A cubic solid is made up of two elements $X$ and $Y$ Atoms of $X$ are present on every alternate corner and one at the enter of cube $Y$ is at $\frac{1}{3} td$ of the total faces The empirical formula of the compound is
To determine the empirical formula of the compound made of elements \( X \) and \( Y \) in the given cubic solid, let's analyze the distribution of atoms as mentioned in the question:
\(\text{Total corner contribution of } X = 4 \times \frac{1}{8} = \frac{1}{2} \text{ atom/unit cell}\)
\(\text{Center contribution of } X = 1 \text{ atom/unit cell}\)
\(\text{Total } X = \frac{1}{2} + 1 = \frac{3}{2} = 1.5 \text{ atoms/unit cell}\)
\(\text{Number of } Y \text{ atoms on faces} = \frac{1}{3} \times 6 = 2 \text{ faces}\)
\(\text{Total face contribution of } Y = 2 \times \frac{1}{2} = 1 \text{ atom/unit cell}\)
Therefore, the ratio of \( X \) to \( Y \) is \( \frac{1.5}{1} : 1 \). To simplify, multiply both by 2:
\(X_{2.5} Y\)
The correct empirical formula of the compound, therefore, is \( X_{2.5} Y \). Hence, the correct answer is:
\(X_{2.5} Y\)