JEE Main 2022 — Chemical Bonding and Molecular Structure Question with Solution

The bond order of a diatomic species can be calculated using molecular orbital theory as follows,

$\mathrm{Bond} \mathrm{order}=\frac{\mathrm{No}.\mathrm{of} \mathrm{BMO} \mathrm{electrons}-\mathrm{No}.\mathrm{of} \mathrm{ABMO} \mathrm{electrons}}{2}$

BMO = bonding molecular orbital

ABMO = Anti bonding molecular orbital.

Based on the total number electrons in bond orders are given in the following table.

Number of ${\mathrm{e}}^{-}$	Bond order
$13$	$2.5$
$14$	$3$
$15$	$2.5$
$16$	$2$

${\mathrm{CN}}^{-} , {\mathrm{NO}}^{+} \mathrm{and} {\mathrm{O}}_2^{2+}$ are having 14 electrons each, hence the bond order is 3.