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Question

According to molecular orbital theory, the correct order of energy for \(\sigma_{2p}\) orbital is :

The correct answer is

\(\mathrm{C_{2} \lt N_{2} \gt O_{2}}\)

In second-period homonuclear diatomics the \(\sigma_{2s}\) and \(\sigma_{2p}\) orbitals have the same symmetry and interact (s–p mixing). The mixing pushes \(\sigma_{2p}\) upward, so its position is judged against the \(\pi_{2p}\) level of the same molecule.

O2: the 2s–2p gap of oxygen is large, s–p mixing is weak, and \(\sigma_{2p}\) lies below \(\pi_{2p}\) (configuration … \(\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p}^{*2}\)).

N2: \(\sigma_{2p}\) is pushed clearly above \(\pi_{2p}\) and is the HOMO; photoelectron spectra show its ionisation energy (15.6 eV) is about 1.1 eV lower than that of the \(\pi_{2p}\) level (16.7 eV).

C2: the ground configuration is … \(\pi_{2p}^{4}\) with \(\sigma_{2p}\) empty, and the state with an electron promoted into \(\sigma_{2p}\) lies very close in energy, so in C2 the \(\sigma_{2p}\) level is only marginally above \(\pi_{2p}\).

So the relative raising of \(\sigma_{2p}\) is greatest in N2, smaller in C2, and in O2 the level drops below \(\pi_{2p}\): C2 < N2 and N2 > O2.

Therefore the order is C2 < N2 > O2.

Hence, \(\mathrm{C_{2} \lt N_{2} \gt O_{2}}\).

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