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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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451 Easy
451 of 705
Remembering +4 | -1
The catalyst used in the manufacture of H2SO4{H_2}S{O_4} by contact process is
View answer and explanation
Correct option: B
V2O5{V_2}{O_5}
Option B (V2O5{V_2}{O_5}) is correct because vanadium(V) oxide is the most commonly used catalyst in the contact process for sulfuric acid production.
452 Easy
452 of 705
Remembering +4 | -1
Bromargyrite is a mineral of:
View answer and explanation
Correct option: D
Br2B{r_2}
Option D (Br2B{r_2}) is correct because Bromargyrite is a mineral of bromine.
453 Easy
453 of 705
Understanding +4 | -1
PCl3PC{l_3} on hydrolysis gives
View answer and explanation
Correct option: D
H3PO3{H_3}P{O_3}
Option B (0) is correct because Sc3+ has no unpaired electrons, resulting in a zero magnetic moment.
454 Easy
454 of 705
Application +4 | -1
Which has the same electronic configuration as of inert gas?
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Correct option: D
Ti4+T{i^{4 + }}
Option D (Ti4+Ti^{4+}) is correct because it has the same electronic configuration as Argon, an inert gas.
455 Medium
455 of 705
Analyzing +4 | -1
Which of the following does not exist?
View answer and explanation
Correct option: B
[KrF3]−[SbF4]+{\left[ {Kr{F_3}} \right]^ - }{\left[ {Sb{F_4}} \right]^ + }
Option B is correct because KrF2 acts as a F- donor and forms complexes with F- acceptors, thus only cationic species or Kr will be present.
456 Easy
456 of 705
Understanding +4 | -1
When ammonium nitrate is heated at 250∘C{250^ \circ }C the gas evolved is
View answer and explanation
Correct option: D
N2O{\mathrm{N}_\mathrm{2}}\mathrm{O}
Option D is correct because ammonium nitrate decomposes upon heating at 250°C to produce nitrous oxide (N2O) and water.
457 Easy
457 of 705
Remembering +4 | -1
Which noble gas is not found in atmosphere?
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Correct option: A
Rn\mathrm{Rn}
Option A (Rn\mathrm{Rn}) is correct because Radon (Rn) is a radioactive noble gas that is not typically found in the atmosphere due to its origin from radioactive decay.
458 Easy
458 of 705
Understanding +4 | -1
The number of sigma bonds in P4O10{P_4}{O_{10}} is:
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Correct option: B
16
Option B (16) is correct because the structure of P4O10P_4O_{10} contains 16 sigma bonds.
459 Easy
459 of 705
Remembering +4 | -1
Which element is extracted commercially by the electrolysis of an aqueous solutions of one of its compounds?
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Correct option: C
Chlorine
Option C (Chlorine) is correct because chlorine is commercially extracted by the electrolysis of an aqueous solution of sodium chloride (brine).
460 Medium
460 of 705
Understanding +4 | -1
Which ion will not be precipitated by aq. (NH4)2S\left({NH}_4\right)_2 {S} ?
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Correct option: D
K+{K}^{+}
Option D (K+{K}^{+}) is correct because potassium ions do not form insoluble sulfides in aqueous ammonium sulfide solution.
461 Easy
461 of 705
Understanding +4 | -1
Which statement is not correct?
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Correct option: D
The most abundant rare gas found in atmosphere is HeHe
Option D is correct because Argon (Ar) is the most abundant rare gas in the atmosphere, not Helium (He).
462 Easy
462 of 705
Understanding +4 | -1
Cu+HNO3 (dil) →a+Cu(NO3)2+H2OZn+HNO3 (dil) →b+Zn(NO3)2+H2O\begin{aligned} & \mathrm{Cu}+\underset{\text { (dil) }}{\mathrm{HNO}_3} \rightarrow \mathrm{a}+\mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{Zn}+\underset{\text { (dil) }}{\mathrm{HNO}_3} \rightarrow \mathrm{b}+\mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \mathrm{O} \end{aligned}

Here a and b are:
View answer and explanation
Correct option: B
NO,N2O\mathrm{NO}, \mathrm{N}_2 \mathrm{O}
Option B is correct because copper reacts with dilute nitric acid to produce nitric oxide (NO), while zinc reacts with dilute nitric acid to produce nitrous oxide (N2ON_2O).
463 Medium
463 of 705
Understanding +4 | -1
Liquid ammonia is used in refrigeration because of its
View answer and explanation
Correct option: B
High heat of vaporisation
Option B (High heat of vaporisation) is correct because liquid ammonia has a high heat of vaporization, making it effective in refrigeration.
464 Medium
464 of 705
Analyzing +4 | -1
Assertion (A) : XeF6\mathrm{XeF}_6 can be used as an oxidising agent.

Reason (R) : Oxidation state of Xee\mathrm{Xe}_{\mathrm{e}} in XeF6\mathrm{XeF}_6 is +6{+ 6}.
View answer and explanation
Correct option: B
Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}.
The correct answer is B (Both A and R are true but R is not the correct explanation of A) because XeF6 can act as an oxidizing agent due to the high electronegativity of fluorine, but the +6 oxidation state of Xe is not the direct reason for its oxidizing ability.
465 Medium
465 of 705
Remembering +4 | -1
An element MM belongs to group- 15 and it occurs as the founder of the apatite family. It is an essential constituent of animal and plant matter. It is also present in bones as well as in living cells. Identify MM.
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Correct option: D
Phosphorus
Option D (Phosphorus) is correct because phosphorus belongs to group 15, is found in the apatite family, and is an essential constituent of animal and plant matter, including bones and living cells.
466 Easy
466 of 705
Understanding +4 | -1
When chlorine water is exposed to sunlight, O2{\mathrm{O}_2} is liberated. Hence:
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Correct option: C
Hydrogen has more affinity to chlorine
Option C (Hydrogen has more affinity to chlorine) is correct because chlorine reacts with water to form hydrochloric acid and oxygen, indicating hydrogen's preference for chlorine.
467 Hard
467 of 705
Understanding +4 | -1
When concentrated H2SO4{H_2}S{O_4} is added to dry KNO3KN{O_3}, brown fumes evolve.These fumes are of:
View answer and explanation
Correct option: C
NO2N{O_2}
Option C (NO2N{O_2}) is correct because the reaction of concentrated sulfuric acid with potassium nitrate produces nitrogen dioxide (NO2N{O_2}) gas, which is brown in color.
468 Easy
468 of 705
Understanding +4 | -1
Which forms new compound in air?
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Correct option: D
Phosphorus in air
Option D (Phosphorus in air) is correct because phosphorus reacts with oxygen in the air to form phosphorus oxides, which are new compounds.
469 Medium
469 of 705
Understanding +4 | -1
Assertion (A): To initiate the reaction between O2\mathrm{O}_2 and a metal, some external heating is required.

Reason (R): Bond dissociation enthalpy of O=O\mathrm{O}=\mathrm{O} bond is high (493.4 kJ mol−1)\left(493.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\right).
View answer and explanation
Correct option: A
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is the correct explanation of AA.
Option A is correct because the high bond dissociation enthalpy of O=O requires initial energy input to start the reaction, even though the reaction is exothermic overall.
470 Easy
470 of 705
Remembering +4 | -1
Hyponitrous acid is:
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Correct option: C
H2N2O2{H_2}{N_2}{O_2}
Option C (H2N2O2{H_2}{N_2}{O_2}) is correct because this is the chemical formula for hyponitrous acid.
471 Easy
471 of 705
Remembering +4 | -1
The word argon means:
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Correct option: D
Lazy
The correct answer is D (Lazy) because the word \"argon\" is derived from the Greek word \"argos,\" which means lazy or inactive.
472 Medium
472 of 705
Analyzing +4 | -1
Which of the following will show disproportionation reaction?
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Correct option: D
NH4Cl⟶Δ\mathrm{NH}_4 \mathrm{Cl} \stackrel{\Delta}{\longrightarrow}
Option C is correct. Fluorine in water undergoes disproportionation, where it is both oxidized and reduced.
473 Easy
473 of 705
Remembering +4 | -1
SO2S{O_2} reacts with Cl2C{l_2} to yield:
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Correct option: C
Sulphuryl chloride
Option C (Sulphuryl chloride) is correct because the reaction of SO2S{O_2} with Cl2C{l_2} produces SO2Cl2S{O_2}C{l_2}, which is sulphuryl chloride.
474 Easy
474 of 705
Remembering +4 | -1
Halogens are:
View answer and explanation
Correct option: B
Electronegative in nature
Option B (Electronegative in nature) is the correct answer because halogens are highly electronegative non-metals.
475 Medium
475 of 705
Understanding +4 | -1
A chloride salt solution acidified with dil. HNO3\mathrm{HNO}_{3} gives a curdy white precipitate, [A], on addition of AgNO3\mathrm{AgNO}_{3}. [A] on treatment with NH4OH\mathrm{NH}_{4} \mathrm{OH} gives a clear solution, B. A and B are respectively
View answer and explanation
Correct option: C
AgCl & [Ag(NH3)2]Cl\mathrm{AgCl} \, \& \, \left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}
Option C (AgCl & [Ag(NH3)2]Cl\mathrm{AgCl} \, \& \, \left[\mathrm{Ag}(\mathrm{NH}_{3})_{2}\right] \mathrm{Cl}) is correct because AgCl forms as a precipitate with AgNO3 and then dissolves in NH4OH to form a complex.
476 Medium
476 of 705
Remembering +4 | -1
In the laboratory H2S{H_2}S gas is prepared by using black lumps and dil. H2SO4{H_2}S{O_4} The black lumps are
View answer and explanation
Correct option: C
FeS
Option C (FeS) is correct because ferrous sulfide reacts with dilute sulfuric acid to produce hydrogen sulfide gas.
477 Easy
477 of 705
Remembering +4 | -1
The product obtained as a result of a reaction of nitrogen with CaC2 Ca{C_2}\, is:
View answer and explanation
Correct option: A
Ca(CN)2\mathrm{Ca(CN}{\mathrm{)}_\mathrm{2}}
The correct answer is A (Ca(CN)2\mathrm{Ca(CN}{\mathrm{)}_\mathrm{2}}) because calcium carbide reacts with nitrogen gas to form calcium cyanamide, which is represented as Ca(CN)2\mathrm{Ca(CN}{\mathrm{)}_\mathrm{2}}.
478 Easy
478 of 705
Remembering +4 | -1
Which one of the following noble gases is not found in the atmosphere
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Correct option: A
Rn{Rn}
Option A (Rn{Rn}) is correct because Radon is a radioactive element produced by the decay of radium and is not typically found in the atmosphere in significant amounts.
479 Easy
479 of 705
Understanding +4 | -1
NH3\mathrm{N}{\mathrm{H}_3} can be collected by the displacement of:
View answer and explanation
Correct option: A
Mercury
Option A (Mercury) is correct because ammonia does not react with mercury, allowing it to be collected by displacement.
480 Medium
480 of 705
Analyzing +4 | -1
Statement-I: There are two types of Cl−P−Cl{Cl}-{P}-{Cl} bond angles of 90∘90^{\circ} and 120∘120^{\circ} and two types of P−Cl{P}-{Cl} single bonds, two weaker than the rest three bonds is PCl5{PCl}_5 molecule.

Statement-II: P{P} in PCls{PCl}_s is sp3d−{sp}^3 {d}-hybridized giving trigonal bipyramid geometry where two P−Cl{P}-{Cl} axial bonds are longer and weaker than the three equatorial bonds.
View answer and explanation
Correct option: A
Statement I and II are correct
The correct answer is A (Statement I and II are correct) because both statements accurately describe the structure and bonding in PCl5.
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