Question

The covalent character of HCl molecule whose observed dipole moment is 1.03 Debye and bond length be $1.275\mathop {\text{A}}\limits^0 $ will be:
(A) 16.82
(B) 83.2
(C) 75%
(D) Data incomplete

Answer 1

Hint: The theoretical dipole moment, charge on the atoms and the bond length is related by the following equation.
\[{\mu _t} = d \times e\]
Ionic character is the ratio of the observed ionic character of the bond to the theoretic ionic character of that bond. We obtain % ionic character if we multiply this ratio by 100.

Complete step by step solution:
Here, we are being asked to find the covalent character of HCl molecule. We are given the observed dipole moment and the distance between H and Cl atoms in HCl molecules.
- There is a relation between the theoretical dipole moment of a molecule and bond length which can be given as below.
\[{\mu _t} = d \times e\] ………(1)
Where ${\mu _t}$ is the theoretical value of dipole moment,
d is the distance between H and Cl molecule or the bond length (in m)
And e is the effective charge on H and Cl atom
- We know that Cl has partial negative charge and H atom has partial positive charge in HCl molecules. That charge is equal to the charge on one electron which is $1.6 \times {10^{ - 19}}C$
So, we can write equation (1) as
\[{\mu _t} = 1.275 \times {10^{ - 10}} \times 1.6 \times {10^{ - 19}}\]
\[{\mu _t} = 2.04 \times {10^{ - 9}}C \cdot m\]
Now, $1C \cdot m = 3 \times {10^{ - 29}}Debye$
So, we obtained that
\[{\mu _t} = 2.04 \times {10^{ - 9}} \times 3 \times {10^{ - 29}}Debye = 6.12Debye\]
 So, we obtained the theoretic dipole moment for the HCl molecule. We are given that the observed dipole moment of this molecule is 1.03 Debye.
- So, now we can find the % ionic character from the given data by the following equation.
$\% {\text{ ionic character = }}\dfrac{{{\mu _o}}}{{{\mu _t}}} \times 100$ …………(2)
Here, ${\mu _o}$ is the observed dipole moment. So, we can put the available values in the equation (2) as
\[\% {\text{ ionic character = }}\dfrac{{1.03}}{{6.12}} \times 100 = 16.83\% \]
Now, we need to find the covalent character of the bond. So, we can simply put the equation that
% covalent character = 100 - % ionic character
% covalent character = 100 – 16.83
% covalent character = 83.2%

Therefore, the correct answer of this question is (B).

Note: Note that there are two units of dipole moment and both are different, so do not assume them as the same as mistakes may occur there. The relation between them is shown below.
$1C \cdot m = 3 \times {10^{ - 29}}Debye$