Magnetic moment of `Ce^(3+)` ion on the basis of 'spin-only' formula will be ______ B.M.

Options

A. 1.232
B. 1.332
C. 1.532
D. 1.732

Explanation

To calculate the magnetic moment of the `Ce^(3+)` ion using the spin-only formula, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Atomic Number of Cerium (Ce)**:
- The atomic number of cerium (Ce) is 58.

2. **Write the Electronic Configuration of Cerium**:
- The electronic configuration of cerium is:
\[ \text{Xe} \, 4f^1 \, 5d^1 \, 6s^2 \]
- Here, "Xe" represents the electron configuration of xenon, which accounts for the first 54 electrons.

3. **Determine the Electronic Configuration of `Ce^(3+)`**:
- The `Ce^(3+)` ion means that cerium has lost 3 electrons. The electrons are lost first from the outermost orbitals:
- 2 electrons from the `6s` orbital and 1 electron from the `5d` orbital.
- Therefore, the electronic configuration of `Ce^(3+)` becomes:
\[ \text{Xe} \, 4f^1 \]

4. **Identify the Number of Unpaired Electrons**:
- In the `4f` subshell, there is 1 electron (from `4f^1`), which is unpaired.
- Thus, the number of unpaired electrons (n) is 1.

5. **Use the Spin-Only Magnetic Moment Formula**:
- The formula for calculating the magnetic moment (μ) based on the number of unpaired electrons is:
\[ \mu = \sqrt{n(n + 2)} \]
- Substitute n = 1 into the formula:
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{1 \times 3} = \sqrt{3} \]

6. **Calculate the Value of the Magnetic Moment**:
- The value of \(\sqrt{3}\) is approximately:
\[ \mu \approx 1.732 \, \text{B.M.} \]

### Final Answer:
The magnetic moment of `Ce^(3+)` ion on the basis of the spin-only formula will be **1.732 B.M.**