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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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301 Medium

301 of 705

Understanding +4 | -1

Which of the following(s) give(s) coloured gases?

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Correct option: D

All of these

Option D (All of these) is correct because all listed reactions produce colored gases: chromyl chloride (reddish-brown), iodine vapor (violet), and nitrogen dioxide (reddish-brown).

302 Easy

302 of 705

Remembering +4 | -1

The mineral clevite on heating gives:

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Correct option: A

Option A (He) is correct because the mineral clevite, when heated, releases Helium gas.

303 Medium

303 of 705

Analyzing +4 | -1

Statement-I: Nitrogen is inactive in normal chemical reactions.

Statement-II: The bond dissociation energy is directly proportional to reactivity.

View answer and explanation

Correct option: C

Statement I is correct but Statement II is wrong

Statement I is correct but Statement II is wrong because nitrogen is inert due to its high bond dissociation energy, and reactivity is inversely proportional to bond dissociation energy.

304 Easy

304 of 705

Remembering +4 | -1

Phosphine is:

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Correct option: A

Basic

Option A (Basic) is correct because phosphine (

\mathrm{P}{\mathrm{H}_3}

) has a lone pair of electrons on the phosphorus atom, making it capable of accepting a proton and acting as a Lewis base.

305 Easy

305 of 705

Understanding +4 | -1

Which of the following pseudohalides best resembles

\mathrm{Cl}^{-}, \mathrm{Br}^{-}

and

\mathrm{I}^{-}

View answer and explanation

Correct option: A

\mathrm{CN}^{-}

Option A (CN⁻) is correct because cyanide ion (CN⁻) shares similar properties with halide ions like Cl⁻, Br⁻, and I⁻, such as forming dimers and disproportionating in alkaline medium.

306 Easy

306 of 705

Remembering +4 | -1

Antimony burns in chlorine to form:

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Correct option: A

SbC{l_3}

Option A (

SbCl_3

) is correct because antimony reacts with chlorine to form antimony(III) chloride (

SbCl_3

307 Medium

307 of 705

Remembering +4 | -1

Ozone reacts with dry iodine to give:

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Correct option: D

{I_4}{O_9}

Option D (

{I_4}{O_9}

) is correct because ozone reacts with dry iodine to produce iodine nonaoxide (

{I_4}{O_9}

308 Medium

308 of 705

Analyzing +4 | -1

In which of the following sets the order of boiling points are incorrect?

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Correct option: B

\mathrm{HBr} > \mathrm{HCl} > \mathrm{HF}

The correct answer is B (HBr > HCl > HF) because HF has the highest boiling point due to hydrogen bonding.

309 Easy

309 of 705

Remembering +4 | -1

Electronegativity of an inert gas is:

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Correct option: D

Zero

Option D (Zero) is correct because inert gases have a stable electronic configuration and thus have no tendency to attract electrons.

310 Easy

310 of 705

Understanding +4 | -1

Order of reactivity of halogen acids towards an alcohol is

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Correct option: C

HI > HBr > HCl

Option C (HI > HBr > HCl) is correct because the acidity of halogen acids increases down the group due to decreasing bond strength.

311 Medium

311 of 705

Analyzing +4 | -1

Assertion:

\mathrm{H}_2 \mathrm{Se}

is more acidic than

\mathrm{H}_2 \mathrm{Te}

Reason: Bond enthalpy decrease with increased in size of the atom.

View answer and explanation

Correct option: D

Assertion is wrong, but reason is correct statement.

Option D (Assertion is wrong, but reason is correct statement) is correct because acidity increases down the group due to decreasing bond enthalpy, making H₂Te more acidic than H₂Se.

312 Medium

312 of 705

Understanding +4 | -1

Statement I:

\mathrm{H}_2 \mathrm{SO}_4

forms only one series of salts

Statement II: The molecule of

\mathrm{H}_2 \mathrm{SO}_4

consists of two

\mathrm{OH}

group

View answer and explanation

Correct option: C

Statement I is correct but statement II is false

Option C (Statement I is correct but statement II is false) is correct because

\mathrm{H}_2\mathrm{SO}_4

forms two series of salts (normal and acid salts), and its molecule has two OH groups.

313 Easy

313 of 705

Analyzing +4 | -1

Potassium chlorate on heating with conc.

{H_2}S{O_4}

gives:

View answer and explanation

Correct option: D

All of these

Option D is correct. Potassium chlorate reacts with concentrated sulfuric acid to produce chlorine dioxide, perchloric acid, and potassium bisulfate.

314 Medium

314 of 705

Analyzing +4 | -1

Which one of the following arrangement does not accurately depict the trend mentioned against it?

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Correct option: D

I_2>{Br}_2>{Cl}_2>{F}_2 \text{ (oxidizing power)}

Option D (I2 > Br2 > Cl2 > F2 (oxidizing power)) is correct because the oxidizing power decreases down the group, with fluorine being the strongest oxidizing agent.

315 Medium

315 of 705

Analyzing +4 | -1

Oxide of a non-metal possesses the following characteristics: (i) It is both a proton donor and proton acceptor. (ii) It is poor conductor of electricity. (iii) It reacts readily with basic and acidic oxides. (iv) It oxidses Fe at boiling point. The oxide is:

View answer and explanation

Correct option: A

{H_2}O

Option A (

{H_2}O

) is correct. Water is amphoteric (proton donor and acceptor), a poor conductor, reacts with both acidic and basic oxides, and can oxidize Fe at boiling point under specific conditions.

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316 Medium

316 of 705

Understanding +4 | -1

The paramagnetic oxides of nitrogen are

View answer and explanation

Correct option: B

Nitrogen monoxide and nitrogen dioxide

The correct answer is B (Nitrogen monoxide and nitrogen dioxide) because both NO and NO2 have unpaired electrons, making them paramagnetic.

317 Easy

317 of 705

Remembering +4 | -1

Chalcogens are:

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Correct option: B

Ore forming elements

Option B (Ore forming elements) is correct because chalcogens, which are group 16 elements (O, S, Se, Te, Po), are often found in ores, especially as sulfides and oxides.

318 Medium

318 of 705

Analyzing +4 | -1

An orange solid

(X)

on heating gives a colourless gas (y) and only a green residue (z). Gas (y) on treatment with

{Mg}

produces a white solid substance:-

View answer and explanation

Correct option: A

{Mg}_3 {N}_2

The correct answer is A (

{Mg}_3 {N}_2

) because the orange solid is ammonium dichromate which decomposes to nitrogen gas, and nitrogen reacts with magnesium to form magnesium nitride.

319 Medium

319 of 705

Analyzing +4 | -1

Which of the following will have the least bond angle?

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Correct option: C

\mathrm{PH}_3

The correct answer is C (PH3) because PH3 has the smallest bond angle due to minimal bond pair-bond pair repulsion, approaching 90 degrees.

320 Easy

320 of 705

Remembering +4 | -1

Noble gases are adsorbed by:

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Correct option: D

All of the above

Option D (All of the above) is correct because finely divided Pd and Pt, colloidal Pd, and coconut charcoal are all known to adsorb noble gases.

321 Easy

321 of 705

Remembering +4 | -1

A radioactive element is:

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Correct option: B

Polonium

Option B (Polonium) is correct because Polonium is a radioactive element, specifically

{{\;}_{84}}Po

is the only radioactive element of group 16.

322 Medium

322 of 705

Analyzing +4 | -1

Assertion (A): Pitch blende, monozite and clevite are not radio active minerals.

Reason (R): Oganesson and Radon are not present in atmosphere.

View answer and explanation

Correct option: D

Both (A) and (R) are false

Option D (Both (A) and (R) are false) is correct because Pitchblende, Monazite, and Clevite are radioactive minerals, and Radon is present in the atmosphere, although Oganesson is not.

323 Easy

323 of 705

Understanding +4 | -1

Which of the following form of interhalogen compounds does not exists?

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Correct option: D

\mathrm{BrC}{\mathrm{l}_7}

Option D (

\mathrm{BrC}{\mathrm{l}_7}

) is correct because bromine cannot achieve a +7 oxidation state to bond with seven chlorine atoms due to its electronic configuration and size.

324 Easy

324 of 705

Remembering +4 | -1

The percentage of available chlorine in a commercial sample of bleaching powder is:

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Correct option: B

0.35

Option B (35%) is correct because the percentage of available chlorine in a commercial sample of bleaching powder is approximately 35%.

325 Easy

325 of 705

Analyzing +4 | -1

Assertion (A) :

\mathrm{SO}_2

from oxidation of sulphur or

\mathrm{H}_2 \mathrm{~S}

contains sulphur and oxygen in the mass ratio of

1: 2

.

Reason (R) : Different samples of a pure chemical substance always contain the same proportions of elements by mass.

View answer and explanation

Correct option: D

\mathbf{A}

is false but

\mathbf{R}

is true.

Option D is correct because the assertion is false (the mass ratio of S:O in SO2 is 1:1), but the reason is true (law of definite proportions).

326 Easy

326 of 705

Analyzing +4 | -1

Arrange the acids (I)

{H_2}S{O_3}

, (II)

{H_3}P{O_3}

, and (III)

HCl{O_3}

in the decreasing order of acidic nature.

View answer and explanation

Correct option: C

\mathrm{III} > I > II

Option C (III > I > II) is correct because the acidic nature of oxoacids increases with the electronegativity of the central atom.

327 Medium

327 of 705

Analyzing +4 | -1

\left[ X \right] + {H_2}S{O_4} \to \left[ Y \right]

a colourless gas with irritating smell.

\left[ Y \right] + {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to

Green solution [X] and [Y] are:

View answer and explanation

Correct option: A

SO_3^{2 - },S{O_2}

Option A (

SO_3^{2 - },S{O_2}

) is correct because sulfite ions react with sulfuric acid to produce sulfur dioxide, a colorless gas with an irritating smell, which then reduces dichromate ions to produce a green solution of chromium(III) sulfate.

328 Easy

328 of 705

Remembering +4 | -1

Which species is not known?

View answer and explanation

Correct option: D

Kr{F_6}

Option D (

Kr{F_6}

) is correct because Krypton is less reactive than Xenon, and

KrF_6

is not a known compound.

329 Easy

329 of 705

Understanding +4 | -1

Which reaction cannot be used for the preparation of the halogen acid?

View answer and explanation

Correct option: A

2KBr + {H_2}S{O_4}\xrightarrow[{\mathrm{Conc}.}]{{}}{K_2}S{O_4} + 2HBr

Option A (

2KBr + {H_2}S{O_4}\xrightarrow[{\mathrm{Conc}.}]{{}}{K_2}S{O_4} + 2HBr

) is correct because HBr is a strong reducing agent and gets oxidized by concentrated H2SO4, which is an oxidizing agent. Therefore, this reaction is not suitable for preparing HBr.

330 Medium

330 of 705

Understanding +4 | -1

Which ion cannot be precipitated from water?

View answer and explanation

Correct option: B

\mathrm{NO}_3^ -

Option B (

\mathrm{NO}_3^ -

) is correct because nitrates of all metals are soluble in water and thus cannot be precipitated.

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