Import Question JSON

$\text{HINT: Dipole moment = charge } \times \text{ distance} \text{Explanation:} \text{First, convert the given dipole moment from Debyes to esu-cm:} $1 \text{ Debye} = 10^{-18} \text{ esu-cm}$ $\mu = 1.2 \text{ D} = 1.2 \times 10^{-18} \text{ esu-cm}$ \text{Next, convert the bond length from meters to centimeters:} $1 \text{ m} = 100 \text{ cm} = 10^2 \text{ cm}$ $\text{d} = 10^{-10} \text{ m} = 10^{-10} \times 10^2 \text{ cm} = 10^{-8} \text{ cm}$ \text{The formula for dipole moment is } \mu = \text{q} \times \text{d} \text{, where q is the charge and d is the distance.} \text{We want to find the charge q (fraction of electronic charge). So, } \text{q} = \frac{\mu}{\text{d}} \text{Step I: Calculate the charge (q) in esu.} $\text{q} = \frac{1.2 \times 10^{-18} \text{ esu-cm}}{10^{-8} \text{ cm}} = 1.2 \times 10^{-10} \text{ esu}$ \text{Step II: The charge of an electron (e) in esu is approximately } 4.8 \times 10^{-10} \text{ esu.} \text{Step III: Calculate the percentage of electronic charge.} $\% \text{charge} = \frac{\text{Calculated value of charge}}{\text{Theoretical value of charge}} \times 100\%$ $\% \text{charge} = \frac{\text{q}}{\text{e}} \times 100\%$ $\% \text{charge} = \frac{1.2 \times 10^{-10} \text{ esu}}{4.8 \times 10^{-10} \text{ esu}} \times 100\%$ $\% \text{charge} = \frac{1.2}{4.8} \times 100\% = 0.25 \times 100\% = 25\%$ \text{So, the fraction of an electronic charge on each atom is } 25\%.}$