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Chemistry · CHEMICAL BONDING AND MOLECULAR STRUCTURE
हिन्दीThe dipole moment of \( \text{HCl} \) is \( 1.07\text{ D} \) and its bond length is \( 127\text{ pm} \). Given that \( 1\text{ D} = 3.33564 \times 10^{-30}\text{ C m} \) and electronic charge \( e = 1.602 \times 10^{-19}\text{ C} \), the fractional electronic charge (\( \delta = q/e \)) residing on each atom is approximately:
Correct answer · Option 4
\( 0.175 \)
Explanation
\( \mu = 1.07 \times 3.33564 \times 10^{-30} = 3.569 \times 10^{-30}\text{ C m} \). \( q = \frac{\mu}{d} = \frac{3.569 \times 10^{-30}}{127 \times 10^{-12}} = 2.81 \times 10^{-20}\text{ C} \). Thus, \( \delta = \frac{2.81 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 0.175 \).
