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Question

The hybrid states of C in diamond and graphite are respectively:


A

sp2,sp3

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B

sp3,sp2

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C

sp2,sp2

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D

sp3,sp3

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Solution

The correct option is B

sp3,sp2


In diamond, each C-atom is sp3 hybridised. C-C bond distance in diamond is 15.4 pm, which is a normal distance of C-C covalent bond. As diamond consists of a number of tetrahedral units, it is a giant molecule. Also since each C-atom form four covalent bonds, it produces a repetitive three-dimensional structure using only covalent bonds.

Graphite is composed of flat two- dimensional sheets or layers of carbon atom. Each sheet or layer consists of flat hexagonal rings of C-atoms and may be regarded as a fused system of benzene rings, in each layer each C-atom is linked to three C-atoms by C-C covalent bond. The C-C bond length in each hexagonal rings is 141.5 pm, as each C-atom is sp2 hybridised, i.e., only three of the valence electrons of each C-atom are involved in forming σ bond (using sp2 hybrid orbitals). The fourth electron forms a π - bond. The π- electrons are delocalized over the whole sheet, and as they are mobile, graphite conducts heat and electricity. Conductance can occur in a sheet (layer) but not from one a sheet to another.

These layers are stacked on one another such that atoms of alternate layers lie vertically beneath one another. The interlayer distance, i.e., the distance between the layers is 340 pm. The interlayer distance is appreciably large, more than twice the covalent radius of carbon. This indicates that the layers are not joined together by covalent bonds but are joined together by weak van der waals force.


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