Effective magentic moment of `Sc^(3+)` ion is

Options

A. `1.73`
B. 0
C. `5.92`
D. `2.83`

Explanation

To find the effective magnetic moment of the `Sc^(3+)` ion, we follow these steps:

### Step 1: Determine the Atomic Number and Electronic Configuration
- The atomic number of Scandium (Sc) is 21.
- The electronic configuration of neutral Sc is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^1 \]

### Step 2: Adjust for the Scandium Ion (Sc^(3+))
- To find the electronic configuration of `Sc^(3+)`, we need to remove three electrons from the neutral Sc configuration.
- The electrons are removed first from the outermost shell (4s) and then from the 3d subshell:
- Remove 2 electrons from 4s: \(4s^2 \rightarrow 0\)
- Remove 1 electron from 3d: \(3d^1 \rightarrow 0\)
- Thus, the electronic configuration of `Sc^(3+)` becomes:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \]

### Step 3: Count the Number of Unpaired Electrons
- In the configuration of `Sc^(3+)`, all the orbitals are filled:
- \(1s^2\), \(2s^2\), \(2p^6\), \(3s^2\), \(3p^6\)
- There are **no unpaired electrons** in this configuration.

### Step 4: Calculate the Effective Magnetic Moment
- The formula for the effective magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{n(n + 2)} \, \text{Bohr magneton} \]
where \(n\) is the number of unpaired electrons.
- Since \(n = 0\) for `Sc^(3+)`, we substitute:
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \, \text{Bohr magneton} \]

### Conclusion
- The effective magnetic moment of `Sc^(3+)` is **0 Bohr magneton**.

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