Arun
Last Activity: 6 Years ago
Dear student
Since oxide ions are fcc so 4O-2 ∴ unit cell
A+2 are at 1/8th of the tetrahedral so 1A+2|unitcell
B+3 occupies ½ of the octahedral voids ∴ 2B+2/unit cell
After removing O-2 ions
Oxide ion / unit cell = 4/8 + 3 = 3.5
A+2 ions/ unit cell = 2/8 + 1 = 10/8 = 1.25
B+3 ions/unit cell = 2
∴ P.F = 3.5 × 4/3 πr3 + 1.25 × 43πr3A+2 + 2 × 4/3 πr3B+3
We know that a = 4r / √2
rA+2 = 0.225, ∴rA, = 0.225r–
rB+3 / r– =0.414 ∴ rB+3 = 0.414r-
Putting all the values
P.F = 0.676
Regards
Arun (askIITians forum expert)