The observed dipole moment of `HCl` is `1.03D`. If the bond length of HCL is `1.3Å`, then the percent ionic character of `H-Cl` bond is

Options

A. 43
B. 21
C. 17
D. 7

Answer

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

### Step 1: Understand the dipole moment
The dipole moment (μ) is defined as the product of the charge (q) and the distance (l) between the charges. The formula is:
\[ \mu = q \times l \]

### Step 2: Given values
- Observed dipole moment of HCl, \( \mu_{\text{observed}} = 1.03 \, D \)
- Bond length of HCl, \( l = 1.3 \, \text{Å} = 1.3 \times 10^{-8} \, \text{cm} \)
- Charge of one unit (ionic character), \( q = 4.8 \times 10^{-10} \, \text{esu} \)

### Step 3: Calculate the dipole moment if HCl were 100% ionic
For a completely ionic bond, the dipole moment can be calculated using:
\[ \mu_{\text{ionic}} = q \times l \]
Substituting the values:
\[ \mu_{\text{ionic}} = (4.8 \times 10^{-10} \, \text{esu}) \times (1.3 \times 10^{-8} \, \text{cm}) \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \mu_{\text{ionic}} = 6.24 \times 10^{-18} \, \text{esu cm} \]
Since \( 1 \, D = 10^{-18} \, \text{esu cm} \), we convert this to Debye:
\[ \mu_{\text{ionic}} = 6.24 \, D \]

### Step 5: Calculate the percent ionic character
The percent ionic character can be calculated using the formula:
\[ \text{Percent Ionic Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{ionic}}} \right) \times 100 \]
Substituting the values:
\[ \text{Percent Ionic Character} = \left( \frac{1.03 \, D}{6.24 \, D} \right) \times 100 \]

### Step 6: Perform the final calculation
Calculating the above expression:
\[ \text{Percent Ionic Character} \approx 16.53\% \]
Rounding this value gives approximately:
\[ \text{Percent Ionic Character} \approx 17\% \]

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

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