Which of the following has highest value for bond order $- C_{6} H_{6}, C O_{3}^{2 -}, N O^{-}$ and $S O_{3}$ ?

C_{6} H_{6}

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C O_{3}^{2 -}

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N O^{-}

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S O_{3}

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Solution

The correct option is D $N O^{-}$
Bond order $(B O)$ can be calculated using following methods:
(i) $B O = \frac{T o t a l n u m b e r o f b o n d s a s s o c i a t e d w i t h t h e c e n t r a l a t o m}{N u m b e r o f σ b o n d s a s s o c i a t e d w i t h t h e c e n t r a l a t o m}$
(or)
$B O = \frac{T o t a l n u m b e r o f b o n d s b e t w e e n t w o a t o m s}{T o t a l n u m b e r o f r e s o n a n c e s t r u c t u r e s}$
(ii) $B O = \frac{N u m b e r o f b o n d i n g e l e c t r o n s - N u m b e r o f a n t i b o n d i n g e l e c t r o n s}{2}$
We use both the concepts as per situation.
Thus,
(i) We use concept (i) for the species $C_{6} H_{6}, C O_{3}^{2 -}$ and $S O_{3}$ and concept (ii) for $N O^{-}$
On the basis of concept (ii)
$B O$ of $N O^{-} = 7 + 8 + 1$
$= 16$ electrons
$∴ B O$ for $N O^{-} = \frac{10 - 6}{2} = 2$
Hence, highest value for $B O$ is of $N O^{-}$ .

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