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 determine which set of ions has magnetic moments equal to \(\sqrt{15}\), \(\sqrt{35}\), \(\sqrt{24}\), and \(0\) respectively, we will calculate the magnetic moments for each ion based on the number of unpaired electrons. The formula for the magnetic moment (\(\mu\)) is given by:

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

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

### Step 1: Calculate the magnetic moment for \( \text{Mn}^{4+} \)

1. **Find the atomic number of Manganese (Mn)**: Mn has an atomic number of 25.
2. **Write the electronic configuration of Mn**:
- Mn: \([Ar] 3d^5 4s^2\)
3. **Determine the configuration for \( \text{Mn}^{4+} \)**:
- Remove 4 electrons: \( \text{Mn}^{4+} \) will have the configuration \([Ar] 3d^3\).
4. **Count the unpaired electrons**:
- In \(3d^3\), there are 3 unpaired electrons.
5. **Calculate the magnetic moment**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \]

### Step 2: Calculate the magnetic moment for \( \text{Fe}^{3+} \)

1. **Find the atomic number of Iron (Fe)**: Fe has an atomic number of 26.
2. **Write the electronic configuration of Fe**:
- Fe: \([Ar] 3d^6 4s^2\)
3. **Determine the configuration for \( \text{Fe}^{3+} \)**:
- Remove 3 electrons: \( \text{Fe}^{3+} \) will have the configuration \([Ar] 3d^5\).
4. **Count the unpaired electrons**:
- In \(3d^5\), there are 5 unpaired electrons.
5. **Calculate the magnetic moment**:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \]

### Step 3: Calculate the magnetic moment for \( \text{Cr}^{2+} \)

1. **Find the atomic number of Chromium (Cr)**: Cr has an atomic number of 24.
2. **Write the electronic configuration of Cr**:
- Cr: \([Ar] 3d^5 4s^1\)
3. **Determine the configuration for \( \text{Cr}^{2+} \)**:
- Remove 2 electrons: \( \text{Cr}^{2+} \) will have the configuration \([Ar] 3d^4\).
4. **Count the unpaired electrons**:
- In \(3d^4\), there are 4 unpaired electrons.
5. **Calculate the magnetic moment**:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \]

### Step 4: Calculate the magnetic moment for \( \text{Cu}^{+} \)

1. **Find the atomic number of Copper (Cu)**: Cu has an atomic number of 29.
2. **Write the electronic configuration of Cu**:
- Cu: \([Ar] 3d^{10} 4s^1\)
3. **Determine the configuration for \( \text{Cu}^{+} \)**:
- Remove 1 electron: \( \text{Cu}^{+} \) will have the configuration \([Ar] 3d^{10}\).
4. **Count the unpaired electrons**:
- In \(3d^{10}\), there are 0 unpaired electrons.
5. **Calculate the magnetic moment**:
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \]

### Conclusion

The magnetic moments calculated are:
- \( \text{Mn}^{4+} \): \(\sqrt{15}\)
- \( \text{Fe}^{3+} \): \(\sqrt{35}\)
- \( \text{Cr}^{2+} \): \(\sqrt{24}\)
- \( \text{Cu}^{+} \): \(0\)

Thus, the correct set of ions with magnetic moments equal to \(\sqrt{15}\), \(\sqrt{35}\), \(\sqrt{24}\), and \(0\) respectively is:

**Answer**: \( \text{Mn}^{4+}, \text{Fe}^{3+}, \text{Cr}^{2+}, \text{Cu}^{+} \)

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