Quick question on the ionization energy and the selection rule

Question

So I am looking through my book and it says ".... the order of the excited states is exactly the same order (3p-4s-3d-4p)".

But now I am looking at a question in the book and it says "Is 3d to 4s transition possible? Why or why not?"

My answer to this question is: No it can't be because it doesn't abide by the selection rule because the difference in the orbital quantum numbers is 2 instead of 1.

Now if this is the case, why does the above statement have the oder of (3p-4s-3d-4p) if we know there cannot be a transition between the 4s and 3d state like this order says?

Answer 1

Order of excited states means ordered by their energy: 3p is lower than 4s is lower than 3d is lower than 4p. As you correctly point out, the transition from 4s to 3d is forbidden due to a selection rule concerning the angular momentum. So, 3d being higher in energy than 4s has nothing to with there being or not being a transition between them.

Note: Nothing is ever really forbidden. You might not have a dipole transition from 4s to 3d, but you might have an electric quadrupole or magnetic dipole or higher multipole transition. However, these have much lower intensity.