A) \[s{{p}^{3}},\] two
B) \[ds{{p}^{2}},\]zero
C) \[ds{{p}^{2}},\]one
D) \[s{{p}^{3}},\]zero
Correct Answer: A
Solution :
\[{{[Ni{{L}_{4}}]}^{2-}}\] i.e, no. of unpaired electron = 2 hybridization \[-s{{p}^{3}}.\]You need to login to perform this action.
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