Question

Fullerene (C₆₀) crystallizes in an FCC unit cell (edge length = 14.14 Å) with one C₆₀ centered at each lattice point. The smallest distance (in Å) between the centers of two C₆₀ molecules is ..........
(Round off to two decimal places)

Hint:

In an FCC unit cell, the smallest distance between centers of molecules is the face diagonal, calculated as \( \sqrt{2}a \).

Answer 1

Correct Answer: 10.1 - 9.9

The FCC (face-centered cubic) unit cell is a type of crystal structure where each corner and face center of the cube is occupied by atoms or molecules. In an FCC lattice, the simplest way to determine the distance between the centers of molecules, such as fullerenes (C₆₀), adhering to the crystal lattice rules is to use the face diagonal of the cube. Each face diagonal passes through the centers of two opposite face-centered spheres and contains two radii of the spheres.

Given edge length a is 14.14 Å.

In FCC, the face diagonal relates to the edge length as follows:

Face diagonal = √2 * a = √2 * 14.14 Å.

The smallest distance between the centers of two C₆₀ molecules is along the face diagonal and equals half of it because it stretches from the center of one face-centered atom to the other:

Distance = (√2 * 14.14) / 2 = (1.414 * 14.14) / 2 = 9.99358 Å.

Rounding off, we get the smallest distance as 9.99 Å.

The smallest distance between the centers of two C₆₀ molecules in an FCC unit cell is therefore 9.99 Å.