Animated Solution for Chemistry - States of Matter: The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is (are)

Welcome to a fascinating exploration of the solid state! Today, we are diving deep into the cubic close packed (ccp) structure.

Before we analyze the specific statements, it is crucial to establish a fundamental geometric truth: the ccp arrangement is geometrically identical to the face-centered cubic (fcc) unit cell. By visualizing an fcc unit cell, we can unlock all the secrets of the ccp structure. Imagine a cube where atoms are present not only at the eight corners but also at the center of each of the six faces. This mental model will be our guiding light.

Let's evaluate the first statement regarding the coordination number. In a standard, bulk ccp lattice, every single atom is surrounded by exactly twelve nearest neighbors. This is a well-known fact.

However, the question sets a clever trap by specifying an atom in the topmost layer. We must think physically about what this means. An atom in the bulk has 6 neighbors in its own layer, 3 in the layer directly above it, and 3 in the layer directly below it ($6+3+3=12$).

For an atom sitting on the absolute surface—the topmost layer—the layer above it simply does not exist! Therefore, we must subtract those 3 missing neighbors. Its coordination number becomes $6+3=9$. Because 9 is not equal to 12, statement (A) is incorrect. Always pay attention to boundary conditions!

Moving on to the second statement, we need to determine the packing efficiency. The packing efficiency tells us what fraction of the total space is actually occupied by atoms.

First, we determine the effective number of atoms ($Z$) in an fcc unit cell. With 8 corner atoms (each contributing $\frac{1}{8}$) and 6 face-centered atoms (each contributing $\frac{1}{2}$), we have $Z=8\times \frac{1}{8}+6\times \frac{1}{2}=4$ effective atoms.

The total volume occupied by these spherical atoms is $4\times \frac{4}{3}\pi r^3$. The volume of the cubic unit cell is $a^3$. As we will prove shortly, the edge length $a$ is related to the radius $r$ by $a=2\sqrt{2}r$.

Substituting this into our efficiency formula:

When we calculate this, we get exactly $74\%$. This represents the maximum possible packing efficiency for identical spheres in three dimensions. Statement (B) is absolutely correct.

Next, let's investigate the empty spaces, or voids, within the lattice. In any close-packed structure, there is a strict mathematical relationship between the number of atoms and the number of voids.

The number of octahedral voids is exactly equal to the effective number of atoms ($Z$). Since $Z=4$ for our fcc unit cell, there are 4 octahedral voids.

The number of tetrahedral voids is exactly double the effective number of atoms ($2Z$). Therefore, there are $2\times 4=8$ tetrahedral voids.

The statement asks for the number of voids per atom. We simply divide the total voids by the total atoms:

This perfectly matches statement (C), making it correct.

Finally, let's verify the relationship between the unit cell edge length ($a$) and the atomic radius ($r$).

In an fcc unit cell, the atoms do not touch along the edges of the cube. Instead, they touch along the face diagonal. If we look at any square face of the cube, the diagonal passes through one corner atom, the face-centered atom, and the opposite corner atom.

The length of this face diagonal, by the Pythagorean theorem, is $a\sqrt{2}$. Because the atoms are touching, this length is also equal to the sum of their radii: one radius from the first corner, two radii (the full diameter) from the face-centered atom, and one radius from the second corner.

This gives us the master equation:

Solving for the edge length $a$, we get:

This confirms that statement (D) is correct.

By systematically breaking down the geometry of the cubic close packed structure, we have successfully navigated the traps and verified the truths. The topmost layer has a reduced coordination number of 9, the packing efficiency is a maximized 74%, there are 1 octahedral and 2 tetrahedral voids per atom, and the edge length is indeed $2\sqrt{2}r$.

Therefore, the correct statements are (B), (C), and (D).