MCQ
The formation of $PH_4^+$ is difficult compared to $NH_4^+$ because
  • A
    lone pair of phosphorus is optically inert
  • B
    lone pair of phosphorus resides in almost pure $p-$ orbital
  • C
    lone pair of phosphorus resides at $sp^3$ orbital
  • lone pair of phosphorus resides in almost pure $s-$ orbital

Answer

Correct option: D.
lone pair of phosphorus resides in almost pure $s-$ orbital
d
$\overset{\,\,\centerdot \,\,\centerdot \,\,\,\,\,\,\,\,}{\mathop{\text{P}{{\text{H}}_{3}}}}\,+{{\text{H}}^{+}}\to \text{PH}_{4}^{+}$

According to Drago's rule lone pair on phosphorus resides in almost pure s-orbital, hence due to non-directionalnature, its overlapping tendency is greatly reduced in comparison to a lone pair present in hybrid orbital, which is directional as present in $\overset{\,\,\centerdot \,\,\centerdot \,\,\,\,\,\,\,\,}{\mathop{\text{N}{{\text{H}}_{3}}}}\,$.

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