Why is[FeF6]3- ion paramagnetic while [Fe(CN)6]4-ion diamagnetic?

In case of [FeF6]3–complex ion, the oxidation state of central iron metal is +3.

The atomic number of iron metal : 26 and that of ion(III)ion : 23. Valence shell electronic configuration of Fe(III) ion : [ Ar ] d5.

Co-ordination number of central metal Fe(III)ion : 6. Arrangement of [FeF6]3– complex ion : Octahedral.

Now ligand F–, which is a weak field ligand. Under the influence of the octahedral

crystal field, the five degenerate d-orbitals of Fe(III) ion are splitted into two sets of energetically different orbitals.

These two sets of orbitals are energetically lower t2g orbital and energetically higher eg orbital.

Since ligand F–, is a weak field ligand, hence the complex is high spin one.

So, under the influence of the octahedral crystal field, the possible electronic arrangement of Fe(III) ion is t2g3, eg2.

The octahedral crystal field splitting of [FeF6]3– complex ion is as follows,

From the above crystal field splitting diagram of Fe(III) ion, it is evidently, shown that, the Fe(III)ion have five unpaired electrons in its outer 3d-orbital.

Hence, hexafluoridoferrate(III) ion, [FeF6]3– is paramagnetic in nature.

On the other hand, in case of [Fe(CN)6]4–ion, the oxidation state of iron is +2 .

The ligand CN– which is a strong field ligand. Now, the valence shell electronic configuration of Fe(II) ion is [ Ar ] d6 .

Again, since the ligand CN– is a strong field ligand, hence the complex is low spin one.

Therefore, under the influence of the octahedral crystal field, the possible electronic arrangement of Fe(II) ion is t2g6, eg0.

The octahedral crystal field splitting of [Fe(CN)6]4–complex ion is as follows,

From the above crystal field splitting diagram of Fe(II) ion, it is evidently shown that, the Fe(II)ion have no unpaired electrons in its outer 3d-orbital.

That is, all electrons are paired, hence [Fe(CN)6]4–complex ion is diamagnetic in nature.

Originally published at https://chemisfast.blogspot.com.