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Answer» `CH_(3)-S-CH_(2)^(+)` `(CH_(3))_(2)NCH_(2)^(+)` `CH_(3)-O-CH_(2)^(+)` `F-CH_(2)^(+)` Solution :The LONE pair of electrons on S is PRESENT in a 3p-orbital while those an N, O and F are present in a 2-p-orbital. Further, the carbon atom bearing the +ve charge has an empty 2p-orbital. Since EFFECTIVE RESONANCE stabilization occurs between orbitals of similar sizes, therefore, least effective resonance will be between S and C atoms. Among N, O and F, since is least electronegative, therefore N, O and F, since N is least electronegative, therefore, most effective resonance will be between N and C and hence `(CH_(3))_(2)NCH_(2)^(+)` carbocations would be the most stable.
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