Spin only magnetic moments of a `d^(8)` ion in octahedral square planar and tetrahedral complexes, respectively are .

Options

A. `2.8BM,0` and `2.8BM`
B. `0,0` and `BM`
C. `2.8,2.8` and `BM`
D. None of these

Answer

To find the spin-only magnetic moments of a d^8 ion in octahedral, square planar, and tetrahedral complexes, we will follow these steps:

### Step 1: Determine the hybridization and electron configuration for the octahedral complex.
- **Hybridization**: In an octahedral complex, the hybridization is typically sp³d².
- **Electron Configuration**: For a d^8 ion, we have 8 electrons to fill in the d orbitals. In an octahedral field, the d orbitals split into two sets: the lower energy t₂g (dxy, dxz, dyz) and the higher energy e_g (dx²-y², dz²) orbitals. The 8 electrons will fill as follows:
- t₂g: 6 electrons (fully filled)
- e_g: 2 electrons (partially filled)
- **Unpaired Electrons**: There are 2 unpaired electrons in the e_g orbitals.

### Step 2: Calculate the magnetic moment for the octahedral complex.
- **Formula**: The magnetic moment (μ) can be calculated using the formula:
\[ μ = \sqrt{n(n + 2)} \]
where n is the number of unpaired electrons.
- **Calculation**: For n = 2 (unpaired electrons):
\[ μ = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83 \text{ Bohr magneton} \]

### Step 3: Determine the hybridization and electron configuration for the square planar complex.
- **Hybridization**: In a square planar complex, the hybridization is dsp².
- **Electron Configuration**: For a d^8 ion in a square planar field, the d orbitals also split, but the arrangement leads to the following filling:
- The d orbitals are filled as follows:
- d_x²-y²: 2 electrons (fully filled)
- d_xy, d_xz, d_yz, d_z²: 6 electrons (fully filled)
- **Unpaired Electrons**: All 8 electrons are paired.

### Step 4: Calculate the magnetic moment for the square planar complex.
- **Calculation**: For n = 0 (unpaired electrons):
\[ μ = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \text{ Bohr magneton} \]

### Step 5: Determine the hybridization and electron configuration for the tetrahedral complex.
- **Hybridization**: In a tetrahedral complex, the hybridization is sp³.
- **Electron Configuration**: For a d^8 ion in a tetrahedral field, the d orbitals remain unpaired:
- The filling will be:
- d_xy, d_xz, d_yz, d_x²-y², d_z²: 8 electrons (distributed among the orbitals)
- **Unpaired Electrons**: There are 2 unpaired electrons.

### Step 6: Calculate the magnetic moment for the tetrahedral complex.
- **Calculation**: For n = 2 (unpaired electrons):
\[ μ = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83 \text{ Bohr magneton} \]

### Conclusion:
- **Octahedral Complex**: 2.83 Bohr magneton
- **Square Planar Complex**: 0 Bohr magneton
- **Tetrahedral Complex**: 2.83 Bohr magneton

### Final Answer:
- The spin-only magnetic moments of a d^8 ion are:
- Octahedral: 2.83 Bohr magneton
- Square Planar: 0 Bohr magneton
- Tetrahedral: 2.83 Bohr magneton