The correct order of sulphur-oxygen bond orde | vedclass

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Question

(C) The bond order is calculated as the total number of bonds divided by the number of resonating structures or the number of terminal atoms.$1$. For

Vedclass · Accepted Answer

$S_2O_3^{2-} < SO_4^{2-} < SO_3$

Vedclass · Answer

(C) The bond order is calculated as the total number of bonds divided by the number of resonating structures or the number of terminal atoms.$1$. For $SO_3$: The structure has $3$ resonance hybrids where $6$ bonds are distributed over $3$ $S-O$ positions. Bond order $= \frac{6}{3} = 2.0$.$2$. For $SO_4^{2-}$: The structure has $4$ equivalent $S-O$ bonds with a total of $6$ bonds in the resonance hybrid. Bond order $= \frac{6}{4} = 1.5$.$3$. For $S_2O_3^{2-}$: The structure involves a central sulphur atom bonded to three oxygen atoms and one sulphur atom. The resonance involves $4$ bonds distributed over $3$ $S-O$ positions. Bond order $= \frac{4}{3} \approx 1.33$.Comparing the values: $1.33 Thus,the correct order is $S_2O_3^{2-} < SO_4^{2-} < SO_3$.

The correct order of sulphur-oxygen bond order in $SO_3$,$S_2O_3^{2-}$,and $SO_4^{2-}$ is:

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