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The structure of diamond

Atomic carbon has an atomic number of 6 and a 1$s^2$2$s^2$2$p^2$ electronic ground state configuration. The carbon atom's electronic configuration is believed to change from its ground state in diamond as follow:

\begin{figure}\centerline{\epsfxsize=12.0cm \epsfbox{spin2.ps}}\end{figure}



If a carbon atom enters into the structure of diamond its two 2$s$ and 2$p$ electrons redistribute into four new equal-energy-level orbitals called 2$(sp^3)$ hybrid orbitals. It require a loss of energy but this effect is compensated by a very profitable covalent bonding. The angular distribution of the wave functions for these four 2$(sp^3)$ orbitals can be illustrated by drawing four lobes whose axes are at 109$^{\circ}$28$'$ to each other, the axes of these lobes thus extend toward the corners of an imaginary tetrahedron centered around the carbon atom [Fig.2.1].

Quantum-mechanical calculations indicate that greater overlap between orbitals results in a stronger covalent bond. The diamond structure represents a three-dimensional network of strong covalent bonds [Fig.2.2], which explains why diamond is so hard.

The diamond structure is cubic with a cube edge length of $a_0=3.567 $ Å and can be viewed as two interpenetrating FCC structures displaced by (1/4,1/4,1/4)$a_0$. The diamond crystal is highly symmetric with a cubic space group $F4_1/d$ $\bar{3}$ $2/m=Fd3m=O^7_h$.

Since all the valence electrons contribute to the covalent bond, they are not free to migrate through the crystal and thus, diamond is a poor conductor with a bandgap of 5.48 eV.


next up previous
Next: The structure of graphite Up: Diamond and graphite Previous: Diamond and graphite
2003-01-02