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Question: The dipole moment of $\ce{HBr}$ is $2.60 \times 10^{-30}$ and the interatomic spacing is $1.41$. What is the percentage ionic character of $\ce{HBr}$?

What I know is that the percentage ionic character is observed dipole moment divided by calculated dipole moment $\times\ 100$.

The calculated dipole moment is $\textrm{charge on electron} \times \textrm{radius of molecule}$.

However, how would I convert interatomic spacing into radius?

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The calculated dipole moment is charge on electron * radius of molecule

That's off a bit, the calculated dipole moment (calculated assuming one electron has been completely transferred from hydrogen to bromine) is dependent on the bond length (interatomic spacing), not the radius. \begin{aligned} \mu_{100\%} &= \mathrm{electron ~charge \times interatomic ~spacing}\\ \mu_{100\%} &= (1.6 \times 10^{-19})\ \mathrm C \times (1.41 \times 10^{-10})\ \mathrm m\\ \mu_{100\%} &= 2.26 \times 10^{-29}\ \mathrm{C\ m}\\ \end{aligned}

As you point out, $$\mathrm{\% ~ionic~ character} = \frac{[\mu_\mathrm{obs}]}{[\mu_{100\ \%}]} = \frac{[2.6 \times 10^{-30}]}{[2.26 \times 10^{-29}]} = 0.12 = 12\ \%$$

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