The calculated spin only magnetic moment of `Cr^(2+)` ion is:

Options

A. `3.87 BM`
B. `4.90 BM`
C. `5.92 BM`
D. `2.84 BM`

Answer

To calculate the spin-only magnetic moment of the Cr²⁺ ion, we can follow these steps:

### Step 1: Determine the electronic configuration of Cr²⁺
Chromium (Cr) has an atomic number of 24. Its ground state electronic configuration is:
\[ \text{Cr: } 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1, 3d^5 \]

When chromium loses two electrons to form Cr²⁺, the electrons are removed first from the 4s orbital and then from the 3d orbital. Therefore, the electronic configuration for Cr²⁺ is:
\[ \text{Cr}^{2+}: 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^4 \]

### Step 2: Identify the number of unpaired electrons
In the 3d subshell of Cr²⁺, we have 4 electrons. The distribution of these electrons in the d-orbitals can be represented as follows:
- The d-orbitals can hold a maximum of 10 electrons (5 orbitals, 2 electrons each).
- For 4 electrons, they will occupy the orbitals singly before pairing occurs (Hund's rule).

Thus, the configuration in the d-orbitals will be:
\[ \uparrow \uparrow \uparrow \uparrow \]
This indicates that there are 4 unpaired electrons.

### Step 3: Use the formula for spin-only magnetic moment
The formula for calculating the spin-only magnetic moment (\( \mu \)) is:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 4: Substitute the value of n
In our case, \( n = 4 \) (the number of unpaired electrons):
\[ \mu = \sqrt{4(4 + 2)} \]
\[ \mu = \sqrt{4 \times 6} \]
\[ \mu = \sqrt{24} \]

### Step 5: Calculate the magnetic moment
Now, we calculate \( \sqrt{24} \):
\[ \mu \approx 4.89 \, \text{BM} \]

### Conclusion
The calculated spin-only magnetic moment of the Cr²⁺ ion is approximately:
\[ \mu \approx 4.9 \, \text{BM} \]