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

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A-1, B-1, C-2) $A-1, B-1, C-2$
$A$. Free radicals are formed by homolytic cleavage and they are trigonal planar.
$B$. Carbocations are formed by heterolytic cleavage,where the central carbon is $sp^2$ hybridized,resulting in a trigonal planar geometry.
$C$. Carbanions have a lone pair of electrons on the central carbon,which is $sp^3$ hybridized,resulting in a pyramidal geometry.

Explore More

Similar Questions

Choose the most reasonable reaction intermediate for the following reaction.

Which of the following is most likely to undergo a favorable hydride shift?

Which of the following is the most stable free radical?

Difficult
View Solution

The correct order of stability for the following carbanions is:
$CH_2=CH^{-}, CH_3-CH_2^{-}, CH\equiv C^{-}$

The stability order of the following carbocations is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo