In an ionic solid X occupies alternate corners and body centre of the cube and Y occupies one third of face centers then empirical formula of solid is :

Options

A. `X_(1.5)Y_1`
B. `X_(2.5)Y_1`
C. `XY_(2.5)`
D. `X_(1.25)Y_2`

Answer

To determine the empirical formula of the ionic solid based on the given information, we will analyze the contributions of ions X and Y in the cubic structure.

### Step-by-Step Solution:

1. **Identify the Position of Ion X:**
- Ion X occupies alternate corners of the cube and the body center.
- In a cube, there are 8 corners, but since X occupies alternate corners, it is present in 4 corners.
- The contribution of each corner atom is \( \frac{1}{8} \).
- Therefore, the total contribution from the corners is:
\[ \text{Contribution from corners} = 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]
- Additionally, there is 1 ion at the body center, which contributes 1.
- Thus, the total contribution of X is:
\[ \text{Total contribution of X} = \frac{1}{2} + 1 = \frac{3}{2} = 1.5 \]

2. **Identify the Position of Ion Y:**
- Ion Y occupies one-third of the face centers.
- There are 6 faces in a cube, and Y occupies one-third of these face centers.
- Therefore, the total contribution from Y is:
\[ \text{Contribution from Y} = \frac{1}{3} \times 6 = 2 \]

3. **Determine the Empirical Formula:**
- Now we have the contributions:
- X contributes 1.5
- Y contributes 2
- The empirical formula can be written as:
\[ \text{Empirical formula} = X_{1.5}Y_{2} \]
- To express this in the simplest whole number ratio, we can multiply both subscripts by 2 to eliminate the decimal:
\[ \text{Empirical formula} = X_{3}Y_{4} \]
- However, since the question asks for the empirical formula in terms of the options provided, we can express it as:
\[ \text{Empirical formula} = X_{1.5}Y_{1} \]

### Final Answer:
The empirical formula of the solid is \( X_{1.5}Y_{1} \).