Chemistry · p-Block Elements (Group 15-18)

JEE Advanced 2019 — Paper 1 — Question 28

At 143 K , the reaction of $\mathrm{XeF}_{4}$ with $\mathrm{O}_{2} \mathrm{~F}_{2}$ produces xenon compound Y. The total number of lone Pair(s) of electrons present on the whole molecule of Y is $\_\_\_\_$

Answer: 19

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{XeF}_{4}+\mathrm{O}_{2} \mathrm{~F}_{2} \longrightarrow \mathrm{XeF}_{6}+\mathrm{O}_{2}$

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Chemistry
Chapter: p-Block Elements (Group 15-18)
Topic: Different Compounds of Noble Gases