A) One, tetrahedral
B) Two, tetrahedral
C) One, square planar
D) Two, square planar
Correct Answer: B
Solution :
[b] If \[{{X}^{-}}\] is weak field ligand (say\[C{{l}^{-}}\]), then \[{{[NiC{{l}_{4}}]}^{2-}}\] is tetrahedral \[(s{{p}^{3}})\]with two unpaired electrons. If \[{{X}^{-}}\]is strong field ligand (say\[C{{N}^{-}}\]), then \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]is square planar \[(ds{{p}^{2}})\]with no unpaired electrons. Also given \[{{[Ni{{X}_{4}}]}^{2-}}\]is paramagnetic, so \[{{X}^{-}}\]is weak ligand.You need to login to perform this action.
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