The hybridisation and unpaired electrons in `[Fe(H_(2)O)_(6)^(2+)]` ion are :

Options

A. `sp^(3)d^(2)`,4
B. `d^(2)sp^(3),3`
C. `d^(2)sp^(3)`,4
D. `sp^(3)d^(2),2`

Answer

To determine the hybridization and the number of unpaired electrons in the complex ion \([Fe(H_2O)_6]^{2+}\), we can follow these steps:

### Step 1: Determine the oxidation state of Iron (Fe)
The complex ion has a charge of \(+2\). The water molecules are neutral ligands, contributing \(0\) charge. Let the oxidation state of iron be \(x\).

\[ x + 6(0) = +2 \implies x = +2 \]

### Step 2: Write the electron configuration of Iron in the +2 oxidation state
Iron has an atomic number of \(26\). The electron configuration of neutral iron is:

\[ [Ar] 4s^2 3d^6 \]

When iron loses \(2\) electrons to form \(Fe^{2+}\), the configuration becomes:

\[ [Ar] 3d^6 \]

### Step 3: Determine the geometry and coordination number
The complex \([Fe(H_2O)_6]^{2+}\) has a coordination number of \(6\), which indicates an octahedral geometry.

### Step 4: Understand the splitting of d-orbitals in octahedral complexes
In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). For \(d^6\) configuration in a weak field ligand like water, we expect a high-spin configuration.

### Step 5: Fill the d-orbitals according to Hund's rule
Since water is a weak field ligand, it does not cause pairing of electrons. Thus, the \(d\) orbitals will be filled as follows:

- The \(t_{2g}\) orbitals will be filled first, followed by the \(e_g\) orbitals.
- The filling will look like this:
- \(t_{2g}\): 3 electrons (1 in each orbital)
- \(e_g\): 3 electrons (1 in each orbital)

This results in the following arrangement:

\[ t_{2g}: \uparrow \uparrow \uparrow \quad e_g: \uparrow \uparrow \]

### Step 6: Count the number of unpaired electrons
From the arrangement above, we can see that there are \(4\) unpaired electrons (3 in \(t_{2g}\) and 1 in \(e_g\)).

### Step 7: Determine the hybridization
In an octahedral complex, the hybridization is \(sp^3d^2\) because it involves the mixing of one \(s\), three \(p\), and two \(d\) orbitals.

### Final Answer
The hybridization and the number of unpaired electrons in \([Fe(H_2O)_6]^{2+}\) are:

\[ \text{Hybridization: } sp^3d^2, \quad \text{Unpaired Electrons: } 4 \]

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