Question

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å.

Short answer

Radius of the Potassium ion will be $1.34\stackrel{0}{A}$.

Step-by-step solution

Define the ionic compound features

In FCC there are total 8 corners and 6 face centres. Therefore, the total number of atoms in this structure is 4. The relation between the lattice parameter and the radius is given by the formula:

$a=\frac{4r}{\sqrt{2}}$

Where$a=$edge or lattice parameter

$r=$ radius of the lattice

 Identify the radius of the molecules.

Now, we are given the edge length $=6.28\stackrel{0}{\mathrm{A}}$ and radius $=1.82\stackrel{0}{\mathrm{A}}$. Now, we will calculate the radius of $C{l}^{-}$ion.

We know that the ratio of the radius of the cation to anion in an FCC lattice is 0.731.

Also, the sum of the radius of cation and anion in FCC is given as equal to half of its edge length. It is represented as:

$r^{+}+r^{-}=\frac{a}{2}$

Where, $r^{+}=$cation and $r^{-}=$anion. In the given salt the cation is $K^{+}$and the anion is $\mathrm{C}{\mathrm{l}}^{-}$.

Putting all the values in the given relation we get:

$\begin{array}{l}{r}_{K^{+}}+r_{C{l}^{-}}=\frac{a}{2}\\ r_{K^{+}}+1.8=\frac{6.28}{2}\\ r_{K^{+}}+1.8=3.14\\ =1.34\\ r_{K^{+}}=1.34\stackrel{0}{A}\end{array}$

Hence the radius of the Potassium ion will be $1.34\stackrel{0}{A}$.