 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/2/a/i/solution

Suppress the filled 1s core pair, which contributes zero net bond order. For N

_{2}

, use the valence ordering

σ_{2 s} < σ_{2 s}^{*} < π_{2 p} < σ_{2 p_{z}} < π_{2 p}^{*} < σ_{2 p_{z}}^{*}

and occupy it as

(σ_{2 s})^{2} (σ_{2 s}^{*})^{2} (π_{2 p})^{4} (σ_{2 p_{z}})^{2}

For O

_{2}

, the ordering of the bonding 2p levels is reversed:

σ_{2 p_{z}} < π_{2 p}

. Occupancy is

(σ_{2 s})^{2} (σ_{2 s}^{*})^{2} (σ_{2 p_{z}})^{2} (π_{2 p})^{4} (π_{2 p_{x}}^{*})^{1} (π_{2 p_{y}}^{*})^{1}

, with parallel spins in the two degenerate antibonding pi orbitals.

_{2}

has

(σ_{2 s})^{2}

; Be

_{2}

has

(σ_{2 s})^{2} (σ_{2 s}^{*})^{2}

in the elementary valence description. Empty 2p-derived levels must also be included to discuss mixing and Be

_{2}

's weak bonding; their precise energies are not set by this qualitative diagram.

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