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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

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

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?

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    $\begingroup$ Your solutions looks fine. May be the answer given is wrong, $\endgroup$ Mar 22 '15 at 8:54
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    $\begingroup$ You need to take into account which atoms are shared with other unit cells. A face centre is shared with one, an edge centre with 4 other cells etc. $\endgroup$
    – matt_black
    Mar 22 '15 at 10:45
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    $\begingroup$ The answer given is wrong and the question is poorly worded; a periodic cubic lattice requires that all the corners in a unit cell be of the same atom. If one of the corners of a cell is made of a different atom, then you're not actually looking at the unit cell. $\endgroup$ Mar 22 '15 at 13:01
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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.

cubic exam question

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

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  • $\begingroup$ There is a way to interpret the unit cell that matches the question. Assume $X$ and $Y$ occur randomly at all corners with on average one corner of a cube having $Y$. Contrived, yes, but workable for cases such as $\ce{Nb(C,N)}$ which appears as a precipitate in the hot rolling of some steels. $\endgroup$ Mar 27 '20 at 12:11

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