Watermelon11
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On a mark scheme it says that the correct electron configuration for Cu+ is [Ar]3d10. It was a multiple choice question and I ticked off [Ar]4s13d9. I’m so confused as to why the mark scheme gives the other answer, can someone please explain why?
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mollyhowarth
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When you get to 3d it like to have either a full shell (10) or half shell (5). When it comes to transition metals especially you tend to take electrons from 4s to to make 3D full/ half full. Cu+ will have one less electron so it will be taken from 4s first if 3d is full/ half full otherwise it is removed from 3d. Hope this helps. Ask your teacher if you still really don’t understand
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sam72016
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(Original post by Watermelon11)
On a mark scheme it says that the correct electron configuration for Cu+ is [Ar]3d10. It was a multiple choice question and I ticked off [Ar]4s13d9. I’m so confused as to why the mark scheme gives the other answer, can someone please explain why?
3d shell fills before 4s shell. So in order to start filling up the 4s shell, the 3d shell has to be full up.

D shell can hold 10 electrons max. 4s shell can hold 2 electrons max.

From your answer of 4s1 3d9, there is space for 1 more electron in the 3d shell (you've filled up 9 instead of 10), so the electron from 4s1 moves to 3d9, so it becomes 4s0 and 3d10 (so just 3d10 basically).
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Watermelon11
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(Original post by sam72016)
3d shell fills before 4s shell. So in order to start filling up the 4s shell, the 3d shell has to be full up.

D shell can hold 10 electrons max. 4s shell can hold 2 electrons max.

From your answer of 4s1 3d9, there is space for 1 more electron in the 3d shell (you've filled up 9 instead of 10), so the electron from 4s1 moves to 3d9, so it becomes 4s0 and 3d10 (so just 3d10 basically).
Thank you!!!
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User135792468
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(Original post by Watermelon11)
On a mark scheme it says that the correct electron configuration for Cu+ is [Ar]3d10. It was a multiple choice question and I ticked off [Ar]4s13d9. I’m so confused as to why the mark scheme gives the other answer, can someone please explain why?
Transition metals fill 4s first then 3d
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Huckipity
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(Original post by sam72016)
3d shell fills before 4s shell. So in order to start filling up the 4s shell, the 3d shell has to be full up.

D shell can hold 10 electrons max. 4s shell can hold 2 electrons max.

From your answer of 4s1 3d9, there is space for 1 more electron in the 3d shell (you've filled up 9 instead of 10), so the electron from 4s1 moves to 3d9, so it becomes 4s0 and 3d10 (so just 3d10 basically).
This is completely wrong. The 4s orbitals always fills up first, and are also removed first when forming an ion. However, there are exceptions to this rule (Copper and Chromium) and we need to be aware of them for the exam. The electron configuration of a copper atom is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10 due to it being more stable this way as one of the 4s electrons jumps to the 3d orbital. Usually it's the other way around (the 4s would fill first) but this is an exception. So if we remove an electron from a copper atom, knowing that the 4s is removed first over the 3d (as it has more energy) we get 1s2, 2s2, 2p6, 3s2, 3p6, 3d10 which is our answer.

(Original post by Watermelon11)
Thank you!!!
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R T
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Huckipity is correct.

This is an infamous "learn but dont learn why" part of a level chemistry. The reason for why this happens is very complicated and not as simple as "half-filled orbitals minimise repulsions" (textbooks say this a lot, it is incorrect, but if your exam board's textbook say this then learn that as the answer). If you want to understand the real answer, wait until your 3rd year chemistry degree / 2nd year physics degree (maybe).

In summary for the d block in period 3:

Sc - 3d1 4s2
Ti - 3d2 4s2
V - 3d3 4s2
Cr - 3d5 4s1
Mn - 3d5 4s2
Fe - 3d6 4s2
Co - 3d7 4s2
Ni - 3d8 4s2
Cu - 3d10 4s1
Zn - 3d10 4s2

Basic rule: for elements you wont get 3d4 or 3d9. If you get 3d4 or 3d9. If you get one of these something went wrong.

for ion formation (e.g. V+, Fe2+), always remove the electron from the 4s orbital as Huckipity rightly points out.
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Pigster
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(Original post by Huckipity)
...The 4s orbitals always fills up first...
Actually, 3d does start to fill before 4s in the d block.
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Huckipity
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(Original post by Pigster)
Actually, 3d does start to fill before 4s in the d block.
Yes I am aware of that it's not strictly true, but we don't need to know it for the AQA exams. Stop trying to inflict confusion on people with knowledge we do not need to know, as we will not be credited for it in the exam, and people will get it mixed up . For what we need to know 4s is always first in and out.
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Pigster
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(Original post by Huckipity)
For what we need to know 4s is always first in and out.
In that case, I am curious to hear how and why for Cr and Cu, it goes 4s first in, but gets bored half way through filling the orbital, so decides to add to the 3d? What is so special about 3d5 or 3d10? Why are V not 3d5 4s0 and Ni 3d10 4s0 if a half-filled or filled shells so stable?

Why can't the rule be "4s first out and its a bit more complex than a simple rule can explain things for which are first in"?

Why teach a 'rule' that is wrong?
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Huckipity
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(Original post by Pigster)
In that case, I am curious to hear how and why for Cr and Cu, it goes 4s first in, but gets bored half way through filling the orbital, so decides to add to the 3d? What is so special about 3d5 or 3d10? Why are V not 3d5 4s0 and Ni 3d10 4s0 if a half-filled or filled shells so stable?

Why can't the rule be "4s first out and its a bit more complex than a simple rule can explain things for which are first in"?

Why teach a 'rule' that is wrong?
I'm by no means doing a chemistry degree so I could be wrong, but I'm pretty sure the 4s in first and out first rule is only correct up to Calcium, after that the 3d fills first as it's more stable or requires less energy. I think it's something to do with the energy of each orbital, and how after Calcium the 3d will have less energy than the 4s, and so the 3d fills first - I could be wrong though. As to why teach a rule that is wrong: it is simply too complex for an A level student to comprehend. I'm pretty sure the explanation relies on 1st year degree level Chemistry knowledge and is just far too in depth for a 16-18 year old.

It's like this with all subjects though, especially at gcse. Like how in gcse chemistry we were told the electron configuration is 2,8,8,8... etc but that's now a complete lie as we know about sub shells. Even at A level, we're really only skimming the surface of a subject as there's way too much to learn. We really only go into detail in organic chemistry at A level, but even then it's just skimming the surface of the topic.
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R T
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(Original post by Huckipity)
I'm by no means doing a chemistry degree so I could be wrong, but I'm pretty sure the 4s in first and out first rule is only correct up to Calcium, after that the 3d fills first as it's more stable or requires less energy. I think it's something to do with the energy of each orbital, and how after Calcium the 3d will have less energy than the 4s, and so the 3d fills first - I could be wrong though. As to why teach a rule that is wrong: it is simply too complex for an A level student to comprehend. I'm pretty sure the explanation relies on 1st year degree level Chemistry knowledge and is just far too in depth for a 16-18 year old.

It's like this with all subjects though, especially at gcse. Like how in gcse chemistry we were told the electron configuration is 2,8,8,8... etc but that's now a complete lie as we know about sub shells. Even at A level, we're really only skimming the surface of a subject as there's way too much to learn. We really only go into detail in organic chemistry at A level, but even then it's just skimming the surface of the topic.
You won't cover any of this in 1st year. This is something you'd do in 3rd year Quantum or 4th year.

I would also point out that 4s is a first-in, first-out model is probably only refuted by something like Sc oxidation to 3+. i.e. For the purposes of A-Level, since 95%+ of reactions can be assumed to use 4s first, it's fine to just say this is always true and ignore special cases.



The first point is to understand that in Chemistry, we don't actually fill orbitals by their energy order, but by their stability. In simple mathematical terms, 4s is higher in energy than 3d because 4>3.

However the 4s orbital is much more stable than the 3d when electrons are occupying lower shells. Textbooks will approximate this to say that their energy is lower, but this is only partially true and indeed this explains why 4s electrons will be removed to or added to first. The thing to take away is that - much like the analogous organic rule - an electron's availability can be considered about equal to its highest possible availability, rather than its average. Since availability is basically just how stable it is this would make sense.

But for the sake of just looking at configurations, we can assume from here on that 3d orbitals are "higher in energy", whereby we are referring to the average energy stability lost by having an electron there instead of in the 4s orbital (or any lower unoccupied orbital).


Understanding why 3d orbitals are filled in the case of 3d5 and 3d10 selectively despite the "lower energy" 4s being available involves explaining exchange energy.


Exchange energy is one of those weird QM things that an A-Level student can't understand because cyclic spin-coupling orbitals aren't taught until 2/3rd year QM. The hand waving argument is that electrons with the same spin in the same energy state with the same angular momentum are able to "swap" with each other to maximise how much space they are able to move in, and therefore minimise the average distance in their coulomb interactions with the atom.

This energy is (as expected) directly proportional to the amount of exchanges which are possible. For 3 d orbital (or p orbital) electrons, it's 2+1 relative units. For 5, it would be 4+3+2+1 (i.e. it's a lot if you can fill it). Lets call this energy drop from exchange A. And let's call the energy gap from 4s to 3d B

If we ignore other coulomb interactions and spin orbit coupling, etc. then we have a very simple energy price to pay for the case of all transition metal candidates (for more hand waving, lets just include that in B and assume we're adding the second electron to the 4s orbital:

If [number of interactions for 1 more d orbital electron] * A > B , then clearly we have a lower average energy and stability config by putting the electron in the d orbital instead of the s. Although we "pay" for the higher energy orbital, we "get back" more energy from electron exchange between the d orbitals.

Since for 1,2,3,4,5 the relative number of exchanges is 0,1,3,6,10 -- this means the gap from d0 to d1, ..., d4 to d5 is 1,2,3,4. It so happens that A ~~ 3.2*B for the 3rd orbital and so while going from d3 to d4 isn't worth it, going from d4 to d5 is worth it (enough to pay for electron promoting to the 3d from the 4s).



In another "just chemistry things" moment, I will note that this doesn't happen with the next row. So 4d and 5s, the energy gap is different (smaller), but the availability of increased space for the larger d orbitals means that the exchange energy is also much smaller. So with 6 electrons in the valence cloud here, we would have 5s2 4d4.
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Pigster
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(Original post by Huckipity)
...Like how in gcse chemistry we were told the electron configuration is 2,8,8,8...
I would add that the GCSE 2.8.8.2 rule is correct. Up to Ca (which is as far as you go at GCSE) you can fit only 2 e- in the first shell before it fills etc. Hopefully teachers says you can get 8 in the 3rd shell before it acts like it is full, rather than is full, but I can't talk for other teachers. The rule is simple and does not need sub-shells for the explanation to work.

Sub-shells are not even hinted at, but if you consider their omission as lying then surely no part of school life involves all of the facts on any matter, so therefore everything is a lie (except, obviously the cake)
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Kookieandpasta
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yh that's the right explaination
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Huckipity
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(Original post by R T)
You won't cover any of this in 1st year. This is something you'd do in 3rd year Quantum or 4th year.

I would also point out that 4s is a first-in, first-out model is probably only refuted by something like Sc oxidation to 3+. i.e. For the purposes of A-Level, since 95%+ of reactions can be assumed to use 4s first, it's fine to just say this is always true and ignore special cases.



The first point is to understand that in Chemistry, we don't actually fill orbitals by their energy order, but by their stability. In simple mathematical terms, 4s is higher in energy than 3d because 4>3.

However the 4s orbital is much more stable than the 3d when electrons are occupying lower shells. Textbooks will approximate this to say that their energy is lower, but this is only partially true and indeed this explains why 4s electrons will be removed to or added to first. The thing to take away is that - much like the analogous organic rule - an electron's availability can be considered about equal to its highest possible availability, rather than its average. Since availability is basically just how stable it is this would make sense.

But for the sake of just looking at configurations, we can assume from here on that 3d orbitals are "higher in energy", whereby we are referring to the average energy stability lost by having an electron there instead of in the 4s orbital (or any lower unoccupied orbital).


Understanding why 3d orbitals are filled in the case of 3d5 and 3d10 selectively despite the "lower energy" 4s being available involves explaining exchange energy.


Exchange energy is one of those weird QM things that an A-Level student can't understand because cyclic spin-coupling orbitals aren't taught until 2/3rd year QM. The hand waving argument is that electrons with the same spin in the same energy state with the same angular momentum are able to "swap" with each other to maximise how much space they are able to move in, and therefore minimise the average distance in their coulomb interactions with the atom.

This energy is (as expected) directly proportional to the amount of exchanges which are possible. For 3 d orbital (or p orbital) electrons, it's 2+1 relative units. For 5, it would be 4+3+2+1 (i.e. it's a lot if you can fill it). Lets call this energy drop from exchange A. And let's call the energy gap from 4s to 3d B

If we ignore other coulomb interactions and spin orbit coupling, etc. then we have a very simple energy price to pay for the case of all transition metal candidates (for more hand waving, lets just include that in B and assume we're adding the second electron to the 4s orbital:

If [number of interactions for 1 more d orbital electron] * A > B , then clearly we have a lower average energy and stability config by putting the electron in the d orbital instead of the s. Although we "pay" for the higher energy orbital, we "get back" more energy from electron exchange between the d orbitals.

Since for 1,2,3,4,5 the relative number of exchanges is 0,1,3,6,10 -- this means the gap from d0 to d1, ..., d4 to d5 is 1,2,3,4. It so happens that A ~~ 3.2*B for the 3rd orbital and so while going from d3 to d4 isn't worth it, going from d4 to d5 is worth it (enough to pay for electron promoting to the 3d from the 4s).



In another "just chemistry things" moment, I will note that this doesn't happen with the next row. So 4d and 5s, the energy gap is different (smaller), but the availability of increased space for the larger d orbitals means that the exchange energy is also much smaller. So with 6 electrons in the valence cloud here, we would have 5s2 4d4.
Really interesting. Thanks for explaining and clearing up my confusion
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