How do we choose last electron & write the quantum numbers for that element?

Question

I was solving questions about writing the quantum numbers of the last electron of an element.

For Ions like Na+ , Na , O , Cl it was easy.

But for elements like Scandium , Yttrium , Iron I am having difficulty.

Electronic configuration respectively of each element = $3d^1$ $4s^2$ , $4d^1$ $5s^2$ , $3d^6$ $4s^2$.

Now, according to the answer in my textbook + online I rechecked. The quantum numbers of the last electron respectively are:

1. For Sc : n = 3 , l = 2 , m = -2 , s = +1/2. This clearly shows that the last electron is not of 4s orbital but of 3d. Why is it that if according to the electronic configuration , $4s^2$ is the last orbital we have written,

Similarly , for other 2 as well:

2. For Fe , n = 3 & not 4

3. For Y , n = 4 & not 5.

I couldn’t find any explanation for it but I am guessing.For example :

We can notice all these elements are exceptions to electronic configuration.

In case of Sc , d orbital is not filled & has only 1 e- but the $4s^2$ orbital is filled. Therefore , we choose the electron of 3d. But still , it is not the last electron but the 1st unpaired electron we can say.

If it is right , then in case when we have $Sc^+$. We will have outer electronic configuration = $4s^1$. In this case , which electron would we choose

The 3d orbital or 4s ?

To brief my questions:

1. Why is the set of quantum numbers in case of elements like Sc,Fe,Y are different to write than Na,O,Na+.

2. What is the reason for such thing to happen?

3. I have made a guess above which is not theoretical but mathematical , if that guess is correct. How would we write set of quantum numbers for $Sc^+$.

Answer 1

Consider the example for Sc where the 3d orbital is not filled but the 4s is.

The fact is that 4s orbitals have a lower energy than the 3d orbitals, which therefore means that 4s orbitals will in fact fill up first. For example, any electrons that are lost from Sc causing it to ionize (Sc$^+$) will come from (most likely) the highest energy level, or 3d.

This is origin of the notation you are using, and you can apply the same reasoning to all the other elements you have listed.