Question

The dipole moment of $HBr$ is $2.60\times {10}^{-30}Cm$ and the interatomic spacing is $1.41\stackrel{\circ }{A}$. What is the percent ionic character of $HBr$?

• A. $50$%
• B. $11.5$%
• C. $4.01$%
• D. $1.19$%

Answer 1

The correct option is B $11.5$%

Theoretical value of dipole moment of $100$% ionic character $=e\times d$

$=(1.6\times {10}^{-19}C)(1.41\times {10}^{-10}m)$

$=2.26\times {10}^{-29}Cm$

Observed value of dipole moment

$=2.60\times {10}^{-30}Cm$

$\therefore$ Percent ionic character $=\frac{observedvalue}{theoreticalvalue}\times 100$

$=\frac{2.60\times {10}^{-30}}{2.26\times {10}^{-29}}\times 100$

$=11.5$%

Hence, the correct option is B.