State True or False for the following statements:
$(i)$ Only carbocation is formed by homolytic cleavage of a bond.
$(ii)$ By heterolytic cleavage of a bond,a carbocation or carbanion is formed.
$(iii)$ The carbon of a carbanion is $sp^2$ and the carbon of a carbocation is $sp^3$.
$(iv)$ The carbon of a carbanion is $sp^3$ and the carbon of a carbocation is $sp^2$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(D) $(i)$ False: Homolytic cleavage results in the formation of free radicals,not carbocations.
$(ii)$ True: Heterolytic cleavage results in the unequal distribution of electrons,leading to the formation of a carbocation $(C^+)$ or a carbanion $(C^-)$.
$(iii)$ False: The carbon of a carbanion is $sp^3$ hybridized (pyramidal geometry),and the carbon of a carbocation is $sp^2$ hybridized (planar geometry).
$(iv)$ True: As explained in $(iii)$,the carbanion carbon is $sp^3$ and the carbocation carbon is $sp^2$.
Final Answer: $(i-F, ii-T, iii-F, iv-T)$

Explore More

Similar Questions

The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical,respectively,are due to

Arrange the following carbocations in the increasing order of stability with respect to their labels:
LabelCarbocation
$1$$CH_3-CH^+-CH_3$
$2$$CH_3^+$
$3$$(CH_3)_3C^+$
$4$$CH_3-CH_2^+$

Chlorine $\left(Cl^{17}\right)$ free radical contains how many electrons around the nucleus?

The most stable carbocation from the following is:

Which of the following carbocations will not undergo rearrangement?

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