JEE Advanced 2019 · Paper 1 · Chemistry Q14

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ChemistryNumerical+3Question 32 of 54

At 143 K, the reaction of $\ce{XeF4}$ with $\ce{O2F2}$ 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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Answer and worked solution

Official answer: 19

$\ce{O2F2}$ is a powerful fluorinating agent:
$\ce{XeF4 + O2F2 -> XeF6 + O2}.$

Lone pairs in Y = $\ce{XeF6}$ .

Xe has one lone pair (6 bond pairs + 1 lone pair, distorted octahedral).
Each F has 3 lone pairs: $6\times3=18$ .

Total: $18+1=19$ .

Answer: 19

Topics:Group 18 VSEPR theory

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JEE Advanced 2019 · Paper 1, Chemistry question 14