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Question Paper from: IIT JEE Main 2021 TestSeries Paper 2

In a compound, XY₂O₄oxide ions are arranged in CCP structure. Cation X is present in the octahedral void and cation Y is equally distributed between octahedral and tetrahedral voids. The fraction of the octahedral voids occupied are

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{6}$

Solution:

Chapter Name: Solid State

Difficulty Level: Moderate

The unit cell of CCP is FCC

$∴$ No of oxide ions = 4

No of T.V = $2 \times 4 = 8$

No of O.V = 4

According to formula, No of cations X present per unit cell = 1

No of cations Y present per unit cell = 2

Since, Y are equally distributed in O.V and T.V (i.e)

One Y cation in O.V and other Y cation in T.V

$∴$ Total number of O.V captured = 1 + 1 = 2

X is also present in O.V fraction of O.V occupied = $\frac{2}{4} = \frac{1}{2}$

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IIT JEE Main 2021 TestSeries Paper 2

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