A) \[s{{p}^{3}},ds{{p}^{2}},ds{{p}^{3}},s{{p}^{2}}\]
B) \[s{{p}^{3}},ds{{p}^{2}},s{{p}^{3}},s{{p}^{3}}\]
C) \[s{{p}^{3}},ds{{p}^{2}},ds{{p}^{3}},s{{p}^{3}}\]
D) \[s{{p}^{3}},s{{p}^{2}},ds{{p}^{3}},s{{p}^{3}}\]
Correct Answer: C
Solution :
\[[N{{H}_{3}}]\to \] Three bond pairs + one lone pair \[\to \]Tetrahedral \[\to s{{p}^{3}}\] hybridisation. \[{{[PtC{{l}_{4}}]}^{2-}}\to \] Four hybrid orbitals \[\to \] square planar \[\to ds{{p}^{2}}\] hybridisation. \[[PC{{l}_{5}}]\to \] Five hybrid orbitals \[\to \] trigonal bipyramid \[\to ds{{p}^{3}}\] hybridisation. \[{{[BC{{l}_{4}}]}^{-}}\to \] Four hybrid orbitals \[\to \] tetrahedral \[\to s{{p}^{3}}\] hybridisation.
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