In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
In ccp structure, each unit cell contains atoms at all the corners and at the centre of all the faces of the cube. Each atom at a corner is shared between eight adjacent unit cells, four unit cells in the same layer and four unit cells of the upper (or lower) layer. Only 1/8th of a particle actually belongs to a particular unit cell. The atom at the faces is also shared with the adjacent unit cell, both sharing half of the atom i.e. only 1/2 of each atom belongs in the unit cell.
To calculate number of atoms per unit cell of ccp,
In a FCC
There are 8 corner atoms × 1/8 atom per unit cell = 8 x 1/8 = 1 atom
There 6 face-centered atoms × 1/2 atom per unit cell = 6 x 1/2 = 3 atoms
Total atoms in unit cell = 1 + 3 = 4 atoms.
The number of tetrahedral voids is twice the number of atoms and octahedral voids is same as the number of atoms.
In cubic closed packed (ccp) lattice, number of atoms per unit cell = 4.
Number of tetrahedral voids present = 2 x number of atoms per unit cell = 2 x 4 = 8.
Given that metal atoms (M) occupy 1/3rd of present tetrahedral voids, M = 1/3 x 8 = 8/3
So the ratio of M and N in a unit cell is M:N = 8/3:4 = 2/3:1
Hence, the formula is M2N3.