Question

If the dipole moment of HBr is \( 2.60 \times 10^{-30} \, \text{Cm} \) and the interatomic spacing is 1.41 Å, then the percent ionic character of HBr is:

• A. 16.23%
• B. 13.21%
• C. 11.50%
• D. 15.81%

Hint:

The ionic character of a molecule can be calculated by comparing its measured dipole moment to the ideal dipole moment assuming full ionic bonding. The formula for the ideal dipole moment uses the charge of the electron and the interatomic separation.

Answer 1

The Correct Option is C

The percent ionic character of a molecule can be calculated using the following formula: \[ \text{Percent ionic character} = \frac{\mu_{\text{measured}}}{\mu_{\text{ideal}}} \times 100 \] where: - \( \mu_{\text{measured}} \) is the measured dipole moment of the molecule, - \( \mu_{\text{ideal}} \) is the ideal dipole moment assuming complete ionic character. The ideal dipole moment \( \mu_{\text{ideal}} \) is given by: \[ \mu_{\text{ideal}} = Q \times r \] where: - \( Q \) is the charge of the electron, \( Q = 1.602 \times 10^{-19} \, \text{C} \), - \( r \) is the interatomic separation, \( r = 1.41 \, \text{Å} = 1.41 \times 10^{-10} \, \text{m} \). Substituting the values: \[ \mu_{\text{ideal}} = 1.602 \times 10^{-19} \times 1.41 \times 10^{-10} = 2.26 \times 10^{-29} \, \text{Cm} \] Now, calculate the percent ionic character: \[ \text{Percent ionic character} = \frac{2.60 \times 10^{-30}}{2.26 \times 10^{-29}} \times 100 = 11.50% \]
Thus, the correct answer is (c).