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KENDRIYA VIDYALAYA SANGATHAN

REGIONAL OFFICE, NEW DELHI

ADVISORS
SHRI S.S. CHAUHAN, DEPUTY COMMISSIONER, KVS RO, DELHI REGION.
SHRI K.C. MEENA, ASSISTANT COMMISSIONER, KVS RO, DELHI REGION.
SHRI GHANSHYAM PANDEY, ASSISTANT COMMISSIONER, KVS RO, DELHI REGION.
SHRI DEVENDER KUMAR, ASSISTANT COMMISSIONER, KVS RO, DELHI REGION.
CONVENOR, MS PRACHI DIXIT, PRINCIPAL KV AGCR COLONY, NEW DELHI.

STUDY MATERIAL PREPARED BY:

NAME OF TEACHER NAME OF KENDRIYA VIDYALAYA

DR. C.SRINIVAS, PGT-CHEMISTRY KV INA COLONY

DR. ANJALI GROVER ,PGT-CHEMISTRY KV SEC-8, R.K PURAM (SHIFT-1)
MS.ARCHANA TIWARI, PGT-CHEMISTRY KV SEC-8, R.K PURAM (SHIFT-1)
DR. RASHMI SRIVASTAVA , PGT-
CHEMISTRY K.V. PASCHIM VIHAR
MS PRANEETA VERMA , PGT-CHEMISTRY KV DR. RAJENDRA PRASAD
MS VARUANA PANDIT, PGT-CHEMISTRY KV DWARKSA SEC-5 (SHIFT-1)
MR ATAL MISHRA, PGT-CHEMISTRY K V NFC VIGYAN VIHAR (SHIFT-2)
MS SAPNA THAKUR , PGT-CHEMISTRY KV ANDREWS GANJ, (SHIFT-1)
MS DEEPSHIKHA, PGT-CHEMISTRY KV AGCR (SHIFT-1)
MS ANJU BALA, PGT-CHEMISTRY KV SEC-8, R.K PURAM (SHIFT-2)
MR. L.N GUPTA ,PGT-CHEMISTRY KV JANAKPURI (SHIFT-1)
MS RUCHIKA, PGT-CHEMISTRY KV MASJID MOTH,( SHIFT-2)
MR. VIKRANT BHARDWAJ ,PGT
CHEMISTRY KV NFC VIGYAN VIHAR(SHIFT-2)

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 Review team for study material
NAME OF TEACHER NAME OF KENDRIYA VIDYALAYA
MS. GEETA GURNANI KV AGCR COLONY (SHIFT-2)

MS. ASHA SINGH KV AGCR COLONY (SHIFT-1)

MS BHAWNA UPADHYAY KV DELHI CANTT NO-1(SHIFT-1)

MS POOJA SHARMA KV DELHI CANTT NO-1(SHIFT-2)

MS ANURADHA KV KHICHRIPUR(SHIFT-1)

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 TABLE OF CONTENTS
PAGE PAGE
S. No. TOPIC S. No. TOPIC
NUMBER NUMBER
1 Revised Syllabus UNIT -6 – Haloalkenes and
3-4 7 40-47
2024-25 Haloarenes
2 UNIT-7– Alcohols, Phenols
UNIT-1– Solution 5-11 8 48-58
and Ethers
3 UNIT-2– UNIT-8– Aldehyde, Ketones
12-18 9 59-70
ElectroChemistry and Carboxylic acid
4 UNIT-3–Chemical
19-25 10 UNIT-9– Amines 71-76
Kinetics
5 UNIT-4– d & f block
26-31 11 UNIT-10– Bio Molecules 77-84
Elements
6 UNIT-5– Coordination
32-39 12 Sample Paper-(1,2,3) 85-107
Compounds

CLASS XII (2024-25) (THEORY)

Time: 3 Hours 70 Marks
S.No. Title No. of Periods Marks
1 Solutions 10 7
2 Electrochemistry 12 9
3 Chemical Kinetics 10 7
4 d -and f -Block Elements 12 7
5 Coordination Compounds 12 7
6 Haloalkanes and Haloarenes 10 6
7 Alcohols, Phenols and Ethers 10 6
8 Aldehydes, Ketones and Carboxylic Acids 10 8
9 Amines 10 6
10 Biomolecules 12 7
Total 108 70
Unit I: Solutions
Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids,
solid solutions, Raoult's law, colligative properties - relative lowering of vapour pressure, elevation of boiling
point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative
properties, abnormal molecular mass, Van't Hoff factor.
Unit II: Electrochemistry
Redox reactions, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical
cells, Relation between Gibbs energy change and EMF of a cell, conductance in electrolytic solutions, specific
and molar conductivity, variations of conductivity with concentration, Kohlrausch's Law, electrolysis and law
of electrolysis (elementary idea), dry cell-electrolytic cells and Galvanic cells, lead accumulator, fuel cells,
corrosion.
Unit III: Chemical Kinetics
Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration, temperature,
catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations
and half-life (only for zero and first order reactions), concept of collision theory (elementary idea, no
mathematical treatment), activation energy, Arrhenius equation.
Unit IV: d and f Block Elements
General introduction, electronic configuration, occurrence and characteristics of transition metals, general
trends in properties of the first-row transition metals – metallic character, ionization enthalpy, oxidation

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 states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation,
preparation and properties of K2Cr2O7 and KMnO4.
Lanthanoids – Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and
its consequences.
Actinoids - Electronic configuration, oxidation states and comparison with lanthanoids.
Unit V: Coordination Compounds
Coordination compounds - Introduction, ligands, coordination number, colour, magnetic properties and
shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, Werner's theory, VBT, and
CFT; structure and stereoisomerism, the importance of coordination compounds (in qualitative analysis,
extraction of metals and biological system).
Unit VI: Haloalkanes and Haloarenes.
Haloalkanes: Nomenclature, nature of C–X bond, physical and chemical properties, optical rotation
mechanism of substitution reactions.
Haloarenes: Nature of C–X bond, substitution reactions (Directive influence of halogen in monosubstituted
compounds only).Uses and environmental effects of - dichloromethane, trichloromethane,
tetrachloromethane, iodoform, freons, DDT.
Unit VII: Alcohols, Phenols and Ethers
Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only),
identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special
reference to methanol and ethanol.
Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol,
electrophilic substitution reactions, uses of phenols.
Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.
Unit VIII: Aldehydes, Ketones and Carboxylic Acids
Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and
chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses.
Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties;
uses.
Unit IX: Amines
Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties,
uses, identification of primary, secondary and tertiary amines.
Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.
Unit X: Biomolecules
Carbohydrates - Classification (aldoses and ketoses), monosaccharides (glucose and fructose), D-L
configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen);
Importance of carbohydrates.
Proteins -Elementary idea of - amino acids, peptide bond, polypeptides, proteins, structure of proteins -
primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of
proteins; enzymes. Hormones - Elementary idea excluding structure.
Vitamins - Classification and functions.
Nucleic Acids: DNA and RNA.

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 UNIT- 1 SOLUTIONS

EXPRESSING CONCENTRATIONS OF SOLUTIONS

𝑴𝒂𝒔𝒔𝒐𝒇𝒔𝒐𝒍𝒖𝒕𝒆
Mass percentage: Mass of solute per 100g of solution 𝑴𝒂𝒔𝒔 % = 𝑻𝒐𝒕𝒂𝒍𝒎𝒂𝒔𝒔𝒐𝒇𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏X 100
Volume percentage: volume of solute per 100ml of solution. Volume % = X 100
Parts per million: parts of a component per million (106) parts of the solution.
.
𝑝𝑝𝑚 = .
𝑋 106
Mole fraction(x): so, 𝑋 = ;𝑋 = In binary solution XA+ XB= 1
Molarity (M) = moles of solute/ vol. of solution in litre M=
Molality(m) = moles of solute/mass of solvent in kg M=
Molality is independent of temperature, whereas molarity is a function of temperature, because volume
depends on temperature and mass does not.
Effect of Temperature Solubility of gases in liquids decreases with rise in temperature. It is due to this
reason that aquatic species are more comfortable in cold waters rather than in warm waters.
HENRY'S LAW (Effect of Pressure) It states that at a constant temperature the solubility of the gas in liquid
is directly proportional to the pressure of the gas above the surface of the liquid.
It also states that the partial pressure (p) of a gas in vapour phase is proportional to the mole fraction of the
gas (x) in the solution. P = KHX where KH is Henry's law constant.
APPLICATION OF HENRY'S LAW
 To increase the solubility of CO2 in soda water and soft drinks the bottle is sealed under high
pressure.
 To avoid bends, toxic effects of high concentration of nitrogen in the blood the tanks used
 Used by scuba divers are filled with air diluted with He.
RAOULT'S LAW :- It states that :
i. For a solution of volatile liquid , the partial vapour pressure of each component of the solution is
directly proportional to its mole fraction of each component
present in solution.
P A = P 0A X A , PB= P0B XB
The total pressure is equal to sum of partial pressure. Ptotal= PA+ PB

ii. For a solution containing non-volatile solute, the lowering in vapour pressure of the solution is
directly proportional to the mole fraction of the solute.
IDEAL SOLUTION The solution which obeys Raoult's law over the entire range of concentration when
enthalpy of mixing and vol. of mixing of pure component to form solution is zero.
Conditions FA- A, FB-B =F A-B I. P = P0 XA = P0B XB II. ΔHmix= 0 III. ΔVmix= 0
This is only possible if A-B interaction is nearly equal to those between A-A and B-B interactions. Ex:-solution
of n-hexane and n-heptane.
NON-IDEAL SOLUTION: The solution which do not obey Raoult's law over the entire range of concentrations.
Conditions: FA- A, FB-B not equal to F A-B
I. PA≠ P0A XAand PB ≠ P0BXB II. ΔHmix≠0 III. ΔVmix≠0
The vapour pressure of such solutions is either higher or lower than that predicted for Raoult's law.

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I. If vapour pressure is higher, the solutions shows positive deviation (A-B
interactions are weaker than those between A-A and B-B ). Ex: mixture of ethanol
and acetone.
PA > P 0AXA ; PB> P0B XB ΔHmix = Positive ;ΔVmix= Positive

II. If vapour pressure is lower, the solution shows negative deviation (A-B
interactions are stronger than those between A-A and B-B). Ex: mixture of
chloroform and acetone.
PA< P0A XA ; PB< P0B XB ΔHmix= negative, ΔVmix= negative

AZEOTROPE: Mixture of liquid having the same composition in liquid and vapour phase and boil at constant
temperature .Azeotropes are of two types :-
(a) Minimum boiling azeotrope :-The solution which shows a large positive deviation from Raoult's law . Ex
Ethanol – water mixture.
(b) Maximum boiling azeotrope :-The solution which shows large negative deviation from Raoult's law. Ex:
nitric acid – water mixture.
COLLIGATIVE PROPERTIES: Properties of ideal solution which depends upon no. of particles of solute but
independent of the nature of the particles are called colligative properties.
1. Relative lowering of vapour pressure: The relative lowering of vapour pressure of a solution containing a
non-volatile solute is equal to the mole fraction of the solute presentin the solution.
–
= X , 𝑋 =
–
For dilute solution, nB<<nA, hence nB is neglected in the denominator. =
2. Elevation in boiling point: Since the addition of a non-
volatile solute lowers the vapour pressure of the solvent, the
vapour pressure of a solution is always lower than that of the
pure solvent and hence it must be heated to a higher
temperature to make its vapour pressure equal to
atmospheric pressure.
ΔTb= kbm Where ,ΔTb= Tb– T0b, 𝐾 = Molal elevation
constant or Ebullioscopic Constant
m= Molality of solution. 𝑀 =

3. Depression in freezing point :Since the addition of a non-

volatile solute lowers the vapour pressure of the solvent, the
vapour pressure of a solution is always lower than that of the pure
solvent and hence it must be freeze to lower temperature to
make its vapour pressure equal to atmospheric pressure.
ΔTf = Kf m where, ΔTf = T0f – Tf, Kf = Molal depression
constant or Cryoscopic constant
m = Molality M= kf 1000 WB /ΔTf WA

M=

4. Osmotic pressure-The excess pressure that must be applied to a solution side to prevent osmosis i.e. to
stop the passage of solvent molecules into it through semi-permeable membrane is called osmotic pressure.
Π= CRT
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 Π= ( n = no. of moles; V= volume of solution(L), R=0.0821L atm mol–1 ; T= temperature in kelvi
ISOTONIC SOLUTION Two solutions having same osmotic pressure and same concentration are called
isotonic solutions. Π1= Π2 also C1 = C2
HYPERTONIC SOLUTION: Hypertonic solution have higher osmotic pressure than the other solution.
HYPOTONIC SOLUTION: hypotonic solution have lower osmotic pressure than the other
solution.0.91%(mass/volume) of sodium
chloride is isotonic with fluid present inside
blood cell.
REVERSE OSMOSIS: The process of movement
of solvent through a semi permeable
membrane from the solution to the pure
solvent by applying excess pressure (P> π) on
the solution, side is called reverse osmosis.
This principle is also applied for Desalination of water and SPM used is cellulose acetate.
VAN'T HOFF FACTOR (i)
i = normal molecular mass / observed molecular mass
i = observed colligative properties / calculated value of colligative properties
i = Total number of moles of particles after association or dissociation / Total number of moles of particles
before association or dissociation i< 1 (for association) i >1 (for dissociation)
MODIFIED FORMS OF COLLIGATIVE PROPERTIES
–
(1) =i (2) ΔTb = iKb m (3) ΔTf = iKf m (4) Π= iCRT
(MULTIPLE CHOICE QUESTION) MCQ
1.Sprinkling of salt helps in clearing the snow-covered roads in hills. The phenomenon involved in the
process is
(a) lowering in vapour pressure of snow (b) depression in freezing point of snow
(c) melting of ice due to increase in temperature by putting salt (d) increase in freezing point of snow
2. The system that forms maximum boiling azeotrope
(a) Acetone-chloroform (b) ethanol-acetone(c) n-hexane-n-heptane (d) carbon disulphide-acetone
3. Which has the lowest boiling point at 1 atm pressure?
(a) 0.1 M KCl (b) 0.1 M Urea (c) 0.1 M CaCl2 (d) 0.1 M A1Cl3
4. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution
is _____________.
(a) the same (b) about twice (c) about three times(d) about six times
5. The values of Van’t Hoff factors for KCl, Al2(SO4)3 and K2SO4, respectively, are
(a) 2, 2 and 5 (b) 2, 5 and 3 (c) 1, 1 and 2 (d) 1, 5 and 3
Answer: 1-(b), 2-(a), 3–(b), 4-(c), 5–(b)

ASSERTION -REASON TYPE QUESTIONS

A statement of assertion is followed by a statement of reason. Mark the correct choice from the options
given below:
(a) Both assertion and reason are true, and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true, but reason is false.
(d) Assertion is false, and reason is true.
1. Assertion : When NaCl is added to water a depression in freezing point is observed.
Reason : The lowering of vapour pressure of a solution causes depression in the freezing point.
2. Assertion : Osmosis does not take place in two isotonic solutions separated by semi-permeablemembrane.
Reason : Isotonic solutions have same osmotic pressure.

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3. Assertion : Azeotropic mixtures are formed only by non-ideal solutions and they may have boiling points
either greater than both the components or less than both the components.
Reason : The composition of the vapour phase is same as that of the liquid phase
of an azeotropic mixture.
4. Assertion : 1 M solution of NaCl has greater osmotic pressure than 1 M solution of glucose atsame
temperature.
Reason : In solution NaCl dissociates to produce more number of particles.
5. Assertion : If one component of a solution obeys Raoult’s law over a certain range of composition, the
other component will not obey Henry’s law in that range.
Reason :Raoult’s law is a special case of Henry’s law.
Answers: 1 – a, 2 – a, 3 – b, 4 – a, 5 – b.
2 MARK QUESTIONS
Q1. Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a
solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
Answer: (i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or
blood cells. When blood cells are placed in this solution, water flows out of the cells and they shrink due to
loss of water by osmosis.
(ii) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium chloride solution or blood cells.
When blood cells are placed in this solution, water flows into the cells and they swell.
Q2. How is that measurement of osmotic pressure more widely used for determining molar masses of
macromolecules than the elevation in boiling point or depression in freezing point of their solutions?
Ans. The osmotic pressure method has the advantage over elevation in boiling point or depression in freezing
point for determining molar masses of macromolecules because
1. Osmotic pressure is measured at the room temperature and the molarity of solution is used instead of
molality.
2. Compared to other colligative properties, its magnitude is large even for very dilute solutions.
Q3. State Henry's l aw. What is the significance of KH ?
Ans. Henry's Law: It states that “the partial pressure of the gas in vapour phase (p) is directly proportional to
the mole fraction of the gas (x) in the solution” , and is expressed as : p=KH X where, KH is the Henry's Law
constant
Significance of KH : Higher the value of Henry's law constant KH ,the lower is the solubility of the gas in the
liquid .
Q4. Calculate the freezing point of a solution containing 60g of
glucose (Molar mass = 180gmol-1) in 250g of water. (Kf of water =
1.86 K kg mol-1).
Ans. Given
Q5.(a) Why is an increase in temperature observed on mixing
chloroform and acetone?
(b) Why does sodium chloride solution freeze at a lower
temperature than water?
Ans: (a) The bonds between chloroform molecules and molecules
of acetone are dipole-dipole interactions but on mixing, the chloroform and acetone molecules, they start
forming hydrogen bonds which are stronger, bonds resulting in the release of energy. This gives rise to an
increase in temperature.
(b) When a non- volatile solute is dissolved in a solvent, the vapour pressure decreases. As a result, the
solution freezes at a lower temperature.

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 3 MARK QUESTIONS
Q1. Non-ideal solution exhibit either positive or negative deviations from Raoult's law. What are these
deviation and why are they caused? Explain with one example for each type.
Ans. When the vapour pressure of a solution is either higher or lower than that predicted by Raoult's law,
then the solution exhibits deviation from Raoult's law. These deviation are caused when solute – solvent
molecular interactions A – B are either weak or stronger than solvent – solvent A – A or solute – solute B – B
molecular interactions.
Positive deviations : When A – B molecular interactions are weaker than A – A and B – B molecular
interaction . For example, a mixture of ethanol and acetone.
Negative deviations: When A – B molecular interaction are stronger than A – A and B – B molecular
interaction. For example, a mixture of chloroform and acetone.
Q3. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water
at 25°C, assuming that it is completely dissociated
SOLUTION

Q4. (i) Identify which liquid will have a higher vapour pressure at 90°C if the boiling point of two liquids A
and B are 140°C and 180°C, respectively.
(ii) Cooking is faster in pressure cooker than in cooking pan.
(ii) Red blood blood cells are RBC shrink when placed in saline water but swell in distilled water. Give
reason.
Ans.
(i) Boiling point of liquid A & B is 140℃& 180℃ respectively.B.PA < B.PB
Liquid A is more volatile in nature.Hence, P0 of A >P0 of B.
So, liquid A has high vapour pressure.
(ii) Food cooks faster in a pressure cooker. Its because the pressure increases inside the cooker, which
also increases the boiling point of water. So, more heat is required to reach the boiling point and is
sufficient to cook food in a reduced time.
(iii) Red blood cells (RBCs) shrink in saline water because the saline solution is more concentrated than the
RBC, causing water to move out of the cell through osmosis. In distilled water, RBCs swell and may
eventually burst because distilled water is hypotonic to RBCs, causing water to move into the cell
through osmosis.
Q5. The molar freezing point depression constant for benzene is 4.90K kgmol–1. Seleniumexists as polymer
Sex.When 3.26 gm of Se is dissolved in 226 gm of benzene, the observed freezing point is 0.1120C lower
than for pure benzene. Decide the molecular formula of Selenium.(At.wt. of selenium is 78.8 g mol-1)
. . . . . .
Ans. 𝛥𝑇 = .
0.112 𝐾 = .
𝑀 = .
= 63𝑔 /𝑚𝑜𝑙

No. of Se atoms in a molecule=631g / mol/78.8 g/mol=8

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Therefore, molecular formula of Selenium = Se8

CASE – BASED QUESTIONS (4 Marks)

Read the given passage and answer the questions
Osmosis is the movement of a solvent across a semipermeable membrane toward a higher concentration of
solute. In biological systems, the solvent is typically water, but osmosis can occur in other liquids,
supercritical liquids, and even gases. When a cell is submerged in water, the water molecules pass through
the cell membrane from an area of low solute concentration to high solute concentration. For example, if
the cell is submerged in saltwater, water molecules move out of the cell. If a cell is submerged in freshwater,
water molecules move into the cell. When the membrane has a volume of pure water on both sides, water
molecules pass in and out in each direction at exactly the same rate. There is no net flow of water through
the membrane. Osmosis can be demonstrated when potato slices are added to a high salt solution. The water
from inside the potato moves out to the solution, causing the potato to shrink and to lose its 'turgor
pressure'. The more concentrated the salt solution, the bigger the loss in size and weight of the potato slice.
Answer the following Question
1. The preservation of fruits by adding concentrated sugar solution protects them against bacterial
action . Give reason? 1
2. Why is osmotic pressure of 1 M KCl solution is higher than 1M urea? 1
3. (i) Give one practical use of the reverse osmosis. 1
(ii) Name one SPM which can be used in reverse osmosis plant. 1
OR
3. A solution containing 15 g urea (molar mass = 60 g mol-1) per litre of solution in water has the same
osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol-1) in water. Calculate the
mass of glucose present in one litre of its solution.
For urea, concentration = 15/60 moles/lt., For glucose, concentration = w/180 moles/lt.
Answer:
1. In sugar solutions , bacterium present in the fruits loses water, shrivels and get destroyed.
2. KCl dissociates to give K+ and Cl- ion while urea is a molecular solid does not dissociates in ions.
3. (i) This can be used for desalination of water. (ii) Film of cellulose acetate
OR
V

so, 45 g of glucose is present in 1 L of solution.

5 MARK QUESTIONS
Q1.(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing
point of 10% glucose in water if the freezing point of pure water is 273.15K.Given:(Molar mass of sucrose
= 342 g mol-1) (Molar mass of glucose = 180 g mol-1)
(b) State Raoult's Law for a solution containing volatile components. How does Raoult's law become a
special case of Henry's Law?
Answer: Mass of sucrose (W)=10g Mass of water =90g

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Molecular weight of sucrose =342g mol−1 Molecular weight of water =18g mol−1
ΔTf=Kfm ΔTf=Tf(solvent)−Tf(solu on) ΔTf=273.15−269.15=4K
m= 10 x 1000/ 90 x 342 m=0.325 So ΔTf=Kf×m =40.325=12.31
For Glucose: ΔTf=Kf×m =12.3×10×1000/180×90=7.6
ΔTf=T (solvent) - T (Solution) So, T(solvent)=273.15−7.6=265.55K
(b) For a solution of volatile liquids ,Raoult's law states that the partial vapour pressure of each component
of the solution is directly proportional to its mole fraction present in solution, i.e.,
PA ∝XA Or PA = P0A X A
According to Henry's Law , the partial pressure of a gas in vapour phase (p) is directly proportional to mole
fraction (x) of the gas in the solution. i.e., p = KHX on comparing it with Raoult's Law it can be seen that partial
pressure of the volatile component or gas is directly proportional to its mole fraction in solutioni.e.; P ∝x
only the proportionality constant K differs from P0 . Thus, it becomes a special case ofHenry's law in which K
= Po.
ASSIGNMENT
Q1. Why is the vapour pressure of a solution of glucose in water lower than that of water?
Q2. State any two characteristics of ideal solutions.
Q3. Some liquids on mixing form “azeotropes”. What are azeotropes ?
Q4.What type of intermolecular attractive interaction exists in the pair of methanol and acetone?
Q5. 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.
Q6. State Raoult's law for the solution containing volatile components. What is the similarity
between Raoult's law and Henry's law ?
Q7. Boiling point of water at 750 mm Hg is 99.63oC. How much sucrose is to be added to 500g of
water such that it boils at 100oC ?
Q8. 18 g of glucose , C6H12O6 (Molar Mass = 180 g mol-1) is dissolved in 1 kg of water in a sauce pan.
At what temperature will this solution boil ? (Kb for water = 0.52 K kg mol-1, boiling point of pure
water = 373.15 K )
Q9. After removing the outer shell of the two eggs in dil. HCl, one is placed in distilled water and the
other in a saturated solution of NaCl . What will you observe and why ?
Q10. Calculate the mass of ascorbic acid (vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to
lower its melting point by 1·5°C. (Kf for CH3COOH) = 3·9 K kg mol-1)

12
 UNIT-2 ELECTROCHEMISTRY
Electrochemistry is that branch of chemistry which deals with the study of production of electricity from
energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-
spontaneous chemical transformation.
GALVANIC CELL
A galvanic cell is an electrochemical cell that
converts the chemical energy of a spontaneous
redox reaction into electrical energy.
Daniell cell having electrodes of zinc and copper
dipping in the solutions of their respective salts.
Function of salt bridge
1. It completes the circuit and allows the flow of current.
2. It maintains electrical neutrality on both sides. Salt-bridge generally contains
solution of strong electrolyte such as KNO3, KCl etc.
Electrode potential/Standardelectrode potential:
A potential difference develops between the electrode and the electrolyte, is called electrode potential.
When the concentrations of all the species involved in a half-cell is unity then the electrode potential is
known as standard electrode potential.
Cell potential is the potential difference between the two electrodes of a galvanic cell.
Electromotive force: The cell potentialis the difference between the electrode potentials (reduction
potentials) of the cathode and anode. It is called the cell electromotive force (e.m.f.) of the cell when no
current is drawn through the cell.
EMF of a cell: Ecell = Eright – Eleft
Nernst equation and its application to chemical cells:
For the electrode reaction: Mn+ (aq) + ne– ---> M(s)
The electrode potential at any concentration measured with
respect to standard hydrogen electrode can be represented by:
but concentration of solid M is taken as unity and we have
Ris gas constant (8.314 JK–1 mol–1), Fis Faraday constant (96487 C
mol–1), T is temperature in kelvin and [Mn+] is the concentration of
the species, Mn+.
For a general electrochemical reaction of the type: Equilibrium Constant from Nernst Equation:
E0cell = (2.303RT/nF) ln Kc

E0cell = (0.059/n) log Kc

Kc= equilibrium constant of the reaction

Electrochemical Cell and Gibbs Energy of the Reaction: ∆rG0 = – n F E0cell

We can calculate equilibrium constant by the equation: ∆rG0 = –RT ln K.
Conductance of Electrolytic Solutions:
Resistance The electrical resistance of any object is directly proportional to its length, l, and inversely
proportional to its area of cross section, A. That is, R ∝l /A or R = ρ (l/A)
The constant of proportionality, ρ (rho), is called resistivity (specific resistance). Its SI units are ohm metre
(Ω m) or ohm centimetre (Ω cm).
Conductance The inverse of resistance, R, are called conductance, G, and we have the relation:
G=1/R = A/ρ l = κ (A/l) [κ = 1/ ρ]
The SI unit of conductance is Siemens, represented by the symbol ‘S’ and is equal to ohm–1 (also known as
mho) or Ω–1.

13
Conductivity The inverse of resistivity, called conductivity (specific conductance) is represented by the
symbol, κ (Greek, kappa). κ = 1/ ρ The SI units of conductivity are S m–1.
Conductivity of Ionic Solutions: In the conductivity cell, the solution confined between the electrodes is a
column of length l and area of cross section A. The resistance of such a column of solution is given by the
equation: R= ρ l/A = l/ κ A
The quantity l/A is called cell constant denoted by the symbol, G*.
The cell constant, G*, is then given by the equation: G* = l/A = R κ κ = cell constant/R = G*/R
Molar conductivity: It is defined asthe conducting power of all the ions produced by dissolving one mole of
an electrolyte in solution. It isdenoted by the symbol Λm(Λ =lambda).
Molar conductivity = Λm = κ/c
units of Λmis in S m2 mol–1.
Variation of Conductivity and Molar Conductivity with Concentration:
Conductivity always decreases with decrease in concentration both, for weak and strong electrolytes. This
can be explained by the fact that the number of ions per unit volume that carry the current in a solution
decrease on dilution.
The conductivity of a solution at any given concentration is the conductance of one unit volume of solution
kept between two platinum electrodes with unit area of cross section and at a distance of unit length.
Molar conductivity increases with decrease in
concentration. This is because the total volume, V, of
solution containing one mole of electrolyte also increases.
At a given concentration, Λm can be defined as the
conductance of the electrolytic solution kept between the
electrodes of a conductivity cell at unit distance but having
area of cross section large enough to accommodate
sufficient volume of solution that contains one mole of the
electrolyte.
When concentration approaches zero, the molar
conductivity is known aslimiting molarconductivity and is represented by the symbol Λm0. The variation in
Λm with concentration is different for strong and weak electrolytes.
Strong Electrolytes For strong electrolytes, Λmincreases slowly with dilution and can be represented by the
equation: Λm = Λm0 – A c ½
If we plot Λm against c1/2, we obtain a straight line with intercept equal Λm0 and slope equalto ‘A’.
Kohlrausch’s law of independent migration of ions.
The lawα states that limitingmolar conductivity of an electrolyte can be represented as the sum ofthe
individual contributions of the anion and cation of the electrolyte. Λm0 = ᶛ+λ°+ + ᶛ–λ°–
Weak Electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations, for such
electrolytes, the change inΛm with dilution is due to increase in the degree of dissociation. At any
concentration c, if α is the degree of dissociation, then α= Λm / Λm0 Ka = c α2/(1-α)
ELECTROLYSIS: It is a process of decomposition of an electrolyte by the passage of electricity through its
aqueous solution or molten (fused) state of electric current.
Quantitative Aspects of Electrolysis Faraday’s Laws of Electrolysis:
(i) Faraday first law of electrolysis: The amount of chemical reaction which occurs at any electrode during
electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution
or melt).
w ∝ Q or w = ZQ, Where Z is called electrochemical equivalent. If a current of I amperes is passed for t
seconds, then Q = It so that, w = ZIt
(ii) Faraday second law of electrolysis: The amounts of different substances liberated by the same quantity
of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights
(Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).

14
For example, when same current is passed through two electrolytic solutions, containing copper sulphate
and silver nitrate connected in series, the weight of copper and silver deposited are:
(Weight of Cu deposited) / (Weight of Ag deposited) = (Eq. wt. of Cu) / (Eq. wt. of Ag)
Products of Electrolysis:
The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic
cell and their standard electrode potentials. Moreover, some of the electrochemical processes although
feasible, are so slow kinetically that at lower voltages these do not seem to occur. For the occurrence of such
reactions some extra potential (called over potential) is required.
(i) Electrolysis of molten sodium chloride: During the electrolysis of molten NaCl, the products of electrolysis
are sodium metal and Cl2 gas.
At cathode: 2Na+ + 2e– → 2Na (reduc on) At anode: 2Cl– →Cl2 + 2e– (oxidation)
The overall reaction: 2NaCl ---> 2Na + Cl2
(ii) Electrolysis of aqueous sodium chloride: During the electrolysis of aqueous sodium chloride solution, the
products are NaOH, Cl2 and H2.
At cathode: Na+ (aq) + e– →Na (s) E0cell = – 2.71 V H+ (aq) + e– →½ H2 (g) E0cell = 0.00 V
0
The reaction with higher value of E is preferred and therefore, the reaction at the cathode during electrolysis
is: H+ (aq) + e– →½ H2 (g)
But H+ (aq) is produced by the dissociation of H2O, i.e., H2O (l) →H+ (aq) + OH– (aq)
Therefore, the net reaction at the cathode: H2O (l) + e– →½H2 (g) + OH– (aq)
At anode: Cl– (aq) →½ Cl2 (g) + e– E0cell= +1.36 V 2H2O (l ) → O2 (g) + 4H+(aq) + 4e– E0cell= +1.23 V
The reaction at anode with lower value of E0 is preferred and therefore, water should be oxidised in
preference to Cl– (aq).
However, on account of over potential of oxygen, the reaction Cl– (aq) →½ Cl2 (g) is preferred.
Thus, the net reactions may be summarized as: NaCl (aq) + H2O (l) →NaOH(aq) + ½H2 (g) + ½Cl2(g)
Dry Cell:
A dry cell consists of a zinc container act
as anode and the cathode is a carbon
(graphite) rod surrounded by powdered
manganese dioxide and carbon. The space
between the electrodes is filled with
moist paste of ammonium chloride and zinc chloride.
Dry cell is commonly used in transistors and
clocks Mercury cell:
It consists of zinc – mercury Amalgam as anode
and a paste of HgO and carbon as the cathode.
The electrolyte is a paste of KOH and ZnO.
The cell potential is approximately 1.35 V and remains constant during its life as the overall reaction does
not involve any ion in solution whose concentration can change during its lifetime.
A secondary cell (lead storage battery) commonly used in automobiles and invertors. It consists of a lead
anode and a grid of lead packed with lead dioxide (PbO2) as cathode. A 38% solution of Sulphuric acid is used
as an electrolyte.
The cell reactions when the
battery is in use are  :

On charging the battery the

reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively.
Fuel cells: To convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into
electrical energy are called fuel cells.
Fuel cell using H2 and O2 produces electricity. In this cell, hydrogen and oxygen are bubbled through porous
carbon electrodes into concentrated NaOH solution. Catalysts like finely divided platinum or palladium metal
are incorporated into the electrodes for increasing the rate of electrodes reactions.
15
The electrode reactions are  :

Corrosion:
In corrosion, a metal is oxidised by loss of electrons to oxygen
and formation of oxides. Corrosion of iron (commonly known as
rusting) occurs in presence of water and air. The chemistry of
corrosion is an electrochemical phenomenon.
x
Corrosion of iron in atmosphere

We can write the reaction

Anode: 2 Fe (s) ----------- 2Fe2+ + 4 e – E0(Fe2+/Fe)= – 0.44 V
Cathode: O2 (g) + 4 H+ (aq) + 4 e– →2 H2O (l) E0 (H+/O2/H2O) = 1.23 V
The overall reaction being:
2Fe(s) + O2 (g) + 4H+ (aq) →2Fe2+ (aq) + 2 H2O (l) E0cell=1.67 V
The ferrous ions are oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of
hydrated ferric oxide (Fe2O3. x H2O) and with further production of hydrogen
MULTIPLE CHOICE QUESTION ANSWER
Q1.What is the observation when the opposing external applied potential to an electrochemical cell is
greater than the cell’s potential?
a) The electrochemical cell behaves like an electrolytic cell b) The electrochemical cell stops functioning
c) Only oxidation reactions occur in the cell d) Only reduction reactions occur in the cell
Q2. If limiting molar conductivity of Ca2+ and Cl– are 119.0 and 76.3 S cm2 mol-1, then the value of limiting
molarconductivityofCaCl2willbe
(a)195.3Scm2 mol-1 (b)271.6Scm2 mol-1
(c)43.3Scm2 mol-1 (d) 314.3 S cm2 mol-1.
Q 3. The emf of the cell: Ni / Ni2+ (1.0 M) // Au3+ (1.0 M) / Au (E° Ni2+/Ni = -0.25 V ; E° Au3+/Au = 1.5 V ) is
(a) 1.25 V (b) -1.25 V (c) 1.75 V (d) 2.0 V
Q 4. The charge required for the reduction of 1 mol of MnO4– to MnO2 is
(a) 1 F b) 3 F c) 5 F d) 6 F
Q5. What changes are observed in the specific conductance and molar conductivity repectively on diluting
the electrolytic solution?
a).Both increase b).Both decrease c).Decrease and increase d).Increase and decrease
ANSWER KEY 1). a 2). b 3). c 4).b 5). c
Given below are two statements labelled as Assertion (A) and Reason (R). Select the most appropriate answer
from the options given below:
a. Both A and R are true, and R is the correct explanation of A
b. Both A and R are true, but R is not the correct explanation of A.
c. A is true, but R is false.
d. A is false, but R is true.
Q1.Assertion:.EMF of the cell is the potential difference between the between electrode potentials of the
cathode and anode when no current is drawn through the cell.
Reason : Anode is kept on the right side and cathode on the left side while representing the galvanic cell.
Q 2.Assertion: A standard hydrogen electrode is also called reversible electrode.
Reason: Standard hydrogen electrode can act both as anode as well as cathode in an electrochemical cell.

16
Q 3. Assertion:Molar conductivity increases with decrease in concentration.
Reason: Conductivity always decreases with decrease in concentration
Q 4. Assertion:Batteries and fuel cells convert chemical energy into electrical energy.
Reason:The reactions carried out electrochemically are not energy efficient and more polluting.
Q 5. Assertion:Tarnishing of silver is not an example of corrosion
Reason:Development of green coating on copper is an example of corrosion
ANSWER KEY
1.c 2. a 3.b 4. C 5. D
SECTION B 2 mark Questions
Q1. How much time would it take in minutes to deposit 1.18 g of metallic copper on a metal object when a
current of 2.0 A is passed through the electrolytic cell containing Cu^(2+) ions?
Cu2+ions?[Cu=63.5g/mol,1F=96,500Cmol-1]
Ans.a 1.18 = 63.5 x 2.0 x t/ 2 x 96500
= 1.18 x2 x 96500/ 63.5 x 2.0 = 1793.23sec =1793.23/ 60 = 29.88min.
Q 2. Value of standard electrode potential for the oxidation of Cl– ions is more positive than that of water,
even then in the electrolysis of aqueous sodium chloride, why are Cl- ions oxidised at anode instead of water?
Ans: The oxidation reactions taking place at anode is: 2Cl– (aq) → Cl2 (g) + 2e– E0(oxid) = -1.36 V
2H2O (l) → O2 (g) + 4H+ (aq) + 4e– E0(oxid) = -1.23 V
Due to the overvoltage of oxygen (O₂), its liberation is kinetically slower than that of Cl– ions.
Therefore, Cl– ions are oxidised to Cl₂ gas.
Q 3. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A
increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Ans: Dilution of the electrolyte mainly decreases interionic attractive force, therefore Λm increases. In case
of strong electrolyte, it dissociated completely in solution so on dilution Λm increases to a small extent. On
the other hand, the electrolyte weak is ionised only to a small extent. Upon dilution, its ionisation or
dissociation increases largely. That is why; Λm for weak electrolyte is more higher on dilution.
From the available information, A is a weak electrolytes and B is a strong electrolyte.
Q 4.Write the cell reaction of a lead storage battery when it is discharged. How does the density of the
electrolyte change when the battery is discharged?
Ans: The cell reaction of a lead storage battery when it is discharged may be given as:
Pb(s) + 4H+(aq) + 2SO42- (aq) → PbO₂(s) + 2PbSO4(s) + 2H₂O(l)
Density of the electrolyte i.e. conc.H2SO4 solution decreases because of the dilution of electrolyte taking
place since water is formed as one of the products.
Q 5. Calculate Ecellfor the reaction given below, if E0cell = 3.17V
Mg (s) + 2Ag+ (0.0001M) Mg2+ (0.130M) + 2Ag (s)
Ans : Ecell = E0cell – (0.059/2) log {[Mg2+] / [Ag+]2 }
= 3.17 – 0.0295 log [(0.130)/ (0.0001)2] = 3.17 V – 0.21V = 2.96 V

SECTION C 03 Mark Questions

Q1. How does fuel cell operate? Why we prefer it over other conventional fuel cells? Write complete
reaction which takes place with respect to hydrogen-oxygen fuel cell.
Ans : Refer the content
Q 2a) Define limiting molar conductivity.
b) The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell
constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1.
17
Answer: a) When the concentration of the solution approaches zero, molar conductivity is called limiting molar
conductivity.
b) Given, Conductivity, κ = 0.146 × 10−3 S cm−1 Resistance, R = 1500 Ω
= Cell constant = κ × R 0.146 × 10−3 × 1500 0.219 cm−1
Q 3. Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and iffor
acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

Answer

Q 4. For the standard cell Cu(s)/Cu2+(aq) || Ag+(aq)/Ag(s)

(i) Identify the cathode and the anode as the current is drawn from the cell.
(ii) Write the reaction taking place at the electrodes.
(iii) Calculate the standard cell potential.
E°(Cu2+/Cu) = +0.34V, E° (Ag+/Ag)= +0.80 V
Ans (i) From the cell representation, Ag/Ag+ electrode is cathode and Cu/Cu+ electrode is anode.
(ii) At anode: Cu(s) ---->Cu2+(aq) + 2e- At Cathode: Ag++ 1e----->Ag (s)
Overall reaction is Cu(s)+ Ag+(aq) → Cu2+(aq)+2 Ag(s)
(iii) E0 =E0 –E0 =E0 –E0 =+0.80 V – (+0.34V) =+0.80V-0.34V =0.46V
cell cathode anode Ag+/Ag Cu2+/Cu

SECTION D
The following question is case-based question. It carries 4 (1+1+2) marks . Read the passage carefully and
answer the questions that follow.
Q 1. E° values for various ions are given in the table
Reaction Std reduction potential (E° value in Volts)
F2(g) + 2e →2F- 2.87
O2+ 4H++ 4e→2H2O 1.23
Zn2++ 2e→Zn(s) -0.76
K + +e →K(s) -2.93
Ni2++ 2e→Ni(s) -0.25
AgCl+e→Ag(s)+Cl- 0.22

When two half cells that differ in their E° values are coupled, redox reaction takes place and the energy
released can be directly used to produce electrical energy. Such cells are known as electrochemical cells. Ions
with more negative reduction potentials can reduce ions of less negative electrode potentials. Charge of one
mole electrons is known a one Faraday (app.96500C). Electroplating is an application of electrolysis, where
a metal is deposited over another metal.
a ) In the given table ,identify the strongest oxidising agent and strongest reducing agent.
b ) Calculate the amount of charge required to deposit 2mol Zinc from zinc chloride solution.
c) Why electrolysis of NaI gives I2whereas NaF gives O2 instead of F2.
Ans:a) Strongest oxidising agent :Fluorine ,Strongest reducing agent in the table :Potassium
b) For 1mol zinc 2mol electrons(2Faraday),that is 2x96500C charge is required .Hence for depositing 2mol
zinc 4F (4 x 96500C)charge is required.
c) I - ions has higher oxidation potentials than water. Hence Iodine gas is released. But F -ions have lower
oxidation potential than water hence water is easily oxidised to O2 gas
18
 SECTION E 5 Mark Questions
Q 1. a)Calculate ∆rG0 and log Kc for the reaction at 298K:
2Cr(s) + 3Fe2+ (aq) ----- 2Cr3+ (aq) +3Fe(s) Given: E0cell =0.30V
0
b) Using the E values of A and B, predict which is better for coating the surface of iron
[ E0( Fe2+/ Fe) = -0.44 V] to prevent corrosion and why? Given: E0( A2+/ A) = -2.37V : E0( B2+/ B) = -0.14V
Ans: a) Given: E0cell = 0.30V , F= 96500C /mol
2Cr(s) --------- 2Cr3+ (aq) + 6 e- 3Fe2+ (aq) + 6e- --------- 3Fe(s)
∆rG0 = – n F E0cell = -6 x 96500 x 0.30 = -173700 J/ mol or -173.7 KJ/mol
0
log Kc = n E cell/ 0.059 or log Kc = 6 x0.30 / 0.059 or log Kc = 30.5
b) A is better for coating the surface of iron .This is because E0 for A is more negative than that of Fe.
Therefore A has greater tendency for oxidation than Fe .Thus A will oxidise in preference to Fe and protect
it from corrosion.
-------------------------------------------------------------------------------------------------------

IMPORTANT FORMULAE OF ELECTROCHEMISTRY

1. NERNST EQUATION
E0cell = (0.059/n) ln Kc

2.Equilibrium constant from Nernst reaction

E0cell = (2.303RT/nF) ln Kc

3.Electrochemical Cell and Gibbs Energy of the Reaction:

∆rG0 = – n F E0cell
We can calculate equilibrium constant by the equation: ∆rG0 = –RT ln K.
4.Resistance R ∝ l /A or R = ρ (l/A)
5. Conductance G=1/R = A/ρ l = κ (A/l) [κ = 1/ ρ]
6.Conductivity κ = 1/ ρ R= ρ l/A = l/ κ A
The quantity l/A is called cell constant denoted by the symbol, G*.
The cell constant, G*, is then given by the equation: G* = l/A = R κ κ = cell constant/R = G*/R
7. Molar conductivity:

8. Kohlrausch’s law of independent migration of ions. Λm0 = ᶛ+λ°+ + ᶛ–λ°–

Weak Electrolytes if  is the degree of dissociation, then  = Λm / Λm0 ,Ka = c 2/(1-)
Faraday’s Laws of Electrolysis: w ∝ Q or w = ZQ, Where Z is called electrochemical equivalent. If a current of
I amperes is passed for t seconds, then Q = It so that, w = ZIt

19
 UNIT-3 Chemical Kinetics
Rate of a reaction ; It is defined as the change in concentration of a reactant or product in a unit time.It
can be expressed in terms of :
(i) The rate of decrease in concentration of any one of the reactants.
(ii) The rate of increase in concentration of any one of the products.
Consider a hypothetical reaction, R ---> P
One mole of the reactant R produces one mole of the product P. If [R]1 and [P]1 are the concentrations of R
and P respectively at time t1 and [R]2 and [P]2 are their concentrations at time t2 then, ∆t = t2 – t1
∆[R] = [R]2 – [R]1 ∆ [P] = [P]2 – [P]1
The square brackets in the above expressions are used to express molar concentration.
Rate of disappearance of R =- =- ….. (1)
Rate of appearance of P = = ………(2)
Since, ∆[R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to make
the rate of the reaction a positive quantity.
Equations (1) and (2) given above represent the average rate of a reaction.
Average rate depends upon the change in concentration of reactants or products and the time taken for that
change to occur.

Units of rate of a reaction: The units will be mol L-1s–1.

In gaseous reactions, the units of the rate equation will be atm s–1.
Rate Law and Rate Constant:
Consider a general reaction: aA + bB ----> cC + dD
Where a, b, c and d are the stoichiometric coefficients of reactants and products.
The rate expression for this reaction is Rate ∝ [A]x [B]y
Where exponents x and y may or may not be equal to the stoichiometric coefficients (a & b) of the reactants.
Above equation can also be written as Rate = k [A]x [B]y -d[R] /dt = k[A] x [B]y
This form of equation is known as differential rate equation, where k is a proportionality constant called rate
constant.
The equation, Rate = k [A]x [B]y which relates the rate of a reaction to concentration of reactants is called rate
law or rate expression.
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.
For example: 2NO (g) + O2 (g) ----> 2NO2 (g) Rate = k [NO]2 [O2] -d[R] /dt = k [NO]2 [O2]
Order of reaction: Rate = k [A]x [B]y
x and y indicate the rate is to the change in concentration of A and B.
Sum of these exponents, i.e., x + y in above equation gives the overall order of a reaction. Where as x and y
represent the order with respect to the reactants A and B respectively.

20
Order of reaction is defined as the sum of powers of the concentration of the reactants in the rate law
expression is called the order of that chemical reaction. . Rate ∝ [A]0
Order of a reaction can be 0, 1, 2, 3 and even a fraction.
A zero-order reaction means that the rate of reaction is independent of the concentration of reactants.
Units of rate constant: Zero order rate of reaction: mol L-1 s-1
First order rate of reaction: s-1
Second order rate of reaction: L mol-1 s-1
Order of a reaction is determined experimentally.
Molecularity of a reaction: The number of reacting species (atoms, ions or molecules) taking part in an
elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called
molecularity of a reaction. Molecularity of a reaction cannot be zero or fractional. Molecularity is a
theoretical concept. For examples:
NH4NO2 -----> N2 + 2H2O Unimolecular reaction
2HI (g) ----> H2 (g) + I2 (g) Bimolecular reaction
2NO (g) + O2 (g) -----> 2NO2 (g) Trimolecular reaction
Integrated rate equations:
Zero order reaction: Zero order reaction means that the rate of the reaction is proportional to zero power
of the concentration of reactants.
Consider the reaction, R -----> P Rate = -d[R] /dt = k [R]0 Rate = -d[R] /dt = k x 1
d[R] = -k dt Integrating both sides [R] = -k t +I (1)
Where, I is the constant of integration
At t = 0 the concentration of reactant [R] = [R]0
Where [R]0 is the initial concentration of reactant.
[R]0 = -k x 0 + I = I, Substituting the value of I in the equation (1)
[R] = -k t + [R]0 k ={[R]0 - [R]}/t
Variation in the concentration Vs time plot for a zero-order reaction
Example of Zero order reaction is the decomposition of gaseous ammonia.
2NH3 (g) ------> N2 (g) + 3 H2 (g)
Rate = k [NH3]0 = k
the rate of the reaction is proportional to the
First Order Reactions: The rate of the reaction is proportional to the first power of the concentration of the
reactant R. For example: R -----> P Rate = -d[R] /dt = k [R] d[R] /[R]= -k dt,
Integrating this equation and we get ln [R] = -kt + I ……..(1) At t = 0,
ln [R]0 = -k x 0 + I, I = ln [R]0, Substituting the value of I in the equation …….(1)
ln [R] = -kt + ln [R]0 k = {ln [R]0 - ln [R]} /t
Remember that, (log a – log b = log(a/b) k = (1/t) ln {[R]0/[R]}
ln {[R]/[R]0} = - kt, taking antilog both sides [R] = [R]0 e –kt
We know that, ln a = 2.303 log a k = (2.303/t) log {[R]0/[R]}
If we plot a graph between log [R]0/[R] Vs t, the slope is k/2.303 for first order
reaction 

First order gas phase reaction: A(g) -----> B(g) + C(g)

Total pressure pt = pA + pB + pC
A(g) -----> B(g) + C(g)
At t = 0 pi atm 0 atm 0 atm
At time t (pi – x) x atm x atm
pt = (pi – x) + x + x = pi + x x = (pt - pi)
pA = pi – x = pi - (pt - pi) = 2pi-pt k= (2.303/t) log [pi/ pA] k= (2.303/t) log [pi/(2pi-pt)]
Half-life of a reaction: The half-life of a reaction is the time in which the concentration of a reactant is
reduced to one half of its initial concentration. It is represented as t1/2.
For a zero-order reaction, rate constant is k ={[R]0 - [R]}/t
21
At t = t1/2, [R] = [R]0 /2 k ={[R]0 - [R]0 /2}/t1/2= [R]0 /2 t1/2 t1/2 = [R]0 /2 k
For a first order reaction, rate constant is k = (2.303/t) log {[R]0/[R]}
At t = t1/2, [R] = [R]0 /2 k = (2.303/t1/2) log {[R]0/[R]0 /2} k = (2.303/t1/2) log 2
k = (2.303/t1/2) 0.3010 t1/2 = 0.693/ k
For zero order reaction t1/2 ∝ [R]0. For first order reaction t1/2 is independent of [R]0.

ORDER UNITS OF RATE CONSTANT INTEGRATED RATE CONSTANT FIRST ORDER GAS PHASE
REACTION
0 -1
Mol L s -1 [ ] – [ ] . [ ]
K= K= log ( )

1 s-1 K=
. [ ]
log [ ]

Collision Theory of Chemical Reactions:

Collision frequency: It is defined as the number of collisions per second per unit volume of the reaction
mixture is known as collision frequency (Z).
Effective collision: The collisions in which molecules collide with proper orientation, breaking of bonds
between reacting species and formation of new bonds to form products are called as effective collisions.
Ineffective collision: The collisions in which molecules collide with improper orientation no products are
formed are called as ineffective collisions.
For example, formation of methanol from Bromoethane
depends upon the orientation of reactant molecules.
The proper orientation of reactant molecules lead to
bond formation whereas improper orientation makes
them simply bounce back and no products are formed.
Diagram showing molecules having proper and
improper orientation: -
Rate of reaction is proportional to
i. The number of collisions per unit volume per second (collision frequency, Z) between the reacting species.
ii. The fraction of effective collisions (properly oriented and possessing sufficient energy), f:
i.e., Rate = - dx/dt = f x Z
Temperature Dependence of the Rate of a Reaction: The rate of a reaction or rate constant becomes almost
double for every 10° rise in temperature. Increase in the rate of reaction with the rise in temperature is
mainly due to the increase in number of effective collisions.
The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius
equation.
Arrhenius equation and calculation of Activation energy:
Where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant
specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol
–1).

Activation energy: The excess energy (over and above the average energy of the reactants) which must be
supplied to the reactants to undergo chemical reactions is called activation energy.

22
It is equal to the difference between the threshold energy needed for the reaction and the average kinetic
energy of all the reacting molecules. That is,
Activation energy = Threshold energy - Average kinetic energy of the reacting molecules
Ea = Threshold energy – Energy of reactants
Low Activation energy: Fast reactions High Activation energy: Slow reactions
It can be understood clearly using the following simple reaction:
H2 (g) + I2 (g) → 2HI (g)

This reaction can take place only when a

molecule of hydrogen and a molecule of
iodine collide to form an unstable
intermediate. The energy required to form
this intermediate, called activated complex,
is known as activation energy (Ea).

Arrhenius equation-

Taking logarithm both side

The plot of ln k Vs 1/T gives a straight line.

Slope = – Ea /R and intercept = ln A.
So we can calculate Ea and A using these values.

Converting to common logarithm ln X = 2.303 log X

2.303log k = 2.303log A – Ea/RT log k = log A – Ea/2.303 RT
Let k1 and k2 are the rate constants for the reaction at two different temperatures T1 and T2 respectively.
log k1 = log A – Ea/2.303 RT1 …(i) log k2 = log A – Ea/2.303 RT2 …(ii)
Subtracting eq. (i) from (ii) log k2 - log k1 = Ea/2.303 R [1/T1 – 1/T2]

MCQ
Q1. Which of the following observations is incorrect about the order of a reaction?
a. Order of a reaction is always a whole number
b. The stoichiometric coefficient of the reactants doesn’t affect the order
c. Order of reaction is the sum of power to express the rate of reaction to the concentration terms of
the reactants in rate law.
d. Order can only be assessed experimentally Ans. (a)
Q2. In the reaction 3A + B → A3B, order of reaction is two with respect to A and one with respect to B, if the
concentration of A is doubled and that of B is halved, then the rate of the reaction will
a. decrease 2 times b. increase 4 times c. increase 2 times d. remain the same Ans. (c)
Q3. when the rate of the reaction is equal to the rate constant, the order of the reaction is
a. zero order b. first order c. second order d. third order Ans. (a)
Q4. A substance ‘A’ decomposes by a first-order reaction starting initially with [A] = 2.00M and after 200min,
[A] becomes 0.15M. For this reaction t1/2 is
a.53.72 min b.50.49 min c.48.45 min d.46.45 min Ans. (a)
Q5. A catalyst alters, which of the following in a chemical reaction?
a. Entropy b. Enthalpy c. Internal energy d. Activation energy Ans. (d)

23
 ASSERTION-REASONING
A statement of assertion is followed by a statement of reason. Mark the correct choice from the options
given below.
a) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
b) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
c) (A) is true but (R) is false
d) (A) is false but (R) is true
1. Assertion (A): A catalyst always lowers the energy of activation.
Reason (R): The catalyst-reactant interaction forms activated adsorbed complex and adsorption is always
exothermic.
2. Assertion (A): Order of reaction can never be fractional for an elementary reaction. Reason (R): An
elementary reaction takes place by one step mechanism.
3. Assertion (A): The elementary reaction is a single step reaction and does not possess a mechanism.
Reason (R): An elementary reaction has order of reaction and molecularity same.
4. Assertion (A): Every collision between molecules does not lead to a chemical reaction. Reason (R): Only
those molecules react during collisions which acquire threshold energy level.
5. Assertion (A): Order and molecularity are same.
Reason (R): Order is determined experimentally.
Answer 1 (a) 2(a) 3(a) 4(a) 5(d)
SECTION B (2 Mark Questions)
Q1. The rate constant for a first order reaction is 60 s–1 . How much time will it take to reduce the initial
concentration of the reactant to its 1/16th value?
Answer- t = 2.303/k log [A0]/[A] = 2.303/ 60 log 1/1/16 = 0.046 s
Q2. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction?
Answer- For first order reaction K75 %=2.303/4 log100/25 =0.6 K50% =2.303/2 log 100/50=0.6
the order of the reaction is 1
Q3. For an elementary reaction A(g) →B(g) + C(g) the half-life period is 10 minute. In what period of time
would the concentration of A be reduced to 10 % of original concentration?
𝟐.𝟑𝟎𝟑 𝟏𝟎𝟎
Answer- The reaction is following first order so k = 2 . 303 log [ R ]
0 K = 𝒕 log[𝟏𝟎]
t [R ]
𝟎.𝟔𝟗𝟑 𝟐.𝟑𝟎𝟑
𝟏𝟎
= 𝒕
log 10 t =2.303 x10/0.693 t=33 minutes

SECTION C (3 Mark Questions)

1. For the reaction A + B →products, the following ini al rates were obtained at various given ini al
concentrations. Determine the half-life period

Answer- For A-Rate =k[A]x For B Rate =k[B]y

From case 1 and 2 2× Rate =k[2A]x x=1
From case 1 and 3 Rate =k[2B] y y=0 Order n=x+y=1+0=1
2. Observe the graph in diagram and answer the following questions.
(i) If slope is equal to -2.0x10-6 sec-1, what will be the value of rate constant?
(ii) How does the half-life of zero order reaction relate to its rate constant?
Answer: (i) Slope = -k = -2.0x10-6 sec-1 Hence k = 2.0x10-6 sec-1
(ii) t ½= R0/2k
3. (a) Consider a certain reaction A → Products with k = 2.0 × 10-2 s-1. Calculate the concentration of A
remaining after 100 s if the initial concentration of A is 1.0 mol l-1

24
 (b) The half-life for radioactive decay of C -14 is 5730 years. An archaeological artifact containing wood
had only 80% of the C -14 found in a living tree. Estimate the age of the sample.
Answer
A

4. (i) What will be the effect of temperature on rate constant?

(ii) State a condition under which a biomolecular reaction is kinetically first order reaction.
(iii) For a zero-order reaction, will the molecularity be equal to zero? Explain.
Answer- a) The rate constant of a reaction is nearly doubled with rise in temperature by 10
b) Bimolecular reactions become kinetically first order when one of the reactants is present in excess
c) No, the molecularity can never be zero or a fractional number. As it represents the total number of
reactants taken in a reaction
SETION D (5-mark question)
1. (a) (i) For the reaction 2X →X2, the rate of reaction becomes three times, when concentration of X is
increased 27 times. What is the order of the reaction?
(ii) Write the rate equation for the reaction 2A + B →2C, if the order of the reaction is zero.
iii) Oxygen is available in plenty in air, yet fuels do not burn by themselves at room Temperature. Explain.
(b) Rate constant for first order reaction has been found to be 2·54 ×10–3 s–1. Calculate its three- fourth
life. [log 2 = 0·3010].
Answer. (a) (i)Rate =k(x)n ; 3Rate =k(27x)n Solving the n =1/3, so order of reaction =1/3
(ii)Rate =k[A]0[B]0=k
(iii)The activation energy for combustion of fuels is generally very high, and not achieved at
room temperature.

(b)

CASE BASED QUESTIONS

Read the message given below and answer the questions :
On the basis of studies of the effect of temperature on the rate constant of any reaction, Arrhenius gave
mathematical relation k = A e-Ea /RT in which Ea is activation energy and A is called frequency factor. The
equation was checked graphically and found to be correct. This equation helps to calculate Activation energy
of a reaction knowing rate constant at two different temperatures or to calculate the rate constant at a
particular temperature if activation energy and rate constant at some other temperature is known.
1. Define activation energy of reaction.
Ans- The excess energy (over and above the average energy of the reactants) which must be supplied to the
reactants to undergo chemical reactions is called activation energy.
2. Define a temperature coefficient of a reaction.
Ans- It is the ratio of two temperature constants differ by 10°C.
3. The plot of ln k versus 1/ T for a reaction has a slope equal to −4500K. What is the ac va on energy of
the reaction?
Ans- The plot of ln k vs 1/T has slope = - Ea/R = -4500 K Ea = 4500 x 8.314 JK-1mol-1 = 37413 JK-1mol-1

25
4. The rate of a reaction doubles when the temperature changes from 170C to 270C. Calculate the activation
energy.
( ) ( )
Ans- log = . Log 2 = . .
Ea = 50140.6 J

26
 UNIT-4 The d- and f- Block Elements
 d- block consists of elements belonging to group 3-12.The general electronic configuration of these
elements is (n-1)d1–10ns1–2.
 Transition elements contain unpaired d electron either in ground state or in ionic state.
 Transition elements exhibit metallic character, high densities, high melting, and boiling point due to
strong inter atomic interaction (metallic bonding) between partially filled d - orbitals.
 Transition elements show variable oxidation state (except Sc in 3d series) due to small energy
difference between (n-1) d & ns orbitals.
 Transition metals show colour due to presence of unpaired d electron in (n-1) d orbital of penultimate
shell which undergoes d-d transition.
 Transition elements are paramagnetic due to presence of unpaired electron. The magnetic moment
can be calculated using the spin only formula μ = √n(n+2) BM.
 Transition elements have high enthalpies of atomization due to presence of unpaired d electrons
resulting in strong inter atomic interaction and metallic bonding.
 Most of transition metals form complex compounds due to
(i)small size (ii)high charge (iii) presence of vacant d-orbital of suitable energy.
 Transition elements have lower value of reduction potential due to high ionization potential, high heat
of sublimation and low enthalpy of hydration.
 Transition elements form interstitial compounds because as small atoms like C, N, O, H occupy the
voids and get bonded to the atom of metals.
Lanthanoids and Actinoids
PROPERTY Lanthanoids Actinoids

Electronic 4f is progressively filled. 5f is progressively filled.

configuration [Xe]4f1–145d0–16s2 [Rn]5f1–14 6d0–1 7s2

Atomic and Lanthanoids show decrease in size of their Actinoids show decrease in size of their
ionic sizes atoms or ions in + 3 oxidation state as we atoms as we go from left to right, it is called
go from left to right. In lanthanoids, the actinoid contraction. 5f electrons have
decrease is called lanthanoid contraction poorer shielding effect than 4f electrons.

Oxidation state: 2, + 3, + 4) out of which + 3 is most +3, +4, +5, +6, +7 oxidation states. This is
common. This is because of large energy because of small energy gap between 5f, 6d
gap between 4f, 5d and 6s orbitals. and 7s orbitals.

Ce, Tb shows +4 and Eu, Yb+2 due to stable

configuration. So, Ce (IV) is good oxidizing.

Chemical Lanthanoids are less reactive than actinoids. Actinoids are highly reactive metals,
reactivity especially when finely divided. Actinoids are
more reactive than lanthanoids due to bigger
atomic size and lower ionisation energy.

Basic character Their compounds are Less basic Their compounds are More basic
Tendency to They do not form Oxo ions They form Oxo ions UO2+, NpO2+, PuO2+
form Oxo ions
Colours Most of their ions are colourless. Most of the actinoids ions are coloured UO2+
(yellow)

27
  Transition elements form alloys due to similar ionic radii as they can mutually substitute their position
in the crystal lattice.
 The oxides of transition metals in lower oxidation states are basic, intermediate oxidation states are
amphoteric, highest oxidation state are acidic.
 The highest oxidation state of an element is equal to no. of unpaired electrons present in (n-1)d & ns
orbitals.
Lanthanide contraction-
Lanthanoid contraction is the steady decrease in atomic and ionic size of lanthanoids as their atomic
number increases.
Causes of Lanthanide contraction: In f-block elements, the positive charge on the nucleus rises by
one unit, and one additional electron enters the same subshell, as the atomic number increases.
Consequences of Lanthanoid contraction-
1.This makes the element's separation of lanthanoids in the pure state difficult.
2. As the size of the lanthanides decreases from the elements La to Lu, the covalent character of the
hydroxides increases, and thus, their basic strength decreases.
3. Atomic size of second and third series members of d-Block is almost same.
Potassium dichromate K2Cr2O7 Preparation: It is prepared from chromite ore (FeCr2O4).
(i) Conversion of chromite ore into sodium chromate
4 FeCr2O4 + 8 Na2CO3 + 7 O2→ 8 Na2CrO4 (yellow) + 2 Fe2O3 + 8 CO2
(ii) Conversion of sodium chromate into sodium dichromate
2Na2CrO4 + 2 H2SO4→Na2Cr2O7 + 2 Na2SO4 + H2O
(iii) Conversion of sodium dichromate into potassium dichromate
Na2CrO4 + 2 KCl→K2Cr2O7 (orange) + 2 NaCl
pH Change-The chromates and dichromates are inter convertible in aqueous solution depending upon
pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.
CrO42– + 2H+→ Cr2O72– + H2O Cr2O72– + 2 OH- → CrO42– + H2O
(yellow) (orange)

Sodium and potassium dichromates acts as strong oxidizing agents in acidic medium.
Cr2O72– +14H+ +6e–→2Cr3+ +7H2O
Oxidation Reaction 6I –→ 3I2 + 6e – 3H2S→6H+ + 3S+ 6e –
6Fe 2+→6Fe 3+ + 6e– 3Sn2+→3Sn4+ + 6e –
Potassium permanganate (KMnO4)
Potassium permanganate is prepared from mineral pyrolusite ore
(MnO2).
(i) Conversion of pyrolusite ore to potassium manganate
2MnO2 + 4KOH + O22K2MnO4 + 2H2O
(ii) Conversion of potassium manganate to potassium
permanganate
2K2MnO4 + Cl2  2K2MnO4 + 2KCl
Disproportionation Reaction: 2KMnO4 K2MnO4 + MnO2 + O2

28
Both KMnO4 &K2Cr2O7 are coloured although they do not have any unpaired d-electron due to charge transfer
from Oxide ion to empty metal d-orbital.
Chemical Properties Potassium permanganate is a powerful oxidizing agent.
In neutral or faintly alkaline solutions:
(a) A notable reaction is the oxidation of iodide to iodate
2MnO4 – + H2O + I– ——> 2MnO2 + 2OH– + IO3 –
(b) Thiosulphate is oxidised almost quantitatively to sulphate:
8MnO4 – + 3S2O3 2– + H2O ——> 8MnO2 + 6SO4 2– + 2OH–
(b) Manganous salt is oxidised to MnO2 ; the presence of zinc sulphate or zinc oxide catalysis the oxidation:
2MnO4 – + 3Mn2+ + 2H2O ——> 5MnO2 + 4H+
In acid solutions:
(a) Iodine is liberated from potassium iodide : 10I– + 2MnO4 – + 16H+ ——> 2Mn2+ + 8H2O + 5I2
(b)Fe ion (green) is converted to Fe (yellow): 5Fe2+ + MnO4 – + 8H+ ——> Mn2+ + 4H2O + 5Fe3+
2+ 3+

(c) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O4 2– + 2MnO4 – + 16H+ ——> 2Mn2+ + 8H2O + 10CO2
(d)Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:
5SO3 2– + 2MnO4 – + 6H+ ——> 2Mn2+ + 3H2O + 5SO4 2–
(e) Nitrite is oxidised to nitrate: 5NO2 – + 2MnO4 – + 6H+ ——> 2Mn2+ + 5NO3 – + 3H2O

SECTION A ( MULTIPLE CHOICE QUESTION ANSWER)

Q.1 Which of the following lanthanoid ions is diamagnetic? (At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70)
a. Ce2+ b. Sm2+ c. Eu2+ d. Yb2+
Q.2 Which of the following statements about the interstitial compounds is incorrect?
a. They retain metallic conductivity b. They are chemically reactive
c. They are much harder than the pure metal d. They have higher melting points than the pure metal
Q3. Least paramagnetic property is shown by
a) Fe b) Mn c)Ni d) Cu

Q4. Which one of the following ions exhibit colour in aqueous solution
a. Sc3+ b.Ni2+ c.Ti4+ d. Zn2+
Q.5 Which one of the following is a diamagnetic ion?
a. Co2+ b.Cu2+ c.Mn2+ d.Sc3+
Answers: 1.d 2.b 3.d 4.b 5.d
ASSERTION -REASON QUESTION ANSWER
A statement of assertion is followed by a statement of reason. Mark the correct choice from the options
given below:
(a)Both assertion and reason are true and reason is the correct explanation of assertion.
(b)Both assertion and reason are true but reason is not the correct explanation of assertion.
(c )Assertion is true but reason is false.
(d) Assertion is false but reason is true.
1. Assertion : Cu2+ iodide is not known.
Reason : Cu2+ oxidizes I- to iodine.
2. Assertion : Mercury is not considered as a transition element.
Reason : Mercury is liquid.
3. Assertion : Cu+ ion is not stable in aqueous solution.
29
 Reason: Large value of I.E. of Cu is compensated by much more negative hydration energy of Cu2+(aq).
4. Assertion :K2Cr2O7 is used as primary standard substance in volumetric analysis.
Reason : K2Cr2O7 has good solubility in water.
5. Assertion : Transition metals used as catalyst.
Reason : Transition metal show variable oxidation state.

Ans- 1) a 2) b 3) a 4) c 5)a

Section B (2 marks questions)

Q1 How would you account for the following ?
(i) Many of the transition elements and their compounds can acts as good catalysts.
(ii) The transition elements have great tendency for complex formation.
Ans.1(i)Because these metals show variable oxidation state and form unstable intermediate which readily
change into product. Free valencies on the surface also favors its catalytic properties.
(ii) Because of small size, high charge on metal ion and availability of vacant d orbitals to accept lone pair of
electrons from ligands.
Q.2Calculate the spin only' magnetic moment of M2+(aq) ion. (Z=27)
Ans. Electronic configuration of the M2+ion (Z=27) would be M2+(aq):(Ar)3d7
It would contain three unpaired d electrons. The' spin only 'magnetic moment is given by the relation
µ= √ n (n +2) BM, µ =√3(3+2)BM=3.87BM
Q.3 i) Why is there a gradual decrease in the atomic sizes of the transition elements in a series with increasing
atomic numbers?
ii) Cu+is not stable in aqueous solution. Why?
Ans3.i) Because as a new electron enters in the d- subshell, each time the nuclear charge increases by unity.
Shielding effect of the d-electrons is not effective, so as the effective nuclear charge increases atomic sizes
of transitions elements decreases.
ii) In aqueous solution Cu+undergoes disproportionation to form a more stable Cu2+ion.
2Cu+ (aq) → Cu2+ (aq) + Cu(s)
The higher stability of Cu2+in aqueous solution may be attributed to its greater negative enthalpy of hydration
than that of Cu+.It compensates the second ionization enthalpy of Cu involved in the formation of Cu2+ions.
Q.4 Which of the following cations are coloured in aqueous solution and why? Sc3+ , V3+ , Ti4+ , Mn2+
Ans4. Sc3+ = 3d0 4s0 V3+ = 3d2 4s0
4+ 0
Ti = 3d 4s 0 Mn = 3d5 4s0
2+
3+ 2+
Thus V and Mn are coloured in their aqueous solution due to presence of unpaired electron and d-d
transition can be observed.
Q.5 i) Why is the highest oxidation state of a transition metal exhibited in its oxides or fluoride?
ii) Arrange the following increasing order of acidic character: CrO3, CrO, Cr2O3
Ans5 i)The highest oxidation state of a transition metal is exhibited in its oxides or fluoride due to its high
electronegativity , low ionization energy and small size.
ii) CrO<Cr2O3<CrO3. Higher the oxidation state, more will be acidic character.

SECTION C (3 Marks Questions)

Q1 Explain the following observations:
(i) Co2+ is easily oxidized to Co3+ in the presence of a strong ligand.
(ii) Oxide of a transition metal in lower oxidation state is basic but in highest oxidation state is acidic. Why?
(iii) Which is a stronger reducing agent–Cr2+or Fe2+and why?
Ans1(i) The electronic configuration of Co3+ is 3d6 4s0. Pairing occurs in the presence of a strong ligand. Thus,
fully filled t2g6 configuration gives stability. However, in Co2+ electronic configuration is 3d7, there is one
unpaired electron even after pairing occurs in the presence of a strong ligand. Hence, Co2+ is oxidized to more
stable Co3+.

30
(ii) The lowest oxide of transition metals is basic in nature because of some valence electrons are not involved
in bonding thus act as a base due to availability of free electron. In the highest oxide of transition metal
electrons of metal are involved in the bonding, but metal can accept electron pair in vacant orbitals. Thus, in
higher oxidation state transition metal oxides are acidic.
(iii) Cr2+is a stronger reducing agent than Fe2+ because 2g
after the loss of one electron Cr2+becomes Cr3+which
has more stable t2g3 (half filled) configuration in medium like water.
Q2. What is meant by the term lanthanoid contraction? What is it due to and what consequences does it
have on the chemistry of elements following lanthanoids in the periodic table?
Ans2.Lanthanoid contraction: Steady decrease in the size of the lanthanoids with increase in the atomic
number across the period. The electrons of 4f orbitals offer imperfect/poor shielding effect in the same
subshell.
Consequence:
Due to this 5d series elements have nearly same radii as that of 4d series.
Decrease in the basic strength from La(OH)3 to Lu(OH)3.
Due to similar atomic size, there is difficulty in separation of lanthanides.
Q3.How would you account for the following:
i)Mn(III) undergoes disproportionation reaction easily.
ii)Chromium is a typical hard metal while mercury is a liquid. Explain why?
iii) Why Cd2+ and Zn2+salts are white?
Ans3 i)Mn3+is less stable and changes to Mn2+which is more stable due to half-filled d-orbital configuration.
That is why, Mn3+undergoes disproportionation reaction.
ii) Chromium has five unpaired d electrons in the d-subshell (3d5 4s1). Hence, metallic bonds are very strong.
In mercury all the d-orbitals are fully filled (3d10 4s2). Hence, the metallic bonding is very weak.
iii Cd2+ and Zn2+salts are white because Cd2+ and Zn2+ have filled d-orbital (d10) and there is no d-d transition.
Q4. Complete the following reactions—
a. MnO4 - + Fe2+ + H+ →
b. MnO4 - + H2O + I ͞ →
c. Cr2O7 2͞ + 2 OH ͞ →

ANS: a. MnO4- +5 Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4 H2O

b. 2MnO4 - + H2O + I ͞ → 2MnO2 + 2OH ͞ + IO3 ͞
c. Cr2O7 2͞ + 2 OH ͞ → CrO4 2͞ + H2O

Q5. What are alloys? Name one important alloy which contains some of the lanthanoid metals. Mention its
use.
Ans. Alloys are homogeneous mixtures of metals with metals or non-metals.
Misch metal (pyrophoric alloy) consists of lanthanoid metal Ce= 40.5%, neodymium 44% iron 4-5% and traces
of S, C, Ca and Al. Misch metal is used to make bullets, shells and lighter flints.
SECTION D (5 Marks Question )
Q1. Give reason.
i)Transition metals and their compounds generally exhibit a paramagnetic behavior.
ii) A transition element scandium ( Z=21 ) does not exhibit variable oxidation state.
iii) Why are Zn, Cd and Hg not regarded as transition elements.
iv) Many of the transition elements are known to form interstitial compounds.
v) Actinoids contraction is greater from element to element than lanthanoids contraction. Why?
Ans1i)Transition metals and their compounds are generally paramagnetic due to unpaired electrons in d-
orbital of transition metal.
ii) Scandium does not exhibit variable oxidation states because it contains three electrons in its valence shell
( 3d1 4s2 configuration) and after losing them, it attains noble gas configuration.

31
 iii)Zn, Cd, Hg neither have partially filled d-orbital in their ground state nor in oxidized state. Thus, they are
not regarded as transition elements.
iv)Transition metals are generally solids and contain lots of interstitial sites. They can trap by small atoms of
other elements such as H, C,N etc., so they are known to form interstitial compounds.
v) This is because of poorer shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons
in lanthanoids.
Q2 i) Which metal in the first transition series(3d series) exhibit +1 oxidation state most frequently and why?
ii) Name a member of lanthanoid series which is well known to exhibit +4 oxidation state.
iii)Why is Cr2+ reducing and Mn3+oxidising when both have d 4 configuration?
iv) Name the oxometal anions of the first transition series of the transition metals in which the
metal exhibits the oxidation state equal to its group number.
v)The highest oxidation state is exhibited in oxides or oxoanion of a transition metal rather than
its fluoride.
Ans2 i) Copper exhibits +1 oxidation state more frequently because it attains fulfilled stability i.e. 3d10.
ii) Cerium
iii)Cr2+ is reducing as its configuration changes from d 4 to d 3, the latter having a half- filled t2g level. On the
other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra
stability.
iv) Cr2O72-andCrO42- (Group number = Oxidation state of Cr = 6)
MnO4– (Group number = Oxidation state of Mn = 7)
v) due to ability of oxygen to form multiple bonds.
CASE BASED QUESTION
1.Elements that have partially filled inner (n-1)d sub-shell in the
atomic state or in the permanent oxidation state are
called transition metal elements. Most of the transition elements
are metal. Transition metals become very hard except Zn, Cd and
Hg. They are very much harder than s-block and p-block elements.
Generally, transition metal elements have very high melting point
and boiling point. The melting point and boiling point of
transition metals gradually increases from left to right along a
particular transition series and reach a maximum value and then
decreases. For example, the melting point of 3d-series gradually
increases from Sc to V and it reach a maximum value for Cr and after
it the melting point slowly decreases. Because of with increase in
atomic mass and atomic number, the number of unpaired electrons of transition metals increases. As the
number of unpaired electrons in the valence shell increases or decreases, the strength of the metallic bond
increases or decreases respectively.
Q1. Why Mn has lower melting point than Cr?
Q2. Why do transition metals of 3d series have lower melting point as compared to 4d series?
Q3 i) Identify the metal with highest melting point in third transition series.
ii) Why Zn, Cd and Hg have lowest m.p and b,p?
Ans 1. The element Cr has the highest melting point because the metallic bonding is strongest due to the
presence of more unpaired electrons
2. 4d series elements have large atomic sizes as compared to 3d series, resulting in greater amount of metallic
bonding. Due to this 4d series elements have higher melting point.
3. i) W (tungsten)

32
ii) Zn, Cd and Hg have no unpaired electron due to which they have weak metallic bonding resulting in lowest
m.p. and b.p.

33
 UNIT-5 COORDINATION COMPOUNDS
Coordination compounds are molecular compounds which are formed from the combination of two or more
simple stable compounds and retain their identity in the solid as well as in the dissolved state.They do not
dissociate completely in the solution . For example- [Ni (NH3)6] Cl2 (Purple)
Double salt-Compounds which completely dissociate in the solution and loose their identity in the solution.
Example-Mohr salt FeSO4.(NH4)2SO4.6H2O
IMPORTANT TERMS
1.Coordination compounds are the compounds in which the central metal atom is bound to a number of
anions or neutral molecules by coordinate bonds.
2.Coordination entity: A coordination entity constitutes a central metal atom or ion bounded to a fix number
of oppositely charged ions or neutral molecules. For example, [Co Cl3 (NH3)3]
3.Coordination sphere: The central metal atom or ion and the molecules or ions bonded to it are enclosed
in a square bracket and a collectively called the Coordination sphere.
4.Central atom or ion: The atom or ion to which a fixed number of neutral molecules or ions are
attached in the coordination entity is called Central atom or ion.
5.Ligands: The neutral molecules or ions bonded to the central atom or ion in the coordination
entity are called ligands. For example, [Ni (NH3)6]2+: Central atom = Ni2+, ligands = NH3 molecules
Types of ligands: Unidentate or monodentate ligands: When a ligand is bound to a metal ion through a single
donor atom, as with Cl–, H2O or NH3, the ligand is said to be Unidentate ligands.
(i) Didentate or Bidentate ligands: When a ligand can bind through two donor atoms as in H2NCH2CH2NH2
(ethane-1, 2-diamine) or C2O42– (oxalate), the ligand is said to be Bidentate ligands.
(ii) Polydentate ligands: Ligands having more the two donor atoms present in the molecule, the ligands
are said to be Polydentate ligands.
Ethylenediaminetetraacetate ion EDTA4– is an important hexadentate ligand. When a di- or Polydentate
ligand uses its two or more donor atoms simultaneously to bind a single metal ion, it is said to be a chelate
ligand..
6. Ambidentate ligand -Ligand which has two different donor atoms and either of the two ligates in the
complex is called ambidentate ligand. Examples of such ligands are the NO2 – and SCN– ions. NO2 – ion
can coordinate either through nitrogen or through oxygen to a central metal atom/ion
7. Homoleptic complex -compounds in which all the ligand are same or identical. Ex- [Ni (NH3)6] Cl2
8. Hetroleptic Complex- compounds in which different type of ligand are attached with central atom
Example- [Co Cl3 (NH3)3]
IUPAC nomenclature of coordination compounds: Rules to write the IUPAC names of the
coordination compounds:
1. The positive part of a coordination compound is named first and is followed by the
negative part.
2. The ligands are named first followed by the central metal. The prefixes di- tri-, tetra-
etc, are used to indicate the number of each kind of ligand present. The prefixes bis
(two ligands) tris (three ligands) etc. are used when the polydentate ligands surround
the central atom.
3. The ligands are named in alphabetical order. Names of the anionic ligands end in O,
those of cationic in ium. Neutral ligands have their regular names except that H2O is
named aqua , NH3 as ammine; NO nitrosyl; and CO carbonyl.
4. There are three types of Monodentate ligand (Negative ) F- , Neutral Water H2O,
Positive Nitronium (NO+)
5. The oxidation state of the central metal is indicated in roman numbers in a bracket.
6. When a complex species has negative charge, the name of the central metal ends in -
ate. For some elements, the name of ion is based on the Latin name of the metal (For
example, argentate for silver).

34
Example of ligands
Name formula Charge
Ammonia NH3 0
Aqua H2O 0
Bromido Br -1
Chlorido Cl -1
Flourido F -1
Hydroxo OH -1
Cyanido CN -1
Nitrito NO2 -1
Ethanediamine en 1, 2-

Some examples of naming-

(a) [Pt (NH3)2 Cl (NO2)] (a)Diamminechloridonitrito-N-platinum (II)
(b) K3 [Cr (C2O4)3] (b)Potassium trioxalatochromate (III)
(c)[Co (NH3)5(CO3)] Cl (c)Pentaamminecarbonatocobalt (III) chloride
(d ) Hg [Co (SCN)4] (d)Mercury (I) tetrathiocyanato-S-cobaltate (III)
FORMULAS OF MONONUCLEAR COORDINATION ENTITIES
The following rules are applied while writing the formulas:
(i) The central atom is listed first.
(ii) The ligands are then listed in alphabetical order. The placement of a ligand in the list does not depend
on its charge.
(iii) Polydentate ligands are also listed alphabetically. In case of abbreviated ligand, the first letter of the
abbreviation is used to determine the position of the ligand in the alphabetical order.
(iv) The formula for the entire coordination entity, whether charged or not, is enclosed in square brackets.
When ligands are polyatomic, their formulas are enclosed in parentheses. Ligand abbreviations are also
enclosed in parentheses.
(v) There should be no space between the ligands and the metal within a coordination sphere.
(vi) When the formula of a charged coordination entity is to be written without that of the counter ion, the
charge is indicated outside the square brackets as a right superscript with the number before the sign. For
example, Co(CN)6]3–
(vii) The charge of the cation(s) must be balanced by the charge of the anion(s).
Werner's theory: The main postulates are
1.In coordination compounds metals show two types of linkages (valences)-primary and secondary.
2.The primary valences are normally ionizable and are satisfied by negative ions.
3.The secondary valences are non-ionizable. These are satisfied by neutral molecules or negative ions.
The secondary valence is equal to the coordination number and is fixed for a metal.
4.The ions/groups bound by the secondary linkages to the metal have characteristic spatial
arrangements corresponding to different coordination numbers.
Valence Bond Theory (VBT) for bonding in Coordination Compounds:
The main assumptions of this theory are listed below:
I. The central metal ion in the complex makes available a number of empty orbitals for the formation
of coordination bonds with suitable ligands.
II. The number of empty orbitals made available for this purpose is equal to coordination number of the
central metal ion.
III. The appropriate atomic orbitals (s, p and d) of the metal hybridize to give a set of equivalent orbitals
of definite geometry such as square planar, tetrahedral and octahedral and so on.
IV. The d-orbitals involved in the hybridization may be either inner d-orbitals i.e. (n-1) d or outer d-
35
 orbitals i.e. nd. For example, in case of octahedral hybridization. The orbitals may be two 3d, one 4s
and three 4p (d2sp3) or one 4s, three 4p and two 4d (sp3d2)
V. Each ligand has at least one orbital (of donor atom)) containing a lone pair of electrons.
VI. The empty hybrid orbitals of metal ion overlap the filled orbitals of the ligand to form metal-ligand
coordinate covalent bonds.
Number of Orbitals, Types of Hybridizations and Magnetic Properties of Coordination Compounds:
Coordination Type of Complex/ion Geometry No. of Magnetic
number hybridization unpaired Character
electrons
4 sp3 [NiCl4]2- Tetrahedral 2 Paramagnetic
4 dsp2 [Ni(CN)4]2- Square planar 0 Diamagnetic
6 sp3d2 [CoF6]3– Octahedral 4 Paramagnetic
6 d2sp3 [Co(NH3)6]3+ Octahedral 0 Diamagnetic

CRYSTAL FIELD THEORY (CFT): The crystal field theory (CFT) is an electrostatic model which considers the
metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and
the ligand.
Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules.
The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate.
This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal
atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of
dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the
dorbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature
of the crystal field.
The conversion of five degenerate d-orbitals of the metal ion into different sets of orbitals having
differentenergies in the presence of electric field of ligand is called crystal field splitting.
Crystal field splitting in octahedral coordination entities:

orbital splitting in an octahedral crystal field

Spectrochemical series: Ligands can be arranged in a series in the order of increasing field strength as givenbelow:
I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O4 2– < H2O < NCS– < edta4– < NH3 < en < CN– < CO
For d4 ions, two possible patterns of electron distribution arise:
(i)If ∆o < P, the fourth electron enters one of the eg orbitals giving the configuration t2g3eg1 .Ligands for
which ∆o < P are known as weak field ligands and form high spin complexes.
(ii) If ∆o > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with
configuration t2g4eg0. Ligands which produce this effect are known as strong field ligands and form low spin
complexes.
36
Crystal field splitting in tetrahedral coordination entities:

Magnetic Properties of Complexes

The complex in which a central transition metal ion has unpaired
electrons is paramagnetic.
The complex in which a central transition metal ion has no unpaired electrons is diamagnetic.
The magnetic moment of a complex is calculated by the spin only formula.
M = √[n(n+2)] BM BM = Bohr Magneton
ISOMERISM IN COORDINATION COMPOUNDS:
Two or more compounds having the same molecular formula but different arrangement of atoms are called
isomers and the phenomenon is called Isomerism.
Two types of isomerism are: (a) Structural isomerism (b) Stereoisomerism
STRUCTURAL ISOMERISM: The isomers which have same molecular formula but different structural
arrangement of atoms or groups of atoms around the central metal ion are called structural isomers.
IONIZATION ISOMERISM: The compounds which have same molecular formula but give different ions in
solution are called ionization isomers. For example: [Co (NH3)5 (SO4)]Br and [Co (NH3)5Br] SO4 .
SOLVATE ISOMERISM: The compounds which have same molecular formula but differ in the number of
solvent molecules present as ligands and as free solvent molecules in the crystal lattice are called solvate
isomers. [Cr (H2O)6]Cl3 (violet) and its solvate isomer [Cr (H2O)5 Cl]Cl2.H2O (grey-green).
COORDINATION ISOMERISM: This type of isomerism arises from the interchange of ligands betweencationic
and anionic entities of different metal ions present in a complex. For example: [Co (NH3)6][Cr(CN)6 ], in which
the NH3 ligands are bound to Co3+ and the CN– ligands to Cr3+. In its coordinationisomer
[Cr (NH3)6][Co(CN)6 ], the NH3 ligands are bound to Cr3+ and the CN– ligands to Co3+.
Linkage isomerism: The compounds which have same molecular formula but differ in the mode of
attachment of a ligand to the metal atom or ion are called linkage isomers. Linkage isomerism arises in a
coordination compound containing ambidentate ligand. For example:
In [Co(NH3)5(NO2)]Cl2 the ligand is bound through nitrogen (NO2).
In[Co (NH3)5(ONO)] Cl2 the ligand is bound through oxygen (–ONO).
STEREOISOMERISM: stereoisomers have the same chemical formula and chemical bonds but they have
different spatial arrangement of atoms.
Two types of isomerism are: (a) Geometrical isomerism (b) Optical isomerism
GEOMETRICAL ISOMERISM: This type of
isomerism arises in heteroleptic complexes due to
ligands occupying different positions around the
central ion. The ligands occupy positions either
adjacent to one another or opposite to one
another. These are referred to cis- form (ligand occupy adjacent position) and trans-form (ligand occupy
opposite position). For example: Geometrical isomers (cis and trans) of Pt [NH3)2 Cl2] Cis- trans-
Square planar complex of the type MABXL (where A, B, X, L are unidentate) shows three isomers-
two cis and one trans. Such isomerism is not possible for a
tetrahedral geometry but similar behavior is possible in
octahedral complexes of formula [MX2L4] in which the two
ligands X may be oriented cis or trans to each other
Geometrical isomers (cis and trans) of [Co(NH3)4Cl2]+ 

37
This type of isomerism also arises when bidentate ligands L–L
are present in complexes of formula [MX2(L–L)2]
Geometrical isomers (cis and trans) of [CoCl2(en)2]
Another type of geometrical isomerism occurs in octahedral
coordination entities of the type [Ma3b3] like [Co(NH3)3(NO2)3]. If three donor atoms of the same ligands
occupy adjacent positions at the corners of an octahedral face, we have
the facial (fac) isomer. When the positions are around the meridian of
the octahedron, we get the meridional (mer) isomer
The facial (fac) and meridional (mer) isomers of [Co(NH3)3(NO2)3] 
(a) OPTICAL ISOMERISM: Optical isomers are mirror images that cannot
be superimposed on one another.These are called as enantiomers. The molecules or ions that
cannot be superimposed are called chiral. The two forms are called
dextro (d) and laevo (l) depending upon the direction they rotate
the plane ofpolarized light in a polarimeter (d rotates to the right,
l to the left).Optical isomerism is common in octahedral
complexes involving Didentate ligands. Optical isomers of [Pt
Cl2(en)2] 2+
Optical isomers (d and l) of cis-[PtCl2(en)2]2+

In a coordination entity of the type [PtCl2(en)2]2+, only the cis-isomer shows optical activity.
Importance and Applications of Coordination Compounds:
1.Hardness of water is estimated by simple titration with Na2EDTA. The Ca2+ and Mg2+ ions form
stable complexes with EDTA.
2.Biological importance: Hemoglobin, the red pigment of blood which acts as oxygen carrier is a coordination
compound of iron. Vitamin B12, cyanocobalamin, the anti– pernicious anaemia factor, is a coordination
compound of cobalt.
3.In catalyst:Coordination compounds are used as catalysts for many industrial process.
a Wilkinson catalyst, is used for the hydrogenation ofalkenes.
4.In medicine: Complexing agents are used for removal of metal poisoning. platinum complex, [Pt
Cl2 (NH3)2] known as cis-platin has been used in cancer therapy.

SECTION A Multiple Choice Questions (MCQ)

Q1. The IUPAC name for K2[PdCl4] is
(a)Potassium tetrachlorinepalladium(II) (b)Potassium tetrachloridopalladate(II)
(c)Potassium tetrachloridopalladium(II) (d)Potassium tetrachlorinepalladate(II)
Q2. The core atom of which of the following biologically significant coordination molecules is magnesium?
(a)Vitamin B12 (b )Hemoglobin (c) Chlorophyll (d) Carboxypeptidase-A
Q3.Which one of the following ligands forms a chelate ring?
a) Acetate b) Oxalate c) Cyanide d) Ammonia
Q4.Which kind of isomerism is exhibited by octahedral [Co(NH3)4 Br2] Cl
a) Geometrical and ionization b) Geometrical only
c) Geometrical and optical d) Optical and ionization
Answers 1 (b) 2(c) 3(b) 4 (a)
ASSERTION -REASON QUESTION ANSWER
A statement of assertion is followed by a statement of reason. Mark the correct choice from the options
given below:
(a)Both assertion and reason are true and reason is the correct explanation of assertion.
(b)Both assertion and reason are true but reason is not the correct explanation of assertion.
38
 (c )Assertion is true but reason is false.
(d) Assertion is false but reason is true.

1. Assertion: [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature.

Reason: Unpaired electrons are present in their d-orbitals.
2. Assertion: The primary valencies are ionizable and are satisfied by positive ions.
Reason: Central atoms/ions in coordination compounds are also referred to as Lewis acids.
3. Assertion: [Ni(CN)4]2- is square planar and diamagnetic.
Reason: It has no unpaired electrons due to the presence of a strong ligand.
4. Assertion: Potassium ferrocyanide is diamagnetic whereas potassium ferricyanide is paramagnetic.
Reason: Crystal field splitting in ferrocyanide ion is greater than that of ferricyanide ion.
5. Assertion: Oxalate ion is a bidentate ligand.
Reason: Oxalate ion has two donor atoms
Answers: 1. B 2. D 3. A 4. C 5. A
B Short answer type questions with answer (2 marks each)
1. Why do compounds having similar geometry have different magnetic moment?
Ans: It is due to presence of weak and strong ligands in complexes, if CFSE is high, the complex will show low
magnetic moment value and vice versa.
2. (a) What type of isomerism is shown by the complex [Co(NH3)5(SCN)]2+?
(b) How does the colour change on heating [Ti(H2O)6]2+ ion??
Answer: (a) Linkage isomerism. Its linkage isomer is [Co(NH3)5(CNS)]2+
(b) On heating, coordinated water molecules are removed. Crystal field splitting vanishes in the absence of
ligands. Hence, no excitation of electrons and no colour.
3. Using IUPAC norms, write the formulae for the following complexes:
(i) Hexaaquachromium (III) chloride
(ii) Sodium trioxalatoferrate (III)
Ans: (i) [Cr(H2O)6]Cl3 (ii) Na3[Fe(C2O4)3]
4.When a coordination compoundNiCl2.6H2O is mixed with AgNO3, 2 moles of AgCl are precipitated per mole
of the compound, Write (i) structural formula of the complex (ii) IUPAC name of the complex
Ans: (i) [Ni(H2O)6]Cl2 (ii) hexaaquanickel(II)chloride
C) Short answer type questions with answer (3 marks each)
1. For the complex [Fe(en)2Cl2], Cl, (en = ethylene diamine), identify
(i) the oxidation number of iron, (ii) the hybrid orbitals and the shape of the complex,
(iii) the magnetic behaviour of the complex
Answer: (i) [Fe(en)2Cl2] Cl or x + 0 + 2 (-1) + (-1) = 0 x + (- 3) = 0 or x = + 3
∴ Oxidation number of iron, x = + 3
(ii) The complex has two bidentate ligands and two monodentate ligands. Therefore, the coordination
number is 6 and hybridization will be d2sp3 and shape will be octahedral.
(iii) In the complex 26Fe3+ = 3d5 4s0 4p0
Due to presence of one unpaired
electrons in d orbitals the complex is
paramagnetic.
2. Write the name, stereochemistry and magnetic behaviour of the following : (At. nos. Mn = 25, Co = 27, Ni
= 28) (i) K4[Mn(CN)6] (ii) [CO(NH3)5 Cl]Cl2 (iii) K2 [Ni(CN)2]
Answer: (i) K4[Mn(CN)6] : IUPAC name : Potassium Hexacyano manganate (II)
Geometry : Octahedral Magnetic behaviour: Paramagnetic (one unpaired electron)

39
 (ii) [CO(NH3)5 Cl]Cl2 :
Name : Pentaammine chlorido cobalt (III) chloride Shape : Octahedral (∵ Coordination number = 6)
Hybridization : d2sp3 Magnetic behaviour : Diamagnetic (no unpaired electrons)
(iii) K2 [Ni(CN)4] :
Name : Potassium tetracyanonickelate (II) Hybridization : dsp2 Magnetic behaviour : Diamagnetic
Shape : Square planar Hybridization : dsp2 (∵ Coordination number = 4)
3. What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic
configuration of d4 in terms of t2g and eg in an octahedral field when (i) Δ0 > P (ii) Δ0 < P
Answer: Crystal field splitting energy: When ligands approach the central metal ion, the degenerate d-
orbitals split into two sets, one with lower energy (t2g) and the other with higher energy (eg). The difference
of energy between these two sets of orbitals is called crystal field splitting energy. (ΔO for octahedral
complexes).
The magnitude of ΔO decides the actual configuration of d-orbitals by the help of mean pairing energy.
● If Δ0 < P then pairing of electrons does not occur and electrons enter in the higher energy e orbitals and
thus form high spin complexes due to weak field ligands. t2g3eg1
● If Δ0 > P then pairing of electrons occurs within the same set and form low spin complexes due to strong
field ligands. t2g4eg0
D) CASE BASED QUESTIONS
CASE I. Read the given passage and answer the questions that follow:
Complex compounds play an important role in our daily life. Werner’s theory of complex compounds says
every metal atom or ion has primary valency (oxidation state) which is satisfied by ligands (+ve, -ve, neutral)
but having lone pair. Primary valency is non –directional, secondary valency is directional. Complex
compound are name according to IUPAC system. Valance bond theory helps in determining shapes of
complexes based on hybridisation, magnetics properties, outer and inner orbital complex. Complex show
ionisation, linkage, solvate, and coordination isomerism also called structural isomerism. Some of them also
shows stereoisomerism i.e. geometrical and optical isomerism. Ambidentate ligand are essential to show
linkage isomerism. Polydentate ligands form more stable complexes then unidentate ligands. There are
called chelating agents. EDTA is used to treat lead poisoning, cis – platin as anticancer agents. Vitamin B12 is
complex of cobalt. Haemoglobin, oxygen carrier is complex of Fe2+ chlorophyll essential for photosynthesis
is complex of Mg2+.
 What is the oxidation state of Ni in [ Ni(CO)4] ?
Ans . Zero.
 One mole of CrCl3.6H2O reacts with excess of AgNO3 to yield 2 mole of AgCI. Write formula of
complex. Write IUPAC name also.
Ans . [Cr(H2O)5CI]CI2.H2O, pentaaqua chlorido chromium (III) chloride .
 Out Cis – [Pt(en)2CI2]2+ and trans (Pt(en)2CI2)2+ which one shows optical isomerism?
Ans . cis – [Pt(en)2CI2]2+ shows optical isomerism.
OR
 What is hybridisation of [CoF6]3- ? [ Co = 27] Give its shape and magnetic properties.
Ans.sp3d2, octahedral , paramagnetic. It is outer orbital complex.

E) Long answer questions (5Mark)

1. (a) Name the following coordination compounds and draw their structures:
(i) [CoCl2(en)2]Cl (ii) [Pt(NH3)2Cl(NO2)] (At. no. Co = 27, Pt = 78)
40
Answer:

(b) Write the name, structure and the magnetic behaviour of each one of the following complexes:
(i) [Pt(NH3)2Cl(NO2)] (ii) Ni(CO)4 (At. nos. Co = 27, Ni = 28, Pt = 78)
Answer:
(i) Diammine chlorido nitrito-N-platinum(II). It is square planar and diamagnetic.
(ii) Tetracarbonyl nickel(0). It is tetrahedral and diamagnetic complex.

41
 UNIT-06 HALOALKANES AND HALOARENES
Haloalkane:- They are compounds having general formula RX . They are halogen derivatives of alkane.
Haloarenes:- They are halogen derivatives of aromatic hydrocarbons (arenes).
Important name reaction-
1. Finkelstein reaction
2. Swarts reaction
3. Sandmeyer’s reaction 4. Fittig Reaction -

5. Wurtz Reaction: R-X + 2Na + R-X  R-R + 2NaX

6. Wurtz-Fittig Reaction 7. Friedal craft reaction

Mechanism of nucleophilic substitution reaction:The nucleophilic substitution reaction can proceed via SN1
mechanism or SN2mechanism.

Unimolecular nucleophilic substitution, SN1 Bimolecular nucleophilic substitution, SN2

This reaction follows first order kinetics This reaction follows second order kinetics
The reactivity order of halo alkanes towards The reactivity of alkyl halide towards SN2 reaction is:
SN1reaction is: 3° R−X < 2° R−X < 1° R−X
1° R−X < 2° R−X < 3° R−X
For the optically active compounds the product is a Here the product have opposite configuration
racemic mixture. (inversion of configuration)
The reaction is favoured in the polar medium. The above reactions are favoured in non-polar
medium

42
This type of nucleophilic substitution takes place in This type of nucleophilic substitution takes place in
two steps, one step via transition state.
+
(1) R:X → R + X - R:X + Nu- → R-Nu + X-
(2) R+ + Nu- →R-Nu

The first step being the rate determining step

The incoming nucleophile interacts with alkyl halide
involves the formation of carbonium ions.
causing the C−X bond to break while forming a new
C−Nu bond. The departure of the leaving group
occurs simultaneously with the backside attack by
the nucleophile.

Preparation of Haloalkanes: Haloalkanes can be prepared by a number of methods

1. By free radical halogenation of alkanes: Chlorination or bromination of alkane usually gives a complex
mixture of isomeric mono and poly halo
alkanes.
Fluorination is a violent reaction so it must be
controlled and iodination is a reversible
reaction so it must be carried out in the presence of strong oxidising agents like nitric acid and periodic acid
as HI acts as a reducing agent. CH4 (excess) + l2 ⇄ CH3l + Hl
2. By addition of HX to alkene:
An alkene is converted to corresponding alkyl
halide by reaction with hydrogen chloride,
hydrogen bromide or hydrogen iodide.
3. From alcohol: The hydroxyl group of
an alcohol is replaced by halogen in
reaction with concentrated halogen
acids, phosphorus halides or thionyl
chloride to give the corresponding alkyl
halide.

Properties of Haloalkanes :-
1. Alkyl halides are colourless when pure but bromides and particularly iodides develop colour when
exposed to light.
2. Halogen compounds have higher boiling points than the corresponding hydrocarbon. This is because the
greater polarity as well as higher molecular mass as compared to the parent hydrocarbon causes the
intermolecular forces of attraction (dipole-dipole and van der Waals) to be stronger in the halogen
derivatives.
3. For monohalogen compounds, the boiling point increases with increasing molecular mass of the halogen
group with a fixed hydrocarbon group, R‒F < R‒Cl < R‒Br < R‒I
4. All halogen derivatives of hydrocarbon are insoluble in water as they are incapable of forming hydrogen
bonds with water but alkyl halides are soluble in non-polar solvents,
5. Reactivity of Haloalkanes towards nucleophilic substitution:
43
For the same alkyl group, as we move from F to I, strength of C−X bond decreases, therefore, the reac vity
order is: R− I > R−Br > R−Cl > R−F

6. Elimination reactions: When a halo alkane with β-hydrogen atom is heated with alcoholic solution of
potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the
αcarbon atom resulting in the formation of an alkene. The reaction follows the Saytzeff rule which states
that “In dehydrohalogenation reactions, the preferred product is that alkene which has the greater
number of alkyl groups attached to the doubly bonded carbon atoms.”

Reactions of Haloarenes
1. Electrophilic Substitution Reaction: The replacement of an atom or group of atoms by an electrophile in
a molecule is called an electrophilic substitution reaction. Haloarenes undergo the usual electrophilic
reactions of the benzene ring such as halogenation, nitration, sulphonation and Friedel-Crafts reactions
 Halogens are ortho and para director, as it shows –I and +R effect but + R effect increases
electron density on ortho and para position.
 Halogens are not electrophilic enough to break the aromaticity of benzenes, which require a
catalyst to activate.
2. Nucleophilic Substitution Reactions in haloarenes
Aryl halides are less reactive towards nucleophilic substitution reaction. Their low reactivity is attributed due
to the following reasons:
(i) Due to resonance, C—X bond has partial double bond
character.
(ii) Stabilization of the molecule by delocalization of
electrons.
(iii) Instability of phenyl carbocation.
However, aryl halides having electron withdrawing groups (like NO2, SO3H etc.) at ortho and para positions
undergo nucleophilic substitution reaction easily.
44
Important Compounds
Trichloromethane, CHCl3 (Chloroform) Chloroform is
slowly oxidized by air in the presence of light to form
poisonous phosgene gas
To avoid this oxidation chloroform is always
stored in dark coloured bottles filled to the
brim to exclude any air. Further bottles are
also filled with small amount of ethyl alcohol to destroy traces of phosgene if formed, to harmless diethyl
carbonate.
Uses of Trichloromethane: 1.Chloroform is employed as a solvent for fats, alkaloids, iodine and other
substances. 2. It mainly used in the production of the freon refrigerant R-22.
Triiodomethane, CHI3 (Iodoform)
Preparation: Any compound containing CH3CO− or CH3CH (OH)− group, when heated with iodine and
aqueous NaOH gives yellow precipitate of iodoform. The reaction is known as iodoform reaction.

Iodoform was used earlier as an antiseptic due to the liberation of free iodine but due to its objectionable
smell, it has been replaced by other formulations containing iodine.
Freons
Preparation: Dichlorodifluoromethane or Freon 12 (CCl2F2) is one of the most common freons in industrial
use and is manufactured from tetrachloromethane by Swarts reaction.
Properties:Freons are extremely stable, unreactive, non-toxic, non-corrosive and easily liquefiable gases.
They lead to the depletion of ozone layer surrounding our planet.
Uses: Freons are used in aerosol propellants, refrigeration and air conditioning purposes.
p,p’-Dichlorodiphenyltrichloroethane (DDT)
Preparation: It is manufactured by the condensation of chloral with chlorobenzene in the presence of H2SO4.
Properties: DDT is a white powder insoluble in water. It is a highly stable fat-soluble compound. Uses:It is
mainly used as an insecticide
Multiple choice questions
1. Racemization occurs in :
(a) SN1 reaction (b) SN2 reaction
(c) Neither SN1 nor SN2 reaction (d) SN2 reaction as well as SN1 reaction
2. The conversion of an alkyl halide into an alkene by alcoholic KOH is Classified as
(a) a substitution reaction (b) an addition reaction
(c) a dehydrohalogenation reaction (d) a dehydration reaction
3. Haloarenes are less reactive towards nucleophilic substitution reactions compared to haloalkanes
due to:
(a) Resonance stabilization b) Higher steric hindrance
c) Lower electronegativity of halogens d) none of the above
4. p-dichlorobenzene has higher melting point than its o- and m- isomers. Why?
a) m-dichlorobenzene is more polar than o-isomer
b) p-isomer has a symmetrical crystalline structure
c) Boiling point of o- isomer is more than p-isomers
d) All of these are correct
5. Phosgene is a common name for a gaseous chemical called.
a) Carbonyl chloride b) Thionyl chloride
c) Carbon dioxide with phosphine d) none of the above
Ans: 1a, 2c,3a, 4b,5a,

45
 ASSERTION AND REASON QUESTIONS
For Questions number 1to 5, two statements are given one labelled as Assertion (A) and the other
labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and
(d) as given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the
Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the
Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
1. Assertion (A): Nucleophilic substitution of iodoethane is easier than chloroethane.
Reason (R): Bond enthalpy of C-I bond is less than that of C-Cl bond.
2. Assertion: SN2 reaction of an optically active aryl halide with an aqueous solution of KOH always
gives an alcohol with opposite sign of rotation.
Reason: SN2 reactions always proceed with inversion of configuration.
3. Assertion: KCN reacts with methyl chloride to give methyl isocyanides
Reason: CN– is an ambident nucleophile.
4. Assertion: Presence of a nitro group at ortho or para position increases the reactivity of haloarenes
towards nucleophilic substitution.
Reason: Nitro group, being an electron withdrawing group decreases the electron density over the
benzene ring.
5. Assertion: The boiling points of alkyl halides decrease in the order : RI > RBr > RCl > RF
Reason : The boiling points of alkyl chlorides, bromides and iodides are considerably higher than
that of the hydrocarbon of comparable molecular mass.
Ans: 1 a, 2d, 3d, 4a, 5c.
SHORT ANSWER TYPE QUESTIONS (2 MARKS)
1. Arrange the following in increasing order of their boiling points:
(i) 1-chloropropane, 2-chloropropane, 1-chlorobutane
(ii) What is an ambident nucleophile? Give one example.
Ans: (i) 2-cholropropane < 1-chloropropane < 1-chlorobutane
(ii) Groups which possess two nucleophilic centres and are called ambident nucleophiles. They can
form one bond with carbon from either side. For example CN-, SCN-
2. Draw the structure of major monohalo
product in each of the following reactions :

Ans: (i)
(ii)

3. Explain why :
46
 (i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) Alkyl halides, though polar, are immiscible with water.
Ans: (i) Chlorobenzene has lower dipole moment than cyclohexyl chloride due to lower magnitude
of -ve charge on the Cl atom and shorter C – Cl distance. Due to greater S-character, a sp2-hybrid
carbon is more electronegative than a sp3-hybrid carbon. Therefore, the sp2-hybrid carbon of C – Cl
bond in chlorobenzene has less tendency to release electrons to Cl than a sp3 hybrid carbon of
cyclohexyl chloride.
(ii) Because they are not able to form H-bond with water.
4. Account for the following:
(i) The C – Cl bond length in chlorobenzene is shorter than that in CH3 – Cl.
(ii) Chloroform is stored in closed dark brown bottles.
Ans: (i) In haloalkanes, the halogen atom is attached to sp3-hybridized carbon while in haloarenes it
is attached to sp2 -hybridized carbon whose size is smaller than sp3 orbital carbon.
Therefore C – Cl bond in chloro-benzene is shorter than alkyl chloride.
(ii) CHCl3 is stored in dark coloured bottles to cut off light because CHCl3is slowly oxidized by air in
presence of light to form an extremely poisonous gas, carbonyl chloride, popularly known as
phosgene.
SHORT ANSWER TYPE QUESTIONS (3 MARKS)
1. Answer the following questions :
(i) What is meant by chirality of a compound? Give an example.
(ii) Which one of the following compounds is more easily hydrolyzed by KOH and why?
CH3CHClCH2CH3 or CH3CH2CH2Cl
(iii) Why Grignard’s reagents should be prepared under anhydrous conditions?
Ans: (i) Chirality: The objects which are non-superimposable on their mirror image are said to be
chiral and this property is known as chirality for Butan-2-ol

C is non-superimposable on its mirror image A.

(iii) Grignard’s reagents are very reactive. They react with alcohol, water, amines etc. to form
corresponding hydrocarbon.
2. Give reasons :
(i) N-Butyl bromide has higher boiling point than t-butyl bromide.
(ii) Racemic mixture is optically inactive.
(iii) The presence of nitro group (-NO2) at o/p positions increases the reactivity of haloarenes
towards nucleophilic substitution reactions.
Ans:(i) n-Butyl bromide has higher boiling point than t-butyl bromide because it has larger surface
area hence have more Van der Waals’ forces.
47
 (ii) Rotation due to one enantiomer is cancelled by another enantiomer.
(iii) The presence of nitro group (-NO2) at ortho and para positions withdraws the electron
density’ from benzene ring and thus facilitating the attack of nucleophile.
3. Write the major product(s) in the following:

Ans:

4. Give reasons for the following :

(i) Chlorine is ortho/para directing in electrophilic aromatic substitution reactions, though
chlorine is an electron withdrawing group.
(ii) Allyl chloride is hydrolyzed more readily than n-propyl chloride
Ans: (i) though chlorine is an electron withdrawing group but it shows +R effect (electron releasing
resonance effect), which increases electron density at ortho/para position. Thus, it behaves as
ortho/para directing group.
(ii) In allyl chloride the carbocation formed by hydrolysis is resonance stabilized, whereas no
such stabilization is there in carbocation of alkyl chloride.
CASE BASED QUESTIONS
Read the passage and answer the following questions:
Saytzeff observed that in dehydrohalogenation reaction, the preferred product is that alkene which
has the greater number of alkyl groups attached to the carbon –carbon double bond. Thus in the
dehydrohalogenation of 2- bromopentane , pentene-2-ene is the preferred product.an alkyl halide
with α-hydrogen atom when reacted with a base or a nucleophile has two competing routes:
substitution and elimination. The actual route taken depends upon the nature of alkyl halide,
strength and size of the base. A bulkier nucleophile will prefer to act as a base and abstract a proton
rather than approach a tetravalent carbon atom for steric reasons. A primary alkyl halide will prefer
SN2 mechanism, a secondary halide SN2 or elimination depending upon the strength of base and a
tertiary halide – SN1 or elimination depending upon the stability of carbocation or the substituted
alkene.
1. What is nucleophile?
2. Write one example of dehydrohalogenation reaction.
48
 3. Vinyl chloride is unreactive to nucleophilic substitution reaction. Why?
OR
(a) Which of the compounds will react faster in SN1 reaction with the “OH” ion? CH3-CH2-Cl or C6H5-CH2-Cl
(b) Predict the order of reactivity of four isomeric bromobutanes in SN1 reaction.
Ans: (i) Nucleophiles are electron-rich species, which have the ability to donate electron pairs.
For eg. –OH, -NH3
(ii)
As a result of resonance,
the carbon-chloride bond (iii)
acquires some double
bond character. Hence,

vinyl chloride does not undergo nucleophillic substitution

reactions.
OR
(a)C6H5-CH2-Cl
(b)

LONG ANSWER TYPE QUESTIONS:

1. (a) An alkyl halide (A) of molecular formula C6H13Cl on treatment with alcoholic KOH gives two isomeric
alkenes (B) and (C) of molecular formula C6H12. Both alkenes on hydrogenation give 2,3-dimethylbutane.
Write the structures of (A), (B) and (C).
(b) Arrange the following in the increasing order of their reactivity towards SN1 reactions :
i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1Bromo-3-
methylbutane.
Ans (a) (A) CH3-C(Cl)(CH3)-CH(CH3)-CH3, (B) CH3-C(CH3)=C(CH3)-CH3, (C) CH2=C(CH3)CH(CH3)
(b) (i)2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
(ii)2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane
(iii)1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-
Bromobutane

49
 UNIT-07
Alcohols and phenols are formed when a hydrogen atom in a hydrocarbon, aliphatic and aromatic
respectively, is replaced by –OH group.
Based on number of -OH (hydroxyl) group attached we have:
Monohydric alcohols: Example: methanol (CH3OH) ethanol (C2H5OH)
Dihydric alcohols: Example:

ethylene-glycol
PHYSICAL PROPERTIES OF ALCOHOL
1. State of existence: Lower members are colorless liquids with characterize smell. Higher members with
more than 12 carbons are solid
2. Boiling point: They have high boiling points due to hydrogen bonding.
.

Boiling point decreases with branching: as surface area gets reduced therefore Vander wall force decrease
and boiling point also decreases. The one degree will have higher boiling point as it has more surface area.

3. Solubility: They are soluble in water due to formation of hydrogen bonds in them. Solubility decreases
with increase in size of the alkyl/aryl group because that would increase the hydrocarbon part of the
compound.

Acidic nature of alcohols and Phenols:

An electron-releasing group increases the electron density on oxygen tending to decrease the polarity of O-
H bond. Acidic strength of alcohol decreases in the order: Primary > Secondary > Tertiary
Alcohols are weaker acids than water.
Phenols are stronger acids than alcohols because of the resonance that takes place. Phenol molecule ionizes
to give phenoxide ion and H+ ion. This becomes possible because phenoxide ion stabilizes itself through
resonance.
Presence of electron withdrawing group in the ring particularly at ortho- and para-positions increases the
acidic character of the phenol. This is because the negative charge on phenoxide ion is dispersed or
delocalized. On the other hand, presence of electron donating groups, like alkyl groups, do not favour the
formation of phenoxide ion. Such compounds will have smaller acidic strength.
IMPORTANT NAME REACTIONS
Kolbe’s Reaction: Sodium phenoxide is generated by treating phenol with sodium hydroxide. It is treated
with carbon dioxide to form ortho-hydroxy benzoic acid (salicylic acid).

50
Reimer-Tiemann Reaction: When phenol is treated with chloroform in the presence of sodium hydroxide,
a −CHO group is introduced at the ortho posi on of the benzene ring. This reaction is known as the Reimer-
Tiemann reaction. The intermediate is hydrolyzed in presence of alkalis to produce salicylaldehyde.

Williamsons Ether synthesis: In this method, an alkyl halide is allowed to react with sodium alkoxide. The
reaction involves SN2 attack of an alkoxide ion on primary alkyl halide. Better results are obtained if alkyl
halide is primary.

Friedel-Crafts alkylation of Anisole: Friedel-Crafts alkylation reaction is the alkylation of aromatic ring using
lewis acid as a catalyst with an alkyl halide.

Friedel-Crafts acylation of Anisole: Friedel-Crafts acylation reaction is the acylation of aromatic ring using
lewis acid as a catalyst with an acyl chloride or acetyl chloride.

51
Distinction between Primary, Secondary and Tertiary alcohols : LUCAS TEST

This test is based upon relative reactivities of various alcohols towards HCl in the presence of anhydrous
ZnCl2. In this test, the alcohol is treated with Lucas reagent which is equimolar mixture of HCl and ZnCl2.
Alcohols are soluble in Lucas reagent and form a clear solution. On reaction, alkyl chlorides are formed which
being insoluble result in cloudiness in the solution.

Neutral Ferric Chloride Test for Phenols: Phenols gives a violet-coloured water soluble complex with neutral
ferric chloride. The complex is a coordination compound in which Iron is hexavalent.
6C6H5-OH + FeCl3 [Fe{OC6H5}6]3- + 3HCl
In general, all compounds containing “enolic group” [ =C-OH] respond to this test. However, the colours of
complexes are different such as green, blue , violet…etc and depend upon the structure of phenols.
Litmus test: Phenol turns blue litmus red, whereas
alcohols have no effect.
Bromine water test: When Phenol is added to Br2water they
give white ppt. Due to formation of 2,4,6-trinitro phenol But ,
alcohol do not give any such test.

Coupling reaction : Azo dye test: Phenol give orange colour, alcohol do not give any reaction.

52
Mechanisms-1: Hydration of ethene to give ethanol :

Mechanism-2: Dehydration of Ethanol to Ethene

Mechanism-3: Dehydration of Ethanol to Ethoxyethane

53
PHENOLS

In phenols, the – OH group is attached to Sp2 hybridised carbon

and thus, the C – O bond acquires a partial double bond
character.

54
Preparation of Phenols:

CHEMICAL PROPERTIES:
Order of reactivity as nucleophile /acidic strength
If we compare acidic strength of primary, secondary and tertiary alcohols we found that more the alkyl
groups attached, more they release electrons and more is the difficulty in loosing H+
Primary Alcohols (Most Acidic), Secondary Alcohols (Acidic) and Tertiary Alcohols (Least Acidic)
Comparison of acidic character of H2O with alcohol.
In alcohols R is electron releasing group which increases
electron density on oxygen and pass of electron from O-H
bond to O becomes difficult due to +I effect. Therefore,
water is more acidic than alcohols.

Chemical properties of phenols

1. Acidic nature of phenols –
o It is sp2 hybridized that is it has 33% s character and 67% p
character that shows it is more electronegative. As a result,
electron density on oxygen decreases and Polarity of O – H bond
increase therefore Hydrogen is not easily removed.
o As compared to alcohol, phenol is more acidic. Because C is sp2 hybridized in phenol and
sp3 hybridized in alcohol.
o The phenoxide ion is more
resonance stabilized than phenol
due to no charge separation as
shown:
In this only one negative charge is getting
dispersed. If electron withdrawing group
is present at ortho or para position like
NO2, CN, Cl etc then they disperse this
negative charge and will enhance acidic
strength, whereas if electron donating
group’s like CH3, alkyl group, NH2 (amine)
or OR etc. are present they decrease the acidic strength. Thus. increasing acidic strength is
55
o-cresol < p-cresol < m-cresol < phenol < o-nitrophenol < 2,4,6-trinitrophenol (picric acid)
Higher Ka and lower pKa value correspond to the stronger acid.
ETHERS:
functional isomers of alcohols. The functional group in them is ROR and are represented as ROR where R is
alkyl group.
IUPAC NOMENCLATURE:
The common name used for them is alkoxy alkane. In this smaller group is written as alkoxy group and bigger
group is written as Alkane.
Depending upon the type of alkyl group on both sides we can classify ethers as:
1. Symmetrical ethers: When it has same R on both sides.
Like: CH3 OCH3 [Methoxymethane] C2 H5 O C2 H5 [ Ethoxyethane]
2. Unsymmetrical ethers: They have different R on both sides.
Like: CH3 O C2H5 Ethyl methyl ether [Methoxyethane]; CH3 O C6H5 [Methoxybenzene]/Anisole
Preparation of ethers
1. By dehydration of alcohol : by this method only primary ethers can be prepared but not tertiary or
unsymmetrical ethers . . Temperature must be maintained at 413k.
2. Williamson’s synthesis - By this method tertiary ether,unsymmetrical ether can be prepared.The
reaction involved is given . If secondary or tertiary ether have to formed than the RX should be
primary.

NOTE : From secondary or tertiary RX we get alkenes instead of ether.

PHYSICAL PROPERTIES :
1. Existence: Lower members are gases and higher members may be liquids or solid due to increase in the
Vander wall force.
2. Dipole moment: The C-O bonds in ethers are polar and thus, ethers have a net dipole moment.
3. Boiling point: Ethers have low boiling point than alcohol because of less polarity.
CH3 CH2 O CH2 CH3 CH3 CH2CH2CH2 OH
Boiling point :307.6K boiling point : 390K
4. Structure is bent. They have larger bond angles than alcohol.
CHEMICAL PROPERTIES:
1.Cleavage by halogen acids: The order of
reactivity of halogen acids towards this
reaction is:
HI > HBr > HCl > HF
The HI bond is broken easily but HF bond
can’t be broken.

In case of unsymmetrical
ether:

56
2.Ring substitution reactions in ethers: (Halogenation, nitration, Friedel craft’s acylation and alkylation)
If we look at resonance structures of ether, we see that attack occur at ortho and para position because on
those position electron density is high so attack of electrophile occurs at ortho and para.

1. Nitration of Anisole:

4. Bromination of Anisole:

MULTIPLE CHOICE QUESTIONS (MCQs)

1. The correct order of boiling point of primary (1°), secondary (2°) and tertiary (3°) alcohols is
(a) 1° > 2° > 3° (b) 3° > 2° > 1° (c) 2° > 1° > 3° (d) 2° > 3° > 1°

2. The correct acidic strength order of the figure

(a) I > II > III (b) III > I > II
(c) II > III > I (d) I > III > II

3. Which compound is predominantly formed

when phenol is allowed to react with bromine in aqueous medium?
(a) Picric acid (b) o-Bromophenol (c) 2, 4, 6-Tribromophenol (d) p-Bromophenol
4. Phenols are more acidic than alcohols because
(a) Phenoxide ion is stabilized by resonance (b) Phenols are more soluble in polar solvents
(c) Phenoxide ion does not exhibit resonance (d) Alcohols do not lose H atoms at all

57
5. Which of the following reagents cannot be used to distinguish between phenol and benzyl alcohol?
(a) FeCl3 (b) Litmus solution (c) Br2/CCl4 (d) All of these

ANSWERS Question 1 2 3 4 5
Answer a b c a C

ASSERTION (A) & REASON (R) BASED QUESTIONS:

(A): Assertion and Reason both are correct statements. And reason is the correct explanation for assertion.
(B): Assertion and reason both are correct statements but reason is not the correct explanation for assertion.
(C): Assertion is correct but reason is wrong statement.
(D): Assertion is wrong but reason is correct statement.
1. Assertion: Phenols are more acidic than aliphatic alcohols.
Reason: The phenoxide ion is more resonance stabilized than alkoxide ion.
2. Assertion: Tertiary alcohols get converted into an alkene instead of a carbonyl Compounds in the
presence of heated metallic copper.
Reason: Tertiary alcohols prefer to undergo dehydrogenation instead of dehydration in the presence of
heated copper
3. Assertion: Ethers are non-polar when they are symmetrical.
Reason: Ethers have a bent structure
4. Assertion: n-Butanol has higher boiling point than 2-methyl propan-2-ol.
Reason: Branching increases the strength of Vander Waals forces
5. Assertion: It is not possible to prepare anisole from chlorobenzene through Willamson’s synthesis.
Reason: The C-Cl bond in chlorobenzene shows partial double bond character due to resonance.

Question 1 2 3 4 5
Answer A C D C A

SHORT ANSWER QUESTIONS: ( 2 MARKS)

Q 1 a). Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular
masses. Explain this fact.
b). Which of the following isomers is more volatile: o-nitrophenol or p-nitrophenol?
Ans: a). Alcohols can form H-bonds with water and break the H-bonds already existing between water
molecules. Therefore, they are in general, soluble in water. On the other hand, hydrocarbons cannot from
H-bonds with water and hence are insoluble in water.
b). o-nitrophenol is more volatile due to intramolecular H-bonding.
Q2. a). Write IUPAC name of CH2(OH)-CH(OH)-CH2(OH).
b). Write the IUPAC name of Aspirin ?
Ans : a). Propane-1,2,3-triol
b). 2-Acetoxybenzoic acid
Q3. a). Arrange the given compounds in the increasing order of acidity:
Phenol, o-nitrophenol and o-Cresol.
b). Alcohols react with active metals like Na, K.. to give corresponding alkoxide. Write down the
decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols ?
Ans: a). o-Cresol < Phenol < o-Nitrophenol
b). Primary >Secondary > Tertiary

58
Q4): Which of the following will have a higher boiling point and why: CH3NH2 or CH3OH?
Ans: CH3OH will have a higher boiling point than CH3NH2. This can be explained as follows. Oxygen is more
electronegative than nitrogen and atomic size of oxygen is less than that of nitrogen. As such, the
intermolecular forces in CH3OH are stronger than in CH3NH2. Therefore, boiling point of CH3OH will be more
than CH3NH2.
Q5) Why primary alcohols are more acidic than secondary alcohol?
Ans. As we move from primary to secondary alcohol, the number of alkyl groups attached to the carbon
having the –OH group increases. Due to +I effect of alkyl group, the electron density on oxygen will be more
in case of 2o alcohols than in case of 1o alcohol. Hence the shared pair of electron of the O-H bond will be
more displaced towards the O atom in case of 1o alcohol than in case of 2o alcohol. Thus the release of proton
(H+) is easier in case of 1o alcohols than in case of 2o alcohol. Therefore, 1o alcohol is more acidic than 2o
alcohol.
SHORT ANSWER QUESTIONS (3 MARKS)
Q 1) a. Explain why phenols do not undergo substitution of the -OH group like alcohols?
b. Explain why nucleophilic substitution reactions are not very common in phenols?
c. Write the IUPAC names of CH3 CH=CH-CH2OH?

Ans: a) The C-O bond in phenols has some double bond character due to resonance and hence cannot be
easily cleaved by a nucleophile. In contrast, the C-O bond in alcohols is a pure single bond and hence can be
easily cleaved by a nucleophile.
b). In phenols, the –OH group is an activating group. It activates the ring particularly at the ortho and
para positions due to greater electron density because of resonance. Hence ring responds to electrophilic
substitution and not nucleophilic substitution.
c). But-2-en-1-ol
Q2) a). Explain why is ortho-nitrophenol more acidic than ortho-methoxyphenol ?
b). Write the product formed when Ethoxy benzene reacts with HI
c). What is the condition of the compound to undergo Iodoform test ?
Ans: a). Due to strong –R and –I-effect of the NO2 group, electron density in the O-H bond decreases and
hence the loss of a proton becomes easy. Further, after the loss of a proton, the-nitro phenoxide ion left
behind is stabilized by resonance, thereby making o-nitrophenol a stronger acid. In contrast, due to +R effect
of the OCH3 group, the electron density in the O-H bond increases thereby making the loss of a proton
difficult. Furthermore, the o-methoxy phenoxide ion left after the loss of a proton is destabilized by
resonance Because the two negative repel each other thereby making o-methoxyphenol a weaker acid. Thus,
o-nitrophenol is more acidic than o-methoxyphenol.
b). Phenol and Ethyl Iodide
c). Presence of terminal methyl group with alcoholic / carbonylgroup [CH3C=O, CH3CH(OH)
Q3) Why is the reactivity of all the three classes of alcohols with conc. HCl and ZnCl2 (Lucas reagent)
different?
Ans: The reaction of alcohols with conc. HCl and ZnCl2 (Lucas reagent) occurs through intermediate
formation of carbocations. Obviously, more stable the carbocation, more reactive is the alcohol. Since the
stability of carbocations follows the order: 3° > 2° >10 , therefore, reactivity of alcohols towards Lucas reagent
follows the same order, i.e. , 3° > 2° > 1°.
CASE-BASED/SOURCE BASED QUESTIONS
Read the passage given below and answer the following questions.
Alcohols and ethers are related to each other as functional isomers but differ widely in characteristics.
Whereas alcohols are completely miscible with water, ethers are nearly immiscible with it. Unlike alcohols
which are extremely reactive, ethers are almost inert chemically. The low reactivity is due to the presence of
C-O-C bond which has which has no active site as is present in case of alcohols(-O-H is polar). However,
59
ethers participate in the formation of co-ordinate complexes by donating the lone electron pairs present
on oxygen and also react with halogen acids under specific conditions involving the cleavage of C-O-C bond.
Answer the following questions:
1. Why is the bond angle in an ether molecule more than the normal tetrahedral bond angle ? 1M
2. What happens when tertiary butyl bromide is heated with sodium ethoxide ? 1M
3. Sodium metal can be used to dehydrate ethers but not alcohols. Why ?
OR
3. Explain Why alcohols and ethers of comparable molecular masses have different boiling points. ? 2M
ANSWERS:
1. Repulsive interactions in the alkyl/Aryl groups on either side of the C-O-C bond are quite large. Therefore,
their bond angles are more than the normal tetrahedral bond angle.
2. 2-Methylpropene or Isobutene is formed as the major product. [Williamson’s ether synthesis must be
performed with primary alkyl halides preferably]. Β-elimination occurs during the above reaction.
3. Since alcohols are weekly acidic, they react with sodium metal to evolve hydrogen. Due to absence of any
C-OH bond in ether, it does not react with sodium metal and the same can be used to prepare anhydrous
ether.
OR
3. Ethers have less boiling points than the isomeric alcohols. For example n-butyl alcohol[C4H9-OH] boils at
390K while diethyl ether [C2H5-O-C2H5] has boiling point of 308K. In alcohols, the molecules are linked by
intermolecular hydrogen bonding which is absent in isomeric ethers. Consequently alcohols boils at higher
temperature.
LONG ANSWER QUESTIONS - 5 MARKS
1. An alcohol A (C4H10O) on oxidation with acidified K2Cr2O7 gives carboxylic acid ‘B’ (C4H8O2). Compound
‘A’ when dehydrated with conc. H2SO4 at 443 K gives compound ‘C’ with aqueous H2SO4 . ‘C’ gives
compound ‘D’ (C4H10O) which is an isomer of ‘A’. Compound ‘D’ is resistant to oxidation but compound
‘A’ can be easily oxidized. Identify A, B, C and D and write their structure.
Ans. A : (CH3)2CHCH2OH and B : CH3CH(CH3)COOH
C : (CH3)2C = CH2 and D : (CH3)3 – C – OH [ Write equations at appropriate places]
ASSIGNMENT:
1) Explain why propanol has higher boiling point than that of the hydrocarbon, butane ?
2) Explain how does —OH group attached to a carbon of benzene ring activate it towards electrophilic
substitution?
3) Suggest a reason for the large difference in the boiling points of butanol and butanal, although they have
the same solubility in water.
4) Give chemical test to distinguish between the following pairs of compounds:
(i) Ethanol and phenol (ii) Methanol and Propan-2-ol.
5) How will you convert: (i) Propene to Propan-l-ol? (ii) Ethanal to Propan-2-ol?
6) Show how will you synthesize:
i)1-phenylethanol from a suitable alkene. ii) Cyclohexyl methanol using an alkyl halide by an SN2
reaction. iii) pentan-1-ol using a suitable alkyl halide?
7) How are the following conversions carried out?
(i) Propene→Propan-2-ol ii) Benzyl chloride →Benzyl alcohol
iii) Ethyl magnesium chloride →Propan-1-ol.
8) Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde. Iii) Brominationofphenolto2,4,6-tribromophenol.
9)..How is aspirin(Acetylsalicylicacid) prepared from salicylic acid?

60
 UNIT-8 ALDEHYDE,KETONE AND CARBOXYLIC ACID
1. Aldehydes and ketones are compounds which contain carbonyl group(>C=O).
2. Carbon of –CHO and ―CO― groups is sp2 hybridized.

3. For IUPAC naming of aldehyde,'-al' is used as suffix and for ketone '-one' is used as suffix.
4. Physical Properties:
HCHO is a gas at normal temperature. Formalin is 40% solution of HCHO. Solubility in water is only
for lower members.
Boiling points: Due to intermolecular dipole-dipole interactions the boiling points of aldehydes and
ketones are comparatively higher than corresponding ethers and hydrocarbons. However, due to
absence of H-bonding, the boiling points of aldehyde/ ketones are lower than those of
corresponding alcohols.
5. Order of reactivity of aldehydes and ketones towards nucleophilic addition is :
(i) HCHO > CH3CHO > CH3CH2CHO.
(ii) HCHO > RCHO > R CO R.
(iii) ArCHO > ArCOR > ArCOAr.
6. Due to presence of a hydrogen atom on the carbonyl group, aldehydes are easily oxidized to the
corresponding carboxylic acids with mild oxidizing agents like Tollens 'reagent, Fehling’s and
Benedict's solution. Thus, aldehydes act as reducing agents.
7. Ketones are not oxidized by mild oxidizing agents but strong oxidizing agents likeHNO3,
KMnO4/KOH, KMnO4/H2SO4, K2Cr2O7/H2SO4 to give mixture of carboxylic acids.
8. Benzaldehyde although reduce Tollens’ reagent, it does not reduce Fehling’s and Benedict’s
solution.
9. Ketones do not give Tollens reagent and Fehling solution test.
10. Aldehydes and ketones with at least one α-H atom get condensed in presence of a base. This is
known as Aldol condensation.
11. Aldol condensation involves carbanion as intermediate.
12. Aldehydes with no α-H atoms undergo Cannizzaro’s reaction.
13. Ketones react with dihydric alcohols to form cyclic ketals.
14. Monocarboxylic acids having (C12C18) carbon atoms, are called fatty acids.
15. Because of stronger hydrogen bonding the boiling points of carboxylic acids are higher than those
of alcohols of comparable molecular mass.
16. Carboxylic acids are weaker than mineral acids, but they are stronger acids than alcohols and
phenols because the conjugate base of carboxylic acid, a carboxylate ion, is stabilised by two
equivalent resonance structures in which the negative charge is at the more electronegative
oxygen atom.
17. Presence of EWGs enhances the acidic character of carboxylic acids.
18. Effect of substituent on the acid strength of aliphatic acids
i) HCOOH.>CH3COOH >(CH3)2CHCOOH.>(CH3)3CCOOH. acidity decreases as the +I effect of alkyl
group increases
ii) FCH2COOH>ClCH2COOH> BrCH2COOH>ICH2COOH>CH3COOH acidity decreases as the –I effect
of halogen decrease
23. -CHO and –COOH group, attached to benzene ring, are deactivating and m-directing.
19. General Methods of Preparation:
(i) Controlled oxidation of 1º alcohols: PCC/CrO3 RCH2OH  RCHO
(ii) Dehydrogenation of 1º alcohols : RCH2OH  RCHO + H2
Cu/573k
61
 iii) Hydrolysis of geminal halides :

20. General Methods of Preparation of Ketones only

(i) Dehydrogenation of 2º alcohols :

(ii) Hydration of alkynes :

Ozonolysis of Alkene O3

(iii) From Grignard's reagents :

HOH/H+
RMgX + R – C ≡ N RCOR+ NH3 + Mg (OH) X
(iv) From acid chlorides :
RCOCl+R2Cd RCOR + CdCl2

CHEMICAL PROPERTIES OF ALDEHYDE AND KETONES

62
63
 (Name Reaction)
H3O+
Stephen H 3C CN + S n C l2 + H C l H 3C CH NH H 3C CHO

CH3 CHO

CrO2Cl2
Etard
H3O+

CHO

Gatterman – CO / HCl
Koch
Anhydrous AlCl3

O H2 O
Rosenmund
C C
reduction H3C Cl Pd / BaSO4 H3C H

O
Clemmensen Zn - Hg
C H 3C CH2 CH3
reduction H 3C CH3 Conc. HCl

Wolff- O
i) NH2-NH2
Kishner C H3C CH2 CH3
reduction H3C CH3 ii) KOH / Ethylene glycol / 

Tollens’ test R-CHO + 2 [Ag(NH3)2]+ + 3 OH- R-COO- + 2Ag + 2H2O + 4 NH3

Fehling’s R-CHO + 2 Cu2+ + 5 OH- R-COO- + Cu2O + 3H2O
test
O
I2 / NaOH
Iodoform C CHI3 + CH3COONa
H3C CH3 OR, NaOI
OH
Aldol dil NaOH 
2 H3C H3C CH CH2 CHO H3C CH CH CHO
condensation CHO

Conc. NaOH
Cannizzaro HCHO
+ HCHO HCOON a + H 3C OH

Hell- i) Cl2 / Red Phosphorus

Volhard- H3C COOH H2C COOH
Zelinsky ii) H2O
Cl
(HVZ)
Distinguish by Chemical Test
1. All aldehydes ( R-CHO) give Tollens’ Test and produce silver mirror.

RCHO + 2 [Ag(NH3)2]+ + 3 OH-  RCOO- + 2 Ag + 2H2O + 4 NH3

Tollens’ Reagent silver ppt

64
 Note: HCOOH(methanoic acid ) also gives this test, ketones(RCOR) do not give this test

2. All aldehydes (R-CHO) and ketones(RCOR) give 2,4-DNP test

RCOR + 2,4-DNP  Orange ppt
R-CHO + 2,4-DNP  Orange ppt

3. Aldehydes and ketones having CH3CO- (keto methyl) group give Iodoform Test. Alcohols
having CH3CH(OH)- group also give Iodoform Test. |

CH3CHO + 3I2 +
4 NaOH  CHI3 +HCOONa + 3 NaI + 3H2O
Yellow ppt
The following compounds give Iodoform Test:
ethanol(C2H5OH), propan-2-ol(CH3CH(OH)CH3), ethanal(CH3CHO), propanone(CH3COCH3), butanone
( CH3COCH2CH3) , pentan-2-one (CH3COCH2 CH2CH3) , acetophenone ( PhCOCH3 )
4. All carboxylic acids ( R-COOH) give Bicarbonate Test
RCOOH + NaHCO3  RCOONa + CO2 + H2O
Brisk effervescence of CO2.
5.FehlingTest-
All aliphatic aldehydes give this test. Aromatic aldehydes do not perform this test. Ketones do not give this
test.
RCHO + Cu2+ + 5OH͞ → RCOO͞ + Cu2O + 3HOH
Red brown ppt of Cu2O is formed on heating aldehyde with Fehling A solution and Fehling B solution.

MCQs
1. Which of the following compounds is most reactive towards nucleophilic addition reactions?

2. The correct order of increasing acidic strength is .

(i) Phenol < Ethanol < Chloroacetic acid < Acetic acid
(ii) Ethanol < Phenol < Chloroacetic acid < Acetic acid
(iii) Ethanol < Phenol < Acetic acid < Chloroacetic acid
(iv) Chloroacetic acid < Acetic acid < Phenol < Ethanol
3. Cannizaro’s reaction is not given by .

65
 4. Through which of the following reactions the number of carbon atoms can be increased in the
chain?
(i) Clemmensen Reduction
(ii) Cannizaro’s reaction
(iii) Aldol condensation
(iv) HVZ reaction
5. The reagent which does not react with both, acetone and benzaldehyde.
(i) Sodium hydrogen sulphite
(ii) Phenyl hydrazine
(iii) Fehling’s solution
(iv) Grignard reagent
6. What is the test to differentiate between penta-2-one and pentan-3-one?
(a) Iodoform test (b)Benedict’s test (c) Fehling’s test (d) Aldol condensation test
7. In order of reactivity of CH3CHO, CH3COC2H5 and CH3COCH3 is
(a) CH3CHO > CH3COCH3 > CH3COC2H5 (b) C2H5COCH3 > CH3COCH3 > CH3CHO
(c) CH3COCH3 > CH3CHO > C2H5COCH3 (d) CH3COCH3 > C2H5COCH3 > CH3CHO
8. To differentiate between pentan-2-one and penta-3-one a test is carried out. Which of the
following isthe correct answer?
(a) Pentan-2-one will give silver mirror test
(b) Pentan-2-one will give iodoform test
(c) Pentan-3-one will give iodoform test
(d) None of these
9. In Clemmensen Reduction carbonyl compound is treated with______.
(i) Zinc amalgam + HCl
(ii) Sodium amalgam + HCl
(iii) Zinc amalgam + nitric acid
(iv) Sodium amalgam + HNO3
10. Which of the following compounds will give butanone on oxidation with alkaline KMnO4 solution?
(i) Butan-1-ol
(ii) Butan-2-ol
(iii) Both of these
(iv) None of these
Ans 1-i ,2-iii,3-i,4-iii,5-iii,6-i,7-i,8-ii,9-i,10-ii
Assertion & Reasoning
NOTE: Here two statements are given - one labelled Assertion (A) and the other labelled Reason (R).
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
66
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true

1 Assertion(A): Hydroxy ketones are not directly used in Grignard reaction.

Reason(R): Grignard reagents react with hydroxyl group.
2 Assertion(A): Aldehydes and ketones have higher boiling point than alcohols.
Reason(R): Alcohols are associated by intermolecular hydrogen bonding while aldehydes and ketones
lack hydrogen bonding.
3 Assertion(A): The order of reactivity of aromatic Aldehydes and ketones towards nucleophilic
addition reaction isC6H5CHO > C6H5 COCH3> C6H5CO C6H5
Reason(R): Aromatic Aldehydes and ketones are less reactive than the corresponding aliphatic aldehydes
and ketones.
4 Assertion (A): Aldehydes and ketones both reacts with Tollen’s reagent .
Reason(R):Both aldehydes and ketones contain a carbonyl group .

Answers : 1-A. 2-D. 3- A .4-D

VERY SHORT ANSWER TYPE QUESTIONS 1 MARK QUESTIONS

1. Suggest a reason for the large difference in the boiling points of butanol and butanal,
although they have same solubility in water.
Ans. The b.p. of butanol is higher than that of butanal because butanol has strong intermolecular H-
bonding while butanal has weak dipole-dipole interaction. However, both of them form H-bonds with
water and hence are soluble.
2. What type of aldehydes undergo Cannizaro reaction?
Ans. Aromatic and aliphatic aldehydes which do not contain α- hydrogens.
3. Out of acetophenone and benzophenone, which gives iodoform test? Write the reaction involved.
(The compound should have CH3CO-group to show the iodoform test.)
Ans. Acetophenone (C6H5COCH3) contains the grouping (CH3CO attached to carbon) and hence gives
iodoform test while benzophenone does not contain this group and hence does not give iodoform test
4. What makes acetic acid a stronger acid than phenol?
Ans. Greater resonance stabilization of acetate ion over phenoxide ion.
5. Arrange the following compounds in increasing order of their acid strength. Benzoic acid, 4-
Nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxy benzoic acid.
Ans. 4-methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid <4, dinitrobenzoic acid.
6. During preparation of esters from a carboxylic acid and an alcohol in the presence of acid catalyst
water or ester formed should be removed as soon as it is formed.
ANS. The formation of esters from a carboxylic acid and an alcohol in the presence of acid catalyst is a
reversible reaction, To shift the equilibrium in forward direction, the water or ester formed should be
removed as fast as it is formed.
7. Cyclohexanone forms cyanohydrin in good yield but 2,2,6 trimethyle cyclo-hexanone does not.
Why?
Ans. In 2,2,6 trimethyl cyclohexanone there is steric hinderance of 3 methyl groups, It does not form
cyanohydrin in good yield.

67
 8. Why formic acid is stronger than acetic acid?
Ans. Due to +I effect, CH3- group in acetic acid increases e- density on carbon atom which makes it weak
acid. While in formic acid no such group is present, hence is more stronger than acetic acid.
9. Arrange the following compounds in an increasing order of their reactivity in nucleophilic addition
reactions : ethanal, propanal, propanone, butanone.
Answer: Butanone < Propanone < Propanal < Ethanal
10. Give a chemical test to distinguish between Benzoic acid and Phenol.
Answer: Benzoic acid forms a brisk effervescence with NaHCO3 solution but phenol does not respond to
this test.
VERY SHORT ANSWER TYPE QUESTIONS 2 MARK QUESTIONS
1. Give the structure and IUPAC name of the product formed when propanone is reacted with
methylmagnesium bromide followed by hydrolysis. Write equation of reaction.
Answer:

2. Name the reagents used in the following reactions :

Answer:
(i) LiAlH4 (Lithium Aluminium Hydride)
(ii) KMnO4 (Alkaline Potassium Permanganate)
3. Write the reagents required in the following reactions :

Answer:

4. Write structures of compounds A, B and C in each of the following reactions:

Answer:A = Ethanal B = 3-Hydroxybutanal (Aldol) C = But-2-enal

5. Arrange the following in the increasing order of their boiling points:
(i) CH3CH2CH2CHO, CH3CH2CH2CH2OH, H5C2-O-C2H5, CH3CH2CH2CH2CH3
(ii) CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
Answer: (i) CH3CH2CH2CH2CH3 < H5C2-O-C2H5, <CH3CH2CH2CHO <CH3CH2CH2CH2OH,
(ii) , CH3OCH3 < CH3CH2CH3 < CH3CHO < CH3CH2OH
6-Draw the structure of the product formed when propanal is treated with zinc amalgam and
concentrated hydrochloric acid.

Answer.

7-a. pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain.
Answer: Due to the presence of a strong electron-withdrawing (-NO2) group in 4-nitrobenzoic acid, it
68
 stabilizes the carboxylate anion and hence strengthens the acid. Therefore, 4-nitrobenzoic acid is more
acidic than benzoic acid and its pKa value is lower.
b. Formaldehyde does not take part in Aldol condensation. Why ?
Answer: Formaldehyde does not contain a-hydrogen atom. Therefore it does not take part in aldol
condensation

Short Answer type questions (03 Marks Question)

Q1 Which acid from each of the following pairs would you expect to be a stronger acid?
(i)CH3COOH or CH2FCOOH
(ii) CH2FCOOH or CH2ClCOOH
(iii) CH2FCH2CH2COOH or CH3CHFCH2COOH
Ans:(i) CH2FCOOH is a stronger acid.
(ii) CH2FCOOH is a stronger acid.
(iii) CH3CHFCH2COOH is a stronger acid.
Q2. How will you convert ethanal into the following compounds?
(i)Butane-1,3-diol
(ii)But-2-enal
(iii)But-2-enoicacid
Ans:

Q3 An organic compound A which has a characteristic odour on treatment with NaOH forms two
compounds B and C. Compound B has a molecular formula C7H8O which on oxidation gives back
compound A. Compound C is sodium salt of acid. C when heated with Sodalime yields an aromatic
hydrocarbon D. Deduce structures A-D.
Answer-: A is C6H5CHO ; B is C6H5CH2OH; C is C6H5COONa ; D is C6H6
Q4 An organic compound A with molecular formula C4H10O undergoes oxidation to give B
C4H8O .B forms an oxime but does not reduce Fehling solution. B reacts with I2 and KOH to
form iodoform. Derive the structures of A and B and write the reactions.
Answer-: B is CH3COCH2CH3 So A is CH3CH(OH)CH2CH3
Q5 An organic compound with molecular formula C5H10O forms 2,4-DNP derivative, but doesnot
answer Tollens test. It was oxidised to a carboxylic acid B with molecular formula C3H6O2 when treated
with alkaline KMnO4 under vigorous conditions. The sodium salt of Bgave a hydrocarbon C on Kolbe’s
electrolytic reaction. Identify A,B,and C and write the reactions involved.
Answer-: A is a ketone B is CH3CH2COOH , C is C2H6 A is CH3CH2CO CH2CH3

69
Q6 Predict the products of the following reactions :

Answer:

LONG ANSWER QUESTION (5 Marks)

1 An organic compound (A) with molecular formula C2Cl3O2H is obtained when (B) reacts with Red P
and Cl2. The organic compound (B) can be obtained on the reaction of methyl magnesium chloride with
dry ice followed by acid hydrolysis.
a. Identify A and B
b. Write down the reaction for the formation of A from B. What is this reaction called?
c. Give any one method by which organic compound B can be prepared from its corresponding acid
chloride.
d. Which will be the more acidic compound (A) or (B)? Why?
e. Write down the reaction to prepare methane from the compound (B).
a. (A): CCl3COOH (B): CH3COOH
b. CH3COOH (i)Red P / Cl2 CCl3COOH, Hell Volhard Zelinsky reaction
(ii)H2O
c. CH3COCl H2O CH3COOH
d. A will be more acidic due to presence of 3 Cl groups (electron withdrawing groups) which increase acidity
of carboxylic acid.
e. CH3COOH (i)NaOH, CaO (ii) heat CH4 + Na2CO3
2(a) Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic
acid is a stronger acid than phenol. Give two reason.
(b) How will you bring about the following converstions?
(i) Propanone to propane (ii) Benzoyl chloride to benzaldehyde (iii) Ethanal to but-2-enal
Answer: (a) Carboxylic acid is a stronger acid than phenol because:
(i) In the resonating structure of phenol and carboxylic acid, the negative charge on the carboxylate ion is
delocalised over two oxygen atoms while they are localized on oxygen atom.

70
 (i) The release of a proton from carboxylic acids is much easier than from phenols

(b) (i) Propanone to Propane

CASE BASED QUESTION ANSWER

Many nucleoplihic additions to carbon-oxygen double bonds are reversible; the overall result of these reactions
depends, therefore, on the position of an equilibrium. This behavior contrasts markedly with most electrophilic
additions to carbon-carbon double bonds and with nucleophilic substitutions at saturated carbon atoms. The
latter reactions are essentially irreversible, and overall results are a functions of relative reaction rates.
i. Arrange following in increasing order of their nucleophilic addition reaction:
Ethanal, Propanal, Propanone, Butanone reaction
Ans - (i) Butanone < Propanone < Propanal < Ethanal
ii. There are two – NH2 groups in semicarbazide. However, only one such group is involved in the formation
of semicarbazones.
(i) Because one of the – NH2 in semicarbazide is involved in the resonance with -CO- group.
iii. Give the structure and IUPAC name of the product formed when propanone is reacted with methyl
magnesium bromide followed by hydrolysis
Answer: IUPAC name : 2-methylpropan-2-ol

71
 OR

iii . Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition
reactions than propanal ? Explain your answer

Ans: Due to electron withdrawing nature of carbonyl group in benzaldehyde the carbon does not attain
positive charge as a result carbon atom of carbonyl is C6H5CHO is less reactive than that of propanal.
(C6H5CHO less polar due to resonance).
(Reasoning Question )
1 Aldehydes (R-CHO) and ketones (R-CO-R) have higher b.pt than hydrocarbon and ether.
Because they are polar so there is dipole-dipole attraction in aldehyde and ketones
2 Lower aldehydes and ketones are miscible with water
Because they form hydrogen bond with water
3 Aldehydes (R-CHO) are more reactive than ketones (R-CO-R) in nucleophilic addition
In ketone the two alkyl groups ( R ) have +I effect so they reduce the electrophilicity co carbonyl
carbon. Also there is steric hindrance in ketone
4 Benzaldehyde ( C6H5CHO ) is less reactive than propanal ( CH3CH2CHO )
Due to resonance the electrophilicity of carbonyl carbon is less in benzaldehyde
4 NaHSO3 is used for separation of aldehydes
It forms a soluble compound with aldehyde which on hydrolysis give back the aldehyde
5 α-H of aldehyde and ketone is acidic in nature
Because the corresponding carbanion is resonance stabilized
6 Carboxylic acids ( R-COOH ) do not give nucleophilic addition reaction like RCHO & RCOR
although it has >C=O
due to resonance the carbonyl carbon looses its electrophilicity
7 Carboxylic acids have higher b.pt than aldehyde, ketones and even than alcohols
There is extensive inter molecular H-bonding in carboxylic acid(RCOOH). Even in vapour
phase it exists as dimer
8 Carboxylic acids are miscible in water
Due to H-bonding with water
9 R-COOH is acidic in nature
Because the conjugate base R-COO- (carboxylate ion) is stable due to resonance
10 Acidic Strength : Cl-CH2-COOH > CH3-COOH > CH3CH2-COOH
Because Cl has –I effect which stabilizes the conjugate base and ethyl gr has +I effect
11 Although phenoxide ion has more no. of resonating structures than carboxylate ion, even though
carboxylic acid is a stronger acid why?

72
 The phenoxide ion has non-equivalent resonance structures in which –ve charge is at less
electronegative C atom and +ve charge as at more electronegative O-atom. In carboxylate ion
–ve charge is delocalized on two electronegative O-atoms hence resonance is more effective.

12 There are two-NH2 group in semicarbazide. However, only one is involved in formation of
semicarbazones. Why?
Due to resonance one NH2 group is involved in resonance and hence can’t participate in the
formation of semicarhazone available for Lone pair of other NH2 group is not involved in
resonance and is nucleophilic attack

73
 UNIT-9 Organic Compounds containing Nitrogen:Amines
Important Terms and Conceptual Areas
 Amines are alkyl or aryl derivates of Ammonia. For example:CH3NH2, C6H5NH2,(CH3)2NH,(CH3)3NH
 Amines have pyramidal structures and act as Lewis bases due to the presence of lone electron
pair on nitrogen atom.
 Nitro compounds, cyanides, amides and oximes upon reduction form primary amines.
H2/Pd/Ethanol
RNO2 → RNH2
Sn/HClOr Fe/HCl
RNO2 → RNH2

 Gabriel-Phthalimde synthesis is very useful for preparing aliphatic primary amines but not aromatic
amines.
 In the primary, secondary and tertiary methyl and ethyl amines, the order of basic strength in aqueous
solution is:

 In non-aqueous solvents such as benzene and also in the gaseous state, the relative basic strength of
methyl amines are in the order of: Tertiary > Secondary > Primary
 Aniline is a weaker base than ethylamine since the electron pair on the nitrogen is involved in
conjugation with the π-electrons of the ring.

 Electron withdrawing substituents decrease the basic strength of aromatic amines and electron
releasing substituents increase the same.
Physical properties of amines:
 Boiling points of amines are less than those of alcohols and carboxylic acids of comparable
molecular masses.
 Amines dissolve in water due to Intermolecular hydrogen bonding. However aromatic amines
do not dissolve in water.
Chemical properties of amines:
 Carbylamine test is given by only primary amines.
 Primary aromatic amines form diazonium salts with nitrous acid in ice cold solution by diazotization
reaction.
 Acylation of amino group with acid chloride or acid anhydride decreases the activation of the ring.
 Aniline give slight yellow precipitate of 2,4,6-tribromoaniline with aq.Bromine. However, in
CS2/CCl4solvent, aniline gives a mixture of ortho and para bromo aniline.
 Nitration of aniline with nitrating mixture gives m-nitroaniline as the major product. However, by
carrying out acylation followed by nitration and then hydrolysis gives a mixture of ortho and para
nitroaniline.
 A coupling reaction is an organic reaction that joins two aromatic rings through a −N=N− bond with
the help of a metal catalyst.
 The diazonium salts have the general formula – R N2 X, where R stands for an aryl group and – X ion
may be Cl– Br, – HSO4, BF4, etc. These are formed by the process of diazotization.
74
 Organic Name Reactions and Tests
01. Hoffmann’s Bromamide Reaction:
R-CO-NH2 + Br2+ 4NaOH R-NH2 + 2NaBr + Na2CO3 + 2H2O
02. Gabriel-phthalimide synthesis:

This is a very useful and convenient method for the preparation of primary amines.
03. Diazotization reaction: -Primary aromatic amines such as aniline react with nitrous acid
[NaNO2+2HCl] under ice cold conditions [273-278k] to form benzene diazonium salt. This reaction is
known as diazotization reaction.
+
C6H5-NH2+ NaNO2+2HClC6H5-N2 -Cl-+ NaCl+ 2H2O
+
CH3-NH2+ NaNO2+2HClCH3-N2 -Cl- → CH3OH + N2 + HCl
Alkyl amines give alcohol in this reaction.
04. Carbylamine Reaction:- Primary aliphatic or aromatic amines upon warming with chloroform and
alc. KOH solution form isocyanides or carbylamines with extremely unpleasant smell. [Note: Secondary
and tertiary amines donot respond to this test]
R-NH2+CHCl3+ 3KOH ----------R-NC +3KCl+ 3H2O[R = alkyl aryl]
Hinsberg’s test :- An amine is shaken with Hinsberg’s reagent [benzenesulphonylchloride] in the presence
of excess of aqueous KOH solution.
a. A primary amine forms N-alkyl benzene sulphonamide which dissolves in aqueous KOH
solution to form potassium salt and upon acidification with dilute HCl regenerates the insoluble
sulphonamide.
b. A secondary amine forms N, N-di alkyl benzene sulphonamide which remains insoluble in
aqueous KOH and even after acidification with dilute HCl.
c. Tertiary amine does not react with benzene sulphonyl chloride and remains insoluble in aqueous
KOH. However, on acidification with dilute HCl it gives a clear solution due to the formation of
the ammonium salt.

75
06. Reactions of diazonium salt:

07. Coupling reaction

MULTIPLE CHOICE

Multiple Choice Questions: [1-Mark questions]

01. The best reagent for converting 2-phenylpropanamide into 2-phenyl propanamine is
a). excess H2 b). Br2 in aqueous NaOH c). Iodine in the presence of red- P4 d). LiAlH4in ether
02. Hoffmann Bromamide [Degradation] reaction is shown by
a). ArNH2 b). ArCONH2 c). Ar-NO2 d). ArCH2NH2
03. Which of the following is a 30amine
a) 1-methylcyclohexylamine c) Triethylamine
b) tert-butylamine d) N-methylaniline
04. The correct IUPAC name for CH2=CHCH2-NH-CH3
a). Allyl methylamine b) 2-amino-4-pentene
c)4-aminopent-1-ene d) N-Methylprop-2-en-1-amine
05. Amongst the following, the strongest base in aqueous medium is
a). CH3-NH2 b) NC-CH2-NH2 c) [CH3]2-NH d) C6H5-NH-CH3
06. The gas evolved when methylamine reacts with nitrous acid is
a). NH3 b) N2 c) H2 d) C2H6
07. Reduction of aromatic nitro compounds using Fe and HCl gives
a). Aromatic oxime b). aromatic hydrocarbon c). Aromatic primaryamine d). Aromatic amide
Cu/HCl
08. The reaction ArN2+Cl- ArCl + N2+ CuCl is named as
76
 a). Sandmeyer’s reaction b) Gatterman reaction c) Claisen reaction d)Carbylamine reaction
09. Best method for preparing primary amines from alkyl halides without changing the
number of carbon atoms in the chain is
a). Hoffmann bromamide reaction
b). Gabriel phthalimide reaction
c). Sandmeyers reaction
d). Reaction with NH3
10. In the nitration of benzene using a mixture of conc. H2SO4and conc. HNO3, the species
which initiates the reaction is
a). NO2 2 b) NO+ c) NO2+ d) NO NaCN H2/Ni Acetic anhydride
Answers

1. d 2. b 3. b 4. d 5. c 6. b 7. c
8. b 9. d 10. c

ASSERTION - REASON QUESTIONS

The questions given below consist of two statements labeled as Assertion(A) and Reason(R). Using the
following key select the correct answer.
a). Both A and R are true and R is the correct explanation of A
b). Both A and R are true but R is not the correct explanation of A
c). A is true but R is false
d). A is false but R is true.
01. Assertion: Acylation of amines gives a mono substituted product whereas alkylation of
amines give poly substituted product.
Reason: Acyl groups technically hinders the approach of further acyl groups
Ans: (c). Correct reason: Alkyl carbocation is a stronger electrophile as compared to acyl
carbocation
02. Assertion: Hoffmann’s degradation reaction is given by primary acid amides only.
Reason : Substituted acid amides do not take part in Hoffmann’s bromamide reaction.
Ans: (a). Reason is the correct explanation for assertion
03. Assertion: N-Ethyl benzene sulphonamide insoluble in alkali.
Reason :Hydrogenattachedtonitrogeninsulphonamideisstrongly acidic.
Ans: ( a ) . Reason is the correct explanation for assertion.
04. Assertion: N,N-Diethylbenzenesulphonamide is insoluble in alkali.
Reason : Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.
Ans: ( b ). Correct reason : N,N-Diethylbenzene sulphonamide does not have acidic hydrogen.
05. Assertion: Only a small amount of HCl is required in the reduction of nitro compounds with Iron
scrap and HCl in the presence of steam.
Reason : FeCl2formed gets hydrolysed to release HCl during the reaction.
Ans: ( a ). Reason is the correct explanation for assertion.
06. Assertion: Acetanilide is less basic than aniline.
Reason :Acetylationofanilineresultsindecreaseofelectrondensityonnitrogen.
Ans: ( a ). Reason is the correct explanation for assertion.

Short answer questions: [2/3-markquestions-answers]

01. Which is more basic, an aqueous solution of [CH3]3N or tetramethyl ammonium hydroxide
Solutionofthe same concentration?
Ans: Trimethylamine is a tertiary amine and is very weakly basic because of steric hindrance and solvation

77
factors which we have discussed under basic character of amines. However, tetramethyl ammonium
hydroxide behave as anionic compound in aqueous solution. Hence, it acts as a strong base.
02. How will you purify aniline containing non-basic impurities?
Ans: Aniline is of basic nature. In order to remove non basic impurities, it is treated with dil. HCl when aniline
forms non-volatile hydrochloride. The impurities can be removed by distillation provided these are low
boiling in nature. From the hydrochloride, aniline can be regenerated by reacting with a base such as sodium
hydroxide.
03. Aniline can’t be prepared by the ammonolysis of chlorobenzene under normal conditions.Discuss.
Ans: In chlorobenzene, the cleavage of C-Cl bond is quite difficult since it has a partial double bond
character due to conjugation. As a result, chlorobenzene does not react with ammonia under normal
conditions and therefore, aniline cannot be prepared by the ammonolysis of chlorobenzene.
04. Why does Aniline even under mild conditions give 2,4,6–tribromoaniline instantaneously when
brominatedwithaqueous solution of Bromine?
Ans: Amino group(-NH2) is highly activating group and activates the ortho and para positions in the ring due
to conjugation. Bromination of aniline involves the electrophilic attack of Br+ ion on the ring and readily
takes place even on shaking aniline with aqueous solution of bromine for a few minutes. As a result, a light-
yellow precipitate of 2,4,6-tribromo aniline is formed.
05. What happens when Aniline reacts with a mixture of nitric acid and sulphuric acid. Explain
Ans: A major portion of aniline is converted into a black tarry mass due to partial oxidation of the ring.
Under strongly acidic conditions, aniline is protonated to form a cation which deactivates the ortho and
para positions in the ring. The nitration occurs mainly at meta position.
06. Primary amines can be alkylated to form secondary amines but tertiary amines &
quaternary ammonium salts are also obtained as the side products. Suggest a way
to convert primary amine to secondary amine only?
Ans: This can be achieved by taking primary amine in excess and alkyl halide in limited amounts.
As a result, only one ‘H ’ atom in-NH2group will be replaced by alkyl group to form a secondary amine.
07. Give plausible explanation for each of the following:
a). Why do primary amines have higher boiling points than tertiary amines?
b). Why are aliphatic amines stronger bases than aromatic amines?
Ans: a). Due to strong intermolecular H-bonding in primary amines.
b). In aromatic amines lone pair is engaged with benzene in resonance.
08. Aniline does not undergo Friedel-Crafts reaction. Justify
Ans: Friedel Crafts reaction is carried out in the presence of Lewis acids such as AlCl3, FeCl3. At the same
time Aniline acts as a Lewis base. Thus, aniline reacts with AlCl3to form asalt.
09. Give a chemical test and reagent to distinguish between ethyl amine and diethylamine.
Ans: Ethylamine reacts with benzene sulphonyl chloride [Hinsberg reagent] and forms a product which is
soluble in alkali. Diethylamine reacts with benzene sulphonyl chloride and forms a product which is
insoluble In alkali.
10. Ethylamine is a stronger base than ammonia. Give reason
Ans: Ethyl amine is more basic than ammonia because the ethyl group is electron releasing and therefore
tend to push the electrons towards nitrogen. This makes the lone pair of ‘N’ morereadily available for sharing
with a proton than NH3in which no electron releasing group is present.

CASE BASED QUESTIONS: 4-Marks

Aniline and other aromatic amines readily take part in the electrophilic substitution reactions such as
halogenations, nitration, sulphonation...etc. Infact, the amino group is an activating group due to the
presence of lone electron pair on the nitrogen atom which is involved in conjugation with the π-electron
pairs in the ring. In certain cases, the ring is so highly activated that it becomes difficult to control the
substitution reaction that is taking place.

78
 1. Why does meta position in aniline not respond to electrophilic substitution?
2. How can you decrease the reactive nature of aniline towards electrophilic
substitution?
3. Out of aniline and anilinium cation which responds more to the electrophilic substitution?
4. Name the product formed when aniline is sulphonated?
Answers:
1. Because amono(-NH2) group activates only the ortho and para positions in the ring.
2. By acylating aniline with acetyl chloride or acetic anhydride to form acetanilide which has
electron withdrawing carbonyl group present.
3. Aniline responds more to the electrophilic substitution.
4. p-amino benzene sulphonic acid, also known as sulphanilic acid is formed.

Assignment Questions
1. Why are alkyl amines stronger bases than aryl amines?
2. Give a chemical test to distinguish between Aniline and Methyl aniline.
3. Why does electrophilic substitutions in aromatic amines takes place more readily than in Benzene?
4. Explain Hinsberg’s test to distinguish between primary, secondary and tertiary amines.
5. How will you convert Aniline into benzene diazonium chloride?
6. Write a chemical test to distinguish between Methylamine and Dimethylamine?
7. Why do primary amines have higher boiling points than tertiary amines?
8. What is Diazotization? Write chemical reaction involved.
9. How will you convert a). Aniline to Chlorobenzene b). Phenol to Aniline
10. How will you convert a). Nitrobenzene to Benzoic acid b). Aniline to Benzyl alcohol
11. Give reasons for the following:
a). o-Toluidine is less basic than Aniline
b). Tertiary amines donot undergo Acylation reactions
12. Write the structures of the following:
a). N-Methyl aniline b). N,N-Dimethyl aniline c). N,N-Dimethylpropan-1 amine
b). Complete the following reaction sequence:
[X] [Y]
R-CO-NH2 R - NH2 R-OH
Identify X and Y.

79
 UNIT-10 BIOMOLECULES

GIST OF THE LESSON

Carbohydrates are optically active polyhydroxy aldehydes or ketones or compounds which
produce such simpler units on hydrolysis.
Classification of Carbohydrates
Monosaccharides Oligosaccharides Polysaccharides

Cannot be hydrolysed to give simpler Yield two to ten Yield a large number of
units of polyhydroxy aldehydes/ ketones monosaccharides on monosaccharides on
Ex : Glucose, fructose, ribose hydrolysis Ex : Sucrose, hydrolysis Ex : Starch,
maltose, lactose cellulose, glycogen

Reducing sugars Non reducing sugars

Reduce Tollen’s and Fehling’s reagent. Do not reduce Tollen’s and Fehling’s reagent.
Sucrose is a non reducing sugar,
All monosaccharides are reducing sugars.

Glucose Preparation
By boiling sucrose with dilute HCl or H2SO4 in alcoholic solution

By boiling starch with dilute H2SO4, at 393 K, under pressure

Structure of Glucose
Glucose has been assigned the above structure based on the following evidences.

 Presence of straight chain of 6 C atoms

 Presence of carbonyl group

a. Reaction with HCN b) Reaction with NH2OH

 Presence of aldehyde group Reaction with Br2

80
  Presence of five -OH groups Reaction with acetic anhydride

 Presence of primary alcohol Reaction with nitric acid

 Evidences of cyclic Structure of Glucose

1. Glucose does not give 2, 4-DNP test, Schiff’s test, and
2. Does not react with NaHSO4 The penta acetate of glucose does not react with hydroxylamine.
This indicates that a free −CHO group is absent from glucose.
3. Glucose exists in two crystalline forms, a and β . The two hemiacetal forms of glucose are
4. called anomers as they differ in the
configuration of –OH group about the 2nd
Carbon atom.The cyclic structure exist in
equilibrium with the open- chain structure.
Haworth structure of glucose 

Glycosidic linkage : Linkage between two monosaccharide units through oxygen atom
Disaccharides
Sucrose Maltose Lactose
On hydrolysis sucrose gives D- (+) On hydrolysis each Commonly called milk sugar,
glucose & D-(-) maltose molecule gives two lactose on hydrolysis gives β D
fructose. α-D – glucose units. galactose & β- D – glucose
The product formed on units.
hydrolysis of sucrose is
called invert sugar as the sign
of rotation changes from. dextro (+)
of sucrose to laevo (−)
Polysaccharides

Starch cellulose Glycogen

It is the main storage It is the It is the Storage

polysaccharide of plants. It is a polymer of a- predominant polysaccharide in
glucose. It consists of two components − constituent of the animal body.
amylase and amylopectin. cell wall of plant It is also Also known as
Amylose has straight chain structure unlike cells. animal starch
amylopectin which is a branched chain It is a Straight-chain because
polymer. polysaccharide, its structure is
Water soluble component of starch is amylose. composed of only β similar to
- D-Glucose. amylopectin

81
Proteins Proteins are polymers of a-amino acids, joined to each other by peptide linkage or peptide bond.
Peptide linkage: Amide formed between −COOH group and −NH2 group of two amino acid molecules.

Amino Acids

Classification of Amino Acids

Based on the relative number of amino and carboxyl groups, they are classified as acidic, basic
and neutral.
Non-essential amino acids Essential amino acids
Amino acids that can be Amino acids that cannot be
synthesized in synthesized in the body, and must
the body be obtained through diet Example −
Example − Glycine, alanine, glutamic Valine, leucine, Isolecuine
acid
Properties of Amino Acids
Exist as dipolar ions, known as zwitter ions, in aqueous solution

In zwitter ion form, amino acids show amphoteric

behaviour. All naturally occurring a-amino acids are optically
active except Glycine which is optically inactive.
Polypeptide − Contains more than ten amino acid molecules

Based on the molecular shape, proteins are classified into two types −
Fibrous proteins Globular proteins

In fibrous proteins, polypeptide chains run parallel and are In this , polypeptide chains coil
held together by hydrogen and disulphide bonds. around, giving a spherical shape.
Fibrous proteins are water insoluble. Ex : Globular proteins are water
Myosin, keratin soluble. Ex : Insulin, albumin

Structures of proteins
Primary structure of proteins: Contains one or more polypeptide chains, and each chain has amino acids
linked with each other by peptide bond in a specific sequence. This sequence of amino acids represents the
primary structure of proteins.
Secondary structure of proteins: It refers to the shape in which a long polypeptide chain can exist; two types
of secondary structures: α helix, β-pleated sheet. Type of interaction occur is peptide bond and hydrogen
bond between C=O and N-H groups at various points. There are two secondary structures of proteins: α-
Helix and β-pleated sheets.
Tertiary structure of proteins: It refers to the overall folding of the polypeptide chains. They exist as fibrous
and globular proteins. The structures are stabilized by hydrogen bonds, disulphide linkages, Vander Waals
forces and electrostatic forces.
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Quaternary structure of proteins: It is the Spatial arrangement of sub-units, each containing two or more
polypeptide chains.
Denaturation of Proteins Loss of biological activity of proteins occurs due to change in temperature or PH.
In this process unfolding of globules and uncoiling of helix occurs. Secondary & tertiary structures of
proteins are destroyed in denaturation. Example − Coagula on of egg white on boiling, curdling of milk.

Vitamins
Organic compounds required in the diet in small amounts to maintain normal health, growth and nutrition
Classified into groups –
Water-soluble vitamins Fat-soluble vitamins
Vitamin C, B-group vitamins (B1, B2, B6, B12) Vitamins A, D, E and K

Vitamin C cannot be stored in our body because it is water soluble. As a result, it is readily excreted in
the urine

Vitamin Source Deficiency disease

Vitamin A Fish liver oil, carrots, Xerophthalmi

butter and milk a, night blindness
Vitamin B1 Yeast, milk, green vegetables and Beri beri
cereals
Vitamin B2 Milk, egg-white, Cheilosis, digestive disorders and burning
liver, kidney sensation of the skin
Vitamin B6 Yeast, milk, egg yolk, Convulsions
cereals and grams
Vitamin B12 Meat, fish, egg and Pernicious anaemia
curd

Vitamin C Citrus fruits, amla Scurvy

and green leafy
vegetables
Vitamin D Exposure to sunlight, Rickets and osteomalacia
fish and egg yolk

Vitamin E Vegetable oils like wheat germ Increased fragility of

oil, sunflower oil RBCs and muscular
weakness
Vitamin k Green leafy vegetables Delay of blood clotting
Nucleic Acids Nucleic acids are polymers of nucleotides which are linked through phosphodiester linkage.
Nucleic acid contains a pentose sugar, phosphoric acid and a base (heterocyclic compound
containing nitrogen).
A Nucleotide is made up of sugar, base and a phosphate group. Nucleosides contain only a sugar & a base.
DNA (Deoxyribonucleic acid) RNA(Ribonucleic acid)
sugar is β-D-2-deoxyribose; sugar is β-D-ribose
Bases in DNA: Adenine (A), guanine (G), Bases in RNA: Adenine (A), guanine (G),
cytosine (C) and thymine (T) cytosine (C) and uracil

83
 In secondary structure, the helices of RNA consists of single-stranded helices.
DNA Types of RNA:
are double-stranded The two strands of Messenger RNA (m-RNA) Ribosomal RNA
DNA are complementary to each other. (r-RNA)
Reason: H-bonds are formed between Transfer RNA (t-RNA) Proteins are
specific pairs of bases. DNA is the synthesized by RNA molecules in the cells
chemical basis of heredity at the command of DNA.

MULTIPLE CHOICE QUESTIONS

1) Which of the following statement is not true about glucose?
(a) it is an aldohexose (b) on heating with HI it forms n-hexane
(c) it is present in furanose form (d) it does not give 2, 4-DNP test
Ans : c
2) In a protein molecule amino acids are linked together by:
(a) peptide bond (b) dative bond (c) glycosidic bond (d) phospodiester bond
Ans: a
3) The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA.
(a) 1st (b) 2nd (c) 3rd (d) 4th
Ans: b
4) Which of the following statement is not true about glucose?
(a) it is an aldohexose (b) on heating with HI it forms n-hexane
(c) it is present in furanose form (d) it does not give 2, 4-DNP test
Ans : c
5) In A protein molecule, amino acids are linked together by:
(a) peptide bond (b) dative bond (c) glycosidic bond (d) phospodiester bond
Ans: a

Assertion and Reason type questions

Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each
of these questions also has four alternative choices, only one of which is the correct answer. You have to
select one of the codes (a), (b), (c), (d) given below :
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false
1. Assertion: D(+)-glucose is dextorotatory in nature.
Reason: ‘D’ represents its dextrorotatory nature.
2. Assertion: Vitamin D can be stored in our body.
Reason: Vitamin D is fat soluble vitamin.
3. Assertion: β-glycodisic linkage is present in maltose.
Reason: Maltose is composed of two glucose units in which C-1 of one glucose unit is linked to C-4 of
another glucose unit.
4. Assertion: In presence of enzyme, substrate molecule can be attacked by the reagent effectively.
Reason: Active sites of enzymes hold the substrate molecule in a suitable position.
5. Assertion: All naturally occurring α-aminoacids except glycine are optically active.
Reason: Most naturally occurring amino acids have L-configuration.
Answers. 1. (c) 2. (a) 3.(d) 4.(a) 5.(b)

84
 2 MARKS QUESTIONS

1. What happens when D-glucose is treated with. the following reagents?

(i) HI (ii) Bromine water
Ans: (i)Forms n- hexane (ii) oxidized to six carbon carboxylic acid (gluconic acid)
2. Differentiate between α- helical and β- pleated sheet structure of secondary protein?
Ans:
α- helical structure β-pleated structure
Intramolecular hydrogen bonding forms . Beta sheets are formed by linking two or
within the polypeptide chain to create a more beta strands by intermolecular
spiral structure hydrogen bonds.
Amino acids exist in the right-handed coiled Amino acids exist in an almost entirely
rod-like structure. extended conformation, i.e. linear or sheet-
like structure

3. What are the hydrolysis products of (i) Maltose (ii) lactose

Ans: i) The product of hydrolysis of maltose is alpha-D-glucose
ii) The product of hydrolysis of Lactose are β-D galactose and β-D glucose.

4. (a)Draw the Pyranose structure of ∝-D-(+)-glucopyranose.

(b) Name the linkage that binds two nucleotides in polunucleotide.

Ans: a) b) Phosphodiester

3 MARKS QUESTIONS
1. (a)What is the biological effect of denaturation of proteins? Which structure of protein is affected by
denaturation?
(b) What are enzymes?
Ans: (a) On denaturation, protein globules unfold and unhelix gets uncoiled and protein looses its
biological activity.Secondary and tertiary structures are affected by denaturation.
(b) Enzymes are biocatalyst which are very specific for a particular reaction and a particular
substrate. Almost all enzymes are globular proteins.
2. List the reactions of glucose which cannot be explained by its open chain structure.
Ans. (i) Despite having the aldehyde group, glucose does not give 2, 4 DNP test or Schiffís test.
(ii)It does not form hydrogen sulphite addition product with NaHSO3.
(iii) The penta acetate of glucose does not react with hydroxylamine indicating the absence of free -
CHO group.
3. How are vitamins classified?
Ans: (i) Fat soluble vitamins – which are soluble in fats and oils. e.g., vitamins A, D, E & K.
(ii) Water soluble vitamins – which are soluble in water e.g., vitamins B& C.

CASE STUDY QUESTIONS (4 MARKS QUESTIONS)

Amino acids contain an —NH2 as well as a —
COOH group. In many non-polar solvents, they
exist in their neutral form, but in aqueous
solution, they exist as dipolar ions (Zwitter ions).
85
This explains their several characteristics
properties, like composition on heating,
solubility in water, large dipole moment. If the
pH is lowered significantly, say to pH 1 or 2,
then carboxylate ion will be protonated,
likewise at a very high pH, the free amino group
is exposed by deprotonation of ammonium ion.

There is a pH corresponding to each amino acid where it

remains neutral and neither moves towards cathode nor
anode when the electric field is applied. This pH of the solution is referred to as isoelectric point. For example,
the isoelectric point of alanine is 6.01, that of isoleucine is 6.02 and so on. Hence, the ionic form of the amino
or carboxylic group is the effect of pH on the functional group in the side chain of amino acid. The side chain
of many amino acids contain a functional group that can also be protonated or deprotonated.
Answer the following questions:
(i) Define zwitter ion.
(ii) What is isoelectric point
(iii) In an amino acid, the carboxyl group ionises at pKa1=2.34 and ammonium ion at pKa2=9.60. The
isoelectric point of the amino acid is at pH ?
OR
(iii). How does the zwitter ion behaves it the pH of solution is significantly high?
Ans: (i) In aqueous solution, amino acids exist as a dipolar ion known as zwitter ion. In aqueous solution, the
carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as
zwitter ion. This is neutral but contains both positive and negative charges.

(ii) Isoelectric point : The pH at which there is no net migration of any ion towards electrode e.g., amino acids
have isoelectric point at pH = 5.5-6.3.
(iii). The isoelectric point of the amino acid is at : pH=pKa1+pKa2/2=2.34+9.60/2=5.97
OR
(iii) at a very high pH, the free amino group is exposed by deprotonation of ammonium ion.

86
 5 MARKS QUESTIONS
1. a) What is the difference between native protein and denatured protein?
b) Which one of the following is a disaccharide? Glucose, Lactose, Amylose ,Frusctose
c) Define the terms nucleotide and nucleoside.
d) What type of linkage is present in saccharides
e) Which vitamin helps in Blood clotting .
Ans. a) Protein found in biological system with unique 3D structure and biological activity is called native
protein. When a protein its native form is subjected to change such as change in temperature, pH,its 20
and 30 structures are destroyed and it loss its biological activity. This protein formed is called denatured
protein.
b) Lactose
c) A unit formed by the combination of only sugar and a nitrogenous base is
called nucleoside.
A unit formed by the combination of sugar, nitrogenous base and a
phosphate group as well is called a nucleotide.
d) Glycosidic linkage
e) Vitamin K

87
 KENDRIYA VIDYALAYA SANGATHAN, DELHI REGION
SAMPLE PAPER -1

CLASS-XII Subject-Chemistry
M.M-70. Time: 3Hours

Read the following instructions carefully.

a) There are 33 questions in this question paper with internal choice.

b) SECTION-A consists of 16 multiple-choice questions carrying 1 mark each.
c) SECTION-B consists of 5 very short answer questions carrying 2 marks each.
d) SECTION-C consists of 7 short answer questions carrying 3 marks each.
e) SECTION-D consists of 2 case- based questions carrying 4 marks each.
f) SECTION-E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed.

SECTION A
The following questions are multiple-choice questions with one correct answer. Each
question carries 1 mark. There is no internal choice in this section.

1. A compound X with the molecular formula C2H8O can be oxidised to another

compound Y whose molecular formulae is C3H6O2. The compound X may be
(a) CH3CH2OCH3 (b) CH3CH2CHO (c) CH3CH2CH2OH (d) CH3CHOHCH3
2. The cell constant of a conductivity cell .
(a) changes with change of electrolyte.
(b) changes with change of concentration of electrolyte.
(c) changes with temperature of electrolyte.
(d) remains constant for a cell.

3. In the first order reaction, the concentration of the reactant is reduced to 1/4th in
60minutes. What will be its half-life?
(a) 120 min (b) 40 min (c) 30 min (d) 25 min
4. What is the ratio of the rate of decomposition of N2O5 to the rate of formation of
NO2+2N2O5(g) ------------> 4NO2(g) + O2(g)
(a) 1:4 (b) 2;1 (c) 4:1 (d) 1:2
5. The potential energy diagram in the reaction R → P is given. ∆H° of the reaction corresponds to the

energy.
(a) a (b) b (c) c (d) a + b

88
 6. Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5 . What
is its atomic number?
(a) 25 (b) 26 (c) 27 (d) 24
7. The hybridization of Co in high spin complex, K3[CoF6] is
(a) sp3d 2 (b) sp3 (c) d2sp3 (d) sp3d
8.The primary and secondary valency of copper in the complex [Cu(NH3)4]SO4 are
(a) 2, 4 (b) 4, 2 (c) 0, 4 (d) 1, 4
9. Which of the following will not give iodoform test?
(a) Ethanol (b) Ethanal (c) Pentan-3-one (d) Pentan-2-one
10. Oxidation of toluene to benzaldehyde by the use of chromyl chloride is called
(a)Wurtz reaction (b) Fittig reaction (c) Etard’s reaction (d) Rosenmund’s reaction
11. In the given sequence, Z is

a. Cyanoethane (c) Ethanamide

b. Methanamine (d) Ethanamine
12. Which of the following is the strongest base?

INSTRUCTIONS: In the following questions a statement of assertion followed by a statement of

reason is given. Choose the correct answer out of the following choices.
a. Both assertion and reason are true and the reason is the correct explanation of assertion.
b. Both assertion and reason are true but the reason is not the correct explanation of
assertion.
c. Assertion is true but reason is false.
d. Assertion is false but reason is true.
13. Assertion: On nitration with dil. HNO3 phenol forms 2- nitrophenol and 4-
nitrophenol.
Reason: The presence of –OH group in phenols deactivates the aromatic ring towards
electrophilic substitution and directs the incoming group to ortho and para positions.
14. Assertion: D (+) -Glucose is dextrorotatory in nature.
Reason: D represents its dextrorotary nature.
15. Assertion: Transition metals show very low degree of para magnetism.
Reason: Transition metal atoms have large number of unpaired electrons.
16. Assertion: Nitration of aniline can be conveniently done by protecting the amino
group by acetylation.
Reason: Acetylation increases the electron-density in the benzene ring.
SECTION B
This section contains 7 questions with internal choice in two questions. The following
89
 questions are very short answer type carry 2 marks each.
17. The rate constant for the decomposition of hydrocarbons is 2.418×10−5 s−1 at 546
K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential
factor? (given: antilog .592=3.90)
18. Give a reaction which:
i)Shows that all the carbon atoms in glucose are linked in a straight chain.
ii)Do not explain its open chain structures
OR
Answer the following questions:
(a) Why are vitamin A and vitamin C essential for us?
(b) What is the difference between a nucleoside and a nucleotide?
19. Explain why
a)the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
b)alkyl halides, though polar, are immiscible with water?
OR
Give the uses each of freon 12 and DDT.
20. Write the IUPAC name and all the geometrical isomers of [Pt(NH3)(Br)(Cl)
(Py)]
21. Write the chemistry of recharging the lead storage battery, highlighting all the materials
that are involved during recharging.
22. Define the following:
a)Order of a reaction
b)Activation energy of a reaction
23. Give correct explanation for each of the following:
i)Cyclohexanone forms cyanohydrin in good yield but 2,2,6-
trimethylcyclohexanone does not.
ii) There are two –NH2 groups in semicarbazide. However, only one is involved in
the formation of semicarbazones.
SECTION C
This section contains 5 questions with internal choice in two questions. The following questions
are short answer type and carry 3 marks each.
24. Write the names of the reagents and equations for the preparation of the following
ethers by Williamson’s synthesis:
a)1-Propoxypropane
b)2-Methoxy-2-methylpropane
c)Ethoxybenzene
25. Explain on the basis of valence bond theory:
(i) [Ni (CN)4]2– ion with square planar structure is diamagnetic and the [NiCl4]2– ion with
tetrahedral geometry is paramagnetic.
ii)Explain [Co (NH3)6]3+ is an inner orbital complex whereas [Ni (NH3)6]2+ is an outer orbital
complex.
26. Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the
decomposition is a first order reaction, calculate the rate constant of the reaction.
OR
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate
constants for a first order reaction?
27. Arrange the following in increasing order of their kb values in gaseous phase:
a)C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, (C2H5)2NH
b)C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2

90
 c)CH3NH2, (CH3)2NH, (CH3)3N, C6HsNH2, C6H5CH2NH2
28( a). Among the isomeric alkanes of molecular formula C5H12, identify the one that on
photochemical chlorination yields
i)A single monochloride.
ii)Three isomeric monochlorides.
iii)Four isomeric monochlorides.
b)Which alkyl halide from the following pairs would you expect to react more rapidly by an
SN2 mechanism? Explain your answer.

OR
How the following conversions can be carried out:
i)Propene to propan-l-ol
ii)l-Bromopropane to 2-bromopropane
iii)Benzene to 4-bromonitrobenzene
SECTION D
The following questions are case-based questions. Each question has an internal choice and
carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that
follow.
29. Proteins are the most abundant biomolecules of the living system. The chief sources of
proteins are milk, cheese, pulses, fish, meat, peanuts, etc. They are found in every part of
the body and form a fundamental basis of the structure and functions of life.Theseare also
required for the growth and maintenance of the body. The word protein is derived from
the Greek word, ‘proteios’ meaning ‘primary’ or of ‘prime importance’. Chemically,
proteins are the polymers in which the monomeric units are the α-amino acids. Amino
acids contain an amino (-NH2) and carboxylic (-COOH) functional groups. Depending
upon the relative position of the amino group with respect to the carboxylic group, the
amino acids can be classified as α, β, and γ- amino acids. Amino acids which are
synthesized by the body are called non-essential amino acids. On the other hand, those
amino acids which cannot be synthesized in the human body and are supplied in the form
of diet (because they are required for proper health and growth) are called essential amino
acids.
Answer the following questions
a) An amino acid may be acidic, alkaline or neutral. How does this
happen?
b) Give one example each of essential and non-essential amino acids in
human food
c) Describe what you understand by primary structure and secondary
structure of proteins.
OR
What is a dipeptide. Name the linkage present.

30. An ideal solution may be defined as the solution which obeys Raoult's law exactly over the
91
entire range of concentration. The solutions for which vapour pressure is either higher or lower
than that predicted by Raoult's law are called non-ideal solutions. Non-ideal solutions can show
either positive or negative deviations from Raoult's law depending on whether the A-B interactions
in solution are stronger or weaker than A - A and B - B interactions.

Answer the following questions

a) Give one example each of an ideal and non-ideal solution.
b) State Raoult’s law.
c) Give two differences between positive deviation and negative deviation.
OR
Name the Azeotropes formed by solutions showing positive deviation and negative
deviation respectively. Give one example each.
SECTION E
This section contains 3 questions with internal choice in two questions. The following
questions are very long answer type and carry 5 marks each.
31. a) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S
cm2mol-1. Calculate the conductivity of this solution.
b) What is the effect of catalyst on:
(i) Gibbs energy (ΔG) and (ii) activation energy of a reaction?
c) The cell in which the following reaction occurs: 2Fe3+ (aq) + 2I– (aq) —> 2Fe2+ (aq)+I2
(s) has E°cell=0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium
constant of the cell reaction.
OR
a). Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this
dilution at298 K. Calculate the electrode potential.
[ E°Zn2+ /Zn = – 0.76 V]

b) Two half cell reactions of an electrochemical cell are given below:

MnO4-(aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e–, E° = + 0.15 V
Construct the redox equation and predict if this reaction is feasible
c) The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1.
Calculate its degree of dissociation and dissociation constant Given λ°(H+)=349.6 S
cm2 mol-1 andλ°(HCOO-) = 54.6 S cm2 mol-1
32 a) Alkenes ( C=C ) and carbonyl compounds (C=0) contain a pie bond, but alkenes show
electrophilic addition reactions, whereas carbonyl compounds show nucleophilic
addition reactions. Explain.
b) Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity
towards HCN)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH,
CH3CH2CH2COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4- Methoxybenzoic
acid (acid strength)

92
 OR
When liquid ′A′ is treated with a freshly prepared ammonical silver nitrate solution, it gives
a bright silver mirror. The liquid forms a white crystalline solid on treatment with
sodium hydrogen sulphite. Liquid ′B′ also forms a white crystalline solid with sodium
hydrogen sulphite, but it does not give a test with ammonical silver nitrate.
a Which of the two liquids is aldehyde?

b Write the chemical equations of these reactions

c.Write down functional isomers of a carbonyl compound with molecular formula C3H6O.
d Which isomer will react faster with HCN and why?
33. Assign a reason for each of the following observations:
(i) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting
and boiling points.
(ii) The ionization enthalpies (first and second) in the first series of the transition elements are
found to vary irregularly.
(iii) The highest oxidation state of a transition metal is usually exhibited in its oxide.
(iv) The oxidising power of the following three oxoions in the series follows the order:
+
VO2 < Cr O 2- <MnO4-
2 7

(v) Co2+. is easily oxidised in the presence of a strong ligand.

MARKING SCHEME
SAMPLE PAPER-01 (2024-25)
CHEMISTRY THEORY(043)
SECTION A
Q1 to 16 each correct answer 1 mark
1 (c) 2 (d) 3 (c) 4(d) 5(c) 6(b) 7(a) 8(a) 9(c) 10(c)
11(d) 12(d) 13(c ) 14(c) 15(d) 16(c)
SECTION B

17. The expression for the preexponentail factor can be obtained from Arrhenius equation.

12
pre -exponential factor, A=3.90×10 /s.

93
18. a

b) glucose + NaHS03 -----> no addition product. Glucose + 2,4-

DNP -------> no reaction.
OR

(i) Because deficiency of vitamin A and vitamin C causes night blindness and scurvy
respectively.
(ii) Nucleoside: A nucleoside contains only two basic components of nucleic acids i.e. a
pentose sugar and a nitrogenous base.
Nucleotide: A nucleotide contains three basic components of nucleic acids i.e a pentose
sugar, a nitrogenous base, a phosphate molecule.

19. (i) In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl
chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. the sp2 hybridized carbon atom
is more electronegative. Therefore, the density of electrons of C−Cl bond near the Cl-atom is less
in chlorobenzene than in cyclohexyl chloride.
(ii) cannot form hydrogen bond with water
OR
Freon-12 (dichlorodifluoromethane, CF2Cl2) is also known as CFC. It’s found in
deodorants, hair sprays. also utilized as a refrigerant in refrigerators and air
conditioners.
DDT’s Applications DDT (p, p-dichlorodiphenyltrichloroethane) is the most widely used
insecticide.

20. Amminebromochloropyridineplatinum(II)

21 . write correct equations.

22. write correct definatins.
23 i) because of presence of three methyl groups at -position w.r.t carbonyl group which
hinder the Nucleophilic attack of CN group due steric hindrance. No such steric
hindrance in Cyclohexanone.
(ii) Semicarbazide has two –NH2 groups but one of these which is directly attached to C=O is
involved in resonance. Electron density on NH2 group decreases hence it does not act as
nucleophile.

(iii) It is a reversible reaction. Therefore, to shift the equilibrium in the forward direction, the
water or the ester should be removed as fast as it is formed.

SECTION C

24. correct reagent and correct equation in each part.

25. (I)[Ni (CN)4]2– square planar geometry and dsp2 hybridization. [NiCl4]2– Tetrahedral
geometry and sp3 hybridization.
(ii)[Co(NH3)6]3+ is an inner orbital complex, d2sp3 hybridization. [Ni (NH3)6]2+ is an
outer orbital complex, sp3d2 hybridization.

94
 26

OR

The unit of rate is: bar min–1 or bar s–1

27.

28. a)i) single mono-chloride is neo-pentane.

ii) The isomer should be n−pentane.
iii) The isomer is 2−methylbutane.
b bromo butane , no steric hindrance
OR
Correct conversion in each part.
SECTION D

29. a) Equal number of amino and carboxyl groups makes it neutral, more amino group
means basic and more carboxylic group means acidic amino acid.
b) ESSENTIAL: Histidine, Isoleucine, Lysine. NON
ESSENTIAL: alanine, arginine, asparagine,
c) The primary structure is comprised of a linear chain of amino acids. The secondary
structure contains regions of amino acid chains that are stabilized by hydrogen bonds
from the polypeptide backbone.
These hydrogen bonds create alpha-helix and betapleated sheets of the secondary
structure.
OR
Peptide composed of two amino acid units. Peptide linkage
30. a) IDEAL SOLUTION: Example- Benzene +Toluene or any other NON IDEAL
SOLUTION: EXAMPLE - CS2+CH3COCH3 or any other.

b correct stat ement nd expression

c any two differences.
OR
Positive deviation: minimum boiling Azeotropes, e.g solution of 95% ethanol in water by
volume
Negative deviation: maximum boiling Azeotropes, e.g solution of 68% nitric acid and 32%
water by mass

95
 SECTION E
31 a) C = 1.5 M, Λm = 138.9 S cm2 mol-1
Λm = K×1000c

∴K = Λm×C1000=138.9×1.51000 = 0.20835 S cm-1

b (i) There will be no effect of catalyst on Gibb’s energy.

(ii) The catalyst provides an alternative pathway by decreasing the activation energy of a
reaction.

OR

The electode reaction is given as

Zn+2 + 2e- → Zn

b) The Net Reaction = 2MnO– (aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ +8H2O
4
Now E°cell = E°cathode – E°anode

= 1.51 – 0.15 = + 1.36 V

∴ Positive value of E°cell favours formation of product c)

96
 32 a). Alkenes undergo electrophilic addition, whereas aldehydes and ketones undergo
nucleophilic addition because, in alkenes, the double bond joins two carbon atoms, and
there is no resultant polarity. While in carbonyl compounds, the double bond joins
atoms having different polarities. The polarity in the carbonyl bond makes it vulnerable
to a nucleophile addition reaction.

b (i) Acetaldehyde > Acetone > Methyl tert-butyl ketone > Di-tert-butyl ketone

(ii) (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH (Br)CH2COOH <

CH3CH2CH(Br)COOH
(iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4- Dinitrobenzoicacid
OR
a) Liquid ′A′ must be Aldehyde because it reacts with both NaHSO3 and ammonical silver
nitrate.
b) write correct equations.
c) Functional isomers of C3H6O containing carbonyl group are CH3CH2CHO (Propanal)
and CH3COCH3 (Propanone).
d) Propanal CH3CH2CHO will react faster with HCN because less steric hindrance and electronic
factors increase its electrophilicity.
33. (i) Because of stronger metallic bonding and high enthalpies of atomization.
(ii) Due to irregulaties in the electronic configuration there is irregularities in the enthalpies
of atomisation. Hence there is irregular variation in I.E. (iii)due to its high
electronegativity, low ionisation energy and small size.
(v)Co2+ ion is easily oxidised to Co3+ ion in presence of a strong ligand because of its
higher crystal field energy which causes pairing of electrons to give inner orbital
complexes (d2sp3).

KENDRIYA VIDYALAYA SANGATHAN, DELHI REGION

SAMPLE PAPER -2

CLASS-XII Subject-Chemistry
M.M-70. Time: 3Hours

Read the following instructions carefully.

97
a) There are 33 questions in this question paper with internal choice.
b) SECTION-A consists of 16 multiple-choice questions carrying 1 mark each.
c) SECTION-B consists of 5 very short answer questions carrying 2 marks each.
d) SECTION-C consists of 7 short answer questions carrying 3 marks each.
e) SECTION-D consists of 2 case- based questions carrying 4 marks each.
f) SECTION-E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed.

SECTION-A
1. An electrochemical cell behaves like an electrolytic cell when
(a) Ecell = Eexternal (c) Ecell = 0
(b) Eexternal > Ecell (d) Eexternal < Ecell
2. Which of the following statements is not correct?
(a) Copper liberates hydrogen from acids.
(b) In its higher oxidation states, manganese forms stable compounds with oxygen and
fluorine.
(c) Mn3+ and Co3+ are oxidizing agents in aqueous solution.
(d) Ti2+ and Cr2+ are reducing agents in aqueous solutions.
3. The number of unpaired electrons in gaseous species of Mn3+, Cr3+ and V3+ respectively are:
(a) 4, 3 and 2
(b) 3, 3 and 2
(c) 4, 4 and 2
(d) 3, 3 and 3
4. The potential Energy diagram for a reaction, R → P is shown below. Enthalpy change of the
reaction corresponds to ----

(a) a (b) a + b
(c) c (d) b
For visually impaired students only

98
 For a reaction X + 2Y → Z The Rate law expression is Rate = k[X][Y]1/2 What is the order of the
reaction

(a) 1 (b) 2 (c) 1.5 (d) 3

5. Which of the following statement is correct in pseudo unimolecular reactions.
(a) Both reactants are present in low concentrations.
(b) Both reactants are present in the same concentrations.
(c) One reactant is present in excess.
(d) One reactant is non-reactive.
6. The table below shows the KH values for some gasses at 293 K and at the same pressure.
KH(kbar) 144.97 69.16 76.48 34.86
Gases Helium Hydrogen Nitrogen Oxygen
In which of the following are the gases arranged in their decreasing order of solubility
(from left to right)?
(a) Helium > Nitrogen > Hydrogen > Oxygen
(b) Hydrogen > Helium > Nitrogen > Oxygen
(c) Nitrogen > Hydrogen > Oxygen > Helium
(d) Oxygen > Hydrogen > Nitrogen > Helium
7. The IUPAC name of [Pt (NH3)2Cl (NO2)] is
(a) Platinumdiaminechloronitrite
(b) Chloronitrito-N-ammine platinum(II)
(c)Diamminechloridonitrito-N-platinum(II)
(d) Diamminechloronitrito-N-platinum(III)
8. Which of the following complexes are homoleptic
(a) [Co(NH3)6] 3+ (c) [Ni(H2O)Cl2]2-
(b) [Co(NH3)4Cl2] + (d) [Ni(NH3)4Cl2]
9. Predict the product of the reaction. CH3-CH2-CH2-O-CH3 + HBr →
(a) Propan-2-ol and Bromomethane
(b) Propan-1-ol and bromomethane
(c) 1-Bromopropane and Methanol
(d) 2-Bromopropane and Methanol
10. Dehydration of alcohol is an example of
(a) addition reaction
(b) elimination reaction
(c) substitution reaction

99
(d) redox reaction
11. Which of the following alcohols will not undergo oxidation?
(a) Butan-1-ol
(b) Butan-2-ol
© 2-Methylbutan-2-ol
(d) 3-Methylbutan-2-ol
12. Which of the following statements concerning methylamine is correct?
(a) Methylamine is stronger base than NH3
(b) Methylamine is less basic than NH3
(c) Methylamine is slightly acidic
(d) Methylamine forms salts with alkali

In the following questions, two statements (Assertion) A and Reason (R) are given. Select the most
appropriate answer from the options given below:
(a) If A and R both are correct and R is the correct explanation of A
(b) If A and R both are correct but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
13. Assertion: Hoffmann’s bromamide reaction is given by primary amines.
Reason: Primary amines are less basic than secondary amines.
14. Assertion: Alkyl halides are more reactive towards nucleophilic substitution reaction as compared
to aryl halides
Reason: In alkyl halides halogen is connected to sp³ hybridised carbon.
15. Assertion: Gabriel phthalimide reaction can be used to prepare aryl and alkyl amines.
Reason: Aryl halides do not undergo nucleophilic substitution reaction easily with phthalimide ion.
16. Assertion: On prolonged heating with HI, glucose forms n-hexane.
Reason: In glucose, all the six carbon atoms are linked in a straight chain.
SECTION –B
17. Calculate the boiling point of solution when 2g of Na2SO4 (M=142g/mol) was dissolved in 50 g of water
assuming that it undergoes complete dissociation.(Kf for water = 0.52Kkg/mol)
18. a) Write chemical reaction involved in Aldol Condensation.
b) Carboxylic acids are not regarded as Carbonyl compounds. Why?
19. Give reason:-
a) SN¹ reaction are accompanied by racemization in optically active alkyl halide.

100
 b) The presence of nitro group -NO2 group at ortho and para positions increases the reactivity of
haloarenes towards nucleophilic substitution reaction.
20. When FeCr2O4 is fused with Na2CO3 in the presence of air it gives a yellow solution of a compound (A).
Compound(A) acidification gives compound (B). Compound (B) on reaction with KCl forms and orange
coloured compound(C). And acidified solution of compound C oxidizes Kl to( D). Identify A, B, C and D
OR
Give the reason for following
a) Ce4+ is a strong oxidising agent in aqueous solution.
b) Actinoids shows a wide range of oxidation state.
21. a)Write reaction of glucose which cannot be explained by its open chain structure.
b) Write one structural difference between RNA and DNA?
SECTION- C

22. i) How much charge is required for the following reduction :-

1 mol of MnO4– to Mn2+
ii) 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?
23. Account for the following.

a) Why is cooking faster in plains than in hilly areas ?

b) Addition of HgI2 to aqueous solution of KI shows increase in vapour pressure. Why?

c) Which colligative property is the best property for determining molecular mass of polymers and why?

24. a) Name the protein and its shape present as oxygen carrier in human body?.

b) Name four forces responsible for stabilisation of 2⁰ and 3⁰ structure of protein.

OR

a) Write any one difference between native protein and denatured protein?

b) On electrolysis in acidic medium, amino acids migrate towards cathode while in basic medium
amino acid migrate towards anode. Why?
25. a) Write down mechanism of dehydration of ethanol to give Diethyl ether.

b) Write down chemical reaction involved in Kolbe’s reaction.

26. a) Predict the major product of acid catalyzed dehydration of butan-1-ol.

b) Write IUPAC name of CH2=CH―CH2―CH2OH
27. An organic compound ‘A’ with molecular formula C4H8O2 was hydrolyzed with dil. H2SO4 to give a
carboxylic acid ‘B’ and an alcohol ‘C’. ‘C’ on dehydration gives ethene and ‘C’ also on oxidation gives back
‘B’. Identify ‘A’, ‘B’ and ‘C’ and write reactions involved.
101
 OR
Name the reagent used in the following reaction:
(a) CH3COCH3  CH3CHOHCH3
(b) CH3CHO  CH3CHOHCH3
(c) C6H5CH2CH3  C6H5COO-K+

28.Complete the following reaction:

(a) C6H5NH2 + CHCl3 + alc. KOH 
(b) C6H5N2Cl + C6H5OH 
(c) C6H5 NH2 + Br2 (aq.) 

SECTION-D
29. Read the passage given below and answer the questions:

Benzene ring in aniline is highly activated. This is due to the sharing of loan pair of nitrogen with the
ring which results in increase in the electron density on the ring and hence facilitates the electrophilic
attack. The substitution mainly takes place at Ortho and para positions because electron density is more
at Ortho and para positions. On reaction with aqueous Bromine all the Ortho and para positions get
substituted resulting in the formation of 2,4,6 tribromoaniline. To get mono bromo compound, the amino
group is protected by reacting it with acetic anhydride before bromination. After bromination the
bromoacetanilide is hydrolysed to give the desired halogenated amine.

Answer the following question

a) Why is pKb of aniline more than benzylamine?

b) How is protection of amino group done in case of aniline?

c) i)Amino group is Ortho and para directing, but nitration of aniline gives meta product along with
Ortho and para product.

ii) Convert Aniline to Benzyl alcohol

OR

i) Why does Aniline not undergo Friedal craft reaction?

ii) Convert Aniline to 2,4,6 –Tribromoflourobanzene.

Q30. Read the passage and answer the questions

Rate of a chemical reaction can be defined as the speed with which reaction proceeds in a forward
direction. The rate of reaction is concerned with decrease in concentration of reactance or increase in
concentration of products per unit time. The rate of a chemical reaction is determined experimentally
and the rate law can not be predicted by balanced chemical equation. The reactions taking place in

102
 one step are called elementary reactions. When a sequence of elementary reaction gives give
products, the reactions are called complex reactions.

Molecularity and order of an elementary reaction are same. Zero order reaction are relatively
uncommon but they occur under special condition. All natural and artificial radioactive decay of
unstable nuclei take place by first order kinetics.

Answer the following question

a) State a condition under which a bimolecular reaction is kinetically a first order reaction.

b) For a reaction A + B —> Products, the rate law is — Rate = k[A][B]3/2.

Can reaction be an elementary reaction? Explain.

c) i)Write two differences between order of reaction and molecularity of reaction.

OR

i) Explain why H2 and O2 do not react at room temperature?

ii) Write the relation between half life and three fourth life of a first order reaction.

SECTION-E
31.a)Write the reactions taking place at anode and cathode of H2-O2 fuel cell.
b) Calculate emf and ΔG° for following cell, at 298K
Mg (s) / Mg2+ (0.001 M) // Cu2+ (0.0001 M) / Cu (s)
(Given Standard reduction potential of Mg Electrode = -2.36 V & standard electrode potential of
Cu Electrode = + 0.34 V)
OR
(a) i) Write the product of electrolysis of NaCl (aq).
ii) Write recharging reaction of lead storage battery.
(b) State Kohlrausch law of independent migration of ions. Calculate Λ°m for acetic acid.
(Given λ°HCl = 426 S cm2 mol-1, λ°NaCl = 126 S cm2 mol-1, λ°AcONa = 91 S cm2 mol-1
32. Attempt any five of the following-

(a) Complete the following reaction

Cr2O72- + I- + H+ →
(b) Why is Cr2+ reducing agent while with same d orbital configuration (d4), Mn3+ is an oxidising
agent?
(c) Zn, Cd & Hg are not considered as transition elements.

(d) Transition metals and their compounds are generally found to be a good catalyst. Why?

103
(e )What is lanthanide contraction?
(f ) Calculate the magnetic moment of Ti3+ ion.
(g) Out of Ti4+ and Fe3+ which ion is coloured and why?
33.a)Account for the followings :
(i) Formaldehyde does not undergo aldol condensation.
(ii) Electrophilic substitution reaction in benzoic acid takes place at m-position.

b) Give a brief account of the following:

(i) Reimer-Tiemann reaction
(ii) Hell-Volhard-Zelinsky reaction
(iii) Clemmensen reduction

OR
(a) Arrange the following set in the increasing order of the property mentioned against them:
(i) Ethanal, propanal, propanone, butanone (reactivity towards nucleophilic addition
reaction)
(ii) Benzoic acid, 3,4-dinitrobenzoic acid, 4-methoxybenzoic acid (acidic strength)
(b) How will you distinguish between the following pairs of compounds ?
(i) Propanal and propanone
(ii) Ethanal and propanal
(iii) Phenol and benzoic acid

104
 KENDRIYA VIDYALAYA SANGATHAN, DELHI REGION

SAMPLE PAPER 3
2024-25

General Instructions:
Read the following instructions carefully.
a) There are 33questions in this question paper with internal choice.
b) SECTION A consists of 16 multiple-choice questions carrying 1 mark each.
c) SECTION B consists of 5 very short answer questions carrying 2 marks each.
d) SECTION C consists of 7 short answer questions carrying 3 marks each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed
SECTION A
The following questions are multiple-choice questions with one correct answer. Eachquestion carries 1 mark. There
is no internal choice in this section.

1 In which of the following pairs both the ions are coloured in aqueoussolution?(a)
Sc3+ , Ti3+ (b) Sc3+, Co2+ (c) Ni2+ , Cu+ (d) Ni2+, Ti3+

2 The difference between the electrode potentials of two electrodes when no current isdrawn
through the cell is called
(a) Cell potential (b) Cell emf (c) Potential difference (d) Cell voltage
3 Methyl ketones are usually characterized through--
(a) Tollen’s reagent (b) Iodoform test (c) Schiff ’s test (d) Benedict solution test.

4 The formation of cyanohydrin from propanone is which type of reaction?

(a) Electrophilic Substitution
(b) Nucleophilic Substitution
(c) Electrophilic Addition
(b) Nucleophilic Addition
5 (CH3)2C = CHCOCH3 can be oxidized to (CH3)2C= CHCOOH by
(a) Chromic acid (b) NaOI (c) Cu at 575 K (d) Tollen’s reagent

105
 6 Correct order of boiling point of isomeric dichlorobenzene is .
(a) para dichlorobenzene> ortho dichlorobenzene> meta dichlorobenzene

(b) meta dichlorobenzene> ortho dichlorobenzene> para dichlorobenzene

(c) ortho dichlorobenzene> meta dichlorobenzene> para dichlorobenzene
(d) para dichlorobenzene> meta dichlorobenzene> ortho dichlorobenzene

7 KMnO4 is not acidified by HCl instead of H2SO4 because

(a) H2SO4 is stronger acid than HCl (b) HCl is oxidized to Cl2 by KMnO4
(c) H2SO4 is dibasic (d) rate is faster in presence of H2SO4
8 The increase in rate constant of a chemical reaction with increasing temperature is due to the
fact that
(a)The activation energy of the reaction decreases with increasing temperature
(b)The concentration of the reactant molecules increases with increasing temperature
(c)The number of reactant molecule acquiring the activation energy increases with increasing
temperature
(d)All of above
9 When a mixture of primary ammine and chloroform is heated with ethanolic KOH to form
Isocyanide which has foul smell. What is the name of reaction involved in the statement
(a)Hinsberg test (b)Diazotization
(c)Gabriel phthalimide synthesis (d)Carbylamine

10 The IUPAC name of m-cresol is .

(a) 3-Methylphenol (b) 3-Chlorophenol (c) 3-Methoxyphenol (d) Benzene-1,3-diol
11 Which of the following compounds does not give Cannizzaro reaction
(a) methanal (b) propanal
(c) benzaldehyde (d) 2,2-dimethyl propenal
12 In a lead storage battery
(a) PbO2 is reduced to PbSO4 at the cathode
(b) Pb is oxidised to PbSO4 at the anode
(c ) Both electrodes are immersed in the same aqueous solution of H2SO4
(d) All the above are true

In the Following questions a statement of Assertion(A) is followed by a statement ofReason(R). Select the most
appropriate answer from the options given below:

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

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 13 Assertion: Conductivity decreases for weak electrolyte and increases for strong
electrolyte with decrease in concentration.
Reason: On dilution, the number of ions per unit volume that carry the current
decreases.
14 Assertion: p-hydroxybenzoic acid is more acidic than phenol.
Reason: carboxyl group helps in the stabilization of the phenoxide ion by dispersal ofnegative
charge due to resonance.
15 Assertion: Compounds containing –CHO group are easily oxidized to corresponding
carboxylic acids.
Reason: Carboxylic acids can be reduced to alcohols by treatment with LiAlH4.
16 Assertion (A): Globular proteins are soluble in water.
Reason(R): Keratin is a fibrous protein.

SECTION B
This section contains 5 questions with internal choice in onequestion. The followingquestions are very short answer type
and carry 2 marks each.

17 Suman took two glasses of water from a water filter. She cools one glass in a fridge and
warms the other glass on a stove. Which glass of water will hold more dissolved oxygen?
Explain using Henry's law.

18 If hydrolysis of cane sugar is completed in acidic medium then its rate of reactiondepends
only on one molecule of cane sugar.
(i) What will be the name of the reaction
(ii) What will be unit of rate constant for this reaction?

19 Which compound in each of the following pairs will react faster in SN2 reaction with
—OH? and why?
(a) CH3Br or CH3I (b) (CH3)3CCl or CH3Cl
20 Arrange the following compounds in increasing order of their property as indicated:
(i) CH3COCH3, C6H5COCH3, CH3CHO (reactivity towards nucleophilic addition
reaction)
(ii) Cl—CH2—COOH, F—CH2—COOH, CH3—COOH (acidic character)
OR
Give reasons:
(i) Oxidation of aldehydes is easier than ketones
(ii) Benzoic acid is a stronger acid than acetic acid
21 (i) Write one structural difference between DNA and RNA..
(ii) Why cannot vitamin C be stored in our body?

SECTION C
This section contains 7 questions with internal choices in one question. The followingquestions are short answer types
and carry 3 marks each.

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 22 A strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2 and a strip of silver metal
is placed in a 1 molar solution of AgNO3. An electrochemical cell is created when the two
solutions are connected by a salt bridge and the two strips
are connected by wires to a voltmeter.
(i) Write the balanced equation for the overall reaction occurring in the cell .
(ii) Calculate the cell potential, E, at 25oC for the cell if the initial concentration of
Ni(NO3)2 is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar.
[E0 2+ = −0.25V; E0 + = 0.80V, log 10−1 = −1]
Ni /Ni Ag /Ag
23 What do you mean by rate constant? The rate constant of a first order reaction increases
from 4 x 10-2 to 24 x 10-2 when the temperature changes from 300 K to 350 K. Calculate
the energy of activation (Ea). (log 2 = 0.3010, log 3 = 0.4771, log4 = 0.6021, log 6 = 0.7782)

24 (a)Why are low spin tetrahedral complexes not formed?

(b)[Fe(H2O)6]3+is strongly paramagnetic where as [Fe(CN)6]3-is weakly
paramagnetic. Explain. (At. no. Fe = 26)
25 Give reasons:
(a) n-Butyl bromide has higher boiling point than t-butyl bromide.
(b) Racemic mixture is optically inactive.
(c) The presence of nitro group (-NO2) at ortho and para positions increases the
reactivity ofhaloarenes towards nucleophilic substitution reactions.
26 Give one chemical test each to distinguish between the following pairs ofcompounds:
(i) Phenol and Benzoic acid (ii)Propan-1-ol and Propan-2-ol
(iii) Methanol andethanol.
OR
Give reasons for the following: -
(a) Alcohols are more soluble in water than the hydrocarbon of comparablemolecular
masses.
(b)Phenoxide ion is more stable than phenol.
(c) Ortho nitro phenol is more acidic than Ortho-methoxyphenol.

27 Give reasons:
(i) Alpha hydrogen of aldehydes and ketones are acidic in nature.
(ii)Propanone is less reactive than ethanal towards addition of HCN.
(iii) Benzoic acid does not give Friedel-Crafts reaction.
28 (i) What are the products of hydrolysis of maltose?
(ii)Name the protein and its shape present in the oxygen carrier in the humanbody.
(iii) Name the linkages which join the amino acids in proteins?

SECTION D
The following questions are case-based questions. Each question has an internal choice andcarries 4 (1+1+2) marks each.
Read the passage carefully and answer the questions that follow.

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 29 Read the passage given below and answer the following questions:
A process in which on passing electricity electrolyte split into ions and ions are deposited at
opposite electrode is called electrolysis.
The amount of substance deposited or liberated during electrolysis is directly proportional
to the quantity of electricity passed through the solution and can be expressed by
relationship w=zIt
(a) Define electrochemical equivalent (Z).
(b) How much electricity in terms of Faraday is required to produce 20⋅0 g of Calcium
from molten CaCl2?
(c) A solution of Ni (NO3)2 is electrolyzed between platinum electrodes using a
current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? (
Atomic mass of Ni is 58.5 u)
OR
(c) A solution of CuSO4 is electrolysed for 10 minutes with a current of 1⋅5 amperes.
What is the mass of substance deposited at the cathode? (Atomic
mass of Cu is 63 u)

30 Read the passage given below and answer the following questions:
In coordination compounds, metal possess two types of valences- primary and secondary
valences. Primary valency is ionizable and secondary valency is non-ionizable. In aseries of
complexes of Co(III) chloride with ammonia, it was found that some of the chloride ions could
be precipitated as AgCl on adding excess amount of AgNO3 solution. The number of ions
furnished by a complex in a solution can be determined by precipitation reactions. The
measurement of molar conductance of solution ofcomplexes helps to estimate the no. of
ions furnished by the complexes in solution.
(a)Write difference between primary and secondary valences.
(b) How many moles of AgCl are formed by the complex [Co(NH3)4Cl2]Cl withexcess
amount of AgNO3 solution.
(c) Calculate number of primary and secondary valences shown in [Co(NH3)5Cl]Cl2.
OR
(c) one mole of NiCl2.6H2O with excess AgNO3 precipitates two moles of AgCl, Write the
structural formula of the compound.

SECTION E
The following questions are long answer type and carry 5 marks each. Two questions havean internal choice.
31 (i) Calculate the freezing point of a solution containing 0.5 g KCl (Molar mass = 74.5g/mol)
dissolved in 100 g water, assuming KCl to be 92% ionized. Kfof water = 1.86K kg / mol.
(ii) Which of the following solutions has higher freezing point?

0.05 M Al2(SO4)3, 0.1 M K3[Fe(CN)6 ] Justify.

OR
(i) State Henry’s law.

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 (ii) Sia’s father is suffering from high blood pressure but he is advised to consume less
quantity of common salt. Why?
(iii) Out of 1M sucrose and 1M urea solution which has more osmotic pressure?

(iv) A compound CuSO4.5H2O undergoes complete dissociation in water. What will be its
value of Van’t Hoff factor?
(v) If molality of dilute solution is doubled, what will be the value of molal elevation
constant(Kb)?

32 Attempt any five of the following- 5

(a) Mn2O7 is acidic whereas MnO is basic.
(b) Among divalent ions of 3d transition series, Mn2+ exhibits maximum
paramagnetism
(c)Actinoids show wide range of oxidation states.
(d)Cu2+ is stable in aqueous solution in spite of having 3d9 configuration. Why?
(e) The E° values of Mn and Zn is more negative. Give reason.
(f) The transition metals are generally paramagnetic in nature why?
(g) Scandium is a transition element but Zinc is not. Why?
33 An aromatic compound A of molecular formula C7H7ON undergoes a series of reactions as
shown below. Write the structures of A, B, C, D and E in the followingreactions :
5

OR
(A) Arrange the following:
(i) In decreasing order of the pKb values:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
(ii) In decreasing order of basic strength: C6H5NH2, C6H5N (CH3)2, (C2H5)2NH andCH3NH2
(B) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms
compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular
formula C6H7N. Write the structures and IUPAC names of compounds A,B and C.

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