Chemical Bonding of Class 11


We can explain the formation of four covalent bonds by an atom of carbon by considering promotion of a 2s electron to a 2p orbital. Let us now consider the formation of a molecule of methane, CH4, by such an excited carbon atom. There will be three C—H bonds formed by overlap of the three 2p-orbitals of carbon with the 1s orbitals of three hydrogen atoms. The sp bonds will be mutually perpendicular to one another. The fourth C—H bond would be formed by overlap of the 2s orbital of carbon with an 1s orbital of hydrogen atom. Since the 2s orbital is spherically symmetrical, the direction of the hydrogen atom held by this bond cannot be directly ascertained. At the same time, one should expect this s—s bond to be of lower strength than the other three s—p bonds. But it is contrary to our experience. We know that all four C—H bonds in methane are all alike and they are arranged symmetrically around the central carbon atom directed along the four corners of a tetrahedron. This necessitates a concept of mixing the s-orbital of carbon with its three p-orbitals before overlap. The resulting equivalent orbitals, each having one-fourth s-character and three –fourth p-character, may now undergo overlap with four hydrogen atoms to form four equivalent C—H bonds. We may reasonably extend this concept to interpret the equivalence of the two bonds in BeF2 or the three bonds in BCl3. This procedure of prior mixing of the orbitals has been given rigorous mathematical formulation.

Hybridization is a concept of mixing different atomic orbitals of comparable energy resulting in an equal number of orbitals with mixed character. The resulting hybrid orbitals undergo better overlap and form stronger bonds than the pure orbitals in conformity with the most stable geometry for a molecule. The formation of four equivalent C—H bonds by carbon in forming methane may then be conceived of in terms of the following successive steps:


At first, a 2s electron of the carbon atom gets unpaired and promoted to a 2p orbital. The 2s and the three 2p orbitals are now hybridized to give four equivalent orbitals, each possessing one part s character to three parts p character in their wave function, directed to the corners of a regular tetrahedron. These sp3 hybrid orbitals now form four equivalent C—H bonds in the methane molecule; the bonds are distributed tetrahedrally around the carbon atom.

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