Which of the following is a low spin (spin-paired) complex ?

Options

A. `[Co(NH_(3))_(6)]^(3+)`
B. `[Ni(NH_(3))_(6)]^(2+)`
C. `[Co(NH_(3))_(6)]^(2+)`
D. `[Fe(C_(2)O_(4))_(3)]^(3-)`

Explanation

To determine which of the following complexes is a low spin (spin-paired) complex, we will analyze the given options based on their hybridization and the number of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Complexes**:
We have the following complexes to analyze:
- Option A: [Co(NH3)6]³⁺
- Option B: [Ni(NH3)6]²⁺
- Option C: [Co(NH3)6]²⁺
- Option D: [Fe(C2O4)3]³⁻

2. **Determine the Oxidation States**:
- For Option A: Co is in the +3 oxidation state.
- For Option B: Ni is in the +2 oxidation state.
- For Option C: Co is in the +2 oxidation state.
- For Option D: Fe is in the +3 oxidation state.

3. **Electron Configuration**:
- Co³⁺: [Ar] 3d⁶
- Ni²⁺: [Ar] 3d⁸
- Co²⁺: [Ar] 3d⁷
- Fe³⁺: [Ar] 3d⁵

4. **Determine Hybridization**:
- Option A: [Co(NH3)6]³⁺ → D²sp³ hybridization (0 unpaired electrons)
- Option B: [Ni(NH3)6]²⁺ → Sp³d² hybridization (2 unpaired electrons)
- Option C: [Co(NH3)6]²⁺ → Sp³d² hybridization (3 unpaired electrons)
- Option D: [Fe(C2O4)3]³⁻ → D²sp³ hybridization (5 unpaired electrons)

5. **Identify Low Spin Complex**:
- A low spin complex is characterized by having a low number of unpaired electrons, which typically occurs in the presence of strong field ligands.
- In Option A, [Co(NH3)6]³⁺ has 0 unpaired electrons, indicating it is a low spin complex.
- The other options have unpaired electrons: Option B has 2, Option C has 3, and Option D has 5.

6. **Conclusion**:
- The complex that is a low spin (spin-paired) complex is **Option A: [Co(NH3)6]³⁺**.

### Final Answer:
**Option A: [Co(NH3)6]³⁺ is a low spin (spin-paired) complex.**