Q. The ionic radii of $Rb^{+}$ and $I^{-}$ are $1.46\ \mathring{\mathrm{A}}$ and $2.16\ \mathring{\mathrm{A}}.$ The most probable type of structure exhibited by it is

A) $CsCl$ type
B) $ZnS$ type
C) $NaCl$ type
D) $CaF_{2}$ type
Explanation

(c)
[table] Radius ratio | Coordination no | Example ; 0.155 $-$ 0.225 | 3 | $B_{2}O_{3}$ ; 0.225 $–$ 0.414 | 4 | $ZnS$ ; 0.414 $–$ 0.732 | 6 | $NaCl$ ; 0.732 $–$ 1 | 8 | $CsCl$ [/table]
In ionic solids the shape of crystal depends upon relative size of ions.
Given, $r_{c^{+}}\left( Rb^{+} \right) = 1.46\ \mathring{\mathrm{A}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r_{a^{-}}\left( I^{-} \right) = 2.16\ \mathring{\mathrm{A}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{c^{+}}}{r_{a^{-}}} = \frac{1.46}{2.16} = 0.676$
$\therefore$ It will have coordination number 6 and structure will be same as of NaCl.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.