Question

The dipole moment of $HBr$ is $1.6\times {10}^{-30}c.m.$ and inter atomic spacing $1A^{\circ }$ . The percentage ionic character of $HBr$ is:
A.$10$
B. $0$
C. $15$
D.$27$

Answer 1

Hint: We have to know that the dipole second emerges in any framework where there is a division of charge. They can, in this way, emerge in ionic bonds just as in covalent bonds. Dipole minutes happen because of the distinction in electronegativity between two artificially fortified molecules.

Complete answer:
We have to see the bond's percent ionic character is the measure of electron dividing among two molecules; restricted electron sharing relates with a high percent ionic character. To decide a bond's percent ionic character, the particles' electronegativities are utilized to foresee the electron dividing among the molecules.
We have to know the bond dipole second is a proportion of the extremity of a synthetic connection between two iotas in an atom. It includes the idea of electric dipole second, which is a proportion of the division of negative and positive charges in a framework.
The bond dipole second is a vector amount since it has both greatness and bearing. An outline depicting the dipole second that emerges in a $HBr$ .
To calculate the percentage ionic character of $HBr$ ,
The charge of an electron = $1.6\times {10}^{-19}C$ .
The inter-atomic spacing = $1A^{\circ }=1\times {10}^{-10}m$ ,
The dipole moment of $HBr$ = $1.6\times {10}^{-30}m$ .
By using the following expression,
$\text{Percentage of ionic character in HBr = }{\displaystyle \frac{Dipole\text{ moment of HBr}\times \text{100}}{Inter\text{ atomic spacing}\times \text{100}}}$
Then, applying all the values in the above equation,
$\text{Percentage of ionic character in HBr = }{\displaystyle \frac{1.6\times {10}^{-30}\times 100}{1.6\times {10}^{-19}\times {10}^{-10}}}$
Therefore, the percentage of ionic character of $HBr$ is $10\mathrm{\%}$ .

Hence, the correct option is (A).

Note:
When, the nuclear separating alludes to the distance between the cores of particles in a material. In strong materials, nuclear separation is depicted by the bond length of its particles. In arranged solids, the nuclear separating between two fortified molecules is by and large around a couple of Angstroms ( $A^{\circ }$ ), which is on the request for ${10}^{-10}$ meters.