JEE Main 2023Chemistryp Block Elements (Group 15, 16, 17 & 18)HardNumerical

JEE Main 2023 — p Block Elements (Group 15, 16, 17 & 18) Question with Solution

JEE Main 2023 (08 Apr Shift 1)

Question

XeF4 reacts with SbF5 to form XeFmn+SbFy2- m+n+y+z=  ?. 

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Show full solutionCorrect answer: 11
Correct answer
11

Step-by-step explanation

Xenon tetrafluoride (XeF4) is square planar and acts as a fluoride donor with SbF5. XeF4 changes to the cation XeF3+ and SbF5 changes to the anion SbF6-.

XeF4+SbF5⟶XeF3+SbF6-

 m = 3, n = 1, y = 6, z = 1

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About this question

This is a previous-year question from JEE Main 2023, covering the p Block Elements (Group 15, 16, 17 & 18) chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.