KVPY-SA 2019 Chemistry Paper with Solutions

Question 1: The hybridizations of N, C and O shown in the following compound respectively, are –

a. sp², sp, sp²
b. sp², sp2, sp²
c. sp², sp, sp
d. sp, sp, sp²

Answer: a

Hybridisation of

O = 1 B.P. (σ bond) + 2 l.p. = 3 ⇒sp²

C = 2 B.P (σ bond) = 2 ⇒ sp

N = 2 B.P. (σ bond) + 1l.P. = 3 ⇒ sp²

So, option (A) is correct

Question 2: The following compounds are

a. geometrical isomers
b. positional isomers
c. optical isomers HCl
d. functional group isomers

Answer: d

a. CH₃ – C ≡ C – CH₂ – CH₂ – CH₂ – CH₃ ⇒ C₇H₁₂ {– yne compound}

b. CH₂ = CH – CH₂ – CH = CH – CH₂ – CH₃⇒ C₇H₁₂ {– diene compound}

Both –yne and –diene compounds are having same molecular formula but different functional group. So, they are functional group isomers of each other.

Question 3: The major product of the following reaction is –

Answer: a

Question 4: IUPAC name of the following compound is –

a. 1-hydroxycyclohex-4-en-3-one
b. 1-hydroxycyclohex-3-en-5-one
c. 3-hydroxycyclohex-5-en-1-one
d. 5-hydroxycyclohex-2-en-1-one

Answer: d

OH → Hydroxy (prefix)

= (double bond) →en (primary suffix)

Ketone →one (secondary suffix)

Root word →cyclo hex

So, IUPAC name: 5–hydroxy cyclohex–2–en–1–one

Question 5: In water-gas shift reaction, hydrogen gas is produced from the reaction of steam with –

a. Methane
b. Coke
c. Carbon Monoxide
d. Carbon Dioxide

Answer: c

The water-gas shift reaction is an equilibrium process. In which carbon monoxide and steam converting into hydrogen and carbon dioxide.

CO + H₂O → CO₂ + H₂

(steam)

Question 6: Treatment with lime can remove the hardness of water caused by –

a. CaCl₂
b. CaSO₄
c. Ca(HCO₃)₂
d. CaCO₃

Answer: c

Lime (Ca(OH)₂) is used to remove the temporary hardness of water causes by salts of Ca and Mg.

Question 7: The most polarizable ion among the following is

a. F^–
b. I^–
c. Na⁺
d. Cl^–

Answer: b

According to Fajan’s rule

Polarization ∝ size of anion ∝ 1/size of cation

So, out of F^–, I^–, Na⁺ ,Cl^–. I^– anion has the largest in size. So, it’s most polarizable.

Question 8: For a multi-electron atom, the highest energy level among the following is –

a. n =5, l = 0, m =0, s = +1/2
b. n =4, l = 2, m =0, s = +1/2
c. n =4, l = 1, m =0, s = +1/2
d. n =5, l = 1, m =0, s = +1/2

Answer: d

Option

(A) n = 5, l= 0, m = 0, s = + ½ ⇒ 5s orbital

(B) n = 4, l = 2, m = 0, s = + ½ ⇒ 4d orbital

(C) n = 4, l = 1, m = 0, s = + ½ ⇒ 4p orbital

(D) n = 5, l = 1, m = 0, s = + ½ ⇒ 5p orbital

So according to (n + l) rule, increasing order of energy level: higher the value of (n+l), higher will be the energy. If (n+l) value is same, then check the value of n. Higher the value of n, higher will be the energy.

4p< 5s < 4d < 5p

So, 5p is highest energy level.

Question 9: The oxide which is neither acidic nor basic is –

a. As₂O₃
b. Sb₄O₁₀
c. N₂O
d. Na₂O

Answer: c

Some non-metal oxides do not show any acidic or basic nature. They are neutral oxide.

Example: H₂O, CO, NO, N₂O etc.

As₂O₃& Sb₄O₁₀ are amphoteric and Na₂O is a basic oxide.

Question 10: The element whose salts cannot be detected by the flame test is –

a. Mg
b. Na
c. Cu
d. Sr

Answer: a

Flame test:

(A) Mg → None of the salts gives [but for burning Mg metal (intense white)]

(B) Na → yellow.

(C)

(D) Sr → Crimson red to scarlet

Question 11: The plot of concentration of a reactant vs. time for a chemical reaction is shown below:

The order of this reaction with respect to the reactant is

a. 0
b. 1
c. 2
d. not possible to determine from this plot

Answer: a

Let, the reaction be zero order

So, according to law of mass action

R ∝ [A]⁰ at time =t

r=k [Since, [A]⁰=1] {k = rate constant}

We know, rate= Change in concentration/change in time

Therefore, -dA/dt =k ⇒ -dA = kdt

On integrating both the side and applying limits

A_t-A₀ =-k_t

A_t = -kt+A₀

On comparing with straight line equation

y=mx+c

A_t = -kt+A₀

So, plot of A_t vs t

So order of reaction with respect to reactant is zero order.

Question 12: During the free expansion of an ideal gas in an isolated chamber,

a. internal energy remains constant
b. internal energy decreases
c. work done on the system is negative
d. temperature increases

Answer: a

Free expansion of an ideal gas in an isolated chamber is irreversible adiabatic free expansion process.

⇒ In an adiabatic process → Heat is constant

So, q = 0 (isolated chamber)

⇒ In irreversible process work done → w = – P^ext. ΔV

For free expansion →P^ext. = 0

So, w = 0

We know, Internal energy (ΔU) = q + w = 0

Hence, internal energy is constant.

Question 13: The number of moles of water present in a spherical water droplet of radius 1.0 cm is [Given the density of water in the droplet = 1.0 g cm^–3]

a. π/18
b. 2π/27
c. 24 π
d. 2π/9

Answer: b

Given:-

Radius of water droplet = 1.0 cm

Density of water droplet = 1.0 g/cm³

We know, density = mass/ volume ⇒ mass = volume

No. of moles = mass/molar mass {mass = volume}

So, moles = volume/molar mass {volume of sphere = (4/3) πr³ }

Mole =

No. of moles = 2π/27

Question 14: Among the following, the correct statement about cathode ray discharge tube is

a. the electrical discharge can only be observed at high pressure and at low voltages
b. in the absence of external electrical or magnetic field, cathode rays travel in straight lines
c. the characteristics of cathode rays depend upon the material of electrodes
d.the characteristics of cathode rays depend upon the gas present in the cathode ray tube

Answer: b

Cathode rays or stream of electrons move in a straight line without any deflection in the absence of electric and magnetic field. But in the presence of the electric field, the cathode rays are deflected towards the positive pole.

Question 15: For a spontaneous process

a. enthalpy change of the system must be negative
b. entropy change of the system must be positive
c. entropy change of the surrounding must be positive
d. entropy change of the system and surrounding (universe) must be positive

Answer: d

For a spontaneous process ΔG < 0

So, we know, ΔG = ΔH – T ΔS

So, for spontaneous Reaction, ΔH = –ve, ΔS = +ve

Therefore, entropy change of the system and surrounding (universe) must be positive.

Question 16: PbO₂ is obtained from

a. the reaction of PbO with HCl
b. thermal decomposition of Pb(NO₃)₂ at 200 °C
c. the reaction of Pb3O4 with HNO3
d. the reaction of Pb with air at room temperature

Answer: c

Question 17: For one mole of a van der Waals gas, the compressibility factor

[Given: “a”, “b” are the standard parameter for van der Waals gas]

a.”b” increases and “a” decreases at a constant temperature
b. “b” decreases and “a” increases at a constant temperature
c. temperature increases at constant “a” and “b” values
d. “b” increases at constant “a” and temperature

Answer: b

a → intermolecular force of attraction

b → volume occupied by molecules.

We know, compressibility factor Z = V_real/V_ideal = PV/RT

To decreases Z value, V_real (numerator) should decrease which decreases when ‘a’ increases & ‘b’ decreases at constant temperature

Question 18: The correct statements among the following

(i) E_2s(H) > E_2s (Li) > E_2s (Na) > E_2s (K)

(ii)The maximum number of electrons in the shell with principal quantum number n is equal to 2n²

(iii)Extra stability of half-filled subshell is due to smaller exchange energy.

(iv)Only two electrons, irrespective of their spin, may exist in the same orbital.

a. i and ii
b. ii and iii
c. iii and iv
d. i and iv

Answer: a

(i) We know that, E = – 13.6 × (z²/n²)ev

So, the higher the value of Z (atomic number) lower is the value of E_2s

(ii) Maximum number of electron in the shell = 2n²

(iii) The extra stability of half-filled subshell is due to higher exchange energy.

(iv) Only two electrons with opposite spin can exist in the same orbital. According to Pauli’s exclusion principle. So, statement (i) and (ii) are correct.

Question 19: An organic compound contains 46.78% of a halogen X. When 2.00 g of this compound is heated with fuming HNO₃ in the presence of AgNO₃, 2.21 g AgX was formed. The halogen X is

[Given: atomic weight of Ag = 108, F = 19, Cl = 35.5, Br = 80, I = 127]

a. F
b. Cl
c. Br
d. I

Answer: c

Given: % if X in organic compound = 46.78%

Weight of organic compound = 2.00g

Weight of AgX formed = 2.21 g

So, weight of X in organic compound =2 x (46.78/100) = 0.94 gm

Let, the molecular weight of X = xg mol^–1

So, by the law of equivalence

No. of eq. of X = no. of eq. of AgX

Mole × valency factor = mole × valency factor

(101.52 + 0.94x) =2.21x

101.52 = 1.27x

x = 79.9 ≈ 80

So, halogen X is Bromine (Br)

Question 20: An organic compound X with molecular formula C₆H₁₀, when treated with HBr, forms a gem dibromide. The compound X upon warming with HgSO₄ and dil. H₂SO₄, produces a ketone which given a positive iodoform test. The compound X is

Answer: d

Given: molecular formula of organic compound (x) = C₆H₁₀ i.e. C_nH_2n–2

So, terminal alkyne reacts with HgSO₄ and dil. H₂SO₄

Produce ketone with gives Iodoform test.

Therefore, Iodoform test is shown by those groups which contain

group in their structure. So, option (d) satisfies all the condition.