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Question

The complex [FeF6]3– is colorless, whereas [CoF6]3– is colored; this is because :

This question was previously asked in
MH SET 2023 Paper 2 Computer Science and Applications Question Paper (26-Mar-2023)
The correct answer is

Spin forbidden transitions in [FeF6]3– and spin allowed transitions in [CoF6]3–

Fluoride is a weak-field ligand, so both hexafluoro complexes are high spin. The colour then depends on whether a d–d transition is spin-allowed (no change in total spin, \(\Delta S=0\)) or spin-forbidden.

In [FeF6]3– the metal is Fe3+, d5 high spin (t2g3eg2), ground term \({}^{6}A_{1g}\). Every excited configuration requires pairing of an electron, and no other sextet state exists, so all d–d transitions are spin-forbidden as well as Laporte-forbidden. The absorptions are extremely weak and the complex looks almost colourless.

In [CoF6]3– the metal is Co3+, d6 high spin (t2g4eg2), ground term \({}^{5}T_{2g}\). Promotion of a t2g electron to eg gives the \({}^{5}E_{g}\) state of the same spin, so the \({}^{5}T_{2g}\rightarrow{}^{5}E_{g}\) band is spin-allowed and the complex is blue.

The fact that both are high spin is true but does not explain the colour difference, the reverse assignment of spin-allowed and spin-forbidden bands is wrong, and the iron complex is not low spin with fluoride ligands.

Hence, Spin forbidden transitions in [FeF6]3– and spin allowed transitions in [CoF6]3–.

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