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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

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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 320
Understanding +4 | -1
Consider the following reaction :

Soda extract+ dilute HNO3+Cl2→CC4\mathrm{HNO}_3+\mathrm{Cl}_2 \xrightarrow{\mathrm{CC}_4} Brown layer. The element confirmed in this test can be:
View answer and explanation
Correct option: B
Br
Option B is correct because the reaction sequence described is a test for bromide ions, where chlorine oxidizes bromide to bromine, which forms a brown layer in CCl4CCl_4.
302 Medium
302 of 320
Analyzing +4 | -1
Among the following which is most likely unhybridized molecule?
View answer and explanation
Correct option: C
PH3{PH}_3
Option C (PH3) is correct because phosphorus in PH3 has a very small electronegativity difference with hydrogen, resulting in minimal hybridization.
303 Medium
303 of 320
Understanding +4 | -1
(X)←BoilH2OH4P2O7→ Strong heating (Y){\text{(X}})\xleftarrow[{{\text{Boil}}}]{{{{\text{H}}_2}{\text{O}}}}{{\text{H}}_4}{{\text{P}}_2}{{\text{O}}_7}\xrightarrow{{{\text{ Strong heating }}}}{\text{(Y)}}
In the sequence of reactions given above, compounds (X)(\mathrm{X}) and (Y)(\mathrm{Y}) are respectively
View answer and explanation
Correct option: D
H3PO4,P2O5\mathrm{H}_3 \mathrm{PO}_4, \mathrm{P}_2 \mathrm{O}_5
Option D is correct because H4P2O7H_4P_2O_7 reacts with water to form H3PO4H_3PO_4, and upon strong heating, it dehydrates to form P2O5P_2O_5.
304 Medium
304 of 320
Understanding +4 | -1
Which among the following is the most reactive?
View answer and explanation
Correct option: D
IClICl
Option D (ICl) is correct because interhalogen compounds like ICl are more reactive than homonuclear diatomic halogens due to the polarity of the bond between the different halogen atoms.
305 Medium
305 of 320
Remembering +4 | -1
Which one of the following reacts with glass?
View answer and explanation
Correct option: B
HF
Option B (HF) is correct because hydrofluoric acid reacts with silica in glass.
306 Medium
306 of 320
Understanding +4 | -1
Which of the following is INCORRECT?
View answer and explanation
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.
307 Medium
307 of 320
Remembering +4 | -1
The function of Fe(OH)3Fe{\left( {OH} \right)_3} in the contact process is
View answer and explanation
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.
308 Medium
308 of 320
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.
309 Medium
309 of 320
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.
310 Medium
310 of 320
Remembering +4 | -1
Antimony dissolves in aquaregia to give:
View answer and explanation
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}.
311 Medium
311 of 320
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.
312 Medium
312 of 320
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.
313 Medium
313 of 320
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:
View answer and explanation
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.
314 Medium
314 of 320
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.
315 Medium
315 of 320
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.
316 Medium
316 of 320
Understanding +4 | -1
Thermally most stable compound is
View answer and explanation
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.
317 Medium
317 of 320
Understanding +4 | -1
Which one is not an acid salt?
View answer and explanation
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.
318 Medium
318 of 320
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
View answer and explanation
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.
319 Medium
319 of 320
Remembering +4 | -1
A hybrid rocket propellant uses
View answer and explanation
Correct option: A
A liquid oxidizer and a solid fuel
The correct answer is A (A liquid oxidizer and a solid fuel) because hybrid rocket propellants typically use a liquid oxidizer and a solid fuel for combustion.
320 Medium
320 of 320
Understanding +4 | -1
HIHI reacts with HNO3HN{O_3} to form:
View answer and explanation
Correct option: D
NO2+I2N{O_2} + {I_2}
Option D is correct. HI is a strong reducing agent and reduces HNO3 to NO2, while HI itself gets oxidized to I2.
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