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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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Showing 121-150 of 320 questions

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121 Medium
121 of 320
Understanding +4 | -1
Assertion: Both the S−O\mathrm{S}-\mathrm{O} bonds in sulphur dioxide are equal.

Reason: Sulphur dioxide molecule exhibits resonance.
View answer and explanation
Correct option: A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
Option A (Both assertion and reason are correct statements, and reason is the correct explanation of the assertion) is correct because the equal bond lengths in SO₂ are due to resonance, which delocalizes the electron density.
122 Medium
122 of 320
Remembering +4 | -1
Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?
View answer and explanation
Correct option: C
Nitric acid
Option C (Nitric acid) is correct because nitric acid is added to the sodium extract to decompose NaCN and Na2S, which would otherwise interfere with the silver nitrate test for halogens.
123 Medium
123 of 320
Analyzing +4 | -1
Which of the following statements are correct?

(i) Arsenic and antimony are metalloids.

(ii) Phosphorus, arsenic and antimony are found mainly as sulphide minerals.

(iii) Covalent redii increases equally from N{N} to Bi.

(iv) Elements of group 15 have extra stability and higher ionisation energy due to exactly half filled ns2np3n s^2 n p^3 electronic configuration.

(v) In group 15 elements only nitrogen is gas whereas all others are solids.
View answer and explanation
Correct option: A
(i), (iv) and (v)
Option A ((i), (iv) and (v)) is correct because arsenic and antimony are metalloids, group 15 elements have extra stability due to half-filled ns2np3ns^2np^3 configuration, and only nitrogen is a gas in group 15.
124 Medium
124 of 320
Analyzing +4 | -1
Which of the following statements are incorrect for XeOF4?{XeOF}_4?
A. It has a square planar structure
B. It has a square pyramidal structure
C. It has pentagonal bipyramidal structure
D. It has 2 lone pair of electrons around Xenon
View answer and explanation
Correct option: C
ACD
Option A is correct because in the given reaction, chlorine undergoes both oxidation and reduction, which defines a disproportionation reaction.
125 Medium
125 of 320
Analyzing +4 | -1
Which of the following chemical reaction does not represents the oxidising behaviour of H2SO4\mathrm{H}_2 \mathrm{SO}_4 ?
View answer and explanation
Correct option: C
NaCl+H2SO4→NaHSO4+HCl\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{HCl}
Option C is correct because the reaction of NaCl with H2SO4 to produce NaHSO4 and HCl does not involve a change in the oxidation state of sulfur, hence it doesn't represent the oxidizing behavior of H2SO4.
126 Medium
126 of 320
Analyzing +4 | -1
The type of hybridisation and number of lone pairs
of electrons on XeXe in XeOF4,XeO{F_4}, respectively, are :
View answer and explanation
Correct option: C
sp3d2s{p^3}{d^2} and 1
Option C (sp3d2s{p^3}{d^2} and 1) is correct because XeOF4XeOF_4 has a square pyramidal geometry with sp3d2s{p^3}{d^2} hybridization and one lone pair on Xe.
127 Medium
127 of 320
Analyzing +4 | -1
Assertion (A): NH3\mathrm{NH}_3 is more polar than NF3.\mathrm{NF}_3.

Reason (R): NF3\mathrm{NF}_3 cannot be hydrolysed.
View answer and explanation
Correct option: B
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
Option B is correct because both the assertion and reason are true, but the reason does not explain the assertion. NH3 is more polar than NF3 due to the direction of dipole moments, and NF3 is difficult to hydrolyze.
128 Medium
128 of 320
Analyzing +4 | -1
Which of the following pair of molecules does not have identical structure?
View answer and explanation
Correct option: C
BF3,ICl3{BF}_3, {ICl}_3
Option C (BF3,ICl3{BF}_3, {ICl}_3) is correct because BF3{BF}_3 is trigonal planar, while ICl3{ICl}_3 is T-shaped.
129 Medium
129 of 320
Remembering +4 | -1
In ordinary Cl2C{l_2} gas Cl35C{l^{35}} and Cl37C{l^{37}} are in the ratio:
View answer and explanation
Correct option: B
0.125694444
Option B (03:01) is correct because the natural abundance ratio of Chlorine-35 to Chlorine-37 is approximately 3:1, which results in the observed atomic weight of 35.5 for chlorine.
130 Medium
130 of 320
Analyzing +4 | -1
Assertion : HOF{HOF} bond angle in HFO{HFO} is higher than HOCl{HOCl} bond angle in HClO{HClO}.

Reason : Oxygen is more electronegative than halogens.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D (If both the Assertion and Reason are incorrect) is correct because the bond angle in HOF is smaller than in HOCl, and oxygen is not more electronegative than fluorine.
131 Medium
131 of 320
Remembering +4 | -1
The solution of phosphine in water
decomposes in the presence of light
giving
View answer and explanation
Correct option: C
Red phosphorous and hydrogen
Option C (Red phosphorous and hydrogen) is correct because phosphine decomposes in the presence of light to form red phosphorus and hydrogen gas.
132 Medium
132 of 320
Understanding +4 | -1
Which of the following(s) give(s) coloured gases?
View answer and explanation
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).
133 Medium
133 of 320
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.
134 Medium
134 of 320
Remembering +4 | -1
Ozone reacts with dry iodine to give:
View answer and explanation
Correct option: D
I4O9{I_4}{O_9}
Option D (I4O9{I_4}{O_9}) is correct because ozone reacts with dry iodine to produce iodine nonaoxide (I4O9{I_4}{O_9}).
135 Medium
135 of 320
Analyzing +4 | -1
In which of the following sets the order of boiling points are incorrect?
View answer and explanation
Correct option: B
HBr>HCl>HF\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.
136 Medium
136 of 320
Analyzing +4 | -1
Assertion: H2Se\mathrm{H}_2 \mathrm{Se} is more acidic than H2Te\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.
137 Medium
137 of 320
Understanding +4 | -1
Statement I: H2SO4\mathrm{H}_2 \mathrm{SO}_4 forms only one series of salts

Statement II: The molecule of H2SO4\mathrm{H}_2 \mathrm{SO}_4 consists of two OH\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 H2SO4\mathrm{H}_2\mathrm{SO}_4 forms two series of salts (normal and acid salts), and its molecule has two OH groups.
138 Medium
138 of 320
Analyzing +4 | -1
Which one of the following arrangement does not accurately depict the trend mentioned against it?
View answer and explanation
Correct option: D
I2>Br2>Cl2>F2 (oxidizing power)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.
139 Medium
139 of 320
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
H2O{H_2}O
Option A (H2O{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.
140 Medium
140 of 320
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.
141 Medium
141 of 320
Analyzing +4 | -1
An orange solid (X)(X) on heating gives a colourless gas (y) and only a green residue (z). Gas (y) on treatment with Mg{Mg} produces a white solid substance:-
View answer and explanation
Correct option: A
Mg3N2{Mg}_3 {N}_2
The correct answer is A (Mg3N2{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.
142 Medium
142 of 320
Analyzing +4 | -1
Which of the following will have the least bond angle?
View answer and explanation
Correct option: C
PH3\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.
143 Medium
143 of 320
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.
144 Medium
144 of 320
Analyzing +4 | -1
[X]+H2SO4→[Y]\left[ X \right] + {H_2}S{O_4} \to \left[ Y \right] a colourless gas with irritating smell. [Y]+K2Cr2O7+H2SO4→\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
SO32−,SO2SO_3^{2 - },S{O_2}
Option A (SO32−,SO2SO_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.
145 Medium
145 of 320
Understanding +4 | -1
Which ion cannot be precipitated from water?
View answer and explanation
Correct option: B
NO3−\mathrm{NO}_3^ -
Option B (NO3−\mathrm{NO}_3^ -) is correct because nitrates of all metals are soluble in water and thus cannot be precipitated.
146 Medium
146 of 320
Remembering +4 | -1
In Haber's process for the manufacture of NH3{NH}_3 :
View answer and explanation
Correct option: B
Finely divided iron is used as a catalyst
Option B (Finely divided iron is used as a catalyst) is correct because iron is used as a catalyst in Haber's process to increase the rate of ammonia production.
147 Medium
147 of 320
Remembering +4 | -1
Which of the following elements is radioactive?
View answer and explanation
Correct option: C
Polonium
Option C (Polonium) is correct because Polonium is a radioactive element.
148 Medium
148 of 320
Analyzing +4 | -1
In which molecule hybrid orbital have only 20%20 \% d- character?
View answer and explanation
Correct option: C
SF4\mathrm{SF}_4
The correct answer is C (SF4) because SF4 has sp3d hybridization, which results in 20% d-character.
149 Medium
149 of 320
Understanding +4 | -1
Which of the following is not a monobasic acid?
View answer and explanation
Correct option: C
H2 S2O3\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_3
Option C (H2 S2O3\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_3) is not a monobasic acid because it has two ionizable hydrogen atoms, making it a dibasic acid.
150 Medium
150 of 320
Application +4 | -1
The number of molecules of water needed to convert one molecule of P2O5  {P_2}{O_{5\;}} into orthophosphoric acid is:
View answer and explanation
Correct option: B
3
Option B (3) is correct because 3 water molecules are required to convert one molecule of P2O5 into two molecules of orthophosphoric acid.
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