IB DP Chemistry - Structure 2.4 From models to materials - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question

Four stable allotropes of carbon are shown in diagrams labelled W, X, Y and Z. Which row correctly matches each structure?

OptionGrapheneFullereneGraphiteDiamond
AZYXW
BWZYX
CXWZY
DYXWZ

▶️ Answer/Explanation

Correct Answer: A

• Z shows a single sheet of hexagons → Graphene.
• Y is a spherical molecule → Fullerene.
• X has stacked layers → Graphite.
• W is a tetrahedral network → Diamond.

Question

What are the hybridizations of the atoms labelled 1, 2 and 3 in the following molecule?

M13/4/CHEMI/HPM/ENG/TZ1/13_01

Option123
Asp2sp2sp
Bsp3sp2sp3
Csp2spsp3
Dsp3sp2sp
▶️ Answer/Explanation

Correct option: B

• Atom 1 (CH3 carbon) forms four σ bonds ⇒ hybridization = sp3.
• Atom 2 (carbonyl carbon) forms three σ bonds and one π bond ⇒ trigonal planar ⇒ sp2.
• Atom 3 (–O–H oxygen) has two lone pairs and two σ bonds ⇒ four electron domains ⇒ sp3.

Question

Which allotropes of carbon show \(sp^2\) hybridization?
I.   Diamond
II.   Graphite
III.   \(C_{60}\) fullerene
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
▶️ Answer/Explanation
Detailed solution

Diamond has a tetrahedral structure where each carbon forms four single bonds → this is \(sp^3\) hybridization. Therefore, diamond is not \(sp^2\)-hybridized.

Graphite has layers of carbon atoms arranged in hexagonal rings. Each carbon is bonded to three others → this is classic \(sp^2\) hybridization.

\(C_{60}\) fullerene (buckminsterfullerene) also has carbon atoms bonded to three neighbors in a spherical structure → this is also \(sp^2\) hybridization.

Therefore, the allotropes showing \(sp^2\) hybridization are:
• Graphite
• \(C_{60}\) fullerene

✅ Answer: (C)

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