Hybridization of Orbitals

Example: Beryllium Hydride (BeH₂)

The atomic number of beryllium is 4. Its electronic configuration is 1s² 2s². In order to form bonds with two hydrogen atoms the valence electrons (2s²) of beryllium atom must overlap with the 1s electrons of the two hydrogen atoms.

Since the valence shell of beryllium atom contains both the electrons in the same orbital (2s), it cannot overlap with the 1s orbital of hydrogen atoms containing one electron because an orbital can contain a maximum of two electrons with opposite spin.

Pauling got over this problem by suggesting that in the process of bond formation an electron from the 2s orbital of beryllium atom gets momentarily excited to the empty 2p orbital. Now, the two valence electrons are in two singly occupied orbitals which can overlap with the 1s orbitals of the two hydrogen atoms and form two bonds.

The two bonds formed by these overlaps would be of different nature. One of these would involve overlapping of 2s orbital of beryllium with 1s orbital of hydrogen while the other would involve overlapping of 2p orbital of beryllium with 1s orbital of hydrogen. However, experimentally the two bonds are found to be equivalent.

This problem is solved with the help of a concept called hybridization of orbitals. According to this, two or more than two non equivalent orbitals (having different energies and shapes) of comparable energies mix or hybridize and give rise to an equal number of equivalent (same energies and shapes) hybrid orbitals.

In case of Beryllium, the two singly occupied orbitals (2s and 2p) hybridize to give two sp-hybrid orbitals. This is called sp-hybridization. These hybrid orbitals can now overlap with the 1s orbitals of hydrogen atoms to give the linear molecule of BeH₂.