Question

In HF molecule, internuclear distance is 0.92 Å and dipole moment approximates 2 D. Estimate the percentage ionic character of HF molecule.

• A. 55
• B. 65
• C. 35
• D. 45

Hint:

The percentage ionic character can be determined by comparing the observed dipole moment with the ideal dipole moment assuming 100% ionic bonding.

Answer 1

The Correct Option is D

The formula for estimating the percentage ionic character is: \[ \text{Percentage Ionic Character} = \frac{\mu_{\text{observed}}}{\mu_{\text{ideal}}} \times 100 \] Where: \( \mu_{\text{observed}} \) is the observed dipole moment (given as 2 D), \( \mu_{\text{ideal}} \) is the ideal dipole moment if the molecule were 100% ionic, calculated as: \[ \mu_{\text{ideal}} = q \times r \] Where:
\( q \) is the charge of the ions \( = 1.6 \times 10^{-19} \, \text{C} \), \( r \) is the internuclear distance \( = 0.92 \, \text{Å} = 0.92 \times 10^{-10} \, \text{m} \).
Now, calculating the ideal dipole moment: \[ \mu_{\text{ideal}} = 1.6 \times 10^{-19} \times 0.92 \times 10^{-10} = 1.472 \times 10^{-29} \, \text{C m} = 4.8 \, \text{D} \] Now, the percentage ionic character is: \[ \text{Percentage Ionic Character} = \frac{2}{4.8} \times 100 = 45 % \]