Question

Xe and F₂ in 1:1 molar ratio when mixed in a closed flask and kept in sunlight for a day, gave white crystals of a compound Q. Two equivalents of Q on reaction with one equivalent of AsF₅ gave an ionic compound X⁺Y^– with the cation having two Xe atoms. The total number of lone pairs present on the cation X⁺ is ___________ (in integer).

Hint:

In xenon fluorides, especially polyatomic ions, lone pairs on central atoms should be evaluated based on VSEPR structures and known bonding schemes. Each Xe in [Xe₂F₃]⁺ typically retains 3 lone pairs.

Answer 1

When xenon and fluorine are combined in a 1:1 molar ratio under sunlight, they form xenon difluoride (XeF₂) as a white crystalline solid.

Upon reacting 2 equivalents of XeF₂ with 1 equivalent of AsF₅, the following ionic compound is formed:

2 XeF₂ + AsF₅ → [Xe₂F₃]⁺ [AsF₆]⁻

The cation [Xe₂F₃]⁺ contains two xenon atoms bridged by a fluorine atom and bonded to terminal fluorines. Each Xe is in the +2 oxidation state.

Each xenon atom has 8 valence electrons. After bonding, each Xe retains 3 lone pairs.
Therefore, total number of lone pairs on both xenon atoms:

2 × 3 = 6

But due to the three-center four-electron bonding in [Xe₂F₃]⁺, each Xe also contributes additional electron density involved in bonding. The correct total count of lone pairs on the cation X⁺ (considering structure and all nonbonding pairs on Xe and F) is:

☑ Answer: 14