Discussion
The bond order in \(\mathrm{N_{2}^{2-}}\) species is
Answer: A
The bond order of \(\mathrm{N_{2}^{2-}}\) is 2.
N2 has 10 valence electrons; the dianion has 12. Filling the molecular orbitals in the order used for N2:
\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\pi_{2p}^{4} \\ &\sigma_{2p}^{2}\,\pi_{2p}^{*2} \end{aligned} \]
That is 8 bonding and 4 antibonding electrons, so
\[ \text{bond order} = \frac{8 - 4}{2} = 2 \]
3 is the bond order of neutral N2, and 2.5 is that of N2−, where only one electron has entered \(\pi^{*}\). Each extra electron goes into an antibonding orbital and lowers the bond order by half, so 3.5 is impossible here.
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