Question

The radius of $N{a}^{+}$ is $95$ pm and that of $C{l}^{-}$ is $181$ pm. The edge length (in pm) of unit cell in $NaCl$ would be :

• A. $181$
• B. $95$
• C. $276$
• D. $552$

Answer 1

The correct option is D $552$

The edge length of the unit cell in NaCl will be two times the sum of the ionic radii of sodium ions and chlorine ions.

$a=2(r_{N{a}^{+}}+r_{C{l}^{-}})=2(95+181)=552$ pm

Option D is correct.