Match column $-I$ and column $-II$ with the correct electronic effect relation:
Column $-I$ Column $-II$
$(i)$ $C_6H_5NH_2$ $(p)$ $-R$ effect
$(ii)$ $C_6H_5OH$ $(q)$ $+R$ effect
$(iii)$ $C_6H_5NO_2$ $(r)$ $+I$ effect
$(iv)$ $CH_3CH_2Cl$ $(s)$ $-I$ effect

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(A) $(i) - (q, s)$; $(ii) - (q, s)$; $(iii) - (p, s)$; $(iv) - (s)$.
Explanation:
$(i)$ $C_6H_5NH_2$: The $-NH_2$ group has a lone pair on nitrogen,showing $+R$ effect,and is more electronegative than carbon,showing $-I$ effect.
$(ii)$ $C_6H_5OH$: The $-OH$ group has lone pairs on oxygen,showing $+R$ effect,and is more electronegative than carbon,showing $-I$ effect.
$(iii)$ $C_6H_5NO_2$: The $-NO_2$ group is electron-withdrawing,showing $-R$ effect,and is highly electronegative,showing $-I$ effect.
$(iv)$ $CH_3CH_2Cl$: The $-Cl$ atom is electronegative,showing $-I$ effect. It does not show resonance with the alkyl group.

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Given below are two statements $:$
Statement $I :$ Hyperconjugation is not a permanent effect.
Statement $II :$ In general,greater the number of alkyl groups attached to a positively charged $C$ atom,greater is the hyperconjugation interaction and stabilization of the cation.
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