Question

In FCC structure of mixed oxide. the
lattice is made up of oxide ions, one
eighth of tetrahedral voids are occupied
by divalent ions (A^2+) while one half of
octahedral voids are occupied by trivalent
ions (B^+). What is the formula of the
oxide?​

Answer 1

Explanation:

• Oxide ion crystal in FCC lattice

O²- = 4 (For FCC Z = 4)

• A²+ in 1/8th tetrahedral void (THV)

A²+ = 1/8 × 8 = 1 (No. of THV = 2Z = 8)

• B²+ in 1/2 of octahedral void (OHV)

B²+ = 1/2 × 4 = 2 (No. OHV = Z = 4)

• Formula = AB2O4