Question

Explain the Pauling's method to determine the ionic radii.

Answer 1

a) Pauling has calculated the radii of the ions on the basis of the observed internuclear distance in four crystals namely.
NaF, KCI, FtbCI and Csl
i) In each ionic crystal the cation and ions are isoelectronic with inert gas configuration.
$K^{+}-2,8,8$
$C{l}^{-}-2,8,8$ Ar type of configuration
ii) The cations and anions of an ionic crystal are assumed to be in contact with each other and the sum of their radii will be equal to the inter nuclear distance between them.
$r\left(K^{+}\right)+r\left(C{l}^{-}\right)=d(K^{+}-C{l}^{-})$
where $\left(K^{+}\right)$ = radius of Cation, $\left(K^{+}\right)$
$r\left(C{l}^{-}\right)$ = radius of anion $r\left(C{l}^{-}\right)$
$d(K^{-}-C{l}^{-})$= inter nucleate distance between and $C{l}^{-}$ ions in $K^{+}C{l}^{-}$ ionic cystal
iii) For a given noble gas configuration the radius of an ion is inversely proportional to its effective nuclear charge.
$i.e.r\left(K^{+}\right)\alpha =\frac{1}{Z^{\ast }\left(K^{+}\right)}$ ....(2)
$r\left(C{l}^{-}\right)\alpha =\frac{1}{Z^{\ast }\left(C{l}^{-}\right)}$ ....(3)
Where $Z^{\ast }\left(K^{\ast }\right)$ and $Z^{\ast }\left(C{l}^{-}\right)$ are the effectivenuclear charge of cation $K^{+}$ and $C{l}^{-}$ respectively. On combining 2) and (3).
$\frac{r\left(K^{\ast }\right)}{r\left(C{l}^{-}\right)}=\frac{Z^{\ast }\left(C{l}^{-}\right)}{Z^{\ast }\left(K^{+}\right)}$ ......(4)
The 2 equations (1) and (4) can be used to evaluated the values of $r\left(K^{+}\right)$ and $r\left(C{l}^{-}\right)$ provided that the value of d $r(K^{+}-C{l}^{-}),Z^{\ast }\left(K^{+}\right)$ and $Z^{\ast }\left(C{l}^{-}\right)$ are known.