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A crystal is made up of metal ions 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure.

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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

1 Answer
1

(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.

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