 Solution

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

A_{1}

2 s

and

2 p_{z}

, giving three

a_{1}

E

functions mix with

(2 p_{x}, 2 p_{y})

, giving a bonding

e

pair and an antibonding

e

pair. Eight valence electrons fill the lower levels as

(2 a_{1})^{2} (1 e)^{4} (3 a_{1})^{2} .

This uses conventional numbering that reserves

1 a_{1}

for the omitted nitrogen core

1 s

orbital. The low

2 a_{1}

and

1 e

levels provide the principal N–H bonding molecular orbitals; the highest occupied

3 a_{1}

level is the mainly nitrogen lone pair, with some bonding admixture permitted by the pyramidal geometry. Empty

4 a_{1}

and

2 e

levels are antibonding. Symmetry fixes which orbitals can mix, rather than their quantitative energy separations.

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