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


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19 Questions MCQ Test 35 Years Chapter wise Previous Year Solved Papers for JEE - Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced

Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced for JEE 2024 is part of 35 Years Chapter wise Previous Year Solved Papers for JEE preparation. The Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced below.
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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 1

In the electrolysis of alumina, cryolite is added to :

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 1

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.

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 2

Nitrogen(I) oxide is produced by :

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 2

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

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 3

The compounds used as refrigerant are

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 3

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

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 4

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

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 5

The material used in the solar cells contains

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 5

Silicon is used in solar cells.

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 6

Sodium nitrate decomposes above  800°C  to give

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 6

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

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

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 7

White phosphorus (P4) has

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 7

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 60o. Each phosphorus has a lone pair of electrons.

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 8

Ammonia, on reaction with hypochlorite anion, can form

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 8

2NH3 + OCl–→ NH2.NH2 + H2O + Cl

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 9

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]

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 10

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 10


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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 11

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 11

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 12

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 12

(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.

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 13

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 13

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.

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 14

For the reaction

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 14

Balanced chemical equation is

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 15

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 15

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.

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 16

The correct statement(s) regarding,

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 16


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)

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 17

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

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 17

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

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 18

The crystalline form of borax has

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 18

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

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Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 19

The nitrogen containing compound produced in the reaction of HNO3 with P4O10

Detailed Solution for Test: MCQs (One or More Correct Option): The p-Block Elements | JEE Advanced - Question 19

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