A crystal is made up of metal ‘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.
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a
+2, +4
b
+1, +3
c
+4, +2
d
+3, +1
answer is A.
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Detailed Solution
If O2−ions form ccp that gives 4.0v’s and 8 T.v’s50% of O.V i.e, 2.0 Vs are occupiiied by M112.5% of T.V i.e 7 T.V is occupied by M2So in one formule unit there is 2M1 1M2 and 4O−2 So M1 must be M1+3 and M2 must be M22+ ⇒M2(M1)2O4