A) \[ds{{p}^{2}},\,\,ds{{p}^{3}},\,\,s{{p}^{2}}\]and \[s{{p}^{3}}\]
B) \[s{{p}^{3}},\,\,ds{{p}^{2}},\,\,ds{{p}^{3}},\,\,s{{p}^{2}}\]
C) \[ds{{p}^{2}},\,\,s{{p}^{2}},\,\,s{{p}^{3}},\,\,ds{{p}^{3}}\]
D) \[ds{{p}^{2}},\,\,s{{p}^{3}},\,\,s{{p}^{2}},\,\,ds{{p}^{3}}\]
Correct Answer: B
Solution :
Hybridization of N in \[N{{H}_{3}}\] is \[s{{p}^{3}}\] that of Pt in \[{{[PtC{{l}_{4}}]}^{2-}}\] is \[ds{{p}^{2}}\] that P in \[PC{{l}_{5}}\] is \[s{{p}^{3}}d\] and that of B in \[BC{{l}_{3}}\] is \[s{{p}^{2}}\].You need to login to perform this action.
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