Explain the following about the complexes `[Fe(CN)_(6)]^(4-) and [Fe(H_(2)O)_(6)]^(2+)`:
(i) What type of shapes do they have ?
(ii) What type of hybridisation is involved in each case ?
(iii) Which of them is outer orbital complex and which one is inner orbital complex ?
(iv) Which of them is low spin complex and which one is high spin complex ?
(v) Compare their magnetic behaviour.
(vi) Why do they have different colours in dilute solution ?
(vii) Write the electronic configuration of metal ion in each case in terms of `t_(2g)` and `e_(g)` orbitals.

Explanation

(i) Both have octahedral shape.
(ii) In `[Fe(CN)_(6)]^(4-)`, `CN^(-)` is a strong ligand. 3d electrons pair up. Hybridisation is `d^(2)sp^(3)`.
In `[Fe(H_(2)O)_(6)]^(2+)` is an outer orbital complex while `[Fe(CN)_(6)]^(4-)` is an inner orbital complex.
(iv) In `[Fe(CN)_(6)]^(4-)`, after pairing up of 3d electrons, no unpaired electron is left while in `[Fe(H_(2)O)_(6)]^(2+)`, 4 unpaired electrons are present in the 3d subshell. Hence, `[Fe(CN)_(6)]^(4-)` is low spin complex while `[Fe(H_(2)O)_(6)]^(2+)` is high spin complex.
(v) As `[Fe(CN)_(6)]^(4-)` has no unpaired electron, it is diamagnetic while `[Fe(H_(2)O)_(6)]^(2+)` has 4 unpaired electrons, it is paramagnetic.
(vi) In both complexes, Fe is in +2 oxidation state and has the configuration `d^(6). CN^(-)` is a strong ligand while `H_(2)O` is a weak ligand. Hence, their crystal field splitting energies are different. Consequently, they absorb different wavelengths from white light and, therefore, the transmitted colours are also different.
(vii) `[Fe(CN)_(6)]^(4-)=t_(2g)^(6)e_(g)^(0), [Fe(H_(2)O)_(6)]^(2+)=t_(2g)^(4)e_(g)^(2)`.