CBSE Physics Class 12 Sample Paper 1 Page

Class XII
Chemistry
Time: 3 Hours Max. Marks: 70
General Instructions:
1. There are 35 questions in this question paper with internal choice.
2. SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
3. SECTION B consists of 7 very short answer questions carrying 2 marks each.
4. SECTION C consists of 5 short answer questions carrying 3 marks each.
5. SECTION D consists of 2 case-based questions carrying 4 marks each.
6. SECTION E consists of 3 long answer questions carrying 5 marks each.
7. All questions are compulsory.
8. Use of log tables and calculators is not allowed.

SECTION-A
Directions (Q. Nos. 1-18) : 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. The boiling points of alcohols are higher than those of hydrocarbons of comparable masses due
to:
(a) ion-dipole interaction (b) dipole -dipole interaction
(c) hydrogen bonding (d) vander Waals forces

2. The role of a catalyst is to change :

(a) enthalpy of reaction
(b) Gibbs’ energy of reaction
(c) equilibrium constant
(d) activation energy of reaction

3. The value of KH for Ar(g), CO (g), HCHO(g) and CH (g) are 40.39, 1.67, 1.83 # 10-5 and 0.413
2 4
respectively. Arrange these gases in increasing order of solubility.
(a) Ar < CO2 < CH4 < HCHO (b) Ar < CH4 < CO2 < HCHO
(c) HCHO < CH4 < CO2 < Ar (d) HCHO < CO2 < CH4 < Ar
4. Out of the following, the strongest base in aqueous solution is:
(a) dimethylamine (b) aniline
(c) methylamine (d) trimethylamine

5. Out of the following transition elements, the maximum number of oxidation states are shown
by:
(a) Cr (Z = 24) (b) Sc (Z = 21)
(c) Fe (Z = 26) (d) Mn (Z = 25)

6. What is the correct IUPAC name of the given compound?

(a) 2-carboxyl-2-methylpropanoic acid

(b) 2-ethyl-2-methylpropanoic acid
(c) 3-methylabutance carboxylic acid
(d) 2, 2-dimethylbutanoic acid

7. For the reaction, 2X + Y $ X2Y

What will be the expression for instantaneous rate of the reaction?
(a)
)+ 1
d (Y (b) 1 d (X 2 Y )
-
2 dt 2 dt
(c) - d
(X )
(d) None of these
2dt

8. For the reaction 2H2O2 $ 2H2O + O 2 , r = k[H2O2]. The reaction is of :

(a) first order (b) second order
(c) third order (d) zero order

9. The compound obtained by the reaction of nitrous acid on aliphatic primary amine is:
(a) alkyl nitrite (b) alcohol
(c) nitroalkane (d) secondary amine
10. A graph was plotted between the molar conductivity of various electrolytes (NaCl, HCl and
NH4 OH) and c (in mol L-1). Which of the following is the correct set?

(a) I (NH4OH), II (HCl), III (NaCl)

(b) I (NaCl), II (HCl), (III) (NH4OH)
(c) I (HCl), II (NaCl), III (NH4OH)
(d) I (NH4OH), II (NaCl), III (HCl)

11. Using valence bond theory, the complex [Cr(NH 3) 6]3+ can be described as :
(a) d 2sp3, inner orbital complex, paramagnetic
(b) d 2sp3, outer orbital complex, diamagnetic
(c) sp3d 2
, outer orbital complex, paramagnetic
(d) dsp2, inner orbital complex, diamagnetic

12. Which of the following compound will not undergo azo coupling reaction with benzene
diazonium chloride?
(a) Phenol (b) Aniline
(c) Nitrobenzene (d) Anisole

13. Major product obtained on reaction of 3-phenyl propene with HBr in presence of organic
peroxide is:
(a) 3-phenyl-2-bromopropane (b) 3-phenyl-1-bromopropane
(c) 1-phenyl-3-bromopropane (d) 1-phenyl-2-bromopropane
14. During dehydration of alcohols to alkenes by heating with concentrated H2SO4, the initiation
step is:
(a) elimination of water
(b) formation of an ester
(c) protonation of alcohol molecule
(d) formation of carbocation

Directions (Q. Nos. 15-18) : Each of the following questions consists of two statements, one is Assertion
and the other is Reason. Give answer :

15. Assertion : Vanadium had the ability to exhibit a wide range of oxidation states.
Reason : The standard potentials Vanadium are rather small, making a switch between oxidation
states relatively easy.
(a) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(c) Assertion is fake but Reason is true.
(d) Assertion is true but Reason is fake.

16. Assertion : DNA has a double strand helix structure.

Reason : The two strands in a DNA molecule are exactly similar.
(a) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(c) Assertion is fake but Reason is true.
(d) Assertion is true but Reason is fake.

17. Assertion : Tertiary butylamine can be prepared by the action of NH3 on tert-butylbromide.
Reason : Tertiary butyl bromide being 3° alkyl halide prefers to undergo elimination on the
treatment with a base.
(a) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(c) Assertion is fake but Reason is true.
(d) Assertion is true but Reason is fake.
18. Assertion : IUPAC name of the compound
CH3 - CH - O - CH2 - CH2 - CH3
y
CH3
is 2-Ethoxy-2-methylethane.
Reason : In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in which a
hydrogen atom replaced by —OR or —OAr group
[where R = alkyl group and Ar = aryl group]
(a) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(b) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(c) Assertion is fake but Reason is true.
(d) Assertion is true but Reason is fake.

SECTION-B
Directions (Q. Nos. 19-25) : This section contains 7 questions with internal choice in two questions. The
following questions are very short answer type and carry 2 marks each.

19. 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.

20. Which one of the following pairs of substances undergoes SN2 substitution reaction faster and
why?

(i)

(ii)

21. A cell is constructed between copper and silver

Cu (s) y Cu2+(aq) Aq+(aq) y Aq (s)
If the two half-cells are working under standard condition, then calculate the emf of the cell.
EcCu /Cu =+ 0.34 V, EcAg /Ag =+ 0.80 V
2+ +

22. Identify compounds (A) and in the following reactions and write the related balanced
chemical equation : (B)
CH3CONH2 P2 4[H]
O5 (A) Sn+HCl
(B)
T
 or
Complete and name the following reaction:
(i) RNH2 + CHCl3 + 3KOH $
(ii) RCONH2 + Br2 + 4NaOH $

23. (i) Sketch the zwitter ionic form of α -amino acetic acid.
(ii) What type of linkage holds together the monomers in DNA?

24. A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at
298K. Calculate the electrode potential.
[Ec(Zn /Zn)
2+ =− 0.76 V]

25. (i) Give the electronic configuration of the d -orbitals of Ti in [Ti(H O) ]3+ ion and explain
2 6
why this complex is coloured ? [At. No. of Ti = 22]
(ii) Write IUPAC name of [Cr(NH3)3 (H2O)3]Cl3.
or
Determine the structure and magnetic behaviour of [CoCl ]42– using valence bond theory.

SECTION-C
Directions (Q. Nos. 26-30) : This section contains 5 questions with internal choice in two questions. The
following questions are short answer type and carry 3 marks each.

26. (i) Draw the structural formulas and write IUPAC names of all the isomeric alcohols with the
molecular formula C5H12O.
(ii) Classify the isomers of alcohols given in part (a) as primary, secondary and tertiary
alcohols.

27. Answer the following questions :(Any three)

(i) What do you mean by depression in freezing point?
(ii) How can the molecular weight of a non-volatile substance be calculated by freezing point
depression method? Only give the formula.
(iii) Measurement of osmotic pressure method is preferred for the determination of molar mass
of macromolecules such as proteins and polymers.
(iv) Elevation of boiling point of 1M KCl solution is nearly double than that of 1 M sugar
solution.
28. (i) Write the IUPAC name of the following complex :
[Co(NH3)4(H2O)Cl]C12
(ii) What is the difference between an Ambidentate ligand and a Bidentate ligand?
(iii) Out of [Fe(NH ) ]3+ and [Fe(C O ) ]3– , which complex is more stable and why ?
3 6 2 4 3

29. What happens when :

(i) N-ethylethanamine reacts with benzenesulphonyl chloride.
(ii) Benzylchloride is treated with ammonia followed by the reaction with Chloromethane.
(iii) Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide.
or
(i) Write the IUPAC name for the following organic compound :

(ii) Complete the following :

C H NO
Sn/HCl A Br2 /KOH B NaNO2 /HCl C HBF4
6 5 2 273 - 278 K T
D

30. How will you convert ethanal to the following compounds?

(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid

SECTION-D
Directions (Q. Nos. 31-32) : The following questions are case-based questions. Each question has an
internal choice and carries 4 marks each. Read the passage carefully and answer the questions that follow.

31. The rate law for a chemical reaction relates the reaction rate with the concentrations or partial
pressures of the reactants. For a general reaction aA + bB $ with no intermediate steps
C
in its reaction mechanism, meaning that it is an elementary reaction, the rate law is given
by r = k [A]x [B]y , where [A] and [B] express the concentrations of A and B in moles per litre.
Exponents x and y vary for each reaction and are determined experimentally. The value of k
varies with conditions that affect reaction rate, such as temperature, pressure, surface area,
etc. The sum of these exponents is known as overall reaction order. A zero order reaction has
a constant rate that is independent of the concentration of the reactants. A first order reaction
depends on the concentration of only one reactant. A reaction is said to be second order when
the overall order is two. Once we have determined the order of the reaction, we can go back
and plug in one set of our initial values and solve for k .
In the context of the given passage, answer the following questions :
 (i) Calculate the overall order of a reaction which has the following rate expression :
Rate = k [A]1/2[B]3/2
(ii) What is the effect of temperature on rate of reaction?
(iii) A first order reaction takes 77.78 minutes for 50% completion. Calculate the time
required for 30% completion of this reaction log 10 = 1, log 7 = 0.8450.

or
(iv) A first order reaction has a rate constant 1 # 10-3 per sec. How long will 5g of this
reactant take to reduce to 3 g?
(log 3 = 0.4771; log 5 = 0.6990)

32. An amino acid is a compound that contains both carboxyl group and an amino group.
Although, many types of amino acids are known, the α -amino acids are the most significant
in the biological world because they are the monomers from which proteins are constructed. A general
structural formula of an α -amino acid is shown in figure below.

Although, figure (a) is a common way of writing structural formulas for amino acids, it is not
accurate because it shows an acid (—COOH) and a base (—NH 2) within the same molecule.
These acidic and basic groups react with each other to form a dipolar ion or internal salt
(figure (b)). The internal salt of an amino acid is given the special name Zwitter ion. Note
that a Zwitter ion has no net charge, it contains one positive charge and one negative charge.
Because they exist as Zwitter ions, amino acids have many of the properties associated with
salts. They are crystalline solids with high melting points and are fairly soluble in water but
insoluble in non-polar organic solvents such as ether and hydrocarbon solvents.
According to the above passage, answer the following questions :
(i) Amino acids are usually colourless, crystalline solids. They behave like salts rather than
simple amines or carboxylic acids. Why amino acids show such a behaviour?
(ii) Amino acids are essential and non-essential depending upon their need. One of the essential
amino acid is lysine. Can you say why lysine is considered an essential amino acid?
(iii) Here are given some amino acids—lysine, Tyrosine, Glycine, Alamine. One of these amino
acids is not optically active. Which one is that amino acid? Also, provide the reason.

or
(iv) The pka , and pka , of an amino acid are 2.3 and 9.7 respectively. What would be the
1 2

isoelectric point of the amino acid? Calculate by defining it.

 SECTION-E
Directions (Q. Nos. 33-35) : The following questions are long answer type and carry 5 marks each. Two
questions have an internal choice.

33. (i) The cell in which the following reaction occurs:

2Fe3+(aq) + 2I−(aq) $ 2Fe2+ (aq) + I2(s)
has EcCell = 0.236 at 298K. Calculate the standard Gibbs energy of the cell reaction.
Volt
(Given : 1F = 96,500 C mol–1)
(ii) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2
hours? (Given : 1F = 96,500 C mol–1)
(iii) Explain the following with reason :
(a) Chlorine can displace iodine from KI solution but iodine can not displace bromine
from KBr solution.
(b) Following reaction is possible or not.
Hg + H2SO4 $ HgSO4 + H2

34. (i) Account for the following :

(a) Transition metals from large number of complex compounds.
(b) The lowest oxide of transition metal is basic whereas the highest oxide is
amphoteric or acidic.
(c) Ec value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compare to
Cr3+/Cr2+.
(ii) Write one similarity and one difference between the chemistry of lanthanoid and actinoid
elements.

or
(i) (a) How is the variability in oxidation states of transition metals different from that of the
p -block elements ?
(b) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why ?
(c) Orange colour of Cr2O2-7 ion changes to yellow colour when treated with an alkali.
Why ?
(ii) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons.
35. (i) Write the product (s) in the following reactions:

(c) CH3 — CH = CH
(a) DIBAL −H
?
(b) H2 O
− CN
(ii) Give simple chemical test to distinguish between the following pairs of compounds :
(a) Butanal and Butan-2-one.
(b) Benzoic acid and Phenol.

or
(i) An organic compound (A) with molecular formula C3H7NO on heating with Br2 and
KOH forms a compound (B), compound (B), on heating with CHCl 3 and alcoholic KOH
produces a foul smelling compound (C) and on reacting with C6H5SO2Cl forms a compound
(D) which is soluble in alkali. Write the structures of (A), (B), (C) and (D).
(ii) Give reasons to support the answer :
(a) Presence of alpha hydrogen in aldehydes and ketones is essential for aldol condensations.
(b) 3-Hydroxy pentan-2-one shows positive result to Tollen’s test.

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Page 1 Sample Paper 1 Solutions CBSE Chemistry Class 12

Sample Paper 3 Solutions

Class XII 2022-23
Chemistry
Time: 3 Hours Max. Marks: 70
General Instructions:
1. There are 35 questions in this question paper with internal choice.
2. SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
3. SECTION B consists of 7 very short answer questions carrying 2 marks each.
4. SECTION C consists of 5 short answer questions carrying 3 marks each.
5. SECTION D consists of 2 case-based questions carrying 4 marks each.
6. SECTION E consists of 3 long answer questions carrying 5 marks each.
7. All questions are compulsory.
8. Use of log tables and calculators is not allowed.

SECTION-A 3. The value of KH for Ar(g), CO2(g), HCHO(g)

and CH 4(g) are 40.39, 1.67, 1.83 # and 0.413
respectively. Arrange these gases in increasing order
10-5
of solubility.
Directions (Q. Nos. 1-18) : The following questions are (a) Ar < < < HCHO
multiple-choice questions with one correct answer. Each CO2 CH4 < HCHO
(b) Ar < CH4 < CO2
question carries 1 mark. There is no internal choice in this
section. (c) HCHO < CH4 < CO2 < Ar
(d) HCHO < CO2 < CH4 < Ar
1. The boiling points of alcohols are higher than those
Ans : (a) Ar < CO2 < CH4 < HCHO
of hydrocarbons of comparable masses due to:
(a) ion-dipole interaction Higher the value of KH , lower will be the solubility
(b) dipole -dipole interaction of gas at given pressure.
(c) hydrogen bonding
4. Out of the following, the strongest base in aqueous
(d) vander Waals forces
solution is:
Ans : (c) hydrogen bonding (a) dimethylamine (b) aniline
Alcohols form intermolecular hydrogen bonds while (c) methylamine (d) trimethylamine
hydrocarbons do not. Ans : (a) dimethylamine
Order of basicity is aqueous solution for amines :
2c 2 1c 2 3c 2 NH3

5. Out of the following transition elements, the

2. The role of a catalyst is to change : maximum number of oxidation states are shown
(a) enthalpy of reaction by:
(b) Gibbs’ energy of reaction (a) Cr (Z = 24) (b) Sc (Z = 21)
(c) equilibrium constant (c) Fe (Z = 26) (d) Mn (Z = 25)
(d) activation energy of reaction
Ans : (d) Mn (Z = 25)
Ans : (d) activation energy of reaction
The role of a catalyst is to change the activation
energy of the reaction. However, catalyst does
not changes Gibbs energy of reaction, enthalpy of
reaction and equilibrium constant.
A catalyst provides an alternate pathway of lower
activation energy.
Transition Element Oxidation States
Sc +3
Cr +2, +3, +4, +5, +6
Mn +2, +3, +4, +5, +6, +7
Fe +2, +3, +4, +6
6. What is the correct IUPAC name of the given R − NH2 + HONO $ ROH + N2 + H2O
1calcohol
compound? 1camine Nitrous acid

10. A graph was plotted between the molar conductivity

of various electrolytes (NaCl, HCl and NH4OH)
and c (in mol L-1). Which of the following is the
correct set?

(a) 2-carboxyl-2-methylpropanoic acid

(b) 2-ethyl-2-methylpropanoic acid
(c) 3-methylabutance carboxylic acid
(d) 2, 2-dimethylbutanoic acid
Ans : (d) 2, 2-dimethylbutanoic acid

7. For the reaction, 2X + Y $ X2Y

What will be the expression for instantaneous rate
of the reaction?
(a) d (Y (b)
)+ 1 1 d (X 2 Y )
-
2 dt 2 dt (a) I (NH OH), II (HCl), III (NaCl)
(c) - d 4
(d) None of these (b) I (NaCl), II (HCl), (III) (NH4OH)
(X)
2dt
- d (X)
Ans : (c) (c) I (HCl), II (NaCl), III (NH4OH)
2dt (d) I (NH4OH), II (NaCl), III (HCl)
The expression for instantaneous rate of the reaction
is as follows: Ans : (c) I (HCl), II (NaCl), III (NH4OH)
d [N2] d [H2] d [NH3] In any graph of molar conductivity vs c , a weak
− =− =
dt 3dt 2dt electrolyte gives a curved decreasing graph and a
strong electrolyte gives straight-line decreasing
graph. Since, graph III is curved, it corresponds to
NH4OH which is a weak electrolyte. Graph I and
II respectively represent HCl and NaCl HCl is a
stronger electrolyte among the two).

11. Using valence bond theory, the complex [Cr(NH3) 6]3+

can be described as :
8. For the reaction 2H2O2 $ 2H2O + (a) d 2sp3, inner orbital complex, paramagnetic
O2, r = k[H2O2]. The reaction is of :
(b) d 2sp3, outer orbital complex, diamagnetic
(a) first order (b) second order
(c) sp3d 2, outer orbital complex, paramagnetic
(c) third order (d) zero order
(d) dsp2, inner orbital complex, diamagnetic
Ans : (a) first order
Ans : (a) d 2sp3 inner orbital complex,
As rate,r = k[H2O2] paramagnetic
It is reaction of first order. Also as it is
decomposition reaction.

9. The compound obtained by the reaction of nitrous

acid on aliphatic primary amine is:
(a) alkyl nitrite (b) alcohol
(c) nitroalkane (d) secondary amine
Ans : (b) alcohol Paramagnetic due to presence of three unpaired
electrons, it is paramagnetic.
12. Which of the following compound will not undergo Step 2 : Formation of carbocation. It is the slowest
azo coupling reaction with benzene diazonium step and hence the rate determining step of the
chloride? reaction.
(a) Phenol (b) Aniline
(c) Nitrobenzene (d) Anisole
Ans : (c) Nitrobenzene
Diazonium cation is a weak electrophile and hence
it reacts with electron-rich compounds containing
electron-donating groups such as –OH, –NH2 and
OCH3 groups and not with compounds containing
electron-withdrawing groups such as –NO2 etc.
Step 3 : Formation of ethene by elimination of a
proton.
13. Major product obtained on reaction of 3-phenyl
propene with HBr in presence of organic peroxide
is:
(a) 3-phenyl-2-bromopropane
(b) 3-phenyl-1-bromopropane
(c) 1-phenyl-3-bromopropane
(d) 1-phenyl-2-bromopropane
Ans : (c) 1-phenyl-3-bromopropane
It is an Anti-Markovnikov addition reaction which
are ass follows : Directions (Q. Nos. 15-18) : Each of the following questions
Organic consists of two statements, one is Assertion and the other
(C6H5) CH2CH = CH2 + peroxide is Reason. Give answer :
HBr
(C 6 H 5 ) CH2CH2CH2Br
15. Assertion : Vanadium had the ability to exhibit a
wide range of oxidation states.
Reason : The standard potentials Vanadium are
rather small, making a switch between oxidation
states relatively easy.
(a) Both Assertion and Reason are true but
Reason is not a correct explanation of
Assertion.
(b) Both Assertion and Reason are true and
14. During dehydration of alcohols to alkenes by heating Reason is the correct explanation of Assertion.
with concentrated H2SO4, the initiation step is: (c) Assertion is fake but Reason is true.
(a) elimination of water
(d) Assertion is true but Reason is fake.
(b) formation of an ester
(c) protonation of alcohol molecule Ans : (b) Both Assertion and Reason are true
and Reason is the correct explanation of
(d) formation of carbocation
Assertion.
Ans : (c) protonation of alcohol molecule
The various steps of the mechanism are given below: 16. Assertion : DNA has a double strand helix structure.
Step 1 : Formation of protonated alcohol Reason : The two strands in a DNA molecule are
exactly similar.
(a) Both Assertion and Reason are true but
Reason is not a correct explanation of
Assertion.
(b) Both Assertion and Reason are true and
Reason is the correct explanation of Assertion.
(c) Assertion is fake but Reason is true.
 (d) Assertion is true but Reason is fake. (d) Assertion is true but Reason is fake.
Ans : (d) Assertion is true but Reason is fake. Ans : (c) Assertion is fake but Reason is true.
Reason is false because the two strands are The IUPAC name of the given compound is
complementary to each other due to the reason that 1-(2-propoxy) propane (or 2-propoxypropane).
the hydrogen bonds are formed between specific Therefore, the assertion is false.
pairs of bases.

17. Assertion : Tertiary butylamine can be prepared

by the action of NH3 on tert-butylbromide. SECTION-B
Reason : Tertiary butyl bromide being 3° alkyl
halide prefers to undergo elimination on the
treatment with a base. Directions (Q. Nos. 19-25) : This section contains 7
questions with internal choice in two questions. The
(a) Both Assertion and Reason are true but
following questions are very short answer type and carry
Reason is not a correct explanation of
2 marks each.
Assertion.
(b) Both Assertion and Reason are true and 19. Time required to decompose SO 2Cl 2 to half of its
Reason is the correct explanation of Assertion.

(c) Assertion is fake but Reason is true. initial amount is 60 minutes. If the decomposition
(d) Assertion is true but Reason is fake. is a first order reaction, calculate the rate constant
of the reaction.
Ans : (c) Assertion is fake but Reason is true.
Ans :
Tertiary butylamine cannot be prepared by the
action of NH3 on tert-butyl bromide. In the basic Time required to decompose half of a substance is
called its half-life (t1/2). So here t1/2 = 60 minutes.
For a first order reaction,
medium, tertiary butyl bromide prefers to undergo
elimination to lose a molecule of HBr to form Rate constant, = 0693
= 60 minutes
0.693
an alkene rather than undergoing nucleophilic t1/2
k
substitution by an amino group. = 1.15 # 10−2 minute−1

20. Which one of the following pairs of substances

undergoes SN2 substitution reaction faster and why?
(i)
18. Assertion : IUPAC name of the compound
CH3 - CH - O - CH2 - CH2 - CH3
y
CH3
is 2-Ethoxy-2-methylethane.
Reason : In IUPAC nomenclature, ether is
regarded as hydrocarbon derivative in which a
hydrogen atom replaced by —OR or —OAr group (ii)
[where R = alkyl group and Ar = aryl group]
(a) Both Assertion and Reason are true but
Reason is not a correct explanation of
Assertion.
(b) Both Assertion and Reason are true and
Reason is the correct explanation of Assertion. Ans :
(c) Assertion is fake but Reason is true.
 r Ans :
(i) —
CH — COOH + + —
CH — COO−
(i) being a primary alkyl halide NH2 2 NH3 2

will react faster as compared to the other (ii) Monomers in DNA are linked by phosphate
substance which is a tertiary alkyl halide. This linkages.
is on account of lesser steric hindrance involved
in the first substance when nucleophile attacks. 24. A zinc rod is dipped in 0.1 M solution of ZnSO 4.
The salt is 95% dissociated at this dilution at 298K.
(ii) will react faster as Calculate the electrode potential.
compared to the other substance since the [Ec(Zn /Zn) =− 0.76 V]
2+

cleavage of C -I bond is easier as compared to

that of C -I bond due to less bond dissociation Ans :
enthalpy. The electrode reaction written at reduction reaction
is,
21. A cell is constructed between copper and silver Zn2+ + 2e− $ (n = 2)
Zn
Cu (s) y Cu2+(aq) Aq+(aq) y Aq (s) Applying Nernst equation, we get

If the two half-cells are working under standard 1 0.0591

EcZn 2+
/Zn = EcZn log
2+
/Zn −
condition, then calculate the emf of the cell. 2 2+
[Zn ]
EcCu /Cu =+ 0.34 V, EcAg /Ag =+ 0.80
2+ +
As 0.1 M ZnSO4 solution is 95% dissociated, this
V
Ans : means that in the solution,
In standard conditions, E cell is called Ec cell and 95
[Zn2+] = 10 # 0.1 = 0.095 M
for this cell, 0
Ec = Ec — Ec Hence, EZn /Zn =− 0.76 − 0.0591 log 1
2+

+ 2+

cell Ag /Ag Cu /Cu 2 0.095

= 0.80 − (+ 0.34) V =+ 0.46 =− 0.76 − 0.02955 (3 −
Volt 1.9777)
=− 0.79021 Volt
22. Identify compounds (A) and in the following
(B)
reactions and write the related balanced chemical
equation :
4[H]
CH3CONH2 P
O (A) Sn+HCl (B)
2

Ans :
4[H]
CH3CONH2 P2 O5
CH CN3
Acetamide T Sn+HCl
(− H 2 Methyl cyanide
O) (A)
CH3CH2NH2
Ethylamine
(B) 25. (i) Give the electronic configuration of the d
-orbitals of Ti in [Ti(H2 O)6 ]3+ ion and explain
or why this complex is coloured ? [At. No. of
Complete and name the following reaction: Ti = 22]
(i) RNH2 + CHCl3 + 3KOH $ (ii) Write IUPAC name of [Cr(NH ) (H O) ]Cl .
(ii) RCONH2 + Br2 + 4NaOH $ 3 3 2 3 3

Ans :
Ans :
(i) Oxidation state of Ti : x + 6 (0) =+ 3
(Warm)
(i) RNH 2 + CHCl3 + 3KOH(alc.) Configuration of Ti3+ ion = (Ar) 3d14s0
RN $= C + 3KCL + Complex is coloured due to the presence of an
3H2O unpaired electron leading to d -d transition.
Alkyl isocyanide (ii) Triamminetriaquachromium (III) chloride.
The reaction is known as carbyl amine reaction.
RCONH + Br + 4NaOH Heat RNH or
(alc.)
2 2 2
(ii) Acid amide Primary amine + Na2CO3 + 2NaBr + 2H2O
4
 De termine the structure and magnetic behaviour of [CoCl ]2– using valence bond theory.
23. (i) Sketch the zwitter ionic form of α -amino acetic Ans :
acid.
(ii) What type of linkage holds together the CO : [Ar] 4s2 3d 7
27

monomers in DNA? Co2+ :[Ar] 4s0 3d 7

Cl– is a weak field ligand, so pairing of electron does
 not occur.

(e)

Due to sp3-hybridisation, its structure is (f)

tetrahedral and presence of three unpaired
electrons make it paramagnetic.

SECTION-C
(g)
Directions (Q. Nos. 26-30) : This section contains 5
questions with internal choice in two questions. The
following questions are short answer type and carry 3
marks each.
(h)
26. (i) Draw the structural formulas and write IUPAC
names of all the isomeric alcohols with the (ii) Primary : (a), (b), (c), (d); Secondary : (e), (f),
molecular formula C5H12O. (h); Tertiary : (g)
(ii) Classify the isomers of alcohols given in part
(a) as primary, secondary and tertiary alcohols.
Ans :
(i) The molecular formula C5H12O represents eight
isomeric alcohols. These are :
5 4 3 2 1
(a) CH3 - CH2 - CH2 - CH2 - CH2 -
OH 27. Answer the following questions : (Any three)
Pentan- 1 - ol (i) What do you mean by depression in freezing
point?
(ii) How can the molecular weight of a non-volatile
substance be calculated by freezing point
depression method? Only give the formula.
(b) (iii) Measurement of osmotic pressure method
is preferred for the determination of molar
mass of macromolecules such as proteins and
polymers.
(iv) Elevation of boiling point of 1M KCl solution is
nearly double than that of 1 M sugar solution.
(c) Ans :
(i) When a non-volatile, non-electrolyte is dissolved
in a solvent, its freezing point decreases. The
decrease in freezing point is called depression
in freezing point. It is directly proportional to
amount (molality) of solute.
Depression in freezing point (TTf )
(d)
 = Freezing point of solvent ambidentate ligand. On the other hand, ligand
Freezing point of which can bind through two donor atoms is
TTf said to O
(iii) [Fe(C be)bidentate
]3– is aligand.
chelate complex and
solution
= TfO − Tf
2 4 3
(ii) The molecular weight of a non-volatile [Fe(NH 3 6) ]3+ is a complex containing
substance can be calculated from depression in unidentate ligand.
freezing point by using the following relation : We know that chelate complexes are more
1000 # Kf # w stable than similar complexes containing
m =
TTf # W unidentate ligands. Thus, [Fe(C2O4)3]3– is more
where, m = molecular weight of solute stable.
Kf = molal depression constant of solvent,
29. What happens when :
w = weight of solute,
(i) N-ethylethanamine reacts with
W = weight of solvent, benzenesulphonyl chloride.
TTf = depression in freezing point. (ii) Benzylchloride is treated with ammonia
(iii) The osmotic pressure method has the advantage followed by the reaction with Chloromethane.
over other methods as pressure measurement (iii) Aniline reacts with chloroform in the presence
is around the room temperature and the of alcoholic potassium hydroxide.
molarity of the solution is used instead of Ans :
molality.
(i) When N-ethylethanamine reacts
(iv) Elevation in boiling point is directly
with benzenesulphonyl chloride, N,
proportional to i , Tb o i . Now as given in
N-diethylbenzenesulphonamide is formed.
the question, elevation of boiling point of 1 M
(ii) When benzylchloride is treated with ammonia,
KCl solution is nearly double than that of
benzylamine is formed which on reaction with
1M sugar solution. It is because KCl being
chloromethane yields a secondary amine,
ionic, dissociates into K+ and Cl- and
N-methylbenzylamine.
therefore. It is van’t Hoff factor, i is 2
(iii) When aniline reacts with chloroform in the
whereas for sugar solution, van’t Hoff
presence of alcoholic potassium hydroxide,
factor is 1 as it does not undergo such a
phenyl isocyanides or phenyl isonitrile is
dissociation.
formed.
or
(i) Write the IUPAC name for the following
organic compound :

(ii) Complete the following :

28. (i) Write the IUPAC name of the following C H NO Sn/HCl A Br /KOH
2 B NaNO2 /HCl
6 5 2
complex : 273 - 278 K

[Co(NH3)4(H2O)Cl]C12 C
HBF4
D
T
(ii) What is the difference between an Ans :
Ambidentate ligand and a Bidentate ligand?
(iii) Out of [Fe(NH 3) ]6 3+ and [Fe(C 2O 4) 3]3– , which (i) N-ethyl-N-methylbenzenamine
complex is more stable and why ? Or N-ethyl-N-ethylaniline.
(ii)
Ans :
(i) Tetraammineaquachloro cobalt III chloride.
(ii) A ligand which has two different donor atoms
but only one of them forms a coordinate bond
at a time with central metal/ion is called
 (iii) Ethanal to But-2-enoic acid

SECTION-D
Directions (Q. Nos. 31-32) : The following questions are
case-based questions. Each question has an internal choice
and carries 4 marks each. Read the passage carefully and
answer the questions that follow.

31. The rate law for a chemical reaction relates the

reaction rate with the concentrations or partial
30. How will you convert ethanal to the following pressures of the reactants. For a general reaction
compounds? aA + bB $ C with no intermediate steps
(i) Butane-1, 3-diol in its reaction mechanism, meaning that it is an
(ii) But-2-enal elementary reaction, the rate law is given by
(iii) But-2-enoic acid r = k [A]x [B]y , where [A] and [B] express the
Ans : concentrations of A and B in moles per litre.
Exponents x and y vary for each reaction and are
(i) Ethanal to Butane-1, 3-diol
determined experimentally. The value of k varies
with conditions that affect reaction rate, such as
temperature, pressure, surface area, etc. The sum
of these exponents is known as overall reaction
order. A zero order reaction has a constant rate
that is independent of the concentration of the
reactants. A first order reaction depends on the
concentration of only one reactant. A reaction is
said to be second order when the overall order is
two. Once we have determined the order of the
reaction, we can go back and plug in one set of our
initial values and solve for k .
In the context of the given passage, answer the
(ii) Ethanal to But-2-enal following questions :
(i) Calculate the overall order of a reaction which
has the following rate expression :
 Rate = k [A]1/2[B]3/2 α -amino acid is shown in figure below.
(ii) What is the effect of temperature on rate of
reaction?
(iii) A first order reaction takes 77.78 minutes for
50% completion. Calculate the time required
for 30% completion of this reaction log 10 =
1, log 7 = 0.8450.
or
(iv) A first order reaction has a rate constant
1# per sec. How long will 5g of this
10-3
reactant take to reduce to 3 g?
(log 3 = 0.4771; log 5 = 0.6990) Although, figure (a) is a common way of writing
structural formulas for amino acids, it is not
Ans :
accurate because it shows an acid (—COOH) and
1 3 4 a base (—NH2) within the same molecule. These
(i) Overall order of reaction = 2 + 2 +2 =
2
(ii) The rate of reaction increases on increasing acidic and basic groups react with each other to
the temperature. form a dipolar ion or internal salt (figure (b)). The
(iii) For first order reaction, internal salt of an amino acid is given the special
t =t
name Zwitter ion. Note that a Zwitter ion has no
=
0.693
50% 1/2
k net charge, it contains one positive charge and one
0.693 negative charge.
77.78 =
k Because they exist as Zwitter ions, amino acids
0.693
k = 77.78 =0.00 have many of the properties associated with salts.
2.303 9 a They are crystalline solids with high melting
and t30% = log points and are fairly soluble in water but insoluble
k (a x) in non-polar organic solvents such as ether and
= 0.009 log(100100
2.303
− 30) hydrocarbon solvents.
According to the above passage, answer the
= 255.89 (log 10 − log following questions :
7) (i) Amino acids are usually colourless, crystalline
= 255.89 (1 − 0.8450) solids. They behave like salts rather than
simple amines or carboxylic acids. Why amino
= 39.66 minutes
acids show such a behaviour?
(iv) Given, or
(ii) Amino acids are essential and non-essential
[A] depending upon their need. One of the essential
0 = 5 g, [A] = 3 g
and = 1 # 10−3 Per amino acid is lysine. Can you say why lysine is
sec considered an essential amino acid?
k (iii) Here are given some amino acids—lysine,
2.303 [A ]0
= log Tyrosine, Glycine, Alamine. One of these
Using, t k A
= 2.303 log 5
1 # 10−3 3 biological world because they are the
= 2.303 # 103(log 5 − log 3) monomers from which proteins are
constructed. A general structural formula of an
= 2.303 (0.6990 − 0.4771)
. 511 sec

32. An amino acid is a compound that contains both

carboxyl group and an amino group. Although,
many types of amino acids are known, the α -
amino acids are the most significant in the
 amino acids is not optically active. Which one is
that amino acid? Also, provide the reason.
or
(iv) The pka , and pka , of an amino acid are 2.3 and
1 2

9.7 respectively. What would be the isoelectric point

of the amino acid? Calculate by defining it.
Ans :
(i) Amino acids behave like salts rather than simple
amines due to the presence of both acidic (—
COON) and basic (—NH2) groups.
(ii) Lysine is considered an essential amino acid as
it cannot be formed in the body and has to be
supplemented to the body through the diet.
(iii) Among the given amino acids, glycine is not
 optically active. It is the only amino acid which Tr Gc =− (2mol) # (96500 C mol−1)
do not have asymmetrical carbon atom and is
the simplest amino acid. # (0.236 V)
or =− 45548 J
(iv) The isoelectric point is the pH at which the Tr Gc =− 45.55
amino acid does not migrate in an electric kJ
(ii) Given,
field. This means that it is the pH at which the Current, I = 0.5 A
amino acid is neutral i.e., the Zwitter ion form
is dominant. Time, t = 2h
pk + pk Quantity of charge (Q) passed = i # t
a1 a2
Isoelectric point = (0.5 A) # (2h)
2
=
= 2.3 + 9.7 = 12.0 = = (0.5A) # (2 # 60 # 60 s)
6.0 2 2
So, the isoelectric point of the amino acid = = 3600 C
6.0 Again, Q = ne−

SECTION-E where, n = number of electrons

e- = charge on electron

Q
Directions (Q. Nos. 33-35) : The following questions are Hence, n =
e−
long answer type and carry 5 marks each. Two questions
= 1.63600 C
have an internal choice. # 10−19 C
= 2250 # 1019
33. (i) The cell in which the following reaction occurs: Thus, number of electrons = 2.250 # 1022
2Fe3+(aq) + 2I−(aq) $ 2Fe2+ (aq) + I2(s) (iii)
has EcCell = 0.236 Volt at 298K. Calculate the (a) Chlorine is a stronger oxidising agent as
standard Gibbs energy of the cell reaction. compared to iodine because its standard
(Given : 1F = 96,500 C mol–1) electrode potential is lower than that of the
(ii) How many electrons flow through a metallic iodine. So, chlorine can displace iodine from
wire if a current of 0.5 A is passed for 2 KI solution.
hours? (Given : 1F = 96,500 C mol–1) 2KI + Cl2 $ 2KCl + I2
(iii) Explain the following with reason : Iodine is a weaker oxidising agent as
(a) Chlorine can displace iodine from KI compared to bromine because the value of
solution but iodine can not displace bromine standard electrode potential of bromine is
from KBr solution. lower than that of the iodine. So, iodine can
(b) Following reaction is possible or not. not displace bromine from KBr solution.
Hg + H2SO4 $ HgSO4 + H2 KBr + I2 $ No reaction.
(b) Hg + H2 SO4 $ HgSO 4
+2 H
Ans :
This reaction is not possible because the
(i) Standard Gibbs free energy is given by the standard electrode potential of Hg is lower
following expressions : (negative value) than that of the hydrogen.
Tr Gc =− nFEcCell where, However, according to electrochemical series,
only those metals can displace hydrogen from
n = number of makes of electrons transferred,
the dilute solution of hydroacids which
F = Faraday’s constant = 96500 C mol−1, have standard electrode potential higher
Ec Cell = cell constant. than that of the hydrogen i.e., which are
Two half-reactions for the given redox reaction placed below hydrogen in the
can be written as : electrochemical series.
2Fe3+ (aq) + 2e− $ 2Fe2+ ...(1)
(aq)
2I-(aq) $ I2(s) + 2e−
2 moles of electrons are involved in the reaction,
hence n = 2
Therefore, by substituting all the values in
equation (1), we get
34. (i) Account for the following : (i)
(a) Transition metals from large number of (a) The variability in oxidation states of transition
complex compounds. metal is due to the incomplete filling of d
(b) The lowest oxide of transition metal is basic -orbitals in such a way, that their oxidation
whereas the highest oxide is amphoteric or states differ from each other by unity, for
acidic. example, Fe2+ and Fe3+ etc. In case of p -block
(c) Ec value for the Mn3+/Mn2+ couple is highly elements, the oxidation state differ by units
positive (+1.57 V) as compare to Cr3+/Cr2+. by two, for example, + and +5. Moreover,
3
(ii) Write one similarity and one difference between in transition elements, the higher oxidation
the chemistry of lanthanoid and actinoid states are more stable for heavier elements in a
elements. group. For example, Mo6+ is more stable than
Ans : Cr4+. In p -block elements, the lower oxidation
states are more stable for heavier members due
(i) to inert pair effect, for example, Pb 2+ is more
(a) Due to the comparatively smaller size of stable than Pb4+.
the metal ions, high ionic charges and the (b) Cu+ ion is unstable in aqueous solution
availability of vacant d -orbitals for bond than Cu2+. This is because, although second
formation, transition metals for a large ionisation enthalpy of copper is large, but
number of complex compounds. hydration enthalpy for Cu2+ is much more
(b) In lower oxidation states, transition metals negative than that for Cu+ and hence, it
behave like metals and metal oxides are is more than compensates for the second
basic in nature. Thus, in lower oxidation ionisation enthalpy of copper. Therefore,
states, transition metal oxides are basic. As many Cu+ compounds are unstable in aqueous
the oxidation state increases, its metallic solution and undergo disproportionation.
character, decreases due to decrease in size, 2Cu+ $ Cu2+ + Cu
thus, it becomes less metallic or more non- (c) Orange colour of Cr 2O 72– ion changes to
metallic. Oxides of a non-metal may be
acidic or neutral. Thus, in higher oxidation yellow colour when treated with an alkali
because of the formation of chromate ions.
states transition metal oxides are amphoteric
Cr2O2−7 + 2OH− $ 2CrO2− 4 + H2 O
or acidic. Dichromatic ion Chromatic ion
(c) The comparatively high Ec value for (orange) (yellow)

Mn3+/Mn2+ is due to the fact that (ii) Chemistry of actinoids is complicated as

Mn2+ (d 5) is quite stable where as compared to lanthanoids because of the
comparatively low value for Cr3+/Cr2+ is following reasons :
because of the extra stability of Cr3+. (a) Actinoids show a wide range of oxidation states
Therefore, Cr3+ cannot be reduced to Cr2+. i.e., +3, +4, +5 and +6 due to small energy
(ii) Similarity : Both lanthanoid and actinoid difference between 5f, 6d and 7s sub-shells
elements show a common oxidation state + 3 of actinoids.
and both are electropositive and very reactive. (b) Actinoids are radioactive due to that, their
Difference : Lanthanoid have less tendency to chemistry is complicated.
form complexes whereas actinoids have greater
tendency to form complexes.

or
(i) (a) How is the variability in oxidation states 35. (i) Write the product (s) in the following reactions:
of transition metals different from that of the
p
-block elements ?
(b) Out of Cu+ and Cu2+, which ion is unstable
in aqueous solution and why ?
(c) Orange colour of Cr2O72- ion changes to
yellow colour when treated with an alkali.
Why ?
(ii) Chemistry of actinoids is complicated as
compared to lanthanoids. Give two reasons.
Ans :
 (c) (a) DIBAL −H to give a buff coloured precipitate of ferric
CH3 — CH = CH − ?
CN
(b) H O 2
benzoate.
(ii) Give simple chemical test to distinguish 3C6H3COOH + FeCl3 " (C6H5COO)3Fe + 3HCl
between the following pairs of compounds : Ferric benzalde
(a) Butanal and Butan-2-one. (buff colour ppt)

(b) Benzoic acid and Phenol. or

Ans : (i) An organic compound (A) with molecular
formula C3H7NO on heating with Br2 and
(i)
KOH forms a compound (B), compound (B),
(a) Cyclohexanone reacts with hydrogen cyanide
(HCN) to form cyclohexanone cyanohydrin. on heating with CHCl3 and alcoholic KOH
produces a foul smelling compound (C) and on
reacting with C6H5SO2Cl forms a compound (D)
which is soluble in alkali. Write the structures
of (A), (B), (C) and (D).
(ii) Give reasons to support the answer :
(a) Presence of alpha hydrogen in aldehydes and
ketones is essential for aldol condensations.
(b) 3-Hydroxy pentan-2-one shows positive
result to Tollen’s test.
(b) The sodium benzoate reacts with soda lime Ans :
to give benzene. O
(i) A $ CH3 - CH2 - C - NH2
B $ CH3 - CH2 - NH2
C $ CH3 - CH2 - NC
D $ CH3 - CH2 - NH - SO2 - C 6 H 5
(ii)
(a) The alpha hydrogen atoms are acidic
in nature due to presence of electron
(c) But-2-en-1-nitrile on reaction with DIBAL-H withdrawing carbonyl group. These can be
followed by water gives but-2-en-1-ol. easily removed by a base and the carbanion
(a) DIBAL −H
CH3 − CH = CH − formed is resonance stabilised.
(b) H2 O
CN (b) Tollen’s reagent is a weak oxidising agent
But− 2 − en− 1 − nitrile CH3CH = CH − not capable of breaking the C—C bond in
CHO Ketones. Thus, ketones cannot be oxidised
But− 2 −en− 1 − ol
(ii)
(a) Butanal being an aldehyde reduces Tollen’s using Tollen’s reagent itself gets reduced to
reagent to shiny silver mirror but butan-2- Ag.
one being a ketone does not reduces Tollen’s
reagent.
—
CH3CH2CH2CHO + 2 [Ag(NH3)2]+ + OH T
🗆🗆🗆🗆🗆🗆🗆
Butanal Tollen's reagent

CH3CH2CH2COO− + 2Ag .+ + 2H2O

4NH
Silver
mirror
O
Tollen's reagent
CH 3 CH 2 CCH3 No silver mirror
(b) Benzoic acid and phenol can be distinguished
by ferric chloride test. Phenol with neutral
FeCl3 to form ferric phenoxide complex
giving violet colouration.
6C6H5OH + FeCl3 " [Fe(OC6H ) H+ + 3Cl
5 6
]3−
Iron −phenol complex
(Violet colour)

But benzoic acid reacts with neutral FeCl3