Which of the following ions has maximum magnetic moment ? ·

Options

A. `Cu^(2+)`
B. `Mn^(2+)`
C. `Ti^(2+)`
D. `Zn^(2+)`

Explanation

To determine which of the given ions has the maximum magnetic moment, we will calculate the magnetic moment for each ion based on the number of unpaired electrons present in their electronic configurations. The formula for calculating the magnetic moment (μ) is:

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

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

### Step-by-step Solution:

1. **Identify the ions and their electronic configurations**:
- **A. Cu²⁺**:
- Atomic number of Cu = 29
- Electronic configuration of Cu: [Ar] 3d¹⁰ 4s¹
- For Cu²⁺, we remove 2 electrons: [Ar] 3d⁹ 4s⁰
- **B. Mn²⁺**:
- Atomic number of Mn = 25
- Electronic configuration of Mn: [Ar] 3d⁵ 4s²
- For Mn²⁺, we remove 2 electrons: [Ar] 3d⁵ 4s⁰
- **C. Ti²⁺**:
- Atomic number of Ti = 22
- Electronic configuration of Ti: [Ar] 3d² 4s²
- For Ti²⁺, we remove 2 electrons: [Ar] 3d² 4s⁰
- **D. Zn²⁺**:
- Atomic number of Zn = 30
- Electronic configuration of Zn: [Ar] 3d¹⁰ 4s²
- For Zn²⁺, we remove 2 electrons: [Ar] 3d¹⁰ 4s⁰

2. **Count the number of unpaired electrons**:
- **Cu²⁺**: 3d⁹ has 1 unpaired electron (n = 1)
- **Mn²⁺**: 3d⁵ has 5 unpaired electrons (n = 5)
- **Ti²⁺**: 3d² has 2 unpaired electrons (n = 2)
- **Zn²⁺**: 3d¹⁰ has 0 unpaired electrons (n = 0)

3. **Calculate the magnetic moment for each ion**:
- **Cu²⁺**:
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73 \text{ magneton} \]
- **Mn²⁺**:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \text{ magneton} \]
- **Ti²⁺**:
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \text{ magneton} \]
- **Zn²⁺**:
\[ \mu = \sqrt{0(0 + 2)} = 0 \text{ magneton} \]

4. **Compare the magnetic moments**:
- Cu²⁺: 1.73 magneton
- Mn²⁺: 5.92 magneton
- Ti²⁺: 2.83 magneton
- Zn²⁺: 0 magneton

5. **Conclusion**:
The ion with the maximum magnetic moment is **Mn²⁺** with a magnetic moment of approximately **5.92 magneton**.

### Final Answer:
**B. Mn²⁺**