What is the state of hybridization of carbon in (a) CO₃^2– (b) diamond (c) graphite?

The state of hybridization of carbon in:

(a) Graphite

1. Each carbon atom in graphite is sp² hybridized and is bound to 3 other carbon atoms.

2. Graphite exists as sheets of hexagonal arrays of carbon. Each carbon atom in graphite thus has a trigonal planar geometry, which implies sp2 hybridization. Moreover, the p orbitals axial to the hexagonal sheets help to bond the layering sheets together.

(b) CO₃^2-

1. C in CO₃²⁻ is sp² hybridized and is bonded to 3 oxygen atoms.

2. No. of single bonds between carbon and other atoms:3

3. No. of lone pairs on carbon atom:0

4. Now sum up the bond pairs and lone pairs = 3 + 0=3. sp² that is, one s and two p orbitals.

(c) Diamond

1. Each carbon in diamond is sp³ hybridized and is bound to 4 other carbon atoms.

2. The electronic configuration of carbon is 1s² 2s² 2p², i.e. with four valence electrons spread in the s and p orbitals.

3. In order to create covalent bonds in diamond, the s orbital mixes with the three p orbitals to form sp³ hybridization.