In a metal oxide, oxide ions are arranged in a hcp array and the metal ion occupies octahedral holes. If one-third of octahedral voids and all tetrahedral voids are empty the metal oxide formula is (M- metal)

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${MO}_{2}$

$M_{2} O_{3}$

$M_{3} O_{2}$

$MO$

answer is B.

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Detailed Solution

It is provided that oxide ions are arranged in $HCP$ and we know that in $HCP$ there are 6 oxide ions per unit cell.

Numbers of octahedral voids = no. of atom per unit cell. $= 6$

So, out of 6 octahedral voids there exists $\frac{1}{3}$ which are empty.

$So, 6 \times \frac{1}{3} = 2 voids are not there$

So, $6 - 2 = 4$ are present

Therefore, the formula of the metal oxide will be $M_{4} O_{6}$ or $M_{2} O_{3}$ .

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