Question

The dipole moment of $HBr$ is $2.6$ $\times$ ${10}^{-30}$ Cm and the inter atomic spacing is $1.41\stackrel{o}{A}$ . The percentage of ionic character in $HBr$ is:

• A. $8.7\%$
• B. $11.5\%$
• C. $12.5\%$
• D. $13.5\%$

Answer 1

Answer: (B)

$11.5\%$

Percentage ionic character = $\frac{\text{Observed dipole moment}}{\text{Theoretical dipole moment}}\times 100$

Theoretical dipole moment = e $\times$d = $(1.6\times {10}^{-19}C)\times (1.41\times {10}^{-10}m)$ = $2.256\times {10}^{-29}Cm$

Observed dipole moment $=2.6\times {10}^{-30}Cm$

Therefore, % ionic character=$\frac{2.6\times {10}^{-30}}{2.256\times {10}^{-29}}\times 100=11.5\%$