PREFACE
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Preparation & printing of study-material is one of the most important aspects of quality
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I feel great pleasure in presenting this study-material for class XII chemistry subject“CHEM-BOOSTER”
CHEM-BOOSTER is basically designed to equip the students with ideal guidance to
improve their grades. It provides “THE EASY WAY TO LEARN CHEMISTRY”
The unique features of this study-material are Adequate number of important terms, definitions & formulae.
 Organic chemistry has been covered under sections like-nomenclature, namedreactions, distinguishing–tests & Q-Answers which will be very useful to the
students.
 Q-answers are provided for every chapter, this will help students to learn & write
the answers correctly in board exams.
 Efforts have been made for thorough coverage of syllabus and make this material
error free.
I hope this material will help students to get through Board exam with flying colours and
teachers will find this book as good assignment material for the students.
I am very thankful to the Deputy Commissioner(Ahmedabad region) – Shri.P.Devakumar
and Assistant Commissioners(Ahmedabad region) – Shri.P.Madan, Shri.Y.P.Singh, for
their motivation and encouragement.
I appreciate the valuable contribution of following team members –
1. Dr. Gunjan Gaur
PGT Chem.
K.V. ONGC Baroda
2. Mrs. Sabiha Shaikh
PGT Chem.
K.V. No.1, Surat
3. Mr. R. R. Singh
PGT Chem.
K.V. EME Baroda
4. Mrs. Samtan Negi
PGT Chem.
K.V. AFS Baroda
My Best wishes to all the users of this material.
Rajni Taneja
PRINCIPAL
KV AFS BARODA
1
INDEX
PHYSICAL CHEMISTRY
1) THE SOLID STATE
2) SOLUTIONS
3) ELECTROCHEMISTRY
4) CHEMICAL KINETICS
INORGANIC CHEMISTRY
1) GENERAL PRINCIPLES & PROCESSES OF ISOLATION OF ELEMENTS
2) THE p-BLOCK ELEMENTS
3) THE d-BLOCK AND f-BLOCK ELEMENTS
4) COORDINATION COMPOUNDS
ORGANIC CHEMISTRY
1) HALOALKANES AND HALOARENES
2) ALCOHOLS,PHENOLS AND ETHERS
3) ALDEHYDES, KETONES AND CARBOXYLIC ACID
4) ORGANIC COMPOUNDS CONTAINING NITROGEN
4
5
5
5
3
8
5
3
4
4
6
4
GENERAL CHEMISTRY
1) SURFACE CHEMISTRY
4
2) BIOMOLECULES
4
3) POLYMERS
3
4) CHEMISTRY IN EVERYDAY LIFE
3
-----------------------X---------------------------------X-----------------------------------X------------ PHYSICAL CHEMISTRY MATERIAL PREPARED BY – MRS. SAMTAN NEGI
 INORGANIC CHEMISTRY MATERIAL PREPARED BY – DR. GUNJAN GAUR
 ORGANIC CHEMISTRY MATERIAL PREPARED BY – MRS.SABIHA SHAIKH
 GENERAL CHEMISTRY – MR. R.R.SINGH
2
PHYSICAL
CHEMISTRY
3
SOLID STATE
Q1.
Ans.
Q2.
Ans.
Q3.
Ans.
Q4.
Ans.
Q5.
Ans.
Q6.
Ans.
Crystalline solids are anisotropic in nature. What does this statement mean?
Crystalline solids are anisotropic in nature means physical properties like
electrical conductivity, refractive index. etc., are show different values when
measured along different directions in the same crystal.
How do metallic and ionic substances differ in conducting electricity?
Metallic substances conduct electricity through electrons while ionic
substances conduct electricity in molten state or in solution through ions.
What type of interactions hold the molecules together in polar molecular
solid?
Dipole – dipole attractions.
What is the total number of atoms per unit cell in a primitive cubic cell body
cetred cubic cell, facecentred cubic cell?
Total number of atoms per unit cell in primitive cubic cell is
1
Total number of atoms per unit cell in body cetred cubic cell is 2
Total number of atoms per unit cell in face centred cubic cell is 4
What is meant by the term 'coordination number'?What is the coordination
number of an atom in:
(a) primitive cubic unit cell
(b) body-centred cubic unit cell.
(c)
face centred cubic unit cell (cubic close-packed structure)
The number of nearest neighbours of any constituent particle present in the
crystallattice is called its coordination number.
The coordination number of an atoms in a
(a) primitive cubic unit cell is 6
(b) in a body-centred cubic structure is 8
(c)
face centred cubic unit cell is 12
Niobium crystallises in body-centred cubic structure.If density is 8.55gcm-3,
calculate atomic radius of niobium using its atomic mass 93 u.
It is given that the density of niobium, d = 8.55 g cm−3
Atomic mass, M = 93 gmol−1
As the lattice is bcc type, the number of atoms per unit cell, z = 2
We also know that, NA = 6.022 × 1023 mol−1
= 3.612 × 10−23 cm3
So, a = 3.306 × 10−8 cm
For body-centred cubic unit cell:
4
=1.432 × 10−8 cm
= 14.32 × 10−9 cm
= 14.32 nm
Q7.
Ans.
Q8.
Ans.
Q9.
Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm
and density is10.5 g cm−3, calculate the atomic mass of silver.
It is given that the edge length, a = 4.077 × 10−8 cm
Density, d = 10.5 g cm−3
As the lattice is fcc type, the number of atoms per unit cell, z = 4
We also know that, NA = 6.022 × 1023 mol−1
Using the relation:
5
Q10.
Ans.
Q11.
Q13
(i)
(ii)
(iii)
(iv)
(v)
(i)
= 107.13 gmol−1
Therefore, atomic mass of silver = 107.13 u
Iron has a bcc unit cell with a cell edge length of 286.65 pm. The density of
iron is 7.874 g cm-3 .To calculate Avogadro’s number.(At. Mass of iron is
56 g mol-1)
Given: Z = 2
A = 286.65 pm
D = 7.874 g cm-3
M = 56 g mol-1
NA =?
d = ZM / (a)3 X NA
7.875 = 2 X 56 / (286.65)3 X 10-30 X NA
NA = 2 X 56 / (286.65)3 X 10-30 X 7.874
NA = 6.054 X 1023
A cubic solid is made of two elements X and Y. Atoms of Y are at the corners
of the cubeand X at the body-centre. What is the formula of the compound?
What are thecoordination numbers ofX and Y?
It is given that the atoms of Y are present at the corners of the cube.
Therefore, number of atoms of Y in one unit cell.
It is also given that the atoms of X are present at the body-centre.
Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of X atoms to the number of Y
atoms, X:Y =1:1
Hence, the formula of the compound is XY.
The coordination number of both X and Y is 8.
Explain the following terms with suitable examples:
Schottky defect
Frenkel defect
Interstitials
F-centres
Ferromagnetism
Schottky defect: Schottky defect is basically a vacancy defect shown by
ionicsolids. In this defect, an equal number of cations and anions are missing
to maintainelectrical neutrality. It decreases the density of a substance. Ionic
substances containing similarsizedcations and anions show this type of
defect. For example: NaCl, KCl, CsCl, AgBr,etc.
6
(ii) Frenkel defect: Ionic solids containing large differences in the sizes of
ions showthis type of defect. When the smaller ion (usually cation) is
dislocated from its normal site to an interstitial site, Frenkel defect is created.
It creates a vacancy defect as well asan interstitial defect. Frenkel defect is
also known as dislocation defect. Ionic solids such
asAgCl, AgBr, AgI, and ZnS show this type of defect.
(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type
of defect is created when some constituent particles (atoms or molecules)
occupy an interstitial siteof the crystal. The density of a substance increases
because of this defect.
(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired
electrons,the ionic sites are called F-centres. These unpaired electrons impart
colour to thecrystals. For example, when crystals of NaCl are heated in an
atmosphere of sodiumvapour, the sodium atoms are deposited on the
surface of the crystal. The Cl ions diffusefrom the crystal to its surface and
combine with Na atoms, forming NaCl. During thisprocess, the Na atoms on
the surface of the crystal lose electrons. These releasedelectrons diffuse into
the crystal and occupy the vacant anionic sites, creating F-centres.
7
Ferromagnetism: The substances that are strongly attracted by a magnetic
fieldare called ferromagnetic substances. Ferromagnetismarises due to
spontaneous alignment of magnetic moments in the same
direction.Ferromagnetic substances can be permanently magnetised even in
the absence of a magnetic field. Some examples of ferromagneticsubstances
are iron, cobalt, nickel, gadolinium, and CrO2.
Q14.
Ans.
Q15.
Ans.
Q16.
Ans.
Define Doping? Classify each of the following as being either a p-type or an
n-type semiconductor:
(i) Ge doped with In (ii) B doped with Si.
Doping: It is a process by which impurity is introduced in semiconductors to
enhance its conductivity.
(i) Ge (a group 14 element) is doped with In (a group 13 element).
Therefore, a hole will be created and the semiconductor generated will be a
p-type semiconductor.
(ii) B (a group 13 element) is doped with Si (a group 14 element). So, there
will be an extra electron and the semiconductor generated will be an n-type
semiconductor.
Why Frenkel defect not found in pure alkali metal halides?
This is because the alkali metal cations have large size which cannot fit into
the interstitial sites.
What changes occur When AgCl is doped with CdCl2?
Cationic vacancy is generated or metal excess defect.
8
SOLUTIONS
Q1.
Ans.
Differentiate between molality and molarity of solution. What is the effect of
change intemperature of a solution on its molality and molarity?
Molality: is the number of moles of solute per thousand grams of solvent
where as Molarity is the number of moles of solute dissolved in one litre of
solution.
Molality is independent of temperature whereas Molarity is
functionoftemperature because volume depends on temperature and the
mass does notor Molarity decreases with increase of temperature.
Q2.
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density
of 20%(mass/mass) aqueous KI is 1.202 g mL-1.
Ans.
(a) Molar mass of KI = 39 + 127 = 166 g mol−1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100
g of
solution.
That is,
20 g of KI is present in (100 − 20) g of water = 80 g of water
Therefore, molality of the solution
= 1.506 m
= 1.506 m
= 1.51 m
(b) It is given that the density of the solution = 1.202 g mL−1
Volume of 100 g solution
= 83.19 mL
= 83.19 × 10−3 L
Therefore, molarity of the solution
= 1.45 M
(c) Moles of KI
9
Moles of water
Therefore, mole fraction of KI
= 0.0263
Q3.
Ans.
Q4.
Ans.
Q5.
Ans.
Q6.
Ans.
Q7.
Concentrated nitric acid used in laboratory is 68 % nitric acid by mass in
aqueous solution.What should be the molarity of such sample of acid if the
density of the solution 1.504 g/ ml?
Given: 68% HNO3 by mass means mass of HNO3= 68 g; d =1.504 g/ ml
Molar mass of HNO3 = 53 g/mol
Molarity ofssolution = % X d X 10 / Molar mass of HNO3
= 68X 1.504 X 10 / 63
= 1022.723 / 63
= 16.23 M
a) State Henry’s law and mention its two important applications.
b) Which of the following has higher boiling point and why?
0.1M NaCl or ) 0.1M Glucose
a) Henry’s law: The solubility of gas in a liquidis directly proportional to
the pressure of the gas.
Applications: 1. Solubility of CO2 is increased at high pressure.
2. Mixture of He and O2 are used by deep sea divers because
Less soluble than nitrogen.
b) 0.1 M NaCl, because it dissociates in solution and furnishes greater
number of particles per unit volume while glucose does not dissociate.
CCl4 and water are immiscible where as ethanol and water are miscible in all
proportions. Correlate this behavior with molecular structure of these
compounds.
CCl4 is non-polar compound, where as water is polar compound. CCl4 can
neither form H- bond with water molecules nor can it break H- bonds between
water molecules, therefore , it is insoluble in water.
Ethanol is polar compound and can form H- bond with water, which is a polar
solvent, therefore it is miscible with water in all proportions.
State the Raoult's law in its general form in reference to solutions. How does
Raoult's law become a special case of Henry’s law?
Raoult's law states that for a solution of volatile liquids, the partial vapour
pressure of each component in the solution is directly proprtional to its mole
fraction in its solution. PA = P0A× XA
PB = P0B × XB
Henry’s law:The partial pressure of the volatile component or gas is directly
proportional to its mole fraction in solution. P = KH × X
Only the proportionality constant KH differs from P0A . Thus, Raoult’s law
becomes a special case of Henry’s law in which KH becomes equal to P0A.
Two liquids A and B boil at 1450C and 1900C respectively. Which of them has
10
Ans.
Q8.
Ans.
higher vapour pressure at 800c?
Liquid A with lower boiling point is more volatile and therefore will have
higher vapour pressure.
Differentiate between Ideal solution, Non- ideal solution, negative and
positive deviations.
Ideal solution
Non- ideal solution Negative
positive
deviations
deviations
i) Those
Those solutions
solutions which
which does not
follows Raoult’s obey Raoult’s law
law under all
for all
conditions of
concentration.
temperature
and pressure.
PA = P0A× XA
PA> P0A× XA
PA<P0A× XA
PB = P0B × XB
PA = P0A× XA
PB> P0B × XB
PB< P0B × XB
0
PB = P B × XB
ii) Hmix = 0
Hmix = 0
Hmix> 0 i.e., +Ve
Hmix< 0 i.e., -Ve
iii) Vmix = 0
Vmix = 0
Vmix> 0 i.e., +Ve
Vmix<0 i.e., -Ve
vi) A-A = B-B =
A-A = B-B = A-B
A-B > A-A and
A-B < A-A and
A-B
B-B
B-B
(intermolecular
forces between
solvent- solvent
and solut-solute
and solvent –
solute are
same)
v)Examples:
Acetone and
Acetone and
HNO3 and water,
methanol and
benzene, CCl4 and benzene, CCl4 and
chloroform and
ethanol,
C6H6, HNO3 and
C6H6
acetone
benzene and
water
toluene
Q9.
Non-ideal solutions exhibit either positive or negative deviations from Raoult’s
law. What are these deviations and why are they caused? Explain with one
example foreach type.
Ans. When the vapour pressure of a non-ideal solution is either higher or lower
than that predicted by Raoult’s law, the solution exhibits deviations.
These deviations are caused because of unequal intermolecular attractive
forcesbetween solute-solvent molecules and solute-solute or solvent-solvent
molecules. Positive deviation eg: mixture of ethanol and acetone, carbondisulphide and acetoneNegative deviation eg: Chloroform and acetone, nitric
acid and water.
Q10. Two liquids A and B on mixing produce a warm solution. Which type of
deviation from Raoult’s law does it show?
Ans. Warming up of the solution means that the processof mixing is exothermic,
i.e Hmix = - Ve. This implies that the solution shows a negative deviation.
Q11. Define the following terms:
(i)
Azeotropess (ii) Colligative properties (iii) Van’t Hoff Factor
11
Ans.
(i)
(ii)
(iii)
(iv)
(iv) Isotonic solution
Azeotrope : the are binary mixtures having same composition in
liquid and vapour phase and boil ay constant temperature.
Colligative Properties : Those properties which depend on the
number of solute particles but not upon their nature are called
Colligative Properties.
Van’t Hoff Factor(i) : It may be defined as the ratio of normal
molecular mass to the observed molecular mass of the solute.
Isotonic solution : Two solutions having same osmotic pressure at
a given temperature are called Isotonic solution.
Q12. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of
using osmotic pressure as compared to other colligative properties for the
determination of molar masses of solutes in solutions?
Ans. The flow of solvent from solution of low concentration to higher concentration
through semipermeable membrane is called osmosis.
The hydrostatic pressure that has to be applied on the solution side to
prevent the entry of the solvent into the solution through the semipermeable
membrane is called the Osmotic Pressure.
Advantage: i) Unlike other colligative properties, osmotic pressure is used to
Determine the Molar mass of macromolecules/polymers like
protein.
ii) Molarity of the solution is used instead of molality.
iii) Compared to other colligative properties, its magnitude islarge
even for very dilute solutions.
Q13. a) Define Reverse Osmosis and state the condition resulting in reverse
osmosis.
b) Mention a large scale use of phenomenon called reverse osmosis?
Ans. (a) Pressure larger than osmotic pressure is applied to the solution side,
i.e, P>𝜋 (osmotic pressure).
b) Desalination of sea water
Q14. What mass of NaCl (molar mass = 58.5 g mol–1) must be dissolved in 65 g of
waterto lower the freezing point by 7.500 C ? The freezing point depression
constant, Kf for water is 1.86 K kg mol–1. Assume van’t Hoff factor for NaCl is
1.87.
Ans. Given: Tf = 7.50C : Weight water = 65 g:
Kf = 1.86 × K kg mol-1 : van’t Hoff factor for NaCl is 1.87
Weknow that: Tf = i Kf m
7.5 = 1.87× 1.86 × w × 1000/ 58.5 × 65
W = 8.2 g
Q15. Derive An equation to express that relative lowering of vapour pressure for a
solution is equal to the mole fraction of the solute in it when the solvent alone
is volatile.
Ans. For a solution of volatile liquids Raoult’s law, is given as
P = PA + PB
If solute (componentB) is nonvolatile then
P = PA = P0A XA = P0A(1 – XB)
(XA + XB = 1 )
0
0
P = P A - P AX B
12
P0AXB = P0A – P
P0A – P / P0A = XB
Thus, relative lowering of vapour pressure is equal to the mole fraction of
non-volatile solute.
Q16.
(i)
(ii)
Ans.
(i)
(ii)
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene
lowered the freezing point by 0.40 K. find the molar mass the solute.
(Kf for benzene is 5.12 K kg mol-1)
Why is freezing point depression of 0.1M sodium chloride solution
nearly twice that of 0.1 M glucose solution?
Given WB = 1.00g; WA = 50g; Kf = 5.12 K kg mol-1 ; Tf = 0.40K
Tf = Kf × WB × 1000 / MB × WA (in grams)
MB = Kf × WB × 1000 / WA × Tf
MB = 5.12 × 1 × 1000 / 0.40 × 50
MB = 256g mol-1
NaCl being an electrolyte, dissociate almost completely to give Na+
and Cl- ions in solutions where as glucose being non-electrolyte, not
does dissociate. Therefore, the number of particles in 0.1 M NaCl
solution is nearly double( i= 2) than that in 0.1 M glucose solution.
Freezing point depression, is a colligative property, therefore
freezing point depression of 0.1M sodium chloride solution nearly
twice that of 0.1 M glucose solution.
Q17. A solution prepared by dissolving 1.25 g of oil of winter green (methyl
salicylate) in 99.0 g of benzene has a boiling point of 80.31 °C. Determine the
molar mass of this compound. (B.P. of pure benzene = 80.10 °C and Kb for
benzene = 2.53 °C kg mol–1)
Ans. Tb = (80.31 – 80.10)0C = 0.210C or 0.21 K
ΔTb = Kb m
0.21 = 2.53 x 1.25g x 1000 / 99 x M
M =152 g mol -1
Where M is molar mass of the solute
Q18. What would be the value of Vant’s Hoff factor for a dilute solution of K2SO4
and K4[Fe(CN)6] in water?
Ans. In dilute solution , For K2SO4  2K+ + SO42
Vant’s Hoff factor(i)
= Number of particles after dissociation / Number of particles before
Dissociation
=3/1
I=3
For K4[Fe(CN)6  4K+ + Fe(CN)64i=5
Q19. A solution prepared by dissolving 8.95mg of a gene fragment in 35.0mL. of
water has an osmotic pressure of 0.335 torr at 25°C. Assuming that the gene
fragment is a non-electrolyte, calculate its molar mass.
Ans. osmotic pressure () = CRT
= n2RT / V
= w2 RT / V M2
M2 = w2 R T /  V
13
M2 = 8.95x10–3 x 0.0821 x 298 x 760 x 1000 / 0.335 x 35
M2 = 14193.3 g mol–1 or 1.42x104g mol-1
Q20. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg
respectively, at 350 K. Find out the composition of the liquid mixture if total
vapour pressure is 600 mm Hg. Also find the composition of the vapour
phase.
Ans. It is given that:
P0A = 450 mm of Hg
P0B = 700 mm of Hg
P Total= 600 mm of Hg
From Raoult’s law, we have
PA = P0A× XA
PB = P0B × XB
Therefore, total pressure, PTotal = PA + PB
Therefore, XB = 1 - XA
= 1 – 0.4
XB = 0.6
Q21. The depression in freezing point of water observed for the same amount of
acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order
given above. Explain briefly.
Ans. Among H, Cl, and F, H is least electronegative while F is most electronegative.
Then, F can withdraw electrons towards itself more than Cl and H. Thus,
trifluoroacetic acid can easily lose H+ ions i.e., trifluoroacetic acid ionizes to
the largest extent. Now, the more ions produced, the greater is the
depression of the freezing point. Hence, the depression in the freezing point
increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid.
Q22. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such
that its osmotic pressure is 0.75 atm at 27°C.
Ans.
We know that,
() = iCRT
= in2RT / V
= iw2 RT / V M2
w2 =  V M2 / i R
Given:  = 0.75 atm; V = 2.5 litre ; i = 2.47 ; T = 273 + 27 = 300K
Here,
R = 0.0821 L atm K-1m0l-1
M2 = 1 40 + 2 × 35.5
= 111g mol-1
w2 = 0.75 × 2.5 × 111 / 2.47 × 0.0821
14
= 3.42 g
Hence, the required amount of CaCl2 is 3.42 g.
15
ELECTROCHEMISTRY
Q1.
Ans.
Define conductivity and molar conductivity for the solution of an
electrolyte. Discuss their variation with concentration.
Conductivity of a solution is defined as the conductance of a solution of 1
cm in length and area of cross-section 1 sq. cm. The inverse of resistivity
is called conductivity or specific conductance. It is represented by the
symbol κ. If ρ is resistivity, then we can write:
Units is S m-1.
Conductivity always decreases with a decrease in concentration, both for
weak and strong electrolytes. This is because the number of ions per unit
volume that carry the current in a solution decreases with a decrease in
concentration.
Molar conductivity of a solution at a given concentration is the conductance
of volume V of a solution containing 1 mole of the electrolyte kept between
two electrodes with the area of cross-section A and distance of unit length.
/\m = 𝐾 × 1000 / C
Units is S cm2 mol-1.
Molar conductivity increases with a decrease in concentration. This is
because the total volume V of the solution containing one mole of the
electrolyte increases on dilution.
The variation of /\m with√𝑐 for strong and weak electrolytes is shown in the
following plot:
Q2.
Ans.
Q3.
Ans.
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?
k = 1/R (l/A)
Where k is conductivity, R is resistance and l/A is cell constant
/\m = k/C
Where /\m is molar conductivity and C is concentration of the solution.
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm−1.
Calculate its molar conductivity.
Given,
16
κ = 0.0248 S cm−1
c = 0.20 M
 Molar conductivity,
Q4.
Ans.
Q5.
Ans.
Q6.
Ans.
Q7.
Ans.
Q8.
Ans.
Q9.
Ans.
= 124 Scm2mol−1
The resistance of a conductivity cell containing 0.001M KCl solution at 298
K is 1500 M. What is the cell constant if conductivity of 0.001M KCl
solution at 298 K is 0.146 × 10−3 S cm−1?
Given,
Conductivity, κ = 0.146 × 10−3 S cm−1
Resistance, R = 1500 M
 Cell constant = κ × R
= 0.146 × 10−3 × 1500
= 0.219 cm−1
What is meant by ‘limiting molar conductivity’ ?
Limiting molar conductivity:It is molar conductivity at infinite dilution or
approaching zero concentration
Given that the standard electrode potentials (Eo) of metals are:
K+ / K = – 2.93 V, Ag+ / Ag = 0.80 V, Cu2+ / Cu = 0.34 V,
Mg2+ /Mg = - 2.37 V, Cr3+ / Cr = – 0.74 V, Fe 2+ / Fe = – 0.44 V.
Arrange these metals in an increasing order of their reducing power.
Ag+ / Ag < Cu2+ / Cu < Fe2+ / Fe < Cr3+ / Cr < Mg2+ / Mg < K+ / K
Two half-reactions of an electrochemical cell are given below:
MnO–4 (aq) + 8 H+ (aq) + 5 e– → Mn2+ (aq) + 4 H2O (l), Eo = + 1.51 V
Sn2+ (aq) → Sn4+ (aq) + 2 e–, Eo = + 0.15 V.
Construct the redox reaction equation from the two half-reactions and
calculate the cell potential from the standard potentials and predict if the
reaction is reactant or product favoured.
Redox Reaction
2MnO4- + 5Sn2+ + 16H+ → 2Mn2+ + 5Sn4+ + 8H2O
E°cell = E°C-E°A
= (+1.51 – 0.15)V = +1.36V
As E°cell is positive, the reaction is product favoured.
Write Nernst equation for the reaction
2Cr + 3Cd2+ →
2Cr3+ + 3Cd
Ecell = E°cell – 0.059 / 6 log[Cr3+]2 / [Cd2+]3
(Here: n = 6)
Determine the values of equilibrium constant (KC) and G0 for the following
reaction:
Ni(s) + 2Ag+ (aq) → Ni2+(aq) + 2Ag(s), E0 = 1.05 V
(IF = 96500 C mol–1)
G0 = -n FE0cell
= -2 x 96500 x 1.05V
= -202650 J mol–1 or -202.6kJ mor–1
log Kc = nE0 / 0.0591
= 2 x 1.05V / 0.0591
17
= 35.53
Kc = 3.412 x 1035
Q10. The cell in which following reaction occur
2Fe+3(aq) + 2I-(aq) → 2Fe+2(aq) + I2(s) has EᴏCell = 0.236 V at 298 K.
Calculate standard Gibbs energy and equilibrium constant.
∆G = -nF EᴏCell = - 2 x 96500 x 0.236 = -45548 J
∆G = -2.303RT log KC = -2.303 x 8.314 x 298 x log KC
- 45548 = -2.303 x 8.314 x 298 x log KC ,
logKC = 7.983, KC = 9.616 x 107
Q11. (a) State Kohlrausch law of independent migration of ions.
(b) Calculate /\m0 for acetic acid.
Given that /\m0 (HCl) = 426 S cm2 mol–1
/\m0 (NaCl) = 126 S cm2 mol–1
/\m0 (CH3COONa) = 91 S cm2 mol–1
Ans. (a) The law states that limiting molar conductivity of an electrolyte can be
represented as the sum of the individual contributions of the Anion and
Cation of the electrolyte.
(b) /\m0 CH3COOH = /\m0 CH3COONa + /\m0 HCI – /\m0 NaCI
= ( 91 + 426 – 126 ) S cm2 mol-1
= 391 S cm2 mol-1
Q12. What type of a battery is lead storage battery? Write the anode and
cathode
reactions and the overall cell reaction occurring in the operation of a lead
storage battery
Ans. It is secondary cell.
Anode Reaction: - Pb + SO42– → PbSO4(s) + 2e–
Cathode. Reaction: - PbO2 + 4H+ + SO42– + 2e– → PbSO4 + 2H2O
Net reaction:- Pb + PbO2 + 2SO42– + 4H+ → 2PbSO4 + 2H2O
Q13. What is H2-O2 fuel cell? Write the reactions occurring at cathode and
anode.
H2-O2 fuel cell : Cells which produce electrical energy directly from the
combustion of fuels such as H2. CO, CH4.
Anode: 2H2(g)
+ 4OH- (aq) 
4H2O(l)
+ 4eCathod:
O2(g) + 2H2O(l) + 4e
4OH- (aq)
Q14. The chemistry of corrosion of iron is essentially an electrochemical
phenomenon. Explain the reactions occurring during the corrosion of iron
in the atmosphere.
Ans. oxidation: Fe (s) → Fe2+ (aq) + 2eReduction: O2 (g) + 4H+ (aq) + 4 e- → 2H2O(1)

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions.
These ferric ions combine with moisture, present in the surroundings, to
form hydrated ferric oxide (Fe2O3.xH2O) i.e., rust.
Q15. Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s)
18
Ans.
= 0.52865 V
= 0.53 V
Q16. 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 = + 0.80 V, Eo cu2+/cu = + 0.34 V)
Ans. The cell reaction : Cu(s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag(s)
Eo cell = 0.80 V – 0.34V = 0.46V
Nernst equation
Ecell = Eocell - 0.059/2 log [ Cu2+] / [ Ag+]2
Ecell = 0.46V - 0.0295 log 0.10/ [ Ag+]2
0.422V = 0.46 V - 0.0295 log 0.10 / [Ag+]2
log 0.10 / [Ag+]2 = 1.2881
[Ag+]2 = 0.0051
[Ag+] = 7.1x10-2 M
Q17. Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its
molar conductivity and if /\m0 for acetic acid is 390.5 S cm2 mol−1, what is
its dissociation constant?
Ans. Given, κ = 7.896 × 10−5 S m−1
c = 0.00241 mol L−1
Then, molar conductivity,
= 32.76S cm2 mol−1
Again, /\m0 = 390.5 S cm2 mol−1
= 0.084
 Dissociation constant,
= 1.86 × 10−5 mol L−1
Q18. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a
current of 5 amperes for 20 minutes. What mass of Ni is deposited at the
cathode?
Ans. Given,
Current = 5A
19
Time = 20 × 60 = 1200 s
Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
20
CHEMICAL KINETICS
Q1.
Ans
.
Q2.
Ans
.
Q3.
Ans
.
Q4.
Ans
.
Define 'order of a reaction'. Identify the order of a reaction if the units of its
rate constant are:
(i) L–1 mol s–1
(ii) L mol–1 s–1
Order of a reaction: The sum of powers of the concentration terms of the
reactants in the rate law expression is called the order of that chemical
reaction.
rate = k[A]P [B]q
Order of reaction = P+q
i) zero order
ii) second order
Distinguish between 'rate law' and 'rate constant' of a reaction.
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, whereas the rate constant is defined as
the rate of reaction when the concentration of the reactant(s) is unity.
A reaction is of first order in reactant A and of second order in reactant B.
How is the rate of this reaction affected when (i) the concentration of B
alone is increased tothree times (ii) the concentrations of A as well as B are
doubled?
Rate = k[A][B]2
(i) When the concentration of B is increased to 3 times, then rate would be
Rate = k[A][3B]2
= 9k[A][B]2
= 9 times the initial Rate1 i.e. rate is increased 9 times
(ii) When the concentration of A as well as B are doubled, then rate would
be
Rate = k[2A][2B]2
= 8k[A][B]2
= 8 times the initial Rate i.e. rate is increased 8 times
(a) Explain the following terms:
(i)
Rate of a reaction
(ii)
Activation Energy
(iii) Molecularity of a reaction
(b) The rate of a reaction increases four times when the temperature
changes from 300 K to 320 K. Calculate the energy of activation of the
reaction, assuming that it does not change with temperature. (R = 8.314 J
K–1 mol–1)
(a)(i) Rate of a reaction- Rate of Change of concentration of reactant or
product with time is called rate of reaction
(ii) Activation Energy – Minimum energy which the reacting molecules
should acquire so that they react to give product is
called activation energy.
(iii)
Molecularity – Number of molecules taking part in rate
determining step of a reaction is called molecularity.
21
(b)
log k2/ k1 = Ea × [T2- T1] / 2.303 R T1 × T2
log 4 = Ea × 320- 300 / 2.303 x 8.314 × 300 x 320
Q5.
Ans
.
Q6.
Ans
.
0.6020 = Ea×20 / 2.303 x 8.314 x 96 x 103
Ea = 55336.7 J mol-1
Ea = 55.33 kJ mol-1
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 ?
[R]t = - kt + [R]0
0.075M = - (0.0030 mol L-1 s-1)t + 0.10M
-0.025M = -(0.0030 mol L-1 s-1)t
t = 8.3s
For a first order reaction, show that time required for 99% completion is
twice the time required for the completion of 90% of reaction.
For a first order reaction, the time required for 99% completion is
For a first order reaction, the time required for 90% completion is
Therefore, t1 = 2t2
Hence, the time required for 99% completion of a first order reaction is
twice the time required for the completion of 90% of the reaction.
Q7.
A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Ans
.
For a first order reaction,
Therefore, t1/2 of the decomposition reaction is
22
Q8.
Ans
.
Q9.
Ans
.
t1/2 = 77.7 min
Dfine the following :
(i)
Elementary step in a reaction
(ii)
Threshold energy of a reaction
(iii)
Effection collision
(i)
Elementary step: Each step of a complex reaction called an
Elementary step.
(ii)
Threshold energy: is the minimum energy which must be
possessed by reacting molecules in order to undergo effective
collision which leads to formation of product molecule.
(iii)
Effection collision: those collisions which leads to formation of
product molecule.
What is meant by pseudo first order reaction? Give an example of a pseudo
first order reaction and write the rate equation for the same.
Pseudo first order reaction: A reaction which is higher order but follows
the kinetics of first order under special conditions is called pseudo first order
reaction.
Example, Acid hydrolysis of ethyl acetate
CH3-COOC2H5 + H2O
CH3COOH + C2H5OH
Here, the rate law is given by expression
Rate = K [CH3-COOC2H5]
The concentration of H2O is so large that its hardly undergoes any change
during the reaction, therefore it does not appear in the rate law.
Q10
.
23
Ans
.
Q11 For the reaction:
.
N2(g) + 3H2(g)  2NH3(g)
If [NH3] / t = 4  10-8 mol L-1 Sec-1, what is the value of - [H2] / t ?
Ans For the reaction:
.
N2(g) + 3H2(g)  2NH3(g)
If [NH3] / t = 4  10-8 mol L-1 Sec-1, what is the value of - [H2] / t ?
- [N2] / t = - 1/3 [H2] / t = +1/2 [NH3] / t
[NH3] / t = - 1/3 [H2] / t
= ½  4  10-8
= 6  10-8 mol L-1 sec-1
Q12 The rate constant for the decomposition of ethyliodide in the reaction
.
C2H5I(g)  C2H4(g) + HI(g)
At 600K is 1.60  10-5 s-1 . Its energy of activation is 209kj/mol. Calculate
the rate constant of the reaction at 700K.
Ans log K2/K1 = Ea/2.303R [1/T1 – 1/T2 ]
.
log K2 = log K1 + Ea/2.303R [1/T1 – 1/T2 ]
= -4.796 + 2.599 = -2.197
K2 = 6.36 X 10-3
Q13 For a certain chemical reaction variation in the concentration in [R] vs. time
.
(s) plot is given below.
For this reaction write / draw
(i) what is the order of the reactions?
(ii) what are the units of rate constant k?
(iii) give the relationship between k and t ½ (half life period)
24
(iv) what does the slope of the above line indicate?
Ans
.
(i)
First Order
(ii)
Time-1(sec-1)
(iii)
K = 0.693 / t1/2
(iv)
Rate constant of the reaction
Q15 For a certain chemical reaction
.
A + 2B → 2C + D
The experimentally obtained information is tabulated below.
Experiment
[A]0
[B]0
Initial rate of reaction
Ans
.
1
0.30
0.30
0.096
2
0.60
0.30
0.384
3
0.30
0.60
0.192
4
0.60
0.60
0.768
For this reaction
(i) derive the order of reaction w.r.t. both the reactants A and B.
(ii) write the rate law.
(iii) calculate the value of rate constant k
(iv) write the expression for the rate of reaction in terms of A and C.
(i) Let order of the reaction w.r.t. is m and w.r.t. B is n. then rate law will be
Rate = k[A]m[B]n
0.096 = k (0.30)m(0.30)n
………..
(i)
m
n
0.384 = k (0.60) (0.3)
………..
(ii)
m
n
0.192 = k (0.30) (0.60)
………..
(iii)
0.768 = k (0.60)m(0.60)n
………..
(iv)
Dividing (ii) by (i), we get
0.384 /0.096 = k (0.60)m(0.3)n / k (0.30)m(0.30)n
4 = 2m ; 22 = 2m ; m= 2
Dividing (iii) by (i), we get
0.192 / 0.096 = k (0.30)m(0.60)n / k (0.30)m(0.30)n
21 = 2n
n=2
Order of the reaction w.r.t. A is 2
Order of the reaction w.r.t. B is 1
(ii)
Rate = k[A]2[B]1
(iv)
K = Rate / [A]2[B]1
= 0.096 / (0.30)2 (0.30) = 3.55 s-1
(v)
Rate = -d[A] / dt = ½ [d] / dt
Q16 Explain the factors which affect the Rate of Chemical Reactions.
.
Ans factors which affect the Rate of Chemical Reactions :
.
(i)Concentration : If a greater concentration of reactant atoms and
molecules is present, there is a greater chance that collisions will occur
among them. More collisions mean a higher reaction rate.Thus, increasing
the concentration of the reactants usually results in a higher reaction rate.
(ii)Temperature: Increasing the temperature causes the particles (atoms
25
or molecules) of the reactants to move more quickly so that they collide
with each other more frequently and with more energy. Thus, the higher the
temperature, the greater the rate of reaction. The rate constant of a
reaction is nearly double with rise in temperature by 100C.
(iii)Surface area: the smaller the particle size, greater the surface area
and faster is the reaction.
(iv)Catalyst: In the presence of catalyst, the rate of reaction generally
increases and the equilibrium state is attained quickly in reversible
reactions.
26
INORGANIC
CHEMISTRY
27
UNIT 6
GENERAL PRINCIPLES AND PROCESS OF ISOLATION OF ELEMENTS
Q.1.Discuss following a. Ore b. Hydraulic washing c. Electromagnetic separation
A.1. a. The mineral from which the metal is conveniently and economically extracted is
called ore. Eg: bauxite, heametite.
b. . The process by which lighter earthly particles are feed from the heavier ore particles
by washing with water.
c. Electromagnetic separation This method
of concentration is employed when either the
ore or the
impurities associated with it are magnetic in
nature.eg haemetite,pyrolusite.
Q.2. Discuss a. Froth floatation.b.
Leaching.
A.2.a.Froth floatation. This method is based
upon the fact that the surface of sulphide ores
is preferentially wetted by oils while that of
gangue is preferentially wetted by water.
Depressant are used to prevent certain types of
particles from forming the froth with the air bubbles. For example, sodium cyanide is used
as depressant.
4NaCN + ZnS -- Na2[Zn(CN)4] + Na2S
b. Leaching. This is a chemical method of concentration
and is useful in the case when the ore is soluble in some
suitable solvent. In this method the powered ore is treated
with certain reagents in which the ore is soluble but the
impurities are not soluble. The impurities left undissolved
are removed by filtration.
Al2O3.2H2O + 2NaOH + 2H2O -2Na[Al(OH)4]
+ 4H2O
2Na[Al(OH)4] + CO2  Al2O3.XH2O
+ 2NaHCO3
Al2O3.XH2O  Al2O3
+ XH2O
Q.3. Discuss calcinations and roasting.
A.3.Calcination. It is a process of heating the ore strongly either in a limited supply of
air or in the absence of air.
For eg:
ZnCO3 
ZnO + CO2
Roasting. It is a process of heating the ore strongly in the presence of excess of
air at a temperature below the melting point of the metal.
2ZnS + 3O2  2ZnO + 2SO2
Fe2O3 + 2Al ------ 2Fe + Al2O3
Cu2S +2 Cu2O------ 6Cu + SO2
Q.4. What do you understand by a. Auto reduction b. Hydrometallurgy
c.Electrometallurgy
A.4 a.Auto reduction. Reduction can be carried out by the auto reduction in which
the anions associated with the help in the reduction. This is used for the
reduction of sulphide ore of Pb, Hg and Cu. In this case, no reducing agent is
28
required. The metal is obtained either by simple roasting or by the reduction of
its partly oxidized form.
HgS + O2 ------ Hg + SO2
PbS + 3O2 ----- 2Pb + 2SO2
b. Hydrometallurgy. The process of extraction of metals by dissolving the ore in
a suitable chemical reagent and the precipitation of the metal by more
electropositive metal is called hydrometallurgy.
Ag2S + 4NaCN -- 2Na[Ag(CN)2] + Na2S
2Na[Ag(CN)2] + Zn – 2Ag + Na2[Zn(CN)4]
Na2[Zn(CN)4] + 4NaOH –- Na2ZnO2 + 4NaCN
c. Electrometallurgy. The process of extraction of metals by electrolysis process
is called electrometallurgy.
At cathode
Na+ + 1e- -- Na
At anode
2Cl- ----------- Cl2 + 2eQ.5. Discuss a. Electrorefining b. Zone refining c. Vapour phase refining d. Van
arkel method.
A.5. a.Electrorefining. This is the most general method for the refining of metals
and based upon the phenomenon of electrolysis. In this method electricity
is
passed through the electrolytic
solution.copper is refined by this method.
At cathode
Mn+ + ne- →
M
At anode
M → Mn+ +
ne
b.Zone refining. This method is used for
the metals which are required in very high
purity e.g Ge.This method is based upon the
principle that the impurities are more soluble in the melt than in the solid state of
the metal.
c. Vapour phase refining.(Mond’s process) This method on the fact that
certain metal are converted to their volatile compounds while the impurities are
not affected during compound formation. The two requirements are :
(a)
The metal should form a volatile compound
(b)
The volatile compound should be easily decopopsed.
Ni + 4CO → Ni(CO)4 → Ni + 4CO
IMPURE
PURE
d. Van arkel method. This method is similar to the above method and used for
getting the ultra pure metal.
Ti + 2I2 → TiI4 → Ti + 2I2
Q.6. Which is the cheapest and most abundant reducing agent which is used in
the
extraction of metal?
A.6. Carbon in form of coke.
Q.7. Why is zinc and not copper used for the recovery of metallic silver from its
cyanide complex [Ag(CN)2]?
A.7. Zinc is more electropositive than silver and therefore, zinc displace silver from its
solution. 2[Ag(CN)2] + Zn → [Zn(CN)4]2- + 2Ag
On the other hand copper is less electropositive than silver and therefore, cannot
displace silver from its solution.
29
Q.8. Why does CaO react with with SiO2 to form a slag?
A.8. CaO is basic oxide whereas SiO is acidic oxide. Their reaction is acid-base reaction
as.
CaO + SiO2 → CaSiO3.
Slag
Q.9. Why is the formation of sulphate in calcinations sometimes advantageous?
A.9. Sulphates are usually water soluble and the gangue remains insoluble. Therefore,
the desired metals is leached away as soluble sulphate from insoluble gangue.
Q.10.
How does NaCN act as a depressant in preventing ZnS from forming
the froth?
A.10.
NaCN forms a layer of zinc complex, Na2 [Zn(CN)4] on the surface of ZnS and
thereby prevents it from the formation of the forth.
Q.11.
At a site, low srade copper ores are available and zinc and iron scraps
are also available. Which of the two scraps would be more suitable for
reducing the leached copper ore and why?
A.11. Since zinc lies above iron in electrochemical series, it is more reactive than iron. As
a result, if zinc scraps are used, the reduction will be fast. However, zinc is a
costlier metal than iron. Therefore, it is advisable and advantageous to use iron
scraps.
Q.12.
Lime stone is used in the manufacture of pig iron from haematite.
Why?
A.12.
Haematite is an ore of iron contain silica (SiO2) as the main impurity. The
purpose of limestone is to remove SiO2 as calcium silicate (CaSiO3) slag.
CaCO3 → CaO + O2
Limestone
CaO +SiO2 → CaSiO3
Basic flux
Q.12.
What is the composition of copper matte?
A.12.
Copper matte contains cuprous sulphide (Cu2S) and iron sulphide (FeS).
Q.13.
A.13.
What is cupellation?
Cupellation is a method used for refining of those metals which contains
impurities of other metals which form volatile oxides. For example, removal of last
traces of lead from silver.
Q.14.
How is the granular zinc obtain?
A.14.
Granular zinc is obtained by pouring molten zinc in cold water.
Q.15.
What is Kroll process?
A.15.
The production of titanium metal at 900K from TiCl4 by reduction with Mg in
argon atmosphere. TiCl4 (g) + 2Mg → Ti (s) + 2MgCl2 (l)
Q.16.
Galena (PbS) and cinnabar (HgS) on roasting often give their
respective metals but zinc blende (zinc) does not. Explain?
A.16.
On roasting sulphides are partly converted to their respective oxides. Since
the oxides of lead and mercury are unstable, these bring about the reduction if their
respective sulphides to the corresponding metals.
PbS + 2PbO → 3Pb + SO2
Unstable
HgS + 2HgO → 3Hg + SO2
Unstable
However, zinc oxide is stable and it does not reduce ZnS to Zn.
Q.17.
What is the stabilizer in froth floatation process? Give examples.
30
A.17.
Stabilizer in froth floatation process stabilizes the forth. For example, aniline,
cresol, etc.
Q.18.
Name three forms of iron. How these three forms do differs?
A.18. Three forms of iron are : cast iron, wrought iron and steel. These differ in their
carbon content.
Q.19.
What is the actual reducing agent of haematite in the furnace?
A.19.
Carbon monoxide, CO
Q.20.
Why do not metals occure as nitrates in the nature?
A.20.
Nitrates of almost all metals are soluble in water.
Q.21.
Why copper matte is put in silica lined converter?
A.21.
Matte has Cu2S and FeS .FeS is oxidized to FeO which combines with silica to
form FeSiO3.
Q.22.
Why alumina can not be reduced by carbon?
A.22.
At high temperature alumina reacts with carbon to form aluminium carbide.
2Al2O3 + 6C → Al4C3 + 3CO2
Q.23.
Name the following.
(a) Any two metals which never occur in native state.
(b) Any two metals which never occur in uncombined state.
A.23.
(a) sodium, calcium
(b) gold, platinum
Q.24.
What is the main difference between red & white bauxite ores?
A.24
Red bauxite contains iron oxides as the main impurity while white bauxite
contains silica as the main impurity.
*****
31
UNIT- 7
P – BLOCK ELEMENTS
GROUP 15
Q. 1. N does not form NCl5.Why?
A. 1. Due to absence of d-orbital, it cannot extend its octet.
Q. 2. NCl3 is easily hydrolysed.Why?
A. 2. Due to presence of vacant of orbital in Cl atom, it can form co-ordinate bond with
H2O.
Q. 3. N2 is highly inert. Why?
A. 3. Due to high bond- dissociation energy.
Q. 4. Why is white P kept under water?
A. 4. Ignition temperature of white P is low (303K); on exposure to air, it spontaneously
catches fire forming P4O10.
Q. 5. H3PO3 act as a reducing agent but H3PO4 does not .Why?
A. 5. H3PO3 has one P−H bond, therefore it acts as reducing agent but H3PO4 does not
have any P−H bond.
Q.6. First element of each group differs in its properties from other elements of
the group .Why?
A.6. Because of: - 1). Small size 2). High ionisation energy 3). Absence of d-orbital.
Q.7. HNH angle is higher then HPH , HAsH & HSbH angle.Why?
A.7. N is highly electropositive it attracts bonded pair of electron more toward itself,
therefore it minimum between lp-bp electros bond. Angle is high.
Q.8. White P is more reactive than red P.
A.8. White P has P4 Tetrahedral Structure and under stain. Red P has polymeric structure
and becomes giant molecule.
Q.9. Bismuth is strong oxidising agent in penta valent state?
A.9. +5 oxidation state is less stable than +3 due to inert pair effect.
Q.10.NO becomes brown when released in air.
A.10. No reacts with O2 to form Brown gas NO2
2NO + O2
2NO2
Q.11. Why does R3P=O exist but R3N=O does not (R= alkyl group)?
A.11. On the other hand P due to the presence of d orbitals forms pπ – dπ multiple bonds
and hence can expands its covalency beyond four. Therefore P forms R3P=O in which
covalency of P is five.
Q.12.PCl5 is ionic in solid state.
A.12. It exists as [PCl4]+ [PCl6]Q.13.NH3 acts as a ligand.
A.13. In NH3, Nitrogen has a lone pair of electron.
32
Q.14.H3PO2 acts as monobasic acids.
A.14. Because it has only one replaceble H-atom
Q.15.NH3 has higher proton affinity than PH3.
A.15. Due to smaller size, high electron density and early availability of electron on N in
NH3.
Q.16. Discuss preparation of HNO3 by Ostwald Process.
A.16. Ostwald process
Pt/Rh
4NH3 + 5O2 → 4NO + 6H2O
2NO + O2↔2NO2
3NO2 + H2O → 2HNO3 +NO
condition : 500 K/9 bar
Q.17. What is the maximum covalence of nitrogen?
A.17. 4
Q.18. PH3 has lower boiling point than NH3. Why?
A18. Due to presence of hydrogen bonds in NH3.
GROUP 16
Q.19. Mention three areas in which sulphuric acid plays an important role.
A.19.1.In the manufacturing of fertilizers 2. It is used in storage batteries
3.It is used in petroleum refining.
Q. 20.
Sulphur disappears when boiled with an aqueous alkaline solution of
Na2SO3. Why?
A.20.
Na2SO3 + S → Na2S2O3.
Q.21.
SF6 is known & not SCl6.
A.21. Because of larger size of Cl atom, 6Cl atoms cannot be accommodated around S.
Q.22.H2S is less acidic then H2Te.Why?
A.22.
Because H −Te bond is weaker then H −S bond.
Q.23.Sulphur in vapour state exhibits paramagnetic behavior?
A.23. In vapour state S exist in S2 form like O2 has two unpaired electrons in antibonding
 * orbital.
Q.24.Thermal stability of H2O is more than H2S.
A.24. Due to H – bonding in H2O.
Q.25.SF6 is well known but SCl6 is not known.
A.25. Due to small size of S, six large Cl atoms can not be accommodated around S atom.
Q.26. Write preparation of sulphuric acid through contact process.
A.26. Contact process S + O2 → SO2
V2O5
2 SO2+ O2 → 2SO3 key reaction ∆rH = -196.6 kJmol-1
SO3 + H2SO4 →H2S2O7 (oleum)
33
H2S2O7 + H2O → 2H2SO4
GROUP 17
Q.2 Why is F2O referred to as a fluoride but Cl2O is an oxide?
7 Because F is more electronegative than O and O is more electronegative than Cl.
A.2
7
Q .28.
Bleaching by Cl2 is permanent & SO2 is temporary.
A.28.
Cl2 bleaching is by oxidation, SO2 bleaching is by reduction, hence the product
can be re-oxidised.
Q.29.
F2 is the strongest oxidizing agent. Why?
A.29.
It is due to low enthalpy of dissociation of F2 & high enthalpy of hydration of
F¯ion.
Q.30.
HI cannot be prepared by heating KI with concentrated H2SO4.Why?
A.30.
HI is strong reducing agent & reduces H2SO4 to SO2 & gets oxidised to I2.
Q.31.ICl is more reactive then I2?
A.31.
I−Cl bond is weaker then I−I bond , therefore ICl is more reactive.
Q.32. S i F62 is known but S i cl 62 is not known,
A.32. Due to smaller size of F, steric repulsion is less in S i F62
Q. 33.
A. 33.
The order of increasing oxidising ability is HIO4<HClO4<HBrO4.Why?
Because standard electrode potential of XO4¯/XO3¯ is highest in HBrO4.
Q. 34.
HClO4 is most acidic amongst HClO, HClO3 & HCO2.Why?
A. 34.
Because it’s conjugate base is most stable due to
resonance
HClO4 ↔ H+ + ClO4¯
Q.35.Bond dissociation energy of F2 is less than that of Cl2.
A.35. Due to smaller size, larger electron – electron repulsion.
Q.36.HF is a weaker acid than Hcl in water.
A.36. Due H – bonding in HF, H+ ions are not formed easily.
Q.37.Addition of Cl2 to KI solution gives a brown colour but excess of Cl2 turns it
colourless.
A.37. Due to the following reactions.
KI + Cl2
Kcl + I2 (Brown)
in excess Cl2 + I2 + H2O
HCl + HIO3 Colourless
34
Q.38 The halogens are coloured.Why?
This
. is Because halogens absorb radiations in the visible region. This results in the
A.38 excitation of valence electrons to a higher energy region. Since the amount of
.
energy required for excitation differs for each halogen, each halogen displays a
different colour
GROUP 18
Q. 39.
Only Xe among noble gases forms compounds. Why?
A. 39.
Because of its low ionization energy.
Q. 40.
Boiling point of noble gases increase with the increase in atomic
number. Why?
A. 40.
Because magnitude of van-der-waal‫ۥ‬s forces increases with increase in atomic
size.
Q. 41.
A. 41.
Ne is generally used in waning signal illumination. Why?
Because it is visible through mist & fog from long distances.
Q.42.Noble gases are Chemically inert.
A.42. Reasons
(i)
Stable configuration
(ii)
High I.E.
(iii) Very low electron affinity.
Q.43. Why is helium used in diving apparatus?
A.43. Because of its very low solubility in blood.
Structures Of Compounds
1) NH3
2) White
Phosphorus
3)
Red Phosphorus
35
4) PCl3
5) PCl5
6)O3
7) S6
8)
S8
10) HOClO
9) SO2
12)HOClO2
11)HOCl
36
13)HOClO3
14)BrF3
15)XeF2
16) XeF4
17) XeOF4
18) XeF6
37
19) XeO3
20) H4P2O7
21)H3PO3
22) H3PO2
23) H3PO4
38
24)H2SO3
25)H2SO4
26) H2S2O8
27) N2O4
28) N2O
29)NO
30)
N2O5
39
31)
NO2
32) H2S2O7
33)ICl4
34)IBr2
-
35)BrO-3
40
UNIT-8
THE d & f BLOCK ELEMENTS
Q.1. Transition metals and their compounds are good catalysts.
A.1. (1) They form reaction intermediates which follow the path of low activation
energy.
(2) They provide suitable large surface area for adsorption of reactant.
Q.2. Enthalpy of atomisation of transition elements is very high.
A.2. Because of large number of unpaired electrons in their atoms.
stronger bonding between the atoms.
Q.3. Explain why transition metals are paramagnetic?
A.3.
They have
The ions of transition metals generally contain one or more unaired electrons
hence the compounds of transition metal are paramagnetic i.e., they are weakly
attracted by magnetic field. The paramagnetic character is directly related to the value
of magnetic moment, m which intern depends upon the number of unpaired electrons.
Q. 4. Why the melting points of transition elements are high ?
A.4. The melting points of transition elements are high due to the presence of strong
intermetallic bonds and covalent bonds.
Q. 5. Why Zn, Cd and Hg are not regarded as transition elements ?
A.5. Because they have the completely filled d-subshell with outer electronic
configuration (n –1)d10 ns2.
Q.6
. Why the compounds of transitions elements are coloured ?
A.6.
The colour of compounds of transition elements depend upon the unpaired
electrons present in d-orbitals fo transition element. If d-orbitals are completely
vacant or completely filled the compounds will be colourless, but if any unpaired
electron is present in d-orbitals, the compound d transition. The will be coloured
due to d → unpaired electron is excited from one energy level to another energy
level with in the same d-sub-shell. For this purpose, the energy is absorbed from
visible region of radiation. The complementary part of the absorbed light i.e.,
reflected light will decide the colour of the compound.
Q.7.
Why the transition elements act as catalyst ? Give two examples.
41
A.7.
(a) Transition metal show variable oxidation states, therefore, they can from
intermediate products of difficult reactant molecules.
(b) Transition elements are capable to form interstitial compounds due to which
they can absorb and activates
the reacting molecules.Example :(a) V2O5 is
used for the oxidation of SO2 to SO3 in contact process of H2SO4.(b) Ni is used as
a catalyst in the hydrogenation of alkenes and alkynes
Q.8. Why transition elements form
(a) Interstitial compounds and
(b) Alloys
A.8. (a) Interstitial compounds : Transition elements form large number of intestitial
compounds. In these compounds small size atoms like hydrogen, carbon,
nitrogen , nitrogen, boron etc. occpy the empty space of metal lattice. The small
entraped atom in the interstics form the bonds with metals due to which
malleability and ductility of the metals decrease, whereas tensile strength
increases.
(b) Alloys : Tranisition element forms alloys with each other because they have
alomost similar sizes. Due to similar sizes atoms of one metal in the crystal
lattice can easily take up the position of the atom of transition elements. Alloys
are more resistant to corrosion than the constituent elements, and usually harder
with higher melting point.
Q. 9. Why is K2Cr2O7 generally preferred over Na2Cr2O7 in volumetric analysis
although both are oxidising agents ?
A.9. Because Na2Cr2O7 is hygroscopic, hence it is difficult to prepare its standard
solution for volumetric analysis, but because of nonhygroscopic nature of K2Cr2O7
its standard solution can be prepared.
Q. 10.
Why do Zr and Hf exhibit similar properties ?
A.10.
Because of the lanthanoid contraction Hf has similar size to Zr, therefore
both Zr and Hf exhibit similar properties.
Q.11.
Give two uses of lanthanoid compounds.
42
A.11.
(i) Misch metal is pyrophoric and used in gas lightness, tracer bullets and
shell.
(ii) Oxides of neodymium and praseodymium are used for making colour glasses.
Q. 12.
All scandium salts are white. Why ?
A.12. Because they have no electron in d-orbital, thus no d - d transition is possible.
Due to this reason all salts of Sc3+ are white.
Q.13.
The first ionisations energies of the 5d-transition elements are
higher then those of 3d and 4d Transition elements.
A.13. Due to lanthanoid contraction.
Q. 14.
Transition metals are known to form complexes.Discuss.
A.14. Because of small size and presence of vacant ‘d’ orbitals.
Q.15. Name a transition metal which does not exhibit variation in oxidation
state in its compounds.
A.15. Zinc in its compounds shows an oxidation state of + 2 only.
Q.16. Name a transition metal which exhibit maximum nuber of oxidation
states in its compounds.
A.16. Mn
Q. 17.
Why do the d-block elements exhibit a larger number of oxidation
states than f-block elements ?
A.17. Because the energy of ns-electron and (n – 1) d-electrons are nearly same,
therefore, ns electrons as well as (n – 1) d-electrons can take part in bond formation in
transition elements. In f-block elements last electron goes to the f-orbitals of second
order outer most shell, thus the difference between the energy of ns-electron and (n –
2) f-electrons increases. Due to this reasons all the (n – 2) f-electrons cannot take
point in bond formation.
Q. 18.
Explain why the first ionisation energies of the elements of the first
transition service do not very much with increasing atomic numbers.
A.18.
With the increasing atomic number, d-electrons add one by one in (n – 1)
shell or penultimate shell. The screening effect of these d- electrons shield the outer selectrons from inward nuclear pull. The effect of the increase in nuclear charge with the
43
increase in atomic number is opposed by the shielding effect of d-electrons . thus due
to these counter effect there is a vary little variations in the values of ionization
energies of first transition series.
Q.19. Discuss extraction of potassium dichromate from chromite ore.
A.19. Preparation:-
Q.20. Draw structure of chromate ion and dichromate ion.
A.20.
Q.21. What is the effect of pH on the colour of the solution of potassium
dichromate?
A.21.
A lower pH, the colour of the solution is orange due to the solution is
orange due to the presence of dichromate ions (Cr2O72–). But in alkaline PH, the colour
of the solution changes to yellow due to the conversion of dichromate ions to chromate
ions.
Q.22. Discuss oxidizing action of potassium dichromate in acidic medium.
A.22.
44
Q.23. Discuss extraction of potassium permanganate from pyrolusite ore.
A.23. Potassium permanganate KMnO4
PREPARATION :-
Q.24. In what way the electronic configuration of transition elements are
different from those of representative elements ?
A.24. In representative elements last electron goes either s-subshell or p-subshell of
last orbit,
while in transition elements last two orbits are incomplete. The outer
electronic configuration of representative element is either ns1–2 or ns2 np1–6, while in
transition element outer electronic configuration is (n – 1) d1–10 ns1–2.
Q. 25. What is lanthanoid contraction ? Discuss its cause and consequences.
A.25. The steady decrease in atomic and as we move from La to Lu, is known as
lanthanide
contraction.
Cause : In lanthanoids, last electron enters the 4f-sub-shell i.e., second last
shell. The shape of
f-orbitals is very much diffused, because of this reason the
mutual shielding of 4f-electrons is very little. The nuclear charge increases at each step
due to the increase in atomic number, while the mutual shielding effect of 4f-electrons
is comparatively negligible. This causes a decrease in size of the 4f sub-shell with the
increase in atomic number.
Consequences : Following are the main consequence of lanthanoid contraction.
(a) Basic strength of oxides and hydroxides of lanthanoids : The basic strength of
oxides
45
and hydroxides of lanthanides decrease with the increase in atomic number the
size of lanthanum ion (Ln3+) decreases due to lanthanide contraction. The
decrease in size of Ln3+,
increases the covalent character i.e., decreases the ionic character between Ln3+
and OH–
ions, consequently the basic strength of oxide and hydroxide will decrease.
(b) Similar sizes of second and third transition series elements : Normally the
atomic radii
increases with the increase in atomic number in the same sub-group, but after
lanthanides
the atomic radii of the elements of second and third transition series are almost
similar.
The expected increase in size from second to third transition series is cancelled
by the decrease
in size due to lanthanide contraction. Due to the similar atomic of radii of second
and third
transition series elements they resemble each other very closely.
(c) Separation of lanthanoids : The properties of lanthanides are very similar,
therefore, it is
difficult to separate them. However due to lanthanoid contraction, decrease in
size of
lanthanoids make the separation possible by ion exchange methods.
Q.26. Write differences between chemistry of Lanthanoids and Actinoids
A.26. Differences between Lanthanoids and Actinoids
1
2
Lanthanoids
Electrons are successively
added in 4f orbitals.
The chemistry of
lanthanoide elements is
fairly similar
1
2
46
Actinoids
Electrons are successively added
in 5f orbitals.
There is a considerable difference
in the chemistry
of actinide elements. This is due
3
4
5
6
7
8
9
10.
because of the large energy
difference between
4f and 3d subshells.
4f electrons have greater
shielding effect.
4f orbitals have higher
binding energies.
Lanthanoids show lesser
number of oxidation states.
They do not have much
tendency for complexes
formation.
Except promethium, all the
lanthanoids are non
radioactive.
Lanthanids do not form
oxocations.
Compounds of lanthanoids
are less basic.
Electronic configuration is
6s2 5d0-1 4f1-14
to very small difference in 5f and
6d energy levels.
3
4
5
6
7
8
5f electrons have poor shielding
effect.
Binding energies of 5f orbitals are
lower.
Actinoids show a variety of
oxidation states.
They have much tendency for
complexes formation.
All the actinoids are radioactive.
Actinoids form oxocations.
9
Compounds of actinoids are more
basic.
10.
Electronic configuration is
7s2 6d0-1 5f1-14
Q.27. For the first row of the transition metals the E° value are:
Metal V Cr Mn Fe Co Ni Cu
E
o
[M+2 /M]
– 1.18 – 0.91 – 1.18 – 0.44 – 0.28 –
2 0.25 +
0.34 volts (V)
Give suitable explanation for the irregular trend in these values
A.27. The irregular trend in the (M+2 /M) E values
for the first row transition
metals is due to the irregular variation in the ionisation and sublimation energies
across the series.
Q.28. Of the ions Co2+, Sc3+ and Cr3+ which ones will give coloured aqueous
solutions and how will each of them respond to a magnetic field and why?
(Atomic numbers: Co = 27, Sc = 21, Cr = 24)
A.28. The electronic configurations of the given ions are,
Co2+ : 1s2 2s2 2p6 3s2 3p6 3d7 No. of unpaired electrons = 3
Sc3+ : 1s2 2s2 2p6 3s2 3p6 3d0 No. of unpaired electrons = 0
47
Cr3+ : 1s2 2s2 2p6 3s2 3p6 3d3 No. of unpaired electrons = 3
Co2+ and Cr3+ ions will give coloured aqueous solutions. Co2+ and Cr3+ are
paramagnetic, and Sc3+ is diamagnetic. Therefore, Co2+ and Cr3+ ions will
get attracted to the magnetic field, whereas Sc3+ ion will be repelled by the
magnetic field.
Q.29. (a) Assign reason for each of the following:
(i) Ce3+ can be easily oxidised to Ce4+.
(ii) E° for Mn3+/Mn2+ couple is more positive than for Fe3+/Fe2+.
(b) Mention two uses of potassium permanganate in the
laboratory.
(Atomic number: Mn = 25, Fe = 26, Ce = 58)
A.29. (i) Ce3+ has only one electron in its 4f orbitals. Due to extra stability of
completely empty orbitals belonging to an energy level as compared to having only one
electron in it, Ce3+ tends to lose its only electron from 4f orbital and get oxidised to
Ce4+.
(ii) Mn3+ has a d 4 configurtion, so it has greater tendency to accept one electron
to acquire d 5 configuration. On the other hand, Fe3+ has a d5 configuration which is
more stable than the d 6 configuration of Fe2+. As a result, reduction of Fe3+ to Fe2+ is
not favoured. Since, E° values reflect the reduction tendency, therefore, E° value for
Mn3+/Mn2+ couple is more positive than Fe3+/Fe2+.
(b)
The uses of potassium permanganate in the laboratory are
(i) As an oxidising agent,
(ii) In volumetric estimation of reducing agents such as Fe2+ salts, oxalic
acid etc.
****
48
UNIT 9
CO-ORDINATION COMPOUNDS
IMPORTANT QUESTIONS
Q. 1. Define coordination compound.
A.1 Coordination compounds are those addition compounds which retain their
identity in dissolved state also like solid state. In coordination compounds
the atoms or group of atoms have been attached to central metal atom or
ion beyond the number possible according to electrovalent or covalent
bonding.
Q. 2. What do you mean by ambidentate ligands ?
A. 2. Any ligand which has two or more donor atoms but only one donor atom is
attached to the metal ion at a time, during complex formation, is known
as ambidentate ligand.
Q. 3. What is the relation between the molar conductivity of the solution
of the complex compound and total number of ions?
A. 3. The total number of ions given by a complex can be predicted with the help
of molar conductivity of the solution of complex compound. Molar
conductivity of the solution of complex compound depends upon number of
ions by a complex in solution.
Q. 4. What do you mean by organometallic compounds ?
A.4 . Any compound which contains at least one metal-carbon bond is called
organometallic
compound.
Q.5. How many isomers are there for octahedral complex [CoCl2(en)
(NH3)2]+.
A. 5. To geometrical isomers i.e., Cis and tans. cis isomer has two optical
isomers i.e., d-cis isomer and l-cis isomer.
Q. 6. Write the IUPAC name for the linkage isomer of [Co (NH3)5.
O.NO]Cl2.
A.6. The linkage isomer will be [Co(NH3)5 NO2]Cl2 and its IUPAC name will be
Pentamminitrito-N-cobalt (III) chloride.
Q. 7. Write the IUPAC name of the ionisation isomer of [Co (NH3)5SO4]
Br.
A.7. The ionisation isomer will be [Co (NH3)5 Br]SO4 and its IUPAC name will be
Pentamminebromidocobalt (III) sulphate.
Q.8. How is ammonia molecule a good ligand ?
A.8. Nitrogen of ammonia has one lone pair of electrons. Because of the small
size of nitrogen, the tendency to donate this electron pair is also high, thus
NH3 is a good ligand.
Q. 9. What do you mean by inner orbital complex and outer orbital
complexes ? Give
examples of each.
49
A. 9. If the d-orbitals used in the hybridization are of lower shell than the s-and
p-orbitals, the (n–1)d2sp3 type of complexes are formed and are called
inner orbital or low spin complexes. For example in K4[Fe (CN)6], 3d, 4s
and 4p orbitals take part in hybridisation, therefore, it is inner orbital
complex. If the d-orbitals used in the hybridization are of the same
principal energy level as that of s-and p-orbitals, the complexes are of
nsp3nd2 type and are called outer orbital complexes or high spin
complexes. For example in [FeF6]3–, 4s, 4p and 4d orbitals take part in
hybridization, therefore, it is outer orbital complex.
Q.10.
It is true that a cyclic complex is usually more stable than an
open one. Substantiate your answer with an example.
A. 10.
Cyclic complexes i.e., chelates are more stable than open
complexes. This is because of reduced strain due to the formation of 5 or 6
membered ring including metal ion. Moreover in cyclic complex, the ligand
is attached with two or more bonds with metal ion, hence more bonds have
to break. Due to this reason cyclic complexes are more stable. The copper
tetraammine complex is less stable than the copper ethylenediamine
complex although in both the cases nitrogen atom is the donor. Many cyclic
complexes like Ni-dmg, Chlorophyll, Haemoglobin are very stable towards
dissociation, while the noncyclic complexes of these metal ions are less
stable anddissociate easily.
Q. 11.
Mention the factors on which stability of the complexes
depends.
A. 11.
Following factors affect the stability of the complexes.
(i) Basic nature of the ligand : Greater the basic nature of the ligand
greater will be the stability. The copper complex with CN– is more stable
than the complex with NH3 (less basic).
(ii) Charge on central metal ion : Greater the charge on central metal ion
greater will be the stability. [Fe (CN)6]3– is more stable than [Fe (CN)6]4–
due to the higher charge on Fe.
(iii) Number of ring structures in complex : If a ring is formed during
complexion, it provides extra stability. That is why chelates are more
stable.
Q. 12.
Define cis and trans isomerism in complexes.
A. 12.
C is and trans isomerism i.e., geometrical isomerism arises due to
the different arrangement of ligands around the central metal ion. In cis
isomer two identical ligand molecules are adjacent to each other i.e., on
same side while in trans isomer two identical ligand molecules are
diametrically opposite to each other. These two isomers differ in physical
and chemical properties of each other and can be separated by some
chemical and physical methods. This type of isomerism is not found in
tetrahedral complexes but is common in square planner and octahedral
complexes.
Q. 13.
What is the importance of coordination compounds in industry
and therapy?
50
A. 13.
In Industry
(i) Photography : In photography, the excess of silver salts has to be
remove to fix the image on the negative and to prevent their further
reduction. For this purpose the negative is immersed in a bath containing
sodium thiosulphate solution, where excess of silver forms the soluble
complex and is washed away.
(ii) In electroplating : Cyanocomplexes of copper, silver, gold etc, are used
for electroplating of these metals. Since the metal complexes release the
metal ions slowly, therefore, a thin and uniform coating of the metal can be
deposited on desired object with the help of electroplating.
In chemotherapy
Many coordination compounds are successfully used as chemotheropeutic
drugs.
(i) Cis-platinum is used in the treatement of cancer.
(ii) Metal complexes of Ni, Fe, Co with some nitrogen and sulphur
containing ligands are very good antituberculosis agents.
Q. 14.
Explain geometrical isomerism with reference to square
planar complexes giving one example. How is that tetrahedral
complexes with simple ligand donot exhibit geometrical isomerism?
A. 14.
Geometrical isomerism arises due to the different geometrical
arrangement of at least two different ligands in square planner complexes.
It is shown by MA2X2, MABX2, MA2XY and MABXY types identical ligands
occupy adjacent position,cis-isomer is formed and when they occupy
opposite positions trans-isomer is formed. In tetrahedral complexes, this
isomerism is not found because the relative position of the ligands with
respect to each other will be the same.
Q.15.
Explain how [Pt (NH3)2 Cl2] and [Pt (NH3)6]Cl4 will differ in
their hybridisation states of Pt in these complexes.
A.15 .
Electrolytic conductance of the complexes depends upon the number
of ions given by the complexes in solution. We know that any ligand
present in the coordinate sphere will not ionize when it is dissolved in
water. Only these groups will form ions which are present in ionic sphere of
the complex, hence; [Pt (NH3) 2Cl2] aq has no ions
[Pt (NH3)]6Cl4 aq has [Pt (NH3 )6 ] +4 (aq) + 4Cl - (aq) ( 5 ions)
threfore, the electrolytic conductivity of [Pt (NH3)6]Cl4 will be higher than
[Pt (NH3)2 Cl2]. Hybridizationof Pt in [Pt (NH3)2 Cl2] is dsp2 and in [Pt (NH3
)6 Cl4 is
d2 sp3.
Q. 16.
Giving a suitable example describe the importance of the
formation of complex compounds in :
(i) the estimation of hardness of water.
(ii) the extraction of a particular metal from its natural source.
A. 16.
(i) Hard water contains magnesium and calcium ion which form
stable complexes with EDTA.
51
Because the values of stability constants of Mg and Ca complexes of EDTA
are different, therefore, the selective estimation of these ions is also
possible. The hardness of water can be detected by titrating it against
EDTA solution.
(ii) Silver can be extracted from its ores by treatment with solution cyanide
solution. Silve forms a soluble complex.
Ag + (aq) + 2NaCN (aq)  Na[Ag (CN2)] aq + Na+ (aq)
Q. 17. Describe each one of the following :
(a) Magentic behaviour of Ni (Co)4
A. 17.
(a) In Ni (Co)4 the oxidation state of Ni is zero. CO ligands create
strong ligand field , thus the unpaired electrons of nickel get paired.
Because there is no unpaired electron in the complex, so it would be
diagmagnetic in nature.
Q.18. Indicate the types of isomerism in the following complexes.
(i) [Co(NH3)5NO2](NO3)2
(ii)
[Co(en)3 ]Cl3
A.18. (i) Ionisation and Linkage (ii) Optical
Q.19.
Define the terms with example (i) Homoleptic complexes (ii)
Ambidentate ligands
A.19. (i) The complexes in which only one kind of ligands are bonded with
central metal atom.
(ii) Monodentate ligands which can co-ordinate with central metal atom
through more than one site.
-NO2 and –ONO, -CN and –NC
Q.20. Write IUPAC name of the following complexesCOMPLEX
[Co(NH3)4Cl(NO2)Cl
Pt(NH3)2Cl(NH2CH3)]Cl
K [Ag(CN)2]
[Ag(NH3)2]Cl
[CrCl2(H2O)4]NO3
[PtCl2(NH3)2]
[PtClNO2(NH3)4]SO4
[Cr(H2O)6]Cl3
IUPAC NAME
Tetraamminechloridonitrito-N-Cobalt(III)chloride
Diamminechlorido(methylamine)Platinum(II)chloride
Potassium dicyanidoargentate(I)
Diamminesilver(I) chloride
Tetraaquadichloridochromium(III) nitrate
Diamminedichloridoplatinum(II)
Tetraamminechloridonitrito-N-platinum(IV) sulphate
Hexaaquachromium(III) chloride
[Cu(NH3)4]SO4
[Ag(CN)2]–
[AIH4]–
[CoCl2(en)2]2SO4
Na3[Co(NO)6]
[CoCl(NO2)(en)2]+
Tetramminecopper(II) sulphate
Dicyanidoargentate(I)
Tetrahydridoaluminium(III)
Dichloridobis(ethylenediamine)cobalt(III) sulphate
Sodium hexanitrosylcobalate(III)
Chloridobis(ethylenediamine)nitrito-Ncobalt(III)
52
Na2[SiF6]
[Co(NO2)3(NH3)3]
[Co(NH3)6]Cl3
[PtCl2(NH3)4]Br2
K2[HgI4]
Na[Au(CN)2]
K4[Fe(CN)6]
K2[PtF6]
K[Ag(CN)2]
K[PtCl5(NH3)]
K3[Fe(CN)6]
K4[Fe(CN)6]
K4[Mo(CN)8]
NH4[Cr(NCS)4(NH3)2]
K3[Fe(C2O4)3]
Na2[Ni(edta)]
[Cu(H2O)2(NH3)4]SO4
[Ni(CO)4]
[Pt(NH3)4][PtCl4]
[CoCl2(NH3)4]3[Cr(CN)6]
Sodium hexafluoridosilicate(IV)
Triamminetrinitrito-N-cobalt(III)
Hexaamminecobalt(III) chloride
Tetraamminedichloridoplatinum(IV) bromide
Potassium tetraiodomercurate(II)
Sodium dicyanidoaurate(I)
Potassium hexacyanidoferrate(II)
Potassium hexafluoridoplatinate(IV)
Potassium dicyanidoargenate(I)
Potassium amminepentachloridoplatinate(IV)
Potassium hexacyanidoferrate(II)
Potassium hexacyanidoferrate(III)
Potassium octacyanidomolybedate(IV)
Ammonium diamminetetrathiocyanato-Nchromate(III)
Potassium trioxalatoferrate(III)
Sodium ethylenediaminetetraacetonickelate(II)
Tetraamminediaquacopper(II) sulphate
Tetracarbonylnickel(0)
Tetraammineplatinum(II)
tetrachloridoplatinate(II)
Tetraamminedichloridocobalt(III)
hexacyanidochromate(III)
Q.21. Draw figure to show splitting of d –orbitals in an a.octahedral crystal field
in tetrahedral field
A.21.
a.
53
b.
b.
Q.22. Write the co-ordination number and d-orbital occupation of central metal ion in
the following complexes.2
(i) [Mn(H2O)6]SO4 (ii) K3[Co(C2O4)3]
Discuss the importance of co-ordination complexes in the following fields.
(i) Medicinal field (ii) Metallurgical field
A.22. (i) CN=6, t2g3 eg2
(ii) CN = 6 , t2g6 eg0
Medicinal- cis platin [PtCl2(NH3)2] ,
Metallurgy- Ag2S + NaCN → Na[Ag(CN)2] + Na2S
Na[Ag(CN)2] + Zn → Na2[Zn(CN)4] + Ag
*****
54
ORGANIC
CHEMISTRY
55
IMPORTANT ORGANIC COMPOUNDS
COMPOUND
HALO
COMPOUNDS
R-X
FORMULA
CH3 -Cl
CH3-CH2-Cl
CH3-CH2-CH2-Cl
(CH3)2 -CH-Cl
CH3-CH2- CH2 -CH2-Cl
(CH3)2 -CH- CH2Cl
CH3-CH2- CH(Cl) –CH3
(CH3)3 –C-Cl
C6H5 –Cl
C6H5 - CH2-Cl
C6H11Cl
COMMON MAME
Methyl chloride
Ethyl chloride
n-Propyl chloride
Isopropyl
chloride
n-Butyl chloride
Isobutyl chloride
sec-Butyl
chloride
ter-Butyl
chloride
Phenyl chloride
Benzyl chloride
IUPAC NAME
Chloromethane
Chloroethane
1-Chloropropane
2-Chloropropane
1-Chlorobutane
1-Chloro-2methylpropane
2-Chlorobutane
2-Chloro-2methylpropane
Chlorobenzene
Benzyl chloride
ALCOHOLS
R-OH
CH3 -OH
CH3-CH2-OH
CH3-CH2-CH2-OH
(CH3)2 -CH-OH
CH3-CH2- CH2 -CH2OH
(CH3)2 -CH- CH2OH
CH3-CH2- CH(OH) –
CH3
(CH3)3 –C-OH
C6H5 -OH
C6H5 - CH2-OH
CH2 = CHOH
Cyclohexyl
chloride
Methyl alcohol
Ethyl alcohol
n-Propyl alcohol
Isopropyl
alcohol
n-Butyl alcohol
Isobutyl alcohol
sec-Butyl alcohol
tert-Butyl
alcohol
Phenol
Benzyl alcohol
Vinyl alcohol
ETHERS
R-O-R’
CH3 –O - CH3
CH3-CH2-O – CH2-CH3
C6H5 - O- CH3
Dimethyl ether
Diethyl ether
Anisole
Methoxymethane
Ethoxy ethane
Methoxybenzene
ALDEHYDES
R- CHO
HCHO
CH3 -CHO
CH3-CH2-CHO
CH3-CH2-CH2-CHO
C6H5 -CHO
Formaldehyde
Acetaldehyde
Propionaldehyde
Butyraldehyde
Benzaldehyde
Methanal
Ethanal
Propanal
Butanal
Benzaldehyde
56
Methanol
Ethanol
Propan-1-ol
Propan-2-ol
Butan-1-ol
2-Methyl propan-1-ol
Butan-2-ol
2-Methylpropan-2-ol
Phenol
Benzyl alcohol
Ethen- 1-ol
KETONES
R-CO - R’
CH3 –CO - CH3
C6H5 –CO - CH3
C6H5 –CO - C6H5
CARBOXYLIC
ACIDS
RCOOH
HCOOH
CH3 -COOH
CH3-CH2-COOH
C6H5 –COOH
ACID
DERIVATIVES CH3 –COCl
ACID
C6H5 –COCl
CHLORIDES
CH3 –CONH2
C6H5 -CONH2
AMIDES
CH3 -COO CH2-CH3
ESTERS
AMINES
RNH2
Primary amines
CH3 - NH2
CH3-CH2- NH2
CH3-CH2-CH2- NH2
C6H5 - NH2
Secondary amines
(CH3)2 -NH
Tertiary amines
(CH3)3 -N
NITRILES
RCN
CH3 - C N
CH3-CH2-CN
C6H5 -CN
NITRO
COMPOUNDS
RNO2
CH3-CH2- NO2
C6H5 -NO2
ISONITRILES
RNC
CH3 - NC
CH3-CH2-NC
C6H5 -NC
Acetone
Methyl phenyl
ketone
Diphenyl ketone
Formic acid
Acetic acid
Propionic acid
Benzoic acid
Propanone
Acetophenone
Benzophenone
Acetyl chloride
Benzoyl chloride
Acetamide
Benzamide
Ethyl acetate
Ethanoyl chloride
Benzoyl chloride
Ethanamide
Benzamide
Ethyl ethanoate
Methyl amine
Ethylamine
n-Propylamine
Aniline
Methanamine
Ethanamine
Propan-1-amine
Benzenamine/Aniline
Dimethyl amine
NMethylmethanamine
Trimethyl amine
Methyl cyanide
Ethyl cyanide
Phenyl cyanide
Methanoic acid
Ethanoic acid
Propanoic acid
Benzoic acid
N,NDimethylmethanamine
Ethane nitrile
Propane nitrile
Benzene nitrile
Nitroethane
Nitrobenzene
Methyl
isocyanide
Ethyl isocyanide
Phenyl
isocyanide
57
Methyl carbylamines
Ehtyl carbylamines
Phenyl carbylamine
NAME REACTIONS
1. Sandmeyer’s reaction
Benzene diazonium salt with cuprous chloride or cuprous bromide results in the
replacement of the diazonium group by –Cl or –Br.
2. Wurtz reaction
Alkyl halides react with sodium in dry ether to give hydrocarbons containing double
the number of carbon atoms present in the halide.
2 RX+
2 Na
R-R +2 NaX
3. Wurtz-Fittig reaction
A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with
sodium in dry ether.
4. Fittig reaction
Aryl halides when treated with sodium in dry ether give diphenyl.
5. Finkelstein reaction
R-X + NaI → R-I + NaX
X=Cl, Br
6. Swarts reaction
H3C-Br +AgF → H3C-F + AgBr
58
7. Kolbe’s reaction
When phenol reacts with sodium hydroxide, phenoxide ion generated
undergoes
electrophilic substitution with carbon dioxide and forms salicylic acid.
8.
Reimer-Tiemann reaction
On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group
is introduced at ortho position of benzene ring.
9. Williamson synthesis
An alkyl halide is treated with sodium alkoxide to form ether.
CH3Br + C2H5ONa
 CH3 – O - C2H5
10. Rosenmund reduction.
Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium
sulphate to give aldehyde..
11. Clemmensen reduction
Aldehydes and ketones are reduced to alkanes on treatment with zinc- amalgam and
concentrated hydrochloric acid.
CH3 –CHO  CH3 - CH3
12. Wolff-Kishner reduction
Aldehydes and ketones are reduced to alkanes on reaction with hydrazine followed by
heating with potassium hydroxide in ethylene glycol.
CH3 –CO - CH3
 CH3 –CH2 - CH3
59
13. Aldol condensation
Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the
presence of dilute alkali form β-hydroxy aldehydes (aldol) or β-hydroxy ketones
(ketol), respectively..
14. Hell-Volhard-Zelinsky reaction
Carboxylic acids having an α-hydrogen are halogenated at the α-position on
treatment with chlorine or bromine in the presence of small amount of red phosphorus
to give α-halocarboxylic acids.
15. Hoffmann bromamide degradation reaction
Primary amines can be prepared by treating an amide with bromine in an aqueous or
ethanolic solution of sodium hydroxide.The amine so formed contains one carbon less
than that present in the amide.
60
Important reactions
i) Chloroform is slowly oxidised by air in the presence of light to an extremely
poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in
closed dark coloured bottles completely filled so that air is kept out.
ii) Phenol is manufactured from the hydrocarbon, cumene (isopropylbenzene)
iii) Carbylamine reaction
Aliphatic and aromatic primary amines on heating with chloroform and ethanolic
potassium hydroxide form isocyanides or carbylamines which are foul smelling
substances. Secondary and tertiary amines do not show this reaction.
iv) Coupling reaction
Benzene diazonium chloride reacts with phenol or aniline to form azo dyes.
61
DISTINCTION TESTS
S
NAME OF
CHEMICAL
NO THE
REAGENT
COMPOUND
1
R-X
Aqueous NaOH
& AgNO3
solution
OBSERVATION PAIRS OF COMPOUNDS THAT CAN
BE DISTINGUISHED
Precipitate
AgCl – White
AgBr – Pale
yellow
AgI – Dark
yellow
1) Ethyl chloride & chlorobenzene
Ethyl chloride gives white ppt
Chlorobenzene does not give
ppt.
2) Cyclohexylbromide &
bromobenzene
Cyclohexylbromiide gives pale
yellow ppt
Bromobenzene does not give the
test.
3) Benzyl chloride & chlorobenzene
Benzyl chloride gives white ppt
Chlorobenzene does not give ppt.
4) Vinyl iodide & Allyl iodide
Vinyl iodide gives pale yellow
ppt.
Allyl iodide does not give the
test.
2
R - OH
Lucas test –
conc HCl & anh
ZnCl2
10 – Does not
react.
20 – Forms
turbidity after
few min
30 - Forms
turbidity
immediately
Yellow ppt of
Iodoform test iodoform
– NaOH &I2
(CHI3)
Note: When
both the
alcohols given
are primary /
sec apply this
62
To distinguish primary , sec & ter
alcohols
Primary – Methanol, Ethanol, n –
alkyl alcohols / alkan – 1-ol, benzyl
alcohol
Secondary – Isopropyl alcohol,
alkan – 2 – ol
Tertiary – ter- buytl alcohol (2methyl – propan – 2- ol)
Alcohols with CH3 – CH – group
give this test
OH
1) Ethyl alcohol & methyl alcohol
(both are 10)
Ethyl alcohol gives yellow ppt
test
Methyl alcohol does not give
this test.
2) Pentan – 2- ol & Pentan – 3- ol
(both are 20)
Pentan – 2- ol gives yellow
ppt
Pentan – 3- ol does not give
this test.
3
Phenol
Neutral ferric
chloride
Violet colour
4
Aldehydes
Tollens test –
ammoniacal
AgNO3 , warm
Silver mirror
Iodoform test Yellow ppt of
– NaOH &I2
iodoform
Note: When
(CHI3)
both the
compounds
given are
aldehydes/
ketones apply
this test
63
1) Phenol & ethyl alcohol
Phenol gives violet colour
Ethyl alcohol does not give this
test
Aldehydes give positive test.
1. Acetaldehyde (Propanal) &
Acetone
(Propanone)
Acetaldehyde (Propanal) gives
Ag mirror
Acetone (Propanone) does not
give this test.
Ald. & ket. with CH3 – C –
group give this test
O
1) Acetaldehyde & benzaldehyde
Acetaldehyde gives yellow
ppt.
Benzaldehyde does not give
this test.
1) Acetaldehyde & formaldehyde
Acetaldehyde gives yellow
ppt.
Formaldehyde does not give
this test.
3) 2 – Pentanone & 3 – Pentanone
2 – Pentanone gives yellow
ppt.
3 – Pentanone does not give
this test.
4) Acetophenone & benzophenone
Acetophenone gives yellow
ppt.
Benzophenone does not give
this test.
5.
Carboxylic
acids
Sodium
bicarbonate
test
Formic acid
( Methanoic Tollens test
acid)
6
Amines
Primary
amines
Aniline
Hinsberg test –
Benzene
sulphonyl
chloride +
NaOH
Effervescence
due to
evolution of
CO2 gas
All carboxylic acids give this
test
Effervescence
due to
evolution of
CO2 gas
HCOOH is the only carboxylic
acid that gives this test
HCOOH and CH3COOH
HCOOH gives this test . CH3COOH
does not give this test
To distinguish primary , sec &
ter amines
Alkyl amine – 10
Dialkylamine -20 (N- alkyl
alkanamine)
Trialkylamine( N,N – dialkyl
alkanamine - 30
10 –Soluble
20 – Insoluble
30 – does not
react with
Hinsberg
reagent
Very
Carbylamine
unpleasant
test – Alcoholic smelling
To distinguish primary amines
KOH + CHCl3
isocyanide gas from other amines
evolves
1)Ethylamine & Diethylamine
Ethylamine gives this test.
Diethylamine does not give this test
2) Aniline & Dimethylamine
Aniline gives this test.
Diethylamine does not give this test
Azodye test –
Reddish
NaNO2 + HCl + orange dye
Aniline & Benzylamine
0
phenol ( 0 to
Aniline gives this test
0
5 C)
Benzylamine does not give this
test.
****************************************************************************************
64
CHAPTER 10
HALOALKANES AND HALOARENES
VSA (one mark each)
1.What are ambident nucleophiles ? Give two eg.
Nucleophiles that possess two nucleophilic centres. Eg CN- & NO2
-
2. What are enantiomers ?
Stereoisomers related to each other as non-superimposable mirror images.
3. What is chirality ?
The property of objects that are non-superimposable on their mirror images.
4. What is called racemisation ?
The process of conversion of enantiomer into a racemic mixture (mixture of d & l
in equal proportions )
5. Why is chloroform stored in closed dark colour bottle ?
In the presence light & air it forms poisonous gas , phosgene.
2CHCl3 +O2
 2COCl2 + 2HCl
6. Alkly halides though polar are immiscible with water. Why ?
Because they do not form intermolecular H bond with water.
7. Chlorobenzene(haloarene) is less reactive than ethyl chloride (haloalkane) towards
nucleophilic substitution reaction. Why ?
In chlorobenzene C – Cl bond acquires partial double bond character due to
resonance. So it is difficult to break the bond.
8. Among isomeric dihalobenzenes, the para-isomers have high melting compared to
their ortho- and meta-isomers. Why?
It is due to symmetry of para-isomers that fits in crystal lattice better as
compared to ortho- and meta-isomers.
9. Arrange the following set of compounds in order of increasing boiling points.
1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane
NOTE : SN1 : 10 < 20 < 30 Tertiary cabocation is more stable
SN2 : 30 < 20 < 10 Steric hindrance is less in primary
65
10. Which one in the following pairs would undergo SN2 reaction faster?
Because iodine is a better leaving group.
11. Which one in the following pairs would undergo SN1 reaction faster?
12. An alkyl halide having molecular formula C4H9Cl is optically active .What is its
name?
Ans.= 2-chlorobutane
13. P-Dichlorobenzene has higher melting point than o and m-isomer.Why?
Ans=The P-isomer being more symmetrical fits closely in the crystal lattice.
SA1 (two marks each)
1. Explane why haloarenes are less reactive towards nucleophilic substation reactions
than Haloalkanes.
(a) Due to resonance in benzene ring there is partial double bond in C-X bond which
is difficult to break.
(b) It is difficult for electron rich nucleophiles to approach electron rich benzene ring
in haloarenes.
66
2. How do the products differ with ethyl bromide react separately with ;
(a) Alc. KCN and Alc. AgCN.
(b) Aq. KOH and Aic. KOH.
Ans.
(a)
Alc.KCN
C2 H5 CN+
KBr
C 2H5Br
Alc.AgCN
C2 H5 NC+ AgBr
(b)
Aq.KOH
C2H5Br
+
KBr
C2H5 Br
Alc.KOH
CH2=CH2 + KBr +H2O
3. Write the structure of major organic product;
C2H5ONa
(a) (CH3 ) 3 CH-CH(Br)CH 2CH 3
Ethanol /Heat
Ans.
(b) CH3CH2Cl + SbF3
(a)
CH3-C=CH-CH2-CH3
Heat
CH3
(b)
CH3CH2F
4. Out of C6H5CH2Cl and C6H5 CHClC6H5 which is more easily hydrolysed by
aqueous KOH and why
Ans. C6H5CHClC6H5 will get hydrolysed easily because carbo cation formed
Will be stabilized by resonance effect of two phenyl groups.
67
5. (a) Give the IUPAC name of the following ;
H3C
CH3
H
H
CH3
Br
(b) Write Wurtz fittig reaction .
Ans. (a)
(b)
4-Bromo-3-methyl –pent-2-ene.
X
R
2Na + X-R
+ 2NaX
Dry ether
6. Arrange the following compounds in order of reactivity towards SN2 displacement:
2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
2-Bromo-2-methylbutane (ter) < 2-Bromopentane(sec) < 1-Bromopentane (pri)
Because of less steric hindrance in 1-Bromopentane
7.
a) How alkyl halides become darker on standing in the presence of sunlight
b) Which is better nucleophile Br‾ or I‾ and why.
ANS:- (a) Alkyl halides are unstable and liberate iodine .
(b) I‾ because of large size and low electronegativity .
68
SA –II (three marks )
1.(a) Write the IUPAC names of organic compounds ‘A’,’B’,’C’ and in the following:Br2
Alc. KOH
CH3 ─CH─CH2─CH2
Alc. KOH
‘B’
‘A’
H2O/H2SO4
‘C’
‘D’
Br
b) Which isomer of C4H9Br has the lowest boiling point?
Ans. a) A=But-2-ene
B=2,3-Dibromobutane
C=But-2-yne
D=Butan-2-one
b)
CH3
‫׀‬
CH3 - C– CH3 has lowest boiling point.
‫׀‬
Br
3. Convert the following:(a) ethanol to propanenitrile
(b) propene to 1-propanol
(c) 1-Bromopropane to 2-Bromopropane
Ans a)
CH3CH2OH
b)CH3CH=CH2
PCl5
KCN
CH3CH2Cl
CH3CH2CN
HBr/peroxide
KOH
CH3CH2CH2Br
c) CH3CH2CH2Br
alc. KOH
CH3CH=CH2
69
HBr
CH3CH2CH2OH
CH3CHBrCH3
4) An organic compound “A” having molecular formula C4H8 on treatment with dil
H2SO4 gives “B” which on treatment with conc. HCl and anhydrous ZnCl2 gives “C” and
on treatment with sodium ethoxide gives back “A”. Identify A, B,C.
Ans.A=But-2-ene
B=Butan-2-ol
C=2-Chlorobutane
5. Write the Structure of major products:(a)
Conc. H2SO4
+ HNO3
Br
NaOH
CH2Br
CH(CH3)2
+ Br2
70
71
CHAPTER 11
ALCOHOLS, PHENOLS AND ETHERS
VSA (one mark each)
Topic:- Reasoning Questions
VSA (one mark each)
1.
Phenol is acidic in nature.
Ans. Phenol is acidic in nature because:
a) phenol , due to resonance, the positive charge rests on oxygen making the
shared
pair of electrons more towards oxygen and hydrogen as H+
b) The carbon attached to OH is SP2 hybridize and is more electronegative, this
decreases the electron density on oxygen, increasing the polarity of O-H bond
and ionization of phenol.
c) The phenoxide ion formed by loss of H+ is more resonance stabilized than
phenol itself.
2.
Phenol has a smaller dipole moment than methanol.
Ans. In phenol due to electron rich benzene ring the C-O bond is less polar whereas in
methanol the C-O bond is highly polar. Therefore the dipole moment of methanol
is higher than phenol.
3.
o- nitrophenol has lower boiling point ( is more volatile ) than p – nitrophenol.
Ans. P- nitrophenol has intermolecular hydrogen bonding which increases the boiling
point while in o- nitro phenol due to presence of intra molecular hydrogen
bonding, there is a decrease in boiling point and increase in volatility.
72
4.
Methanol is miscible with water while iodomethane is not.
Ans.
Methanol can form intermolecular hydrogen
bonding with water but there is no hydrogen
bonding in iodomethane and water. Therefore
methanol in miscible in water.
5.
Alcohols have higher boiling points than isomeric ethers.
Ans. Alcohols can form intermolecular hydrogen bonds due to their high polarity
whereas, ether cannot. Therefore alcohols have higher boiling points than
isomeric ethers.
6.
Ethers are soluble in water alkanes are not.
Ans. Ethers can form H- bonding with water molecule whereas alkenes cannot.
Therefore ethers are soluble in water and alkanes are not.
7.
The order of acidic strength in alcohols is R CH2OH > R2 CHOH > R3 COH
Ans. In alcohols, the acidic strength is due to polar nature of O-H bond. An electron
releasing group e.g., alkyl groups, increases electron density on oxygen tending
to decrease the polarity of O-H bond. This decreases the acid strength. Therefore
the order of acid strength is
73
8. During preparation of ester from alcohol and acid, water has to be removed as soon as it
is formed.
Ans.
9.
Ethers cannot be prepared by dehydration of secondary or tertiary alcohols.
Ans. For secondary and tertiary alcohols, elimination competes over substitution and
alkenes are formed on acidic dehydration as the reaction follows Sn1 mechanism.
Therefore the acidic dehydration of secondary or tertiary alcohols does not give
ethers.
10. Reaction of anisole with HI gives methyl iodide and phenol.
Ans.
11. tert butyl alcohol has lower b.pt than n-butyl alcohol. Why ?
Ans. In tert butyl alcohol there is more branching and so less van der Waals forces
due to less surface area.
12.
Alcohols are more soluble in water than hydrocarbons of comparable molecular
mass. Why ?
Ans. Because alcohols form hydrogen bond with water.
74
13. o-nitrophenol is more acidic than o-methoxyphenol. Why ?
Ans. Because of electron withdrawing nitro group in o-nitrophenol there is
effective delocalisation negative charge.
14. Alcohols are weaker acids than water. Why?
Ans. Water is a better donor than alcohols.
RO - + H2O
ROH + OH –
15. Phenol is more acidic than alcohols. Why?
Ans. Phenoxide is more stabilised by resonance than alkoxide ion.
16. What is denaturation of alcohols ?
Ans. Commercial alcohol is made unfit for drinking by adding copper sulphate/
pyridine.
17.
Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, phenol, 4-methylphenol
Ans. Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 2,4, 6-trinitrophenol.
18.
Write the names of reagents and equations for the preparation of the following
ether by Williamson’s synthesis: 2-Methoxy-2-methylpropane
Ans. Reagents : Ethyl bromide & sodium tert butoxide
C2H5Br + (CH3)3CONa
(CH3)3CO C2H5 + NaBr
****************************************************************************************
75
ALCOHOLS, PHENOLS AND ETHERS
Topic:- Conversions
SA-I (two marks each)
1.
Ethene to 1,2 -ethanediol
Ans.
2.
Phenol to Salicyldehyde
Ans.
3.
Butanol to Butanoic acid
4.
Ethanol to propanone
5.
Phenol to salicylic acid
Ans.
76
6.
Methanol to Ethanol
7.
Ethanol to propanol
8.
Phenol to Benzyl Alcohol
Ans.
9.
Ethanal to propan -2- ol
Ans.
10.
Ans.
l – propanol to 2 – bromo propane
77
ALCOHOLS, PHENOLS AND ETHERS
Topic:- Identification Question
SA-II (three marks each)
1.
An organic compound ‘ A ‘ having molecular formula C3 H6 on treatment with aq.
H2SO4 give ‘B’ which on treatment with Lucas reagent gives ‘C’. The compound
‘C’ on treatment with ethanolic KOH gives back ‘ A’ .Identify A, B , C .
Ans.
2.
An organic compound A (C6H6O) gives a characteristic colour with aq. FeCl3
solution. (A) On reacting with CO2 and NaOH at 400k under pressure gives (B)
which on acidification gives a compound (C) .The compound (C) reacts with
acetyl chloride to give (D) which is a popular pain killer. Deduce the structure of
A,B,C & D.
Ans.
3.
An organic compound (X) when dissolved in ether and treated with magnesium
metal forms a compound Y. The compound, Y, on treatment with acetaldehyde
and the product on acid hydrolysis gives isopropyl alcohol. Identify the
compound X. What is the general name of the compounds of the type Y.
Ans. The compound X is CH3Br and Y is CH3MgBr The compounds of the type ‘Y’ are
called Grignard reagent.
78
4.
A compound ‘A’ with molecular formula C4H10O on oxidation forms compound ‘B’
gives positive iodoform test and on reaction with CH3MgBr followed by hydrolysis
gives (c). Identify A, B & C.
Ans. The compound ‘B’ is obtained by oxidation of C4H10O and gives positive iodoform
test and also reacts with CH3MgBr , it must be methyl Ketone , it must be methyl
ketone having four carbon atoms i.e, CH3COCH2CH3 .
This can be obtained by oxidation of 2 – butanol i.e , CH3 CH CH2 CH3 Therefore ,
the
reactions are:
5.
An aromatic compound (A) having molecular formula C6H6O on treatment with
CHCl3 and KOH gives a mixture two isomers ‘B’ and ‘C’ both of ‘B’ & ‘C’ give same
product ‘D’ when distilled with Zn dust. Oxidation of ‘D’ gives ‘E’ of formula
C7H6O2. The sodium salt of ‘E’ on heating with soda lime gives ‘F’ which may also
be obtained by distilling ‘A’ with zinc dust. Identify compounds ‘A’ to ‘F’ giving
sequence of reactions.
79
6.
Compound ‘A’ of molecular formula C5H11Br gives a compound ‘B’ of molecular
formula C5H12O when treated with aq. NaOH. On oxidation the compound yields a
mixture of acetic acid & propionic acid. Deduce the structure of A, B & C.
Ans.
80
The reactions are:
Ans.
****************************************************************************************
81
CHAPTER 12
ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
VSA (one mark each)
1.
The b.pts of aldehydes & ketones are higher than ethers of comparable molecular
masses. Why ?
Ans. Due to stronger dipole – dipole interactions in aldehydes & ketones.
2.
Acetaldehyde (ethanal) and acetone (propanone) are miscible with water. Why?
Ans. Because they form hydrogen bond with water.
3.
Arrange the following set of compounds in order of increasing boiling points.
CH3CH2CH2CHO, CH3CH2CH2CH2OH, CH3CH2 OCH2 CH3 ,CH3CH2CH2 CH3
Ans. CH3 ,CH3CH2CH2 CH3 < CH3CH2 OCH2 CH3 < CH3CH2CH2CHO <
CH3CH2CH2CH2OH
4.
Ketones are less reactive than aldehydes towards nuclephilic addition reaction.
Why?
Ans. Two electron releasing alkyl groups in ketones reduce the electrophilicity of
carbonyl. Also due to steric hindrance nucleophilic attack is not easier.
5.
Benzaldehyde is less reactive than propanal towards nucleophilic addition
reaction. Why ?
Ans. Resonance in benzaldehyde reduces the electrophilicity of carbonyl carbon.
6.
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6trimethylcyvlohexanone does not.
Ans. The three methyl groups reduce the elecrophilicity of carbonyl and also offer
steric hindrance to the nucleophilic attack
(ii) In semicarbazide ,the NH2 group directly attached to the -C = O group is not
involved in the formation of semicarbazones.Give reason.
The lone pair of electrons of the – NH2 group directly attached to –C = O group is
involved in resonance. Hence it does not act nucleophile
7.
Arrange the following compounds in increasing order of their reactivity towards
nuclephilic addition reaction.
Methanal, propanone, ethanol
Ans. Propanone < Ethanal < Methanal
8.
Acetaldehyde undergoes aldol condensation. Why ?
Ans. Because it contains alpha hydrogen.
9.
Formaldehyde & benzaldehyde do not undergo aldol condensation. Why ?
Ans. Because they do not contain alpha hydrogen.
82
10. Formaldehyde & benzaldehyde show Cannizaro reaction. Why ?
Ans. Because they do not contain alpha hydrogen.
11. What is formalin ? Mention its uses.
Ans. 40% formaldehyde is formalin. It is used to preserve biological specimen and to
prepare bakelite.
12. Carboxylic acids are higher boiling liquids than alcohols. Why ?
Ans. Carboxylic acids form more extensive hydrogen bond than alcohols and exisi as
dimer.
13. Lower members of carboxylic acids are miscible with water. Why ?
Ans. Because they form hydrogen bond with water.
14.
Carboxylic acids are more acidic than phenol. Why ?
OR
Ka of carboxylic acid is more than that of phenol.Why ?
OR
pKa of carboxylic acid (acetic acid) is more than that of phenol. Why?
Ans. Carboxylate ion is more stabilised than phenoxide ion due to resonance with
negative charge on more electronegative oxygen atom.
15. Carboxylic acids do not show the reactions of carbonyl group. Why ?
Ans. Due to resonance the double bond character is reduced in carboxylic acids.
16. What is decarboxylation reaction ?
Ans. Carboxylic acids loose CO2 to form hydrocarbons when their sodium salts are
heated with sodalime.
sodalime
CH3COONa
CH4 + CO2 + Na2CO3
17.
During the preparation of esters from a carboxylic acid and an alcohol in the
presence of an acid catalyst, the water or the ester should be removed as soon
as it is formed. Why ?
Ans. The preparation of esters from a carboxylic acid and an alcohol in the presence of
an acid catalyst is a reversible reaction. So, the water or the ester should be
removed as soon as it is formed to shift the equilibrium towards the ester
formation (forward) according to Le Chatelier’s principle.
18.
Exercise 12.12 page no 378
(i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
(reactivity towards HCN)
(ii) (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH <
CH3CH2CH(Br)COOH
(acid strength)
(iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4Dinitrobenzoic acid (acd strength)
83
19. Why aldehydes and ketones have lower boiling point than alcohols.
Ans. Alcohol molecules are associated with intermolecular H-bonding whereas they are
not.
20.
Compare the acid strength of the following acidsHCOOH , CH3COOH , C6H5COOH
Ans. HCOOH > C6H5COOH > CH3COOH
21. Why chloroacetic acid stronger than acetic acid.
Ans- Cl is electron withdrawing group thus chloroacetate ion is more stable than
acetate ion.
22. Which of the aldehyde undergo Cannizaro reaction.
Ans. Aldehydes which do not have alpha hydrogen.e.g. HCHO.
SA-I (two marks each)
1.
Although phenoxide ion has more number of resonating structures than
carboxylate ion, carboxylic acid is a stronger acid than phenol.
Ans- In carbooxlate ion –ve charge is delocalized over two oxygen atoms which are
higly electronegative whereas in phenoxide ion –ve charge is delocalized over
only one oxygen atom. Carboxylate ion is more stable phenoxide ion that is why
carboxylic acids are more acidic than phenols.
2. State
(i) Why benzaldehyde does not undergo aldol condensation.
(ii) How an acid amide may be converted to the parent acid.
Ans (i) Because of absence of ɑ - Hydrogen
O
O
H
+
(ii) CH3 – C –NH2 + H2O
CH3 – C- OH + NH3
3. Complete the following
(i)
COOH
SO3/H2SO4
(ii)
COOH
Conc HNO3
Conc H2SO4
84
Ans (i)
COOH
COOH
+ H2O
SO3/H2SO4
(ii)
SO3
COOH
COOH
+
Conc HNO3
H2O
Conc H2SO4
NO2
4.. What are nucleophilic addition reactions- How would you account –Aldehydes are more
reactive than ketones towards nucleophiles?
Ans. Reactions which are intiated by nucleophiles are called nucleophilic addition reactions.
O
OH
CH3-C-H + HCN
CH3-CH -CN (acetaldehyde
Cynohydrin)
Aldehydes are more reactive because they are more polar than ketones due to lesser no. of
electron releasing alkyl groups, also steric hinderance in aldehydes is less than in ketones.
SA-II (three marks each)
1. Write the major products of the following reactions:
(i) Nitration of benzaldehyde
(ii) Sulphonation of benzaldehyde
Ans
(i) Nitration Nitration of benzaldehyde with a mixture of conc. H2SO4 and HNO3 forms 3nitro- benzaldehyde and 3, 5-dinitrobenzaldehyde.
CH
O
HNO3 + H2SO4
CH
O
HNO3 + H2SO4
CH
O
273 – 283 K
Benzaldehyde
NO2
3-nitro- benzaldehyde
85
NO2
NO2
3, 5-dinitrobenzaldehyde
(ii) Sulphonation. Sulphonation of benzaldehyde with conc. H2SO4 forms 3- benzaldehyde
CH
sulphonic acid and then 3- benzaldehyde-1, 5-disulphonic acid.
O
CH
O
CH
O
H2SO4
H2SO4
SO3H
H3OS
SO3H
3- benzaldehyde sulphonic
acid
Benzaldehyde
3- benzaldehyde-1,
5-disulphonic acid
1. What happens whena. Acetophenone is treated with bromine in the presence of anhdy AlCl3.
b. Acetophenone is nitrated.
AnsCOCH3
COCH3
Br2/anhyd AlCl3
a.
Br
b.
COCH3
COCH3
COCH3
HNO3/H2SO4
HNO3/H2SO4
NO2
3-nitroacetophenone
3.
Complete the following reactionsO
a. R-C-H
Zn/HCl,4[H]
------O
b. CH3-C-NH2
HNO2
------
86
NO2
3,5-dinitroacetophenone
NO2
AnsO
a. . R-C-H
Zn/HCl,4[H]
O
b. . CH3-C-NH2
4.
HNO2
CH3COOH + N2 + H2O
Carry out the following conversion—
a. Acetylene to Acetaldehyde
b. Toluene to Benzaldehyde
Ans—a. CHΞCH + H2O
b.
R-CH3 + H2O
H2SO4/HgSO4
CH2=CHOH
CH3CHO
CH3
[O]
CH
O
CrO2Cl2
87
+ H2O
CHAPTER 13
ORGANIC COMPUNDS CONTAINING NITROGEN
VSA (one mark each)
1.
Ammonolysis is not preferred for preparation for primary amines.Why ?
Ans. Because it gives mixture of amines.
2.
Alkyl amines have higher b.pts than alkanes of comparable molecular mass.
Why?
Ans. Because alkyl amines form intermolecular hydrogen bond.
3.
Alcohols have higher b.pts than alkyl amines of comparable molecular mass.
Why?
Ans. Because in alcohols the strength of H – bond is more than that in amines since O
is more electronegative than N.
4.
Arrange the following set of compounds in order of increasing boiling points.
n-Butylamine(10), Diethylamine(20),Ethyl dimethylamine(30) [isomeric amines]
Ans. Ethyl dimethylamine(30) < Diethylamine(20) < n-Butylamine(10)
Reason - 30 amine – No H bond
In primary amine 2 hydrogen atoms are available for H- bond. In sec amine only
one H atom available for H- bond.
5.
(i) Ethylamine (aliphatic amines) is a stronger base than aniline (aromatic
amine). Why ?
Ans. In ethyl amine the electron releasing alkyl group increases the electron density
on nitrogen. In aniline the electrons on N is involved in resonance.
(ii) pKb of aniline is more than that of methyl amine. Why?
Ans. In methyl amine the electron releasing alkyl group increases the electron density
on nitrogen. In aniline the electrons on N is involved in resonance.
6.
Amines are less acidic than alcohols . Why ?
Ans. RO - is more stable than RNH –
7.
Why cannot primary amines be prepared by Gabriel phthalimide ?
Ans. Aryl halides do not undergo nucleophilic substitution with the anion formed by
phthalimide.
8.
Methylamine in water reacts with ferric chloride to precipitate hydrated ferric
oxide. Why ?
Ans. Methylamine is more basic than water. So it forms OH- with water. The OH- ions
react with ferric chloride and gives ppt.
88
9.
Aniline does not undergo Friedel- Crafts reaction. Why ?
Ans. In Friedel- Crafts reaction AlCl3 which is used a catalyst is a Lewis acid. It reacts
with aniline and forms salt.
10.
Nitration of aniline gives a substantial amount of m-nitroaniline besides ortho
and para derivatives. Why ?
Ans. During nitration, in presence of strong acidic medium aniline is protonated to
form the anilinium ion which is meta directing.
11.
Diazonium salts of aromatic amines are more stable than those of aliphatic
amines. Why ?
Ans. Arenediazonium ion is stabilised by resonance.
12. Arrange the following: (13.4 page no 400)
(i) (C2H5)2NH > (C2H5 )3 N > C2H5NH2 (Basic strength in aqueous medium)
(ii) (CH3)2NH >
CH3NH2 > (CH3)3 N > NH3(Basic strength in aqueous medium)
(iii) C6H5NH2 >
(C2H5 )3 N > (C2H5)2NH > C2H5NH2 (Basic strength in gaseous
phase)
(iv) p-nitroaniline < Aniline < p-toluidine (Basic strength)
(v) Diethylamine(20) < n-Butylamine(10) < n- butylalcohol (boiling point)
NOTE
 Hydrocarbons < Ethers < Ald & Ketones < Amines < Alcohols < Carboxylic acid
(boiling pts)

Electron withdrawing group increases the acidic strength
Higher Ka, lower pKa higher is the acidic strength

Electron releasing group increases the basic strength
Higher Kb, lower pKb higher is the basic strength
13.
Arrange the following In decreasing order of basic strength:
C6H5NH2,C6H5(CH3)2,(C2H5)2NH and CH3NH2
Ans- (C2H5)3N > CH3NH2 > C6H5NH2 > C6H5N(CH3)2 .
14. Why do amines react as nucleophiles ?
Ans. It is because they have lone pair of electrons .
15. Give reason why secondary amines are more basic than primary amines ?
Ans. In secondary amines , there are two alkyl groups which increase electron
density on N more than “N” present in primary amine in which there is one alkyl group.
89
TWO MARKS QUESTIONS : Q1.
A1.
(b).
Write chemical equations for the following conversations :
(a). CH3-CH2-Cl- into CH3- CH2 –NH2
(b). C6H5-CH2-Cl into C6H5-CH2-CH2-NH2
(a) CH3-CH2-Cl ethanolic NACN CH3-CH2-C Ξ N reduction CH3-CH2-CH2-NH2
C6H5-CH2-Cl
ethanolic NACN C6H5-CH2-CΞ N
H2/Ni
C6H5-CH2-CH2-NH2
Q2.
Give reasons :
(a). Electrophilic substitution in case of aromatic amines takes place more
readily than benzene.
(b). Nitro compounds have higher boilings points than hydrocarbons having
almost same molecular mass.
Ans. (a) It is because –NH2 group is electron releasing, i.e. activating and increase
electron density at o-and-p-positions . Electrophilic substitution reactions takes
place faster.
(b).Nitro compounds are more polar than hydrocarbons , therefore, have more
van der Waals” forces of attraction than hydrocarbons, hence higher boiling
point.
5.How will you convert Aniline to Benzonitrile?
AnsN2+Cl-
NH2
CΞN
NaNO2+HCl
0 – 5 0C
KCN
CuCN
Aniline
Benzonitrile
90
THREE MARK QUESTION1.
a. Explain why is so that methyl bromide reacts with KCN to give mostly methyl
isocyanides?
b. Why nitro alkanes have higher boiling points than the corresponding
hydrocarbons?
c. Before nitration, aniline is converted to Acetanilide?
Ans. a. KCN is an ionic compounds it generates CN- which replace Br-.AgCN is covalent
comp,it forms bond through N forming isocyanides.
b. Nitro alkanes are more polar than the hydrocarbons.
c. Aniline get oxidized with HNO3,therefore it is converted in to acetanilide
before nitration.
2. Account for the following :(i) Pkb of aniline is morethan that of methylamine.
(ii) Ethylamine is soluble in water whreas aniline is not.
(iii)Why has aniline a weak basic nature than aliphatic amines.
Ans- (i) aniline has an electron withdrawing phenyl group so it is the weaker base
than ammonia. Methyl group in methylamine is electron donating group so it is
stronger base than ammonia.
(ii) ethyl group in ethylamine is comparatively a small group and causes no
hindrance in formation of hydrogen bonding and hence it is soluble in water.
(iii) Due to resonance in aniline.
4. Arrange the following
(i)
in decreasing order of the pKb value:
C2H5NH2, C6H5NHCH3,(C2H5)2NH AND CH3NH=
(ii)
in decreasing order of basic strength: C6H5NH2, C6H5N(CH3)2,(C2H5)2NH and
CH3NH2
(iii)
in increasing order of basic strength:
(a)
aniline,p-nitroaniline and p-toylidine
(b)
C6H5NH2,C6H5NHCH3,C6H5CH2NH2
Ans)
(i) C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH
(ii) (C2H5)2NH>CH3NH2>C6H5NH2>C6H5N(CH3)2
(iii)-(a) p-nitroaniline<aniline<p-toulidine
(b)C6H5NH2<C6H5NCH3<C6H5CH2NH2
5.
(a) Give possible explanation for each of the following:(i) the presence of a base is needed in the ammonolysis of alkyl halides.
(ii)aromatic primary amines cannot be prepared by Gabriel phtaliminde
synthesis
(b)Write the IUPAC name of CH3-N-C-CH3 - C2H5
91
Ans-
(a)(i) To remove the HX produced during the reaction.
(ii)Aryl halides do not undergo nucleophilic substitution with POTASSIUM
PHTHALIMIDE
(b) N-Ethyl-N-methylethanamide.
92
GENERAL
CHEMISTRY
93
UNIT 5 SURFACE CHEMISTRY
Surface Chemistry – The branch of chemistry which deals with the study of nature of
surfaces and the changes occurring on the surfaces is called surface chemistry.
The phenomenon of attracting and retaining the molecules of a substance on the
surface of a liquid or a solid resulting into higher concentration of the molecules on the
surface is called adsorbate and the substance on which it is adsorbed is called
adsorbent.
The reverse process i.e. removal of the adsorbed substance from the surface is called
desorbtion (which can be brought about by heating or reducing the pressure). The
adsorption of gases on the surface of metals is called occlusion.
DIFFERENCE BETWEEN ADSORPTION AND ABSORPTION
ADSORPTION
ABSORPTION
It is a surface phenomenon.
It is a bulk property.
Conc. of Adsorbate is more on the
The adsorbate/substance is uniformly
surface than in the bulk.
distributed throughout.
It is rapid in beginning and its rate
It occurs at a uniform rate.
slowly decreases until equilibrium is
attained.
E.g. Silica Gel. Aluminum Gel,
e.g. CaCl2 etc.
Colloidal etc.
Types of Adsorption – Depending upon the nature of forces between particles of
Adsorbate and Adsorbent, Adsorption is of two types:
1. Physical Adsorption or Vander Waal’s Adsorption or Physiorption :
When a gas is held on the surface of solid by Vander Waal’s forces without resulting
in the formation of any chemical bond between Adsorbate and Adsorbent called
Physical Adsorption.
2. Chemical or Chemisorption (Langmuir Adsorption) :
When a gas is held on to the surface of the solid by forces similar to those of a
chemical bond (covalent or ionic), the type of Adsorption is called chemical
Adsorption.
DIFFERENCE BETWEEN PHYSICAL AND CHEMICAL ADSORPTION
PHYSICAL
CHEMICAL
Definition:
Definition:
Reversible process.
Irreversible process.
Rate of Adsorption decreases with
Rate of Adsorption initially increases
increase in temperature.
then decreases by increase in
temperature.
It is not specific in nature.
It is highly specific in nature.
No compounds formed on its surface. Surface compounds are formed.
Forms multimolecular layer.
Forms unimolecular layer.
Enthalpy of Adsorption is the order of Enthalpy of Adsorption is the order of
94
20-40 KJ mol-1.
Increase with increase surface area
of adsorbent.
200-400 KJ mol-1.
It also increases with increase
surface area of adsorbent.
Adsorption of Gases on Solids:The extent of adsorption of a gas on a solid surface depends upon the following
factors:i.
Nature of Gas: - The easily liquefiable gases are adsorbed more than the
permanent gases (H2, N2, O2, He, Ne).
Liquifiable Gases – NH3, SO2, SO3, HCL etc.
Liquefaction of gas
ii.
iii.

TC (critical temperature).
 The minimum temperature above which a gas cannot be liquefied even
by applying very high pressure is called critical temperature (TC).
Nature of Adsorbent: - The extent of adsorption also depends upon the nature
of adsorbent. E.g. Activated charcoal or animal charcoal can be adsorbed
those gases easily which are easy to liquefy.
 Permanent gases can be adsorbed on the surface of transition metal.
Effect of Pressure: Isotherm – A graph between the amount of gas adsorbed per gram of
the adsorbent (x/m) and the equilibrium pressure of the adsorbate at
constant temperature is called Adsorption.
At
value of Ps, of Equilibrium
pressure, x/m reaches its
maximum value & then remain
constant even pressure is
increased. It is saturation state
pressure called saturation
pressure.
and
x = amount of as adsorbate
m = mass of adsorbent
P = pressure, P0 = independent pressure.
 P = Kp.
At high pressure x/m  P0 = Kp0 = K.
a. At low pressure x/m
b.
c. At intermediate x/m = Kp1/n
1/n = 0 to 1 (0.1 to 0.5) probably
95
log(x/m) = log(K) + 1/n(log(P))
log(x/m) = 1/n(log(P)) + log(K)
y = mn + c
A graph between x/m VS log(P), a straight line with slope equal to 1/n and
ordinate intercept (at y-axis) equal to log(K).
PHYSICAL
CHEMICAL
96
Catalysis – A substance which accelerates the rate of a chemical reaction without
undergoing any change in its chemical composition or mass during the reaction called
catalyst.
 The Phenomenon of increasing the rate of a reaction with the help of a catalyst is
known as catalysis.
2KClO3
2KCl + 3O2
By adding small amount of MnO2, decomposition becomes faster and occurs at
lower temperature.
2KClO3
2KCl + 3O2
Types of Catalysis:A. Homogeneous Catalysis – When a catalyst mixes homogenous with the reactants
& this form a single phase, catalyst called homogenous & this type of catalysis
called Homogenous Catalysis.
I.
II.
SO2 oxidised to SO3 in presence of NO(g).
2SO2(g) + O2(g)
2SO3(g)
Hydrolysing Ester
CH3COOC2H5(l) + H2O(l)
III.
Hydrolysis of pure sucrose.
C12H22O11 + H2O
IV.
CH3COOH + C2H5OH
C6H1206 + C6H12O6
{ Glucose + Fructose }=Soln
Preparation of Di-ethyl ether in the presence of conc. H2SO4.
2C2H5OH
C2H5-O-C2H5 + H2O
All reactants and catalyst are in same phase.
B. Heterogeneous Catalysis – When a catalyst exist in different phase then that of
reactants called heterogeneous catalyst and process called heterogeneous
catalysis.
2SO2(g) + O2(g)
2SO3 [Contact process of H2SO4]
97
In such type of catalysis, the catalyst is generally solid while reactant are gases
and reaction from surface of solid catalyst. Hence, it is also called surface
catalysis.
e.g. N2(g) + 3H2(g)
2NH3
Oxidation of NH3
4NH3(g) + 5O2(g)
Hydrogenation of Vegetable Oil
Oil + H2
4NO + 6H2O
Vegetable Ghee
Enzyme Catalysis:-Enzyme are the biological catalyst when the rate of bio-chemical
reaction.
Enzymes are proteins. There are certain substances which may increase the reactivity
of enzyme; they are called activator or co-enzyme.
These co-enzymes are generally metal ions like zinc, magnesium, sodium, potassium,
etc.
Inversion of consumption
C12H12O11 + H2O
C6H12O6
2(C6H10O5)n + nH2O
C6H12O6(aq) + C6H12O6(aq)
2C2H5OH(aq) + 2CO2(g)
nC12H22O11(aq)
C12H22O11 + H2O
2C6H12O6
(NH2)2CO + H2O
Milk
2NH3 + CO2
Curd
Proteins
Amino acids
Characteristics:a. Enzymes have very high efficiency(106)
b. Very small amount of enzyme is reactive.
c. Very specific in nature.
d. Enzymes are active at optimum temperature and PH.
 Temperature at which enzyme activity is maximum called optimum temp [37
0C] around PH =7.
e. The activity of most enzyme catalyst is closely regulated.
Mechanism of enzyme Activity:The reactant molecule (substrate) binds itself to the active site on the surface of
enzyme. The molecules of reactants fit into these cavities (active site) like key fit in a
98
specific lock. This active binding results in the formation of enzyme substrate complex
(activated complex).
I. E + S
ES
II.
ES
EP
III.
EP
E+P
Product
1. Binding of enzyme with substrate to form enzyme substrate.
E+S
ES
2. Enzyme substrate converted into enzyme product.
E+S
ES
3. Product gets released from surface of enzyme.
EP
E+P
Applications of enzymes:Enzymes are used in the Fermentation of Carbohydrates
Sucrose
Glucose + Sucrose
Glucose
Ethyl Alcohol + CO2
DIFFERENCE BETWEEN A TRUE SOLN, COLLOIDAL AND SUSPENSION
TRUE SOLUTION
COLLOIDAL
SUSPENSION
Particle size <1nm
Between 1nm – 1000nm >100nm. i.e. >10-6
( <10 A0 )
(10-9 – 10-6)
Particles are invisible
Scattering of light by
Visible to naked or
particles is observed
under a microscope.
under ultraMay or may not show
microscope(Tyndale
Tyndale effect.
effect)
Particles do not settle
Do not settle down
Settle down on standing
down
Appear clear and
Translucent
Opaque
transparent
Do not show Brownian
Show Brownian
Show Brownian
movement
movement
movement
Homogeneous
Appear homogeneous
Heterogeneous
but heterogeneous
99
Brownian movement: - Movement of particle in zigzag form when they collide with
solvent particles.
Tyndale effect: - When a beam of light passes through colloidal solution then scattering
of light takes place, it is called Tyndale effect.
Classification of colloidal: -
True
Solution






Solute
Collodial
System
Solution
Dispersion
Dispersal
Medium
Dispersed phase means the substance means the substance distributed in the
dispersion medium in the form of colloidal particles.
Dispersion medium in which the substance is dispersed in the colloidal particles.
The colloidal system thus obtained is sometimes called a colloidal dispersion or
disperse phase.
Depending upon the nature of interaction between the particles of dispersed
phase and dispersion medium, Colloids are of two types:
1. Lyophilic colloids
2. Lyophobic Solution
Lyophilic colloids:- (liquid loving) – Substances like gum, gelatin, starch, rubber etc.
which when mixed with suitable liquids as the dispersion medium directly form colloidal
solution called lyophilic and solutions formed called lyophilic solutions of water take as
dispersion medium, solution called hydrophilic solution.
Lyophobic solution:- (liquids hating) – In this type of solutions, dispersion phase has
very little affinity for the dispersion medium. These solutions are less stable than
hydrophilic solutions.
E.g. Solutions of metals and their insoluble compounds like sulphides and oxides.
If water is taken as dispersion medium, solution is called hydrophilic.
100
DIFFERENCE BETWEEN LYOPHOBIC &
LYOPHOBIC
They are irreversible solutions
Less stable and get coagulated, by
heating or by agitating
Not much hydrated
Cannot prepared directly, prepared
by special method only
Viscosity is almost same as that of
the medium
Surface tension is nearly same as
that of dispersion medium
Solution generally formed by metals,
their sulphides etc.
(inorganic substances)
They carry either +ve or-ve charge
HYOPHILIC SOLUTION
HYOPHILIC
Reversible solutions
Quite stable, are not easily
coagulated by electrolysis
Highly hydrated
Prepared easily by directly mixing
with the liquid dispersion medium
Viscosity is higher than that of
medium
Surface tension is usually lower than
that of the dispersion medium
Usually formed by organic
substances like starch, gum, proteins
etc.
Carry no charge
CLASSIFICATION – DEPENDING UPON MOLECULAR SIZE
Types of colloids:a. Multi molecular colloids:-When on dispersion of a substance in the dispersion
medium, a large number of atoms or smaller molecules of the substance(< 1mm)
aggregate together to form species having size in colloidal range, the species thus
formed are called multimolecular colloids. E.g. a gold solution consists of particles
of various sizes which are clusters of various gold atoms.
A solution of sulphur consists of colloidal particles which are aggregates of S8
molecules.
b. Macro molecular colloids:- These are the substances which on dissolution form
solution in which the dispersed phase have the size of particles of colloidal range.
Such substances are called macromolecular colloids. E.g. colloidal solution of
starch, protein, cellulose, carbohydrates etc. These solutions are quite stable &
versatile with True solution.
Manmade: Polythene, Nylon, synthetic rubber, etc.
c. Associated colloids:- These are substances which when dissolved in the medium
behave as normal electrolyte at low concentration but behave as colloid at higher
concentration.
Soap solution at low concentration act as electrolyte but at higher concentration
is colloid.
The aggregated particles thus formed are micelles. The formation of micelles takes
place only above a particular temp called Kraft Temp (Tk) and above a particular
concentration called critical micelle concentration (CMC).
101
Kraft Temperature:-The minimum temperature at which micelle formation takes
place called Kraft Temperature.
CMC (Critical Micelle Concentration):- The concentration at which micelle
concentration takes place.
Mechanism of micelle formation:- Micelle are generally formed by amphiphilic
molecules which have both lyophilic as well as lyophobic ends.[e.g. Soap solution].
Such molecules are thrown as surface active molecules or surfactant molecules.
 Soap sodium stearate C11H35COO-Na+
C17H35 - long hydrocarbon part is hydrophobic tail.
COO- is hydrophilic head.
Concentration below CMC [ 3 x 10-3 mol L-1] act as good electrolyte & ionize as
C17H35COO- & Na+ ions.
As concentration increases CMC, hydrophobic part
starts receding away from the solvent and are
made to approach each other & COO- part interact
to water molecule. This ultimately leads to the
formation of cluster a large having dimension of
colloidal particles. In each such cluster a large
number of stearate gaps are clumped together in a
spherical manner, so that their hydrocarbon parts
interact with one another but COO- part remain
projected in water Na+.
Cleansing action of soap:Suppose any oil patch is sticking of cloth. When soap dissolves in water, it dissociated
into two parts. The stearate ions array around oily patch and hydrophilic parts project
outside the grease/oily patch/droplet.
As hydrophilic part is polar can interact with water molecules, present around oily
droplet as a result oily droplet pulled away from surface of cloth into water to form
ionic micelle which is then washed away with the excess of water.
Bredig’s Method [Electro-disintegration]
This method is mainly
used for the preparation of
metal solution. The metal
whose solution to be
prepared is made as two
electrodes immersed in
dispersion medium.
Dispersion medium kept in
dispersion medium.
Dispersion medium kept in
ice cold bath. An electric spark is introduced between the electrodes. The metal
102
vapourises and condensed immediately to form colloidal solution. The colloidal solution
is unstable therefore a small amount of stabilizer is added.
Peptization Method:A process of converting a freshly prepared ppt. into colloidal form by shaking it with
dispersion medium in the presence of small amount of electrolyte.
The electrolyte used for this purpose is called peptizing agent.
Fe(OH)3 + FeCl3
Fe(OH)3Fe3+ (colloidal solution)
Fe(OH)3 + FeCl3
Colloidal particles of Fe(OH)3
Cause of Peptization:- As
electrolyte is added to a freshly
precipitated substance,
the particles of ppt. preferentially adsorb
one particular type of ions of electrolyte (ions common with the precipitated
substance). As a result they get dispersed due to repulsion. It gives particles colloidal
size.
Electrophoresis:- The phenomenon/process
of moment of the colloidal particles towards
the opposite electrode by passing electric
current is called Electrophoresis.
Emulsions:- Emulsions are kind of colloidal
solutions in which both dispersion phase and
dispersion medium are liquids. Any two
immiscible liquids form an emulsion.
If two immiscible liquids do not mix well the
emulsion is unstable and it require a substance for stability called Emulsification.
Types of Emulsification:i.
Oil in water O/W: In this type of emulsion oil acts
as
dispersed phase and water act as dispersion
medium. E.g. milk is emulsion of soluble fat in
water.
ii.
Water in Oil W/O: In the case water acts as
dispersed phase and oil act as dispersion medium.
Identification of Emulsion:I. Dye Test :- To the emulsion some oil soluble dye added, if background
becomes coloured, the emulsion is O/W.
II.
Dilution Test :- If emulsion can be diluted with H2O, it shows that H2O is
dispersion medium & emulsion is O/W. If water form a separate layer
emulsion is W/O.
Applications
 Cleansing action of soap
 Digestion of fats, (fat emulsify by alkaline solution)
 Metallurgical process(both floatation process)
 Disinfectants.
103
 Building roads. Emulsion asphalt & H2O.
De-Emulsification:- It is a process of decomposing an emulsion back into its
constituent’s liquids. It can be done by different methods like centrifugation, filtration,
boiling, freezing and by addition of some chemicals.
Gels:- Gel is a colloidal system in which a liquid is dispersed in a solid. The process of
gel formation called gelation. Gels are of two categories.
a. Elastic Gels:- Gels which have the property of elasticity
e.g. gelatin, agar-agar, starch.
b. Non-Elastic Gel:- Gels which do not have property of elasticity
e.g. silica gel.
Application of Colloids: Purification of drinking water:- The water obtained from natural sources often
contains suspended impurities. Alum is added to such water to coagulate the
suspended impurities and make water for drinking purpose.
 Rubber Industry:- Latex is a colloidal solution of rubber particles which are –vely
charged. Rubber is obtained by coagulation of latex.
 Industrial products:- Paints, inks synthetic plastics, rubber, graphite lubricants,
cement etc. are all colloidal solutions.
Important Questions
1. Why substances like Pt. & Pd. are used to carry electrolysis of aqueous
solution?
Ans: They absorb H2 gas on the electrolysis of aqueous solution.
2. Why Physiorption decrease with increase temperature?
Ans:
3.
Why
powdered substances are more effective adsorbed than crystalline forms?
Ans: Due to more surface area. Surface area  adsorption.
4. Why hydrolysis of ester slow at beginning & than become fast?
Ans: RCOOR’ + H2O
RCOOH + R’OH
Acid produced
in the reaction acts as a catalyst (autocatalyst). In the case of homogenous auto-catalyst , rate increases with passage
of time. Rate=K[CH3COOC2H5][CH3COOH].
5. What is the role of desorption in the process of catalysis?
Ans: In heterogeneous catalysis, the products remain adsorbed on the
of catalyst.
104
surface
Desorption makes the surface of the solid catalyst free for fresh
adsorption of the reactants on the surface.
6. What modification can you suggest in Hardy-Schulz law?
Ans: According to this law, greater is valency of ion, stronger is its power to
coagulate colloidal solution.
-Here not only valency of ion but concentration of ion should be considered.
7. Why adsorption always exothermic?
Ans: In adsorption surface energy of adsorbent decreases which appears as heat.
Therefore, adsorption is always an exothermic process.
When a gas adsorbed on the surface of a solid, its energy decreases.  ∆S=-ve,
i.e. ∆G=T∆S, for spontaneous process ∆G=-ve.
Here if ∆S=-ve. -T∆S is +ve. Therefore if ∆H=-ve, then ∆G=-ve.
Hence adsorption is always exothermic.
8. Explain what is observed:
i.
When a beam of light is passed through a colloidal solution.
Ans: It appears bright at night angle from the incident beam. It is called
Tyndall effect, it takes place due to scattering of light by colloidal particles.
ii.
An electrolyte NaCl is added to ferric oxide solution.
Ans: The =ve charged particles of Fe(OH)3 get coagulated by the
oppositely charged Cl- ion of NaCl.
iii.
Electric current is passed through a colloidal solut5ion.
Ans: By passing electric current colloidal solution then charged colloidal
particles more towards oppositely charged electrodes, this phenomenon
called Electrophoresis.
105
Biomolecules
Biomolecules may be defined as complex lifeless chemical substances which form the
basis of life i.e. they not only build up living systems (creatures) but are also
responsible for their growth, maintenance and their ability to reproduce. E.g.
carbohydrates, fat, protein, nucleic acid, starch, glycogen etc.
Carbohydrates/Saccharides: General formula is Cx (H2O) y
Polyhydroxy aldehydes and ketones are called carbohydrates.
CARBOHYDRATES
Sugar (soluble in water & sweet in taste)
Non-Sugar
Oligosaccharide
Monosaccharide
(Reducing &
Non
hydrolysable e.g.
Glucose &
Fructose
Reducing
Maltose
Hydrolyzed
into 2
moles of DGlucose
Non-Reducing
Lactose
Hydrolyzed
into DGlucose & DGalactose
Polysaccharide
e.g. Starch & Cellulose
(Non-Reducing),
Insoluble in water and
do not have sweet taste
e.g. Sucrose
Hydrolyzed into Dglucose & D-Fructose
Mutarotation: Spontaneous change in specific rotation of an optically active
compound.
D-glucose
D-mannose
D-fructose
Anomers: Pairs of isomers having different configuration at C-1 e.g. α- and β- glucose
Inversion of Sugar
Sucrose on hydrolysis gives dextro-rotatory glucose and laevo-rotatory fructose. The
laevo rotation of fructose is more than dextro rotation of glucose; hence the resulting
solution becomes laevo-rotatory. Since, the hydrolysis of Sucrose is accompanied by a
change in the sign of optical rotation from dextro rotatory to laevo rotatory, the overall
106
process called inversion of Sugar and equimolar mixture of D-(+) glucose and D-(-)
fructose called invert sugar.
Protein:- High molecular mass complex biopolymers of amino acids.
Protein
Hydrolysis
hydrolysis
Polypeptides
α – aminoacids
Structure of 𝛂 –amino acids:
(R- Different side chains for different amino acids)
Zwitter ion – Dipolar ion carrying positive as well as negative charge
CH3
C
C COO
NH3+
Amino acids
R
Essential amino acids: Amino acids which cannot be synthesised in the body and must
therefore be obtained through diet are known as essential amino acids.
Example: - Lysine, Valine, leucine etc.
Non-essential amino acids: amino acids which can be synthesised in the body and are
not required in our diet are known as non-essential amino acids.
Example: - Glycine, Alanine, Proline etc.
Isoelectric Point: pH at which all amino acid molecules exist as neutral ion and do
not migrate towards any electrode.
pH< isoelectric point – Positive ion predominates which moves towards cathode
pH > isoelectric point – Negative ion predominates and it moves towards anode.
Peptide bond: Amide bond (-CONH-) which combine two amino acid molecules.
Structure of protein:
a)
Primary structure determine the sequence of amino acids
b)
Secondary structure :
𝛂- helix: H-bonding exist between >NH and >C=O groups of different
polypeptides
Chain gives right handed helical structure.
β – Structure: H – bonding exist between >NH and >C=O groups of different
polypeptide chains which may be parallel or antiparallel to each other.
c)
Tertiary structure: Overall folding of polypeptide chains gives two molecular
shapes
i.e. fibrous or globular.
107
Functions of proteins: 1.As enzymes catalyze biochemical reactions
2. As hormones regulate metabolic process
3. As antibodies protect the bodies against toxic substances
Nucleic Acids: Polynucleotides which are of two types
1) DNA
2) RNA
Nucleotide: Nucleoside + Phosphate
Nucleoside: Base + Sugar
Sugar
: Deoxyribose & Ribose
Denaturation of proteins: Disruption of the native conformation of protein due to
changes in
pH, temp and presence of salts. It disrupts higher structure of protein
without affecting the primary structure
Native Protein: Energetically most stable form of protein with definite configuration
and
biological activity.
Structure of DNA:
1. Double helical structure made up of two polynucleotide strands which are
antiparallel to each other.
2. Both stands are held together by H-bonds. A combine with T (A=T) and C with G
(C=G)
Genetic code: Hereditary information in an organism is encoded in form of special
sequence of bases on polynucleotide chain. It is coma less & universal. It is
degenerate.
Codon: Combination of 3 nitrogenous bases which code a particular amino acid
sequence in protein synthesis.
Replication: Duplication of DNA
Transcription: Process of Synthesis of RNA from DNA
Translation: Synthesis of protein by RNA
Mutation: Sudden chemical changes in DNA molecule which lead to synthesis of
protein with altered amino acids sequence.
Distinguish Between
DNA
RNA
1
contains deoxyribose sugar.
Contains ribose sugar
2
Contains A,T,G,C as bases.
Contains A,U,G,C as bases
3
Has double helix structure
Has single helix structure
108
4
Can replicate
Can not replicate
Globular Proteins
They consists of polypeptide chains,
partly of helical structure folded
around itself in 3-dimensions so as to
give a spherical shape.
Soluble in water.
Sensitive to small change in temp &
pH. e.g. Elbumines in egg, insulin.
Fibrous Proteins
They have long and || polypeptide
chains half together in linear thread
like form (fibers) which is held by
intermolecular H-bond & disulphide
bonds.
Insoluble.
Stable to moderate change in temp &
pH.
Enzymes
1
Co-Enzymes
They are biological catalyst require to catalyze
They are non protein part
biochemical reactions
which increases the activity
of enzymes.
2.
All enzymes are globular proteins e.g. Lactase,
Coenzymes are non
Invertase,
proteinous
e.g. Metal ions like Zn2+,
Mg2+,Mn2+ etc
Starch
Cellulose
1.
It is a white amorphous powder
It is a colourless amorphous powder
2.
It is a mixture of two polysaccharides ,
It is a straight chain polysaccharide
amylase(straight chain) & amylopectin
(branched chain)
3.
4.
Its solution in water gives blue colour with
Its solution in water does not give blue
iodine solution.
colour with iodine solution.
It has D-Glucose unit joined together by
It has D-Glucose unit joined by β-
α-glycosidic linkages
glycosidic linkages.
Amylose
1.
Amylopectin
Straight chain polymer madeup of α – D
Branched chain polymer of α – D
glucose units.
glucose units
109
2.
Gives blue colour with I2 solution
Does not give blue colour with I2
solution
3.
Having 100 – 300 D-glucose units
Having only 25 – 30 D-glucose units
Important reaction of Glucose
Evidence against open chain structure:i.
Aldehyde group present but glucose does not react with NaHSO3 & NH3.
ii.
Glucose does not give the Schiff’s Test & 2,4-DNP test for aldehyde.
iii.
Glucose does not react with Grignard reagents.
iv.
Glucose penta-acetate does not react hydroxyl amine, which shows that
aldehyde group is absent in glucose penta-acid.
v.
Glucose exist in two sterioisomeric forms (  & ).
vi.
An aqueous solution of glucose shows Mutarotation i.e. specific rotation
gradually decreases from +1100 to 52.50 in case of α-glucose and increase
from 19.70 to 52.50 in case of β-glucose.
 All these observations indicate that free aldehydic group is not present in glucose.
(II) The penta-acetate of glucose does not react with hydroxylamine. It indicate that
free aldehyde group is not present in glucose.
Avitaminosis: Condition of vitamin deficiency.
110
Hypervitaminosis: Excess intake of vitamins.
Fat soluble vitamin: Vitamin A, D, E, & K
Water soluble vitamins: Vitamin B & C
Hypoglycemic Factor: Insulin is also known as hypoglycemic factor & decrease
glucose
concentration in blood
Saturated Fatty acids: e.g. Palmitic acid & stearic acid
Unsaturate Fatty acid: e.g. Oliec acid, linolenic acid
S
Name of vitamin
Source
Deficiency disease
No.
1
Vitamin A
Fish oil particularly shark liver
Xerophthalmia i.e cornea hardening
(bright eye
oil, liver
of eye
Vitamin B1
Yeast, milk, green vegetables
Beri-beri (a disease of nervous
(thiamin)
etc
system)
Vitamin B2
Yeast, vegetables, milk, egg
(riboflavin)
white liver and kidney
Dark red tongue (glossitis),
dermatitis cheilosis (fissuring at
corners of mouth and lips)
Vitamin B6
Cereal, grams, molasses, yeast,
egg yolk and meat
Severe dermatitis, convulsions
vitamin)
2
3
4
(pyridoxine)
5
Vitamin H ( Biotin)
Yeast, liver, kidney and milk
6
Vitamin B12
Liver of ox, sheep, pig, fish etc
Dermatitis, loss of hair and
paralysis
Pernicious anaemia
7
Vitamin C
Citrus fruits, green vegetables
Scurvy
8
Vitamin E
Wheat germ oil, cotton seed oil
Sterility
9
Vitamin K
Cereals, leafy vegetable
Haemorrhagic conditions
10
Coenzyme Q10
Chloroplasts of green plants
Low order of immunity of body
against many diseases
Some Important Quetsions
Q1.How are vitamins classified? Name the vitamin responsible for coagulation of blood.
Ans: - vitamins are classified into two groups depending upon their solubility in water
or fat.
1. Water soluble vitamins: - These include vitamin B-complex and vitamin C.
111
2. Fat soluble vitamins: - These include vitamins A,D,E and K. They are stored in liver
and adipose. Vitamin K is responsible for coagulation of blood.
Q2.What are nucleic acids? Mention their two important functions.
Ans: - Nucleic acids are biomolecules which are found in the nuclei of all living cells in
form of nucleoproteins or chromosomes.
Nucleic acids are of two types; deoxyribonucleic acid (DNA) and ribonucleic acid
(RNA).The two important functions of nucleic acids are as follows:
1. DNA is responsible for the transmission of hereditary effects from one generation to
another. This due to unique property of replication, during cell division when two
identical DNA strands are transferred to the daughter cells.
2.DNA and RNA are responsible for synthesis of all proteins needed for the growth and
maintenance of our body.
Q3.What is meant by reducing "reducing sugars "?
Ans:- Carbohydrates which reduce Tollen's reagent or Fehling solution are called
reducing sugars. This is due to the presence of free aldehydic group e.g.,glucose.
Q4. What is invert sugar?
Ans :- Invert sugar is a equimolar mixture of (+) glucose (+52.50) and (-) fructose (92.40).It is obtained by the hydrolysis of sucrose .Since sucrose is dextrorotatory , but
after hydrolysis the mixture is laevorotatory, the mixture is called invert sugar.
Q5. Describe denaturation of protein.
Ans :- Denaturation :- When the natural conformation of a protein is disturbed in which
it does not change in primary structure but loses some or all of its secondary, tertiary
or quaternary structure is called denaturation of protein. As a result of denaturation,
the bioactivity of protein is lost. Protein denaturation is caused by change of
temperature, change of pH, addition of soluble salts like ammonium sulphate,
magnesium sulphate, addition of solvents like water, alcohol etc.
Q6. Describe Peptide linkage or Peptide bond.
Ans :- Peptide bond :- Proteins are condensation polymers of α-amino acids which are
the same or different α-amino acids are connected by peptide linkage. Chemically, a
peptide bond is an amide linkage formed between -COOH group of one α-amino acid NH3 group of other α-amino acid by loss of a water molecule.
Q7. What is the difference in structures of α-D(+) glucose and β-D(+) glucose?
Ans: These are Anomers and differ in the configuration of H-atom and –OH group
about C1 atom.
Q8. What is difference between a nucleotide and nucleoside?
Ans: A nucleoside contains a pentose sugar and base (purine or pyrimidine) while in
nucleotide a phosphoric acid component is also present.
Q9. Name two RNA molecules found in the cells of organism.
Ans: Messenger RNA (m-RNA) and Ribosomal RNA (r-RNA).
Q10. Give one use of enzyme streptokinase in medicines.
112
Ans: This enzyme can dissolve blood clots. It is useful medicine for checking heart
attacks due to blood clotting.
113
POLYMERS
Very short answer type questions:
1. Define the term homopolymerisation giving an example.
Ans. Polymerisation process involving only single monomeric species is known as
homopolymerisation. For example formation of polyethene from ethane.
2. Define the term Polymerisation.
Ans. It is a process of formation of a high molecular mass polymer from one or
more monomers by linking together a large number of repeating structural units
through covalent bonds.
3. Write the name and structure of the monomer of polymer: PVC.
Ans. The monomer of PVC is vinyl chloride.
CH2=CH-Cl
4. Write the monomer of polyethene.
Ans. Ethene (CH2=CH2)
5. What does the designation 6,6 means in the name of Nylon-6,6?
Ans. 6,6 means that both the monomers are of 6 carbon each.
6. Give an example of elastomers.
Ans. Natural rubber of Buna-S.
7. What is the primary feature for a molecule to be a monomer in a condensation
polymerisation reaction?
Ans. Monomer should have more than one functional groups.
8. Give the chemical name of Teflon.
Ans. Polytetrafluoroethylene.
9. What are the monomer units of Bakelite?
Ans. Phenol and Formaldehyde.
10.
What does SBR stands for?
Ans. Styrene Butadiene Rubber.
11.
Give an example of step growth polymer.
Ans. Terylene or Dacron.
12.
Name a polymer used to make cups for hot drinks.
Ans. Polystyrene is used to make disposable cups for hot drinks as it does not
become soft at a temperature near boiling point of water.
13.
Write the name and structure of monomer of natural rubber.
Ans. Isoprene or 2-Methylbuta-1,3-diene.
14.
What are the monomer units of terylene?
Ans. Ethylene glycol and terepthalic acid.
15.
What does PMMA stands for?
Ans. Polymethylmetaacrylate.
114
SHORT ANSWER TYPE QUESTIONS:
1. Write the structure of the monomer of each of the following polymers:
(i)
Nylon-6
(ii)
Teflon
(iii)
Neoprene
2. Differentiate between thermoplastic and thermosetting polymers. Give one
example of each.
Ans.
Thermoplastic polymer
Thermosetting polymer
1. They are linear polymer
1. They have three dimensional
without cross-linking.
network of covalent bonds
with cross links.
2. Upon heating these polymers
2.Upon heating they retain
become soft and becomes
their strength and does not
hard again on cooling.
become soft.
3. These
plastics
can
be 3.These plastics cannot be recycled.
recycled. They can be melted
and reused.
4.Example:
PVC,
Nylon, 4.Example: Phenol formaldehyde
Polystrene.
resin, urea formaldehyde resin.
3. Explain each of the following giving suitable example of each:
(i) Elastomers
(ii) Condensation polymers
(iii)Addition polymers
Ans. (i) Elastomers: Polymers which are rubber like with elastic properties are
called Elastomers.
(ii) Condensation polymers: The polymer formed by the repeated condensation
reaction between two same or different bifunctional monomeric units with the
elimination of simple molecules like water, ammonia etc., are called as condensation
polymers.
(iii)Addition polymers: The polymers formed by the repeated addition of monomer
molecules possessing double or triple bond are called addition polymers.
4. Mention two important uses of each of the following polymers.
(i)
Bakellite
(ii)
Nylon-6,6
(iii)
PVC
Ans. (i) Bakellite: For making electrical switches, sockets, plugs etc.
(ii) Nylon-6,6: For making bristles for brushes, ropes, in textile industries.
(iii)PVC For making insulation for electrical wires, for making foot wear, vinyl
flooring, buckets, toys etc.
115
5. Name the sub-groups into which the polymers are classified on the basis of
magnitude of intermolecular forces.
Ans. On the basis of magnitude of intermolecular forces polymers can be classified
as: a. Elastomers b. Fibers c. Thermoplastic polymers d. thermosetting polymers.
6. Write the name and structures of the monomers of the following polymers:
Buna-S, Dacron and Neoprene
Ans.
Polymer
Monomer
Buna-S
CH2=CH-CH=CH2 (Butadiene), CH2=CH-C6H5 (Styrene)
Dacron
HOCH2-CH2OH ( Ethylene Glycol), HOOC-C6H4-COOH
(Terepthalic acid)
Neoprene
CH2=C(Cl)-CH-CH3 (Chloroprene)
7. What are biodegradable polymers? Give two examples.
Ans. Polymers which can be easily degraded by micro-organisms present in nature
are called biodegradable polymers. Common examples are poly-β-hydroxybutyrateco-β-hydroxyvalerate (PHBV), Nylon-2-nylon-6.
8. Distinguish between ‘Chain growth polymerisation’ and ‘Step growth
polymerisation’ and give one example of each.
Ans.
Chain growth polymerisation
Step growth polymerisation
Only
addition
reaction
adds Any two molecular species present
repeating unit one at a time.
can react.
Monomers
are
unsaturated Monomers are generally bifunctional
compounds like alkenes, alkadienes organic compounds.
etc.
This type of polymerisation occurs Loss of simple molecule like water
without the loss of any small or ammonia takes place.
molecule.
Example: polymerisation of ethene Example: formation of nylon-6,6
to form polyethene.
from hexamethylene diamine and
adipic acid.
9. What is the structural difference between LDPE and HDPE?
Ans. LDPE (Low density polyethylene) has a branched structure hence does not
packs well. It is a transparent polymer with moderate tensile strength and
toughness. It is widely used as packing materials thin plastic bags, flexible pipes,
squeeze bottles etc.
HDPE (High density polyethylene) has a linear structure and hence packs well. It is
a translucent polymer with greater toughness, higher tensile strength than LDPE. It
is used in the manufacture of Buckets, mugs, toys, housewares etc.
116
10.
Give the differences between natural rubber and vulcanised rubber.
OR
How does vulcanisation change the character of the natural rubber?
Ans.
Natural rubber
Vulcanised rubber
It becomes soft at temperature It has a much wider useful
above
335K
and
brittle
at temperature range.
temperature below 283K
It shows high water absorption
It absorbs water to a much lesser
capacity.
extent.
It is highly soluble in non-polar
It is less soluble in non-polar
solvents.
solvents.
It is readily attacked by oxidising
It is not readily attacked by
agents.
oxidising agents.
11.
Why are the numbers 6,6 and 6 put in the names of nylon-6,6 and nylon6?
Ans. Nylon-6,6 is obtained from hexamethylene diamine and adipic acid. As each of
these monomers contains six carbon each. Nylon-6 is obtained by polymerisation of
caprolactum of its hydrolysis product, aminocaproic acid. As there is only one
monomer with 6 carbon atoms, the nylon so obtained is known as nylon-6.
12.
Distinguish between homopolymer and copolymer. Give example of each.
Ans. Homopolymer: A polymer obtained by polymerisation of only one type of
monomer is called as homopolymer. For example Nylon-6 is obtained from
caprolactum only.
Copolymer: A polymer obtained by polymerisation of two types of monomer is
called as copolymer. For example, Nylon-6,6 is obtained by copolymerisation of
hexamethylene diamine and adipic acid.
13.
How is Dacron obtained from its monomers? Give one use of Dacron.
Ans. Dacron is obtained by condensation polymerisation of ethylene glycol and
terepthalic acid with the elimination of water molecules. The reaction is carried out
at 420-440K in the presence of a catalyst consisting of mixture of zinc acetate and
antimony trioxide.
440K , Zn ( OCOCH 3) 2 , Sb 2 O 3



nHOCH2CH2OH + nHOOC-C6H4-COOH 420
[-O-CH2-CH2-O-CO-C6H4-CO-]
(Dacron)
14.
What are elastomers? Explain giving an example?
Ans. These polymers have weakest intermolecular forces of attractions. These are
highly elastic. These can be stretched due to weak intermolecular forces. They
regain their original shape when stress is removed due to cross links. For example:
Buna S.
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15.
How do you explain the term functionality of a monomer?
Ans. For a substance to act as a monomer, it must have at least two reactive sites
or bonding sites. The number of reactive sites in a monomer is referred to as its
functionality. In a olefin, the double bond can be considered as a site for two free
valencies. When a double bond is broken two single bonds become available for
bonding. Other bifunctional monomers are aminocaproic acid, pthalic acid, ethylene
glycol etc.
16.
What are natural and synthetic polymers? Give examples of each type.
Ans. Natural Polymers: Polymers which are found in natural living organisms, i.e.,
animals and plants are called natural polymers. For example: proteins, cellulose,
nucleic acids, resins etc.
Synthetic polymers: Man made polymers are called as synthetic polymers. For
example: Plastics, PVC, Nylon-6,6 etc.
17.
Describe the preparation of Buna-S frim its monomers.
Ans. Buna-S is a synthetic polymer obtained by polymerisation of butadiene and
styrene. Sodium is used as a polymerising agent.
heat

nCH2=CH-CH=CH2 + nC6H5CH=CH2 Na,
[-CH2-CH=CH-CH(C6H5)-CH2-]n
18. Explain the role of initiator in the free radical polymerisation of an alkene.
Ans. In free radical polymerisation of alkenes an initiator like benzoyl peroxide,
acetyl peroxide etc are used. The initiator undergoes homolytic fission in the
presence of heat or light to produce free radicals.
ission
Homolyticf

 2C6H5COO.
(benzoyl free radical)
As this redical reacts with another molecule of ethene, another bigger size free
radical is formed and chain propagates to form polyethene.
C6H5COO-OOCC6H5
19.
Explain the following terms with suitable examples:
(i)
Polyamides
(ii)
Polyesters
(iii)
Macromolecules
Ans. (i) Polyamides: Polyamides are polymers with amide linkage in their
chains. For example silk, wool, nylon-6, nylon-6,6.
(ii) Polyesters: Polymers with ester linkages in their chains are called as
polyesters. For example terylene, glyptal, etc.
Macromolecules:
Molecules of large size are generally called as
macromolecules. Most of these are polymers formed by small repeating structural
units.
20.
Write the uses of the following polymers:
(i)
Melamine
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(ii)
Teflon
(iii)
PAN
Ans. Melamine: it is used in the manufacture of unbreakable crockery.
Teflon: it is used in making oil seals, gaskets and for non-stick coated utensils.
PAN ( Polyacrylonitrile): it is used as a substitute for wool and for making
commercial fibres.
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CHEMISTRY IN EVERYDAY LIFE
VERY SHORT ANSWER TYPE QUESTIONS:
1. What is meant by narrow spectrum antibiotics?
Ans. Antibiotics which are mainly effective against gram positive of gram negative
bacteria are called as narrow spectrum antibiotics. For example Penicillin G is a
narrow spectrum antibiotic.
2. What is meant by broad spectrum antibiotics?
Ans. These are medicines which are effective against several different types of
micro-organisms. For example tetracycline, chloramphenicol etc.
3. What do you understand by the term ‘Chemotherapy’?
Ans. Use of chemicals for therapeutic effect is called as Chemotherapy.
4. Name a drug used in case of mental depression?
Ans. Equanil, barbituric acid derivatives such as seconal, neuronal are used as
antidepressents.
5. Give an example of Narcotic which is used as analgesic.
Ans. Morphine is a narcotic which is used as an analgesic i.e. to reduce pain.
6. What is the role of Bithional in toilet soaps?
Ans. Bithional acts as antiseptic when added to soap. Thus, it increases the
antiseptic properties of the soap.
7. Define antipyretic. Give two examples.
Ans. These are the chemicals which are used to bring down the body temperature
during high temperature. For example: Asprin, Paracetamol.
8. What is tincture of iodine? For what purpose it is used?
Ans. Alcohol water solution containing 2-3% iodine is known as tincture of iodine. It
is used as antiseptic.
9. What precautions will you take before administering penicillin to a patient?
Ans. The patient should be tested for sensitivity to penicillin before it is
administered.
10.
Give an example of non-ionic detergent.
Ans. Lauryl alcohol ethoxylate.
11.
Name the chemical responsible for antiseptic properties of Dettol.
Ans. Chloroxylenol and terpineol in a suitable solvent.
12.
To which class of drugs cimetidine and ranitidine belongs?
Ans.
Anti-histamines.
13.
Name a food preservative which is most commonly used by food
producers.
Ans. The most common food preservative used by food producers is sodium
benzoate.
14.
Describe antiseptics giving suitable examples.
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Ans. Antideptics are the chemical substances which prevent the growth of microorganism and may even kill them. Boric acid, hydrogen peroxide, bithional, is
common antiseptics.
15.
How are transparent soaps manufactured?
Ans. Transparent soaps are manufactured by dissolving soap in ethanol and
evaporating the excess solvent.
16.
What are artificial sweetening agents? Give two examples.
Ans. These are the chemical substances which are sweet in taste but do not add to
calorie intake. Some common examples are aspartame saccharin, alitame etc.
17.
Why the use of aspartame is is limited to cold food and drinks?
Ans. This is because aspartame decomposes at baking and cooking temperatures.
Hence aspartame can be used only with cold drinks and foods.
18.
Name the additive added to shaving soap to prevent rapid drying.
Ans. Glycerol.
19.
Name an analgesic which is also antipyretic.
Ans.
Asprin.
20.
Name an artificial sweetener which appear and taste like sugar and is
stable at cooking temperature.
Ans. Sucrolose.
SHORT ANSWER TYPE QUESTIONS:
1. What are biodegradable and non-biodegradable detergents? Give one example
of each.
Ans. Detergents which are easily degraded by micro-organisms present in the
rivers, ponds etc., are called as biodegradable polymers. For example sodiumdodecylbenzene sulphonate.
Non-biodegradable detergents are not easily degraded by mico-organisms.
For example sodium-1,1-dimethyldecylbenzene sulphonate.
2. State the reason for each of the following:
a. Soap do not work in hard water.
b. Synthetic detergents are better than soaps.
Ans. a. ordinary soaps are sodium or potassium salts of higher fatty acids like
stearic acid, palmitic acid. Hard water contains calcium and magnesium ions. These
ions form insoluble scum with soap and useless as cleansing agent.
b. soaps do not work in hard water but synthetic detergents are effective in
cleaning even in hard water.
3. Explain the following giving one example of each type:
a. Antacids
b. Disinfectants c. Enzymes
Ans. (i) Antacids: The chemicals which control the release of acid in gastric juices or
neutralises the excess acid in gastric juices are called antacids.
(ii) Disinfectants: The chemical substances which are used to kill microorganisms but
they cannot be applied on living tissues are called as disinfectants. For example 1%
solution of phenol in water is a disinfectant.
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(iii) Enzymes: Proteins which perform the role of biological catalysts and catalyse
biological reactions are called enzymes. For example, an enzyme maltase catalyses the
hydrolysis of maltose to glucose.
4. Explain the following giving one example of each type:
a. Tranquilizers
b. Food preservatives
Ans. a. Tranquilizers: These are chemicals used for the treatment of stress and mild
or even severe mental diseases. For example, veronal, serotonin, derivatives of
barbituric acid etc.
b. Food preservatives: Chemicals which are used to protect food items against
micro-organisms are called food preservatives. For example, sodium benzoate, sodium
metabisulphite etc.
5. Mention one use of the each of the following:
a. Ranitidine
b. Paracetamol c. Tincture of iodine
Ans. Ranitidine: Antihistamine /Antacid
Paracetamol: Analgesic, antipyretic
Tincture of iodine: Antiseptic
6. Describe the following giving one example:
a. Antifertility drugs
b. Antioxidants
Ans. Antifertility drugs: These drugs are used to prevent unwanted pregnancies. For
example Norethindrone
Antioxidants: Chemicals which prevent the oxidation of food stuff during
storage etc., are called as antioxidants. Vitamin C and E are natural oxidants. BHA (
Butylated Hydroxy Anisole) is a synthetic antioxidant.
7. what type of drug is Chloramphenicol, Equanil and asprin?
Ans. Chloramphenicol is an antibiotic, Equanil is a tranquilizer and asprin is an
analgesic.
8. Describe and illustrate with an example- a detergent.
Ans. Chemical substances which help in removing dirt by emulsifying oil of grease are
called as detergents. Detergents help in lowering the surface tension of the water.
Sodium salt of p-dodecylbenzene sulphonate is the most common example of synthetic
detergent.
9. How is antibiotic different from antiseptics? Give one example of each.
Ans. Antibiotic can be taken orally or injected into the body while an antiseptic can only
be applied to the skin, outside the body. Erythromycin is an antibiotic while
chloroxylenol is an antiseptic.
10. Sleeping pills are recommended by doctors to the patients suffering form
sleeplessness, but it is not advisable to take its doses without consultation with the
doctor. Why?
Ans. Not only the sleeping pills, in fact most of the drugs taken in the doses higher
than recommended may cause harmful effect and act as poison. Therefore, a doctor
should always be consulted before taking medicine.
11. How antiseptics differ from disinfectants. Give one example of each.
Ans.
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Antiseptic
1. Antiseptic can kill or prevent the
growth of micro-organism.
2. Antiseptics do not harm the living
tissues. Therefore they can be
applied on the skin.
3. Antiseptics are used for dressing
of wounds, cuts and in the
treatment of skin diseases.
4. Examples: Furacine, boric acid,
tincture of iodine.
Disinfectant
1. Disinfectants can kill of prevent
the growth of micro-organism.
2. Disinfectants are toxic to the
living tissues. Therefore they cannot
be applied on the living tissues.
3. Disinfectants are used for
disinfecting floors, toilets, drains,
instruments etc.
4. Examples: 1% aqueous solution
of phenol.0.4ppm chlorine.
12. Write the IUPAC name of Asprin, also write its usage. Why it not be taken on an
empty stomach?
Ans. IUPAC name of Asprin is 2-Acetoxybenzoic acid. In the stomach, it gets
hydrolysed to give salicylic acid , which on empty stomach can damage the lining of
the stomach.
13. Why are Cimitidine and ranitidine better antacid than sodium hydrogen carbonate?
Ans. Excessive intake of sodium hydrogen carbonate can make stomach alkaline and
trigger the production of even more acid. Also these treatments controls only the
symptoms of hyperacidity and not the cause. Cimetidine, ranitidine are better antacids
because they control the cause of hyperacidity. It is observed that hormone Histamine
stimulates the production of pepsin and hydrochloric acid in the stomach. Drugs like
cimetidine and ranitidine prevent the interaction of histamine with the receptors
present in the stomach wall. This results in production of lesser amount of acids and
controlling hyperacidity.
14. What are analgesic medicines? How they are classified when they are commonly
recommended for use?
Ans. The chemical substance which are used to relieve pain are called analgesics.
These are of two types:
(i)
Non-narcotic
(ii)
Narcotic
Non-narcotic drugs are effective in relieving skeletal pain preventing heart attack, viral
inflammation etc. narcotic drugs are recommended for the post-operative pains,
cardiac pain, terminal cancer etc.
15. Describe the following with an example:
(i) Edible colours
(ii) cationic detergents
Ans: Edible colours: These are synthetic or natural dyes used to impart attractive
colours to food, drugs etc. common examples are carotene, diazonium salt of
sulphanilic acid.
Cationic detergents: cationic detergents are quaternary ammonium salts of
amines with acetates/ chlorides/ bromides as anions. For example cetyltrimethyl
ammonium bromide.
16. Explain why diabetic patients are advised to take artificial sweetners instead of
natural sugar?
123
Ans. Artificial sweeteners like Aspartame, alitame are either not metabolised by the
body or do not produce carbohydrates like glucose when metabolised. Thus these
artificial sweeteners help in controlling the blood glucose level.
17. Account for the following:
(i) Asprin drug helps in preventing the heart attack.
(ii) Detergents are non-biodegradable while soaps are biodegradable.
Ans. (i) Asprin is found to help in prevention of heart attack because of its anti-blood
clotting action. The blood clots if formed, are mainly responsible for heart attack.
(ii) Common detergents which contain branched chain hydrocarbon part in their
molecules are non-biodegradable. This is because micro-organisms present in the
waste water cannot metabolise branched chain hydrocarbons. On the other hand soaps
contain linear chain hydrocarbons, which can be biodegraded more easily and water
pollution is prevented.
18. In chemotherapy, what are target molecules?
Ans. Drugs usually interact with biological macromolecules such as carbohydrates,
proteins, lipids, nucleic acids. These are called target molecules. Drugs are designed to
interact with specific targets so that these have the least chance of affecting other
targets. This minimises the side effects and localises the action of drug.
19. What are antihistamines? Explain how they act on human body?
Ans. Antihistamines are drugs which interfere with the natural action of hormone
histamine by competing with histamine for binding sites of receptors where histamines
exert its effect. They are also called as anti-allergic, drugs used to treat allergy.
Allergic reactions like skin rashes are caused due to liberation of histamine in the body
that is why these drugs are also called as anti-histamines.
20. Can sulphanilamide be classified as an anti-biotic? Give suitable reasons.
Ans. Yes, according to the modern definition of antibiotic, an antibiotic is a substance
produced wholly or partially by chemical synthesis, which in low concentration inhibits
the growth or destroy micro-organisms by intervening in their metabolic processes.
Sulphanilamide fits in the definition as it is chemically synthesized and destroys microorganisms.
*****************************
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HOTS
CHAPTER 10
HALOALKANES AND HALOARENES
1. Draw the structure of most stable carbocation of the formula C5H11+ .
CH3
Ans: CH3—CH2—C+--CH3.
2.(a)Convert Nitrobenzene to m-Bromoiodobenzene.
(b)Draw the structure of the product of the Bromination of following compounds in
presence of FeBr3.
(b)Because –NO2 deactivates the ring but OCH3 activates.
3(a) Identify ‘A’ and ‘B’
i)
CH3-C≡CH
NaNH2
Na/NH3
‘A’
‘B’
ii) CH3I
(b) Distinguish between CHCl3 and CHI3
Ans (a) A=But-2-yne
125
B= But-2-ene
(b) Dissolve them in 50% alcohol solution.Iodoform will give yellow ppt with
AgNo3 because C-I bond is weaker than C-Cl bond whereas CHCl3 does not react.
4. Write the major products:F
CH3ONa
(a)
NO2
CH3
‫׀‬
(b) CH3-C-CH2Br
‫׀‬
CH3
C2H5OH
(c)
COOC6H5
Br2
Ans (a)
FeBr3
OCH3
NO2
(b)
CH3
‫׀‬
CH3 - C - CH2 – CH3
‫׀‬
OC2H5
126
(c)
COOC6H5
Br
5. Two isomeric compounds ‘A’ and ‘B’ have same formula C11H13OCl. Both are
unsaturated give the same compound ‘C’ on catalytic hydrogenation and produce 4chloro-3-exthoxy benzoic acid on vigorous oxidation. ’A’ exits in two geometrical
isomers ‘D’ and ‘E’ but B does not. Identify A to E. Which has higher melting point out
of D and E
Ans
CH = CH – CH3
OC2H5
Cl
CH2 - CH = CH2
B=
OC2H5
Cl
127
C=
CH2CH2CH3
OC2H5
Cl
D=
H
Cl
H
C=C
CH3
OC2H5
(Cis)
E=
H
Cl
CH3
C=C
H
OC2H5
( trans)
E has higher Melting point due to symmetry.
6. Convert the following:(a) ethanol to propanenitrile
(b) propene to 1-propanol
(c) 1-Bromopropane to 2-Bromopropane
Ans a)
CH3CH2OH
PCl5
CH3CH2Cl
KCN
HBr/peroxide
KOH
128
CH3CH2CN
b)CH3CH=CH2
CH3CH2CH2Br
alc KOH
c) CH3CH2CH2Br
CH3CH2CH2OH
HBr
CH3CH=CH2
CH3CHBrCH3
7) An organic compound “A” having molecular formula C4H8 on treatment with dil
H2SO4 gives “B” which on treatment with conc. HCl and anhydrous ZnCl2 gives “C” and
on treatment with sodium ethoxide gives back “A”.Identify A, B,C.
Ans.A=But-2-ene
B=Butan-2-ol
C=2-Chlorobutane
129
CHAPTER 11
ALCOHOLS, PHENOLS AND ETHERS
1.
An organic compound ‘ A ‘ having molecular formula C3 H6 on treatment with aq.
H2SO4 give ‘B’ which on treatment with Lucas reagent gives ‘C’. The compound
‘C’ on treatment with ethanolic KOH gives back ‘ A’ .Identify A, B , C .
Ans.
2.
An organic compound A (C6H6O) gives a characteristic colour with aq. FeCl3
solution. (A) On reacting with CO2 and NaOH at 400k under pressure gives (B)
which on acidification gives a compound (C) .The compound (C) reacts with
acetyl chloride to give (D) which is a popular pain killer. Deduce the structure of
A,B,C & D.
Ans.
3.
An organic compound (X) when dissolved in ether and treated with magnesium
metal forms a compound Y. The compound, Y, on treatment with acetaldehyde
and the product on acid hydrolysis gives isopropyl alcohol. Identify the
compound X. What is the general name of the compounds of the type Y.
Ans. The compound X is CH3Br and Y is CH3MgBr The compounds of the type ‘Y’ are
called Grignard reagent.
130
4.
A compound ‘A’ with molecular formula C4H10O on oxidation forms compound ‘B’
gives positive iodoform test and on reaction with CH3MgBr followed by hydrolysis
gives (c). Identify A, B & C.
Ans. The compound ‘B’ is obtained by oxidation of C4H10O and gives positive iodoform
test and also reacts with CH3MgBr , it must be methyl Ketone , it must be methyl
ketone having four carbon atoms i.e, CH3COCH2CH3 .
This can be obtained by oxidation of 2 – butanol i.e , CH3 CH CH2 CH3 Therefore ,
the
reactions are:
5.
An aromatic compound (A) having molecular formula C6H6O on treatment with
CHCl3 and KOH gives a mixture two isomers ‘B’ and ‘C’ both of ‘B’ & ‘C’ give same
product ‘D’ when distilled with Zn dust. Oxidation of ‘D’ gives ‘E’ of formula
C7H6O2. The sodium salt of ‘E’ on heating with soda lime gives ‘F’ which may also
be obtained by distilling ‘A’ with zinc dust. Identify compounds ‘A’ to ‘F’ giving
sequence of reactions.
131
6.
Compound ‘A’ of molecular formula C5H11Br gives a compound ‘B’ of molecular
formula C5H12O when treated with aq. NaOH. On oxidation the compound yields a
mixture of acetic acid & propionic acid. Deduce the structure of A, B & C.
Ans.
132
The reactions are:
Ans.
133
CHAPTER 12
ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
1.
How are formalin and trioxane related to methanal?
Ans- Formalin is 40% aq solution of methanal whereas trioxane is trimer of methanal.
2.
How is acetone obtained from ethanol?
Ansi) CH3MgBr
Cu
Cu
CH3CH2OH
CH3CHO
CH3CHCH3
CH3CCH3
+
ii) H2O/H
575 K
575 K
OH
O
3.
Ans
Carboxylic acids do not give characterstics reactions of carbonyl group,explain
why?
In carboxylic acids,carbonyl group is involved in resonance
O
O-
R-C- O- H
R-C=O+-H
Therefore it is not a free group. But no resonance is possible in aldehydes and
ketones.
4.
Name on reagent used to convert phenol into salicyladehyde?
Ans. CHCl3/NaOH at 343 K.
5.
Why does benzoic acid not undergo Friedel Crafts reaction?
Ans. Due to the deactivation of the benzene ring by electron withdrawing effect of the
-COOH group.
134
CHAPTER 13
ORGANIC COMPUNDS CONTAINING NITROGEN
1.
Ans-
Why methylamine is a stronger base than trimethylamine?
Due to steric strain.Addition of protonincreases the crowing on the nitrogen
which increases the steric strain. Trimethylamine have the more
strain,consequently lesser tendency of protonation,so basicity is reduced.
2.
Why do amines dissolve in mineral acids?
Ans- The nitrogen atom in amines contains a lone pair of electron which it can donate.
Thus they accept a proton from mineral acid to form a salt which is water
soluble.
:
R-N H2 + H+Cl[ R-NH3]+ClWater soluble salt.
3.
Electrophilic substation in case of aromatic amines takes place more readily than
benzene?
Ans- Amino group( -NH2) in aniline is electron pushing group.It activates the benzene
ring due to +I effect, and it stabilized by resonance .Thus aromatic amines
undergo electrophilic substation at ortho and para position more readily than
benzene.
4.
How would convert methylamine into ethylamine?
AnsHI
CH3NH2 + HNO2
CH3OH
CH3I+KCN
methanol
iodomethane
CH3CH2NH2
Ethanamine
CH3CN
Ethanenitrile
LiAlH4
5.
Amines are basic substances while amides are neutral?
Ans- Due to presence of lone pair of electronson nitrogen of amines,they are basic in
nature where as nitrogen of amides acquires positive charge due to resonance with
carbonyl group which makes it neutral.
135
136