A crystal is made up of metal ions 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1' occupies 50% of octahedral voids and the cation 'M2' occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1' and 'M2' are, respectively:
(1) + 1, +3
(2) +3, +1
(3) +2, +4
(4) +4, +2
(3) +2, +4
Explanation:
In the ccp lattice of oxide ions effective number of O–2 ions = 8 × 1/8 + 6 x 1/2 = 4
In the ccp lattice,
No. of octahedral voids = 4
No. of tetrahedral voids = 8
Given M1 atoms occupies 50% of octahedral voids and M2 atoms occupies 12.5 of tetrahedral voids
No. of M1 metal atoms = 4 x 50/100 = 2
No. of M2 metal atoms = 8 x 12.5/100 = 1
∴ Formula of the compound = (M1)2(M2)O4
∴ Oxidation states of metals M1 & M2 respectively are +2 and +4.