# Can the regions of 1s and 2s subshells overlap? I was studying atomic orbitals and always had this question lingering.

As we can see in the image, the 1s and 2s subshells both are known to closely surround nucleus. The overlapping peaks in the graph (of probability vs radius) lead me to believe that the part region of 2s before the node should overlap with the region under 1s.

And by 1s and 2s I mean their contours ( regions which contain 90% probability of electron occurring).

Am I committed a blunder here or do subshells overlap?

Additionally, does the same logic apply when comparing the spatial distribution of say 2p and 2s aswell? Do they have overlapping regions of high electron probability density as well?

There is nothing physically special about 90%. And if we don't think of 90%, then of course all orbitals of all atoms overlap, since each of them is infinite and spreads over all space. True, most of them overlap in the regions of extremely low probability, like $10^{-100}$, so for all practical purposes we might just as well consider they don't. But orbitals of one atom (pretty much all of them) would overlap in the areas of significant probability.
One might wonder why wouldn't they interact the way the orbitals of different atoms would. Well, there is a reason to that. Being eigenfunctions of the same operator, they are orthogonal, that is, the overlap integral equals zero if we consider the signs of $\psi$.
I'm not sure that thinking about sub-shells is very useful at all. They are not distinct entities but just part of the shape of the wave-function which in turn depends on the quantum numbers. You can see from your graph and figures (which are different ways of displaying the same thing) that there seems to be a chance that electrons in 1s, 2s & 3s occupy the same region of space. However, the overlap integral between total wavefunctions (radial and angular parts) from the same atom e.g. $\int\psi_{1s}\psi_{2s}d\tau =0$ shows that they are orthogonal. If n, l, m are the quantum numbers for one orbital and n', l', m' for the other orthogonality is found from the product $\delta_{nn'}\delta_{ll'}\delta_{mm'}$ where $\delta_{ab}$ is the delta function, = 1 if $a=b$ else zero.