Arrange the following free radicals in the correct order of their stability:
$(i)$ $CH_2=\dot{C}H$
$(ii)$ $\dot{C}H_3$
$(iii)$ $CH_3-\dot{C}H-CH_3$
$(iv)$ $(CH_3)_3\dot{C}$

The most stable carbanion among the following is

State True or False for the following statements:
$(i)$ $\mathop{C}\limits^{+}H_3$ and $CH_3\mathop{C}\limits^{+}H_2$ both are primary carbocations.
$(ii)$ $(CH_3)_3\mathop{C}\limits^{+}$ is a tertiary carbocation and all carbons are $sp^3$ hybridized.
$(iii)$ $\mathop{C}\limits^{+}H_3$ is trigonal planar.
$(iv)$ $CH_4$ is not trigonal planar.

Consider the following compound $(X)$:
$H-C \equiv C-CH_2-CH(CH_3)-CH_3$
The most stable and least stable carbon radicals,respectively,produced by homolytic cleavage of the corresponding $C-H$ bond are:

Arrange the following carbocations in increasing order of stability:
$1. \, (CH_3)_2 - \overset{+}{C} - CH_2 - CH_3$
$2. \, (CH_3)_3 - \overset{+}{C}$
$3. \, (CH_3)_2 - \overset{+}{C} H$
$4. \, CH_3 - \overset{+}{C} H_2$
$5. \, \overset{+}{C} H_3$

Which of the following carbocations is least stable?