Chapter 2 - Atomic Structure and Interatomic Bonding - Questions and Problems - Page 48: 2.17

$0.155 nm$

We have $r_{0}=r_{C}+r_{A}$ with $r_{C}$ and $r_{A}$, the radii of the cation and anion, respectively. Using equation 2.14, we find $r_{0}$ with $Z_{C}=+2 $ and $Z_{A}=-2 $ $r_{0}=\sqrt {\frac{(2.31\times{10}^{-28}N.m^{2})(|+2|)(|-2|)}{1.67\times{10}^{-8}N}}=0.235\times{10}^{-9}m=0.235nm $ Thus, we obtain: $r_{A}=r_{0}-r_{C}=0.235nm-0.080nm=0.155nm$