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