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The correct order of basicity of below is


$ \begin{array}{1 1}(a)\;(I) > (II) > (III)&(b)\;(III) > (I) > (II)\\(c)\;(III) > (II) > (I)&(d)\;(II) > (I) > (III)\end{array}$

1 Answer

In this compound the lone pair is in an $SP^3$ hybrid orbital and there is no resonance (no $\pi$ system).So more basic.
$SP^2$ hybrid orbital-So more basic than the 1-lone pair in resonance.
However the order is $(III) > (I) > (II)$
Hence (b) is the correct answer.
answered Mar 3, 2014 by sreemathi.v