Question

A solid AB has Zns type structure. If tge edge length of the unit cell is 400 pm and tge radius of B- ion is 0.130 nm then the radius of A+ ion is

Answer 1

Answer:

The answer is 0.47 nm

Explanation:

Hello , welcome to brainly.Below is the solution for your problem.

ZnS has FCC structure.

For FCC structure, a=2r+ + 2r-

Where, r+ : radius of cation

            r- : radius of anion

            a= edge length

So,

r+ +r- = a/2

Given that, a=400 pm=400×0.001 nm =0.4 nm

                  r- = 0.130 nm

Putting these value in above equation, we get,

r+ +0.130 = 0.4/2

∴ r+ = 0.2 - 0.130

∴ r+ = 0.70 nm

Hope this explanation clears your doubt.Thanks