in a cubic packed structure of mixed oxides,the lattice is made up of oxide ions,one fifth of tetrahedral voids are occupied by divalent ions,while one half of the octahederal voids are occupied by trivalent ions then the formula of the oxide is

To determine the formula of the mixed oxide based on the given information about the cubic packed structure, we need to analyze the arrangement of ions and the occupancy of the voids. Let's break this down step by step.

Understanding the Structure

In a cubic close-packed (CCP) structure, the oxide ions (O^2-) form the lattice. In this arrangement, there are two types of voids: tetrahedral and octahedral. The tetrahedral voids are smaller and can accommodate cations, while the octahedral voids are larger and can also hold cations.

Counting the Ions

In a unit cell of a cubic close-packed structure, there are:

• 4 oxide ions (O^2-) per unit cell.
• 8 tetrahedral voids.
• 4 octahedral voids.

Occupancy of the Voids

According to the problem:

• One fifth of the tetrahedral voids are occupied by divalent ions (let's denote them as M²⁺).
• One half of the octahedral voids are occupied by trivalent ions (denote these as N³⁺).

Calculating the Number of Ions

Now, let's calculate how many divalent and trivalent ions are present:

• For tetrahedral voids: One fifth of 8 voids = 8/5 = 1.6. Since we cannot have a fraction of an ion, we interpret this as 1 divalent ion (M²⁺) effectively contributing to the formula.
• For octahedral voids: One half of 4 voids = 4/2 = 2. Thus, there are 2 trivalent ions (N³⁺).

Formulating the Compound

Now, we can summarize the contributions of each type of ion:

• 4 oxide ions (O^2-)
• 1 divalent ion (M²⁺)
• 2 trivalent ions (N³⁺)

Balancing the Charges

Next, we need to ensure that the overall charge of the compound is neutral. The charges contributed by the ions are:

• From oxide ions: 4 × (-2) = -8
• From divalent ions: 1 × (+2) = +2
• From trivalent ions: 2 × (+3) = +6

Now, summing these charges gives:

-8 + 2 + 6 = 0

This indicates that the charges balance out, confirming the formula is correct.

Final Formula

Putting it all together, the formula of the mixed oxide can be expressed as:

MO₄N₂

Where M represents the divalent cation and N represents the trivalent cation. This formula reflects the stoichiometry of the ions in the cubic packed structure of the mixed oxide.