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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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601 Medium

601 of 705

Understanding +4 | -1

Statement-I: White phosphorous react with sulphuryl chloride gives

{PCl}_5

and

{SO}

Statement-II:

{PCl}_5

react with finally divided

{Sn}

gives

{SnCl}_4

and

{PCl}_3

View answer and explanation

Correct option: A

Statement I and II are correct

Option A (Statement I and II are correct) is correct because both statements accurately describe chemical reactions involving phosphorus compounds.

602 Easy

602 of 705

Understanding +4 | -1

Assertion (A): HF is a liquid (boiling point

=293 \mathrm{~K}

).

Reason (R): It is due to strong hydrogen bonding.

View answer and explanation

Correct option: A

Both

\mathrm{A}

and

\mathrm{R}

are correct;

\mathrm{R}

is the correct explanation of

A

Option A is correct because HF is a liquid at room temperature due to strong hydrogen bonding between HF molecules, which significantly increases its boiling point.

603 Medium

603 of 705

Remembering +4 | -1

The formula of iodine acetate is:

View answer and explanation

Correct option: B

I{\left( {C{H_3}COO} \right)_3}

Option B (

I{\left( {C{H_3}COO} \right)_3}

) is correct because iodine acetate is an ionic compound with iodine having a +3 oxidation state and acetate having a -1 charge.

604 Easy

604 of 705

Analyzing +4 | -1

Assertion: Oxygen has less negative electron gain enthalpy than sulphur.

Reason: Oxygen is more electronegative than sulphur.

View answer and explanation

Correct option: B

Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.

The correct answer is B (Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.) because both statements are true, but electronegativity doesn't directly explain the electron gain enthalpy difference.

605 Easy

605 of 705

Understanding +4 | -1

Oxygen differs from sulphur in:

View answer and explanation

Correct option: D

Nature of hydrides

Option D (Nature of hydrides) is correct because oxygen forms a neutral hydride (

{H_2}O

), while sulfur forms an acidic hydride (

{H_2}S

606 Easy

606 of 705

Remembering +4 | -1

Which of the following isotopes is present in largest amount?

View answer and explanation

Correct option: A

{O^{16}}

Option A (

{O^{16}}

) is correct because Oxygen-16 is the most abundant isotope of oxygen, making up the vast majority of naturally occurring oxygen.

607 Easy

607 of 705

Remembering +4 | -1

{\mathrm{F}_2}

combines with all non-metals directly except:

View answer and explanation

Correct option: A

{\mathrm{N}_2}

Option A (

{\mathrm{N}_2}

) is correct because nitrogen is relatively inert and does not readily react directly with fluorine under normal conditions.

608 Medium

608 of 705

Analyzing +4 | -1

The perhalate ion with maximum oxidizing power is:

View answer and explanation

Correct option: B

BrO_4^ -

Option B (

BrO_4^ -

) is correct because the oxidizing power of perhalate ions increases down the group:

BrO_4^ - > ClO_4^ - > IO_4^ -

609 Easy

609 of 705

Remembering +4 | -1

{CaCl}_2

and

{Ca}({OCl})_2

are components of:

View answer and explanation

Correct option: D

Bleaching powder

Option D (Bleaching powder) is correct because Calcium hypochlorite (

Ca(OCl)_2

) and Calcium chloride (

CaCl_2

) are the main components of bleaching powder.

610 Medium

610 of 705

Remembering +4 | -1

Phosphine is prepared by the reaction of

View answer and explanation

Correct option: C

P and NaOH

Option C (P and NaOH) is correct because phosphine is produced by the reaction of phosphorus with sodium hydroxide and water.

611 Easy

611 of 705

Remembering +4 | -1

Nuclear fusion produces

View answer and explanation

Correct option: C

Helium

Option C (Helium) is correct because nuclear fusion involves the combination of light nuclei, such as deuterium, to form heavier nuclei, such as helium, releasing a large amount of energy.

612 Medium

612 of 705

Remembering +4 | -1

Statement I: Ammonia is a colourless gas with a pungent order.

Statement II: Its freezing and boiling points are 198.4 and

239.7 \mathrm{~K}

respectively.

View answer and explanation

Correct option: A

Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.

Option A (Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because both statements about ammonia's properties and its freezing/boiling points are accurate, and the latter explains the former due to hydrogen bonding.

613 Easy

613 of 705

Understanding +4 | -1

In the preparation of

{O_2}

from

KCl{O_3},\;Mn{O_2}

acts as:

View answer and explanation

Correct option: B

Catalyst

Option B (Catalyst) is correct because

Mn{O_2}

speeds up the decomposition of

KCl{O_3}

without being consumed in the reaction.

614 Medium

614 of 705

Application +4 | -1

In the reaction

4 \mathrm{HNO}_3+\mathrm{P}_4 \mathrm{O}_{10} \rightarrow 4 \mathrm{HPO}_3+\mathrm{X}

, the product

\mathrm{X}

View answer and explanation

Correct option: A

\mathrm{N}_2 \mathrm{O}_5

Option A (

\mathrm{N}_2 \mathrm{O}_5

) is correct because it balances the chemical equation with the given reactants.

615 Medium

615 of 705

Analyzing +4 | -1

Assertion : Iodine is more soluble in water than in carbon tetrachloride.

Reason : Iodine is a polar compound.

View answer and explanation

Correct option: D

If both the Assertion and Reason are incorrect.

The correct answer is D because iodine is non-polar and thus more soluble in carbon tetrachloride (a non-polar solvent) than in water (a polar solvent).

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616 Medium

616 of 705

Analyzing +4 | -1

Statement I:

\mathrm{SeCl}_4

does not have a tetrahedral shape

Statement II:

\mathrm{Se}

\mathrm{SeCl}_4

has two lone pairs

View answer and explanation

Correct option: C

Statement I is correct but statement II is false

Option C is correct because Se in

SeCl_4

has one lone pair and seesaw shape, making statement I correct and statement II false.

617 Medium

617 of 705

Analyzing +4 | -1

A" obtained by Ostwald's method involving air oxidation of

\mathrm{NH}_{3}

, upon further air oxidation produces "B". "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test.

Identify

\mathrm{A}

and

\mathrm{B}

, respectively.

View answer and explanation

Correct option: A

\mathrm{NO}, \mathrm{NO}_{2}

Option A (

\mathrm{NO}, \mathrm{NO}_{2}

) is correct because ammonia oxidation yields NO (A), which further oxidizes to NO2 (B). NO2 then reacts with water and oxygen to form HNO3, which gives NO and a positive brown ring test.

618 Medium

618 of 705

Understanding +4 | -1

Assertion (A) : Anhydrous sulphuric acid has very high viscosity.

Reason (R) :

\mathrm{H}

-bonding makes it viscous.

View answer and explanation

Correct option: A

Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

Option A (Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

.) is correct because the high viscosity of anhydrous sulfuric acid is due to extensive hydrogen bonding between molecules.

619 Medium

619 of 705

Understanding +4 | -1

The correct order of acidic nature of oxides is in the order

View answer and explanation

Correct option: B

{N_2}O < \mathrm{ }NO < \mathrm{ }{N_2}{O_3} < \mathrm{ }N{O_2} < {N_2}{O_5}

Option B (

{N_2}O < \mathrm{ }NO < \mathrm{ }{N_2}{O_3} < \mathrm{ }N{O_2} < {N_2}{O_5}

) is correct because the acidic nature of oxides increases with the oxidation number of the central atom.

620 Easy

620 of 705

Remembering +4 | -1

Sulphur is soluble in:

View answer and explanation

Correct option: D

C{S_2}

Option D (

C{S_2}

) is correct because sulphur is soluble in carbon disulfide (

C{S_2}

) due to similar intermolecular forces.

621 Medium

621 of 705

Understanding +4 | -1

Which of the following is the least reactive interhalogen compound?

View answer and explanation

Correct option: B

\mathrm{BrF}

Option B (

\mathrm{BrF}

) is correct because as the electronegativity difference between the halogens decreases, the reactivity decreases.

\mathrm{BrF}

has the smallest difference among the options.

622 Easy

622 of 705

Analyzing +4 | -1

A compound ‘X’ upon reaction with

{H_2}O

produces a colourless gas ‘y’ with rotten fish smell. Gas ‘y’ is absorbed in a solution of

CuS{O_4}

and

C{u_3}{P_2}\,

as one of the products, predict the compound ‘X’

View answer and explanation

Correct option: A

C{a_3}{P_2}

Option A (

C{a_3}{P_2}

) is correct because it reacts with water to produce phosphine (

P{H_3}

), which has a rotten fish smell and reacts with

CuS{O_4}

to form

C{u_3}{P_2}

623 Medium

623 of 705

Understanding +4 | -1

Chlorine on reaction with hot and concentrated sodium hydroxide gives

View answer and explanation

Correct option: B

C{l^ - }\mathrm{ }\,\mathrm{and}\,\,\mathrm{ }ClO_3^ -

The correct answer is B (

C{l^ - }\mathrm{ }\,\mathrm{and}\,\mathrm{ }ClO_3^ -

) because chlorine reacts with hot and concentrated sodium hydroxide to form chloride and chlorate ions.

624 Easy

624 of 705

Analyzing +4 | -1

S-I: Oxygen is more electronegative than sulphur, yet

{H}_2{S}

is acidic, while

{H}_2{O}

is neutral.

S-II:

{H}-{S}

bond is weaker than

{O}-{H}

bond.

View answer and explanation

Correct option: A

Both S-I and S-II are true

Option A (Both S-I and S-II are true) is correct because oxygen is more electronegative than sulfur, yet H2S is acidic while H2O is neutral, and the H-S bond is weaker than the O-H bond.

625 Medium

625 of 705

Analyzing +4 | -1

Which of the following halogen oxides is ionic?

View answer and explanation

Correct option: A

{\mathrm{I}_4}{\mathrm{O}_9}

The correct answer is A (

{\mathrm{I}_4}{\mathrm{O}_9}

) because it exists as

\mathrm{I}{\left( {\mathrm{I}{\mathrm{O}_3}} \right)_3}

, which is an ionic compound composed of

I^+

and

IO_3^-

ions.

626 Easy

626 of 705

Analyzing +4 | -1

Assertion:

Cl,Br

and

I

show disproportionation, but

F

doesn't.

Reason: This is because it is the most electronegative element of the group.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because fluorine's high electronegativity prevents it from disproportionating, as it is already in its most oxidized state.

627 Medium

627 of 705

Analyzing +4 | -1

{P}-{O}-{P}

bond is present in

View answer and explanation

Correct option: B

{H}_4 {P}_2 {O}_5

Option B is correct because pyrophosphorous acid (H4P2O5) contains a P-O-P bond, while hypophosphoric acid (H4P2O6) does not. The structure of H4P2O5 has a direct P-P bond, and the structure of H4P2O6 has a P-O-P bond.

628 Easy

628 of 705

Understanding +4 | -1

Which of the following is most volatile?

View answer and explanation

Correct option: D

\mathrm{HI}

Option D (

\mathrm{HI}

) is correct because HI has the weakest intermolecular forces (van der Waals forces) among the given hydrogen halides, leading to the highest volatility.

629 Medium

629 of 705

Remembering +4 | -1

The highest industrial consumption of molecular hydrogen is to produce compounds of element:

View answer and explanation

Correct option: B

Nitrogen

Option B (Nitrogen) is correct because the largest industrial use of hydrogen is in the Haber-Bosch process to produce ammonia (NH₃) from nitrogen and hydrogen.

630 Easy

630 of 705

Understanding +4 | -1

Which of the following will not release

\mathrm{NO}_2

upon reaction with

\mathrm{HNO}_3

(in respective conditions).

View answer and explanation

Correct option: A

\mathrm{Cu}

in dilute

\mathrm{HNO}_3

acid

Option A (

\mathrm{Cu}

in dilute

\mathrm{HNO}_3

acid) is correct because copper reacts with dilute nitric acid to produce nitric oxide (NO), not nitrogen dioxide (

\mathrm{NO}_2

Revision notes and key takeaways

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