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Question

(C) The stability of carbanions increases with an increase in the $s-$character of the hybrid orbitals of the carbon atom bearing the negative charge.

Vedclass · Accepted Answer

$sp^3 < sp^2 < sp$

Vedclass · Answer

(C) The stability of carbanions increases with an increase in the $s-$character of the hybrid orbitals of the carbon atom bearing the negative charge.This is because a higher $s-$character makes the orbital more electronegative,which better stabilizes the negative charge.The $s-$character in different hybridizations is: $sp$ $(50\%)$,$sp^2$ $(33.3\%)$,and $sp^3$ $(25\%)$.Therefore,the correct order of stability is: $sp^3 < sp^2 < sp$.

With a change in hybridisation of the carbon bearing the charge,the stability of a carbanion increases in the order:

Similar Questions

Which of the following species $(A)$,$(B)$,$(C)$,and $(D)$ contain an $sp^2$ hybridized carbon atom? $(A) CH_3^+$,$(B) CH_3^-$,$(C) dot{C}H_3$,$(D) :CH_2$

Arrange the following carbocations in decreasing order of their stability:
$\stackrel{+}{C}H_3, (CH_3)_3\stackrel{+}{C}, CH_3\stackrel{+}{C}H_2, (CH_3)_2\stackrel{+}{C}H$

What is the order of stability of carbon free radicals among ${3^\circ}$,${2^\circ}$,${1^\circ}$,and $\mathop C\limits^\bullet {H_3}$?

$CH_3-C(CH_3)_2-CH_2-CH_2^+ \to CH_3-C^+(CH_3)-CH(CH_3)-CH_3$
How many $H^-$ shifts are involved in the above rearrangement?

What is the increasing order of stability for the following free radicals?

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