JEE Chemistry 2023 Question with Solution

Given: The species are $\mathrm{ClO_3^-}$, $\mathrm{XeF_4}$, $\mathrm{SF_4}$ and $\mathrm{I_3^-}$.

Find: The maximum number of lone pairs on the central atom.

Analyse each species using its Lewis structure:

1. $\mathrm{ClO_3^-}$: the central chlorine atom has 1 lone pair.
2. $\mathrm{XeF_4}$: the central xenon atom has 2 lone pairs.
3. $\mathrm{SF_4}$: the central sulfur atom has 1 lone pair.
4. $\mathrm{I_3^-}$: the central iodine atom has 3 lone pairs.

Comparing these values, the maximum number of lone pairs on the central atom is $3$.

Therefore, the required numerical answer is $3$.

Given: Determine the number of lone pairs on the central atom in each species.

Find: Which species has the maximum value.

For $\mathrm{ClO_3^-}$:

The Lewis structure shows that the central Cl atom has 1 lone pair.

For $\mathrm{XeF_4}$:

The Lewis structure shows that the central Xe atom has 2 lone pairs.

For $\mathrm{SF_4}$:

The Lewis structure shows that the central S atom has 1 lone pair.

For $\mathrm{I_3^-}$:

The Lewis structure shows that the central I atom has 3 lone pairs.

Thus, among all the given species, the largest number of lone pairs on the central atom is $3$, found in $\mathrm{I_3^-}$.