A cubic solid is made up of two elements X an | vedclass

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(D) Total corners in a cube are $8$. Atoms of $X$ are at alternate corners,so number of $X$ atoms at corners = $8 	imes \frac{1}{2} 	imes \frac{1}{8

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$X_{1.5} Y_2$

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(D) Total corners in a cube are $8$. Atoms of $X$ are at alternate corners,so number of $X$ atoms at corners = $8 	imes \frac{1}{2} 	imes \frac{1}{8} = 0.5$. One $X$ atom is at the center of the cube,so total $X = 0.5 + 1 = 1.5$. Total faces in a cube are $6$. Atoms of $Y$ are at $\frac{1}{3}$ of the total faces,so number of $Y$ atoms = $6 	imes \frac{1}{3} 	imes \frac{1}{2} = 1$. Wait,let us re-evaluate: $Y$ is at $\frac{1}{3}$ of the total faces,meaning $6 	imes \frac{1}{3} = 2$ faces. Each face contributes $\frac{1}{2}$ to the unit cell. So,$Y = 2 	imes \frac{1}{2} = 1$. The ratio $X:Y = 1.5:1 = 3:2$. The formula is $X_3 Y_2$.

$A$ cubic solid is made up of two elements $X$ and $Y$. Atoms of $X$ are present on every alternate corner and one at the center of the cube. $Y$ is at $\frac{1}{3}$ of the total faces. The empirical formula of the compound is

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