Question

To determine the radii of a univalent ion, Pauling gave a method as follows:
i. In an ionic crystal of the type $M^{+}{X}^{-}$, cations and anions are in contact with each other. Therefore, the sum of their ionic radius is equal to the interionic distance.
ii. $r_M^{+}+r_X^{-}=\frac{c}{Z_{eff}\left(M^{+}\right)}+\frac{c}{Z_{eff}\left(X^{-}\right)}pm$ $Z_{eff}$ is the effective nuclear charge $Z_{eff}=Z-\sigma$ ; where $\sigma$ is the screening constant.
For $\text{Ne},\sigma =4.5$
$\therefore r_{\text{M}}^{+}=\frac{c}{Z_{\text{eff}}\left(M^{+}\right)}$ and $r_X^{-}=\frac{C}{Zeff\left(X^{-}\right)}$

In $\text{NaF}$ crystal,$r_{Na}^{+}+r_F^{-}=230$ pm therefore, the radius of $F^{-}$ is

• A. 136 pm
• B. 94.5 pm
• C. 112 pm
• D. 115 pm

Answer 1

The correct option is A 136 pm

$r_M^{+}+r_X^{-}=\frac{c}{Z_{eff}\left(M^{+}\right)}+\frac{c}{Z_{eff}\left(X^{-}\right)}pm$
$230=\frac{c}{(11-4.5)}+\frac{c}{(9-4.5)}pm$
$230=\frac{11c}{(6.5\times 4.5)}pm$
$c=\frac{6.5\times 4.5\times 230}{11}pm$
$r_{F^{-}}=\frac{6.5\times 4.5\times 230}{11}\times \frac{1}{4.5}$
$r_{F^{-}}=136pm$