Structure of a mixed oxide is CCP. The unit cell is composed of oxide ion. One fourth of the tetrahedral voids are occupied by divalent metal A & the octahedral voids are occupied by monovalent metal B. What will be the formula of the solid?
Asked by Kalita Padmanath | 11th May, 2017, 01:24: PM
Number (N) of atoms in ccp = 4=O2-
Hence no. of tetrahedral holes = 2 x N = 2 x 4= 8
No. of A2+ ions = 8 x ¼ =2
No. of octahedral voids = No. of B+ ions = n = 4
Ratio of O2- : A+2: B+ =4: 2: 4=2: 1: 2
Hence the formula of oxide = AB2O2
Number (N) of atoms in ccp = 4=O2-
Hence no. of tetrahedral holes = 2 x N = 2 x 4= 8
No. of A2+ ions = 8 x ¼ =2
No. of octahedral voids = No. of B+ ions = n = 4
Ratio of O2- : A+2: B+ =4: 2: 4=2: 1: 2
Hence the formula of oxide = AB2O2
Answered by Vaibhav Chavan | 11th May, 2017, 04:26: PM
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