Question

Which of the following options are correct for $ {{[Fe{{(CN)}_{6}}]}^{3-}} $ complex? (i) $ {{d}^{2}}s{{p}^{3}} $ hybridization(ii) $ s{{p}^{3}}{{d}^{2}} $ hybridization (iii) Para magnetic (iv) Diamagnetic

• A. (i) and (iii)
• B. (i) and (iv)
• C. (ii) and (iii)
• D. (ii) and (iv)

Answer 1

Answer: (A)

In $ {{[Fe{{(CN)}_{6}}]}^{3-}}, $ Fe is present as $ F{{e}^{3+}} $ $ F{{e}^{3+}}=[Ar]3{{d}^{5}}4{{s}^{0}} $ $ C{{N}^{-}} $ being strong field ligand pair up these unpaired electrons so that now the complex have only one unpaired electron. The two 3d, one 4s and three 4p orbitals hybridise to give six $ {{d}^{2}}s{{p}^{3}} $ hybridised orbitals which are occupied by electrons of $ C{{N}^{-}} $ . Since, the complex contain one unpaired electron, it is paramagnetic in nature.