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Which of the following has the minimum bond length?
A
$O_2$
B
$O_2^+$
C
$O_2^-$
D
$O_2^{2-}$
Detailed Solution
MO configuration of $O_2$ (16 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \sigma2p_z^2\ \pi2p_x^2 = \pi2p_y^2\ \pi^*2p_x^1 = \pi^*2p_y^1$
Bond order = $\frac{N_b - N_a}{2}$. For $O_2$: $\frac{10 - 6}{2} = 2$
$O_2^+$ (one antibonding electron removed): $\frac{10 - 5}{2} = 2.5$
$O_2^-$ (one antibonding electron added): $\frac{10 - 7}{2} = 1.5$
$O_2^{2-}$ (two antibonding electrons added): $\frac{10 - 8}{2} = 1$
Bond length is inversely proportional to bond order.
$O_2^+$ has the highest bond order (2.5) and therefore the minimum bond length.
Bond order = $\frac{N_b - N_a}{2}$. For $O_2$: $\frac{10 - 6}{2} = 2$
$O_2^+$ (one antibonding electron removed): $\frac{10 - 5}{2} = 2.5$
$O_2^-$ (one antibonding electron added): $\frac{10 - 7}{2} = 1.5$
$O_2^{2-}$ (two antibonding electrons added): $\frac{10 - 8}{2} = 1$
Bond length is inversely proportional to bond order.
$O_2^+$ has the highest bond order (2.5) and therefore the minimum bond length.