Compare the complexes ${[F e {(H_{2} O)}_{6}]}^{2 +}$ and ${[F e {(C N)}_{6}]}^{4 -}$ with the help of crystal field theory.

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Solution

${[F e {(H_{2} O)}_{6}]}^{2 +}$ , Here ${F e}^{+ 2}$ is surrounded by $6$ $H_{2} O$ ligands. Since they are weak ligands, it does not cause pairing up of $e^{-}$ s and other orbital is formed with ${S P}^{3} d^{2}$ hybridisation.
${[F e {(C N)}_{6}]}^{4 -}$ ${C N}^{-}$ is a strong ligand. Hence, it forms an inner orbital complex with $d^{2} {s p}^{3}$ hybridisation.

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Give the electronic configuration of the following complexes on the basis of crystal field splitting theory. $[C o F_{6}]^{3 -}, [F e (C N)_{6}]^{4 -}$ and $[C u (N H_{3})_{6}]^{2 +} .$

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following
(a) $[C o F_{6}]^{3 -}, [C o (H_{2} O)_{6}]^{2 +}, [C o (C N)_{6}]^{3 -}$
(b) $F e F_{6}^{3 -}, [F e (H_{2} O)_{6}]^{2 +}, [F e (C N)_{6}]^{4 -}$