What is the increasing order of stability for the carbanions $P: \bar{C}H_3$,$Q: C_6H_5\bar{C}H_2$,$R: (CH_3)_2\bar{C}H$,and $S: H_2\bar{C}^{-}CH=CH_2$?

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.

Which one of the following is most stable?

Arrange the following in decreasing order of stability.
$(i)$ $CH_3^{\bullet}$,$(CH_3)_3C^{\bullet}$,$(CH_3)_2CH^{\bullet}$ and $CH_3CH_2^{\bullet}$
$(ii)$ $CH_3CH_2^{+}$,$(CH_3)_3C^{+}$,$(CH_3)_2CH^{+}$ and $CH_3^{+}$

Due to the presence of an unpaired electron,free radicals become .......

Rank the following radicals in order of increasing stability (least $ < < < $ most).