x0.4260 /0.693
½
1
½
½
½
½
½
½
= 8494 years
½