Question

The dipole moment of $HBr$ is $0.78\times {10}^{-18}$ esu-cm and bond length of $HBr$ is $1.41A^{\circ }$. The ionic character of $HBr$ is:

• A. $7.5$
• B. $11.52$
• C. $15$
• D. $27$

Answer 1

Answer: (B)

$11.52$

Dipole moment of $HBr$ $\cdot 78\times {10}^{-18}$ esu-cm. Molecule is 100% ionic if $e=4\cdot 803\times {10}^{-18}$ esu-cm. Percentage of ionic character of $HBr$ is $\frac{\cdot 781}{4\cdot 803\ast 1.41}\times 100$ = $11.53$% approx.