Question 1

Why does the 4s orbital fill before the 3d orbital ( in general)?

Accepted Answer

The Aufbau principle states that electrons are allocated to atomic orbitals in order of increasing energetic content. To calculate the order of energy we use the Madelung rule:
a) In H atoms, all subshells of same n have the same energy.
b) in atoms of 2+ electrons, subshells increase in energy as the value of n+l increases, and for subshells of same n+l value, subshells with a lower n are lower in energy.
In this example, 4s orbitals will have an energy of n+l=4 and 3d orbitals have an energy of n+l=5, that is why 4s orbitals fill before 3d orbitals.
Instead of calculating everything, you can refer yourself to the Madelung diagram, that will give you the full order of orbitals filling. There are only a few expectations to these rules, with more complex reasons.

Question 2

Hi, 
I just read somewhere that in the electron configuration of Cu the 4s subshell is filled after the 3d subshell, what can be the reason for that?

Accepted Answer

We observe that 3d4 and 3d9 do not exist as electronic configurations due to instability issues. In general any d4 and d9 configuration will be changed to d5 and d10 respectively by an electron from the next s subshell being excited to the d subshell . This is what happens with Chromium as well. 
Chromium's electronic configuration is supposed to be :
1s2 2s2 2p6 3s2 3p6 3d4 4s2    . But d5 is a situation when the entire d subshell is half filled and this is a more stable scenario than d4. So an electron get's excited from the 4s subshell to the 3d subshell to achieve the half filled subshell stability. Finally, this is the actual electronic configuration of Chromium :
1s2 2s2 2p6 3s2 3p6 3d5 4s1 . 
So as you can see this special scenario repeats with Copper when a 3d9 gets changed to 3d10 . 
So instead of 
1s2 2s2 2p6 3s2 3p6 3d9 4s2 
we have
1s2 2s2 2p6 3s2 3p6 3d10 4s1

Question 3

Why is the electron configuration of Calcium [Ar]4s2 and not [Ar] 3d2? Shouldn't the electrons in Calcium first fill energy level 3?

Accepted Answer

Electrons fill into atomic orbitals according to the aufbau principle, from the lowest energies to the highest energies. When comparing the energies of the 4s and 3d orbitals, the 4s is lower in energy than the 3d. Even though the 3d is in a lower shell number, it still possesses slightly more energy so it is filled after the 4s. Hope that helps.

Question 4

Does it matter the order the notation is written in? At 5:17 3d is written before 4s, is it ok to switch the order of those?

Accepted Answer

There is some debate which way is better, either 4s3d or 3d4s, for describing the electron configurations of transition metals. Short answer, technically the more correct way would be to write 3d4s, but it is still acceptable to write 4s3d.

The difference arises due to the energy levels of the two subshells. When writing electron configurations the aufbau principal says that electron subshells with lower energy should be filled first. So this would mean writing the lower energy subshells further to the left and the higher ones toward the right. So the two ways of writing the electron configurations are having a disagreement about which subshell is lower in energy.

The problem is when we start filling them for transition metals. So when the 4s and 3d subshells are unoccupied by electrons the 4s is lower than the 3d subshell so we would fill the 4s first and write it first giving us 4s3d. However once filled, the 3d subshell actually becomes less energetic and has lower energy than the filled 4s orbital. Now when filled the 3d subshell would be written first since it has lower energy giving us 3d4s.

So technically the electron configurations for elements like scandium or titanium should be written like 3d4s since they are filling those subshells causing the new energy difference as compared to the subshells being unoccupied. However I find it easier to write them as 4s3d since it follows the order of the pyramid Sal has in the upper left, so I use that format instead.  The important part is just to be aware of the different energy levels of the subshells when they're unoccupied versus occupied.

Hope that helps.

Question 5

I'm really confused about the Aufbau principle. Since it basically says electrons fill lower energy shells first then does that mean electrons will always go from 1 to 2 to 3 shells rather than skipping 3d and going to the 4th shell like shown with Sc in the video? Does that mean Sc "breaks" this principle? And how does this principle apply to naming electron configurations?

Accepted Answer

The aufbau principal says that atoms start filling orbitals with the lowest energy and progressively fill higher energy orbitals. This does not necessarily mean that we fill them in order simply based on which electron shell they are in, ie. 1, 2, 3, then 4. We fill electrons according to the energy levels of the subshells. For example, the 4s subshell is lower in energy than the 3d so we fill the 4s before the 3d. Despite the third electron shell not being completely filled, the only thing that matters are the relative energy levels of the subshells. The actual order we fill subshells in follows something called the diagonal rule in chemistry which you can look up.

Hope that helps.

Hope that helps.

Question 6

Why does argon's electron configuration have [Ne] but neon's electron configuration not have [He]?

Accepted Answer

Anshul, the reason why Neon's electron configuration does *not* have `[He]` is most likely because its configuration is so short that it can be easily written and read without using the shorthand version.

Once you get to elements further down in the periodic table, that's when using the Noble gas notation as a shorthand makes more sense (like with Argon - as Sal does in the video).

You could still write the `[He]` shorthand for Neon => `Ne: [He] 2s2 2p6` ( if you wanted to show only its valence electrons for example ), and it would be correct as well, but it's not necessary.

I hope this answered your question. :-)

Question 7

At 5:22, why does it go to 3d and then 4s? It doesn't follow the diagonal.

Accepted Answer

The issue with the energy levels of the subshells is that they can differ from element to element. The diagonal pattern assumes we’re adding electrons to an element with less protons like calcium. An element like scandium has more electrons and this effects the effective nuclear charge felt by the electrons and the relative energy levels of the 4s and 3d subshells.

For an element like calcium, the diagonal rule applies and the 4s subshell is filled before the 3d subshell because the 4s is lower in energy. But for scandium, the 3d subshell is now lower in energy than the 4s so it is filled first. It also explains why scandium cations involve the removal of 4s electrons first since higher energy electrons are removed before lower energy ones.

The aufbau principal has its fair share of exceptions from the expected rule. Learning chemistry involves learning general rules and the exceptions to those rules.

Hope that helps.

Question 8

Hi! So only eight electrons are needed in every energy level to make it stable, but these energy levels are able to hold more electrons? Thanks!

Accepted Answer

So the eight electron idea you're referring to is known as the octet rule which is a general rule that says certain elements like having a valence shell with eight electrons to be stable.   The valence shell being the shell with electrons which participate in chemical reactions and is often the outermost electron shell. The octet rule applies best to second period elements (second row on the periodic table). The second electron shell holds a maximum of eight electrons and its filling creates a stability which elements like to attain.

The rest of the periodic table is full of exceptions to this octet rule. The simplest being the first period elements (hydrogen and helium). They only really make use of the first electron shell which holds a maximum of two electrons. So their valence shell is considered full and stable with just two electrons and so follow what is known as the duet rule.

The third period elements follow the octet for the most part, but you'll find plenty of examples of elements in that row exceeding eight valence electrons. Chlorine in the chlorate ion for example has twelve valence electrons. Elements in the second period don't exceed that eight valance electron limit because their valence shell only holds a maximum of eight electrons. However elements in the third period have an electron shell which can potentially accommodate eighteen electrons. So third period elements would like to follow the octet rule, but they're not limited by it like the second period elements are.

Hope that helps.

Question 9

I am super confused. Aufbau principle is confusing. Why don't we fill 3d? I'd imagine it would be 1s - 2s2p - 3s3p3d - 4s4p4d4f. But instead it is 1s - 2s2p - 3s3p - 4s4p3d (3d4s4p). What is the matter? What is the actual pattern then? Does it mean if we could keep going longer and longer, eventually we could fill up the 1p? So many questions...
Oh, and one more question - does it mean that 3d orbital is higher on energy than 4s orbital? I'd assume yes.

Accepted Answer

The aufbau principal says that we fill orbitals first with lowest energy and build up from there, increasing in energy. So the order we fill orbitals in goes according to the orbital’s energy relative to the other orbitals. We fill the 1s first because it is lowest in energy, then the 2s because it is next higher in energy, and then the 2p is even higher, and so on. The order does not simply follow the electron shell number which means we often backfill lower shells with lower energy subshells, like filling the 4s before the 3d. And we do this again before the 4s subshell is lower in energy than the 3d subshell.

There wouldn’t be a 1p subshells because the first electron shell only contains a 1s subshell.

Hope that helps.

There wouldn’t be a 1p subshells because the first electron shell only contains a 1s subshell.

Hope that helps.

Question 10

Does it matter how your write the configuration?
For example: Sc:[Ag]3d1 4s2 ; can it also be written as Sc: [Ag]4s2 3d1 ?

Accepted Answer

Yes, it actually matters, the reason being that in the case of Scandium, the 4s2 is written afterwards as those electron as Scandium has three valence electrons. The aufbau principle tells us they should add to the lowest available energy sublevel. Therefore, the first electron goes into 3d.

However, it turns out to be more favorable for the second to enter 4s. This is because when there are 20 protons or more, the 3d orbitals are more compact than 4s. The higher repulsion energy of a second electron entering 3d makes it unfavorable compared with the second electron entering 4s. Likewise, the third electron goes into 4s because it is the lowest cost energy option.

Course: AP®︎/College Chemistry > Unit 1

The Aufbau principle

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Video transcript