Arrange the bonds in order of increasing ionic character in the molecules. $\mathrm{LiF}$, $\mathrm{K}_{2}\mathrm{O}$, $\mathrm{N}_{2}$, $\mathrm{SO}_{2}$ and $\mathrm{ClF}_{3}$.

(C) $\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{ClF}_{3}<\mathrm{K}_{2}\mathrm{O}<\mathrm{LiF}$

(A) $\mathrm{ClF}_{3}<\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{K}_{2}\mathrm{O}<\mathrm{LiF}$

(B) $\mathrm{LiF}<\mathrm{K}_{2}\mathrm{O}<\mathrm{ClF}_{3}<\mathrm{SO}_{2}<\mathrm{N}_{2}$

(D) $\mathrm{N}_{2}<\mathrm{ClF}_{3}<\mathrm{SO}_{2}<\mathrm{K}_{2}\mathrm{O}<\mathrm{LiF}$

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Chemical Bonding and Molecular Structure

Arrange the bonds in order of increasing ionic character in the molecules.

LiF

K_{2} O

N_{2}

SO_{2}

and

ClF_{3}

Answer: (C)

Ionic character of a bond increases with the electronegativity difference between the bonded atoms. N

_{2}

(identical atoms) has zero ionic character; SO

_{2}

has a small difference; ClF

_{3}

has a somewhat larger one; K

_{2}

O (a metal oxide) has a large difference; LiF (an alkali-metal halide) has the largest difference among these. So the increasing order is:

N_{2} < SO_{2} < ClF_{3} < K_{2} O < LiF

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option