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Structure of a mixed oxide is cubic close - packed (c.c.p). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is:

Option 1)
A₂BO₂

Option 2)
A₂B₃O₄

Option 3)
AB₂O₂

Option 4)
AB O₂

Answers (1)

best_answer

As we learnt in

No. of atoms(z) for end centered unit cell -

Lattice point: at corners and one pair of opposite face centers.

For end centered cubic, z=2

- wherein

Given that its a c.c.p and unit cell is composed of oxide ions.

So, $O^{2}^{-} ions=4$

Octahedral voids $=\frac{1}{2} tetrahedral\ voids= \frac{1}{2}\times 8=4$

$\therefore A=\frac{1}{4}\times 8=2$

B = 4

Summing up, the formula is $A B_{2}O_{2}$

Option 1)

A₂BO₂

Incorrect option

Option 2)

A₂B₃O₄

Incorrect option

Option 3)

AB₂O₂

Correct option

Option 4)

AB O₂

Incorrect option

Posted by

prateek

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