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).
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Correct Answer: 14
Solution and Explanation
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.
Total lone pairs on xenon atoms:
2 × 3 = 6
Each fluorine atom has 3 lone pairs. There are 3 fluorine atoms in the cation:
3 × 3 = 9
Total number of lone pairs on the cation (Xe and F combined):
Final Answer: 14
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