Animated Solution for Chemistry - States of Matter: For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one face are shown for clarity) , the packing fraction is approximately (packing fraction = $\frac{\text{Packing efficiency}}{100}$)

When tackling solid state problems, the visual representation is everything. At first glance, you might assume the large Y anions form a standard Face-Centered Cubic (FCC) lattice. However, look closely at the provided image. The Y atoms are located at the corners of the unit cell, and crucially, they are touching each other along the edges.

This is the defining characteristic of a Simple Cubic lattice for the Y atoms. Because they touch along the edge, the relationship between the edge length $a$ and the radius of the anion $r_{-}$ is simply:

Now, let's locate the cation X. The image shows it sitting perfectly in the center of the face, nestled between the four corner Y atoms. This specific 2D arrangement creates a square planar void.

To find the size of this void, we analyze the face diagonal. The X atom touches the Y atoms along this diagonal. The length of the face diagonal is $a\sqrt{2}$. Therefore, the sum of the radii along the diagonal is:

Substituting $a=2r_{-}$, we get:

Solving for the radius of the cation $r_{+}$, we find the classic radius ratio for a square planar void:

Before we can calculate the packing fraction, we need to know exactly how many atoms of each type are effectively inside one unit cell.

For the simple cubic lattice of Y, there are 8 corner atoms, and each is shared by 8 adjacent unit cells:

For the X cations, they are located at the face centers. A cube has 6 faces, and an atom at a face center is shared between 2 unit cells:

So, the effective formula of our unit cell is $X{Y}_3$.

The packing fraction (P.F.) is the ratio of the volume occupied by the atoms to the total volume of the unit cell. Let's set up the master equation:

Substitute our known values ($Z_Y=1$, $Z_X=3$, and $a=2r_{-}$):

Now, we factor out the common terms and substitute the radius ratio $r_{+}=0.414r_{-}$:

Notice how elegantly the $r_{-}^3$ terms cancel out, leaving us with a purely numerical expression:

Calculating the final value:

The packing fraction is approximately 0.63, making option (B) the correct choice. This problem beautifully demonstrates why you must never assume a lattice type without carefully analyzing the contact points in the given diagram!