With a change in hybridisation of the carbon | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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: $sp^3 < sp^2 < sp$.

Column $-I$	Column $-II$
$A$. Free radical	$1$. Trigonal planar
$B$. Carbocation	$2$. Pyramidal
$C$. Carbanion	$3$. Linear

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 is the most stable carbocation?

Match the intermediates given in Column $-I$ with their probable structure in Column $-II$.

Column $-I$ Column $-II$

$A$. Free radical $1$. Trigonal planar

$B$. Carbocation $2$. Pyramidal

$C$. Carbanion $3$. Linear

Some pairs of ions are given below. In which pair is the first ion more stable than the second?

What is the decreasing order of stability (most stable $\to$ least stable) of the following carbocations?

The stability of the following carbocations decreases in the order:

Vedclass Test Series

Exam Paper Generator

Online Exam Module