The dipole moment of HBr is $2.60 \times 10^{- 30} Cm$ and the inter-atomic distance is $1.41 Å$ . What is the percent ionic character of HBr? Give your answer after multiplying with 10.

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answer is 115.

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Detailed Solution

$%$ ionic character $\begin{aligned} = \frac{2.6 \times 10^{- 30}}{1.6 \times 10^{- 19} \times 1.41 \times 10^{- 10}} \times 100 = 11.5 \end{aligned}$

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