In an ionic solid, `B^(x-)` ions constitute a ccp lattice, if `A^(y+)` ions occupy 25%, of the tetrahedral voids, the possible ions in the solid are

Options

A. `A^(+)` and `B^(-)`
B. `A^(2+)` and `B^(4-)`
C. `A^(2+)` and `B^(-)`
D. `A^(4+)` and `B^(-)`

Answer

To solve the problem, we need to analyze the information given about the ionic solid composed of ions \( B^{x-} \) and \( A^{y+} \).

### Step-by-Step Solution:

1. **Identify the Structure**:
- The \( B^{x-} \) ions form a cubic close-packed (ccp) lattice. In a ccp lattice (also known as face-centered cubic, FCC), there are 4 atoms per unit cell (denoted as \( Z = 4 \)).

2. **Calculate the Number of Tetrahedral Voids**:
- In a ccp lattice, the number of tetrahedral voids is twice the number of atoms in the lattice. Therefore, if \( Z = 4 \), the number of tetrahedral voids \( = 2Z = 2 \times 4 = 8 \).

3. **Determine the Number of \( A^{y+} \) Ions**:
- According to the problem, \( A^{y+} \) ions occupy 25% of the tetrahedral voids. Since there are 8 tetrahedral voids, the number of \( A^{y+} \) ions occupying these voids is:
\[ \text{Number of } A^{y+} = 0.25 \times 8 = 2 \]

4. **Establish the Ratio of Ions**:
- From the previous steps, we have:
- Number of \( A^{y+} \) ions = 2
- Number of \( B^{x-} \) ions = 4 (from the ccp lattice)
- The simplest ratio of \( A \) to \( B \) is:
\[ A : B = 2 : 4 = 1 : 2 \]

5. **Determine the Charges**:
- Let the charge of \( B \) be \( -x \). Therefore, the total negative charge contributed by \( B^{x-} \) is:
\[ 4 \times (-x) = -4x \]
- Let the charge of \( A \) be \( +y \). Therefore, the total positive charge contributed by \( A^{y+} \) is:
\[ 2 \times (+y) = +2y \]
- For the compound to be electrically neutral, the total positive charge must equal the total negative charge:
\[ 2y = 4x \]
- Simplifying gives:
\[ y = 2x \]

6. **Possible Values for \( x \) and \( y \)**:
- The simplest integer values that satisfy \( y = 2x \) can be \( x = 1 \) and \( y = 2 \). Thus, the possible ions in the solid are:
- \( A^{2+} \) and \( B^{-} \).

### Conclusion:
The possible ions in the solid are \( A^{2+} \) and \( B^{-} \), leading to the formula \( A_2B_4 \) or simplified to \( AB_2 \).