Question

The dipole moment of $HBr$ is 2.60D and the interatomic spacing is 1.41pm. What is the percentage ionic character of $HBr$?

• A. 12
• B. 0.12
• C. 1.2
• D. None of the above

Answer 1

Answer: (A)

Dipole moment occur in between ions in a ionic bond and between atoms in a covalent bond dipole moments arises due to difference in the electronegativities of the constituent atoms.

Dipole moment is measured in de bye units. which is equal to the distance between the charges.

$1Debye=3.34\times {10}^{-30}Cm$

Given Data $\Rightarrow$

Dipole moment of $HBr=2.601030D$

inter atomic spacing $=1.41pm.$

% ionic moment of $HBr=2.60\times {10}^{-30}Cm$

% Ionic character is the ratio between observed dipole moment and the dipole moment of the molecule in case of full electron transfer.

% ionic character $=\frac{\mu obs}{\mu 100}$

charge of electron is $1.6\times {10}^{-19}$ Coloumbs,

$\mu 100$% is the charge of electron times the inter atomic distance.

$\mu 100$%=$q\gamma r$ $q=$ charge of electron

$r=$ interatomic spacing

$=1.6\times {10}^{-19}coloumbs\times 14\times {10}^{-12}\left(m\right)$

$\mu 100$% $=2.256\times {10}^{-29}$ Coloum - metre

% ionic character $=\frac{2.601030\times {10}^{-30}coloumb-metre}{2.256\times {10}^{-29}coloumb-metre}$

$=0.115$ or $11.5$%

$\approx 12$%

$\therefore$ The % ionic character is $12$%