KVPY-SX 2018 Chemistry Paper with Solutions

Question 1: The amount (in mol) of bromoform (CHBr₃) produced when 1.0 mol of acetone reacts completely with 1.0 mol of bromine in the presence of aqueous NaOH is

a. 1/3
b. 2/3
c. 1
d. 2

Answer:(a)

The reaction of acetone with Br₂ in the presence of NaOH follow as

Here 3 moles of Br₂ reacts with 1 mole acetone, produce 1 mole of CHBr₃

So 1 moles of Br₂ = 1/3 mole of CHBr₃

Hence, the moles of CHBr₃ = 1/3 moles

Therefore, the correct option is (a).

Question 2: The following compound can readily be prepared by Williamson ether synthesis by reaction between

Answer:(b)

Williamson ether synthesis ->

Order of reactivity of halide->

R–I > R–Br > R–Cl > R–F (1⁰ > 2⁰ > 3⁰)

Hence, in the following reaction, we can readily prepare the above ether compound.

Because the stability of carbocation (C⁺) increases the rate of SN² decreases.

Therefore, the correct option is (b).

Question 3: X and Y

are:

a. enantiomers
b. diastereomers
c. constitutional isomers
d. conformers

Answer:(d)

(1) and (2) are identical.

Hence, they are conformers.

Therefore, the correct option is (d).

Question 4: The higher stabilities of tert-butyl cation over isopropyl cation, and trans-2-butene over propene, respectively, are due to orbital interactions involving

a. σ -> π and σ -> π *
b. σ -> vacant p and π -> π *
c. σ -> σ * and σ -> π
d. σ -> vacant p and σ -> π *

Answer:(d)

Hint:- See the key point and definition of hyperconjugation

Hyperconjugation structure = (Number of αH + 1)

As the α Hydrogen increases the number of Hyperconjugation (σ bond vacant p orbital overlapping) structure increases and stability of carbocation also increases.

Hence, tert-butyl carbocation is more stable compared to isopropyl carbocation.

Hyperconjugation structure = (Number of αH + 1)

As the α Hydrogen increase in alkene number of Hyperconjugation (σ bond π* Interaction) structure increase and stability of alkene also increase.

Hence, 2 butene is more stable compare to propene.

Therefore, the correct option is (d).

Question 5: Benzaldehyde can be converted to benzyl alcohol in concentrated aqueous NaOH solution using

a. acetone
b. acetaldehyde
c. formic acid
d. formaldehyde

Answer:(d)

It is an example of Cannizzaro reaction.

The reaction mechanism follows as

Reactivity order towards Nucleophilic addition -> HCHO >PhCHO

Therefore, the correct option is (d).

Question 6: The major product of the following reaction

is:

Answer:(a)

Solution -> NaBH₄ ideal reducing agent for reduction of the carbonyl group to corresponding alcohol without affecting other groups.

The reaction is as follows;

Therefore, the correct option is (a).

Question 7: Among the following species, the H–X–H angle (X = B, N or P) follows the order

a. PH₃< NH₃< NH₄⁺< BF₃
b. NH₃< PH₃< BF₃
c. BF₃< PH₃< NH₃
d. BF₃< NH₃< PH₃

Answer:(a)

1. PH₃-> no hybridisation (Drago’s rule). Bond angle – 93⁰

2.NH₃-> sp³ hybridisation (with one lone pair) Bond angle – 107⁰

Due to lone pair-bond pair repulsion, the bond angle will reduce from 109⁰28’ to 107⁰.

3.NH₄⁺ -> sp³ hybridisation. Bond angle -109⁰28’

4. BF₃-> sp² hybridisation. Bond angle – 120⁰

Question 8: The ionic radii of Na⁺, F^–, O^2–, N^3– follow the order

a. O^2–> F^–> Na⁺> N^3–
b. N^3–> Na⁺> F^–> O^2–
c. N^3–> O^2–> F^–> Na⁺
d. Na⁺> F^–> O^2–> N^3–

Answer:(c)

Ionic size increases with the addition of electron and size decrease with loss of an electron.

Nitride ion has 10 electrons in contrast with only 7 protons due to the above-given reason. It has the highest inter electronic repulsions. Hence its size or radius is maximum.

Note: For isoelectronic species.

Ionic size is proportional to the negative charge on the anion.

Hence, correct order follow as N^3–> O^2–> F^–> Na⁺

Therefore, the correct option is (c).

Question 9: The oxoacid of phosphorus having the strongest reducing property is:

a. H₃PO₃
b. H₃PO₂
c. H₃PO₄
d. H₄P₂O₇

Answer:(b)

Reducing Hydrogen => P–H (Bounded Hydrogen)

-> Hence, in H₃PO₂ have (2) reducing H due to this it has strongest reducing properties.

Therefore, the correct option is (b).

Question 10: Among C, S and P, the element(s) that produce(s) SO₂ on reaction with hot conc. H₂SO₄ is/are:

a. only S
b. only C and S
c. only S and P
d. C, S and P

Answer:(d)

Reaction of C, S and P with H₂SO₄ follows as

C + H₂SO₄-> CO₂ + H₂O + SO₂

S + H₂SO₄-> SO₂ + H₂O

P + H₂SO₄-> SO₂ + H₃PO₄ + H₂O

=> Hence, here all produced SO₂ gas

So, the correct option is (d).

Question 11: The complex that can exhibit linkage isomerism is

a. [Co(NH₃)₅(H₂O)]Cl₃
b. [Co(NH₃)₅(NO₂)]Cl₂
c. [Co(NH₃)₅(NO₃)](NO₃)₂
d. [Co(NH₃)₅Cl]SO₄

Answer:(b)

Linkage Isomerism->

Linkage isomerism is the existence of coordination compounds that have the same composition differing from the connectivity of metal to ligand.

=> Here, in case (B) NO₂ is a ambidentate ligand So it can bind with metal as N(NO₂) and O(ONO).

Hence, compound [Co(NH₃)₅(NO₂)]Cl₂ show linkage as follow:

Therefore, the correct option is (b).

Question 12: The tendency of X in BX₃ (X = F, Cl, OMe, NMe) to form a π bond with boron follows the order

a. BCl₃< BF₃<B(OMe)₃< B(NMe₂)₃
b. BF₃< BCl₃< B(OMe)₃< B(NMe₂)₃
c. BCl₃<B(NMe₂)₃< B(OMe)₃< BF₃
d. BCl₃< BF₃< B (NMe₂)₃< B(OMe)₃

Answer:(a)

In all, highest back bonding in the case of B(NMe₂)₃ due to less E.N & small size and N have good electron density for back bonding due to +I effect of attached group.

=> E.N increases back bonding decreases.

=> electron density increases back bonding increases.

=> Size increases back bonding decreases.

=> Hence, the overall order of back bonding follow as

B (NMe₂)₃> B (OMe)₃> BF₃> BCl₃

Therefore, the correct option is (a).

Question 13: Consider the following statements about Langmuir isotherm:

(i) The free gas and adsorbed gas are in dynamic equilibrium

(ii) All adsorption sites are equivalent

(iii) The initially adsorbed layer can act as a substrate for further adsorption

(iv) The ability of a molecule to get adsorbed at a given site is independent of the occupation of neighbouring sites

The correct statements are

a. (i), (ii), (iii) and (iv)
b. only (i), (ii) and (iv)
c. only (i), (iii), and (iv)
d. only (i), (ii) and (iii)

Answer:(b)

According to Langmuir isotherm ->

=> ∆H of adsorption of each binding site is same because the surface is homogenous.

=> There is a dynamic equilibrium between free gas and adsorbed gas.

=> The ability of a molecule to get adsorbed at a given site is independent of the occupation of the neighbouring site.

Therefore, the correct option is (b).

Question 14: Among the following, the plot that correctly represents the conductometric titration of 0.05 M H₂SO₄ with 0.1M NH₄OH is:

Answer:(b)

-> Reaction of H₂SO₄ with NH₄OH follow as.

H₂SO₄+NH₄OH -> (NH₄)₂SO₄ (salt)+H₂O

Initial -> H₂SO₄ is a strong acid so dissociation 100%

Hence, due to the small size of (H⁺) conductivity is maximum.

=> Upon addition of NH₄OH in H₂SO₄, salt [(NH₄)₂SO₄] will form and small H⁺ ion is replaced by larger Ion (NH₄⁺) so conductivity decreases.

After salt formation, if we add Excess of NH₄OH, there is no change in conductivity because NH₄OH is a week base. Hence, the dissociation is very less.

Therefore, the correct option is (b).

Question 15: The correct representation of the wavelength-intensity relationship of ideal blackbody radiation at two different temperatures T₁ and T₂ is:

Answer:(a)

From Wien’s displacement law

λmT =constant

λ proportional to 1/T

=> T increases λ decreases. [At higher temperature λ will be small]

Therefore, the correct option is (a).

Question 16: The pressure (P)-volume (V) isotherm of a van der Waals gas, at the temperature at which it undergoes gas to liquid transition, is correctly represented by

Answer:(b)

Vapour -> liquid (volume decreases )

At critical temperature ( l, in equilibrium)

Therefore, the correct option is (b).

Question 17: A buffer solution can be prepared by mixing equal volumes of

a. 0.2 M NH₄OH and 0.1 M HCl
b. 0.2 M NH₄OH and 0.2 M HCl
c. 0.2 M NaOH and 0.1 M CH₃COOH
d. 0.1 M NH₄OH and 0.2 M HCl

Answer:(a)

-> No buffer formed.

Therefore, the correct option is (a).

Question 18: The plot of total vapour pressure as a function of mole fraction of the components of an ideal solution formed by mixing liquids X and Y is:

Answer:(b)

Since we know that P proportional to x(mole fraction of solute)

For an ideal solution, according to Raoult’s law

Therefore, the correct option is (b).

Question 19: On complete hydrogenation, natural rubber produces

a. polyethene
b. ethylene-propylene copolymer
c. polyvinyl chloride
d. polypropylene

Answer:(b)

Natural rubber formed by the polymerization of isoprene and on hydrogenation provides copolymer of ethylene and propylene.

Copolymer -> Polymer which is synthesized from more than one species of monomers.

Ex. [A–B–A–B–A–B–A–B ——–]_n

Therefore, the correct option is (b).

Question 20: The average energy of each hydrogen bond in A-T pair is x kcal mol^–1 and that in G-C pair is y kcal mol^–1. Assuming that no other interaction exists between the nucleotides, the approximate energy required in kcal mol–1 to split the following double-stranded DNA into two single strands is

a. 10x + 9y
b. 5x + 3y
c. 15x + 6y
d. 5x + 4.5 y

Answer:(a)

Hint -> See the hydrogen bounded structure of A–T (Adenine –Thymine) and –

G–C(Guanine– Cytosine)

Solution

->No OH bond between A–T = 2

Average energy of each H bond in A–T pair = X Kcal mol^–1

Total no of A–T pair in structure = 5

Hence, total energy = 2x ×5 = 10 x

-> No of H bond between G–C = 3

Average energy of each H-bond in G–C pair = Y Kcal mol^–1

Total no of G–C pair in structure = 3

Hence total energy = 3y×3 = 9y

-> Now, the total energy required to split all double-stranded DNA into two single strands.

=10 x + 9y

Therefore, the correct option is (a).

Question 21: For the electrochemical cell shown below

Pt|H₂ (P = 1 atm)|H⁺(aq., x M)||Cu²⁺ (aq., 1.0 M)|Cu(s)

The potential is 0.49 V at 298 K. The pH of the solution is closest to

Gas constant, R is 8.31 J K^–1mol^–1

Faraday constant, F is 9.65 × 104 J V^–1mol^–1]

a. 1.2
b. 8.3
c. 2.5
d. 3.2

Answer:(c)

Since reduction potential of E⁰_H+/H2 = 0 V & E⁰_Cu+2/Cu = 0.34V

Hence, anode and cathode reaction is as follows

E⁰_cell = E⁰_cell – (0.0591/n) log Q {Q = [H⁺]²/[Cu²⁺], [H⁺] = x, [Cu²⁺] = 1M}

=> 0.49 = 0.34-(0.06/2) log [H⁺]²

=> 0.15 = +2×(0.06/2) [-log [H⁺]]/pH

=> pH = 0.15/0.06

{ RT/f = 0.0591, n = 2}

=> pH = 2.5

Therefore, the correct option is (c).

Question 22: Consider the following reversible first-order reaction of X at an initial concentration [X]₀. The values of the rate constants are k_f = 2s^–1 and k_b = 1s^–1

A plot of concentration of X and Y as function of time is

Answer:(b)

K_eq = K_f/K_b = α/x₀– α = 2

{ K_eq = K_f/K_b =2/1 = 2}

K_f -> rate constant for forward reaction

K_b -> rate constant for backward reaction

=> α = 2x₀ – 2α

=> 3α = 2x₀

=> α = 2/3 x₀

At equilibrium

Hence, here concentration ‘y’increases equal to concentration ‘x’decreases.

So, graph (B) best represents this.

Therefore, the correct option is (b).

Question 23: Nitroglycerine (MW = 227.1) detonates according to the following equation:

2C₃H₅(NO₃)₃(l) -> 3N₂(g) + ½O₂(g) + 6CO₂(g) + 5H₂O(g)

The standard molar enthalpies of formation of ΔH_f⁰ for all the compounds are given below:

ΔH_f⁰ [C₃H₅(NO₃)₃] = –364 kJ/mol

ΔH_f⁰ [CO₂(g)] = –393.5 kJ/mol

ΔH_f⁰ [H₂O(g)] = –241.8 kJ/mol

ΔH_f⁰ [N₂(g)] = 0 kJ/mol

ΔH_f⁰ [O₂(g)] = 0 kJ/mol

The enthalpy change when 10 g of nitroglycerine is determined is

a. –100.5 kJ
b. –62.5 kJ
c. –80.3 kJ
d. –74.9 kJ

Answer:(b)

Reaction

2C₃H₅(NO₃)₃(l) -> 3N₂(g) + ½O₂(g) + 6CO₂(g) + 5H₂O(g)

We know that ΔH_f⁰ = (ΔH_f⁰)_P– (ΔH_f⁰)_Reactant

= 0 + 0 + 6 × (–393.5) + 5 × (–241.8) – 2 × (–364)

ΔH_f⁰ = –2842 kJ

Therefore for one mole nitroglycerine ΔH_f⁰ = -2812/2= –1421 kJ mole^–1

Wt of nitroglycerine = 10 g

Mw of nitroglycerine = 227

Moles of nitroglycerine = 10/227 moles

Hence, (ΔH_f⁰ ) for 10/227 moles = 1421 × 10/227.1 = 62.5 kJ

Therefore, the correct option is (b).

Question 24: The heating of (NH₄)₂Cr₂O₇ produces another chromium compound along with N₂ gas. The change of the oxidation state of Cr in the reaction is

a. +6 to +2
b. +7 to +4
c. +8 to +4
d. +6 to +3

Answer:(d)

Heating reaction of (NH₄)₂Cr₂O₇ follow as

Hence, change in the oxidation state of Cr -> +6 to +3

Therefore, the correct option is (d).

Question 25: The complex having the highest spin-only magnetic moment is

a. [Fe(CN)₆]^3–
b. [Fe(H₂O)₆]²⁺
c. [MnF₆]^4–
d. [NiCl₄]^2–

Answer:(c)

(1)[Fe(CN)₆]^3–

Fe³⁺ -> d⁵

(d²sp³-> Hybridisation)

CN^– -> Strong field ligand, hence, pairing occur.

Unpaired electron = 1

MM = √(n(n+2)) = 1.732 BM

(2)[Fe(H₂O)₆]²⁺

Fe²⁺-> d⁶

(sp³d²-> Hybridisation)

H₂O->Weak field ligand, hence, the pairing does not occur.

Unpaired electron = 4

MM = 4.90 BM

(3)[MnF₆]^4–

Mn²⁺-> d⁵

F^–-> Weak field ligand

Hence, pairing not occur

Unpaired electron = 5

MM = 5.92 BM

(4)[NiCl₄]^2–

Cl^–-> Weak field ligand

Hence, paring not occur

Ni²⁺-> d⁸

MM = 2.83 BM

Therefore, the correct option is (c).

Question 26: Among Ce(4f¹ 5d¹ 6s²), Nd(4f⁴ 6s²), Eu(4f⁷ 6s²) and Dy(4f¹⁰ 6s²), the elements having highest and lowest 3^rd ionization energies, respectively, are

a. Nd and Ce
b. Eu and Ce
c. Eu and Dy
d. Dy and Nd

Answer:(b)

Ionisation energy -> Minimum amount of energy required to remove the most loosely bonded electron of an isolated neutral gaseous atoms or molecules.

Therefore, the correct option is (B).

Question 27: The major product of the following reaction sequence is:

Answer:(c)

Therefore, the correct option is (c).

Question 28: Among the following reactions, a mixture of diastereomers is produced from

Answer:(a)

Diastereomers-> Stereoisomers that are not mirror images, different compounds with different physical properties.

Therefore, the correct option is (a).

Question 29: Reaction of phenol with NaOH followed by heating with CO₂ under high pressure, and subsequent acidification gives compound X as the major product, which can be purified by steam distillation. When reacted with acetic anhydride in the presence of a trace amount of conc. H₂SO₄, compound X produces Y as the major product. Compound Y is

Answer:(a)

Therefore, the correct option is (a).

Question 30: A tetrapeptide is made of naturally occurring alanine, serine, glycine and valine. If the C-terminal amino acid is alanine and the N-terminal amino acid is chiral, the number of possible sequences of the tetrapeptide is

a. 12
b. 8
c. 6
d. 4

Answer:(d)

Possible sequence:

Ser – Val – Gly – Ala

Ser – Gly – Val – Ala

Val – Ser – Gly – Ala

Val – Gly – Ser – Ala

Total 4.

Therefore, the correct option is (d).