What fraction of charge is present on I in covalently bonded $ H^{{\delta ^ + }} - I^{{\delta ^ - }} $ if the dipole moment of HI is 0.38 D and the bond length is 1.61 $ A^ \circ $?
A. 5 %
B. 16 %
C. 33 %
D. 79 %

Hint:Ionic compounds include salts, halides, hydroxides etc. ionic solids are held together by the electrostatic attraction between the positive and negative ions. Plainly there will be repulsion if the ions of the same charge are adjacent , and attraction when positive ions are surrounded by negative ions.

Complete answer:
Molecules like H-X having two polar ends are known as dipoles are polar molecules. The measure of polarity in a polar compound is given by its dipole moment ($\mu$).
Dipole moment is defined as the product of the net positive or negative charge and distance between the two charged ends, i.e. the dipole length.
Dipole moment $\mu = e \times d$(where e is electric charge and d is the distance between the charges)
It is measured in debye unit(D). Let us now understand the concept of percentage ionic character, every ionic compound has some percentage of covalent character according to Fajan's rule. The percentage of ionic character in a compound having some covalent character can be calculated by the following equation
The percent ionic character=(observed dipole moment/calculated dipole moment assuming 100% ionic bond)$ \times 100$
In the given compound of HI we have been asked to find “$\delta = e$ ” in this case is charge percentage on I. Therefore we will put the given values in the equation above which will be
$
\delta = \dfrac{{dipole(D)}}{d} \\$
$\Rightarrow \delta = \dfrac{{0.38 \times {{10}^{ - 18}}esu - cm}}{{1.61 \times {{10}^{ - 10}}cm}} \;(1A^\circ = {10}^{-10}\,m) \\$
$\Rightarrow \delta = \dfrac{{0.38 \times {{10}^{ - 18}}esu - cm}}{{1.61 \times {{10}^{ - 10}}cm}} = 2.4 \times {10^{ - 11}}esu
$
We know that 1D=$1 \times {10^{ - 18}}esu - cm$ so
$0.38D=0.38 \times {10^{ - 18}}esu - cm$
Now let us calculate the fraction of charge where the total charge is present on both H and I atoms which are the same in magnitude and hence on adding the total charge becomes $4.8 \times {10^{ - 10}}esu$. Therefore we get,
$\Rightarrow \%\, \delta = \dfrac{{2.4 \times {{10}^{ - 11}}esu}}{{4.8 \times {{10}^{ - 10}}esu}} \times 100 = \dfrac{{100}}{{20}} = 5\% $

Hence the correct option is A.

Note:
The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no bond or compound is either completely covalent or ionic. Further dipole moment is a vector quantity and is depicted by a small arrow with tail on the positive centre and head pointing towards the negative centre. In the case of polyatomic molecules the dipole moment not only depends upon the individual dipole moments of bond but on spatial arrangement as well.