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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.

Similar Questions

Identify the correct statement.

The stability order of the following resonance structures is:
$I. CH_2=CH-CH=O$
$II. CH_2^+-CH=CH-O^-$
$III. CH_2^--CH=CH-O^+$

Write the resonance structures for $SO_{3}$,$NO_{2}$ and $NO_{3}^{-}$.

Explain the structure of $CO_{3}^{2-}$ ion in terms of resonance.

Which of the following has the highest resonance energy?

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