Question

Write the oxidation state and hybridisation of the central metal in the following complex : \[ [Fe(H_2O)_6]^{3+} \] [Atomic number of Fe = 26]

Hint:

For octahedral complexes: \[ d^2sp^3 \rightarrow \text{Inner orbital complex} \] \[ sp^3d^2 \rightarrow \text{Outer orbital complex} \] Always calculate the oxidation state first and then write the electronic configuration of the metal ion before determining hybridisation.

Answer 1

Step 1: Find the oxidation state of Fe.
$H_2O$ is a neutral ligand (charge $= 0$). For $[Fe(H_2O)_6]^{3+}$: $x + 6(0) = +3$, giving oxidation state of Fe $= +3$.
Step 2: Write the configuration of $Fe^{3+}$.
Fe ($Z = 26$): $[Ar]\,3d^6\,4s^2$. Removing 3 electrons gives $Fe^{3+}$: $[Ar]\,3d^5$ (five singly occupied $d$-orbitals, half-filled configuration).
Step 3: Determine the hybridisation.
For an inner orbital octahedral complex, two inner $3d$ orbitals combine with one $4s$ and three $4p$ orbitals giving $d^2sp^3$ hybridisation and octahedral geometry. \[ \boxed{\text{Oxidation State} = +3,\quad \text{Hybridisation} = d^2sp^3} \]