Among the following, the molecule expected to be stabilized by anion formation is:
$\mathrm{C}_{2}, \mathrm{O}_{2}, \mathrm{NO}, \mathrm{F}_{2}$
Correct Option: 1
Configuration of $C_{2}$
$=\sigma 1 s^{2} \sigma * 1 s^{2} \sigma 2 s^{2} \sigma * 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}$
Configuration of $\mathrm{C}_{2}^{-}$
$=\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \sigma 2 p_{z}^{1}$
Bond order
$=\frac{\text { No. of bonding } \mathrm{e}^{-}-\text {No. of antibonding } \mathrm{e}^{-}}{2}$
$\mathrm{C}_{2}$ has $s-p$ mixing and the HOMO is $\pi 2 p_{x}=\pi 2 p_{y}$ and LUMO is $\sigma 2 p_{z^{*}}$. So, the extra electron will occupy bonding molecular orbital and this will lead to an increase in bond order.$\mathrm{C}_{2}^{-}$has more bond order than $\mathrm{C}_{2}$.