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Question

(A) The $\pi -$ bond order is calculated by dividing the total number of $\pi -$ bonds by the total number of $S-O$ bonds in the resonance hybrid stru

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The $\pi -$ bond order is calculated by dividing the total number of $\pi -$ bonds by the total number of $S-O$ bonds in the resonance hybrid structure.						$Species$			$S-O$ $\pi -$ Bond order							$SO_3$			$1$							$SO_4^{2-}$			$0.5$							$SO_3^{2-}$			$0.33$			Thus,the decreasing order of $\pi -$ bond order is $SO_3 > SO_4^{2-} > SO_3^{2-}$.

Choose the $CORRECT$ comparison of decreasing $\pi -$ bond order of $S-O$ bond.

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