A substance is found to hav a magnetic moments of `3.9BM`.How many unpaired electrons does it contain?

Answer

To solve the problem of determining the number of unpaired electrons in a substance with a magnetic moment of 3.9 Bohr magnetons (BM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Moment**:
The magnetic moment (μ) is related to the number of unpaired electrons (n) by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
Given that the magnetic moment is 3.9 BM, we can set up the equation:
\[ 3.9 = \sqrt{n(n + 2)} \]

2. **Square Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ (3.9)^2 = n(n + 2) \]
This simplifies to:
\[ 15.21 = n(n + 2) \]

3. **Rearrange the Equation**:
Rearranging gives us a standard quadratic equation:
\[ n^2 + 2n - 15.21 = 0 \]

4. **Use the Quadratic Formula**:
The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 2\), and \(c = -15.21\). Plugging in these values:
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-15.21)}}{2 \cdot 1} \]

5. **Calculate the Discriminant**:
First, calculate the discriminant:
\[ b^2 - 4ac = 4 + 60.84 = 64.84 \]

6. **Find the Square Root**:
Calculate the square root of the discriminant:
\[ \sqrt{64.84} \approx 8.06 \]

7. **Substitute Back into the Formula**:
Now substitute back into the quadratic formula:
\[ n = \frac{-2 \pm 8.06}{2} \]

8. **Calculate Possible Values for n**:
This gives us two potential solutions:
\[ n_1 = \frac{6.06}{2} \approx 3.03 \quad \text{(take the positive root)} \]
\[ n_2 = \frac{-10.06}{2} \quad \text{(discard as it is negative)} \]

9. **Round to the Nearest Whole Number**:
Since the number of unpaired electrons must be a whole number, we round \(3.03\) to \(3\).

### Final Answer:
The substance contains **3 unpaired electrons**.