Q. Calculate the ionic radius of a$\ Cs^{+}$ ion, assuming that the cell edge length for CsCl is 0.4123 nm and that the ionic radius of a $CI^{-}$ ion is 0.181 nm
A) 0.352 nm
B) 0.116 nm
C) 0.231 nm
D) 0.176 nm
Explanation
(d)
$CsCl$ has a bcc lattice. So, $d_{body} = a\sqrt{3}$
or $d_{body} = \sqrt{3} \times \ 0.4123\ nm = 0.7141\ nm$
The sum of the ionic radii of $Cs^{+}$ and $Cl^{-}$ ions is half this distance $ie$
$r_{Cs^{+}} + r_{Cs^{-}} = \frac{d_{body}}{2} = \frac{0.7141}{2}\ nm$
= 0.3571 nm
Ionic radius of $Cs^{+} = 0.3571 - 0.181 = 0.1761$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.