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**Hint:**The formula to find the % ionic character of the bond is

\[\% {\text{ Ionic character = }}\dfrac{{{\text{Observed dipole moment}}}}{{{\text{Theoretical dipole moment}}}} \times 100\]

**Complete step by step solution:**

We need to find the % ionic character in a bond where the actual dipole moment and its bond length is given.

- There is a formula which relates the charge on the dipoles of a bond, its dipole moment and the bond length. The formula can be given as

\[{\text{Dipole moment = Charge}} \times {\text{Bond length}}\]

We are given that the bond length of the H-X bond is 2.00$\mathop A\limits^0 $ which is equal to $2.00 \times {10^{ - 10}}m$ as $1\mathop A\limits^0 = {10^{ - 10}}m$ .

We know that H-X bond is an ionic bond and H atom is positive charge and X atom has negative charge. The charge on them is +1 and -1 respectively.

- That means the charge on them is of one electron only and we know that the charge of one electrons is $1.6 \times {10^{ - 19}}C$

So, we can say that the dipole moment of H-X bond will be $1.6 \times {10^{ - 19}} \times 2 \times {10^{ - 10}} = 32 \times {10^{ - 30}}Cm$

So, we obtained that the theoretical dipole moment of the compound is $32 \times {10^{ - 30}}Cm$. We are given that the observed dipole moment is $5.12 \times {10^{ - 30}}Cm$. From this, we can find the % ionic character of the molecule by the formula below.

\[\% {\text{ Ionic character = }}\dfrac{{{\text{Observed dipole moment}}}}{{{\text{Theoretical dipole moment}}}} \times 100\]

So, we can write that

\[\% {\text{ Ionic character = }}\dfrac{{5.12 \times {{10}^{ - 30}}}}{{32 \times {{10}^{ - 30}}}} \times 100 = 0.16 \times 100 = 16\% \]

Thus, we can say that the ionic character of this H-X bond will be 16%.

**So, the correct answer is (B).**

**Note:**Do not get confused between the observed and the theoretic dipole moment of a bond. Theoretical dipole moment is the value we obtain from the equation by calculation. Observed or actual dipole moment is the dipole moment measured practically.

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