A) \[{{[PdC{{l}_{4}}]}^{2-}}\]
B) \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]
C) \[Pd{{(CN)}_{4}}{{]}^{3-}}\]
D) \[{{[NiC{{l}_{4}}]}^{2-}}\]
Correct Answer: D
Solution :
\[Ni(28)=3{{d}^{8}}\,\,4{{s}^{2}}\] \[N{{i}^{2+}}=3{{d}^{8}}\,\,4{{s}^{0}}\] Here \[\times \,\times \] represents electron pairs donated by \[C{{l}^{-}}\]ion. \[{{[NiC{{l}_{4}}]}^{2-}}\] shows paramagnetic characteristics due to the presence of two unpaired electrons. The 3d orbitals remain undisturbed because \[C{{l}^{-}}\] ligand is weak field ligand therefore, \[N{{i}^{2+}}\] ion undergoes \[s{{p}^{3}}\]hybridisation resulting in tetrahedral structure of the complex.You need to login to perform this action.
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