Calculate the magnetic moment of a high-spin octahedral complex that has six electrons in 3d-orbitals

Answer

To calculate the magnetic moment of a high-spin octahedral complex with six electrons in the 3d orbitals, we can follow these steps:

### Step 1: Identify the electron configuration
The complex has six electrons in the 3d orbitals, which corresponds to the electron configuration of 3d^6.

### Step 2: Understand octahedral splitting
In an octahedral field, the 3d orbitals split into two sets:
- **T2g** (lower energy level)
- **Eg** (higher energy level)

For a high-spin complex, electrons will fill the orbitals according to Hund's rule, meaning they will occupy the degenerate orbitals singly before pairing up.

### Step 3: Fill the orbitals
For 3d^6 in a high-spin octahedral complex:
- The first three electrons will fill the T2g orbitals singly.
- The next three electrons will fill the remaining T2g orbitals, leading to a configuration of:
- T2g: ↑ ↑ ↑ (3 unpaired electrons)
- Eg: ↑ ↑ (2 paired electrons)

### Step 4: Count the number of unpaired electrons
From the filling:
- T2g has 4 unpaired electrons (3 in T2g and 1 in Eg).
- Therefore, the number of unpaired electrons (n) = 4.

### Step 5: Use the formula for magnetic moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
Where n is the number of unpaired electrons.

### Step 6: Substitute the value of n
Substituting n = 4 into the formula:
\[ \mu = \sqrt{4(4 + 2)} \]
\[ \mu = \sqrt{4 \times 6} \]
\[ \mu = \sqrt{24} \]

### Step 7: Calculate the magnetic moment
Calculating the square root:
\[ \mu \approx 4.89 \text{ Bohr magneton} \]

### Step 8: Round off the value
Rounding off gives us approximately:
\[ \mu \approx 5 \text{ Bohr magneton} \]

### Final Answer
The magnetic moment of the high-spin octahedral complex with six electrons in the 3d orbitals is **5 Bohr magneton**.

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