The bond length of the S=O bond is maximum in | vedclass

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(B) According to $Bent's$ rule,more electronegative atoms prefer to occupy hybrid orbitals having less $s$-character,while less electronegative atoms

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$SOBr_2$

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(B) According to $Bent's$ rule,more electronegative atoms prefer to occupy hybrid orbitals having less $s$-character,while less electronegative atoms prefer orbitals with more $s$-character.In $SOX_2$ molecules,the $S=O$ bond has more $s$-character as the electronegativity of the halogen atom $(X)$ increases.As the electronegativity of $X$ decreases $(F > Cl > Br)$,the $s$-character in the $S-X$ bonds increases,which means the $s$-character in the $S=O$ bond decreases.Since bond length is inversely proportional to $s$-character,the $S=O$ bond length increases as the $s$-character in the $S=O$ bond decreases.Therefore,the $S=O$ bond length is maximum in $SOBr_2$ because $Br$ is the least electronegative among $F, Cl,$ and $Br$.

The bond length of the $S=O$ bond is maximum in which of the following compounds?

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