Which of the following correctly matched?

Options

A. `[Fe(CN)_(6)]^(4-) and [Fe(CN)_(6)]^(3-)`-both are octahedral and diamagnetic with `d^(2)sp^(3)`. Hybridisation.
B. `Ni(CO)_(4) and [Ni(CN)_(4)]^(2-)`-both are tetrahedral and diamagnetic with `sp^(3)`-hybridisation
C. `Ni(CO)_(4) and [Co(CO)_(4)]^(-)`-both are tetrahedral and diamagnetic
D. `[Co(H_(2)O)_(6)]^(3+) and [Cr(H_(2)O)_(6)]^(3+)`-both are paramagnetic annd metal is `d^(2)sp^(3)`-hybridised

Answer

To determine which of the given coordination complexes is correctly matched, we will analyze each complex, calculate their oxidation states, electron configurations, and hybridizations, and then identify their magnetic properties.

### Step 1: Analyze the first complex - \( \text{Fe(CN)}_6^{4-} \) and \( \text{Fe(CN)}_6^{3-} \)

1. **Calculate Oxidation States**:
- For \( \text{Fe(CN)}_6^{4-} \):
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
- For \( \text{Fe(CN)}_6^{3-} \):
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

2. **Electron Configurations**:
- \( \text{Fe}^{2+} \): Argon (Ar) \( 4s^0 3d^6 \)
- \( \text{Fe}^{3+} \): Argon (Ar) \( 4s^0 3d^5 \)

3. **Hybridization**:
- Both complexes use strong field ligands (CN⁻), leading to pairing of electrons.
- \( \text{Fe}^{2+} \) will undergo \( d^2sp^3 \) hybridization (octahedral).
- \( \text{Fe}^{3+} \) will undergo \( d^2sp^3 \) hybridization (octahedral).

4. **Magnetic Properties**:
- \( \text{Fe}^{2+} \) is **diamagnetic** (all electrons paired).
- \( \text{Fe}^{3+} \) is **paramagnetic** (one unpaired electron).

### Step 2: Analyze the second complex - \( \text{Ni(CO)}_4 \) and \( \text{Ni(CN)}_4^{2-} \)

1. **Calculate Oxidation States**:
- For \( \text{Ni(CO)}_4 \):
\[ x + 0 = 0 \implies x = 0 \]
- For \( \text{Ni(CN)}_4^{2-} \):
\[ x + 4(-1) = -2 \implies x - 4 = -2 \implies x = +2 \]

2. **Electron Configurations**:
- \( \text{Ni}^0 \): Argon (Ar) \( 4s^2 3d^8 \)
- \( \text{Ni}^{2+} \): Argon (Ar) \( 4s^0 3d^8 \)

3. **Hybridization**:
- \( \text{Ni(CO)}_4 \) has \( sp^3 \) hybridization (tetrahedral).
- \( \text{Ni(CN)}_4^{2-} \) has \( dsp^2 \) hybridization (square planar).

4. **Magnetic Properties**:
- Both complexes are **diamagnetic** (all electrons paired).

### Step 3: Analyze the third complex - \( \text{Co(CN)}_4^{-} \) and \( \text{Co(CO)}_4^{0} \)

1. **Calculate Oxidation States**:
- For \( \text{Co(CN)}_4^{-} \):
\[ x + 4(-1) = -1 \implies x - 4 = -1 \implies x = +3 \]
- For \( \text{Co(CO)}_4^{0} \):
\[ x + 0 = 0 \implies x = 0 \]

2. **Electron Configurations**:
- \( \text{Co}^{3+} \): Argon (Ar) \( 4s^0 3d^6 \)
- \( \text{Co}^0 \): Argon (Ar) \( 4s^2 3d^7 \)

3. **Hybridization**:
- \( \text{Co(CN)}_4^{-} \) will undergo \( sp^3 \) hybridization (tetrahedral).
- \( \text{Co(CO)}_4^{0} \) will also undergo \( sp^3 \) hybridization (tetrahedral).

4. **Magnetic Properties**:
- \( \text{Co}^{3+} \) is **diamagnetic** (all electrons paired).
- \( \text{Co}^0 \) is **diamagnetic** (all electrons paired).

### Conclusion
- **Option A**: Incorrect (both are paramagnetic).
- **Option B**: Incorrect (one is tetrahedral, one is square planar).
- **Option C**: Correct (both tetrahedral and diamagnetic).
- **Option D**: Incorrect (different hybridizations).

### Final Answer
**Option C is the correct match.**