Chemistry » Chemical Bonding » Covalent Bonding

# Formation of Covalent Bonds

## Formation of Covalent Bonds

Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, H2, contains a covalent bond between its two hydrogen atoms. The figure below illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line.

Along the x-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the x-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms.

The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.

The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.

It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of H2, the covalent bond is very strong; a large amount of energy, 436 kJ, must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:

$${\text{H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{H}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H=436\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Conversely, the same amount of energy is released when one mole of H2 molecules forms from two moles of H atoms:

$$\text{2H}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{\text{2}}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H=-436\phantom{\rule{0.2em}{0ex}}\text{kJ}$$