In HCI molecule, expected value of dipole moment is 6.12 D but experimental value is 1.03 D. Then, the percentage ionic character will be

Options

A. 16.13
B. `15.14`
C. 6.02
D. 18.9

Answer

To determine the percentage ionic character of HCl, we can use the formula:

\[ \text{Percentage Ionic Character} = \left( \frac{\mu_{\text{experimental}}}{\mu_{\text{expected}}} \right) \times 100 \]

Where:
- \(\mu_{\text{experimental}}\) is the experimental dipole moment.
- \(\mu_{\text{expected}}\) is the expected dipole moment.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Expected dipole moment, \(\mu_{\text{expected}} = 6.12 \, \text{D}\)
- Experimental dipole moment, \(\mu_{\text{experimental}} = 1.03 \, \text{D}\)

2. **Substitute the Values into the Formula:**
\[ \text{Percentage Ionic Character} = \left( \frac{1.03 \, \text{D}}{6.12 \, \text{D}} \right) \times 100 \]

3. **Calculate the Fraction:**
\[ \frac{1.03}{6.12} \approx 0.168 \]

4. **Multiply by 100 to Get the Percentage:**
\[ \text{Percentage Ionic Character} \approx 0.168 \times 100 \approx 16.8\% \]

5. **Final Result:**
The percentage ionic character of HCl is approximately \(16.8\%\).