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Bond Order by MOT

The ion that ...

Question

The ion that is isoelectronic with $C O$ and has same bond order is:

C N^{-}

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O_{2}^{+}

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{O_{2}}^{-}

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N_{2}^{+}

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Solution

The correct option is A $C N^{-}$
The number of electrons are:
$C O = 6 + 8 = 14$ electrons, BO $= 3$
$C N^{-} = 6 + 7 + 1 = 14$ electrons, BO $= 3$
$O_{2}^{+} = 8 + 8 - 1 = 15$ electrons, BO $= 2.5$
${O_{2}}^{-} = 8 + 8 + 1 = 17$ electrons, BO $= 1.5$
Hence, $C N^{-}$ is isoelectronic with $C O$ .

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