In a standard -type structure, the cation is present in a cubic void. If the edge length of the crys — Solid State Chemistry Question
Question
In a standard $CsCl$ -type structure, the cation is present in a cubic void. If the edge length of the crystal unit lattice is exactly $390\text{ pm}$ and the radius of the cation ($r_+$) is $150\text{ pm}$, what is the radius of the anion ($r_-$) in $\text{pm}$? (Round to the nearest integer)
💡 Solution & Explanation
In a cubic void arrangement (like $CsCl$), the cation and anion touch along the body diagonal of the simple cubic unit cell formed by the anions. The body diagonal length is $a\sqrt{3} = 2(r_+ + r_-)$. Therefore, $r_+ + r_- = \frac{a\sqrt{3}}{2} = \frac{390 \times 1.732}{2} = 337.74\text{ pm}$. The anion radius $r_- = 337.74 - 150 = 187.74\text{ pm}$. Rounded to the nearest integer, it is $188\text{ pm}$.