Structure of a mixed oxide is cubic close-packed (c.c.p.). the cubic unit cell of mixed oxide is composed of oxide ions. one fourth of the tetrahedral voids are occupied by divalent metal a and the octahedral voids are occupied by a monovalent metal
b. the formula of the oxide is
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Actually the bigger molecules like oxygen occupy the lattice points.In ccp unit cell the effective number of atoms occupying lattice points is 8(1/8)+6(1/2)=4 and in ccp the number of tetrahedral voids is 8 so divalents ions will be 8/8 and number of octahedral voids are 4 so one half would be 2 finally,XY2O4.
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Answer: AB2O2
Explanation :
In ccp structure the oxide anion will occupy the lattice point and cations will occupy the voids .
Let there be N no of O2− in a crystal .
Therefore octahedral voids =N tetradral voids = 2N
A2+ ion occupies the tetrahedral voids so number of atoms A=1/4×2N=N/2
B2+ ion occupies the octahedral voids so number of atoms B=1×N=N
Hence the fomula is AN/2BNON=AB2O2
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