The p-Block Elements

Class 12

21 previous-year questions from this chapter every option and the correct answer, free · where the marks are

1. Group 13 and 14 elements

Exam focus: The inert pair effect makes the lower oxidation state more stable going down groups 13 and 14 — Tl⁺ is more stable than Tl³⁺, Pb²⁺ more stable than Pb⁴⁺ — the reverse of what "higher group, higher favoured state" intuition suggests. Diborane is the other signature trap: B₂H₆ has too few electrons for conventional 2-electron bonds at every position, forcing two 3-centre-2-electron bridge bonds at the bridging hydrogens.

p-Block Elements: Group 13 and 14 — Boron and Carbon Families

Vedantu NEET English

Covers Group 13 (boron family) and Group 14 (carbon family) NCERT line by line for NEET, including the anomalous behaviour of the first member and important compounds — a direct, on-target match.

ChemistryNEET 2023show ▾

Taking stability as the factor, which one of the following represents the correct relationship?

A. $TlC l_{3} > TlCl$
B. $In I_{3} > InI$
C. $AlCl > AlC l_{3}$
D. $TlI > Tl I_{3}$ ✓

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ChemistryNEET 2023show ▾

Match List I with List II.

List I	List II
A. Coke	I. Carbon atoms are $s p^{3}$ hybridised
B. Diamond	II. Used as a dry lubricant
C. Fullerene	III. Used as a reducing agent
D. Graphite	IV. Cage-like molecules

Choose the correct answer from the options given below:

A.A-II, B-IV, C-I, D-III
B.A-IV, B-I, C-II, D-III
C.A-III, B-I, C-IV, D-II✓
D.A-III, B-IV, C-I, D-II

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ChemistryNEET 2022show ▾

Choose the correct statement:

A.Both diamond and graphite are used as dry lubricants
B.Diamond and graphite have two-dimensional networks
C.Diamond is covalent and graphite is ionic
D.Diamond is $s p^{3}$ hybridised and graphite is $s p^{2}$ hybridised✓

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ChemistryNEET 2022show ▾

Which of the following statements is not correct about diborane?

A.Both the boron atoms are $s p^{2}$ hybridised✓
B.There are two 3-centre-2-electron bonds
C.The four terminal B–H bonds are two-centre two-electron bonds
D.The four terminal hydrogen atoms and the two boron atoms lie in one plane

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ChemistryNEET 2021show ▾

B F_{3}

is a planar and electron-deficient compound. The hybridisation and number of electrons around the central atom, respectively, are:

A. $s p^{3}$ and 4
B. $s p^{3}$ and 6
C. $s p^{2}$ and 6✓
D. $s p^{2}$ and 8

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ChemistryNEET 2019show ▾

Which of the following species is not stable?

A. $[Si F_{6}]^{2 -}$
B. $[GeC l_{6}]^{2 -}$
C. $[Sn (OH)_{6}]^{2 -}$
D. $[SiC l_{6}]^{2 -}$ ✓

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ChemistryNEET 2019show ▾

Which of the following is an incorrect statement?

A. $Pb F_{4}$ is covalent in nature✓
B. $SiC l_{4}$ is easily hydrolysed
C. $Ge X_{4}$ (X = F, Cl, Br, I) is more stable than $Ge X_{2}$
D. $Sn F_{4}$ is ionic in nature

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2. Group 15 and 16: nitrogen, oxygen families

Exam focus: N₂'s triple bond makes it exceptionally unreactive, while P4 in the same group is so reactive it's stored underwater — a contrast NEET uses to test whether "same group, similar reactivity" has been over-applied. Catenation is actually stronger for sulfur than oxygen despite O sitting above S, because lone-pair repulsion between the small oxygen atoms weakens the O–O single bond.

p-Block Elements: Group 15 and 16 — Nitrogen and Oxygen Families

Vedantu JEE

A focused lecture specifically on Group 15 (pnictogens) and Group 16 (chalcogens) trends with worked numericals — a tight, direct match for this exact subtopic rather than a bundled chapter video.

ChemistryNEET 2025show ▾

Given below are two statements: Statement I: Like nitrogen, which can form ammonia, arsenic can form arsine. Statement II: Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below:

A.Both Statement I and Statement II are correct
B.Both Statement I and Statement II are incorrect
C.Statement I is correct but Statement II is incorrect✓
D.Statement I is incorrect but Statement II is correct

Solution

Arsenic forms arsine, $As H_{3}$ (I correct). Antimony does form the pentoxide $S b_{2} O_{5}$ (II incorrect).

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ChemistryNEET 2024show ▾

Among Group 16 elements, which one does not show the –2 oxidation state?

A.Se
B.Te
C.Po✓
D.O

Solution

Polonium, the heaviest and most metallic member, does not show the −2 state; O, S, Se and Te all form −2 ions or hydrides.

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ChemistryNEET 2024show ▾

Given below are two statements: Statement I: The boiling points of the hydrides of Group 16 elements follow the order

H_{2} O > H_{2} Te > H_{2} Se > H_{2} S

. Statement II: On the basis of molecular mass,

H_{2} O

is expected to have a lower boiling point than the other members of the group, but due to the presence of extensive H-bonding in

H_{2} O

, it has a higher boiling point. In the light of the above statements, choose the correct answer from the options given below:

A.Both Statement I and Statement II are false
B.Statement I is true but Statement II is false
C.Statement I is false but Statement II is true
D.Both Statement I and Statement II are true✓

Solution

Boiling points rise with molar mass down the group ( $H_{2} S < H_{2} Se < H_{2} Te$ ), but water sits above all of them because of extensive hydrogen bonding. Both statements are true.

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ChemistryNEET 2023show ▾

Match List I with List II.

List I (Oxoacids of sulphur)	List II (Bonds)
A. Peroxodisulphuric acid	I. Two S–OH, four S=O, one S–O–S
B. Sulphuric acid	II. Two S–OH, one S=O
C. Pyrosulphuric acid	III. Two S–OH, four S=O, one S–O–O–S
D. Sulphurous acid	IV. Two S–OH, two S=O

Choose the correct answer from the options given below:

A.A-I, B-III, C-II, D-IV
B.A-III, B-IV, C-I, D-II✓
C.A-I, B-III, C-IV, D-II
D.A-III, B-IV, C-II, D-I

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ChemistryNEET 2021show ▾

In which one of the following arrangements is the given sequence not strictly according to the property indicated against it?

A.HF < HCl < HBr < HI : increasing acidic strength
B. $H_{2} O < H_{2} S < H_{2} Se < H_{2} Te$ : increasing $p K_{a}$ values✓
C. $N H_{3} < P H_{3} < As H_{3} < Sb H_{3}$ : increasing acidic character
D. $C O_{2} < Si O_{2} < Sn O_{2} < Pb O_{2}$ : increasing oxidising power

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ChemistryNEET 2020show ▾

Which of the following oxoacids of sulphur has a

- O - O -

linkage?

A. $H_{2} S O_{3}$ , sulphurous acid
B. $H_{2} S O_{4}$ , sulphuric acid
C. $H_{2} S_{2} O_{8}$ , peroxodisulphuric acid✓
D. $H_{2} S_{2} O_{7}$ , pyrosulphuric acid

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ChemistryNEET 2019show ▾

Which is the correct thermal stability order for

H_{2} E

(E = O, S, Se, Te, Po)?

A. $H_{2} S < H_{2} O < H_{2} Se < H_{2} Te < H_{2} Po$
B. $H_{2} O < H_{2} S < H_{2} Se < H_{2} Te < H_{2} Po$
C. $H_{2} Po < H_{2} Te < H_{2} Se < H_{2} S < H_{2} O$ ✓
D. $H_{2} Se < H_{2} Te < H_{2} Po < H_{2} O < H_{2} S$

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ChemistryNEET 2019show ▾

Match the following industrial processes with their principal product, and choose the correct option:

Column I	Column II
(a) Pure nitrogen	(i) Chlorine
(b) Haber process	(ii) Sulphuric acid
(c) Contact process	(iii) Ammonia
(d) Deacon's process	(iv) Sodium/barium azide

A.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
B.(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
C.(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
D.(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)✓

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3. Group 17 and 18: halogens, noble gases

Exam focus: F–F bond dissociation enthalpy is anomalously low, even lower than Cl–Cl, because the small fluorine atoms pack their lone pairs close enough to repel each other strongly — bond strength doesn't track atomic size monotonically down this group. Among noble gases, only xenon forms an established range of compounds; helium, neon and argon's very high ionisation enthalpy keeps them essentially compound-free, a fact often tested as assertion-reason.

Halogens and Noble Gases (Group 17 and 18)

Unacademy NEET

Single lecture covering both group 17 (halogens) and group 18 (noble gases). Not a primary-preferred channel, but a well-established national NEET coaching brand.

ChemistryNEET 2025show ▾

Match List I with List II.

List I	List II
A. $Xe O_{3}$	I. $s p^{3} d$ ; linear
B. $Xe F_{2}$	II. $s p^{3}$ ; pyramidal
C. $XeO F_{4}$	III. $s p^{3} d^{3}$ ; distorted octahedral
D. $Xe F_{6}$	IV. $s p^{3} d^{2}$ ; square pyramidal

Choose the correct answer from the options given below:

A.A-II, B-I, C-IV, D-III✓
B.A-II, B-I, C-III, D-IV
C.A-IV, B-II, C-III, D-I
D.A-IV, B-II, C-I, D-III

Solution

$Xe O_{3}$ : $s p^{3}$ with one lone pair, pyramidal (II). $Xe F_{2}$ : $s p^{3} d$ with three lone pairs, linear (I). $XeO F_{4}$ : $s p^{3} d^{2}$ with one lone pair, square pyramidal (IV). $Xe F_{6}$ : $s p^{3} d^{3}$ , distorted octahedral (III).

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ChemistryNEET 2023show ▾

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Helium is used to dilute oxygen in diving apparatus. Reason R: Helium has high solubility in

O_{2}

. In the light of the above statements, choose the correct answer from the options given below:

A.Both A and R are true and R is the correct explanation of A
B.Both A and R are true and R is NOT the correct explanation of A
C.A is true but R is false✓
D.A is false but R is true

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ChemistryNEET 2022show ▾

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): ICl is more reactive than

I_{2}

. Reason (R): The I–Cl bond is weaker than the I–I bond. In the light of the above statements, choose the most appropriate answer from the options given below:

A.(A) is not correct but (R) is correct
B.Both (A) and (R) are correct and (R) is the correct explanation of (A)✓
C.Both (A) and (R) are correct but (R) is not the correct explanation of (A)
D.(A) is correct but (R) is not correct

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ChemistryNEET 2021show ▾

Noble gases are so named because of their inertness towards reactivity. Identify an incorrect statement about them.

A.Noble gases are sparingly soluble in water
B.Noble gases have very high melting and boiling points✓
C.Noble gases have weak dispersion forces
D.Noble gases have large positive values of electron gain enthalpy

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ChemistryNEET 2021show ▾

Statement I: Acid strength increases in the order HF ≪ HCl ≪ HBr ≪ HI. Statement II: As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases. In the light of the above statements, choose the correct answer from the options given below:

A.Both Statement I and Statement II are true✓
B.Both Statement I and Statement II are false
C.Statement I is correct but Statement II is false
D.Statement I is incorrect but Statement II is true

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ChemistryNEET 2019show ▾

Match the xenon compounds in Column I with their molecular geometry in Column II:

Column I	Column II
(a) $Xe F_{4}$	(i) Pyramidal
(b) $Xe F_{6}$	(ii) Square planar
(c) $XeO F_{4}$	(iii) Distorted octahedral
(d) $Xe O_{3}$	(iv) Square pyramidal

A.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
B.(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)✓
C.(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
D.(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

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