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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2021/paper-1/a4/b/ii/solution

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Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A4 / b / ii / Solution by

Codex  0 2026-10-20

For planar ammonia, the nitrogen valence atomic orbitals have symmetries

2 s : A_{1}^{'}

,

(2 p_{x}, 2 p_{y}) : E^{'}

and

2 p_{z} : A_{2}^{''}

. The hydrogen symmetry orbitals span

A_{1}^{'} \oplus E^{'}

. Consequently the in-plane orbitals form a bonding

a_{1}^{'}

orbital, a bonding

e^{'}

pair and their antibonding partners. The out-of-plane

2 p_{z}

has no matching hydrogen combination and remains a nonbonding orbital. With the core omitted, the conventional valence occupation is

(2 a_{1}^{'})^{2} (1 e^{'})^{4} (1 a_{2}^{''})^{2} .

The preceding ammonia molecular orbital diagram shows the correlation. On pyramidalization,

A_{1}^{'}

and

A_{2}^{''}

both correlate with

A_{1}

of the C3v point group, permitting s–p orbital mixing. The pure planar lone pair can acquire bonding character, lowering its energy and helping stabilize the pyramid. Symmetry alone does not provide a numerical inversion barrier.^[ref]

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