The correct order of S-O bond length is | vedclass

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(B) The bond length is inversely proportional to the bond order $(B.L. \propto \frac{1}{B.O.})$. First,calculate the bond order $(B.O.)$ for each spec

Vedclass · Accepted Answer

$SO_3^{2-} > SO_4^{2-} > SO_2 > SO_3$

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(B) The bond length is inversely proportional to the bond order $(B.L. \propto \frac{1}{B.O.})$. First,calculate the bond order $(B.O.)$ for each species: $1$. For $SO_3^{2-}$: The resonance hybrid has $3$ $S-O$ bonds with a total of $4$ bonds,so $B.O. = \frac{4}{3} \approx 1.33$. $2$. For $SO_4^{2-}$: The resonance hybrid has $4$ $S-O$ bonds with a total of $6$ bonds,so $B.O. = \frac{6}{4} = 1.5$. $3$. For $SO_2$: The resonance hybrid has $2$ $S-O$ bonds with a total of $3$ bonds,so $B.O. = \frac{3}{2} = 1.5$. $4$. For $SO_3$: The resonance hybrid has $3$ $S-O$ bonds with a total of $4$ bonds,so $B.O. = \frac{4}{3} \approx 1.33$. Comparing $SO_2$ and $SO_3$ using Bent's rule or hybridization,$SO_2$ $(sp^2)$ has a higher $s$-character in the $S-O$ bond than $SO_3$ ($sp^2$ but with different electronic environment),leading to shorter bonds in $SO_3$ compared to $SO_2$. Comparing $SO_3^{2-}$ and $SO_4^{2-}$,$SO_3^{2-}$ has a lower bond order $(1.33)$ than $SO_4^{2-}$ $(1.5)$,so $SO_3^{2-}$ has a longer bond length. The overall order of bond length is $SO_3^{2-} > SO_4^{2-} > SO_2 > SO_3$.

The correct order of $S-O$ bond length is

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