The dipole moment of KCl is $3.336\times {10}^{-29}$ Cm. The interatomic distance ${\mathrm{K}}^{+}\mathrm{and} {\mathrm{Cl}}^{-}$ ion in KCl is 260 pm. Calculate the dipole moments of KCl, if there were opposite charges of the fundamental unit located at each nucleus. The percentage ionic character of KCl (XY%) is ___

From the given data q = 1.602 × 10^–19C , 
$\mathrm{r}=260\mathrm{pm}=260\times {10}^{-12}\mathrm{m}=2.6\times {10}^{-10}\mathrm{m}$
Magnitude of dipole moment for 100% ionic 
$\text{ character }=\mathrm{qr}=\left(1.602\times {10}^{-19}\right)\left(2.6\times {10}^{-10}\right)$

$=4.165\times {10}^{-29}\mathrm{Cm}$

Actual dipole moment $=3.336\times {10}^{-29}\mathrm{Cm}$ 
$\therefore \mathrm{\%}\text{ of ionic bond }=\frac{3.336\times {10}^{-29}}{4.165\times {10}^{-29}}\times 100$
$=80.1\mathrm{\%}$
The bond is 80.1% ionic.