Question

As per Pauling, the radius of an isoelectronic ion in a given crystal is inversely proportional to the effective nuclear charge, $Z^{\ast }$.
$Z^{\ast }=Z-\sigma$; where $\sigma$ represents the screening constant. The crystal structure of $\text{NaF}$ is shown as below :

The radius of $F^{-}$ in pm is $(130+x)$ . The value of $x$ in nearest possible integer is

Answer 1

Screening constant of $N{a}^{+}and{F}^{-}$ are same
$N{a}^{+}=1s^2{s}^{2}2p^6$
$\sigma =0.35\times 8+2\times 0.85$
$=2.8+1.7=4.5$
$\frac{r_{N{a}^{+}}}{r_{F^{-}}}=\frac{Z_{F^{-}}^{\ast }}{Z_{N{a}^{+}}^{\ast }}$
$=\frac{(Z_F^{-}-{\sigma }_F^{-})}{(Z_N{a}^{+}-{\sigma }_{N{a}^{+}}}$
$=\frac{4.5}{6.5}=\frac{9}{13}$
$r_{N{a}^{+}}=\frac{9}{13}{r}_{F^{-}}$
$r_{N{a}^{+}}+r_{F^{-}}=2.35\stackrel{o}{A}$
$r_{F^{-}}=\frac{13}{22}\times 2.35\stackrel{o}{A}$
$=1.38\stackrel{o}{A}1.388\times {10}^{-10}m=138.8pm\approx 139pmx=9$