Magnetic momentum of an ion `M^(+3)` is `sqrt( 35)` BM. Then, number of electron in d orbital of M element will be `"___________"`

Answer

To find the number of electrons in the d orbital of the element M, given that the magnetic moment of the ion M^(+3) is √35 BM, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Moment**:
The magnetic moment (μ) is given by the formula:
\[ μ = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Set Up the Equation**:
Given that the magnetic moment is √35 BM, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{35} \]

3. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ n(n + 2) = 35 \]

4. **Rearrange the Equation**:
Rearranging gives us a standard quadratic equation:
\[ n^2 + 2n - 35 = 0 \]

5. **Factor the Quadratic**:
We need to factor the quadratic equation:
\[ n^2 + 7n - 5n - 35 = 0 \]
This can be factored as:
\[ (n + 7)(n - 5) = 0 \]

6. **Solve for n**:
Setting each factor to zero gives us:
\[ n + 7 = 0 \quad \text{or} \quad n - 5 = 0 \]
Thus:
\[ n = -7 \quad \text{or} \quad n = 5 \]

7. **Determine Valid Solutions**:
Since the number of electrons cannot be negative, we discard \( n = -7 \). Therefore, we have:
\[ n = 5 \]

8. **Conclusion**:
The number of electrons in the d orbital of the element M is:
\[ \text{5} \]