Electronegativity Chemistry Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

medium
In the molecule HF, the electronegativity of H is 2.1 and F is 4.0. Describe the bond polarity and indicate the direction of the dipole.

Solution

  1. 1
    Calculate ΔEN=∣4.0−2.1∣=1.9\Delta\text{EN} = |4.0 - 2.1| = 1.9.
  2. 2
    This large difference indicates a highly polar covalent bond (borderline ionic).
  3. 3
    The bonding electrons are pulled toward fluorine, creating a partial negative charge (δ−\delta^-) on F and a partial positive charge (δ+\delta^+) on H.
  4. 4
    The dipole points from H (δ+\delta^+) toward F (δ−\delta^-).

Answer

Hδ+–Fδ−,ΔEN=1.9\text{H}^{\delta+}\text{–F}^{\delta-},\quad \Delta\text{EN} = 1.9
The greater the electronegativity difference, the more polar the bond. In HF, fluorine strongly attracts the shared electrons, creating a significant molecular dipole.

About Electronegativity

A dimensionless measure of how strongly an atom attracts the shared electrons in a covalent bond toward itself, quantified on the Pauling scale from 0.7.

Learn more about Electronegativity →

More Electronegativity Examples

Example 1 easy

Arrange the following elements in order of increasing electronegativity: Na, Cl, F, O.

Example 3 easy

Which bond is more polar: H–Cl or H–Br? (EN values: H = 2.1, Cl = 3.0, Br = 2.8)

Example 4 medium

Which bond is more polar, C–H or O–H? Use these electronegativity values: C = 2.5, H = 2.1, O = 3.5.

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