question_answer

The unit cell cube length for Lid (just like \[NaCl\] structure) is \[5.14\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]. Assuming anion- anion contact, the ionic radius for chloride ion is-

A) \[1.815\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

B) \[2.8\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

C) \[3.8\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]      

D) \[4.815\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

Correct Answer: A

Solution :

\[2{{r}_{Cl-}}\,=\,\,\frac{\sqrt{2}}{2}\,\,\times \,\,a\,\,=\,\frac{\sqrt{2}}{4}\,\,\times \,\,5.14\,\,=\,\,1.815\]