How many of the following metal ions have similar value of spin only magnetic moment in gaseous state_______________(Given : Atomic number : V, 23 , Cr, 24 , Fe, 26 , Ni, 28)
`V^(3+), Cr^(3+), Fe^(2+), Ni^(3+)`

Answer

To determine how many of the given metal ions have similar values of spin-only magnetic moment in the gaseous state, we will first need to calculate the number of unpaired electrons for each ion. The spin-only magnetic moment (μ) can be calculated using the formula:

\[ \mu = \sqrt{n(n + 2)} \]

where \(n\) is the number of unpaired electrons.

### Step 1: Determine the electronic configuration of each metal ion.

1. **Vanadium (V) - Atomic number 23**
- Ground state configuration: \([Ar] 3d^3 4s^2\)
- For \(V^{3+}\): Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: \([Ar] 3d^2\)
- Unpaired electrons: 2

2. **Chromium (Cr) - Atomic number 24**
- Ground state configuration: \([Ar] 3d^5 4s^1\)
- For \(Cr^{3+}\): Remove 3 electrons (1 from 4s and 2 from 3d)
- Configuration: \([Ar] 3d^3\)
- Unpaired electrons: 3

3. **Iron (Fe) - Atomic number 26**
- Ground state configuration: \([Ar] 3d^6 4s^2\)
- For \(Fe^{2+}\): Remove 2 electrons (2 from 4s)
- Configuration: \([Ar] 3d^6\)
- Unpaired electrons: 4 (in a high-spin state)

4. **Nickel (Ni) - Atomic number 28**
- Ground state configuration: \([Ar] 3d^8 4s^2\)
- For \(Ni^{3+}\): Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: \([Ar] 3d^7\)
- Unpaired electrons: 3 (in a high-spin state)

### Step 2: Calculate the spin-only magnetic moment for each ion.

1. **For \(V^{3+}\) (2 unpaired electrons)**:
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83 \, \mu_B \]

2. **For \(Cr^{3+}\) (3 unpaired electrons)**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \, \mu_B \]

3. **For \(Fe^{2+}\) (4 unpaired electrons)**:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \, \mu_B \]

4. **For \(Ni^{3+}\) (3 unpaired electrons)**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \, \mu_B \]

### Step 3: Compare the values of spin-only magnetic moments.

- \(V^{3+}\): 2.83 μB
- \(Cr^{3+}\): 3.87 μB
- \(Fe^{2+}\): 4.89 μB
- \(Ni^{3+}\): 3.87 μB

### Conclusion:

From the calculations, we observe that \(Cr^{3+}\) and \(Ni^{3+}\) have the same spin-only magnetic moment value of approximately 3.87 μB.

Thus, **2 metal ions** have similar values of spin-only magnetic moment.