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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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661 Medium
661 of 705
Remembering +4 | -1
Which one of the following reacts with glass?
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Correct option: B
HF
Option B (HF) is correct because hydrofluoric acid reacts with silica in glass.
662 Easy
662 of 705
Remembering +4 | -1
Ordinary oxygen contains:
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Correct option: C
A  mixture  of  O16,  O17  and  O18  isotopes\mathrm{A\;mixture\;of\;}{\mathrm{O}^{16}},\mathrm{\;}{\mathrm{O}^{17}}\mathrm{\;and\;}{\mathrm{O}^{18}}\mathrm{\;isotopes}
Option C is correct because ordinary oxygen exists as a mixture of isotopes O-16, O-17, and O-18.
663 Easy
663 of 705
Analyzing +4 | -1
A gas does not turn lime water milky, supports the combustion of burning magnesium. It has no smell and is colourless. It extinguishes a glowing splint but under some circumstances reacts with oxygen and hydrogen. It is not poisonous. The gas is likely to be :
View answer and explanation
Correct option: B
Nitrogen
Option B (Nitrogen) is correct because nitrogen is a colorless, odorless gas that does not turn lime water milky, supports the combustion of magnesium, extinguishes a glowing splint, and can react with oxygen and hydrogen under certain conditions.
664 Medium
664 of 705
Understanding +4 | -1
Which of the following is INCORRECT?
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Correct option: D
Sb2O3……{Sb}_2{O}_3 \ldots \ldots Purely Basic Oxide
Option D (Sb2O3……{Sb}_2{O}_3 \ldots \ldots Purely Basic Oxide) is correct because Sb2O3{Sb}_2{O}_3 is amphoteric, not purely basic.
665 Easy
665 of 705
Remembering +4 | -1
Concentrated nitric acid on heating decomposes to give:
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Correct option: D
NO2N{O_2} and O2{O_2}
The correct answer is D (NO₂ and O₂) because concentrated nitric acid decomposes upon heating to produce nitrogen dioxide (NO₂), water, and oxygen (O₂).
666 Easy
666 of 705
Understanding +4 | -1
Assertion (A) : In aqueous solution, SO2\mathrm{SO}_2 reacts with H2S\mathrm{H}_2 \mathrm{S} forming sulphur.

Reason (R): SO2\mathrm{SO}_2 is an effective reducing agent.
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 is correct because both the assertion and reason are true, but the reason does not explain the assertion.
667 Medium
667 of 705
Remembering +4 | -1
The function of Fe(OH)3Fe{\left( {OH} \right)_3} in the contact process is
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Correct option: A
To remove arsenic impurity
Option A (To remove arsenic impurity) is correct because Fe(OH)3Fe(OH)_3 is used to adsorb and remove arsenic impurities in the contact process for sulfuric acid production.
668 Medium
668 of 705
Analyzing +4 | -1
Assertion: H2O\mathrm{H}_2 \mathrm{O} is a liquid while H2 S\mathrm{H}_2 \mathrm{~S} is a gas

Reason: There is stronger intermolecular hydrogen bonding in H2 S\mathrm{H}_2 \mathrm{~S}.
View answer and explanation
Correct option: C
Assertion is correct, but reason is wrong statement.
Option C (Assertion is correct, but reason is wrong statement) is correct because water is a liquid due to strong intermolecular hydrogen bonding, while hydrogen sulfide is a gas due to weaker van der Waals forces.
669 Easy
669 of 705
Remembering +4 | -1
In Birkeland and Eyde process, the temperature of the electric arc is about:
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Correct option: C
3000∘C{3000^ \circ }C
Option C (3000∘C{3000^ \circ }C) is correct because the Birkeland-Eyde process requires a very high temperature (around 3000°C) to dissociate nitrogen molecules.
670 Easy
670 of 705
Remembering +4 | -1
Oleum is
View answer and explanation
Correct option: C
Fuming H2SO4{H_2}S{O_4}
Option C (Fuming H2SO4{H_2}S{O_4}) is correct because Oleum is a solution of sulfur trioxide in sulfuric acid, also known as fuming sulfuric acid.
671 Hard
671 of 705
Analyzing +4 | -1
Assertion: Fluorine oxidises H2O\mathrm{H}_2 \mathrm{O} to O2\mathrm{O}_2 but Cl2\mathrm{Cl}_2 does not.

Reason: Electronegativity of fluorine is greater than oxygen but EN of chlorine is lesser than oxygen.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option D is correct because the assertion is incorrect (the hydrogen atoms are not in one plane), but the reason is correct (the carbon atoms are sp2 and sp hybridized).
672 Easy
672 of 705
Remembering +4 | -1
Noble gases are:
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Correct option: D
All of the above
Option D (All of the above) is correct because noble gases are indeed colorless, odorless, tasteless, and non-inflammable.
673 Medium
673 of 705
Analyzing +4 | -1
Assertion (A) : Structure of SO42−\mathrm{SO}_4^{2-} is identical for S2O32−\mathrm{S}_2 \mathrm{O}_3^{2-}.

Reason (R) : Sulphur atoms in S2O32−\mathrm{S}_2 \mathrm{O}_3^{2-} are in different oxidation states.
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 is correct. Both the assertion and reason are true, but the reason does not explain why the structures are identical. The identical structure arises from similar bonding and hybridization.
674 Medium
674 of 705
Remembering +4 | -1
Antimony dissolves in aquaregia to give:
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Correct option: C
SbCl5\mathrm{SbC}{\mathrm{l}_5}
Option C (SbCl5\mathrm{SbC}{\mathrm{l}_5}) is correct because Antimony dissolves in aquaregia to give SbCl5\mathrm{SbC}{\mathrm{l}_5}.
675 Medium
675 of 705
Analyzing +4 | -1
Assertion (A): NO3−{NO}_3^{-} is planar while NH3{NH}_3 is pyramidal.

Reason (R): N{N} in NO3−{NO}_3^{-} is sp2{sp}^2 hybridized but in NH3{NH}_3 it is sp3{sp}^3-hybridized.
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 the planar structure of NO3−{}_3^{-} is due to sp2{}^2 hybridization, while the pyramidal structure of NH3{}_3 is due to sp3{}^3 hybridization.
676 Medium
676 of 705
Understanding +4 | -1
Assertion (A): As to Bi\mathrm{Bi}, there is only a small increase in covalent radius.

Reason (R): dd and/or ff-orbitals are completely filled in heavier elements.
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 (Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is the correct explanation of AA) is correct because the small increase in covalent radius from As to Bi is due to the poor shielding effect of the filled d and f orbitals in Bi.
677 Medium
677 of 705
Remembering +4 | -1
On passing ammonia gas through a reagent ' xx ', brown precipitate is observed that confirms the presence of NH4+\mathrm{NH}_4^{+}ion. The reagent ' x\mathrm{x} ' is:
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Correct option: A
Nessler's reagent
Option A (Nessler's reagent) is correct because Nessler's reagent reacts with ammonia gas to form a brown precipitate, confirming the presence of NH4+ ions.
678 Hard
678 of 705
Remembering +4 | -1
The compound is used in the estimation of carbon monoxide. Here, AA refers to
View answer and explanation
Correct option: A
I2O5\mathrm{I}_2 \mathrm{O}_5
Option A (I2O5\mathrm{I}_2 \mathrm{O}_5) is correct because iodine pentoxide (I2O5) is used to estimate carbon monoxide by oxidizing it to carbon dioxide.
679 Easy
679 of 705
Analyzing +4 | -1
Which of the following statements regarding properties of halogens are correct?

(i) Due to small size electron gain enthalpy of fluorine is less than that of chlorine.

(ii) Iodine has same physical state but different colour as compare to other members of the group.

(iii) Fluorine shows no positive oxidation state.

(iv) In X2(g)+H2O(l)→HX(aq)+HOX(aq)X_2(g)+H_2O(l) \to HX (aq) + HOX (aq)

(where X2=ClX_2=Cl or Br)Br)

(v) F2F_2 is the strongest oxidising halogen.
View answer and explanation
Correct option: B
(i), (iii), (iv) and (v)
Option B ((i), (iii), (iv) and (v)) is correct because it accurately describes the properties of halogens, including the electron gain enthalpy of fluorine, the absence of positive oxidation states for fluorine, the reaction of halogens with water, and the oxidizing strength of fluorine.
680 Easy
680 of 705
Understanding +4 | -1
Which one is most stable to heat -
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Correct option: D
HClO4{HClO}_4
Option D is correct because as the oxidation state of the central halogen atom increases, the halogen-oxygen bond becomes more covalent, increasing thermal stability.
681 Medium
681 of 705
Analyzing +4 | -1
Statement I: The coordination number of ' NN ' is seldom, if ever 5 where as that of PP, which is in the same group, is 5 in a large number of compounds

Statement II: The first IE of nitrogen is more than that of P\mathrm{P}
View answer and explanation
Correct option: B
Statement I and statement II are correct but statement II is not the correct explanation of statement I
Option B (Statement I and statement II are correct but statement II is not the correct explanation of statement I) is correct because both statements are factually true, but the ionization energy of nitrogen does not explain its lower coordination number compared to phosphorus.
682 Easy
682 of 705
Remembering +4 | -1
Ammonium salts are oxidized in the soil to nitrites by:
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Correct option: B
Nitrifying bacteria
Option B (Nitrifying bacteria) is correct because nitrifying bacteria are responsible for the oxidation of ammonium salts to nitrites in the soil.
683 Medium
683 of 705
Analyzing +4 | -1
In the reaction, X+I2+2HCl→SnCl4+2HIX + I_2 + 2HCl \to {SnCl}_4 + 2HI

The correct option regarding XX is/are
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because SnCl2 is a strong reducing agent, an angular molecule, and used as a reagent in the test of Hg2+ radical.
684 Easy
684 of 705
Understanding +4 | -1
The noble gases are unreactive because they:
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Correct option: D
Have stable electronic configuration or closed valency shell
Option D (Have stable electronic configuration or closed valency shell) is correct because noble gases are unreactive due to their complete octet, resulting in a stable electronic configuration.
685 Easy
685 of 705
Analyzing +4 | -1
An inorganic salt (A) is decomposed on heating to give two products (B) and (C). Compound (C) is a liquid at room temperature and is neutral to litmus while the compound (B) is a colourless neutral gas. Compounds (A), (B) and (C) are
View answer and explanation
Correct option: A
NH4NO3,N2O,H2O{NH}_4{NO}_3,{N}_2{O},{H}_2{O}
Option A (NH4NO3,N2O,H2O{NH}_4{NO}_3,{N}_2{O},{H}_2{O}) is correct because ammonium nitrate decomposes into dinitrogen monoxide and water, matching the description.
686 Medium
686 of 705
Understanding +4 | -1
Thermally most stable compound is
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Correct option: A
HClO4\mathrm{HClO}_4
Option A (HClO4\mathrm{HClO}_4) is correct because the thermal stability of oxyacids of chlorine increases with the oxidation state of chlorine; HClO4\mathrm{HClO}_4 has chlorine in the +7 oxidation state, making it the most stable.
687 Easy
687 of 705
Remembering +4 | -1
First compound of intert gases was prepared by scientist Neil Barthleta in 1962. This compound is
View answer and explanation
Correct option: A
XePtF6{XePtF}_6
Option A (XePtF6{XePtF}_6) is correct because the first noble gas compound prepared by Neil Bartlett in 1962 was Xenon hexafluoroplatinate (XePtF6{XePtF}_6).
688 Medium
688 of 705
Understanding +4 | -1
Which one is not an acid salt?
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Correct option: A
NaH2PO2Na{H_2}P{O_2}
Option A (NaH2PO2Na{H_2}P{O_2}) is correct because H3PO2H_3PO_2 is a monobasic acid, meaning it only has one replaceable hydrogen atom, and thus cannot form an acid salt with two H+H^+ ions.
689 Medium
689 of 705
Understanding +4 | -1
Choose the correct statements about the hydrides of group 15 elements.

A. The stability of the hydrides decreases in the order NH3>PH3>AsH3>SbH3>BiH3\mathrm{NH}_3 > \mathrm{PH}_3 > \mathrm{AsH}_3 > \mathrm{SbH}_3 > \mathrm{BiH}_3

B. The reducing ability of the hydride increases in the order NH3<PH3<AsH3<SbH3<BiH3\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3

C. Among the hydrides, NH3\mathrm{NH}_3 is strong reducing agent while BiH3\mathrm{BiH}_3 is mild reducing agent.

D. The basicity of the hydrides increases in the order

NH3<PH3<AsH3<SbH3<BiH3\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3

Choose the most appropriate from the option given below
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Correct option: B
A and B only
Option B (A and B only) is correct because the stability and reducing ability trends are accurate, while basicity decreases down the group and NH3 is not a strong reducing agent.
690 Easy
690 of 705
Remembering +4 | -1
Bone black is polymorphic form of
View answer and explanation
Correct option: A
Phosphorus
Option A (Phosphorus) is correct because bone black is a polymorphic form of phosphorus.
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