JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

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The observed dipole-moment of HCl molecule is 1.03 D . If

\mathrm{H}-\mathrm{Cl}

bond distance is

1.275 \, \mathop {\text{A}}\limits^ \circ

and electronic charge is

4.8 \times 10^{-10}

esu. What is the percent polarity of HCl?

View answer and explanation

Correct option: C

\frac{1.03 \times 100 \times 10^{-18}}{4.8 \times 10^{-10} \times 1.275 \times 10^{-8}}

The correct answer is

\frac{1.03 \times 100 \times 10^{-18}}{4.8 \times 10^{-10} \times 1.275 \times 10^{-8}}

. Percent polarity is calculated by dividing the observed dipole moment by the calculated dipole moment and multiplying by 100.

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