Question

Use electronegativity values to rank the bonds in the following compounds from least ionic to the most ionic character: IF, ICl, ClF, BrCl and ${\mathbf{Cl}}_{\mathbf{2}}$?

Short answer

IF is the most ionic while ${\text{Cl}}_{\text{2}}$has the least ionic character

$C{l}_2⟨BrCl⟨ICl⟨ClF⟨IF$

Step-by-step solution

Introduction

Electronegativity is the measure with which an atom attracts the shared pair of electrons towards itself. If an atom is more electronegative it is more likely to attract the shared pair of electrons than does the less electronegative.

Explanation:

The differences in the electronegativities of the two atoms describe the polarity of the bond or a molecule. The one having highest electronegativity difference will be the most ionic and the one having least electronegativity difference will be least ionic.

Electronegativity of F= 3.98

Electronegativity of Cl=3.16

Electronegativity of Br=2.96

Electronegativity of I=2.66

From the electronegativity values given above, we can find out the electronegativity difference and thus polar character of the molecule.

IF= 1.32

ICl= 0.5

ClF= 0.82

BrCl=0.2

=0

So the ionic character follows-

$C{l}_2⟨BrCl⟨ICl⟨ClF⟨IF$i.e., IF is the most ionic while $C{l}_2$has the least ionic character.