# 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?

$$\ce{X2YZ24}$$

### 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?

• Your solutions looks fine. May be the answer given is wrong, Mar 22 '15 at 8:54
• 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. Mar 22 '15 at 10:45
• 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. Mar 22 '15 at 13:01

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.

• 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. Mar 27 '20 at 12:11