JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Chemical Bonding and Molecular Structure

The lone pairs on the central atom follow from the valence electrons left after bonding. In $\mathrm{I}{\mathrm{F}}_7$ iodine spends all seven valence electrons on seven bonds, leaving no lone pair. In $\mathrm{IC}{\mathrm{l}}_4^{-}$ iodine has $7+1=8$ electrons; four go into bonds and the remaining four form two lone pairs. In $\mathrm{Xe}{\mathrm{F}}_6$ xenon has eight valence electrons, six used in bonding and one pair left over, a single lone pair. In $\mathrm{Xe}{\mathrm{F}}_2$ xenon keeps three lone pairs after two bonds. Thus $\mathrm{I}{\mathrm{F}}_7$ has zero, $\mathrm{IC}{\mathrm{l}}_4^{-}$ has two, $\mathrm{Xe}{\mathrm{F}}_6$ has one and $\mathrm{Xe}{\mathrm{F}}_2$ has three lone pairs on the central atom.