Question

The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is:
A. 288 pm
B. 398 pm
C. 618 pm
D. 144 pm

Answer 1

Hint:Face centered cubic cell is a type of non-primitive unit cell that is it contains one or more lattice points at other than 8 corner positions in the unit cell and as the name suggests beside the 8 corners there are unit cell at each face which makes it 6 lattice points and total number of atoms per unit cell is:
$\dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 6 = 1 + 3 = 4$

Complete answer:
In all types of packaging there is always some free space in the form of voids or vacant spaces. Packing efficiency is the percentage of total space filled by the particles. Similarly FCC packing has voids too. Therefore two types of commonly occurring voids are octahedral and tetrahedral voids. In this number of octahedral voids present in the lattice is equal to the number of lattice points and the number of tetrahedral voids is twice the number of lattice points.
Since it is an ionic substance that is its voids are also filled with ions that are smaller in size. Pattern of arrangement and type of voids both depend upon ionic size where ${r^ + }$ is the radius of the cation and ${r^ - }$is the radius of anion. For FCC the relation is
$2({r^ + } + {r^ - }) = a$ where a is the edge length of the unit cell, substituting the values we get,
$
\Rightarrow 2(110 + {r^ - }) = 508 \\
\Rightarrow 220 + 2{r^ - } = 508 \\
\Rightarrow 2{r^ - } = 508 - 220 = 288pm \\
\Rightarrow {r^ - } = \dfrac{{288}}{2} = 144pm
 $

Hence the correct option is D.

Note:
In ionic solids that are bigger in size generally, i.e anions form FCC and smaller ions i.e cations occupy the voids. In a given compound some fraction of tetrahedral and octahedral voids are occupied by cations.