A crystal is made up of metal ions ' M 1 ' and ' M 2 ' and oxide ions. Oxide ions form a ccp lattice structure. The cation ' M 1 ' occupies 50 % of octahedral voids and the cation ' M 2 ' occupies 12 . 5 % of tetrahedral voids of oxide lattice. The oxidation numbers of ' M 1 ' and ' M 2 ' are respectively :

Step-by-step solution

In the ccp latice of oxide ions effective number of O - 2 ions = 8 × 1 8 + 6 × 1 2 = 4 In the ccp latice, No. of octahedral voids = 4 No. of tetrahedrla voids = 8 Given M 1 atoms occupies 50 % of octahedral voids and M 2 atoms occupies 12 . 5 of tetrahederal voids No. of M 1 metal atoms = 4 × 50 100 = 2 No. of M 2 metal atoms = 8 × 12 .5 100 = 1 ∴ Formula of the compound = M 1 2 M 2 O 4 ∴ Oxidation states of metals M 1 & M 2 respectively are + 2 and + 4