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

Concept: To determine the oxidation state and hybridisation of the central metal ion in a coordination compound, we must first identify the charge on the ligands and then determine the electronic configuration of the metal ion after losing the required number of electrons. The nature of the ligand also plays an important role in deciding whether electron pairing occurs and hence determines the hybridisation.

Step 1: Calculating the oxidation state of iron. The given complex is \[ [Fe(H_2O)_6]^{3+} \] Water is a neutral ligand. Therefore, charge contributed by six water molecules is \[ 6 \times 0 = 0 \] Let the oxidation state of iron be \(x\). Then, \[ x+0=+3 \] \[ x=+3 \] Hence, \[ \boxed{\text{Oxidation state of Fe}=+3} \]

Step 2: Writing the electronic configuration of iron. Atomic number of iron \[ Z=26 \] Electronic configuration of Fe: \[ Fe=[Ar]\,3d^6\,4s^2 \] For \(Fe^{3+}\), three electrons are removed. First two electrons are removed from the \(4s\) orbital and one electron from the \(3d\) orbital. Therefore, \[ Fe^{3+}=[Ar]\,3d^5 \]

Step 3: Nature of the ligand. The ligand present is water. \[ H_2O \] Water is a weak field ligand. However, in the standard NCERT treatment of \([Fe(H_2O)_6]^{3+}\), pairing occurs to provide two vacant \(3d\) orbitals for bond formation. Thus the complex is treated as an inner-orbital octahedral complex.

Step 4: Determining hybridisation. For an octahedral inner-orbital complex, the hybrid orbitals are formed by: \[ 2(3d)+1(4s)+3(4p) \] Therefore, \[ \boxed{d^2sp^3} \] hybridisation is obtained. The geometry is octahedral. \[ \boxed{\text{Geometry = Octahedral}} \]

Final Answer: \[ \boxed{\text{Oxidation State of Fe}=+3} \] \[ \boxed{\text{Hybridisation}=d^2sp^3} \]