JEE Chemistry 2026 Question with Solution

Given: The molecules are $\mathrm{H_2S}$, $\mathrm{H_2O}$, $\mathrm{NF_3}$, $\mathrm{NH_3}$ and $\mathrm{CHCl_3}$.

Find: The number of lone pairs on the central atom of the molecule having the lowest dipole moment.

Step 1: Compare dipole moments of the given molecules. Approximate dipole moments are

Among these, $\mathrm{NF_3}$ has the lowest dipole moment.

Step 2: Identify the central atom and lone pairs in $\mathrm{NF_3}$. In $\mathrm{NF_3}$, the central atom is nitrogen. Nitrogen has $5$ valence electrons. It forms three $\mathrm{N-F}$ bonds and retains one lone pair.

Step 3: Conclude. Therefore, the molecule with the lowest dipole moment is $\mathrm{NF_3}$, and the number of lone pairs on its central atom is $1$. The correct option is A.

Given: The listed molecules include bent, pyramidal and tetrahedral species.

Find: Which one has the smallest net dipole moment and then count lone pairs on its central atom.

In molecules like $\mathrm{NF_3}$, bond dipoles can oppose the lone pair dipole, resulting in an unusually low net dipole moment. This makes the resultant molecular dipole much smaller than in $\mathrm{NH_3}$, $\mathrm{H_2O}$ or $\mathrm{CHCl_3}$.

Once $\mathrm{NF_3}$ is identified, count electrons on nitrogen:

• Nitrogen has $5$ valence electrons.
• Three electrons are used in forming three bonds with fluorine atoms.
• One pair remains non-bonding.

Hence the central atom has one lone pair, so the answer is A.