JEE Main 2024 — Chemical Bonding And Molecular Structure Question with Solution

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:

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

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

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

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