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The edge length of the unit cell of LiCl (NaCl-like structure) is 520 pm. Assuming that the lithium Solid State Chemistry Question

Question

The edge length of the unit cell of LiCl (NaCl-like structure) is 520 pm. Assuming that the lithium ion is small enough that the chloride ions are in contact, the ionic radius of chloride ion (in pm) is $( \frac{1}{\sqrt{2}} = 0.7 )$

Answer: 0182

💡 Solution & Explanation

In an NaCl structure where the cations are small enough that the anions touch, the anions are in contact along the face diagonal. Thus, $4r_{Cl^-} = \sqrt{2}a$. So $r_{Cl^-} = \frac{\sqrt{2}a}{4} = \frac{a}{2\sqrt{2}} = \frac{520 \times 0.7}{2} = \frac{364}{2} = 182 \text{ pm}$. Therefore, correct answer is 0182.

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