Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. Th

Q: Why do we take alternate tetrahedral voids?

The question specifically states 'alternate tetrahedral voids'. In a standard FCC unit cell, there are 8 tetrahedral voids. Occupying alternate ones means occupying exactly half of them, which is 4.

Q: Why do the corner atom and tetrahedral void atom touch?

Since both positions are occupied by the same atom X, they will pack as closely as possible. The shortest distance between any two sites in this specific arrangement is between a corner and its nearest tetrahedral void.

Q: What is the distance from the corner to the tetrahedral void?

The tetrahedral voids are located at one-fourth the distance along the body diagonal from each corner. Since the body diagonal is $\sqrt{3}a$, the distance is $\frac{\sqrt{3}a}{4}$.

Q: Why is the packing efficiency so low (34%) compared to regular FCC (74%)?

In this structure, the atoms in the tetrahedral voids push the corner and face-centered atoms apart. They no longer touch along the face diagonal. The new touching condition along the body diagonal significantly increases the unit cell volume relative to the atom size, dropping the efficiency.

Q: Is this structure related to any real-world material?

Yes! This is exactly the crystal structure of Diamond, where carbon atoms occupy FCC sites and alternate tetrahedral voids. Silicon and Germanium also crystallize in this structure.

Question

Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

Options:

(A) 25

(B) 35

(C) 55

(D) 75

Sigma Percentile · Accepted Answer

### Visualizing the Lattice

Imagine you are shrinking down to the atomic scale and stepping inside a crystal lattice. The problem tells us that Atom X occupies the Face-Centered Cubic (FCC) lattice sites. This means there is an atom at every corner of the cube and one at the center of every face.

But that's not all! Atom X also occupies the **alternate tetrahedral voids**. In a standard FCC unit cell, there are 8 tetrahedral voids located along the body diagonals (one near each corner). Occupying *alternate* voids means exactly half of them are filled.

Let's count the total number of atoms, $Z$, in this unit cell. The FCC lattice points contribute $8 	imes \frac{1}{8} + 6 	imes \frac{1}{2} = 4$ atoms. The alternate tetrahedral voids contribute another $4$ atoms entirely within the cell. Therefore, the total effective number of atoms is $Z = 4 + 4 = 8$.

### The Touching Condition

To find the packing efficiency, we need to relate the edge length of the unit cell, $a$, to the radius of the atom, $r_X$. In a regular FCC lattice, atoms touch along the face diagonal. However, because we have stuffed extra atoms into the tetrahedral voids, the lattice expands. The new closest contact point is between a corner atom and the atom in the nearest occupied tetrahedral void.

The distance from a corner to a tetrahedral void is exactly one-fourth of the body diagonal. Since the body diagonal has a length of $\sqrt{3}a$, this distance is $\frac{\sqrt{3}a}{4}$.

Because both positions are occupied by the identical Atom X, they must touch each other. This gives us our master equation:

$$2r_X = \frac{\sqrt{3}a}{4}$$

Rearranging this to solve for the edge length $a$, we get:

$$a = \frac{8r_X}{\sqrt{3}}$$

### Calculating Packing Efficiency

Now, let's bring it all together. Packing efficiency $\eta$ is the percentage of the total unit cell volume that is actually occupied by the atoms. The formula is:

$$\eta = \frac{Z \cdot V_{	ext{atom}}}{V_{	ext{cell}}} 	imes 100$$

Substitute our known values: $Z = 8$, $V_{	ext{atom}} = \frac{4}{3}\pi r_X^3$, and $V_{	ext{cell}} = a^3$.

$$\eta = \frac{8 	imes \frac{4}{3}\pi r_X^3}{\left(\frac{8r_X}{\sqrt{3}}ight)^3} 	imes 100$$

Let's carefully expand the denominator. Cubing $\frac{8}{\sqrt{3}}$ gives $\frac{512}{3\sqrt{3}}$. Notice how beautifully the $r_X^3$ terms cancel out, leaving us with pure numbers:

$$\eta = \frac{\frac{32}{3}\pi}{\frac{512}{3\sqrt{3}}} 	imes 100 = \frac{32\pi \cdot 3\sqrt{3}}{3 \cdot 512} 	imes 100$$

Simplifying the fraction, we get:

$$\eta = \frac{\sqrt{3}\pi}{16} 	imes 100$$

Plugging in the values $\sqrt{3} \approx 1.732$ and $\pi \approx 3.14$, we find that $\eta \approx 34\%$. Looking at our options, the closest value is **35%**.

### The Diamond Connection

Here is a fascinating piece of trivia: the structure we just analyzed is mathematically identical to the **Diamond cubic crystal structure**! In diamond, carbon atoms occupy the exact same positions (FCC + alternate tetrahedral voids). Because the atoms are forced to sit relatively far apart to maintain their tetrahedral bonding geometry, the packing efficiency drops drastically from the standard FCC's $74\%$ down to a remarkably open $34\%$.

Animated Solution for Chemistry - States of Matter: Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

Select Answer:

Visualized Solution

$Z_{eff}$

$Z = Z_{FCC} + Z_{TV}$

$a \leftrightarrow r_{X}$

$a = \frac{8 r _{X}}{3}$

\eta = \frac{Z \cdot V_{\text{atom}}}{V_{\text{cell}}} \times 100

\eta = \frac{8 \times \frac{4}{3}\pi r_X^3}{\left(\frac{8r_X}{\sqrt{3}}\right)^3} \times 100

\eta \approx 34\%

$Z = 8$

The Sigma Insight: Solid State

Visualizing the Lattice

The Touching Condition

Calculating Packing Efficiency

The Diamond Connection

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