Question

What is the correct electronic configuration of the central atom in $K_{ 4 }[Fe(CN)_{ 6 }]$ on crystal field theory?
(A.) ${ t }_{ 2g }^{ 4 }{ e }_{ g }^{ 2 }$
(B.) ${ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }$
(C.) ${ e }^{ 3 }{ t }_{ 2 }^{ 3 }$
(D.) ${ e }^{ 4 }{ t }_{ 2 }^{ 2 }$

Answer 1

Hint : To find the correct answer first try to find the oxidation state of Fe in this complex and then write its electronic configuration. Here you have to remember that ${ CN }^{ - }$ is a strong ligand that would pair up the electrons present in the d-orbitals.

Complete step by step solution :
Let’s find the correct answer to this question,
First, we need to calculate the oxidation state of iron -
4(+1) + x + 6(-1) = 0
+4 + x -6 = 0
x = +2
In $K_{ 4 }[Fe(CN)_{ 6 }]$,
Electronic configuration of Fe in ground state = [Ar]$3d^{ 6 }4s^{ 2 }$
Electronic configuration of Fe in excited state = ${ Fe }^{ 2+ }$= [Ar]$3d^{ 6 }4s^{ 0 }$
We can represent them as:

In this compound ${ CN }^{ - }$ is a strong field ligand that would pair up the electrons present in the d-orbitals.

We can see there are 3 pairs of electrons in ${ t }_{ 2g }$ and 0 pairs of electrons in ${ e }_{ g }$.
So, the correct electronic configuration of the central atom in $K_{ 4 }[Fe(CN)_{ 6 }]$ on crystal field theory is ${ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }$

So, the correct answer is “Option B”.

Additional Information: ${ t }_{ 2g }$ and ${ e }_{ g }$ refer to symmetry classes to which the d-orbitals belong in octahedral complexes.
In the crystal field model, the normally degenerate 3d orbitals split into two groups, a doubly degenerate set of two orbitals and lower energy triply degenerate set. When analyzed by mathematical group theory, the wave functions for these two groups fall into different symmetry groups, designated ${ e }_{ g }$ and ${ t }_{ 2g }$ respectively.

Note : Solid potassium ferrocyanide, both as the hydrate and anhydrous salts, has a complicated polymeric structure. We can also draw the structure of this complex $K_{ 4 }[Fe(CN)_{ 6 }]$ as,