Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2023 (Series HFGIE/5) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2023 (Series HFGIE/5) with Solutions
General Instructions:
- This Question Paper contains 35 questions. All questions are compulsory.
- Question paper is divided into Five sections-A, B, C, D and E.
- In Section-A: Question Numbers 1 to 18 are Multiple Choice (MCQs) type questions carrying 1 mark each.
- In Section-B: Question Numbers 19 to 25 are Very Short Answer (VSA) type questions carrying 2 marks each.
- In Section-C: Question Numbers 26 to 30 are Short Answer (SA) type questions carrying 3 marks each.
- In Section-D: Question Numbers 31 & 32 are case based questions carrying 4 marks each.
- In Section-E: Question Numbers 33 to 35 are Long Answer (LA) type questions carrying 5 marks each.
-
There is no overall choice. However, an internal choice has been provided in 2 questions in
Section-B, 2 questions in Section-C, 2 questions in Section-D and 2 questions in Section-E. - Use of calculators is NOT allowed.
Set – I Code No.
56/5/1
Section-A
Question 1.
Which of the following molecules has a chiral centre correctly labelled with an asterisk (*)?
(a) CH
3
C*HBrCH
3
(b) CH
3
C*HClCH
3
Br
(c) HOCH
2
C*H(OH)CH
3
OH
(d) CH
3
C*Br
2
CH
3
Answer:
(c) HOCH
2
C*H(OH)CH
3
OH
All the 4 groups attached to C* are different.
Question 2.
Which of the following alcohols will not undergo oxidation?
(a) Butanol
(b) Butan-2-ol
(c) 2-Methylbutan-2-ol
(d) 3-Methylbutan-2-ol
Answer:
(c) 2-Methylbutan-2-ol
2-Methylbutan-2-ol is a tertiary alcohol which doesn’t undergo oxidation.
Question 3.
A voltaic cell in made by connecting two half cells represented by half equations below:
Sn
2+
(aq) + 2e
–
→ Sn(s) E° = -0.14 V
Fe
3+
(aq) + e
–
→ Fe
2+
(aq) E° = + 0.77 V
Which statement is correct about this voltaic cell? 1
(a) Fe
2+
is oxidised and the voltage of the cell is -0.91 V
(b) Sn is oxidised and the voltage of the cell is 0.91 V
(c) Fe
2+
is oxidised and the voltage of the cell is 0.91 V
(d) Sn is oxidised and the voltage of the cell is 0.63 V
Answer:
(b) Sn is oxidised and the voltage of the cell is 0.91 V
Since the E° of Sn/Sn
2+
is negative, hence it will undergo oxidation easily than Fe
3+
/Fe
2+
whose E° is +0.77 V.
E°
cell
= E°
cathode
– E°
anode
= 0.77 – (-0.14) = 0.77 + 0.14 = +0.91 V
Question 4.
Four half reactions I to IV are shown below:
I. 2Cl
–
→ Cl
2
+ 2e
–
II. 40H
–
→ O
2
+ 2H
2
O + 2e
–
III. Na
+
+ e
–
→ Na
IV. 2H
+
+ 2e
–
→ H
2
Which two of these reactions are most likely to occur when concentrated brine is electrolysed?
(a) I and III
(b) I and IV
(c) II and III
(d) II and IV
Answer:
(b) I and IV
The net reaction for electrolysis of brine is as follows:
NaCl + H
2
O → Na
+
+ OH
–
+ \(\frac{1}{2}\)H
2
+ \(\frac{1}{2}\)Cl
2
Question 5.
Which property of transition metals enables them to behave as catalysts?
(a) High melting point
(b) High ionisation enthalpy
(c) Alloy formation
(d) Variable oxidation states
Answer:
(d) Variable oxidation states
Due to variable oxidation states transition metals can form complexes which make them a good catalyst.
Question 6.
In the two tetrahedral structures of dichromate ion
(a) 4 Cr —O bonds are equivalent in length.
(b) 6 Cr —O bonds are equivalent in length.
(c) All Cr —O bonds are equivalent in length.
(d) All Cr —O bonds are non-equivalent.
Answer:
(b) 6 Cr —O bonds are equivalent in length.
Question 7.
1 mole of liquid A and 2 moles of liquid B make a solution having a total vapour pressure 40 torr. The vapour pressure of pure A and pure B are 45 torr and 30 torr respectively. The above solution
(a) is an ideal solution.
(b) shows positive deviation.
(c) shows negative deviation.
(d) is a maximum boiling azeotrope.
Answer:
(b) shows positive deviation.
Total moles = 3
According to Raoult’s law
P
A
= x
A
× \(\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}\) = \(\frac{1}{3}\) × 45 = 15 torr ; P
B
= x
B
× \(\mathrm{P}_{\mathrm{B}}^{\circ}\) = \(\frac{2}{3}\) × 30 = 20 torr Expected pressure = 15 + 20 = 35 torr ; Actual pressure = 40 torr
∴ Actual pressure > Expected pressure
As It does not obey Raoult’s law, it shows positive deviation.
Question 8.
Which of the following would not be a good choice for reducing nitrobenzene to aniline?
(a) LiAlH
4
(b) H
2
/Ni
(c) Fe and HCl
(d) Sn and HCl
Solution:
(a) LiAlH
4
LiAlH 4 is not used for reduction of nitrobenzene to aniline as it is dangerous for the reaction.
Question 9.
If molality of a dilute solution is doubled, the value of the molal elevation constant (K
b
) will be
(a) halved
(b) doubled
(c) tripled
(d) unchanged
Answer:
(d) unchanged
The value of molal elevation constant (K b ) is constant for a particular solvent. Thus, it will be unchanged when molality of the dilute solution is doubled.
Question 10.
Hydrolysis of sucrose is called
(a) inversion
(b) hydration
(d) saponification
(c) esterification
Answer:
(a) inversion
Hydrolysis of sucrose takes place in presence of invertase enzyme, hence known as inversion.
Question 11.
Which one of the following has lowest pK
a
value? [1]
(a) CH
3
-COOH
(b) O
2
N-CH
2
-COOH
(c) Cl-CH
2
-COOH
(d) HCOOH
Answer:
(b) O
2
N-CH
2
-COOH
As NO
2
is a stronger electron withdrawing group than —Cl so it makes the acid
stronger with least pK
a/sub> value.
Question 12.
Which of the following cell was used in Apollo space programme? [1]
(a) Mercury cell
(b) Daniel cell
(c) H
2
— O
2
Fuel cell
(d) Dry cell
Answer:
(c) H
2
— O
2
Fuel cell
H 2 – O 2 Fuel cell
Question 13.
The following experimental rate data were obtained for a reaction carried out at 25 °C:
A
(g)
+ B
(g)
→C
(g)
+ D
(g)
What are the orders with respect to A
(g)
and B
(g)
?
Answer:
The rate law for A + B → C + D is
Rate = K[A]
p
[B]
q
The initial rate becomes
(Rate)
o
= K[A]
p
[A]
q
Comparing Experiment II and III, we get
(Rate)
2
= K(3.0 × 10
-2
)
p
(4.0 × 10
-2
)
q
= 1.89 × 10
-4
……..(i)
(Rate)
3
= K(6.0 × 10
-2
)
p
(4.0 × 10
-2
)
q
= 7.56 × 10
-4
…….(ii)
Dividing equation (ii) with equation (i)
Question 14.
The magnetic moment of [NiCl
4
]
2-
(a) 1.82 BM
(b) 2.82 BM
(c) 4.42 BM
(d) 5.46 BM
[Atomic number: Ni = 28]
Answer:
(b) 2.82 BM
Ni
2+
= 3d
8
4s
0
For questions number 15 to 18, 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 Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
Question 15.
Assertion (A) : Proteins are polymers of a-amino acids connected by a peptide bond.
Assertion (R) : A tetrapeptide contains 4 amino acids linked by 4 peptide bonds. [1]
Answer:
(c) Assertion (A) is true, but Reason (R) is false.
A tetrapeptide contains 4 amino acids linked by 3 peptide bonds.
Question 16.
Assertion (A) : For a zero order reaction the unit of rate constant and rate of reaction are same.
Assertion (R) : Rate of reaction for zero order reaction is independent of concentration of reactant. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Question 17.
Assertion (A) : Acetic acid but not formic acid can be halogenated in presence of red P and Cl
2
.
Assertion (R) : Acetic acid is a weaker acid than formic acid. [1]
Answer:
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the Correct explanation of Assertion (A).
Formic acid (HCOOH) does not contain a-hydrogen atom for substitution by Cl atom.
— CH
3
group of acetic acid causes +I effect on — COOH group which reduces its acidic character.
Question 18.
Assertion (A) : Trans [CrCl
2
(ox)
2
]
3-
shows optical isomerism.
Assertion (R) : Optical isomerism is common in octahedral complexes involving didentate ligands.
Answer:
(d) Assertion (A) is false, but Reason (R) is true.
Trans-form does not show optical isomerism as the molecule has a plane of symmetry.
Section – B
Question 19.
(a) (i) What should be the signs (positive/negative) for E
°
cell
and ∆G° for a spontaneous redox reaction occurring under standard conditions? [2 × 1 = 2]
Answer:
E°
cell
must be positive and ∆G° must be negative for a spontaneous redox reaction occurring under standard conditions.
(ii) State Faraday’s first law of electrolysis.
Answer:
Faraday’s first law of electrolysis states that the amount of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.
Or
(b) Calculate the emf of the following cell at 298 K: [2]
Fe
(s)
| Fe
2+
(0.01 M) || H
+
(1M)
| H
2
(g) (1 bar), Pt
(s)
Given: E°
Cell
= 0.44V. [2]
Answer:
Question 20.
What happens to the rate constant k and activation energy E
a
as the temperature of a chemical reaction is increased? Justify. [2]
Answer:
The rate constant K increases with temperature as it becomes almost double for every 10° rise in temperature.
The activation energy is not affected by the change in temperature.
Question 21.
(a) Which of the following species cannot act as a ligand? Give reason. [2 × 1 = 2]
OH
–
, \(\mathrm{NH}_4^{+}\), CH
3
NH
2
, H
2
O
Answer:
\(\mathrm{NH}_4^{+}\) cannot act as a ligand because it is electron deficient group and is unable to donate e
–
pair.
(b) The complex [Co(NH
3
)
5
(NO
2
)]Cl
2
is red in colour. Give IUPAC name of its linkage isomer.
Answer:
The complex [Co[NH
3
]
5
(NO
2
)]Cl
2
contain 4 unpaired electrons which get paired after ligand coordination, so the colour is due to charge transfer spectra.
The linkage isomer is [Co(NH
3
)
5
(ONO)] Cl
2
.
IUPAC name. Pentaamminenitrito -(O)-Cobalt (III) chloride
Question 22.
Why is boiling point of o-dichlorobenzene higher than p-dichlorobenzene but melting point of para isomer is higher than ortho isomer.
Answer:
The boiling point of o-dichlorobenzene is greater than the p-isomer due to its greater polar nature but the melting point of p-dichlorobenzerte is greater than o-isomer due to more symmetric nature of p-isomer where it fits in crystal lattice better as compared to o-isomer. As a result of this, intermolecular forces of attraction become stronger in para-isomer and therefore greater energy is required to break its lattice and it melts at higher temperature.
Question 23.
For the pair phenol and cyclohexanol, answer the following: [2 × 1 = 2]
(a) Why is phenol more acidic than cyclohexanol?
Answer:
Phenol is more acidic than cyclohexanol because the conjugate base of phenol is much more stable than conjugate base of cyclohexanol. Moreover the conjugate base of phenol can delocalize the negative charge throughout the ring through resonance and stabilize it effectively, whereas conjugate base of cyclohexanol has no resonance structures to stabilize the charge and so is less stable.
(b) Give one chemical test to distinguish between the two.
Answer:
Phenol reacts with neutral FeCl
3
solution and gives violet colour solution whereas cyclohexanol does not react with FeCl
3
and doesn’t give any colour.
Question 24.
(a) (i) Draw the zwitter ion structure for sulphanilic acid. [2 × 1 = 2]
Answer:
Zwitter ion structure of suiphanilic acid:
(ii) How can the activating effect of -NH
2
group in aniline be controlled?
Answer:
The activating effect of — NH
2
group can be controlled by protecting the —NH
2
group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. The lone pair of electrons on the nitrogen of acetanilide interacts with oxygen atom due to resonance.
Or,
(b) (i) Complete the reaction with the main product formed: [2 × 1 = 2]
(ii) Convert Bromoethane to Propanamine.
Answer:
Question 25.
Give the reaction of glucose with hydrogen cyanide. Presence of which group is confirmed by this reaction? [2]
Answer:
Section – C
Question 26.
(a) For the reaction:
2N
2
O
5(g)
→ 4NO
2(g)
+ O
2(g)
at 318 K
Calculate the rate of reaction if rate of disappearance of N
2
O
5(g)
is 1.4 × 10
-3
ms
-1
.
Answer:
(b) For a first order reaction derive the relationship t
99%
= 2t
90%
. [1+2=3]
Answer:
Question 27.
(a) On the basis of crystal field theory write the electronic configuration for d
5
ion with a strong field ligand for which ∆
0
> P.
(b) [Ni(CO)
4
] has tetrahedral geometry while [Ni(CN)
4
]
2-
has square planar yet both exhibit dimagnetism. Explain. [Atomic number: Ni = 28] [1 + 2 = 3]
Answer:
Question 28.
(a) Illustrate Sandmeyer’s reaction with an equation.
Answer:
Sandmeyer’s equation:
(b) Explain, why (CH
3
)
2
NH is more basic than (CH
3
)
2
N in aqueous solution. [1 + 2 = 3]
Answer:
Secondary amines (CH
3
)
2
NH are more basic than tertiary amines (CH
3
)
3
N due to steric hindrance of 3 bulky group in 3° amine which hinders the protonation at nitrogen atom and hence reduces its basicity.
Question 29.
Give reasons for any three of the following observations: [3 × 1 = 3]
(a) Penta-acetate of glucose does not react with hydroxylamine.
Answer:
Penta-acetate of glucose does not have a free — OH group at Cl and therefore, cannot be converted to the open chain form to give — CHO group, hence it does not form the oxime.
(b) Amino acids behave like salts.
Answer:
Amino acids behave like salts due to the presence of both acidic and basic groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton, an amino group can accept a proton and giving rise to a dipolar ion known as zwitter ion.
(c) Water soluble vitamins must be taken regularly in diet.
Answer:
Water soluble vitamins dissolve in water and are not stored by the body. Moreover, they are eliminated from the body in the form of urine, so we require a continuous supply of these vitamins in our diet.
(d) The two strands in DNA are complimentary to each other.
Answer:
DNA has double helical structure, contains two strands which are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms H-bond with thymine. Thus, the sequence of bases in one strand automatically determines that of the other. Hence, the 2 strands of DNA are complementary to each other.
Question 30.
(a) (i) Why is the C—O bond length in phenols Less than that in methanol’
Answer:
The C —O bond length in phenol is slightly less than that in methanol due to partial double bond character of C —O bond because of the conjugation of lone pair of electrons of oxygen with the aromatic ring.
(ii) Arrange the following in order of increasing boiling point:
Ethoxyethane, Butanal, Butanol, n-butane
Answer:
n-butane < Ethoxyethane < Butanal < Butanol
(iii) How can phenol be prepared from anisole? Give reaction. [3 × 1 = 3]
Answer:
Or,
(b) (i) Give mechanism of the following reaction:
(ii) Illustrate hydroboration-oxidation reaction with an example.
Answer:
Section – D
The following questions are Case Based questions. Read the passage carefully and answer the questions that follow:
Question 31.
Nucleophilic Substitution
Nucleophilic Substitution reaction of haloalkane can be conducted according to both S
N
1 and S
N
2 mechanisms. S
N
1 is a two step reaction while S
N
2 is a single step reaction. For any haloalkane which mechanism is followed depends on factors such as structure of haloalkane, properties of leaving group, nucleophilic reagent and solvent.
Influences of solvent polarity: In S
N
1 reaction, the polarity of the system increases from the reactant to the transition state, because a polar solvent has a greater effect on the transition state than the reactant, thereby reducing activation energy and accelerating the reaction. In S
N
2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (S
N
1) of tertiary chlorobutane at 25° C in water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (S
N
2) of 2-Bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. Hence the level of solvent polarity has influence on both S
N
1 and S
N
2 reaction, but with different results. Generally speaking weak polar solvent is favourable for S
N
2 reaction, while strong polar solvent is favourable for S
N
1. Generally speaking the substitution reaction of tertiary haloalkane is based on S
N
1 mechanism in solvents with a strong polarity (for example ethanol containing water).
Answer the following questions:
(a) Why racemisation occurs in S
N
1? [1]
Answer:
In S
N
1 reaction, a carbocation intermediate is formed which is a planar molecule and can lead to form d- and l[- products. Hence racemisation occurs.
(b) Why is ethanol less polar than water? [1]
Answer:
Ethanol is less polar than water because in water, two hydrogen atoms are bonded to an oxygen while in ethanol, one hydrogen atom and one alkyl group are bonded to an oxygen.
(c) Which one of the following in each pair is more reactive towards S
N
2 reaction?
(i) CH
3
—CH
2
—I or CH
3
CH
2
—Cl
Answer:
CH
3
CH
2
— I (as -I is a better leaving group)
(ii)
Answer:
Or
(c) Arrange the following in the increasing order of their reactivity towards S
N
1
reactions: [2 × 1]
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
Answer:
1-Bromopentane < 2-Bromopentane < 2-Bromo-2-methylbutane
(ii) 1 -Bromo-3-methylbu tane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
Answer:
1-Bromo-3-methylbutane < 2-Bromo-3-methylbutane < 2-Bromo-2-methylbutane
Question 32.
Rahul set-up an experiment to find resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a.c. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from null point he also calculated the conductivity Λ
m
and molar conductivity and recorded his readings in tabular form.
Answer the following:
(a) Why does conductivity decrease with dilution? [1]
Answer:
Conductivity decreases with dilution because the number of ions responsible for carrying current decreases in a unit volume.
(b) If Λ
m
° of KCl is 150.0 S cm
2
mol
-1
, calculate the degree of dissociation of 0.01 M KCl. [1]
Answer:
Given: Λ
m
° = 150.0 S cm
2
mol
-1
Concentration of KCl = 0.01 M
Molar conductivity, Λ
m
= \(\frac{K}{C}\) = 141 S cm
2
mol
-1
α = \(\frac{\Lambda_m}{\Lambda_m^o}\) = \(\frac{141}{150}\) = 0.94
(c) If Rahul had used HCl instead to KCl then would you expect the Am values to be more or less than those per KCl for a given concentration. Justify. [2 × 1]
Answer:
If Rahul had used HCl instead of KCl of same concentration then molar conductivity Λ
m
of HCl will be higher. Since smaller the cation, higher the molar conductivity. Here H
+
ions are smaller than K
+
ions.
Or
(c) Amit, a classmate of Rahul repeated the same experiment with CH
3
COOH solution instead of KCI solution. Give one point that would be similar and one that would be
different in his observations as compared to Rahul. [2 × 1]
Answer:
Similarity is for both KCl and CH
3
COOH, the molar conductivity Λ
m
will increase with dilution or decrease in concentration.
Difference in KCl being a strong electrolyte shows a linear relation between molar conductivity and concentration, as it is completely ionised while CH
3
COOH will not show linear relation due to its weak nature.
Section – E
Question 33.
(a) (i) Why is boiling point of 1M NaCl solution more than that of 1M glucose solution?
Answer:
Since NaCl is ionic and completely dissociates to give 2 ions per molecule which is not possible in glucose. So NaCl requires more heat to break its strong crystal lattice and thus have high boiling point. Moreover, both have same molality but Van’t Hoff factor of NaCl is more than glucose.
(ii) A non-volatile solute ‘X’ (molar mass = 50 g mol
-1
) when dissolved in 78 g of benzene reduced its vapour pressure to 90%. calculate the mass of X dissolved in the solution.
Answer:
Let p be the vapour pressure of pure benzene.
The vapour pressure of solution, p’ = \(\frac{90}{100}\)p = 0.9 p.
Molar mass of solute, M = 50 g mol
-1
Molar mass of benzene, m = 78 g mol
-1
(iii) Calculate the boiling point elevation for a solution prepared by adding 10 g of MgCl
2
to 200 g of water assuming MgCl
2
is completely dissociated.
(K
b
for Water 0.512 K kg mol
-1
, Molar mass MgCI
2
= 95 g mol
-1
) [1 + 2 + 2 = 5]
Answer:
Given: W
A
= 10 g, W
B
= 200 g, K
b
= 0.512 K kg mol
-1
, M
B
= 95 g mol
-1
Or,
(b)
(i) Why is the value of Van’t Hoff factor for ethanoic acid in benzene close to 0.5?
Answer:
Ethanoic acid undergoes association in benzene.
(ii) Determine the osmotic pressure of a solution prepared by dissolving 2.32 × 10
-2
g of KMsub>2SO
4
in 2L of solution at 25° C, assuming that K
2
SO
4
is completely dissociated.
(R = 0.082 L atm K
-1
mol
-1
, Molar mass K
2
SO
4
= 174 g mol
-1
)
Answer:
(iii) When 25.6 g of Sulphur was dissolved in 1000 g of henzene, the freezing point lowered by 0.512 K. Calculate the formula of Sulphur
(K
f
for benzene = 5.12 K kg mol
-1
, Atomic mass of Sulphur = 32 g mol
-1
) [1 + 2 + 2 = 5]
Answer:
Question 34.
(a)
(i) Write the reaction involved in Cannizaro’s reaction. [1 + 1 + 3 = 5]
Answer:
Cannizaro’s Reaction. Aldehydes which do not contain any α-hydrogen atom undergo self oxidation and reduction on treatment with cone, alkali to form Carboxylic acid salt and alcohol.
(ii) Why are the boiling point of aldehydes and ketones lower than that of corresponding carboxylic acids?
Answer:
The boiling point of aldehydes and ketones are lower than that of corresponding carboxylic acids because of their polar nature due to which they have sufficient intermolecular dipole-dipole interaction between the opposite ends of C = O dipoles but this dipole-dipole interaction becomes weaker than intermolecular H-bonding in carboxylic acids.
(iii) An organic compound ‘A’ with molecular formula C
5
H
8
O
2
is reduced to n-pentane with hydrazine followed by hearing with NaOH ana Glycol. ‘A’ forms n dioxime with hydroxylamine and gives a positive lodoform and Toflen’s test. Identify ‘A’ and give its reaction for lodoform and Tollen’s test.
Answer:
Here the compound A is
Or,
(b) (i) Give a chemical test to distinguish between ethanal acid and ethanoic acid.
Answer:
Distinction between ethanal and ethanoic acid:
Ethanal gives iodoform test whereas ethanoic acid does not.
(ii) Why is the Œ-hydrogens of aldehydes and ketones are acidic in nature?
Answer:
The α-hydrogen atoms of aldehydes and ketones are slightly acidic in nature due to strong electron withdrawing inductive (-I) effect of the carbonyl group. The acidity is also due to resonance stabilization of the conjugate base. Due to —I effect it withdraws electrons from the adjacent C—C bond which makes α-carbon e
–
deficient and it in turn withdraws e
–
from C
α
—H bond. As a result, the electron density in C
α
— H bond decreases and α-hydrogens are weakly held and therefore, can easily be abstracted by strong base as:
(iii) An organic compound ‘A’ with molecular formula C
4
H
8
O
2
undergoes acid hydrolysis to form two compounds ‘B’ and ‘C’. Oxidation of ‘C’ with acidified potassium permanganate also produces ‘B’. Sodium salt of ‘B’ on heating with soda lime gives methane.
(1) Identify ‘A’, ‘B’ and ‘C’.
(2) Out of ‘B’ and ‘C’, which will have higher boiling point? Give reason. [1 + 1 + 3 = 5]
Answer:
(1) The organic compound ‘A’ is an ester CH
3
COOCH
2
CH
3
(Ethyl ethanoate) and undergoes esterification reaction in presence of dil. H
2
SO
4
.
(2) Out of ethanoic acid and ethanol, ethanol will have higher boiling point due to intermolecular hydrogen bonding.
Question 35.
(a) Why is chemistry of actinoids complicated as compared to lanthanoids? [1 + 2 + 2 = 5]
(b) Complete the following reaction and justify that it is a disproportionation reaction:
3Mn\(\mathrm{O}_4^{2-}\) + 4H
+
→ ……. + …… + 2H
2
O.
(c) The given graph shows the trends in melting points of transition metals.
Explain the reason why Cr has highest melting point and manganese (Mn) a lower melting point.
Answer:
(a) Chemistry of actinoids is complicated as compared to lanthanoids due to its more radioactive elements. They are present in small quantities and their half life is extremely small. Moreover, they shows more variable oxidation states than lanthanoids and their 5f orbital is more exposed to outer environment while 4f orbital of lanthanoids are deeply buried.
(b)
Manganate ion \(\left(\mathrm{MnO}_4^{2-}\right)\) is unstable in acidic solution and disproportionate to give Permanganate \(\left(\mathrm{MnO}_4^{-}\right)\) via oxidation and MnO
2
by reduction.
(c) The highest melting point of Cr is attributed to the involvement of greater number of unpaired electrons (d
5
) from 3d in addition to 4s electrons in interatomic bonding. Therefore metallic bonding is maximum and has high melting point.
Mn has exceptionally low melting point due to stable half-filled 3d orbital which is more tightly held by the nucleus. This reduces the de-localization of electrons resulting in weaker metallic bonding due to low interatomic forces of attraction and is easier to break.
Set – II Code No. 56/5/2
Except for the following questions, all the remaining questions have been asked in Set I.
Question 3.
Consider the following standard electrode potential values:
Fe
3+
(aq) + e
–
→ Fe
2+
(aq)
, E° = +0.77 V
Mn\(\mathrm{O}_{4(\mathrm{aq})}^{-}\) + 8H
+
+ 5e
–
→ Mn
2+
(aq)
+ 4 H
2
O
(l)
, E° = +1.51 V
What is the cell potential for the redox reaction?
(a) -2.28 V
(b) -0.74 V
(c) +0.74 V
(d) +2.28 V
Answer:
(c) +0.74 V
Mn\(\mathrm{O}_4^{-}\) undergoes reduction due to its high positive electrode potential value than Fe
3+
/Fe
2+
system.
\(\mathrm{E}_{\text {cell }}^o\) = \(\mathrm{E}_{\text {o }}^{\mathrm{ot} h o d e}\) – \(\mathrm{E}_{\text {anode }}^{\circ}\) = 1.51 – (+0.77) = +0.74 V
Question 6.
Which of the following ions has the electronic configuration 3d
6
?
(Atomic number: Mn = 25, Co = 27, Ni 28) [1]
(a) Ni
3+
(b) Co
3+
(c) Mn
2+
(d) Mn
3+
Answer:
(b) Co
3+
\({ }_{27} \mathrm{Co}^{3+}\) = [Ar]3d 6 4s 0
Question 7.
Which of the following aqueous solution will have highest boiling point? [1]
(a) 1.0 M KCl
(b) 1.0 M K
2
SO
4
(c) 2.0 M KCl
(d) 2.0 M K
2
SO
4
Answer:
(d) 2.0 M K
2
SO
4
The elevation in boiling point is proportional to the number of ions in solution or on Van’t Hoff factor.
We can compute the product of Van’t Hoff factor and the molality for the following:
Question 9.
Amides can be converted into amines by the reaction named [1]
(a) Hoffmann degradation
(b) Ammonolysis
(c) Carbylamine
(d) Diazotisation
Answer:
(a) Hoffmann degradation
Question 10.
Which of the following statements is not true about glucose? [1]
(a) It is an aldohexose.
(b) On heating with HI it forms n-hexane.
(c) It is present in pyranose form.
(d) It gives 2, 4 DNP test.
Answer:
(d) It gives 2, 4 DNP test.
For questions number 15 to 18, 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 Assertion (A).
(c) Assertion (A) is true, hut Reason (R) is false.
(d) Assertion (A) is false, hut Reason (R) is true.
Question 15.
Assertion (A) : Vitamin C cannot be stored in our body.
Reason (R) : Vitamin C is fat soluble and is excreted from the body in urine. [1]
Answer:
(c) Assertion (A) is true, hut Reason (R) is false.
Vitamin C is water soluble and is excreted from the body in the form of urine.
Question 16.
Assertion (A) : The half life of a reaction is the time in which the concentration of the reactant
is reduced to one half of its initial concentration.
Reason (R) : In first order kinetics when concentration of reactant is doubled, its half life is
doubled. [1]
Answer:
(c) Assertion (A) is true, hut Reason (R) is false.
When initial concentration of a reactant in first order kinetic is doubled, its half life is not affected as t
1/2
is independent of the initial concentration of reactant.
∴ t
1/2
= \(\frac{0.693}{K}\)
Question 17.
Assertion (A) : Bromination of benzoic acid gives m-bromobenzoic acid.
Reason (R) : Carboxyl group increases the electron density at the meta position. [1]
Answer:
(c) Assertion (A) is true, hut Reason (R) is false.
Carboxyl group is an electron withdrawing group and it decreases electron density at ‘O’ and ‘P’ position but not increase the e density at ‘m’ position while it directs the electrophile at (m) (meta position) due to its meta directing nature.
Question 18.
Assertion (A) : EDTA is a hexadentate ligand.
Reason (R) : EDTA has 2 nitrogen and 4 oxygen donor atoms. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Question 22.
Write equations for the following: [2 × 1 = 2]
(a) Oxidation of chloroform by air and light.
Answer:
Oxidation of chloroform by air and light:
(b) Reaction of chlorohenzene with CH
3
Cl/anhyd. AlCl
3
Answer:
Friedel Craft’s alkylation:
Question 25.
Give the reaction of glucose with acetic anhydride. Presence of which is confirmed by this
reaction? [2]
Answer:
Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five —OH groups which are attached to different ‘C’ atoms.
Question 27.
(a) On the basis of crystal field theory write the electronic configuration for d
5
ion with a weak ligand for which ∆
0
< P.
Answer:
Electronic configuration for d
5
with weak ligand.
(b) Explain [Fe(CN)
6
]
3-
is an inner orbital complex whereas [FeF
6
]
3-
is an outer orbital complex.
[Atomic number: Fe = 26] [1 + 2 = 3]
Answer:
In [Fe(CN)
6
]
3-
, Hexacyanoferrate (III) ion, CN is a strong field ligand, so the pairing of e
–
occurs and d
2
sp
3
hybridization takes place and forms inner orbital complex.
Fe
3+
= [Ar] 3d
5
4s
0
4p
0
In [FeF
6
]
3-
, Hexafluoridoferrate (III) ion, F is a weak field ligand and no pairing of electrons takes place, so sp
3
d
2
hybridization occurs and forms outer orbital complex. Fe
3+
= [Ar] 3d
5
4s
0
4p
0
4d
0
Question 35.
(a) A transition element X has electronic configuration [Ar]4s
2
3d
3
. Predict its likely oxidation states.
Answer:
X = [Ar] 4s
2
3d
3
The atomic number is 23 and the element will be Vanadium. Its oxidation states are +2, +3, +4 and
(b) Complete the reaction mentioning all the products formed:
Answer:
On heating KMnO
4
disproportionates into K
2
MnO
4
and MnO
2
.
(c) Account for the following:
(i) In the 3d transition series, zinc has the lowest enthalpy of atomisation.
Answer:
Zn has 3d
10
4s
2
configuration and does not have any unpaired electrons in its atomic and ionic state. So the interatomic electronic bonding is the weakest in zinc which contributes to its least enthalpy of atomisation in the 3d series.
(ii) Cu
+
ion is unstable in aqueous solution.
Answer:
Cu
+
ion is unstable in aqueous solution because it undergoes disproportionation. It gets oxidised into cupric ion and is reduced into copper metal. Although energy is required to remove one electron from Cu
+
to Cu
2+
, high hydration energy of Cu
2+
compensates it.
(iii) Actinoids show more number of oxidation states than lanthanoids. [1 + 1 + 3 = 5]
Answer:
Actinoids show more number of oxidation states than lanthanoids because its 5f, 6d and 7s energy levels are of comparable energies and they all participate in bonding. Moreover, poor shielding of 5f orbitals than 4f orbitals is also responsible for this.
Set – III Code No. 56/513
Except for the following questions, all the remaining questions have been asked in Set I and Set II.
Question 3.
Consider the following standard electrode potential values:
Sn
2+
(aq)
+ 2e
–
+ Sn
(S)
; E° = -0.91 V
Fe
3+
(aq)
+ e
–
→ 2 Fe
2+
(aq); E° = +0.77 V
What is the cell reaction and potential for the spontaneous reaction that occurs?
(a) 2 Fe
2+
(aq)
+ Sn
2+
(aq) → 2 Fe
3+
(aq)
+ Sn
(s)
; E° = -0.91 V
(b) 2 Fe
3+
(aq)
+ Sn
(s)
→ 2 Fe
2+
(aq)
+ Sn
2+
(aq)
; E° = +0.91 V
(c) 2 Fe
2+
(aq)
+ Sn
2+
(aq)
→ 2 Fe
3+
(aq)
+ Sn(s); E° = +0.91 V
(d) 2 Fe
3+
(aq)
+ Sn
(s)
→ 2 Fe
2+
(aq)
+ Sn
2+
(aq); E° = +1.68 V
Answer:
(b) 2 Fe
3+
(aq)
+ Sn
(s)
→ 2 Fe
2+
(aq)
+ Sn
2+
(aq)
; E° = +0.91 V
E° of Fe
3+
/Fe
2+
has positive value due to which it undergoes reduction while Sn undergoes oxidation. So their
E°
cell
= E°
cathode
– E°
anode
= 0.77 – (-0.14) = 0.77 + 0.14
∴ E°
cell
= +0.91 V
2Fe
3+
+ Sn → 2Fe
2+
+ Sn
2+
Question 6.
The unit of molar conductivity is
(a) S cm
-2
mol
-1
(b) S cm
2
mol
-1
(c) S
-1
cm
2
mol
-1
(d) S cm
2
mol
Answer:
(b) S cm
2
mol
-1
Question 7.
Out of the following 1.0 M aqueous solutions, which one will show largest freezing point depression?
(a) NaO
(b) Na
2
SO
4
(c) C
6
H
12
O
6
(d) Al
2
(SO
4
)
3
Answer:
(d) Al
2
(SO
4
)
3
As Al 2 (SO 4 ) 3 (2Al 3+ + 3S\(\mathrm{O}_4^{2-}\)) ionizes to give total 5 number of ions than the others, so will show largest freezing point depression.
Question 9.
In the reaction
C
6
H
5
NH
2
+ CHCl
3
+ 3KOH → A + 3B + 3C the product A is
(a) Ç
6
H
5
NC
(b) C
6
H
5
CN
(c) C
6
H
5
Cl
(d) C
6
H
5
NHCH
3
Answer:
(a) Ç
6
H
5
NC
Carbylamine reaction forms phenyl isocyanide.
C
6
H
5
NH
2
+ CHCl
3
+ 3KOH → C
6
H
5
NC + 3KCl + 3H
2
O
Question 10.
β-pleated sheet structure in proteins refers to
(a) primary structure
(b) secondary structure
(c) tertiary structure
(d) quaternary structure
Answer:
(b) secondary structure
For questions number 15 to 18, 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) arc true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
Question 15.
Assertion (A) : The backbone of DNA and RNA molecules is a chain consisting of heterocyclic base, pentose sugar and phosphate group.
Reason (R) : Nucleotides and nucleosides mainly differ from each other in presence of phosphate group.
Answer:
(b) Both Assertion (A) and Reason (R) arc true, but Reason (R) is not the correct explanation of Assertion (A).
Question 16.
Assertion (A) : Order of reaction is applicable to elementary as well as complex reactions.
Reason (R) : For a complex reaction molecularity has no meaning. [1]
Answer:
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Since a complex reaction proceeds through several steps and each step has its own molecularity which may be different. So, molecularity of overall complex reaction is meaningless. On the other hand, order of a complex reaction is determined by the slowest step in its mechanism and is not meaningless even in the case of elementary reactions.
Question 17.
Assertion (A) : The final product in Aldol condensation is always α, β-unsaturated carbonyl
compound.
Reason (R) : α, β-unsaturated carbonyl compounds are stahilised due to conjugation. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Question 18.
Assertion (A) : [Co(NH
3
)
5
SO
4
] Cl gives a white precipitate with silver nitrate solution.
Reason (R) : The complex dissociates to give Cl- and \(\mathrm{SO}_4^{2-}\) ions. [1]
Answer:
(c) Assertion (A) is true, but Reason (R) is false.
Question 22.
Why haloarenes are not reactive towards nucleophilic substitution reaction? Give two reasons.
[2]
Answer:
Haloarenes are not reactive towards nucleophilic substitution reaction due to following reasons:
(i) The e” pairs of halogen atom are in conjugation with the benzene ring through resonance and C —Cl bond acquires a partial double character which makes its cleavage very difficult for S N reaction.
(ii) The sp 2 hybridised carbon of haloarene has greater s character and is more electronegative due to which it holds the e – pair more tightly which reduces its (C — X) bond length and imparts its stability and hence the cleavage of bond becomes difficult.
Question 27.
(a) On the basis of crystal field theory, write the electronic configuration for d
4
with a strong field ligand for which ∆
0
> p.
Answer:
d
4
with strong field ligand. Here pairing occurs
(b) A solution of [Ni(H
2
O)
6
]
2+
is green but a solution of [Ni(CO)
4
] is colourless. Explain.
[Atomic number: Ni = 28] [1 + 2 = 3]
Answer:
[Ni(H
2
O)
6
]
2+
. Here H
2
O is a weak field ligand and does not pair up the unpaired electrons. Hence due to d-d transition, red light is absorbed and complimentary green light is emitted.
In [Ni(CO)
4
], here CO is strong field ligand which causes pairing up of electrons leaving no unpaired electrons for absorption of light and hence the complex is colourless.
Ni = [Ar] 3d
8
4s
2
4p
0
Question 35.
(a) Write the number of unpaired electrons in Cr
3+
. [Atomic number of Cr = 24)
Answer:
(b) Complete the reaction mentioning all the products formed:
Cr
2
\(\mathrm{O}_7^{2-}\) + 3H
2
S + 8H + →
Answer:
Cr
2
\(\mathrm{O}_7^{2-}\) + 3H
2
S + 8H
+
→ 2Cr
3+
+ 3S + 7H
2
O
(c) Account for the following:
(i) Mn
2+
is more stable than Fe
2+
towards oxidation to +3 state.
Answer:
25
Mn
2+
= [Ar] 3d
5
4s
0
;
26
Fe
2+
= [Ar] 3d
6
4s
0
Mn
2+
is more stable due to half filled d-orbitais and does not get oxidised easily while Fe
2+
has 6 electrons in their 3d orbital, so they tend to lose one electron to form Fe
3+
through oxidation and get stable d
5
configuration.
(ii) Copper has exceptionally positive \(\mathrm{E}_{\mathrm{M}}^{\mathrm{o}}{ }^{2+} / \mathrm{M}\) value.
Answer:
The \(\mathrm{E}_{\mathrm{M}^{\prime 2+} / \mathrm{M}}\) value for metal depends on atomization enthalpy, ionization enthalpy and hydration enthalpy. Copper has a high value of atomisation enthalpy and low hydration enthalpy as a result of which, the overall \(\mathrm{E}^0 \mathrm{M}^{2+} / \mathrm{M}\) for Copper becomes positive.
(iii) Eu
2+
with electronic configuration [Xe]4f
7
6s
2
is a strong reducing agent. [1 + 1 + 3 = 5]
Answer:
Eu
2+
= \([\mathrm{Xe}] 4 f^7 6 s^2\). In +2 oxidation state it has stable half filled f
7
configuration but to acquire stable common oxidation state it oxidises readily into Eu
3+
and acts as a strong reducing agent.