Electronegativity and Bond Polarity Example Problem

Determining Covalent or Ionic Bonds

Nitrogen molecule
Nitrogen molecule. Getty Images/PASIEKA

This example problem demonstrates how to use electronegativity to determine bond polarity and whether or not a bond is more covalent or more ionic.

Problem:

Rank the following bonds in order from most covalent to most ionic.

a. Na-Cl
b. Li-H
c. H-C
d. H-F
e. Rb-O

Given:
Electronegativity values
Na = 0.9, Cl = 3.0
Li = 1.0, H = 2.1
C = 2.5, F = 4.0
Rb = 0.8, O = 3.5

Solution:

The bond polarity, δ can be used to determine if a bond is more covalent or more ionic.

Covalent bonds are not typically polar bonds so the smaller the δ value, the more covalent the bond. The reverse is true for ionic bonds, the greater the δ value, the more ionic the bond.

δ calculated by subtracting the electronegativities of the atoms in the bond. For this example, we are more concerned with the magnitude of the δ value, so the smaller electronegativity is subtracted from the larger electronegativity.

a. Na-Cl:
δ = 3.0-0.9 = 2.1
b. Li-H:
δ = 2.1-1.0 = 1.1
c. H-C:
δ = 2.5-2.1 = 0.4
d. H-F:
δ = 4.0-2.1 = 1.9
e. Rb-O:
δ = 3.5-0.8 = 2.7

Answer:

Ranking the molecule's bonds from most covalent to most ionic shows

H-C > Li-H > H-F > Na-Cl > Rb-O