Question

A solid AB has NaCl type closed packed structure. If the anion ($B^{-}$) has a radius of $250$ pm, what should be the ideal radius (in pm) of the cation?

• A. $78.2$
• B. $103.4$
• C. $156.2$
• D. None of the above

Answer 1

Answer: (B), (C)

The correct options are
B $103.4$
C $156.2$

For face centered cubic structure, the radius ratio $\frac{r_c}{r_a}$ is between 0.414 and 0.732. For the options A, B and C, we will calculate the radius ratio and see if it fits in the above range. If it fits in the above range, then it is the ideal radius of the cation.
(A) $\frac{r_c}{r_a}=\frac{78.2}{250}=0.313$
(B) $\frac{r_c}{r_a}=\frac{103.4}{250}=0.414$
(C) $\frac{r_c}{r_a}=\frac{156.2}{250}=0.625$
As can be seen the radius ratio of options B and C fits in the range 0.414 to 0.732. Hence, the ideal radius of the cation will be either 103.4 p or 156.2 pm.