Question

The bond length of HCl bond is $2.29 \times {10^{ - 10}}$m. The percentage ionic character of HCl, if measured dipole moment is $6.226 \times {10^{ - 30}}C - m$, is :
a.) 8%
b.) 20%
c.) 17%
d.) 50%

Answer 1

Hint : The percentage ionic character can be defined as the ratio of observed dipole moment to calculated dipole moment multiplied by 100.
Percentage ionic character = $\dfrac{{Observed{\text{ dipole moment}}}}{{{\text{Calculated dipole moment}}}} \times 100$
Where Observed dipole moment is the practically measured dipole moment and
Calculated dipole moment is the one theoretically calculated by assuming that one electron has been completely transferred.

Complete answer :
First, let us write what is given to us in question and what we have to find out.
We have been given, Bond length of HCl bond = $2.29 \times {10^{ - 10}}$m
Measured dipole moment = $6.226 \times {10^{ - 30}}C - m$
We have to find out : Percentage ionic character of HCl
Now, what is the percentage ionic character?
It may be defined as the ratio of observed dipole moment to calculated dipole moment multiplied by 100.
Percentage ionic character = $\dfrac{{Observed{\text{ dipole moment}}}}{{{\text{Calculated dipole moment}}}} \times 100$
The observed dipole moment is given to us while the calculated one is also needed.
So, let us calculate that one first.
Theoretically dipole moment can be calculated as-
Dipole moment = Charge on electron * bond length of the molecule
Dipole moment = $1.6 \times {10^{ - 19}} \times 2.29 \times {10^{ - 10}}$
Dipole moment = $3.664 \times {10^{ - 20}}$
As we have calculated the dipole moment. So, now let us calculate percentage ionic character as -
Percentage ionic character = $\dfrac{{6.226 \times {{10}^{ - 19}}}}{{3.664 \times {{10}^{20}}}} \times 100$
Percentage ionic character = 16.9 %
This value is approximately equal to 17 %.

So, the option c.) is the correct answer.

Note :
It must be noted that we calculate the theoretical dipole moment by assuming that one electron of hydrogen atom has been completely transferred to the bromine atom.
Further, move step by step in numericals to get the correct answer.