Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ____________.

Solution

Chemistry ma’am · 4 steps

Count the fcc lattice points first, then add the carbons sitting in half of the tetrahedral voids.

1. FCC lattice points. An fcc unit cell has 8 corner atoms, each shared by 8 cells, and 6 face-centre atoms, each shared by 2 cells. That gives 8 × 1/8 + 6 × 1/2 = 1 + 3 = 4 carbon atoms.
2. Tetrahedral voids. Every close-packed arrangement has 2 tetrahedral voids per atom. With 4 atoms in the cell, there are 2 × 4 = 8 tetrahedral voids, all lying fully inside the cube.
3. Half are filled. Carbon occupies 50% of them, so 1/2 × 8 = 4 more carbon atoms sit inside the cell.

   Z = 4 + 4 = 8

4. So. A diamond unit cell holds 8 carbon atoms.

The common slip: 4. Forgetting the void atoms and counting only the fcc lattice points gives 4, not 8.

The idea to keep. In any close-packed cell with N atoms there are 2N tetrahedral and N octahedral voids. Count the lattice atoms, then add whatever fraction of the voids is filled.

Answer from the NTA final answer key. Solution by PrepWiser.