In a crystalline solid, anions B arranged in ccp lattice and cations A occupy 50% of the octahedral voids. And 50% of the tetrahedral voids. What is the formula of the solid?

Options

A. AB
B. `A_(3)A_(2)`
C. `A_(2)B_(2)`
D. `A_(2)B_(3)`

Answer

To determine the formula of the crystalline solid described in the question, we will follow these steps:

### Step 1: Identify the arrangement of anions
The question states that anions B are arranged in a CCP (cubic close-packed) lattice. This is equivalent to a face-centered cubic (FCC) structure.

### Step 2: Calculate the number of anions (B)
In a face-centered cubic (FCC) lattice:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculating the total number of anions B:
\[ \text{Total B} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, there are 4 anions B in the unit cell.

### Step 3: Determine the number of octahedral and tetrahedral voids
In a face-centered cubic (FCC) lattice:
- The number of octahedral voids is equal to the number of atoms, which is 4.
- The number of tetrahedral voids is double the number of octahedral voids, which is \( 2 \times 4 = 8 \).

### Step 4: Calculate the number of cations (A) in the voids
According to the question:
- Cations A occupy 50% of the octahedral voids.
- Cations A occupy 50% of the tetrahedral voids.

Calculating the number of cations A:
- From octahedral voids:
\[ \text{Cations from octahedral voids} = 50\% \times 4 = 2 \]
- From tetrahedral voids:
\[ \text{Cations from tetrahedral voids} = 50\% \times 8 = 4 \]

### Step 5: Total number of cations (A)
Adding the contributions from both types of voids:
\[ \text{Total A} = 2 + 4 = 6 \]

### Step 6: Write the empirical formula
Now, we can write the formula of the solid based on the number of cations and anions:
\[ \text{Formula} = A_6B_4 \]

### Step 7: Simplify the formula
To simplify the formula, divide both subscripts by 2:
\[ \text{Simplified Formula} = A_3B_2 \]

### Final Answer
The formula of the solid is \( A_3B_2 \).

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