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Although Strontium is in group 2, reducing the number of electrons as it becomes ionized makes it group 18, period 4 in terms of electrons. Therefore, ionized Strontium (Sr2+) is in the same period as Cu+. Cu+ has four shells and Sr2+ has four as well. The deciding factor for which ion has the greater ionic radius is the number of protons. The fewer protons given that the comparing ions are in the same period means that it has a greater radius. Sr2+ has more protons whilst being in the same period as Cu+, therefore the increase in nuclear charge provides a greater force of attraction between the nucleus and the electrons, decreasing the ionic radius. Does this mean that Cu+ has a greater ionic radius than Sr2+? Why is it said in my workbook that Sr2+ has a larger radius than Cu+?

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  • $\begingroup$ The wikipedia table of ionic radii says r(Cu+)=91 nm, r(Sr62+)=132 nm. Not the decreasing radius trens along the periods link and that ions for the first two groups are relatively big. $\endgroup$
    – Poutnik
    Dec 11, 2023 at 9:54

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I think I got it. I did not consider the amount of shells the ions concluded for some reason, which is the obvious thing to do. Cu+ has an electron configuration of 1s2,2s2,2p6,3s2,3p6,3d10 and Sr2+ has an electron configuration of 1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6. Putting it in a way that looks better in order to visualize the shells, it would look like this respectively: (1s2),(2s2,2p6),(3s2,3p6,3d10) and (1s2),(2s2,2p6),(3s2,3p6),(4s2,3d10,4p6). As you can see, Cu+ only has 3, whilst Sr2+ has 4. As Sr2+ has more inner shells, Sr2+ is larger.

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