With the help of VBT explain the geometry of `K_4 [Fe(CN)_6` and predict its magnetic property.

Explanation

(i) The co-ordination number of `Fe^(2+) ` is 6 .
(ii) The geometry is octahedron.
(iii) It is diamagnetic .
`Fe` is Present as `Fe^(2+)` in `K_4 [Fe(CN)_6]`.
Fe : 26 : [Ar]
`Fe^(2+) :24 :[Ar]`

Hybrid orbitals in `[Fe(CN)]_6^(4-)`: [Ar]
When strong `CN^(-)` ligands approach the `Fe^(2+)` ion, the four unpaired electrons present in 3d-orbitals are forced to pair up against Hund.s rule. To accommodate the 12 electrons donated by 6 `CN^(-)` ligands `Fe^(2+)` must provide 6 vacant orbitals. The two 3d-orbitals, one 4s-orbital and three 4p-orbitals undergo `d^2 sp^3` hybridization giving 6 hybrid orbitals. They align themselves at 6 corners of an octahedron with `Fe^(2+)` at the centre. The 6 pairs of electrons of 6 `CN^-` ligands overlap with 6-hybrid vacant orbitals forming 6 co-ordinate bonds. So, co-ordination number of `Fe^(2+)` is 6.
After overlapping and bond formation, there are no unpaired electrons in the complex. So, it is diamagnetic.
Since (n-1) d-orbitals take part in bonding it is an inner orbital complex.