Question

in certain solid the oxide ions are arranged in CCP. A occupy 1/6 of the tetrahedral voids and cation B occupy 1/3 of the octahedral void .
The probable formula of the compound is

Answer 1

Oxides- 8×1/8 + 6×1/2 = 4

A- 8× 1/6 = 8/6= 4/3

B- 4×1/3= 4/3

therefore,

O4A4/3B4/3 = O12A4B4

Formula- O3AB

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Answer 2

Answer:

The probable formula of the compound is $ABO_3$.

Explanation:

Let the number of atoms in CCP arrangement be n

Number of tetrahedral voids in CCP = 2n

Number of octahedral voids in CCP = n

number of oxide ion = n

If 1/6 of the tetrahedral voids are occupied by atom A.

Number A atom = $\frac{1}{6}\times 2n=\frac{1}{3}n$

If 1/3 of the octahedral voids are occupied by atom B.

Number B atom = $\frac{1}{3}\times n=\frac{1}{3}n$

Ratio of oxide ion to  atom A to atom B;$n:\frac{1}{3}n:\frac{1}{3}n=3:1:1$

The probable formula of the compound is $ABO_3$.