which of the following sets of ions has the magnetic moment equal to `sqrt15`, `sqrt35`, `sqrt24` and 0 respectively?

Options

A. `Mn^(4+), Fe^(3+), Cr^(2+), Cu^+`
B. `Fe^(2+), Ti^(2+), Co^(3+), Cr^(2+)`
C. ``Zn^(2+), Mn^(4+), Cr^+,Cu^+`
D. `Fe^(3+), Mn^(4+), Cr^(2+), Ti^(2+)`

Answer

To solve the problem of identifying which set of ions has the magnetic moments equal to \(\sqrt{15}\), \(\sqrt{35}\), \(\sqrt{24}\), and \(0\) respectively, we will analyze each ion's electronic configuration and determine the number of unpaired electrons. The magnetic moment (\(\mu\)) can be calculated using the formula:

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

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

### Step 1: Identify the Ions and Their Configurations

1. **Manganese (Mn\(^{4+}\))**
- Atomic number: 25
- Ground state configuration: \([Ar] 4s^2 3d^5\)
- For Mn\(^{4+}\), we remove 4 electrons (2 from 4s and 2 from 3d):
- Excited state configuration: \([Ar] 3d^3\)
- Number of unpaired electrons: \(3\)

2. **Iron (Fe\(^{3+}\))**
- Atomic number: 26
- Ground state configuration: \([Ar] 4s^2 3d^6\)
- For Fe\(^{3+}\), we remove 3 electrons (2 from 4s and 1 from 3d):
- Excited state configuration: \([Ar] 3d^5\)
- Number of unpaired electrons: \(5\)

3. **Chromium (Cr\(^{2+}\))**
- Atomic number: 24
- Ground state configuration: \([Ar] 4s^1 3d^5\)
- For Cr\(^{2+}\), we remove 2 electrons (1 from 4s and 1 from 3d):
- Excited state configuration: \([Ar] 3d^4\)
- Number of unpaired electrons: \(4\)

4. **Copper (Cu\(^{+}\))**
- Atomic number: 29
- Ground state configuration: \([Ar] 4s^1 3d^{10}\)
- For Cu\(^{+}\), we remove 1 electron (from 4s):
- Excited state configuration: \([Ar] 3d^{10}\)
- Number of unpaired electrons: \(0\)

### Step 2: Calculate the Magnetic Moments

1. **For Mn\(^{4+}\)**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

2. **For Fe\(^{3+}\)**:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \]

3. **For Cr\(^{2+}\)**:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \]

4. **For Cu\(^{+}\)**:
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \]

### Step 3: Match the Magnetic Moments with the Given Values

- Mn\(^{4+}\): \(\sqrt{15}\)
- Fe\(^{3+}\): \(\sqrt{35}\)
- Cr\(^{2+}\): \(\sqrt{24}\)
- Cu\(^{+}\): \(0\)

### Conclusion

The correct set of ions corresponding to the magnetic moments \(\sqrt{15}\), \(\sqrt{35}\), \(\sqrt{24}\), and \(0\) respectively is:
- Mn\(^{4+}\) → \(\sqrt{15}\)
- Fe\(^{3+}\) → \(\sqrt{35}\)
- Cr\(^{2+}\) → \(\sqrt{24}\)
- Cu\(^{+}\) → \(0\)

Thus, the answer is the first option.