Read the passage given below and answer the following question
In the simplest cases of monatomic metals such as Cu and `alpha-`Fe, mentioned above, the arrangement of metal atoms in the structure is simply the same as the arrangement of lattice points. In more complex structures such as NaCl. The lattice point represents an ion pair. This is still a very simple example, however, and in most inorganic structures the lattice point represents a considerable number of atoms In crystals of organic molecules such as proteins, the lattice point represents an entire protein molecule. Obviously, the lattice point gives no information whatsoever as to the atoms and their arrangements Which it represents, what the lattice does show is how these species are packed together in 3D. The combination of crystal system and lattice type gives the Bravais lattice of a structure. There are 14 possible Bravais lattices.
Source: West, Anthony R. 'Crystal Structures and Crystal Chemistry'. In Solid State Chemistry. 1-81, Wiley, 2014.
A mixed oxide has ccp arrangement in which the cations 'X' occupy `1//3`rd of octahedral voids and the cations 'Y' occupy `1//3`rd of tetrahedral voids. The formula of oxide is:

Options

A. `X_2Y_3O_2`
B. `XY_3O`
C. `X_2YO_3`
D. `XY_2O_3`

Answer

To find the formula of the mixed oxide based on the given information, we will follow these steps:

### Step 1: Determine the effective number of atoms in the CCP structure.
In a cubic close-packed (CCP) structure, there are 6 atoms at the corners and 1 atom in the body center. The contribution from the corner atoms is \( \frac{1}{8} \times 8 = 1 \) and from the body center is 1. Therefore, the total effective number of atoms in the CCP structure is:

\[ \text{Total effective atoms} = 6 \times \frac{1}{8} + 1 = 4 \]

### Step 2: Calculate the number of octahedral voids.
In a CCP structure, the number of octahedral voids is equal to the number of atoms, which is 4. Since cations 'X' occupy one-third of the octahedral voids, we can calculate the number of 'X' cations:

\[ \text{Number of } X = \frac{1}{3} \times 4 = \frac{4}{3} \]

### Step 3: Calculate the number of tetrahedral voids.
In a CCP structure, the number of tetrahedral voids is twice the number of atoms, which is \( 2 \times 4 = 8 \). Since cations 'Y' occupy one-third of the tetrahedral voids, we can calculate the number of 'Y' cations:

\[ \text{Number of } Y = \frac{1}{3} \times 8 = \frac{8}{3} \]

### Step 4: Determine the number of oxide ions.
In a CCP structure, the number of oxide ions (O) is equal to the effective number of atoms, which is 4. Therefore, the number of oxide ions is:

\[ \text{Number of } O^{2-} = 4 \]

### Step 5: Write the ratio of cations and anions.
Now we have:
- \( X = \frac{4}{3} \)
- \( Y = \frac{8}{3} \)
- \( O^{2-} = 4 \)

To find the simplest whole number ratio, we can express this as:

\[ \text{Ratio} = X : Y : O^{2-} = \frac{4}{3} : \frac{8}{3} : 4 \]

### Step 6: Simplify the ratio.
To simplify the ratio, we can multiply each term by 3 to eliminate the fractions:

\[ \text{Ratio} = 4 : 8 : 12 \]

Now, dividing each term by 4 gives:

\[ 1 : 2 : 3 \]

### Step 7: Write the formula of the oxide.
From the simplified ratio, we can write the formula of the oxide as:

\[ \text{Formula} = X Y_2 O_3 \]

Thus, the formula of the mixed oxide is:

\[ \text{Final Answer: } XY_2O_3 \]