A) tetrahedral
B) octahedral
C) square planar
D) linear
Correct Answer: B
Solution :
\[{{K}_{4}}[Fe{{(CN)}_{6}}]4{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{4-}}\] The oxidation number of Fe in\[{{[Fe{{(CN)}_{6}}]}^{4-}}\]is +2. \[F{{e}^{2+}}-1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{6}},4{{s}^{0}}4p\] Since\[C{{N}^{-}}\]is a strong field ligand, pairing occurs and the hybridisation of\[{{[Fe{{(CN)}_{6}}]}^{4-}}\]is \[{{d}^{2}}s{{p}^{3}}\]and structure is octahedral.You need to login to perform this action.
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