UPSEE Solved Model Paper 5

Given oxide ions are in cubic close packed (ccp) arrangement or in face centered cubic arrangement (fcc).
A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube.
Each atom located at the face-centre is shared between two adjacent unit cells and only ½ of each atom belongs to a unit cell. Thus, in a face-centred cubic (fcc) unit cell:
8 corners atoms ×1/8 atom per unit cell=8×1/8 =1 atom 6 face-centred atoms ×1/2 atom per unit cell = 6 ×1/2= 3 atoms
Therefore total number of atoms per unit cell = 4 atoms.
Therefore, number of O2−
∴ Number of O2– ions = 4 {8 ×1/8 + 6 ×1/2 = 4}
Number of octahedral voids per particle = 1
∴ Total number of octahedral voids = 1 x 4 = 4
Since M ions occupy 2/3 of the octahedral voids,
Number of metal ions , (M) = 4 ×

2

3

=

8

3

∴ Formula is

M2

3

O4 = M

2

3

O = M2O3
Therefore the formula of the oxide is M2O3.