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JEE Main levelSingle correct optionDifficulty 2 of 3ss029

In rock salt the ions touch along the cell edge. If the unit cell edge is $564 pm$ and the radius of $Cl X^{-}$ is $181 pm$ , the radius of $Na X^{+}$ is

A141 pm
B101 pm
C383 pm
D55 pm

Answer: B

Along an edge the ions are in the sequence

Cl X^{-} - Na X^{+} - Cl X^{-}

, so

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

. Hence

r_{+} = a /2 - r_{-} = 282 - 181 = 101 pm

.

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