The observed dipole moment of `HCl` is `1.03D`. If the bond length of HCL is `1.275Å`, then the percent ionic character of `H-Cl` bond is

Options

A. 0.6
B. 0.39
C. 0.29
D. 0.17

Answer

To find the percent ionic character of the H-Cl bond in HCl, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Observed dipole moment (P) of HCl = 1.03 D
- Bond length (d) of HCl = 1.275 Å

2. **Convert Bond Length to Meters:**
- 1 Å = \(1 \times 10^{-10}\) m
- Therefore, \(1.275 \, \text{Å} = 1.275 \times 10^{-10} \, \text{m}\)

3. **Calculate the Theoretical Dipole Moment:**
- The dipole moment (P) is given by the formula:
\[ P = Q \times d \]
- Where \(Q\) is the charge and \(d\) is the distance (bond length).
- The charge of an electron (Q) is approximately \(1.6 \times 10^{-19}\) C.
- Substitute the values:
\[ P = (1.6 \times 10^{-19} \, \text{C}) \times (1.275 \times 10^{-10} \, \text{m}) \]

4. **Calculate the Theoretical Dipole Moment:**
- Performing the multiplication:
\[ P = 2.04 \times 10^{-29} \, \text{C m} \]

5. **Convert Dipole Moment to Debye:**
- 1 Debye = \(3.335 \times 10^{-30} \, \text{C m}\)
- To convert the theoretical dipole moment to Debye:
\[ P_{\text{theoretical}} = \frac{2.04 \times 10^{-29} \, \text{C m}}{3.335 \times 10^{-30} \, \text{C m/Debye}} \approx 6.1169 \, \text{D} \]

6. **Calculate the Percent Ionic Character:**
- The formula for percent ionic character is:
\[ \text{Percent Ionic Character} = \left(\frac{\text{Observed Dipole Moment}}{\text{Theoretical Dipole Moment}}\right) \times 100 \]
- Substitute the values:
\[ \text{Percent Ionic Character} = \left(\frac{1.03 \, \text{D}}{6.1169 \, \text{D}}\right) \times 100 \]

7. **Perform the Calculation:**
- Calculate the fraction:
\[ \text{Percent Ionic Character} \approx 16.83\% \]
- Rounding gives approximately \(17\%\).

### Final Answer:
The percent ionic character of the H-Cl bond is approximately **17%**.