Which of the following is inner orbital complex as well as diamagnetic in nature ?

Options

A. `[Ir(H_(2)O)_(6)]^(3+)`
B. `[Ni(NH_(3))_(6)]^(2+)`
C. `[Cr(NH_(3))_(6)]^(3+)`
D. `[Co(NH_(3))_(6)]^(2+)`

Answer

To determine which of the following complexes is an inner orbital complex and is also diamagnetic in nature, we can follow these steps:

### Step 1: Understand the Definitions
- **Inner Orbital Complex**: A complex where the d-orbitals involved in bonding are from the inner d-subshell (n-2). This typically occurs with transition metals from the 4d and 5d series.
- **Diamagnetic**: A substance that has all its electrons paired, resulting in no net magnetic moment.

### Step 2: Analyze Each Complex
1. **[Edium(H2O)6]³⁺**:
- Edium (likely referring to a transition metal like Lutetium) in the +3 oxidation state has a d-electron configuration of 5d⁶.
- Water (H2O) is a strong field ligand, which means it will cause pairing of the d-electrons.
- The hybridization will be d²sp³, indicating that the inner d-orbitals are involved.
- Since all electrons are paired, this complex is diamagnetic.

2. **[Ni(H2O)6]²⁺**:
- Nickel in the +2 oxidation state has a d-electron configuration of 3d⁸.
- Water is a weak field ligand for Ni²⁺, leading to no pairing of electrons.
- This complex is paramagnetic due to unpaired electrons.

3. **[Cr(NH3)6]³⁺**:
- Chromium in the +3 oxidation state has a d-electron configuration of 3d³.
- Ammonia (NH3) is a weak field ligand, leading to unpaired electrons.
- This complex is also paramagnetic.

4. **[Co(NH3)6]²⁺**:
- Cobalt in the +2 oxidation state has a d-electron configuration of 3d⁷.
- Again, ammonia is a weak field ligand, leading to unpaired electrons.
- This complex is paramagnetic.

### Step 3: Conclusion
From the analysis, the only complex that is both an inner orbital complex and diamagnetic is **[Edium(H2O)6]³⁺**.

### Final Answer
The correct answer is **[Edium(H2O)6]³⁺**.

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