Assertion : `[Fe(H_(2)O)_(6)]^(2+)` is `sp^(3) d^(2)` hybridised and paramagnetic complex ion .
Reason : `[Fe(H_(2)O)_(6)]^(2+)` has four unpaired electrons as `H_(2)O` is a weak field ligand .

Options

A. If both Assertion and reason are true and reason is the correct explanation of assertion
B. If both assertion and reason are true but reason is not the correct explanation of assertion
C. If assertion is true but reason is false
D. if both assertion and reason are false

Explanation

To solve the given question, we need to analyze both the assertion and the reason provided regarding the complex ion \([Fe(H_2O)_6]^{2+}\).

### Step-by-Step Solution:

1. **Identify the oxidation state of Iron in the complex**:
- The complex ion is \([Fe(H_2O)_6]^{2+}\).
- Since \(H_2O\) is a neutral ligand, the charge of the complex is solely due to the iron ion.
- Therefore, the oxidation state of iron (Fe) in this complex is +2.

2. **Determine the electronic configuration of Iron**:
- The atomic number of iron (Fe) is 26.
- The electronic configuration of neutral iron is:
\[ [Ar] 3d^6 4s^2 \]
- For \([Fe]^{2+}\), we remove two electrons (from the 4s orbital):
\[ [Fe]^{2+} = [Ar] 3d^6 \]

3. **Analyze the ligand field strength**:
- \(H_2O\) is classified as a weak field ligand.
- Weak field ligands do not cause pairing of electrons in the d-orbitals.

4. **Fill the d-orbitals according to Hund's rule**:
- Since \(H_2O\) is a weak field ligand, the six electrons in the \(3d\) subshell will occupy the available orbitals singly before pairing occurs.
- The distribution of the \(3d^6\) electrons will be:
- \(d_{xy}^1, d_{xz}^1, d_{yz}^1, d_{x^2-y^2}^1, d_{z^2}^1\) (5 unpaired electrons) and one electron will pair up in one of the orbitals.
- Therefore, there will be **4 unpaired electrons** in total.

5. **Determine the hybridization**:
- The coordination number of the complex is 6 (due to 6 water molecules).
- For a coordination number of 6, the hybridization is \(sp^3d^2\).
- This involves the mixing of one s orbital, three p orbitals, and two d orbitals.

6. **Check the paramagnetism**:
- A complex is paramagnetic if it has unpaired electrons.
- Since \([Fe(H_2O)_6]^{2+}\) has 4 unpaired electrons, it is indeed paramagnetic.

### Conclusion:
- The assertion that \([Fe(H_2O)_6]^{2+}\) is \(sp^3d^2\) hybridized and paramagnetic is **correct**.
- The reason that it has four unpaired electrons due to \(H_2O\) being a weak field ligand is also **correct**.
- Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.