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JEE Main 20238 AprilMorning Shift

Exam: JEE Main
Year: 2023
Date: 8 April
Shift: Morning Shift
Subject: Chemistry
Chapter: p Block Elements (Group 15, 16, 17 & 18)

JEE Main 2023 · 8 April · Morning Shift · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

XeF _ 4 reacts with SbF _ 5 to form [ XeF _ m ] ^ n + [ SbF _ y ] ^...

{XeF}_{4}

reacts with

{SbF}_{5}

to form

{\left[{XeF}_{m}\right]}^{n +}{\left[{SbF}_{y}\right]}^{2 -}m + n + y + z = ?

.

Correct answer

11

Step-by-step solution

Xenon tetrafluoride ( ${XeF}_{4}$ ) is square planar and acts as a fluoride donor with ${SbF}_{5}$ . ${XeF}_{4}$ changes to the cation ${\left[{XeF}_{3}\right]}^{+} and$ {SbF}_{5} $changes to the anion$ {\left[{SbF}_{6}\right]}^{-} $.</p><p>$ {XeF}_{4} + {SbF}_{5} ⟶ {\left[{XeF}_{3}\right]}^{+}{\left[{SbF}_{6}\right]}^{-} $</p><p>$ m = 3, n = 1, y = 6, z = 1

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