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A FCC crystal...

Question

A FCC crystal of $O^{2 -}$ (oxide) ion has ${M n}^{n +}$ ions in all octahedral voids. Then value of n is
(1) 4
(2) 1
(3) 2
(4) 3

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Solution

In FCC crystal, no. of oxide ions per unit cell.
In FCC, Number of close packed spheres = number of octahedral voids
$∴$ Total no. of octahedral voids = 4
Number of $M^{n +}$ particles = 4
As all O.V. are occupied by $M^{n +}$ .
$∴$ Formula of compound is $M_{4} O_{4}$
4n + (-2)4 = 0
n = $\frac{8}{4}$
n = 2
Hence, option (3) is correct.

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