Chemical formula of a compound crystallized in a cubic lattice Question


  In a cubic lattice of $\ce{XYZ},$ $\ce{X}$ atoms are present at all corners except one corner which is occupied by $\ce{Y}$ atoms. $\ce{Z}$ atoms are present at face centres. What is the formula of compound?


Answer


   $\ce{X2YZ24}$


My solution

Let the number of atoms in a unit cell be $N.$ Then

$$
\begin{align}
N(\ce{X}) &amp;= 7 × \frac{1}{8} = \frac{7}{8}\\
N(\ce{Y}) &amp;= 1 × \frac{1}{8} = \frac{1}{8}\\
N(\ce{Z}) &amp;= 6 × \frac{1}{2} = 3
\end{align}
$$

Therefore, the ratio would be

$$N(\ce{X}):N(\ce{Y}):N(\ce{Z}) = \frac{7}{8}:\frac{1}{8}:3,$$

which is the same thing as $7:1:24.$ So, the formula is $\ce{X7YZ24}.$

What am I doing wrong?
I would like to say your answer is right, but the question is wrong. 

Remember that a lattice is "infinitely" repeated units in 3 dimensional space so you should be able to expand the unit cell in x, y and z axes, indefinitely.

Now imagine your cubic lattice with an $\ce{Y}$ atom in just 1 (out of 8) corners. 

Expanding the unit cell in the x-axis will necessarily duplicate the $\ce{Y}$ atom in the x-axis because all expanded unit cells must be identical in composition. Similarly, expanding the unit cell in the y-axis will duplicate the $\ce{Y}$ atom in the y-axis, and expanding the unit cell in the z-axis will duplicate the $\ce{Y}$ atom in the z-axis. 

In the end you will find the having one corner as $\ce{Y}$ atom will necessarily have ALL corners as $\ce{Y}$ atoms. In other words, $\ce{X}$ IS $\ce{Y}$, and the formula is necessarily $\ce{XZ_3}$ or $\ce{YZ_3}$

Unfortunately, I don'k think the designer of the question saw his fatal contradiction because this very question is seen in the (mock) Joint Entrance Examination (JEE) in India.

  

And $\ce{X_2YZ_24}$ is not in the options even in the wrong question.