Q. 114.4( 9 Votes )

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

Answer :

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.


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