CAIE A-Level Chemistry A2 29.3 Shapes of Aromatic Organic Molecules and Bonds Questions

Practise explaining aromatic geometry through sp² hybridisation, σ bonding and sideways p-orbital overlap.

Exam points

state that benzene is a planar hexagonal molecule with 120° bond angles and sp²-hybridised carbon atoms
describe σ bonds as end-on overlap involving carbon sp² orbitals and hydrogen 1s orbitals where applicable
explain the delocalised π system as sideways overlap of parallel p orbitals above and below the ring

Covered syllabus points

29.3.1—Shape of benzene and other aromatic molecules

[Maximum number: 1]

Cumene is an aromatic hydrocarbon used in the synthesis of other useful chemicals.

cumene

Complete Table 5.1 to show the number of $\mathrm{sp}^{2}$ and $\mathrm{sp}^{3}$ hybridised carbon atoms that are present in a molecule of cumene.

Table 5.1

$\mathrm{sp}^{2}=6 / \mathrm{six} \mathrm{sp}^{3}=3 /$ three

[Maximum number: 2]

Ruthenium and osmium are transition metals below iron in Group 8 of the Periodic Table.

Fig. 4.1 shows another ruthenium complex.

Fig. 4.1

This complex contains the neutral ligand pyrazine.

Figure for Question 2 — CAIE A-Level Chemistry A2

Pyrazine is an aromatic compound. The bonding and structure of pyrazine is similar to that of benzene.

Describe and explain the shape of pyrazine.
In your answer, include:
- the hybridisation of the nitrogen and carbon atoms
- how orbital overlap forms $\pi$ bonds between the atoms in the ring.

shape is (hexagonal ring) planar / (trigonal) planar / $120^{\circ}$

- carbons and nitrogens are $\mathrm{sp}^{2}$ hybridised

- a p orbital (from each atom) overlaps sideways/laterally (with each other above and below the ring forming $\pi$ bonds)

mark as •

[Maximum number: 4]

Describe the structure of a benzene molecule, $\mathrm{C}_{6} \mathrm{H}_{6}$ .
Your answer should include:
- the shape of the molecule
- the relative lengths of the C-C bonds
- bond angles
- the hybridisation of the carbon atoms
- the overlap between orbitals that produces each type of bond present.

Any four from the following points:

- (regular) hexagon OR planar

- all C-C bonds same length [1]

- all bond angles $120^{\circ}$

- all carbon atoms $\mathrm{sp}^{2}$ hybridised [1]

- C-H bonds are $\mathrm{s}-\mathrm{sp}^{2}$ overlap [1]

- C-C bonds have $\mathrm{sp}^{2}-\mathrm{sp}^{2}$ overlap [1]

- C-C bonds have p-p overlap

- $\pi$ used correctly and $\sigma$ used correctly once each

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Covered syllabus points

Question 1

Question 2

Question 3