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 : [2012 M]a)ABO2b)A2BO2c)A2B3O4d)AB2O2Correct answer is option 'D'. Can you explain this answer?

Question

Answer

According to ccpccp, structure,
Number of O2− ions = 4

tetrahedral void = 8
and octahedral void = 4
∵ A ions occupy l/4th of tetrahedral voids
∴∴ Number of A ions = 1/4 × 8 = 2
Again B ions occupy all the tetrahedral voids

∴∴ Number of B ions = 4

A : B : O = 2 : 4 : 4 = 1 : 2 : 2

∴∴ the formula of oxide = AB2O2

Answer

Answer

Answer:

To determine the formula of the mixed oxide, we need to analyze the information provided about the structure and the occupancy of the voids by different metal ions.

Crystal Structure:
The mixed oxide has a cubic close-packed (c.c.p) structure. In a c.c.p structure, the oxide ions (O2-) form a cubic lattice in which each corner of the cube is occupied by an oxide ion and the face-centered positions are also occupied by oxide ions. This arrangement results in a unit cell with a total of 4 oxide ions.

Occupancy of Voids:
In a c.c.p structure, there are two types of voids - tetrahedral voids and octahedral voids.

- Tetrahedral Voids:
One-fourth of the tetrahedral voids are occupied by divalent metal A. This means that for every 4 oxide ions, there is 1 metal A ion occupying a tetrahedral void. The charge on the metal A ion is 2+.

- Octahedral Voids:
The octahedral voids are occupied by a monovalent metal B. Since the charge on the metal B ion is 1+, it can occupy an octahedral void without disrupting the charge balance of the crystal structure.

Formula of the Oxide:
Based on the occupancy of the voids, we can determine the formula of the oxide.

- Metal A occupies 1/4th of the tetrahedral voids. So, the ratio of metal A ions to oxide ions is 1:4.

- Metal B occupies the octahedral voids. So, the ratio of metal B ions to oxide ions is 1:2.

Therefore, the formula of the oxide is AB2O2, which matches with option 'D'.

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