Question

The dipole moment of LiH is $1.964 \times {10^{ - 29}}Cm\;$ and the intermolecular distance between Li and H in this molecule is . The percentage ionic character in molecule is (write the value to the nearest integer):

Answer 1

To solve this question, we need to find the dipole moment of $100\%$ ionic LiH. This value will be our theoretical value of the dipole moment which can be determined by using the given intermolecular distance. Then with the help of this theoretical value and given experimental value of the dipole moment, we will determine the percentage ionic character in the molecule.

Complete step by step answer:
Our first step is to find the theoretical value of the dipole moment when LiH is $100\%$ionic. This can be determined by multiplying one electronic charge to the intermolecular distance between Li and H in this molecule.And we know that one electronic charge is $1.602 \times {10^{ - 19}}C$.

Thus, the dipole moment of $100\%$ ionic LiH $= \left( {1.596 \times {{10}^{ - 10}}m} \right) \times \left( {1.602 \times {{10}^{ - 19}}C} \right) = 2.556 \times {10^{ - 29}}Cm$
Now, we will find the fractional ionic character which is the ratio of the actual or experimental value to the theoretical value.
Hence, the fractional ionic character $= \dfrac{{1.964 \times {{10}^{ - 29}}Cm\;}}{{2.556 \times {{10}^{ - 29}}Cm}} = 0.76838$
And the percentage ionic character in molecule $= 0.76838 \times 100 = 76.838\% \approx 77\%$

Hence, the percentage ionic character in molecules is $77\%$.

Note: We have seen the concept of dipole moment. Dipole moment is defined as the measure of net molecular polarity, which is the magnitude of the charge at either end of the molecular dipole times the distance between the charges. Dipole moments tell us about the charge separation in a molecule. The larger the difference in electronegativities of bonded atoms, the larger the dipole moment. For example, NaCl has the highest dipole moment because it has an ionic bond which is highest charge separation.