Question

The crystal structure of diamond is fcc, and the atom coordinates in the unit cell are (0, 0, 0), ( ½, ½ ,0) , ( ½ , 0, ½ ) , (0, ½ , ½ ) , ( ¼ , ¼ , ¼ ) , ( ¾ , ¼ , ¾ ), ( ¾ , ¾ , ¼ ), and ( ¼, ¾ , ¾ ). The lattice parameter is a= 3.57 Å. What is the c-c bond distance in diamond?

Short answer

The C-C bond length in diamond is $1.54A°$.

Step-by-step solution

FCC crystal

The face centred cubic unit cell consists of eight atoms in the corners and are shared by eight other unit cells. In the center of each face one more atom is present and is shared by two unit cells. Overall there will be four atoms in the unit cell of fcc. But for diamond there will be eight atoms.

Calculation

Given, the lattice parameter$a=3.57A°$.

Each atom in a diamond cubic crystal is tetrahedrally bonded to 4 other atoms, so there are 4 nearest neighbors (NNs).

In other words, the coordination number (CN) is 4. The NN distance is between the two coordinates (0,0,0) and ( ¼, ¼ , ¼ ) is

$\sqrt{{\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2}=\frac{\sqrt{3}}{4}a$

The bond length of the diamond crystal(distance between two nearest neighbors):

$\begin{array}{c}\mathrm{Bond}\mathrm{length}=\frac{\sqrt{3}\mathrm{a}}{4}\\ =\frac{\sqrt{3}\times 3.57\mathrm{A}°}{4}\\ =1.54\mathrm{A}°\end{array}$.