KVPY-SX 2016 Chemistry Question Paper with Solutions PDF Download

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KVPY SX 2016 Chemistry Paper with Solutions

KVPY SX 2016 – Chemistry

Question 1. The shape of SCl₄ is best described as a

a) Square
b) tetrahedron
c) square pyramid
d) see-saw

Answer: (d)

SCl₄

\(\begin{array}{l}S \rightarrow \frac{1s^{2}}{2} \frac{2s^{2}2p^{2}}{8} \frac{3s^{2}3p^{4}}{6}\end{array} \)

\(\begin{array}{l}Hybridization = \frac{Valence\, electron\, of\, central\, atom \pm charge + number\, of\, single\, bonded\, atom}{ 2}\end{array} \)

\(\begin{array}{l}\frac{6+0+4}{2} = 5 (sp^{3}d)\end{array} \)

5 = lp + BP

⇒ lp = 5 – BP = 5 – 4 = 1

Hence, the shape is K shape or see-saw shape.

KVPY SX 2016 Chemistry Question 1 Solution

Therefore, the correct option is (d).

Question 2. Among the following atomic orbital overlaps, the non-bonding overlap is

KVPY SX 2016 Chemistry Question 2

Answer: (a)

Overlapping is possible when axis of two orbitals is in same direction but in case of option(A), axis is not in the same direction.

Possible types of overlapping are as follows;

KVPY SX 2016 Chemistry Question 2 Solution

Therefore, the correct option is (a).

Question 3. Among the following complexes, the one that can exhibit optical activity is

a) [CoCl₆]^3–
b) [Co(en)Cl₄]^–
c) cis-[Co(en)₂Cl₂]⁺
d) trans-[Co(en)₂Cl₂]⁺

Answer: (c)

a) All of the ligands are same. Hence, it is not optically active.

b) [Co(en)Cl₄]^Θ

Due to the presence of plane of symmetry in elements, it is not optically active.

KVPY SX 2016 Chemistry Question 3 Solution

Question 4. The pK_a of oxoacids of chlorine in water follows the order

a)HClO< HClO₃< HClO₂< HClO₄
b) HClO₄< HClO₃< HClO₂<HClO
c) HClO₄< HClO₂< HClO₃<HClO
d) HClO₂<HClO< HClO₃< HClO₄

Answer: (b)

Oxidation state ↑, Acidic strength ↑

[H⁺] ↑, Acidic strength ↑, K_a↑, pK_a ↓

Order of acidic strength

HClO₄

(+7)

> HClO₃

(+5)

> HClO₂

(+3)

>HClO

(+1)

∴ Order of pK_a⇒ HClO> HClO₂> HClO₃< HClO₄

Therefore, the correct option is (b).

Question 5. The packing efficiency of the face centered cubic (fcc), body centered cubic (bcc) and simple/primitive cubic (pc) lattices follows the order

a)fcc> bcc > pc
b) bcc >fcc> pc
c) pc > bcc >fcc
d) bcc > pc >fcc

Answer: (a)

Unit cell	FCC	BCC	SC
Packing efficiency	74%	68%	52.4%

Hence, the order of packing efficiency is FCC > BCC > SC.

Therefore, the correct option is (a).

Question 6. The ratio of root mean square velocity of hydrogen at 50 K to that of nitrogen at 500 K is closest to

a) 1.18
b) 0.85
c) 0.59
d) 1.40

Answer: (a)

∵ We know

\(\begin{array}{l}v_{rms} = \sqrt{\frac{3RT}{M}}\end{array} \)

R → gas constant

M → Molecular weight of gas (g mol^–1)

T → Temperature (K)

⇒ M for H₂ = 2 g mol^–1

T = 50 K

⇒ M for N₂ = 28 g mol^–1

T = 500 K

\(\begin{array}{l}\frac{(v_{rms})_{H_{2}}}{(v_{rms})_{N_{2}}} = \sqrt{\frac{\frac{3\times R\times 50}{2}}{\frac{3\times R\times 50}{28}}} = 1.18\end{array} \)

Therefore, the correct option is (a).

Question 7. The molecule with the highest dipole moment among the following is

a) NH₃
b) NF₃
c) CO
d) HF

Answer: (d)

Molecule

NH₃

NF₃

CO

HF

Dipole moment (µ)

1.46

D

0.24

D

0.122

D

1.82

D

• ∆EN ↑, µ ↑ [CO, HF]

• µ ↑, EN of central atom ↑ (NH₃)

• µ ↓, EN of surrounding atoms ↑ (NF₃)

Therefore, the correct option is (d).

Question 8. The most stable Lewis acid-base adduct among the following is

a) H₂O → BCl₃
b) H₂S → BCl₃
c) H₃N → BCl₃
d) H₃P → BCl₃

Answer: (c)

Greater the tendency to donate lp, more stable will be the Lewis acid base adduct.

Here, B binds especially with nonmetals N and O [due to good overlapping], so option (a) and (c) have this advantage.

But electronegativity of N is less as compared to O.

(Electronegativity ↑ electron-donating tendency ↓)

Hence, here H₃N → BCl₃ is most stable acid base adduct.

Therefore, the correct option is (c).

Question 9. The reaction of D-glucose with ammoniacal AgNO₃ produces

KVPY SX 2016 Chemistry Question 9 Options

Answer: (c)

Question 10. The reagent (s) used for the conversion of benzene diazonium hydrogen sulfate to benzene is/are –

a) H₂O
b) H₃PO₂ + H₂O
c) H₂SO₄ + H₂O
d) CuCl/HCl

Answer: (b)

KVPY SX 2016 Chemistry Question 10 Solution

Question 11. The major product obtained in the reaction of toluene with 1-bromo-2-methyl propane in the presence of anhydrous AlCl₃ is

Answer: (c)

Reaction of Toluene with 1-Bromo-2-methyl propane follows as

KVPY SX 2016 Chemistry Question 11 Solution

Note: Here, both O and P products are possible but due to steric hindrance Ortho is minor and Para is the major product.

Therefore, the correct option is (c).

Question 12. The major product in the following reaction is

KVPY SX 2016 Chemistry Question 12 Options

Answer: (b)

Acetylation reaction of salicylic acid follows;

KVPY SX 2016 Chemistry Question 12 Solution

Therefore, the correct option is (b).

Question 13. The compounds containing sp hybridized carbon atom are

KVPY SX 2016 Chemistry Question 13 Options

a) (i) and (ii)
b) (iii) and (iv)
c) (ii) and (iii)
d) (i) and (iv)

Answer: (b)

Hence, here compound (iii) and (iv) contain ‘sp’ carbon atom.

Therefore, the correct option is (b).

Question 14. Upon heating with acidic KMnO₄ an organic compound produces hexan-1,6-dioic acid as the major product. The starting compound is

a)Benzene
b) cyclohexene
c) 1-methylcyclohexene
d) 2-methylcyclohexene

Answer: (b)

KVPY SX 2016 Chemistry Question 14 Solution

Therefore, the correct option is (b).

Question 15. It takes 1h for a first order reaction to go to 50% completion. The total time required for the same reaction to reach 87.5% completion will be

a) 1.75 h
b) 6.00 h
c) 3.50 h
d) 3.00 h

Answer: (d)

Given: t_½ = 1 hr

∵ We know kt = ln

\(\begin{array}{l}\left ( \frac{a}{a-x} \right )\end{array} \)

………(1) {a = initial concentration}

\(\begin{array}{l}k = \frac{1}{t} 1n \left ( \frac{a}{a-x} \right ) = \frac{1}{1}1n\left ( \frac{a}{a/2} \right )\end{array} \)

{at t_½ = a/2}

\(\begin{array}{l}k = \frac{1}{1}1n(2)\end{array} \)

Time required for 87.5% completion = ?

Now, from equation (1)

\(\begin{array}{l}t= \frac{1}{1n(2)} \left ( \frac{a}{a-0.875a} \right )\end{array} \)

\(\begin{array}{l}t= \frac{1}{1n(2)} \left ( \frac{a}{a/8} \right )\end{array} \)

{t = time, k = rate constant}

\(\begin{array}{l}t= \frac{1n8}{1n2} = \frac{2.079}{0.693} = 3hrs\end{array} \)

Therefore, the correct option is (d).

Question 16. A unit cell of calcium fluoride has four calcium ions. The number of fluoride ions in the unit cell is

a) 2
b) 4
c) 6
d) 8

Answer: (d)

Formula of calcium fluoride: CaF₂→ Ca²⁺ + 2F^–

Number of Ca²⁺ ion in unit cell = 4

So, to maintain electrical neutrality, the number of F^– present in unit cell

Let the oxidation number of F = x {Oxidation number of Ca = +2 and F = –1}

Then, 4(+2) + x(–1) = 0

x = 8

So, the number of F^– ion in unit cell = 8

Therefore, the correct option is (d).

Question 17. The equilibrium constant of a 2-electron redox reaction at 298 K is 3.8 × 10^–3. The cell potential E° (in V) and the free energy change ∆G° (in kJ mol^–1) for this equilibrium respectively, are

a) –0.071, –13.8
b) –0.071, 13.8
c) 0.71, –13.8
d) 0.071, –13.8

Answer: (b)

Given equilibrium constant (K) = 3.8 × 10^–3

Temperature = 298 K

Number of electron (n) = 2

∵ We know that ∆ G = –2.303RTlogK = –nF E°_cell

∆G = –2.303 × 8.314 × 298 × log(3.8 × 10^–3)

⇒∆G = 13809.387 J = 13.809 kJ

∆G = –nF E°_cell

13809.387 = –2 × 96500 × E°_cell

⇒E°_cell = –0.071 V

Therefore, the correct option is (b).

Question 18. The number of stereoisomer possible for the following compound is CH₃–CH=CH–CH(Br)–CH₂–CH₃

a) 2
b) 3
c) 4
d) 8

Answer: (c)

KVPY SX 2016 Chemistry Question 16 Solution

n = 2 (Number of steriogenic-centre)

Total stereoisomers = 2ⁿ (when symmetry absent)

= 2² = 4

Therefore, the correct option is (c).

Question 19. In the radioactive disintegration series

\(\begin{array}{l}_{232}^{90}\textrm{Th}\rightarrow _{208}^{82}\textrm{Pb}\end{array} \)

, involving α and β decay, the total number of α and β particles emitted are

a) 6 α and 6 β
b) 6 α and 4 β
c) 6 α and 5 β
d) 5 α and 6 β

Answer: (b)

Let, the numbers of α and β particles involved are n₁and n₂ respectively.

₉₀Th²³²→ ₈₂Pb²⁰⁸ + n₁α + n₂β^– {α → [₂He⁴], β^–→ _–1e⁰}

α-decay

_PX^A→ _P–2X^A–4 + ₂He⁴

β-decay

_PX^A→ _P+1X^A + _–1e⁰

Mass Balance

⇒ Mass of α particle = 4

232 = 208 + n₁ × 4

⇒ n₁ = 6

Balancing of proton and atomic number

90 = 82 + 2n₁ – n₂

n₂ = 4

Therefore, the correct option is (b).

Question 20. In the following compressibility factor (Z) vs pressure graph at 300 K, the compressibility of CH₄ at pressure < 200 bar deviates from ideal behavior because

KVPY SX 2016 Chemistry Question 20

a) The molar volume of CH₄ is less than its molar volume in the ideal state.
b) The molar volume of CH₄ is same as that in its ideal state.
c) Intermolecular interactions between CH₄ molecules decrease.
d) The molar volume of CH₄ is more than its molar volume in the ideal state.

Answer: (a)

Compressibility factor (z) =

\(\begin{array}{l}\frac{PV}{RT}=\frac{V_{real}}{V_{ideal}}\end{array} \)

…(1)

∵ Z < 1 at P < 200 bar

∴ from equation (1)

\(\begin{array}{l}\frac{V_{real}}{V_{ideal}}< 1\end{array} \)

⇒ V_r<V_i

Hence, the molar volume of CH₄ is less than its molar volume in the ideal state.

Therefore, the correct option is (a).

Question 21. X, Y and Z in the following reaction sequence are

KVPY SX 2016 Chemistry Question 21 Options

Answer: (d)

KVPY SX 2016 Chemistry Question 21 Solution

Therefore, the correct option is (d).

Question 22. The reagent required for the following two step transformation are

KVPY SX 2016 Chemistry Question 20

a) (i) HBr, benzoyl peroxide; (ii) CH₃CN
b) (i) HBr, (ii) NaCN
c) (i) Br₂, (ii) NaCN
d) (i) NaBr, (ii) NaCN

Answer: (b)

KVPY SX 2016 Chemistry Question 22 Solution

Therefore, the correct option is (b).

Question 23. In the reaction sequence

Answer: (a)

Therefore, the correct option is (a).

Question 24. In the following reactions

KVPY SX 2016 Chemistry Question 24 Options

Answer: (a)

KVPY SX 2016 Chemistry Question 24 Solution

Therefore, the correct option is (a).

Question 25. Copper (atomic mass = 63.5) crystallizers in a FCC lattice and has density 8.93 g.cm^–3. The radius of copper atom is closest to

a) 361.6 pm
b) 511.4 pm
c) 127.8 pm
d) 102.8 pm

Answer: (c)

Given Atomic mass of Cu = 63.5 g

Lattice FCC→ N = 4

d = 8.93 g/cm³

∵ We know that

\(\begin{array}{l}d = \frac{N\times M}{N_{A}\times a^{3}}\end{array} \)

\(\begin{array}{l}\Rightarrow a^{3} = \frac{N\times M}{N_{A}\times d}\end{array} \)

\(\begin{array}{l}\Rightarrow a^{3} = \frac{4\times 63.5}{6.023\times 10^{23}\times 8.93}\end{array} \)

a³ = 47.2 × 10^–24

⇒ a = (47.2 × 10^–24)^1/3

a = 3.61 × 10^–8 cm

a = 3.61 × 10^–10 m

a = 361 × 10^–12 m

a = 361 pm

∵ a = r (for FCC)

\(\begin{array}{l}r = \frac{a}{2\sqrt{2}} = \frac{361}{2\sqrt{2}} = 127.8 pm\end{array} \)

Therefore, the correct option is (c).

Question 26. Given the standard potentials E°(Cu²⁺/Cu) and E°(Cu⁺/Cu) as 0.340 V and 0.522 V respectively, the value of E°(Cu²⁺/Cu⁺) is

a) 0.364 V
b) 0.158 V
c) –0.182 V
d) –0.316 V

Answer: (b)

KVPY SX 2016 Chemistry Question 26 Solution

Question 27. For electroplating, 1.5 amp current is passed for 250 s through 250 mL of 0.15 M solution of MSO₄. Only 85% of the current was utilized for electrolysis. The molarity of MSO₄ solution after electrolysis is closest to. [Assume that the volume of the solution remained constant]

a) 0.14
b) 0.014
c) 0.07
d) 0.035

Answer: (a)

By Faraday’s law we know that

Equivalent =

\(\begin{array}{l}r = \frac{I\times t}{96500}\end{array} \)

{where, I = current and t = time}

Equivalent =

\(\begin{array}{l}\frac{wt.}{eq.wt.}\end{array} \)

Moles (n) =

\(\begin{array}{l}= \frac{I\times t}{96500\times v.f.}\end{array} \)

∵ According to question only 85% current utilized for electrolysis.

∴ Deposited moles (n) =

\(\begin{array}{l}\frac{1.5\times 250}{96500\times 2} \times \frac{85}{100} = 0.00135 moles\end{array} \)

\(\begin{array}{l}Molarity = \frac{moles\, of\, solute}{volume (L)}\end{array} \)

\(\begin{array}{l}0.15 = \frac{n\times 1000}{250}\end{array} \)

⇒ n = 0.375

Hence, the initial moles = 0.375 moles

Deposited moles = 0.00165 moles

Number of moles left in solution = 0.375 – 0.00165 = 0.03585 (moles)

Molarity =

\(\begin{array}{l}\frac{0.03585\times 1000}{250} = 0.143 M\end{array} \)

Therefore, correct option is (a).

Question 28. The hybridization of the central atom and the shape of [IO₂F₅]^2– ion respectively, are–
KVPY SX 2016 Chemistry Solved Questions

Answer: (d)

[IO₂F₅]^2–→ Central atom = I

\(\begin{array}{l}Hybridization = \frac{Valence\, electron\, of\, central\, atom \pm charge + number\, of\, single\, bonded\, atom}{ 2}\end{array} \)

Hybridization (n) =

\(\begin{array}{l}\frac{7+5+2}{2} = 7 (sp^{3}d^{3})\end{array} \)

Hence, the shape is pentagonal bipyramidal.

∵ Double bond causes more repulsion, so they would be on axial position (180° angle to each other).

Therefore, the correct option is (d).

Question 29. 2.33 g of compound X (empirical formula CoH₁₂N₄Cl₃) upon treatment with excess AgNO₃ solution produces 1.435 g of a white precipitate. The primary and secondary valences of cobalt in compound X, respectively, are; [Given: Atomic mass: Co = 59, Cl = 35.5, Ag = 108]

a) 3, 6
b) 3, 4
c) 2, 4
d) 4, 3

Answer: (a)

Weight of the compound (x) = 2.33 g

Molecular weight of the compound x, (CoH₁₂N₄Cl₃)

= 59 + 12 × 1 + 14 × 4 + 35.5 × 3

= 233.5 gmol^–1

Moles of compound (x) =

\(\begin{array}{l}\frac{2.33}{233.5} \approx 0.01\, moles\end{array} \)

Weight of AgCl = 1.435 g

Molecular weight of AgNO₃ = 143.5 g mol^–1

Moles of AgNO₃ =

\(\begin{array}{l}\frac{1.435}{143.5} = 0.01\, moles\end{array} \)

KVPY SX 2016 Chemistry Question 29 Solution

Oxidation state of Co in complex = x + 4 × 0 + 2 × (–1)

⇒ x = +3

Coordination number of Co = 6

Hence, the primary valence is 3 and secondary valence is 6.

Therefore, the correct option is (a).

Question 30. The specific conductance (k) of 0.02 M aqueous acetic acid solution at 298 K is 1.65 × 10^–4 S cm^–1. The degree of dissociation of acetic acid is [Given: equivalent conductance at infinite dilution of H⁺ = 349.1 S cm²mol^–1 and CH₃COO^– = 40.9 S cm²mol^–1]

a) 0.021
b) 0.21
c) 0.012
d) 0.12

Answer: (a)

Given:

K = 1.65 × 10^–4Scm^–1

Molarity of CH₃COOH = 0.02 M

\(\begin{array}{l}\lambda _{H}^{\alpha }\end{array} \)

= 349.1 Scm²mol^–1, CHCOO^ = 40.9 Scm²mol^–1

∵ We know

\(\begin{array}{l}\alpha = \frac{\lambda }{\lambda _{H}^{\alpha }}\end{array} \)

…(1)

\(\begin{array}{l}\lambda_{m} = \frac{1000\times k}{M} = \frac{1000\times 1.65\times 10^{-4}}{0.02}\end{array} \)

λ_m = 8.25 Scm²mol^–1

Therefore, the correct option is (a).

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