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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced

19 Questions MCQ Test Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper | Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced

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Attempt Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced | 19 questions in 40 minutes | Mock test for JEE preparation | Free important questions MCQ to study Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper for JEE Exam | Download free PDF with solutions

*Multiple options can be correct

QUESTION: 1

In the electrolysis of alumina, cryolite is added to :

A.
lower the melting point of alumina
B.
increase the electrical conductivity
C.
minimise the anode effect
D.
remove impurities from alumina

Solution:

Because of high melting point (2050ºC), pure alumina cannot be electrolysed. Hence a mixture of alumina, cryolite (m.p. 1000ºC) and calcium fluoride (to lower the temperature of the melt) is electrolysed at about 900ºC.
NOTE : The function of cryolite is to increase the electrical conductivity of the electrolyte, and to lower the temperature of the melt.

*Multiple options can be correct

QUESTION: 2

Nitrogen(I) oxide is produced by :

A.
thermal decomposition of ammonium nitrate
B.
disproportionation of N₂O₄
C.
thermal decomposition of ammonium nitrite
D.
interaction of hydroxylamine and nitrous acid.

Solution:

NH₄NO₃ ——→ N₂O + 2H₂O
NH₂OH.HCl + NaNO₂ ——→ N₂O + NaCl + 2H₂O

*Multiple options can be correct

QUESTION: 3

The compounds used as refrigerant are

A.
NH₃
B.
CCl₄
C.
CF₄
D.
CF₂Cl₂
E.
CH₂F₂

Solution:

NH₃ and CF₂Cl₂ (freon-12) are used as refrigerants.

*Multiple options can be correct

QUESTION: 4

The major role of fluorspar (CaF₂), which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na₃AlF₆), is

A.
as a catalyst
B.
to make the fused mixture very conducting
C.
to lower the temperature of the melt
D.
to decrease the rate of oxidation of carbon at the anode.

Solution:

To make the fused mixture very conducting and to reduce the temperature of the melt.

*Multiple options can be correct

QUESTION: 5

The material used in the solar cells contains

A.
Cs
B.
Si
C.
Sn
D.
Ti

Solution:

Silicon is used in solar cells.

*Multiple options can be correct

QUESTION: 6

Sodium nitrate decomposes above 800°C to give

A.
N₂
B.
O₂
C.
NO₂
D.
Na₂O

Solution:

Sodium nitrate on decomposition upto 500^oC gives NaNO₂ and oxygen.

While at higher temperature (i.e. above to 800°C), NaNO₂ further decomposes into Na₂O, N₂ and O₂.

*Multiple options can be correct

QUESTION: 7

White phosphorus (P₄) has

A.
six P-P single bonds
B.
four P-P single bonds
C.
four lone pairs of electrons
D.
PPP angle of 60°

Solution:

The four atoms in a P₄ molecule are situated at the corners of a tetrahedron. There are six P - P single bonds with PPP bond angle equal to 60^o. Each phosphorus has a lone pair of electrons.

*Multiple options can be correct

QUESTION: 8

Ammonia, on reaction with hypochlorite anion, can form

A.
NO
B.
NH₄Cl
C.
N₂H₄
D.
HNO₂

Solution:

2NH₃ + OCl–→ NH₂.NH₂ + H₂O + Cl^–

*Multiple options can be correct

QUESTION: 9

A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are)

A.
NH₄NO₃
B.
NH₄NO₂
C.
NH₄Cl
D.
(NH₄)₂SO₄

Solution:

When ammonium salt NH₄NO₃ or NH₄NO₂ (ammonium salts are colourless) is boiled with excess of NaOH, ammonia (NH₃) gas is evolved as follows:

NH₄NO₂ + NaOH → NaNO₂ + NH₃ + H₂O
NH₄NO₃ + NaOH → NaNO₃ + NH₃ + H₂O

The NH₃ gas evolved is non-flammable gas.
When the gas evolution ceases we are left with NaNO₂ or NaNO₃ in solution.
These salts get reduced when Zn is added to this solution containing salt (NaNO₂ or NaNO₃) and excess NaOH and NH₃ gas is evolve.

Thus the colourless salt [H] is either NH₄NO₂ or NH₄NO₃.
Thus (a) and (b) are correct answers. [NOTE : NaCl formed has no reaction with NaOH]

*Multiple options can be correct

QUESTION: 10

The nitrogen oxide(s) that contain(s) N-N bond(s) is(are)

A.
N₂O
B.
N₂O₃
C.
N₂O₄
D.
N₂O₅

Solution:

*Multiple options can be correct

QUESTION: 11

Which of the following halides react(s) with AgNO₃(aq) to give a precipitate that dissolves in Na₂S₂O₃(aq)?

A.
HCl
B.
HF
C.
HBr
D.
HI

Solution:

*Multiple options can be correct

QUESTION: 12

With respect to graphite and diamond, which of the statement(s) given below is (are) correct ?

A.
Graphite is harder than diamond.
B.
Graphite has higher electrical conductivity than diamond.
C.
Graphite has higher thermal conductivity than diamond.
D.
Graphite has higher C–C bond order than diamond.

Solution:

(a) Diamond is harder than graphite.
(b) Graphite is good conductor of electricity as each carbon is attached to three C-atoms leaving one valency free, which is responsible for electrical conduction, while in diamond, all the four valencies of carbon are satisfied, hence insulator.
(c) Diamond is better thermal conductor than graphite.
Whereas electrical conduction is due to availability of free electrons; thermal conduction is due to transfer of thermal vibrations from atom to atom. A compact and precisely aligned crystal like diamond thus facilitates fast movement of heat.
(d) In graphite, C – C bond acquires double bond character, hence higher bond order than in diamond.

*Multiple options can be correct

QUESTION: 13

The correct statement(s) about O₃ is(are)

A.
O—O bond lengths are equal
B.
Thermal decomposition of O₃ is endothermic
C.
O₃ is diamagnetic in nature
D.
O₃ has a bent structure

Solution:

Ozone is diamagnetic in nature (due to presence of paired electron) and both the O – O bond length are equal. It has a bent structure.

*Multiple options can be correct

QUESTION: 14

For the reaction

A.
Stoichiometric coefficient of HSO₄^- is 6
B.
Iodide is oxidized
C.
Sulphur is reduced
D.
H₂O is one of the products

Solution:

Balanced chemical equation is

*Multiple options can be correct

QUESTION: 15

The correct statement(s) for orthoboric acid is/are

A.
It behaves as a weak acid in water due to self ionization.
B.
Acidity of its aqueous solution increases upon addition of ethylene glycol
C.
It has a three dimensional structure due to hydrogen bonding
D.
It is a weak electrolyte in water

Solution:

H₃BO₃ does not undergo self ionization. However, it acts as a weak acid in water (hence it is a weak electrolyte in water).

Addition of cis-diols (e.g., ethylene glycol) to aqueous solution of orthoboric acid leads to complex formation, thus acidity of aqueous solution of orthoboric acid is increased.

*Multiple options can be correct

QUESTION: 16

The correct statement(s) regarding,

(i) HClO, (ii) HClO₂, (iii) HClO₃ and (iv) HClO₄, is(are)

A.
The number of Cl=O bonds in (ii) and (iii) together is two
B.
The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
C.
The hybridization of Cl in (iv) is sp³
D.
Amongst (i) to (iv), the strongest acid is (i)

Solution:

Number of Cl = O bonds in (ii) and (iii) together is 3 Number of lone pairs on Cl in (ii) and (iii) together is 3 Hybridisation of Cl in all the four is sp³
Strongest acid is HClO₄ (iv)

*Multiple options can be correct

QUESTION: 17

Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are

A.
CH₃SiCl₃ and Si(CH₃)₄
B.
(CH₃)₂SiCl₂ and (CH₃)₃SiCl
C.
(CH₃)₂SiCl₂ and CH₃SiCl₃
D.
SiCl₄ and (CH₃)₃SiCl

Solution:

(CH₃)₂SiCl₂ form linear polymer on hydrolysis and (CH₃)₃SiCl is a chain terminator.

*Multiple options can be correct

QUESTION: 18

The crystalline form of borax has

A.
tetranuclear [B₄O₅(OH)₄]^2– unit
B.
all boron atoms in the same plane
C.
equal number of sp² and sp³ hybridized boron atoms
D.
one terminal hydroxide per boron atom

Solution:

Structure of borax

Correct formula of borax is Na₂[B₄O₅(OH)₄].8H₂O
(A) Borax has tetranuclear. [B₄O₅(OH)₄]^2– unit
(B) Only two 'B' atom lie in same plane
(C) two Boron are sp² and two are sp³ hybridised.
(D) one terminal hydroxide per boran atom