Question

A fcc crystal of o-2 (oxide) ions has m +n ions in 50% of octahedral voids then the value of n is

Answer 1

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

hope u understand

^_^

Answer 2

Answer:

The value of n for $M^{n+}$  in FCC crystal is equal to 4.

Explanation:

We know the number of atoms in FCC crystal = 4

Therefore number of oxide ions = 4

Octahedral voids  in crystal lattice= 4

$M^{n+}$ ions occupied 50% octahedral voids.

Therefore the number of $M^{n+}$ ions  in the crystal = 2

The formula of compound $=M_{2}O_{4}$ or  $MO_{2}$

The charge on M $=n+2(-2)=0$

            ∴          $n=4$