Question

The dipole moment of HBr is $2.60\times {10}^{-30}cm$ and the interatomic spacing is $1.41$ . What is the per cent ionic character of $HBr$ ?

• A. $10.11$%
• B. $9.11$%
• C. $11.5$%
• D. $15$%

Answer 1

The correct option is C $11.5$%

Dipole moment assuming 100 % ionic character would be the product of its charge and the inter atomic distance.

It would be $4.8\times {10}^{-10}esu\times 1.41\times {10}^{-8}cm=6.77\times {10}^{-18}esu.cm$
But $1esu.cm=3.3356\times {10}^{-12}Cm$

Hence, the dipole moment assuming 100% ionic character is
$6.77\times {10}^{-18}esu.cm\times 3.3356\times {10}^{-12}=2.26\times {10}^{-29}C.m$

Percent ionic character is the ratio of the observed dipole moment to the dipole moment assuming 100% ionic character. This ratio is multiplied by 100.

Percent ionic character $=\frac{2.60\times {10}^{-30}C.m}{2.26\times {10}^{-29}C.m}\times 100=11.5$.