# Ionic radius of Al[3+] compared to K[+], Ca[2+], S[2-], and Cl[-]

I came across this question:

Which of the following is bigger?

$$\ce{K+, Ca^2+, Al^3+, S^2-, Cl-}$$

Now, I know that the radius decreases as the period progresses to the right and increases as we go down on the P.T. So, I get to this:

$$\ce{S^2- > Cl- > K+ > Ca^2+}$$

Also, I know that $\ce{Al^3+ < Al^0}$ and that $\ce{Al^0 > S^0}$. But I can't figure out how to connect that to the inequality above.

• $\ce{Al{3+}}$ is a 2nd group element, and should be even smaller than the rest. – Pritt says Reinstate Monica Apr 27 '17 at 16:01
• Think of $\ce{Sc^3+}$; it should be easy to compare it to $\ce{Al^3+}$ and also to the rest. – Ivan Neretin Apr 27 '17 at 16:11
• Well $Sc^{3+}<Sc$ and $Al^{3+}<Al$. Of course $Sc>Al$. If $Sc^{3+}>Al^{3+}$, then yes, I can compare them. And then $Al^{3+}$ is the smallest. But why does $Sc>Al\rightarrow Sc^{3+}>Al^{3+}$ hold? – George Apr 27 '17 at 16:22