Animated Solution for Chemistry - States of Matter: Lithium forms body-centred cubic structure. The length of the side of its unit cell is $351\text{ pm}$. Atomic radius of the lithium will be

Imagine you are shrinking down to the atomic level and stepping inside a crystal of Lithium. What do you see? Lithium crystallizes in a Body-Centred Cubic (BCC) structure. This means if you look at a single unit cell—the fundamental building block of the crystal—you will find an atom at every single one of the $8$ corners of the cube, and one solitary atom sitting right in the dead center of the body.

Now, here is the crucial physical reality: these atoms are not just floating in space; they are packed together. In a BCC structure, the atoms at the corners do not touch each other along the edges of the cube. The atom in the center pushes them apart! Instead, the atoms touch each other diagonally right through the center of the cube. This line is called the body diagonal.

Let's translate this physical picture into mathematics. If the edge length of our cubic unit cell is $a$, we can use the 3D Pythagorean theorem to find the length of the body diagonal. The face diagonal is $\sqrt{2}a$, and combining that with the vertical edge $a$, the body diagonal becomes $\sqrt{3}a$.

Because the atoms touch along this exact line, we can also express this length in terms of the atomic radius, $r$. Starting from one corner, we travel through one radius $r$ of the corner atom, then completely through the center atom which contributes its full diameter $2r$, and finally through the radius $r$ of the opposite corner atom.

Adding these up, the total length is $r+2r+r=4r$. Equating our two expressions for the body diagonal gives us our master equation:

We are given that the edge length $a$ is $351\text{ pm}$. Our goal is to find the atomic radius $r$. Let's rearrange our master equation to isolate $r$:

Now, we substitute the known value of $a$ into the equation. We know that the square root of $3$ is approximately $1.732$. Let's plug that in:

Multiplying the numerator gives us approximately $607.932$. Now, we just need to divide by $4$:

Rounding off our result to the nearest integer, we get $152\text{ pm}$.

This elegant geometric relationship perfectly predicts the atomic radius of Lithium based purely on the macroscopic measurement of its unit cell edge length. The correct option is indeed (d).