# Why is strontium(II) ion bigger than krypton atom?

$$\ce{Sr^2+}$$ is exactly the same as $$\ce{Kr}$$, in terms of electrons and orbitals. The only difference between the two, is that $$\ce{Sr^2+}$$ has a couple of extra protons in the nucleus (and probably a couple of extra neutrons too, but these don't influence ionic/atomic size).

Considering the two extra protons in the nucleus of $$\ce{Sr^2+}$$, and the overall 2+ charge on the ion, shouldn't this draw the valence electrons in closer to the nucleus in $$\ce{Sr^2+}$$ in comparison to $$\ce{Kr}$$?

(Krypton size= 88 pm, while Sr2+ is 132 pm.)

• These are different, incomparable kinds of bigness. – Ivan Neretin May 14 at 7:29
• No size data are quoted, making the context of this question difficult to trace. Do tell. – Oscar Lanzi May 14 at 9:38
• References for size data being considered would be very useful as well – Ian Bush May 14 at 11:11
• Incomparable kinds of bigness? I see a textbook problem comparing the radius of an sodium atom to its ion (see Zumdahl & Zumdahl, 9th Edition, Ch. 7, Exercise 109). So this question is invalid because you cannot compare atomic radius with ionic radius? Notwithstanding my last sentence, I need to say that I never saw a question on a standardized test asks me to compare atomic radius with ionic radius. – L. Cang May 14 at 13:47
• Gas phase strontium(II) has a smaller ionic radius than krypton's atomic radius. The thing is that in a crystal lattice it isn't quite the same as in the gas phase. – Buck Thorn May 15 at 12:01

According to Ptable and LumenLearning, $$\ce{Kr}$$ has an atomic radius of 88 pm, while $$\ce{Sr^2+}$$ has an atomic radius of 132 pm. Although it is true that $$\ce{Kr}$$ and $$\ce{Sr^2+}$$ have the same electron orbital configuration and $$\ce{Sr^2+}$$ has two more protons than $$\ce{Kr}$$, $$\ce{Sr^2+}$$'s greater nuclear attraction is likely countered by the nature of metal ions being more willingly to lose electrons, which implies the electrons are more loosely held by the nucleus than the noble gas $$\ce{Kr}$$.
Another reason I conjecture is that $$\ce{Sr^2+}$$ has unfilled 5s orbitals to readily accept promoted electrons, while $$\ce{Kr}$$'s electrons have less tendency to fill the 5s orbitals because the property of being a noble gas.