Question

In a crystal, oxide ions are arranged in FCC and A2+ ions are at 1/8 of the terahydralhyderal void. ions of B3+ occupied 1/2 of octahyderal voids. calculate the packing fraction of the crystal, if o2- ions are removed from aletrnate corners and A2+ is being placed at two of the corners.

Answer 1

Answer:

The answer is 3) 4.

In FCC crystal the number of octahedral voids = 4

50% of them are occupied by the metal ion so no of voids occupied =2, hence the number of metal ions =2

In FCC lattice the total number of O2- atoms are =4 ( 8 at corners and 6 at face centers)

Therefore the formula of the compound becomes M2O4 = MO2.

Now since the cahrge on a compond is zero so 1 x n +( 2 x -2) = 0

=> n=4

Answer 2

Answer:

n=4

Hope it's helpful to you