Question

Calculate the percentage ionic character of the HCl molecule. The bond length is 1.275$A{}^\circ$ and the observed dipole moment is 1.03 D.

Answer 1

For calculation of ionic character in a molecule, we need to determine the dipole moment within the molecule. The dipole moment is the physical property which is a measure of the polarity of a chemical bond between two atoms in a molecule. The higher the electronegativity difference between the two atoms, the more will be the ionic character to the bond and vice versa.

Complete answer:
- The dipole moment is the mathematical product of the total amount of positive charge or negative charge and the distance between the center of the charge distribution.
-For calculating the percent ionic character in a molecule, we need to follow the steps below-
(a) We start by calculating the dipole moment by taking the product of the magnitude of charge and separation between charges, i.e
$\mu =q\times r$
where $\mu$ is the dipole moment
q is the separated charge
r is the distance between them
Here, q= $1.6\times {{10}^{-19}}C$ ; r= $1.275A{}^\circ \times {{10}^{-10}}m$
So, dipole moment ($\mu$ ) $=1.6\times {{10}^{-19}}\times 1.275\times {{10}^{-10}}m$
$=2.04\times {{10}^{-29}}cm$
(b) Now, we have to convert the centimeter into the standard unit of dipole moment, i.e Debye.
We know that, $1D=3.335\times {{10}^{-30}}cm$
$=\dfrac{2.04\times {{10}^{-29}}}{(3.335}$ $=\dfrac{2.04\times {{10}^{-29}}}{\begin{aligned} & 3.335\times {{10}^{-30}} \\\ & =\dfrac{20.4}{3.335} \\\ & =6.1169D \\\ \end{aligned}}$
(c) Now, percentage ionic character $=\dfrac{\exp value}{theoretical value}\times 100$
$=\dfrac{1.03D}{6.1169D}\times 100$
$=16.83$

Note: For calculating the dipole moment you need to remember the charge of an electron i.e,$1.6\times {{10}^{-19}}$ . For the calculation of percentage ionic character, remember the conversion units i.e,$1D=3.335\times {{10}^{-30}}cm$ .