The edge length of a face centred cubic cell of an ionic substance is 508pm. If the radius of the cation is 110pm, the radius of the anion is

144 pm

In face centred cubic (FCC) structure, 8 atoms are at the 8 corners of the cubic structure and 6 atoms are at the 6 faces of the structure. The cross-section of the FCC use is as shown below,

Each face is tightly packed with corner atoms. Atoms have a radius equal to ‘r’.If seen diagonally, one complete atom and two parts of an atom are situated at the diagonal.

Let's consider that, the anion is at the corners of the cubic structure and the cation is at the center of the face. Then the face diagonal having a length of a is equal to twice the sum of the radius of the cation and anion.

That is the formula that related the radius of the cation, the radius of the anion with the edge length,

\(2 r^+ + 2 r^− = a\) --------- (1)

Where \(r^+\) is the radius of the cation, \(r^−\) is the radius of the anion, and ‘a’ is the edge length.

Here we are given,

The edge length of the FCC cell is, \(508\) pm

The radius of the cation is given as, \(110\) pm

We are interested to find the radius of the anion.

Let's substitute the values in the above equation (1). we have,

\(2 r^+ + 2 r^− = a\)

⇒ \(r^− = \frac{a − 2 r^+}{2}\)

⇒ \(r^− = \frac{508 − 2 (110)}{2}\)

∴ \(r^− = 144\) pm

Therefore, the radius of the anion is equal to 144 pm.