Question

Write the hybridization and magnetic behaviour of \([\mathrm{Fe(CN)_6}]^{4-}\).

(Atomic number : Fe = 26)

Hint:

\(\mathrm{Fe^{2+}}\) is \(3d^6\); strong field \(\mathrm{CN^-}\) pairs all electrons, so \(d^2sp^3\), diamagnetic.

Answer 1

Step 1: Understanding the Concept:
In valence bond theory, the hybridization of the central metal depends on the oxidation state of the metal and on whether the ligand is strong field or weak field. A strong field ligand pairs up the d electrons.

Step 2: Oxidation state and electronic configuration:
In \([\mathrm{Fe(CN)_6}]^{4-}\), the charge on each \(\mathrm{CN^-}\) is \(-1\), so \(x + 6(-1) = -4\) and \(x = +2\).
\(\mathrm{Fe}\,(Z = 26) = [\mathrm{Ar}]\,3d^6 4s^2\), so \(\mathrm{Fe^{2+}} = 3d^6\).

Step 3: Effect of the ligand:
\(\mathrm{CN^-}\) is a strong field ligand. It forces the six d electrons to pair up in the three \(t_{2g}\) orbitals, \(3d^6 = t_{2g}^6\). This leaves two d orbitals empty.

Step 4: Hybridization and magnetic behaviour:
The two empty \(3d\) orbitals, one \(4s\) and three \(4p\) orbitals mix to give six \(d^2sp^3\) hybrid orbitals, which accept six electron pairs from the six \(\mathrm{CN^-}\) ligands. The complex is octahedral and inner orbital.
There is no unpaired electron, so the complex is diamagnetic.

Final Answer:
Hybridization: \(d^2sp^3\) (octahedral, inner orbital complex). Magnetic behaviour: diamagnetic (\(n = 0\)).