Animated Solution for Chemistry - States of Matter: Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?

The problem of calculating packing efficiency often feels like a standard plug-and-chug exercise, but this specific question from JEE Main 2019 introduces a beautiful twist. We are not just dealing with a standard lattice; we are dealing with a modified, expanded lattice where a massive atom has been introduced into the body center. Let's break down the geometry and see how this affects the packing fraction.

Imagine you are looking at two distinct crystal structures. Solid 1 is your textbook Body-Centered Cubic (BCC) lattice. It is composed entirely of atoms of type A. There are A atoms at all eight corners and one A atom sitting perfectly in the body center.

Solid 2, on the other hand, is a bit of a hybrid. It still has the A atoms at the eight corners, but the body center is now occupied by a different atom, type B. The catch? Atom B is huge. Its radius is exactly twice that of atom A ($r_B=2r_A$). Because of this massive atom in the center, the entire unit cell has to expand. The problem tells us that the edge length of Solid 2 is 50% larger than that of Solid 1.

To find the packing efficiency of Solid 2, we need its volume, which means we need its edge length ($a_2$). But we only know $a_2$ in terms of $a_1$. So, our first mission is to find $a_1$.

Let's look closely at Solid 1. In a standard BCC lattice, the atoms touch each other along the body diagonal. The length of this body diagonal is $\sqrt{3}{a}_1$. Since it passes through two corner atoms (radius $r_A$) and one central atom (diameter $2r_A$), the total length is $4r_A$.

Equating these gives us our first master relationship:

Now, let's transition to Solid 2. We are given that its edge length ($a_2$) is 50% more than $a_1$. Mathematically, this means:

Substituting our expression for $a_1$ into this equation, we get $a_2$ purely in terms of $r_A$:

Packing efficiency is simply the ratio of the volume actually occupied by the atoms to the total volume of the unit cell, expressed as a percentage.

Let's figure out the occupied volume in Solid 2. The unit cell contains atoms of type A at the 8 corners. Since each corner atom is shared by 8 adjacent cells, their effective contribution is $8\times \frac{1}{8}=1$ atom of A. The body center contains exactly 1 full atom of B.

The total volume occupied is the sum of their individual volumes:

We know that $r_B=2r_A$. Let's substitute this in. Notice how the volume scales with the cube of the radius!

Next, we need the total volume of the unit cell for Solid 2. This is simply the cube of its edge length:

We have all the pieces of the puzzle. Let's plug them into the packing efficiency formula:

The $r_A^3$ terms beautifully cancel out, leaving us with a pure geometric constant:

Using the approximations $\pi \approx 3.14$ and $\sqrt{3}\approx 1.732$:

Rounding to the nearest given option, we get 90%.

This is a fascinating result! A standard BCC lattice has a packing efficiency of only 68%. By introducing a massive atom into the interstitial void and allowing the lattice to expand slightly, the packing efficiency shot up to an incredible 90%. It perfectly illustrates how doping and interstitial defects can drastically alter the physical properties of a crystal.