Electron configurations for d-block and p-block

Question

Question 1: Why is it that when you get to the transition metals (or the d-block), the energy of the $\ce{3d}$ orbitals becomes slightly less than that of $\ce{4s}$, making $\ce{4s}$ have the electrons with the most energy?

Also, is it correct to write electron configurations for transition metals with the $\ce{4s}$ orbital being the outermost one? For example like this $\ce{Sc = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$ instead of this $\ce{Sc = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$

Question 2: If my first thought is true, then do you write electron configurations for the p-block just like the d-block, with the $\ce{4s}$ orbital being the outermost one? Or do you follow the regular pattern of letting the $\ce{3d}$ block comes after the $\ce{4d}$ block?

For example, would it be like this $\ce{Ga = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^1}$ or would it be like this $\ce{Ga = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^1}$

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According to google, if you search up "scandium electron configuration" it would say my first one). If you search up "gallium electron configuration", you would see my second one)

Answer 1

I did not completely understand you question here, but I am gonna give it a try, so maybe I might answer your question.

I do not think that it matters how you write them, but its just depends on how you fill them. The order for filling orbitals is as follows: $$\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s}$$

In high school, I just used to write the $(n+1)s$ before the $nd$. Because with some exceptions, the rule says that you will always fill the $(n+1)s$ orbital before you fill the $nd$ orbital. Meaning that you fill $\ce{4s}$ before you fill $\ce{3d}$, $\ce{5s}$ before you fill $\ce{4d}$, and so on.

The reason is that the $\ce{s}$ actually has a lower energy than the $\ce{d}$ orbital. You wrote that wrong in your question. That is the only you fill the s before the d, and when you make ions, you will have to remove the electrons from s before you do from $\ce{d}$.

But like I said, you can write them like you want as long as you have filled them correctly in the right order. Likewise once you have completely filled your $ns$, only then you can move to fill the $(n-1)d$ orbital and then again proceed to fill the $\ce{np}$ orbital. Its just how we fill them.

In some books, you will find that they write the s as the outermost but some write d as the outermost. Frankly, in my book and myself have learned to write $\ce{d}$ at the end. But I do not think that writing it matters.

For example, take a look at this site. They write $s$ orbital at the end of the configuration and also note the energy diagram. But in the following Wikipedia website, they have written the $\ce{d}$ at the end instead of $\ce{s}$. So basically, the writing of the orbitals doesnt matter, but the order of filling matters.

Answer 2

The 4s sublevel in an atom have a lower energy than the 3d sublevel if the atom only has 4s electrons and no 3d electrons, so 4s "fills first" as you progress across period 4 of the P.T., and group 1 and 2 metals will have the s sublevel at the top, but if you have an element in the d block, the ELECTRONS IN the 4s orbital have higher energy than the electrons in the 3d orbitals (they flip), so the 4s electrons leave first and fill last.

http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html

So, in other words, 1) Order of filling depends on what you mean by order of filling.
a) As you progress along a period on the P.T. each successive element will (with exceptions) have electrons added to their d orbitals compared to previous elements.

b) As you fill up a given element in the d-block, the electrons of known sublevels will fill in the 3d sublevel before the 4s (4d before 5s etc.) however you can not predict the electron configuration by using the rule that the electrons will go into the sublevels of a given d-block atom, by first entering nd and then into the (n-1)p because, for example, the entire 3d sublevel is not at a lower energy than the entire 4s sublevel for Gallium, ONLY the 3d electrons that Gallium HAS are lower energy than Gallium's 4s electrons.

Google now lists the "b" answer for both Scandium and Gallium

2b) Ga = $1s^22s^22p^63s^23p^63d^{10}4s^24p^1$

1b) Sc = $1s^22s^22p^63s^23p^64s^23d^1$

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=scandium%20electron%20configuration

xs electrons are higher than (x-1)d electrons that are present in a given atom

only the (x-1)d electrons that are present in an atom are lower energy than the xs present in that atom. You can't know for sure how many electrons will be in the (x-1)d and xs, but ones with 2 electrons in the xs will have the rest in the (x-1)p, its just that the xs electrons are higher.

d-orbitals still get "fuller" as you move across the P.T.