Magnetic moment (spin only) of octahedron complex having CFSE=-`0.8Delta_(0)` and surrounded by weak field ligands can be:

Options

A. `sqrt(15)` BM
B. `sqrt(8)`BM
C. (a) and (b) both
D. none of these

Answer

To solve the problem of determining the magnetic moment (spin only) of an octahedral complex with a given crystal field stabilization energy (CFSE) of -0.8Δ₀ and surrounded by weak field ligands, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Crystal Field Stabilization Energy (CFSE):**
- In an octahedral complex, the d-orbitals split into two sets: the lower energy \( t_{2g} \) (three orbitals) and the higher energy \( e_g \) (two orbitals).
- The CFSE can be calculated using the formula:
\[ \text{CFSE} = (-0.4 \Delta_0 \times n_{t_{2g}}) + (0.6 \Delta_0 \times n_{e_g}) \]
where \( n_{t_{2g}} \) is the number of electrons in the \( t_{2g} \) orbitals and \( n_{e_g} \) is the number of electrons in the \( e_g \) orbitals.

2. **Setting Up the Equation:**
- Given CFSE = -0.8Δ₀, we can set up the equation:
\[ -0.4 \Delta_0 \times n_{t_{2g}} + 0.6 \Delta_0 \times n_{e_g} = -0.8 \Delta_0 \]
- Dividing through by Δ₀ (assuming Δ₀ ≠ 0):
\[ -0.4 n_{t_{2g}} + 0.6 n_{e_g} = -0.8 \]

3. **Considering Electron Configuration:**
- Since the complex is surrounded by weak field ligands, electrons will fill the orbitals singly before pairing occurs.
- Let's consider two scenarios for \( d \)-electron configurations.

4. **Scenario 1: d⁶ Configuration**
- If we assume a \( d^6 \) configuration, we can have:
- \( n_{t_{2g}} = 4 \) and \( n_{e_g} = 2 \)
- Plugging into the CFSE equation:
\[ -0.4 \times 4 + 0.6 \times 2 = -1.6 + 1.2 = -0.4 \quad \text{(not valid)} \]
- Let's try \( n_{t_{2g}} = 3 \) and \( n_{e_g} = 3 \):
- This gives:
\[ -0.4 \times 3 + 0.6 \times 3 = -1.2 + 1.8 = 0.6 \quad \text{(not valid)} \]

5. **Scenario 2: d⁷ Configuration**
- For \( d^7 \) configuration, we can have:
- \( n_{t_{2g}} = 5 \) and \( n_{e_g} = 2 \)
- Plugging into the CFSE equation:
\[ -0.4 \times 5 + 0.6 \times 2 = -2 + 1.2 = -0.8 \quad \text{(valid)} \]
- This means we have 3 unpaired electrons (5 in \( t_{2g} \) and 2 in \( e_g \)).

6. **Calculating the Magnetic Moment:**
- The formula for the spin-only magnetic moment (μ) is:
\[ \mu = \sqrt{n(n + 2)} \quad \text{where } n \text{ is the number of unpaired electrons.} \]
- Here, \( n = 3 \):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Final Answer:
The magnetic moment (spin only) of the octahedral complex is \( \sqrt{15} \) Bohr magnetons.