The increasing order of basicity of $R C N, R C H = N R$ and $R {N H}_{2}$ is:

R C N < R C H = N R < R H_{2} N

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R {N H}_{2} < R C N < R C H = N R

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R C H = N R < R {N H}_{2} < R C N

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R H_{2} N < R C H = N R < R C N

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Solution

The correct option is A $R C N < R C H = N R < R H_{2} N$
In $R C N$ the $N$ atom is $s p$ hybrid, in $R C H = N R$ it is $s p^{2}$ hybrid and in $R N H_{2}$ is $s p^{3}$ hybrid.
The $s p$ hybrid $N$ atom is most electron negative due to maximum S character. More electron negative has less tendency to donate electrons and therefore is least basic. Thus the order of basicity is $R C N < R C H = N R < R H_{2} N$ .

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