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(B) In $RNH_2$,the $N$ atom is $sp^3$ hybridized. In $RN=CHR^1$,the $N$ atom is $sp^2$ hybridized,whereas in $RC \equiv N$,the $N$ atom is $sp$ hybrid

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$III > I > II$

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(B) In $RNH_2$,the $N$ atom is $sp^3$ hybridized. In $RN=CHR^1$,the $N$ atom is $sp^2$ hybridized,whereas in $RC \equiv N$,the $N$ atom is $sp$ hybridized.The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom. As the $s$-character of the hybrid orbital increases,the electronegativity of the nitrogen atom increases,making the lone pair less available for donation.The $s$-character increases in the order: $sp^3 Therefore,the electronegativity of the $N$ atom increases in the order: $sp^3 Consequently,the basicity decreases as the $s$-character increases.The order of decreasing basicity is: $RNH_2 (III) > RN=CHR^1 (I) > RC \equiv N (II)$.

Arrange the following in the order of decreasing basicity:
$I. RN=CHR^1$ $II. RC \equiv N$ $III. RNH_2$

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Arrange the following in the order of decreasing basicity:$I. RN=CHR^1$$II. RC \equiv N$$III. RNH_2$