A diatomic molecule has a dipole moment of $1.2 \mathrm{~D}$. If the bond distance is $1 \mathrm{~A}^{\circ}$, then fractional charge on each atom is _________ $\times 10^{-1}$ esu.

(Given $1 \mathrm{~D}=10^{-18}$ esucm)

<p>To find the fractional charge on each atom in a diatomic molecule, we can use the relationship between dipole moment (μ), charge (q), and distance (d):</p> <p>$\mu = q \times d$</p> <p>Where:</p> <ul> <li>μ is the dipole moment.</li> <br/><li>q is the magnitude of the charge on each atom.</li> <br/><li>d is the distance between the charges.</li> </ul> <p>First, convert units to be consistent with esu:</p> <ul> <li>Bond distance in cm: $1 \mathrm{~A}^{\circ} = 1 \times 10^{-8} \mathrm{~cm}$</li> <br/><li>Dipole moment in esu·cm: $1.2 \mathrm{~D} = 1.2 \times 10^{-18} \mathrm{~esucm}$</li> </ul> <p>We are given that the bond distance ($ d $) is $ 1 \mathrm{~A}^{\circ} $ and that the dipole moment $( \mu $) is $ 1.2 \mathrm{~D} $. Using these values, we can calculate the charge ($ q $) as follows:</p> <p>$q = \frac{\mu}{d}$</p> <p>Now plug in the given values:</p> <p>$q = \frac{1.2 \times 10^{-18} \mathrm{~esucm}}{1 \times 10^{-8} \mathrm{~cm}}$</p> <p>$q = 1.2 \times 10^{-10} \mathrm{~esu}$</p> <p>$q = 1.2 \times 10^{-9} \times 10^{-1} \mathrm{~esu}$</p> <p>$q = 0.000000012 \times 10^{-1} \mathrm{~esu}$</p>