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Bond Length

The correct o...

Question

The correct order of increasing $C - O$ bond length in $C O, C O_{2}$ and $C O_{3}^{2 -}$ is:

A

C O_{3}^{2 -} < C O_{2} < C O

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B

C O < C O_{2} < C O_{3}^{2 -}

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C

C O_{3}^{2 -} < C O < C O_{2}

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D

C O = C O_{2} = C O_{3}^{2 -}

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Solution

The correct option is B $C O < C O_{2} < C O_{3}^{2 -}$
Method (1) Bond-multiplicity increases bond-length.
Thus,
(i) $C \equiv) \to$ has least bond-length.
(ii) $O = C + O \to$ next to $C = O$ .
(iii)
Hence, increasing order of bond-length is $C O < C O_{2} < C O_{3}^{2 -}$
Method (2)
(i) More be the value of bond-order smaller be the bond-length and more be the bond-energy.
(ii) Bond-order $(B O) = \frac{T o t a l n u m b e r o f b o n d s a s s o c i a t e d b e t w e e n t w o a t o m s i n a l l s t r u c t u r e}{T o t a l n u m b e r o f c a n o n i c a l f o r m s}$
(a) $C O \to C \equiv O \to B O = \frac{3}{1} = 3$
(b) $C O_{2} \to O = C = C \to B O = \frac{4}{2} = 2$
Hence, order of bond-length $C O < C O_{2} < C O_{3}^{2 -}$
Epimer is used to designate diastereomers that differ in configuration at only one chiral centre.

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