A) \[~ds{{p}^{2}},\text{ }s{{p}^{2}},\text{ }s{{p}^{3}},\text{ }ds{{p}^{3}}\]
B) \[~s{{p}^{3}},\text{ }ds{{p}^{3}},\text{ }ds{{p}^{3}},\text{ }s{{p}^{2}}\]
C) \[~ds{{p}^{2}},\text{ }sp\text{ },\text{ }s{{p}^{2}},\text{ }s{{p}^{3}}\]
D) \[~ds{{p}^{3}},\text{ }ds{{p}^{3}},\text{ }s{{p}^{2}},\text{ }s{{p}^{3}}\]
Correct Answer: B
Solution :
N in \[N{{H}_{3}}\]is \[s{{p}^{3}}\]hybridized, Pt in \[{{[Pt\,C{{l}_{4}}]}^{2-}}\] is \[ds{{p}^{2}}\]hybridized, P in\[PC{{l}_{3}}\]is\[s{{p}^{3}}d\] hybridized, whereas B in\[BC{{l}_{3}}\]is \[s{{p}^{2}}\]hybridized. Hence, the correct option is [b].
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