A solid `AB` has `NaCl` structure. If the radius of the cation `A` is `100` pm, what is the radius of anion `B`?

Explanation

The ionic solid is expected to have an octahedral structure in which the ratio `(r^(+)//r^(-))` lies between `0.414` to `0.732`.
Radius of cation `(r^(+))` = 100 pm
`therefore` Radius of anion `(r^(-))` may be calculated as follows:
`r^(+)//r^(-)=0.414" to "0.732`
`r^(-)=(r^(+))/(0."414 to 0.732")=(("100 pm"))/("0.414 to 0.732 ")="241.54 pm to 136.61 pm".`
Thus, the radius of anion will be in the range 241.54 to 136.61 pm.