Question

$AX,BY,BX$ and $BY$ have rock salt type structure with following internuclear distances:

Salt	Anion-anion distance in ${A^o}$	Cation-anion distance in ${A^o}$
$AX$	2.40	1.70
$AY$	1.63	1.15
$BX$	2.66	1.88
$BY$	2.09	1.48

Ionic radii of ${A^ \oplus }$ and ${B^ \oplus }$ respectively are:
A. $0.35$ and $0.68{A^o}$
B. $0.68$ and $0.35{A^o}$
C. $1.20$ and $0.80{A^o}$
D. $0.80$ and $1.20{A^o}$

Answer 1

We are given the internuclear distances between constituents of the packed structure. As cation-anion distance is readily available to us, if we know the radii of the anion in each case, we can easily obtain the cation radius.

Complete answer:
From the data, we can see that the cation will not be in contact with all the anions in $AX$. This is due to the difference in internuclear distances we observe from $AX$ and $BX$. Hence there must be contact between the anions in this compound. If we consider anion-anion distance as $2{r^ - }$ , from the table we can calculate the radius of ${X^ - }$ as half of this value.
$\Rightarrow$ Radius of ${X^ - } = \dfrac{{2.40}}{2} = 1.2{A^o}$
We know that the sum of radii of the cation and anion is equal to the cation-anion distance.
So, if we subtract this radius from the cation-anion distance of $BX$, we can get the radius of the cation, that is, radius of ${B^ \oplus }$
$\Rightarrow$ Radius of ${B^ \oplus } = 1.88 - 1.20 = 0.68{A^o}$
Subtracting this value from the cation-anion distance of $BY$ will give us the radius of the anion in this case, that is, radius of ${Y^ - }$
$\Rightarrow$ Radius of ${Y^ - } = 1.48 - 0.68 = 0.80{A^o}$
Subtracting this from the cation-anion distance of $AY$ will give us the radius of the cation ${A^ \oplus }$
$\Rightarrow$ Radius of ${A^ \oplus } = 1.15 - 0.80 = 0.35{A^o}$
Hence the radius of ${A^ \oplus }$ and ${B^ \oplus }$ are $0.35{A^o}$ and $0.68{A^o}$ respectively. Therefore, the correct option is option A.

Note:
Rock salt is sodium chloride ($NaCl$) and the anions (chloride ions) are arranged in a face centred cubic cell structure, with the cations (sodium ions) occupying the octahedral voids, and the tetrahedral voids left empty. The coordination number of both ions in this type of structure is 6. Note that although the cation-anion distance is representative of the size of the total molecule, cations may not be always touching the anion, like in this question. In this question, we were able to take the size of the anion ${X^ - }$ as half of the anion-anion distance because we found that they were in contact with each other.