JEE Chemistry 2023 Question with Solution

Given: The interhalogens are $IF_5$ and $IF_7$.

Find: The sum of lone pairs present on the central atom in both molecules.

For $IF_5$:

Total valence electrons are $42$.

So, $10$ bonding electrons are used in $5$ $I-F$ bonds.

These electrons are used to complete the octet of the fluorine atoms, and $2$ electrons remain on iodine. Hence iodine has 1 lone pair in $IF_5$.

For $IF_7$:

Total valence electrons are $56$.

So, $14$ bonding electrons are used in $7$ $I-F$ bonds.

These electrons are used to complete the octet of the fluorine atoms, and no electrons remain on iodine. Hence iodine has 0 lone pairs in $IF_7$.

Therefore,

The total number of lone pairs on the central atoms in $IF_5$ and $IF_7$ is 1.

The extracted solution image indicates that the central iodine atom in $IF_5$ has 1 lone pair, while the central iodine atom in $IF_7$ has zero lone pairs. The text in the second solution says "0.1.00", which is inconsistent with the working and diagram, so the worked solution is taken as authoritative.

Thus, the required sum is 1.