Question

The Valence Bond Theory (VBT) explains the formation, magnetic behaviour and geometry of coordination compounds. The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields.

Answer the following question :

(c) Write hybridization and magnetic behaviour of the complex \( \mathrm{[Fe(CN)_6]^{3-}} \).

[Atomic No. : Fe = 26]

Hint:

Fe3+ is d5 and CN- is a strong ligand, so electrons pair up to t2g5 (1 unpaired). The complex uses d2sp3 hybrid orbitals and is paramagnetic.

Answer 1

Step 1: Find the oxidation state of iron:
Let the oxidation number of Fe be x. Cyanide is -1.
x + 6(-1) = -3, so x = +3.
Iron (Z = 26) is \( \mathrm{[Ar]\,3d^6\,4s^2} \), so \( \mathrm{Fe^{3+}} \) is \( 3d^5 \).

Step 2: Decide the ligand strength:
CN- is a strong field ligand. It forces the d electrons to pair as far as possible.

Step 3: Rearrange the electrons:
The five 3d electrons go into the three lower orbitals as \( t_{2g}^{5} \). Two pairs fill two orbitals and one electron stays alone in the third.
Two 3d orbitals are now empty.

Step 4: Hybridisation:
The two empty 3d orbitals, one 4s orbital and three 4p orbitals mix to give six \( d^2sp^3 \) hybrid orbitals.
Each hybrid orbital takes a lone pair from one CN- ligand. This is an inner orbital, octahedral complex.

Step 5: Magnetic behaviour:
There is one unpaired electron, so the complex is paramagnetic.
\[ \mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \text{BM} \]

Final Answer:
Hybridisation is \( d^2sp^3 \) (octahedral, inner orbital) and the complex is paramagnetic with one unpaired electron.
\[ \boxed{d^2sp^3,\ \text{paramagnetic (1 unpaired electron)}} \]