$\ce{Sr^2+}$ 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.)