Assertion: `K_(4)[Fe(CN_(6)]` is diamagnetic and `[Fe(H_(2)O)_(6)]Cl_(3)` is paramagnetic.
Reason: Hybridisation of central metal in `K_(4)[Fe(CN)_(6)]" is "sp^(3) d^(2)`, which in `[Fe(H_(2)O)_(6)]Cl_(3)" is "d^(2) sp^(3)`.

Options

A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion
B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
C. Assertion is true but Reason is false
D. Both Assertion and Reason are false

Explanation

To solve the question regarding the assertion and reason about the complexes \( K_4[Fe(CN)_6] \) and \( [Fe(H_2O)_6]Cl_3 \), we will analyze both statements step by step.

### Step 1: Analyze the Assertion
The assertion states that \( K_4[Fe(CN)_6] \) is diamagnetic.

1. **Determine the oxidation state of Iron in \( K_4[Fe(CN)_6] \)**:
- The cyanide ion \( CN^- \) has a charge of -1, and there are 6 cyanide ions.
- Therefore, the total negative charge from the cyanide ions is \( 6 \times (-1) = -6 \).
- To balance this, the oxidation state of Iron (Fe) must be +2.
- Thus, \( Fe^{2+} \) is present in the complex.

2. **Electronic configuration of \( Fe^{2+} \)**:
- The electronic configuration of neutral Iron (Fe) is \( [Ar] 3d^6 4s^2 \).
- For \( Fe^{2+} \), it loses two electrons, resulting in \( [Ar] 3d^6 \).

3. **Identify the ligand field strength**:
- Cyanide \( CN^- \) is a strong field ligand, which causes pairing of electrons in the d-orbitals.

4. **Determine the electron pairing**:
- In the presence of a strong field ligand, the 6 electrons in \( 3d \) will pair up, resulting in no unpaired electrons.
- Therefore, \( K_4[Fe(CN)_6] \) is indeed diamagnetic.

### Step 2: Analyze the Reason
The reason states that the hybridization of the central metal in \( K_4[Fe(CN)_6] \) is \( sp^3d^2 \) and in \( [Fe(H_2O)_6]Cl_3 \) it is \( d^2sp^3 \).

1. **Hybridization of \( K_4[Fe(CN)_6] \)**:
- Since \( Fe^{2+} \) has 6 electrons in the \( 3d \) orbitals and is coordinated by 6 cyanide ligands, the hybridization involves 2 \( d \) orbitals, 1 \( s \) orbital, and 3 \( p \) orbitals.
- Therefore, the correct hybridization is \( d^2sp^3 \) (not \( sp^3d^2 \)).

2. **Hybridization of \( [Fe(H_2O)_6]Cl_3 \)**:
- In this complex, \( Fe \) is in the +3 oxidation state, which means it has \( [Ar] 3d^5 \) configuration.
- Water \( H_2O \) is a weak field ligand, and the hybridization will involve 3 \( p \) orbitals and 2 \( d \) orbitals.
- Thus, the hybridization is also \( d^2sp^3 \) (not \( sp^3d^2 \)).

### Conclusion
- The assertion is **true**: \( K_4[Fe(CN)_6] \) is diamagnetic.
- The reason is **false**: The hybridization descriptions provided are incorrect.

### Final Answer
The correct option is **C**: The assertion is true, but the reason is false.