A) \[Ni{{(CO)}_{4}}\]
B) \[{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
C) \[{{[NiC{{l}_{4}}]}^{2-}}\]
D) \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]
Correct Answer: D
Solution :
The electronic configuration of\[Ni\]is\[[Ar]\text{ }3{{d}^{8}},4{{s}^{2}}\]. \[CO\]and\[C{{N}^{-}}\]being strong field ligand cause pairing while\[{{H}_{2}}O\]and\[Cl\]do not. Thus, \[Ni{{(CO)}_{4}};\]tetrahedral, no unpaired electron \[{{[Ni{{(CN)}_{4}}]}^{2-}};\]octahedral, two unpaired electrons. \[{{[NiC{{l}_{4}}]}^{2-}};\]tetrahedral, two unpaired electrons. \[{{[Ni{{(CN)}_{4}}]}^{2-}};\]square planar, no unpaired electrons.
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