Question

What is the state of hybridization of carbon in (a) CO32– (b) diamond (c) graphite​

Answer 1

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) CO32-

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.

Answer 2

I don't know the answer mate sorry

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