In a CsCl-type crystal the edge length of the cubic cell is 412 pm and the radius of the anion is 181 pm. Using sqrt3 = 1.732 and assuming cation-anion contact along the body diagonal, find the radius of the cation in pm (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3ss174

In a CsCl-type crystal the edge length of the cubic cell is $412 pm$ and the radius of the anion is $181 pm$ . Using $3 = 1.732$ and assuming cation-anion contact along the body diagonal, find the radius of the cation in pm (nearest integer).

Numerical answer type, answer in pm, tolerance 0.5.

Along the body diagonal

3 a = 2 (r_{+} + r_{-})

, so

r_{+} + r_{-} = \frac{1.732 \times 412}{2} = 356.79 pm

. Then

r_{+} = 356.79 - 181 = 175.79

, which is about 176 pm.