KF crystallizes in the NaCl type structure. If the radius of ion is and ion is , what is the shortes — Solid State Chemistry Question

💡 Solution & Explanation

In an NaCl structure, the edge length $a = 2(r^+ + r^-)$. Thus $a = 2\left(\frac{186}{\sqrt{2}} + \frac{214}{\sqrt{2}}\right) = 2\left(\frac{400}{\sqrt{2}}\right) = 400\sqrt{2} \text{ pm}$. The shortest distance between two cations ($\text{K}^+$ ions) is half the face diagonal, which equals $\frac{a}{\sqrt{2}} = \frac{400\sqrt{2}}{\sqrt{2}} = 400 \text{ pm} = 4.00 \text{ \AA}$. Therefore, correct answer is 4.