Which of the following system has maximum value of `mu` (only spin magnetic moment) ?

Options

A. `d^(5)(Delta_(0)gtP)`
B. `d^(8)` (tetrahedral)
C. `d^(6)` (high spin)
D. `d^(9)` (octahedral)

Answer

To determine which of the given systems has the maximum value of the spin magnetic moment (μ), we will follow these steps:

### Step 1: Understand the Formula for Spin Magnetic Moment
The spin magnetic moment (μ) is calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons. The more unpaired electrons there are, the greater the value of μ.

### Step 2: Analyze Each Configuration
We will analyze the given configurations one by one to find the number of unpaired electrons.

**Option A: d5 Configuration (Low Spin)**
- Given that the splitting energy (Δo) is greater than the pairing energy (P), this is a low-spin configuration.
- In a low-spin d5 configuration, the electrons will pair up in the lower energy orbitals first.
- The configuration will have 1 unpaired electron.
- Calculation of μ:
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73 \text{ Bohr magnetons} \]

**Option B: d8 Configuration (Tetrahedral)**
- Tetrahedral complexes are always high-spin.
- In a d8 tetrahedral configuration, the electrons do not pair up.
- The configuration will have 2 unpaired electrons.
- Calculation of μ:
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83 \text{ Bohr magnetons} \]

**Option C: d6 Configuration (High Spin)**
- In a high-spin d6 configuration, the electrons will occupy the higher energy orbitals before pairing.
- The configuration will have 4 unpaired electrons.
- Calculation of μ:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \text{ Bohr magnetons} \]

**Option D: d9 Configuration (Octahedral)**
- In a d9 configuration, there is typically 1 unpaired electron.
- Calculation of μ:
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73 \text{ Bohr magnetons} \]

### Step 3: Compare the Values of μ
Now we compare the values of μ calculated for each configuration:
- d5 (Low Spin): μ ≈ 1.73 Bohr magnetons
- d8 (Tetrahedral): μ ≈ 2.83 Bohr magnetons
- d6 (High Spin): μ ≈ 4.89 Bohr magnetons
- d9 (Octahedral): μ ≈ 1.73 Bohr magnetons

### Conclusion
The configuration with the maximum value of μ is the **d6 high-spin configuration**, which has 4 unpaired electrons and a magnetic moment of approximately 4.89 Bohr magnetons.

**Final Answer: Option C (d6 High Spin)**

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