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Number of molecules having bond order $2$ from the following molecule is. . . . . . . $\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2$
$1$
$2$
$5$
$8$
Solution
$\left(12 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2\left[\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right]$
$\text { B.O. }=\frac{8-4}{2}=2$
$\mathrm{O}_2$
$\left(16 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2, \sigma 2 \mathrm{pz}^2$
${\left[\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right]\left[\pi^* 2 \mathrm{p}_{\mathrm{x}}^1=\pi^* 2 \mathrm{p}_{\mathrm{y}}^1\right]}$
$\text { B.O. }=\frac{10-6}{2}=2$
$\mathrm{Be}_2$
$\left(8 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2$
$\text { B.O. }=\frac{4-4}{2}=0$
$\text { Li }_2$
$\left(6 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2$
