On the basis of valence bond theory explain the shape and magnetic character of `K_4[Fe(CN)_6]` or `[FeCN_6]^(4-)` ion.

Explanation

Outer electronic configuration of iron (Z = 26). Its ground state is `3d^6 4s^2`. Eron in this complex is in +2 oxidation state. Iron achieves the +2 oxidation state by the loss of two 4s electrons. The resulting `Fe^(2+)` ion has outer electronic, configuration of `3d^6`. It has been experimentally observed that this complex is diamagnetic and as such has no unpaired electrón: To account for this, two unpaired electrons in the 3d-subshell pair up, thus leaving two 3d-orbitals empty. These two vacant 3d-orbitais alongwith one 4s-orbital and three 4p-orbitals hybridise to give six equivalent `d^2sp^3` hybridised orbitals. Six pairs of electrons, one from each cyanide ion, occupy the six vacant hybrid orbitals so produced. The resulting complex ion has an octahedral geometry and is diamagnetic due to the absence of unpaired electrons. The hybridisation and the formation of the complex is illustrated ahead.