For two elements forming a bond, `|X_A –X_B| = 2.0`. What is percent ionic character for this covalent molecule?

Options

A. 46
B. 50
C. 20
D. 30

Answer

To find the percent ionic character of a covalent molecule given the electronegativity difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electronegativity Difference**:
We are given that the electronegativity difference \( |X_A - X_B| = 2.0 \).

2. **Use the Formula for Percent Ionic Character**:
The formula to calculate the percent ionic character is:
\[ \text{Percent Ionic Character} = 16 \times \Delta E_n + 3.5 \times (\Delta E_n)^2 \]
where \( \Delta E_n \) is the electronegativity difference.

3. **Substitute the Value into the Formula**:
Substitute \( \Delta E_n = 2.0 \) into the formula:
\[ \text{Percent Ionic Character} = 16 \times 2.0 + 3.5 \times (2.0)^2 \]

4. **Calculate Each Term**:
- First term:
\[ 16 \times 2.0 = 32 \]
- Second term:
\[ 3.5 \times (2.0)^2 = 3.5 \times 4 = 14 \]

5. **Add the Results**:
Now, add the two results together:
\[ 32 + 14 = 46 \]

6. **Conclusion**:
Therefore, the percent ionic character of the covalent molecule is \( 46\% \).