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 calculate the percentage ionic character of HCl based on the given dipole moments, we can follow these steps:

### Step 1: Understand the formula for percentage ionic character
The formula to calculate the percentage ionic character is given by:

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

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

### Step 2: Identify the values given in the question
From the question, we have:
- \(\mu_{\text{observed}} = 1.03 \, \text{D}\) (experimental value)
- \(\mu_{\text{calculated}} = 6.12 \, \text{D}\) (expected value)

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the formula:

\[ \text{Percentage Ionic Character} = \left( \frac{1.03 \, \text{D}}{6.12 \, \text{D}} \right) \times 100 \]

### Step 4: Perform the calculation
First, calculate the fraction:

\[ \frac{1.03}{6.12} \approx 0.168 \]

Now, multiply by 100 to find the percentage:

\[ 0.168 \times 100 \approx 16.8 \]

### Step 5: Final result
Thus, the percentage ionic character of HCl is approximately:

\[ \text{Percentage Ionic Character} \approx 16.8\% \]