# How can an atom have 2 outer shells?

For example, my textbooks says how many electrons in the outer shell in Nickel? and it says the answer is the electrons in both the d shell and s shell. Doesn't shell equal= n quantum number? And n=4 is the outermost shell so shouldn't the answer exclude the electrons in d shell (since they are in the n=3 shell? Also the book says that Iodine has 7 electrons in its outer shell why did they skip over the de shell in this situation?

• Electron shells is very obsolete concept, easily misleading to incorrect ideas. See also atomic orbitals Jun 23 '20 at 19:43

First of all, I would like to thank @john for the question. Though this looks like a homework question, but this question actually tests the principles of quantum mechanics we understand in atomic levels. I would also like to thank @drmoishe-pippik for his simple answer, but I would like to dig deeper into quantum mechanics.

While solving hydrogen atom in quantum mechanics, the total Hamiltonian contains the kinetic energy terms and the nuclear-electron attraction term. There are no terms corresponding to electron-electron repulsion in the Hamiltonian. This gives rise the fact that we can separate the wavefunction into coordinates as $$\psi(r,\theta,\phi)=\psi_n(r)Y_l^m(\theta,\phi)$$, and the eigenvalue (energy) of the Hamiltonian is dependent only on the nuclear part quantum number $$(n)$$. This is true for any hydrogenic system ($$\ce{H,He+,Li^2+})$$, i.e. a system with one electron and one nucleus. The numbering of shells are based on hydrogenic orbitals only. So if we say an electron is in "$$3p$$" orbital, it means the electron (which is the only electron in the system) has an wave function corresponding to $$n=3,l=1$$ and $$-1\le m \le 1$$.

However, as we go to systems with more than one electrons, electron-electron repulsion comes to play and we are unable to separate the wave function into coordinates. Till date, exact solution of Helium atom is an extremely difficult challenge in quantum mechanics. However, even for multi-electron systems, one imagines hydrogenic orbitals while understanding the electronic structure of atoms and that is why we say, "$$\ce{C}$$ has $$1s^22s^22p^4$$ electronic configuration". The truth is, however, the orbitals we know of is hydrogenic in nature and we fill up electrons there without even trying to account for electron-electron repulsion and correlation. In facr, the orbitals you imagine may (and most probably will) not look like the orbitals in C atom at all.

So, there is nothing such as outer shell of electrons. There are some available approximate methods like Hartree Fock to estimate the orbitals using discrete basis, however, there are some issues in each of the method we know of today. In other words, we do not know which ones are the "outer shell" yet. However, my best bet in this case will be to take up any one of the standard methods for approximating multiple electron atoms and call the orbital(s) with highest energy level as outer orbitals.

However, if you are up to solving a homework problem only, I will say to go with the answer of @drmoishe-pippik in this case.