Question

Discuss the nature of bonding in $[Fe(CN{)}_6{]}^{-4}$ on the basis of valence bond theory. What is the structure and magnetic characteristics? (At number Fe = 26).

Answer 1

Hint:From the given complex it can be seen that there is one Fe ion and six cyanide ions. The oxidation number of cyanide ions is +1 and the total charge of the complex is -4, so the oxidation state of the Fe ion can be calculated. Cyanide ion is a strong field ligand.

Complete answer:
The atomic number of iron (Fe) is 26. The outer electronic configuration of Fe is $3d^{6}4s^2$. The iron present in the complex ion is in +2 oxidation state, thus the iron loses its two electrons to form $F{e}^{2+}$ ion. The outer electronic configuration of $F{e}^{2+}$ is $3d^6$.
The ligand present in the complex is cyanide ion which is a strong field ligand which results in pairing.

The two electrons of 3d orbital jump and pair up with the two more unpaired electrons in 3d orbital thus leaving two empty d-orbital, one s-orbital and three p-orbital. Thus the resulting hybridization will be $d^{2}s{p}^3$. The empty orbital is filled with six pairs of electrons from $C{N}^{-}$ ion.
As, no unpaired electron is present in any orbital, therefore the compound is diamagnetic.
The complex ion formed $[Fe(CN{)}_6{]}^{-4}$, has octahedral geometry.

Note:
During the bonding of electrons, the pairing is done according to Hund’s rule which states that first every orbital will be filled with single electrons only then pairing of electrons is done.