Which of the following is a pair of diamagnetic complexes ?

Options

A. `[Co(NH_(3))_(6)]^(3+) , [Fe(CN)_(6)]^(4-)`
B. `[Co(ox)_(3)]^(3-) , [FeF_(6)]^(3-)`
C. `[Fe(ox)_(3)]^(3-) , [FeF_(6)]^(3-)`
D. `[Fe(CN)_(6)]^(3-) , [CoF_(6)]^(3-)`

Explanation

To determine which of the given complexes are diamagnetic, we need to analyze their electronic configurations and oxidation states. A diamagnetic complex is one that has no unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Complexes**: The question provides two complexes, which we need to analyze for their magnetic properties.

2. **Determine Oxidation States**:
- For the first complex, Co(NH₃)₆³⁺:
- Ammonia (NH₃) is a neutral ligand, so the oxidation state of cobalt (Co) can be calculated as follows:
\[ x + 6(0) = +3 \implies x = +3 \]
- Therefore, cobalt is in the +3 oxidation state.

- For the second complex, Fe(CN)₆²⁻:
- Cyanide (CN) has a charge of -1. Therefore, the oxidation state of iron (Fe) can be calculated as:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
- Thus, iron is in the +2 oxidation state.

3. **Determine Electronic Configurations**:
- For Co³⁺ (Cobalt in +3 state):
- The ground state electronic configuration of cobalt (Co) is [Ar] 3d⁷ 4s².
- In the +3 state, two electrons are removed from the 4s and one from the 3d, resulting in:
\[ \text{Co}^{3+}: [Ar] 3d⁶ \]

- For Fe²⁺ (Iron in +2 state):
- The ground state electronic configuration of iron (Fe) is [Ar] 3d⁶ 4s².
- In the +2 state, two electrons are removed from the 4s, resulting in:
\[ \text{Fe}^{2+}: [Ar] 3d⁶ \]

4. **Filling the d-Orbitals**:
- Both complexes have strong field ligands (NH₃ and CN⁻), which cause pairing of electrons.
- For Co³⁺ (3d⁶):
- The 6 electrons will fill the d-orbitals as follows:
\[ \uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow\downarrow \]
- This results in 0 unpaired electrons, confirming that Co(NH₃)₆³⁺ is diamagnetic.

- For Fe²⁺ (3d⁶):
- The 6 electrons will fill the d-orbitals similarly:
\[ \uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow\downarrow \]
- This also results in 0 unpaired electrons, confirming that Fe(CN)₆²⁻ is diamagnetic.

5. **Conclusion**:
- Both complexes Co(NH₃)₆³⁺ and Fe(CN)₆²⁻ are diamagnetic as they have no unpaired electrons.

### Final Answer:
The pair of diamagnetic complexes is **Co(NH₃)₆³⁺ and Fe(CN)₆²⁻**.