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:
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:
Option 1 -
+2, +4
Option 2 -
+1, +3
Option 3 -
+3, +1
Option 4 -
+4, +2
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1 Answer
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Correct Option - 1
Detailed Solution:In the ccp lattice of oxide ions effective number of O? ² ions = 8 ×? + 6 × ½ = 4
In the ccp lattice,
No. of octahedral voids = 4
No. of tetrahedral voids = 8
Given M? atoms occupies 50% of octahedral voids and M? atoms occupies 12.5 of tetrahedral voids
No. of M? metal atoms = 4 ×? /? = 2
No. of M? metal atoms = 8 × ¹²·? /? = 1
∴ Formula of the compound = (M? )? (M? )O?
∴ Oxidation states of metals M? & M? respectively are +2 and +4 .
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