The spin only magnetic of `Cr^(3+)` in aqueous solution would be `:`

Options

A. `sqrt(3) B.M.`
B. `sqrt(15)B.M.`
C. `sqrt(8)B.M.`
D. `sqrt(35)B.M.`

Explanation

To find the spin-only magnetic moment of `Cr^(3+)` in aqueous solution, we can follow these steps:

### Step 1: Determine the Atomic Number and Electron Configuration of Chromium
- Chromium (Cr) has an atomic number of 24.
- The electron configuration of neutral chromium is:
\[ [Ar] 4s^1 3d^5 \]

### Step 2: Identify the Oxidation State
- In the question, chromium is in the +3 oxidation state (`Cr^(3+)`).
- This means that it has lost 3 electrons.

### Step 3: Adjust the Electron Configuration for `Cr^(3+)`
- When chromium loses 3 electrons, it loses the 1 electron from the 4s orbital and 2 electrons from the 3d orbital.
- The electron configuration for `Cr^(3+)` will be:
\[ [Ar] 3d^3 \]
- This indicates that there are 3 electrons in the 3d subshell.

### Step 4: Determine the Number of Unpaired Electrons
- In the `3d^3` configuration, all three electrons are unpaired.
- Therefore, the number of unpaired electrons (n) is 3.

### Step 5: Use the Formula for Spin-Only Magnetic Moment
- The formula for calculating the spin-only magnetic moment (μ) is:
\[ \mu = \sqrt{n(n + 2)} \]
where n is the number of unpaired electrons.

### Step 6: Substitute the Value of n into the Formula
- Substituting n = 3 into the formula:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Step 7: Conclusion
- The spin-only magnetic moment of `Cr^(3+)` in aqueous solution is:
\[ \mu = \sqrt{15} \, \text{BM} \]

### Final Answer
- The correct option is the one that states the magnetic moment is \(\sqrt{15} \, \text{BM}\).

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