Question

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

• A. $X _2 Y _{1.5}$
• B. $X _{2.5} Y$
• C. $XY _2, 5$
• D. $X _{1.5} Y _2$

Answer 1

The Correct Option is B

1. Atoms of X: - X is present on alternate corners of the cube, which means it occupies 4 out of the 8 corners. - Each corner contributes $\frac{1}{8}$ to the unit cell. \[ \text{Contribution from corners} = 4 \times \frac{1}{8} = \frac{1}{2}. \] - Additionally, there is one X atom at the center of the cube, contributing 1 atom. \[ \text{Total contribution from X} = \frac{1}{2} + 1 = 1.5 \text{ atoms}. \] 2. Atoms of Y: - Y atoms are present at $\frac{1}{4}$ of the 6 faces of the cube. - Each face contributes $\frac{1}{2}$ atom to the unit cell. \[ \text{Contribution from faces} = 6 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{2} \text{ atoms}. \] 3. Empirical Formula: - The ratio of atoms of X to Y is: \[ \text{X : Y} = 1.5 : 1.5 = 1 : 1. \] - However, the empirical formula accounts for their fractional contributions, giving: \[ \text{Empirical formula} = X_{2.5}Y. \] 1. Atoms at corners contribute $\frac{1}{8}$ to the unit cell, while atoms at the center contribute fully.
2. Atoms at faces contribute $\frac{1}{2}$ of their count to the unit cell.
3. By calculating individual contributions of X and Y, the correct empirical formula is derived as X$_{2.5}$Y.