The dipole moment of HX molecule is $1.92\mathrm{D}$ and bond distance is $1.2{ \mathrm{A}}^0$. What is the percentage ionic character of HX ?

$$\begin{gathered} \mathrm{Dipole}\mathrm{moment}(\mathrm{theoretical}) \mathrm{\mu }=\mathrm{Q}\times \mathrm{d}\mathrm{coulomb}-\mathrm{metre} \\ \mathrm{Q} \mathrm{is}\mathrm{the}\mathrm{charge}\mathrm{of}\mathrm{electron}{ =1.6\times 10}^{-19}\mathrm{C} \\ \mathrm{d} \mathrm{isthebondlengthinmetres} \\ \mathrm{\mu }=1.6\times {10}^{-19}\times 1.2\times {10}^{-10}=1.92\times {10}^{-29}\mathrm{C}-\mathrm{m} \\ =19.2\times {10}^{-30}\mathrm{C}-\mathrm{m}=5.75\mathrm{D} \\ \left(1 \mathrm{debye} =3.34\times {10}^{-30}\mathrm{C}-\mathrm{m}\right) \\ \% \mathrm{ioniccharacter} = \frac{\mathrm{actualdipole}}{\mathrm{theorticaldipole}} \times 100 \\ =\frac{1.92}{5.75}\times 100=33.4\mathrm{\%} \end{gathered}$$

Hence, the correct option is (A).