The Department of Technology & Science, Government of India conducts the KVPY exam to inspire the students to take up basic science courses and research careers in science by giving scholarships. KVPY-SX 2019 Chemistry paper solutions are given on this page. BYJU’S provides step by step solutions so that students can easily understand them. Students are advised to revise these solutions so that they will get an idea of the exam pattern and difficulty level.

**Question 1:** The major product of the following reaction

are:

Solution:

**Answer: (b)**Option (b) is correct.

**Question 2:** Among the following,

The compounds which can undergo an S_{N}1 reaction in an aqueous solution, are

- (a) I and IV only
- (b) II and IV only
- (c) II and III only
- (d) II, III and IV only

Solution:

**Answer: (c)**So, option (c) is correct.

**Question 3:** The major product of the following reaction

Solution:

**Answer: (a)**[DIBAL–H] -> (i–Bu)

_{2}Al–H -> Di isobutyl aluminium hydride.

**Question 4:** Permanent hardness of water can be removed by

- (a) heating
- (b) treating with sodium acetate (CH
_{3}CO_{2}Na) - (c) treating with Ca(HCO
_{3})_{2} - (d) treatment with sodium hexametaphosphate (Na
_{6}P_{6}O_{18})

Solution:

**Answer: (d)**The permanent hardness of water can be removed by using calgon process. The hardness of water causes by salts of Ca and Mg.

**Question 5:** Alkali metals (M) dissolve in liquid NH_{3} to give

- (a) MNH
_{2} - (b) MH
- (c) [M(NH
_{3})x]^{+}+ [e(NH_{3})y]^{– } - (d) M
_{3}N

Solution:

**Answer: (c)**All alkali metals like lithium, sodium, potassium etc. dissolves in liquid ammonia to give deep blue coloured solution. The blue colour is due to presence of solvated electrons.

**Question 6:** The absolute configurations of the following compounds

Respectively are

- (a) R and R
- (b) S and S
- (c) R and S
- (d) S and R

Solution:

**Answer: (d)**Draw the Fischer project i.e. 2–D representation and then assign priorities.

According to Cahn Ingold Prelog (CIP) priority rules –

So, it should be ‘R’ but here, lower priority group is on horizontal line. Therefore, Configuration will be reversed. i.e. correct configuration ‘S’.

It should be ‘S’ lower priority group is on

horizontal line i.e. ‘R’.

So, option (d), S, R is correct.

**Question 7:** The diamagnetic species among the following is

- (a) O
_{2}^{+} - (b) O
_{2}^{-} - (c) O
_{2 } - (d) O
_{2}^{2-}

Solution:

**Answer: (d)**According to MOT –

Number of unpaired e

^{–s}= 0 => diamagneticOption (d) is correct.

**Question 8:** Among the following transformations, the hybridization of the central atom remains unchanged in

- (a) CO
_{2}-> HCOOH - (b) BF
_{3}-> BF_{4}^{-} - (c) NH
_{3}-> NH_{4}^{+} - (d) PCl
_{3}-> PCl_{5}

Solution:

**Answer: (c)**

**Question 9:** For an octahedral complex MX_{4}Y_{2} (M = a transition metal, X and Y are monodentate achiral ligands), the correct statement, among the following, is

- (a) MX
_{4}Y_{2}has 2 geometrical isomers one of which is chiral - (b) MX
_{4}Y_{2}has 2 geometrical isomers both of which are achiral - (c) MX
_{4}Y_{2}has 4 geometrical isomers all of which are achiral - (d) MX
_{4}Y_{2}has 4 geometrical isomers two of which are chiral

Solution:

**Answer: (b)**MX

_{4}Y_{2}-> Octahedral complexNote -> If there is plane of symmetry (POS) or centre of symmetry (COS) in a complex. Then it will be achiral and optically inactive. Geometrical isomers exist is cis and trans forms.

i.e. two geometrical isomers both of which are achiral.

So, option (b) is correct.

**Question 10:** The values of the Henry's law constant of Ar, CO_{2}, CH_{4} and O_{2} in water at 25^{0}C are 40.30, 1.67, 0.41 and 34.86 kbar, respectively. The order of their solubility in water at the same temperature and pressure is

- (a) Ar > O
_{2}> CO_{2}> CH_{4} - (b) CH
_{4}> CO_{2}> Ar > O_{2} - (c) CH
_{4}> CO_{2}> O_{2}> Ar - (d) Ar > CH
_{4}> O_{2}> CO_{2}

Solution:

**Answer: (c)**According to Henry’s law, the partial pressure of a gas in vapour phase (P) is proportional to the mole fraction of the gas (X) in the solution.

P proportional to X

P = K

_{H}XK

_{H}= Henry’s constantP = Partial pressure of gas above liquid surface

so, according to Henry’s law solubility of gas in liquid proportional to partial pressure of gas above liquid surface.

**Question 11:** Thermal decomposition of N_{2}O_{5} occurs as per the equation below

2N_{2}O_{5}-> 4NO_{2} + O_{2}

The correct statement is

- (a) O
_{2}production rate is four times the NO_{2}production rate - (b) O
_{2}production rate is the same as the rate of disappearance of N_{2}O_{5} - (c) rate of disappearance of N
_{2}O_{5}is one-fourth of NO_{2}production rate - (d) rate of disappearance of N
_{2}O_{5}is twice the O_{2}production rate.

Solution:

**Answer: (d)**Rate of disappearance of N

_{2}O_{5}= twice the O_{2}production rateSo, option (d) is correct.

**Question 12:** For a 1^{st} order chemical reaction.

- (a) the product formation rate is independent of reactant concentration
- (b) the time taken for the completion of half of the reaction (t
_{1/2}) is 69.3% of the rate constant (k) - (c) the dimension of Arrhenius pre-exponential factor is reciprocal of time
- (d) the concentration vs time plot for the reactant should be linear with a negative slope.

Solution:

**Answer: (c)**For I

^{st}order reaction –t = (2.303/K) log

_{10}[A_{0}/A_{t}]And t

_{1/2}= 0.693/KArrhenius equation: K = Ae

^{-Ea/RT}[A = pre-exponential factor]

In a first order reaction, unit of rate constant is s

^{–1}. So pre-exponential factor are also reciprocal seconds. Because exponential factor depends on frequency of collision. Its related to collision theory and transition state theory.

**Question 13: **The boiling point of 0.001 M aqueous solutions of NaCl, Na_{2}SO_{4}, K_{3}PO_{4} and CH_{3}COOH should follows the order

- (a) CH
_{3}COOH < NaCl < Na_{2}SO_{4}< K_{3}PO_{4} - (b) NaCl < Na
_{2}SO_{4}< K_{3}PO_{4}< CH_{3}COOH - (c) CH
_{3}COOH < K_{3}PO_{4}< Na_{2}SO_{4}< NaCl - (d) CH
_{3}COOH < K_{3}PO_{4}< NaCl < Na_{2}SO_{4}

Solution:

**Answer: (a)**We know that, ∆T

_{b}= i ×K_{b}× mmolarity or molality = 0.001 m {at very diluted solution}

So, ∆T

_{b}proportional to i ;{molality is constant for all}i increases as Boiling point increases

For NaCl => strong electrolyte i = 2 {Na

^{+}+ Cl^{–}}For Na

_{2}SO_{4}=> strong electrolyte i = 3 {2Na^{+}+ SO_{4}^{2–}}For K

_{3}PO_{4}=> strong electrolyte i = 4 {3K^{+}+PO_{4}^{3-}}For CH

_{3}COOH => weak acid i < 2 (calculated from i= 1+ (n-1)α , where α<1 )So, order of Boiling Point: K

_{3}PO_{4}> Na_{2}SO_{4}> NaCl > CH_{3}COOHSo, option (a) is correct.

**Question 14:** An allotrope of carbon which exhibits only two types of C–C bond distance of 143.5 pm and 138.3 pm, is

- (a) charcoal
- (b) graphite
- (c) diamond
- (d) fullerene

Solution:

**Answer: (d)**Fullerene a soccer ball shaped molecule has 60 vertices with a carbon atom at each vertex. It contains both single and double bond with C–C at a distance of 143.5 pm and 138.3 pm.

Where as in diamond C–C bond length 154 pm and in graphite 140 pm.

**Question 15: **Nylon-2-nylon-6 is a co-polymer of 6-aminohexanoic acid and

- (a) glycine
- (b) valine
- (c) alanine
- (d) leucine

Solution:

**Answer: (a)**Nylon–2–Nylon–6 is a copolymer of glycine and amino caproic acid (6-amino hexanoic acid)

**Question 16:** A solid is hard and brittle. It is an insulator solid state but conducts electricity in molten state. The solid is a

- (a) molecular solid
- (b) ionic solid
- (c) metallic solid
- (d) covalent solid

Solution:

**Answer: (b)**Ionic solid is hard and brittle eg. NaCl

It is an insulator in solid state but in molten state it conducts electricity as it has free ions in molten state.

**Question 17:** The curve that best describes the adsorption of a gas (x g) on 1.0 g of a solid substrate as a function of pressure (p) at a fixed temperature is:

- (a) 1
- (b) 2
- (c) 3
- (d) 4

Solution:

**Answer: (b)**Given: mass = 1g

According to freundlich adsorption isotherm

x/m = KP

^{1/n}(n>1)x = KP

^{1/n}x/K = P

^{1/n}(x/K)

^{n}= POn comparing parabola equation y

^{n}= x. If n = 2So, graph (b) is correct.

**Question 18:** The octahedral complex CoSO_{4}Cl.5NH_{3} exists in two isomeric forms X and Y. Isomer X reacts with AgNO_{3} to give a white precipitate, but does not react with BaCl_{2}, Isomer Y gives white precipitate with BaCl_{2} but does react with AgNO_{3}.

Isomers X and Y are

- (a) Ionization isomers
- (b) Linkage isomers
- (c) Coordination isomers
- (d) Solvate isomers

Solution:

**Answer: (a)**CoSO

_{4}Cl.5NH_{3}Isomers of CoSO

_{4}Cl.5NH_{3}are (I) [Co(NH_{3})5SO_{4}]Cl gives white ppt with AgNO_{3}but not with BaCl_{2}and (II) [Co(NH_{3})5Cl]SO_{4}gives white ppt with BaCl_{2}but not with AgSO_{4}.(I) [Co(NH

_{3})_{5}SO_{4}]Cl + AgNO_{3 }-> AgCl + [Co(NH_{3})_{5}SO_{4}]NO_{3}[white ppt]

(II) [Co(NH

_{3})_{5}Cl]SO_{4}+ BaCl_{2 }-> BaSO_{4}+ [Co(NH_{3})_{5}Cl]Cl_{2}[white ppt]

So, isomer (I) and (II) are ionization isomers.

**Question 19:** The correct order of basicity of the following amines is:

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

Solution:

**Answer: (b)**So, order of basicity - I > III > II > IV

**Question 20:** Electrolysis of a concentrated aqueous solution of NaCl results in

- (a) increase in pH of the solution
- (b) decrease in pH of the solution
- (c) O
_{2}liberation at the cathode - (d) H
_{2}liberation at the anode

Solution:

**Answer: (a)**Electrolysis of aqueous NaCl leads to the formation of NaOH, which is a base. So, its pH will be increases.

H

_{2}liberation will be at cathode due to (H^{+}-> H_{2}) gain of electron at cathode.So, option (a) is correct.

**Question 21:** The product of which of the following reaction forms a reddish brown precipitate when subjected to Fehling's test?

Solution:

**Answer: (d)**Fehling reagent react with aldehyde which has α–H such aldehyde form an enolate and thus give a positive Fehling test and a reddish brown precipitate is obtained.

Reaction: R–CHO + 2Cu

^{2+}+ 5OH^{–}->RCOO^{–}+ Cu_{2}O + 3H_{2}O (reddish brown)Benzaldehyde gives tollen’s as well as schiff’s test but not gives fehling's solution test because benzaldehyde does not contain Alpha hydrogen and cannot form intermediate enolate to proceed for further and hence it does not react with fehling's solution test however aliphatic aldehydes gives fehling's solution test.

So, option (d) product gives Fehling test.

**Question 22:** The major products X, Y and Z in the following sequence of transformations are

Solution:

**Answer: (b)**So, option (b) is correct.

**Question 23:** In the following reaction, P gives two products Q and R, each in 40% yield.

If the reaction is carried out with 420 mg of P, the reaction yields 108.8 mg of Q.

The amount of R produced in the reaction is closest to

- (a) 97.6 mg
- (b) 108.8 mg
- (c) 84.8 mg
- (d) 121.6 mg

Solution:

**Answer: (c)**Mass of R = mole × m.wt. = (4/5 )m mole × 106 g mol

^{–1}= 84.4 mgSo, option (c) is correct.

**Question 24:** Solubility products of CuI and Ag_{2}CrO_{4} have almost the same value (~4 × 10^{–12}).

The ratio of solubilities of the two salts (CuI: Ag_{2}CrO_{4}) is closest to

- (a) 0.01
- (b) 0.02
- (c) 0.03
- (d) 0.10

Solution:

**Answer: (b)**S

_{1}/S_{2}= 0.02So, option (b) is correct.

**Question 25: **Given that the molar combustion enthalpy of benzene, cyclohexane, and hydrogen are x, y, and z, respectively, the molar enthalpy of hydrogenation of benzene to cyclohexane is

- (a) x – y + z
- (b) x – y + 3z
- (c) y – x + z
- (d) y – x + 3z

Solution:

**Answer: (b)**So, equation (1) + equation (3) × 3 – equation (2); to get equation (4)

C

_{6}H_{6}+ 3H_{2}-> C_{6}H_{12}x + 3z – y = x – y + 3z

So, molar enthalpy of hydrogenation of benzene to cyclo hexane = x – y + 3z

Option (b) is correct.

**Question 26:** Among the following, the pair of paramagnetic complexes is

- (a) K
_{3}[Fe(CN)_{6}] and K_{3}[CoF_{6}] - (b) K
_{3}[Fe(CN)_{6}] and [Co(NH_{3})_{6}]Cl_{3} - (c) K
_{4}[Fe(CN)_{6}] and K_{3}[CoF_{6}] - (d) K
_{4}[Fe(CN)_{6}] and [Co(NH_{3})_{6}]Cl_{3}

Solution:

**Answer: (a)**Number of unpaired e

^{–s}= 1 => paramagneticNumber of unpaired e

^{–s}= 4 => paramagneticSo, option (a) is correct.

**Question 27:** The major products X and Y in the following sequence of transformations

are:

Solution:

**Answer: (d)**So, option (d) is correct.

**Question 28:** 3.0 g of oxalic acid [(CO_{2}H)_{2}.2H_{2}O] is dissolved in a solvent to prepare a 250 mL solutions. The density of the solution is 1.9 g/mL. The molality and normality of the solution, respectively, are closest to

- (a) 0.10 and 0.38
- (b) 0.10 and 0.19
- (c) 0.05 and 0.19
- (d) 0.05 and 0.09

Solution:

**Answer: (c)**Given: Mass of oxalic acid = 3.0 g

Mol. wt. of oxalic acid (COOH)

_{2}.2H_{2}O = 126 g mol^{–1}Density = 1.9 g/mL and volume of solution = 250 mL

d = m/v

m = d × v

Molality = mass of solute/m.mass of solute× Mass of solvent(kg)

Molality = m/m.m×(d×v) = 3/(126×10.9×250×10

^{-3})Molality = 0.05

We know normality = molality ×n

_{factor}Molarity = mole/volume(L)

n

_{factor}of oxalic acid = 2 (because it lose 2H^{+})N = (n/VL) ×n

_{factor}Normality = ((3/126)/(250 ×10

^{-3}))2= 0.19

So, option (c) is correct.

**Question 29:** In a titration experiment, 10 mL of a FeCl_{2 }solution consumed 25 mL of a standard K_{2}Cr_{2}O_{7} solution to reach the equivalent point. The standard K_{2}Cr_{2}O_{7 }solution is prepared by dissolving 1.225 g of K_{2}Cr_{2}O_{7} in 250 mL water. The concentration of the FeCl_{2} solution is closest to [Given: molecular weight of K_{2}Cr_{2}O_{7} = 294 g mol^{–1}]

- (a) 0.25 N
- (b) 0.50 N
- (c) 0.10 N
- (d) 0.04 N

Solution:

**Answer: (a)**So, by law of equivalence,

Number of eq. of FeCl

_{2}= number of eq. of K_{2}Cr_{2}O_{7}N

_{1}V_{1}= N_{2}V_{2}Given: volume of FeCl

_{2}= 10 mLVolume of K

_{2}Cr_{2}O_{7}= 25 mLWt. of K

_{2}Cr_{2}O_{7}= 1.225 gVolume of solution = 250 mL

m. wt. of K

_{2}Cr_{2}O_{7}= 294 g mol^{–1}N

_{1}×10 = M×nf ×25N

_{1}×10 = ((1.225/294)/(250×10^{-3}))25×6Molarity = mole/volume(L)

N

_{1}= 1.225×25×6/(294×250×10^{-3}×10)= 0.25N

**Question 30:** Atoms of an element Z form hexagonal closed pack (hcp) lattice and atoms of element X occupy all the tetrahedral voids. The formula of the compound is

- (a) XZ
- (b) XZ
_{2} - (c) X
_{2}Z - (d) X
_{4}Z_{3}

Solution:

**Answer: (c)**Let, number of atom in HCP = a

We know, the number of tetrahedral voids will be twice as that of number of atoms present in a unit cell.

So, no. of tetrahedral void = 2a

Z = a, X = 2a

Z : X

a : 2a

1 : 1

So, formula of compound = X

_{2}ZTherefore, option (c) is correct.

KVPY-SX 2019 Chemistry Paper with Solutions