Chemistry - Percentage ionic character when electronegativity is given

Solution 1:

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference $\Delta \chi$.

Percent ionic character $= (1-e^{-(\Delta \chi/2)^2} )\times 100$

But I'd like to correct the definition of percent ionic character in your question using dipole moment $\mu$ (not Observed value of ionic character):

Percent ionic character = $\Large\frac{\mu_{\text{observed}}} {\mu_{\text{calculated} }}$ $\times 100 \%$

Where $\mu_{\text{calculated}}$ is calculated assuming a 100% ionic bond.

For more details please see this page.

Solution 2:

$$\text{% of ionic character} = 16\times ∆\mathrm{EN} + 3.5\times (∆\mathrm{EN})^2$$

where $∆\mathrm{EN}$ is electronegativity difference. For example, in $\ce{H-F}$ $∆\mathrm{EN} = 2$:

$$ \begin{align} \text{% of ionic character} &= 16\times 2 + 3.5\times 2^2 \\ &= 32 + 14 \\ &= 46~(\%) \end{align} $$