Question

if the ionic radius of k+ and f- are about 1.34 amstrong each,then the expected values of atomic radii of k and f should be what?

Answer 1

Since, the ionic radio of both K + and F- is the same, hence the atomic radius of K+ will be greater than that of F-. Because the cationic radius is smaller than it's atomic radius and anionic radius is greater than it's atomic radius.

Answer 2

Explanation:

A we known that $K^{+}$ ion is formed by loss of an electron whereas $F^{-}$ ion is formed by gain of an electron.

In case of a cation, its atomic radii is smaller that its neutral state atomic radii. And, in case of an anion its atomic radii is larger than its atomic radii in neutral state.

So, when atomic radii of both these ions become equal to 1.34 amstrong then it means that in their neutral state also they have the same atomic radius that is why, on reaching excited state only they have same radii.

Therefore, we can conclude that if the ionic radius of $K^{+}$ and $F^{-}$ are about 1.34 amstrong each,then the expected values of atomic radii of K and F should be the same.