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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Mixed difficulty

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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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391 Medium
391 of 705
Remembering +4 | -1
Bromine is obtained on a commercial scale from:
View answer and explanation
Correct option: B
Carnallite
Option B (Carnallite) is correct because bromine is commercially extracted from carnallite, a hydrated potassium magnesium chloride mineral.
392 Medium
392 of 705
Analyzing +4 | -1
S-I: Solubility of noble gases in water decreases with increasing size of the noble gas.

S-II: Solubility of noble gases in water is due to dipole induced dipole interaction.
View answer and explanation
Correct option: D
S-I is false, S-II is true
Option D is correct because the solubility of noble gases in water increases with increasing size due to increased polarizability and dipole-induced dipole interactions are responsible for their solubility.
393 Easy
393 of 705
Remembering +4 | -1
Interhalogen compounds are:
View answer and explanation
Correct option: D
Covalent compounds
Option D (Covalent compounds) is correct because interhalogen compounds are formed by covalent bonding due to relatively small electronegativity differences between the halogens.
394 Medium
394 of 705
Analyzing +4 | -1
A white precipitate was formed when BaCl2\mathrm{BaCl}_2 was added to water extract of an inorganic salt. Further, a gas ' X\mathrm{X} ' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl\mathrm{HCl}. The anion present in the inorganic salt is
View answer and explanation
Correct option: B
SO32−\mathrm{SO}_3^{2-}
Option B (SO32−\mathrm{SO}_3^{2-}) is correct because BaCl2 precipitates sulfite ions as BaSO3, which releases SO2 gas with a characteristic odor upon reaction with HCl.
395 Easy
395 of 705
Remembering +4 | -1
Argon is used
View answer and explanation
Correct option: C
In high temperature welding
Option C (In high temperature welding) is correct because argon is an inert gas and is used to provide a non-oxidizing atmosphere in high-temperature welding to prevent oxidation of the metals.
396 Easy
396 of 705
Analyzing +4 | -1
Minimum bond length will be in:
View answer and explanation
Correct option: B
HF
Option B (HF) is correct because fluorine has the smallest atomic size among the given options, leading to the shortest bond length when bonded with hydrogen.
397 Medium
397 of 705
Understanding +4 | -1
Assertion (A) : Both H3PO3\mathrm{H}_3 \mathrm{PO}_3 and H3PO4\mathrm{H}_3 \mathrm{PO}_4 have the same number of hydrogen atoms but H3PO4\mathrm{H}_3 \mathrm{PO}_4 is a tribasic acid and H3PO3\mathrm{H}_3 \mathrm{PO}_3 is a dibasic acid.

Reason (R) : 1 mole of H3PO3\mathrm{H}_3 \mathrm{PO}_3 is neutralised by 2 moles of NaOH\mathrm{NaOH} while 1 mole of H3PO4\mathrm{H}_3 \mathrm{PO}_4 is neutralised by 3 moles of NaOH\mathrm{NaOH}.
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}.
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}) is correct because both the assertion and the reason are true statements, but the reason doesn't directly explain why H3PO3\mathrm{H}_3\mathrm{PO}_3 is dibasic and H3PO4\mathrm{H}_3\mathrm{PO}_4 is tribasic.
398 Easy
398 of 705
Remembering +4 | -1
Atomicity of phosphorus is:
View answer and explanation
Correct option: D
4
Option D (4) is correct because phosphorus exists as a tetra-atomic molecule, P4P_4.
399 Easy
399 of 705
Remembering +4 | -1
Which oxide of N is neutral?
View answer and explanation
Correct option: D
N2O{N_2}O
Option D (N2O{N_2}O) is correct because N2O{N_2}O (nitrous oxide) is a neutral oxide of nitrogen.
400 Easy
400 of 705
Analyzing +4 | -1
In which of the following reaction, oxidation number of N{N} does not change
View answer and explanation
Correct option: C
2NO2→N2O42 {NO}_2 \rightarrow {N}_2 {O}_4
Option C is correct. In the reaction 2NO2 -> N2O4, the oxidation number of N does not change. It remains +4 on both sides of the equation.
401 Medium
401 of 705
Understanding +4 | -1
Which of the following is mixed anhydride?
View answer and explanation
Correct option: D
N2O4\mathrm{N}_2 \mathrm{O}_4
Option D (N2O4\mathrm{N}_2 \mathrm{O}_4) is correct because N2O4\mathrm{N}_2 \mathrm{O}_4 is a mixed anhydride as it gives a mixture of HNO2\mathrm{HNO}_2 and HNO3\mathrm{HNO}_3 upon hydrolysis.
402 Medium
402 of 705
Understanding +4 | -1
Which of the following is/are CORRECT from the given below?

(I) Brown ring test for nitrate depends on the ability of Fe2+{Fe}^{2+} to reduce nitrate to nitric oxide
(II) Red phosphorus is much less reactive than white phosphorus

(III) Red phosphorus readily catches fire in air to give white fumes

(IV) White phosphorus is a translucent white waxy solid
View answer and explanation
Correct option: D
I, II and IV
Option D (I, II and IV) is correct because statements I, II, and IV are correct regarding the brown ring test, reactivity of red phosphorus, and the nature of white phosphorus.
403 Easy
403 of 705
Analyzing +4 | -1
Of the following, which has three electron bond in its structure?
View answer and explanation
Correct option: B
Nitric oxide
Option B (Nitric oxide) is correct because nitric oxide (NO) has an odd number of electrons, resulting in a three-electron bond in its Lewis structure.
404 Medium
404 of 705
Understanding +4 | -1
NH3\mathrm{NH}_3 reacts with HCl\mathrm{HCl}. The correct reasons are

I. NH3\mathrm{NH}_3 can donate a pair of electrons.

II. The Cl−\mathrm{Cl}^{-} ion formed has a stable configuration.

III. The N\mathrm{N}-atom of NH3\mathrm{NH}_3 gains electrons.

IV. A proton in HCl\mathrm{HCl} can accept an electron pair from NH3\mathrm{NH}_3.

The correct option is
View answer and explanation
Correct option: B
I, II and IV
Option B (I, II and IV) is correct because ammonia donates an electron pair, chloride ion has a stable configuration, and a proton accepts the electron pair.
405 Medium
405 of 705
Understanding +4 | -1
A gas that cannot be collected over water is
View answer and explanation
Correct option: A
SO2S{O_2}
Option A (SO2S{O_2}) is correct because sulfur dioxide is highly soluble in water, reacting to form sulfurous acid, and therefore cannot be collected over water.
406 Medium
406 of 705
Remembering +4 | -1
When copper reacts with dilute nitric acid the gaseous product obtained is
View answer and explanation
Correct option: C
NO{NO}
Option C (NO{NO}) is correct because when copper reacts with dilute nitric acid, nitric oxide (NO{NO}) is the primary gaseous product.
407 Easy
407 of 705
Understanding +4 | -1
The following statements concern elements in the
periodic table. Which of the following is true?
View answer and explanation
Correct option: C
Elements of Group 16 have lower ionization
enthalpy values compared to those of Group
15 in the corresponding periods
Option C is correct because Group 16 elements have lower ionization enthalpy than Group 15 elements due to the extra stability of the half-filled p-orbitals in Group 15.
408 Easy
408 of 705
Remembering +4 | -1
Which inert gas has the highest boiling point?
View answer and explanation
Correct option: A
Xe\mathrm{Xe}
Option A (Xe) is correct because the boiling point of noble gases increases with increasing atomic mass due to stronger London dispersion forces, and Xenon (Xe) has the highest atomic mass among the options.
409 Easy
409 of 705
Understanding +4 | -1
NCl3\mathrm{NC}{\mathrm{l}_3} on hydrolysis yields:
View answer and explanation
Correct option: C
NH3  and HOCl\mathrm{N}{\mathrm{H}_\mathrm{3}}\mathrm{\;and}\,\mathrm{ HOCl}
Option C (NH3  and HOCl\mathrm{N}{\mathrm{H}_\mathrm{3}}\mathrm{\;and}\,\mathrm{ HOCl}) is correct because the hydrolysis of NCl3 results in the formation of ammonia and hypochlorous acid.
410 Medium
410 of 705
Analyzing +4 | -1
What will be the A{A} and B{B} in the following equations.

8NH3+3Cl2(A)→6NH4Cl+N2NH3+3Cl2(B)→NCl3+3HCl\begin{gathered} 8{{N}}{{{H}}_3} + \mathop {3{{C}}{{{l}}_2}}\limits_{(A)} \to 6{{N}}{{\text{H}}_4}{{Cl}} + {{{N}}_2} \quad \\ {{N}}{{{H}}_3} + \mathop {3{{C}}{{{l}}_2}}\limits_{(B)} \to {{NC}}{{{l}}_3} + 3{{HCl}} \quad \\ \end{gathered}
View answer and explanation
Correct option: A
A={A}= Excess, B={B}= Excess
Option B (N2O5\mathrm{N}_2 \mathrm{O}_5) is correct because when N₂O₅ is hydrolyzed, it forms only HNO₃, where nitrogen remains in the +5 oxidation state, thus no disproportionation occurs.
411 Easy
411 of 705
Remembering +4 | -1
P2O5{P_2}{O_5} is used extensively as a:
View answer and explanation
Correct option: A
Dehydrating agent
Option A (Dehydrating agent) is correct because P2O5{P_2}{O_5} has a strong affinity for water and is widely used to remove water from other substances.
412 Medium
412 of 705
Remembering +4 | -1
What is the correct order of occurrence (% by weight) in air of Ne, Ar and Kr?
View answer and explanation
Correct option: B
Ar>Ne>KrAr > Ne > Kr
Option B (Ar>Ne>KrAr > Ne > Kr) is correct because Argon has the highest percentage by weight in air, followed by Neon, and then Krypton.
413 Hard
413 of 705
Analyzing +4 | -1
Which of the following dissolves in water but does not give any oxyacid solution?
View answer and explanation
Correct option: B
OF2O{F_2}
Option B (OF2O{F_2}) is correct because OF2O{F_2} reacts with water to form HFHF and O2O_2, not an oxyacid.
414 Easy
414 of 705
Remembering +4 | -1
Ozonization of water is carried out to remove:
View answer and explanation
Correct option: A
Bacterial impurities
Option A (Bacterial impurities) is correct because ozonization is a water treatment process that effectively disinfects water by killing bacteria and viruses.
415 Medium
415 of 705
Analyzing +4 | -1
Among the compounds given in the options, identify the one which does not gives a precipitate with excess of both NaOH\mathrm{NaOH} solution and NH3\mathrm{NH}_3 solution?
View answer and explanation
Correct option: B
Zn(NO3)2\mathrm{Zn}\left(\mathrm{NO}_3\right)_2
The correct answer is B (Zn(NO₃)₂) because zinc hydroxide initially precipitates with both NaOH and NH₃, but then dissolves in excess due to the formation of soluble complexes.
416 Medium
416 of 705
Analyzing +4 | -1
Which of the following is not correct?
View answer and explanation
Correct option: D
π∗2px<π2py<σ2pz\pi *2px < \pi 2py < \sigma 2pz (no. of nodal plane)
Option D (π∗2px<π2py<σ2pz\pi *2px < \pi 2py < \sigma 2pz (no. of nodal plane)) is correct because the number of nodal planes increases in the order σ2pz\sigma 2pz (0 nodal planes) < π2py\pi 2py (1 nodal plane) < π∗2px\pi *2px (2 nodal planes).
417 Medium
417 of 705
Understanding +4 | -1
Assertion (A): Phosphoric acid has no reducing properties.

Reason (R): Phosphoric acid does not contain P−H{P}-{H} bonds.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because phosphoric acid does not have P-H bonds, which are responsible for reducing properties in other phosphorus acids.
418 Medium
418 of 705
Understanding +4 | -1
Observe the following statements
I. Bleaching powder is used in the preparation of chloroform.
II. Bleaching powder decomposes in the presence of CoCl2\mathrm{CoC}{\mathrm{l}_\mathrm{2}} to liberate O2{\mathrm{O}_2}
III. Aqueous KHF2\mathrm{KH}{\mathrm{F}_\mathrm{2}} is used in the preparation of fluorine.
View answer and explanation
Correct option: D
Only I and II are correct
Option C (Only I and III are correct) is incorrect, so Option D (Only I and II are correct) is correct because aqueous KHF2 is NOT used in the preparation of fluorine.
419 Hard
419 of 705
Understanding +4 | -1
Assertion: Ba(N3)2\mathrm{Ba}\left(\mathrm{N}_3\right)_2 give highly pure N2\mathrm{N}_2

Reason: Atomic weight of Ba\mathrm{Ba} is very high
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because both the assertion and the reason are true, but the reason doesn't explain why barium azide gives pure nitrogen.
420 Medium
420 of 705
Analyzing +4 | -1
Assertion (A): When H2S(g){H}_2 {S}({g}) is passed into aqueous ZnCl2{ZnCl}_2, solution, ZnS{ZnS} is precipitated.

Reason (R) : ZnS{ZnS} is insoluble in dil. HCl{HCl}.
View answer and explanation
Correct option: D
A\mathbf{A} and R\mathbf{R} both are false.
Option D (A\mathbf{A} and R\mathbf{R} both are false) is correct because ZnSZnS does not precipitate when H2SH_2S is passed into ZnCl2ZnCl_2 solution, and ZnSZnS is soluble in dilute HClHCl.
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