Which of the following pairs have identical values of bond order?

N_{2}^{+}

and

O_{2}^{+}

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F_{2}

and

N e_{2}

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O_{2}^{2 -}

and

B_{2}

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C_{2}

and

N_{2}

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Solution

The correct options are
A $N_{2}^{+}$ and $O_{2}^{+}$
C $O_{2}^{2 -}$ and $B_{2}$
$N_{2}^{+} = \frac{9 - 4}{2} = 2.5$ $O_{2}^{+} = \frac{10 - 5}{2} = 2.5$
$F_{2} = \frac{10 - 8}{2} = \frac{2}{2} = 1$ ${N e}_{2} = \frac{10 - 10}{2} = 0$
$O_{2}^{2 -} = \frac{10 - 8}{2} = 1$ $B_{2} = \frac{6 - 4}{2} = 1$
$C_{2} = \frac{8 - 4}{2} = 2$ $N_{2} = \frac{10 - 4}{2} = 3$
So the molecular pairs with similar Bond orders are as follows:-
(a) $N_{2}^{+}$ and $O_{2}^{+}$ (Bond order=2.5)
(b) $O_{2}^{2 -}$ and $B_{2}$ (Bond order=1)

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(i). $N_{2} \to N_{2}^{+} + e^{-}$

(ii) $O_{2} \to O_{2}^{+} + e^{-}$

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