Question

A binary solid (AB) has a rock salt structure. If the edge length is 500 pm, and radius of cation is 80 pm. Find the radius of an anion.
(A) 100 pm
(B) 120 pm
(C) 250 pm
(D) 170 pm

Answer 1

Hint: Use the relation between radius of cation and radius of anion for FCC lattice structure as rock salt (NaCl) has a FCC structure. The structure of NaCl is formed by repeating the face centered cubic unit cell. It has a 1:1 stoichiometry ratio of Na: Cl.

Formula Used: Radius of cation + Radius of anion = edge length/2

Complete Step By Step Solution
Given, edge length = 500 pm
Radius of cation = 80 pm
Using the formula, Radius of cation $ + $ Radius of anion = edge length/2 we get,
80 pm + Radius of anion = 500 pm/2
Therefore, radius of anion = 250 pm $ -80 $ pm
= 170 pm.
So, the correct option is (D).

Additional Information
FCC structure: In the FCC arrangement, there are eight atoms at corners of the unit cell and one atom at the center of each face. The atom in the face is shared with the adjacent cell. FCC unit cells thereby consist of four atoms, eight eighths contribution at the corners and six halves contribution in the faces. Two simple regular lattices have highest average density namely face-centered cubic and hexagonal close-packed lattices.

Note
Take care that radius of cation, radius of anion and edge length all must be in the same units. If not, convert them into the same units and then solve for the unknown quantity. Also check for the type of structure of the compound given and then use the formula accordingly.
The relation between edge length (a) and radius of atom (r) for BCC lattice is 3a =4r.