QJEEAdv19Ch7a94: JEE Advanced 2019 Chemistry - Class 12 JEEAdv19Ch7a94 Solution: At 143 K, the reaction of XeF 4 with O 2 F 2 produces a xenon…

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QJEEAdv19Ch7a94: JEE Advanced 2019 Chemistry - Class 12 JEEAdv19Ch7a94 Solution: At 143 K, the reaction of XeF 4 with O 2 F 2 produces a xenon…

At 143 K, the reaction of XeF₄ with O₂F₂ produces a xenon compound Y. The total number of lone pair(s) of electrons present on the whole molecule of Y is .................

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Accepted Answer

XeF₄ reacts with O₂F₂ to form XeF₆.O₂F₂ is fluoronating reagent.

X e F_{4} + O_{2} F_{2} \overset{143 K}{⟶} \underset{(Y)}{X e F_{6}} + O_{2}

The structure of XeF₆ is

JEE Advanced 2019 Paper 1 Offline Chemistry - p-Block Elements Question 45 English Explanation

Y compound (XeF₆) has 3 lone pair in each fluorine and one lone pair in xenon.

Hence, total number of lone pairs electrons is 19.

QJEEAdv19Ch7a94: JEE Advanced 2019 Chemistry - Class 12 JEEAdv19Ch7a94 Solution: At 143 K, the reaction of XeF 4 with O 2 F 2 produces a xenon…

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