Which of the following is the most stable carbocation?

Arrange the following carbanions in the order of their stability:
$i$. $RC \equiv C^{\ominus}$
$ii$. $C_6H_5^{\ominus}$
$iii$. $R_2C = CH^{\ominus}$
$iv$. $R_3C - CH_2^{\ominus}$

Compare the stability of the following free radicals:
$(1)$ $CH_3 - \dot{CH} - CH_3$
$(2)$ $C_6H_5 - \dot{CH_2}$
$(3)$ $\dot{CH_2} - CH(CH_3)_2$
$(4)$ $\dot{CH_2} - CH_3$

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

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 of the following is the correct decreasing order of stability of carbanions?
$(1) (CH_3)_3C^-$
$(2) (CH_3)_2CH^-$
$(3) CH_3CH_2^-$
$(4) C_6H_5CH_2^-$