Which of the following complex compound is low spin, inner orbital , diamagnetic complex ?

Options

A. `[Ni(NH_3)_6]Cl_2`
B. `K_3 [Fe(CN)_6]`
C. `K_2 [PtCl_6]`
D. `[Cr(H_2 O)_6 ] Cl_3`

Explanation

To determine which of the given complex compounds is a low spin, inner orbital, diamagnetic complex, we need to analyze each option based on the definitions of these terms:

1. **Low Spin**: This refers to complexes where the pairing of electrons occurs in the lower energy orbitals before occupying the higher energy orbitals. This typically happens in the presence of strong field ligands.

2. **Inner Orbital**: This means that the complex uses the inner d-orbitals (n-1 d-orbitals) for bonding, which occurs in transition metals with a higher oxidation state.

3. **Diamagnetic**: A complex is diamagnetic if it has no unpaired electrons.

Let's analyze each option step by step:

### Step 1: Analyze the first complex (Ni(NH₃)₆Cl₂)
- **Oxidation State**: Nickel (Ni) in this complex has an oxidation state of +2 (Ni²⁺).
- **Electron Configuration**: Ni has the configuration [Ar] 3d⁸ 4s². For Ni²⁺, it becomes [Ar] 3d⁸.
- **Ligand Field**: Ammonia (NH₃) is a weak field ligand, leading to a high spin configuration. Therefore, this complex is paramagnetic (has unpaired electrons).
- **Conclusion**: This complex is **not** low spin, inner orbital, or diamagnetic.

### Step 2: Analyze the second complex ([Fe(CN)₆]³⁻)
- **Oxidation State**: Iron (Fe) in this complex has an oxidation state of +3 (Fe³⁺).
- **Electron Configuration**: Fe has the configuration [Ar] 3d⁶ 4s². For Fe³⁺, it becomes [Ar] 3d⁵.
- **Ligand Field**: Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons. Thus, all 5 electrons will pair in the lower d-orbitals.
- **Conclusion**: This complex is **not** diamagnetic (it has unpaired electrons).

### Step 3: Analyze the third complex ([PtCl₆]²⁻)
- **Oxidation State**: Platinum (Pt) in this complex has an oxidation state of +4 (Pt⁴⁺).
- **Electron Configuration**: Pt has the configuration [Xe] 4f¹⁴ 5d⁹ 6s². For Pt⁴⁺, it becomes [Xe] 4f¹⁴ 5d⁶.
- **Ligand Field**: Chloride (Cl⁻) is a weak field ligand, but due to the high oxidation state of Pt, pairing occurs. All 6 electrons will occupy the lower d-orbitals.
- **Conclusion**: This complex is low spin, uses inner d-orbitals, and has no unpaired electrons, making it **diamagnetic**. This is the correct answer.

### Step 4: Analyze the fourth complex ([Cr(H₂O)₆]³⁺)
- **Oxidation State**: Chromium (Cr) in this complex has an oxidation state of +3 (Cr³⁺).
- **Electron Configuration**: Cr has the configuration [Ar] 3d⁵ 4s¹. For Cr³⁺, it becomes [Ar] 3d³.
- **Ligand Field**: Water (H₂O) is a weak field ligand, leading to a high spin configuration. Therefore, this complex has unpaired electrons.
- **Conclusion**: This complex is **not** diamagnetic.

### Final Conclusion:
The only complex that fits all the criteria of being low spin, inner orbital, and diamagnetic is **[PtCl₆]²⁻**.

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