How many of the following ions have spin mangnetic moment more than `4 B.M`.
`Ti^(3+), Cu^(+), Ni^(2+),Fe^(3+),Mn^(2+), Co^(2+)`

Explanation

To determine how many of the given ions have a spin magnetic moment greater than 4 Bohr Magnetons (B.M.), we will calculate the spin magnetic moment for each ion using the formula:

\[ \text{Spin Magnetic Moment} = \sqrt{n(n + 2)} \]

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

Let's analyze each ion step by step:

1. **Titanium Ion (Ti³⁺)**:
- Electronic configuration: \( \text{Ti}^{3+} \) has an electronic configuration of \( [Ar] 3d^1 \).
- Number of unpaired electrons (\( n \)): 1 (since there is one electron in the 3d subshell).
- Spin magnetic moment:
\[ \sqrt{1(1 + 2)} = \sqrt{3} \approx 1.73 \, \text{B.M.} \]

2. **Copper Ion (Cu⁺)**:
- Electronic configuration: \( \text{Cu}^{+} \) has an electronic configuration of \( [Ar] 3d^{10} \).
- Number of unpaired electrons (\( n \)): 0 (since all 10 electrons are paired).
- Spin magnetic moment:
\[ \sqrt{0(0 + 2)} = 0 \, \text{B.M.} \]

3. **Nickel Ion (Ni²⁺)**:
- Electronic configuration: \( \text{Ni}^{2+} \) has an electronic configuration of \( [Ar] 3d^8 \).
- Number of unpaired electrons (\( n \)): 2 (in the 3d subshell, there are 2 unpaired electrons).
- Spin magnetic moment:
\[ \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \, \text{B.M.} \]

4. **Iron Ion (Fe³⁺)**:
- Electronic configuration: \( \text{Fe}^{3+} \) has an electronic configuration of \( [Ar] 3d^5 \).
- Number of unpaired electrons (\( n \)): 5 (since all 5 electrons are unpaired).
- Spin magnetic moment:
\[ \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \text{B.M.} \]

5. **Manganese Ion (Mn²⁺)**:
- Electronic configuration: \( \text{Mn}^{2+} \) has an electronic configuration of \( [Ar] 3d^5 \).
- Number of unpaired electrons (\( n \)): 5 (all 5 electrons are unpaired).
- Spin magnetic moment:
\[ \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92 \, \text{B.M.} \]

6. **Cobalt Ion (Co²⁺)**:
- Electronic configuration: \( \text{Co}^{2+} \) has an electronic configuration of \( [Ar] 3d^7 \).
- Number of unpaired electrons (\( n \)): 3 (there are 3 unpaired electrons).
- Spin magnetic moment:
\[ \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \]

### Summary of Results:
- \( \text{Ti}^{3+} \): \( \sqrt{3} \approx 1.73 \, \text{B.M.} \) (not > 4)
- \( \text{Cu}^{+} \): \( 0 \, \text{B.M.} \) (not > 4)
- \( \text{Ni}^{2+} \): \( \sqrt{8} \approx 2.83 \, \text{B.M.} \) (not > 4)
- \( \text{Fe}^{3+} \): \( \sqrt{35} \approx 5.92 \, \text{B.M.} \) (is > 4)
- \( \text{Mn}^{2+} \): \( \sqrt{35} \approx 5.92 \, \text{B.M.} \) (is > 4)
- \( \text{Co}^{2+} \): \( \sqrt{15} \approx 3.87 \, \text{B.M.} \) (not > 4)

### Conclusion:
The ions with spin magnetic moments greater than 4 B.M. are \( \text{Fe}^{3+} \) and \( \text{Mn}^{2+} \). Therefore, **2 ions** have a spin magnetic moment greater than 4 B.M.