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Chemistry · Chemical Bonding

JEE Main 2024 — 1 February, Shift 1 — Question 69

Arrange the bonds in order of increasing ionic character in the molecules. LiF,K2O,N2,SO2\mathrm{LiF}, \mathrm{K}_{2} \mathrm{O}, \mathrm{N}_{2}, \mathrm{SO}_{2} and CIF3\mathrm{CIF}_{3}.

  1. Option A:

    CIF3<N2<SO2<K2O<LiF\mathrm{CIF}_{3}<\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{K}_{2} \mathrm{O}<\mathrm{LiF}

  2. Option B:

    LiF<K2O<CIF3<SO2<N2\mathrm{LiF}<\mathrm{K}_{2} \mathrm{O}<\mathrm{CIF}_{3}<\mathrm{SO}_{2}<\mathrm{N}_{2}

  3. Option C:

    N2<SO2<CIF3<K2O<LiF\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{CIF}_{3}<\mathrm{K}_{2} \mathrm{O}<\mathrm{LiF}

    Correct
  4. Option D:

    N2<CIF3<SO2<K2O<LiF\mathrm{N}_{2}<\mathrm{CIF}_{3}<\mathrm{SO}_{2}<\mathrm{K}_{2} \mathrm{O}<\mathrm{LiF}

Answer: C

Step-by-step solution

Increasing order of ionic character N2<SO2<ClF3<K2O<LiF\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{ClF}_{3}<\mathrm{K}_{2} \mathrm{O}<\mathrm{LiF}

Ionic character depends upon difference of electronegativity (bond polarity).

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 1
Subject
Chemistry
Chapter
Chemical Bonding
Topic
Introduction to Bonding and Lewis Octet Theory

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