The ionic radius of the $\ce{Sr^2+}$ ion is $\mathrm{132\,pm}$, while the ionic radius of the $\ce{K^+}$ ion is $\mathrm{152\,pm}$. Why is this the case? I would have thought that since $\ce{K^+}$ has one less shell of electrons than $\ce{Sr^2+}$, its radius would be significantly smaller.

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    $\begingroup$ See also Libretexts.org - Periodic_Trends_in_Ionic_Radii $\endgroup$
    – Poutnik
    Apr 23, 2020 at 6:08
  • $\begingroup$ If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed. $\endgroup$
    – Poutnik
    Apr 23, 2020 at 6:13
  • $\begingroup$ Some quick analysis: $\ce {Rb^{+}}$ has an ionic radius of 166 pm. What this means is that we cannot attribute the effect to d-orbital contraction, i.e. poorer shielding effect provided by the 3d orbitals. Going from $\ce {K^{+}}$ to $\ce {Ca^{2+}}$, there is a significant drop from 152 to 114 pm. Similarly, there is a big drop from $\ce {Rb^{+}}$ to $\ce {Sr^{2+}}$, from 166 to 132 pm. It may well be the case that the addition of another proton cause a significant reduction of the ionic radius, such that it outweighs the increase in radius due to presence of an additional electron shell. $\endgroup$ Apr 23, 2020 at 9:57
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/64335/…. $\endgroup$ Apr 24, 2020 at 5:07


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