QJEEM24Ch0694: JEE Main 2024 Chemistry - Class 11 JEEM24Ch0694 Solution: A diatomic molecule has a dipole moment of 1.2   D . If…

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):

$$\mu =q\times d$$

Where:

• μ is the dipole moment.


• q is the magnitude of the charge on each atom.


• d is the distance between the charges.

First, convert units to be consistent with esu:

• Bond distance in cm: $$1{\mathrm{A}}^{\circ }=1\times {10}^{-8}\mathrm{cm}$$


• Dipole moment in esu·cm: $$1.2\mathrm{D}=1.2\times {10}^{-18}\mathrm{esucm}$$

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:

$$q=\frac{\mu }{d}$$

Now plug in the given values:

$$q=\frac{1.2\times {10}^{-18}\mathrm{esucm}}{1\times {10}^{-8}\mathrm{cm}}$$

$$q=1.2\times {10}^{-10}\mathrm{esu}$$

$$q=1.2\times {10}^{-9}\times {10}^{-1}\mathrm{esu}$$

$$q=0.000000012\times {10}^{-1}\mathrm{esu}$$