Question

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

Hint:

Weak field ligands (H\(_2\)O, F\(^-\)) → high spin → outer orbital complex (\(sp^3d^2\)).

Answer 1

Step 1: Find oxidation state of Fe. Water is a neutral ligand. Let oxidation state of Fe = \( x \) \[ x + 6(0) = +3 \Rightarrow x = +3 \]
Step 2: Electronic configuration of Fe. Fe (Z = 26): \[ [Ar] \, 3d^6 4s^2 \] Fe\(^{3+}\): \[ 3d^5 \]
Step 3: Determine hybridisation. Water is a weak field ligand → high spin complex. For octahedral high spin \(d^5\): \[ \text{Hybridisation} = sp^3d^2 \]
Answer:

• Oxidation state = +3
• Hybridisation = \( sp^3d^2 \)