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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.

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  • $\begingroup$ $\ce{Al{3+}}$ is a 2nd group element, and should be even smaller than the rest. $\endgroup$
    – Pritt says Reinstate Monica
    Apr 27, 2017 at 16:01
  • $\begingroup$ Think of $\ce{Sc^3+}$; it should be easy to compare it to $\ce{Al^3+}$ and also to the rest. $\endgroup$
    – Ivan Neretin
    Apr 27, 2017 at 16:11
  • $\begingroup$ 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? $\endgroup$
    – George
    Apr 27, 2017 at 16:22

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