Give reasons for the two $S = O$ bonds in ${S O}_{2}$ are identical.

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Solution

Electronic configuration of $S = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$
During the formation of $S O_{2}$ , one $e^{-}$ from $3 p$ orbital goes to $3 d$ due to the presence of vacant d-orbital.
Two of these orbitals form sigma bonds with two oxygen atoms & the third contains a lone pair, p-orbital & d-orbital contains an unpaired $e^{-}$ each. One of these $e^{-} s$ forms $p Π - p Π$ bond with one oxygen atom and the other form $p Π - d Π$ bond with the other oxygen. This is the reason that $S O_{2}$ has a bent structure and are identical.

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