Spin only' magnetic moment of Ni in `[Ni(dmg)_2]` is same as that found in

Options

A. Ni in `[NiCl_2(PPh_3)_2]`
B. Mn in `[MnO_4]^(2-)`
C. Co in `[CoBr_4]^(2-)`
D. Pt in `[Pt (H_2O)_2 Br_2]`

Explanation

To determine the 'spin only' magnetic moment of Ni in `[Ni(dmg)_2]` and find a compound with the same magnetic moment, we will follow these steps:

### Step 1: Identify the oxidation state of Nickel in `[Ni(dmg)_2]`
- The ligand `dmg` (dimethylglyoxime) has a charge of -2. Therefore, in `[Ni(dmg)_2]`, Nickel (Ni) must be in the +2 oxidation state to balance the -4 charge from the two `dmg` ligands.

### Step 2: Determine the electron configuration of Ni²⁺
- The atomic number of Nickel (Ni) is 28. Its electron configuration is:
\[ \text{Ni: } [Ar] 3d^8 4s^2 \]
- For Ni²⁺, we remove two electrons (the 4s electrons are removed first):
\[ \text{Ni}^{2+}: [Ar] 3d^8 \]

### Step 3: Identify the geometry of `[Ni(dmg)_2]`
- The complex `[Ni(dmg)_2]` is a square planar complex due to the nature of the `dmg` ligand, which is a bidentate ligand. In square planar complexes, the d-orbitals are split in such a way that the electrons pair up.

### Step 4: Calculate the spin-only magnetic moment
- The formula for the spin-only magnetic moment (\( \mu_s \)) is given by:
\[ \mu_s = \sqrt{n(n+2)} \]
where \( n \) is the number of unpaired electrons.
- In the case of Ni²⁺ in a square planar complex, all 8 d-electrons will pair up, resulting in 0 unpaired electrons:
\[ n = 0 \]
- Therefore, the magnetic moment is:
\[ \mu_s = \sqrt{0(0+2)} = \sqrt{0} = 0 \]

### Step 5: Find a compound with the same magnetic moment
- We need to find another compound that also has a magnetic moment of 0.
- The options provided in the question include:
- NiCl₂·PbH₃
- MnO₄²⁻
- COBr₄²⁻
- Pt₂H₂·Br₂

- **Nickel in NiCl₂·PbH₃**: Nickel is in +2 oxidation state and is likely to form a square planar complex, hence it will also have a magnetic moment of 0.
- **Manganese in MnO₄²⁻**: Manganese in this case has unpaired electrons, so it will not have a magnetic moment of 0.
- **Carbon in COBr₄²⁻**: This is a tetrahedral complex, which typically has unpaired electrons, hence not 0.
- **Platinum in Pt₂H₂·Br₂**: Platinum in +2 oxidation state forms a square planar complex and will also have a magnetic moment of 0.

### Conclusion
The spin-only magnetic moment of Ni in `[Ni(dmg)_2]` is the same as that found in:
- **NiCl₂·PbH₃** and **Pt₂H₂·Br₂**.