Question

$HBr$ has dipole moment $2.6\times {10}^{-30}$ Cm. If the ionic character of the bond is $11.5$ %, the interatomic spacing(in $\stackrel{o}{A}$) is (write the value to the nearest integer) :

Answer 1

Let the inter atomic spacing be x cm.

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

It would be $4.8\times {10}^{-10}esu\times xcm=4.8x\times {10}^{-10}esu.cm$
But $1esu.cm=3.356\times {10}^{-12}C.m$

Hence, the dipole moment assuming 100% ionic character is
$4.8x\times {10}^{-10}\times 3.3356\times {10}^{-12}=1.601x\times {10}^{-21}Cm$

But actual dipole moment is $2.6\times {10}^{-30}Cm$

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.

$11.5=\frac{2.6\times {10}^{-30}Cm}{1.601x\times {10}^{-21}Cm}\times 100$

$x=1.4\times {10}^{-8}cm=1.4A^o$