The correct statement about the magnetic properties of ${[Fe (CN)_{6}]}^{3 -}$ and ${[{FeF}_{6}]}^{3 -}$ is $(Z = 26)$ .

A
both are paramagnetic.
B
both are diamagnetic
C
${[Fe (CN)_{6}]}^{3 -}$ is diamagnetic, ${[{FeF}_{6}]}^{3 -}$ is paramagnetic.
D
${[Fe (CN)_{6}]}^{3 -}$ is paramagnetic, ${[{FeF}_{6}]}^{3 -}$ is diamagnetic.

Solution:

The electronic configuration of free ${Fe}^{3 +}$ ion is $3 d^{5}$ , i.e

In ${[Fe (CN)_{6}]}^{3 -}$ , the configuration is

In ${[{FeF}_{6}]}^{3 -}$ , the configuration is

Both the complexes contain unpaired electron (s). Hence, these are paramagnetic.

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