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

JEE Advanced 2024 — Paper 2 — Question 28

The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are

  1. Option A:

    [PF4]+[PF6]−\left[\mathrm{PF}_{4}\right]^{+}\left[\mathrm{PF}_{6}\right]^{-}and [PCl4]+[PF6]\left[\mathrm{PCl}_{4}\right]^{+}\left[\mathrm{PF}_{6}\right]

  2. Option B:

    [PCl4]+[PCl4 F2]−\left[\mathrm{PCl}_{4}\right]^{+}\left[\mathrm{PCl}_{4} \mathrm{~F}_{2}\right]^{-}and [PCl4]+[PF6]−\left[\mathrm{PCl}_{4}\right]^{+}\left[\mathrm{PF}_{6}\right]^{-}

    Correct
  3. Option C:

    PF3\mathrm{PF}_{3} and PCl3\mathrm{PCl}_{3}

  4. Option D:

    PF5\mathrm{PF}_{5} and PCl3\mathrm{PCl}_{3}

Answer: B

Step-by-step solution

On fluorination of PCl5\mathrm{PCl}_{5} in polar organic solvent ionic isomers are formed, i.e. [PCl4]+[PCl4 F2]−→\left[\mathrm{PCl}_{4}\right]^{+}\left[\mathrm{PCl}_{4} \mathrm{~F}_{2}\right]^{-} \rightarrow

(Colourless crystal) [PCl4]+[PF6]−→(\left[\mathrm{PCl}_{4}\right]^{+}\left[\mathrm{PF}_{6}\right]^{-} \rightarrow( White crystal)

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Chemistry
Chapter
p-Block Elements (Group 15-18)
Topic
Compounds of other Group 15 elements - preparation and properties
The species formed on fluorination of phosphorus pentachloride in a… | JEE Advanced 2024 PYQ with Solution · DhiX AI