Question

The observed dipole moment of $HBr$ is $0.79D$. Calculate the percentage ionic character of the bond in $HBr$ if bond distance is $1.42$ $\dot{A}$.

• A. $16.9\%$
• B. $12.0\%$
• C. $5.0\%$
• D. $20\%$

Answer 1

Answer: (B)

$12.0\%$

$\text{Ionic character}=\frac{\text{Observed dipole moment}}{\text{Theoretical dipole moment}}\times 100\text{Theoritical dipole moment}=(1.6\times {10}^{-19}C)\times (1.42\times {10}^{-10}m)=2.272\times {10}^{-29}Cm1D=3.33564\times {10}^{-30}Cm\text{Percentage ionic character}=\frac{0.79\times 3.33564\times {10}^{-30}}{2.272\times {10}^{-29}}\times 100=12\%$