Sp3 hybridization

Sp³ hybridization

This type of hybridization can be explained by taking the example of CH₄ molecule in which three is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp³ hybrid orbital of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp³ hybrid orbital. The four sp³ hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp³ hybrid orbital is 109.5 as shown below.

The structure of NH₃ and H₂O molecules can also be explained with the help of sp³ hybridization.  In NH₃, the valence shell (outer) electronic configuration of nitrogen in the grounds state is 2s²  2p¹_x  2p¹_y  2p¹_z having three unpaired electrons in the sp³ hybrid orbitals and a lone pair of electrons is present in the fourth one. These three hybrid orbitals overlap with 1s orbitals of hydrogen atoms to form three N-H sigma bonds.  As we know that the force of repulsion between a lone pair and a bond pair is more than the force of repulsion between two bond pairs of electrons. The molecule thus gets distorted and the bond angle is reduced to 107 from 109.5. The geometry of such a molecule will be pyramidal as shown below.

In the case of H₂O molecule, the four oxygen orbitals (one 2s and three 2p) undergo sp³ hybridization forming four sp³ hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These four sp³ hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by hydrogen atoms while the other two by the lone pairs. The bond angle in this case is reduced to 104.5 from 109.5 and the molecule thus acquires a V-shape or angular geometry.