In the complex $F e K_{2} [F e {(C N)}_{6}]$ ,

Both Fe atoms are in the same oxidation state.

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Both Fe atoms are in the different oxidation state.

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The coordination number of ion is 4.

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The complex is a high spin complex.

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Solution

The correct option is A Both Fe atoms are in the same oxidation state.
In $F e^{+ 2} K_{2}^{+ 2} [F e^{+ 2} (C N)_{6}]$ , both Fe atoms are in the same oxidation state.
As $C N^{-}$ is a strong ligand so pairing of electron happens and low spin complex forms.
As here Fe is connected with 6 ligands so coordination number of ion is 6.

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