daviem
Badges: 14
Rep:
?
#1
Report Thread starter 6 years ago
#1
Why does the 4s sub-level come before the 3d sub-level, but not the 3p sub-level? And if 4s comes before 3d, why is 3s not before 2p?
Thanks
0
reply
username913907
Badges: 10
Rep:
?
#2
Report 6 years ago
#2
(Original post by daviem)
Why does the 4s sub-level come before the 3d sub-level, but not the 3p sub-level? And if 4s comes before 3d, why is 3s not before 2p?
Thanks
Short answer... Shielding.

Long answer ... What level you at? Not gonna waste time if you're not uni level
0
reply
zacharyuk
Badges: 0
Rep:
?
#3
Report 6 years ago
#3
From what I understand, the higher the principle quantum number (shell number), the higher the energy level of that shell. Within these, the different sub shells have varied energy levels, with s being the lowest, p being slightly higher and so on.

There is a gap between the most energised sub shell within the first shell (1s) and the least energised within the second shell (2s). Same thing between the second and third shells; 2p is of a lower energy value than 3s so it is filled first.

However, as the third shell is nearly filled, the energy level of 3d is so high that for some reason the energy level of 4s is lower, and so they overlap and 4s is filled first. I'm guessing it has something to do with how large the atomic radius in a d shell is and that the 4s is so much smaller, but I could be mistaken.

So basically you just have to memorise that 4s is usually filled before 3d - however there is, of course, an exception to this exception, in the cases of chromium and copper where 4s is only has one electron rather than two so that the 3d sub shell may be more stable.

Something else to remember is that the energy level of the sub shells changes once they become full, such that once they are full 4s actually is of higher energy level than 3d and so it loses its electrons first. Fe has the configuration 2,8,14,2 in order of shells, however Fe2+ has the configuration 2,8,14 - the 3d sub shell has some electrons in even though 4s does not.

I think all of the above is roughly correct and I hope it helps, but if I've gone wrong somewhere, someone please let me know!
0
reply
Pigster
Badges: 20
Rep:
?
#4
Report 6 years ago
#4
In the case of d-block elements, the d subshell (partly) fills before the s subshell, but when e- are added to the a d orbital it raises the energy level of all d orbitals, eventually making them higher in energy than the s subshell, so they start to fill.

In the case of Cr, 4 e- go into the d subshell before the d subshell is raised above the s subshell, so the 5th e- goes into the s subshell.
0
reply
Infraspecies
Badges: 16
Rep:
?
#5
Report 6 years ago
#5
(Original post by Pigster)
In the case of d-block elements, the d subshell (partly) fills before the s subshell, but when e- are added to the a d orbital it raises the energy level of all d orbitals, eventually making them higher in energy than the s subshell, so they start to fill.

In the case of Cr, 4 e- go into the d subshell before the d subshell is raised above the s subshell, so the 5th e- goes into the s subshell.
No.
Orbital energy is never a function of occupancy.
0
reply
Infraspecies
Badges: 16
Rep:
?
#6
Report 6 years ago
#6
(Original post by zacharyuk)
Something else to remember is that the energy level of the sub shells changes once they become full, such that once they are full 4s actually is of higher energy level than 3d and so it loses its electrons first. Fe has the configuration 2,8,14,2 in order of shells, however Fe2+ has the configuration 2,8,14 - the 3d sub shell has some electrons in even though 4s does not.

I think all of the above is roughly correct and I hope it helps, but if I've gone wrong somewhere, someone please let me know!
You were doing well until here, and I guess that this is an -acceptable- explanation for A-Level (by which I mean it isn't correct but it could be worse).

However.
The energy of a subshell is not a function of its occupancy. In this case, it is a function of the atomic number. 4s is filled before 3d in potassium and calcium because, for Z=19 or 20, 4s is lower energy than 3d.
For iron, however, at Z=26, 3d is now lower in energy than 4s, hence 4s empties first.
I know; it's unlikely to have an element where you would have such atomic numbers without total occupancy of 4s, so you may ask how your explanation is different. However; your explanation doesn't take into account the partial occupancy in chromium and copper, nor is it consistent with quantum mechanical approaches that determine energies of orbital wave functions.

So, orbital energies are not a function of occupancy, but of atomic number (among other things-but this is the important one to demonstrate the trend).
0
reply
username913907
Badges: 10
Rep:
?
#7
Report 6 years ago
#7
Am I gonna have to write out the long answer? I said it was shielding and everyone else has written all sorts of wacko occupancy stuff!!!
0
reply
daviem
Badges: 14
Rep:
?
#8
Report Thread starter 6 years ago
#8
(Original post by JMaydom)
Short answer... Shielding.

Long answer ... What level you at? Not gonna waste time if you're not uni level
AS level
0
reply
daviem
Badges: 14
Rep:
?
#9
Report Thread starter 6 years ago
#9
(Original post by zacharyuk)
From what I understand, the higher the principle quantum number (shell number), the higher the energy level of that shell. Within these, the different sub shells have varied energy levels, with s being the lowest, p being slightly higher and so on.

There is a gap between the most energised sub shell within the first shell (1s) and the least energised within the second shell (2s). Same thing between the second and third shells; 2p is of a lower energy value than 3s so it is filled first.

However, as the third shell is nearly filled, the energy level of 3d is so high that for some reason the energy level of 4s is lower, and so they overlap and 4s is filled first. I'm guessing it has something to do with how large the atomic radius in a d shell is and that the 4s is so much smaller, but I could be mistaken.

So basically you just have to memorise that 4s is usually filled before 3d - however there is, of course, an exception to this exception, in the cases of chromium and copper where 4s is only has one electron rather than two so that the 3d sub shell may be more stable.

Something else to remember is that the energy level of the sub shells changes once they become full, such that once they are full 4s actually is of higher energy level than 3d and so it loses its electrons first. Fe has the configuration 2,8,14,2 in order of shells, however Fe2+ has the configuration 2,8,14 - the 3d sub shell has some electrons in even though 4s does not.

I think all of the above is roughly correct and I hope it helps, but if I've gone wrong somewhere, someone please let me know!
Oh ok, that sorta makes sense, thanks
0
reply
daviem
Badges: 14
Rep:
?
#10
Report Thread starter 6 years ago
#10
Thank you for answering everyone, you all seem to have varying answers however haha
0
reply
username913907
Badges: 10
Rep:
?
#11
Report 6 years ago
#11
(Original post by daviem)
AS level
OK.... sorry to have to say this but the real answer is just a bit too advanced for you in all likelihood. It's basically down to differences in the structure of the orbitals meaning that the 4s is shielded less than the 3d.

(Original post by daviem)
Oh ok, that sorta makes sense, thanks
No, NO it doesn't..... sorry to the poster but they clearly don't understand it.
0
reply
zacharyuk
Badges: 0
Rep:
?
#12
Report 6 years ago
#12
(Original post by JMaydom)
No, NO it doesn't..... sorry to the poster but they clearly don't understand it.
In that case, would it trouble you to explain it properly so that I do?

Everything I posted was correct according to the AS textbook, although I understand that doesn't mean it is necessarily true or free of large simplifications.

Please elaborate on what actually happens so that I won't be wrong in the future.
0
reply
langlitz
Badges: 17
Rep:
?
#13
Report 6 years ago
#13
(Original post by JMaydom)
OK.... sorry to have to say this but the real answer is just a bit too advanced for you in all likelihood. It's basically down to differences in the structure of the orbitals meaning that the 4s is shielded less than the 3d.



No, NO it doesn't..... sorry to the poster but they clearly don't understand it.
I'd be interested if you'd care to give a full explanation (in the context of quantum mechanics etc). I'm second year uni
0
reply
username913907
Badges: 10
Rep:
?
#14
Report 6 years ago
#14
(Original post by zacharyuk)
In that case, would it trouble you to explain it properly so that I do?

Everything I posted was correct according to the AS textbook, although I understand that doesn't mean it is necessarily true or free of large simplifications.

Please elaborate on what actually happens so that I won't be wrong in the future.
Sorry, it's not meant to be a putdown. I certainly didn't know this at A-level.

Before I do you need to go look at radial distribution functions. They are core to explaining this at a uni level.
0
reply
username913907
Badges: 10
Rep:
?
#15
Report 6 years ago
#15
(Original post by langlitz)
I'd be interested if you'd care to give a full explanation (in the context of quantum mechanics etc). I'm second year uni
Don't think I could do it in the 'context of quantum mechanics' if that means a proper mathematical explanation. I've only ever discussed it in terms of a qualitative analysis of it - from inorganic chemistry - but the radial distribution functions all fall out of QM.

The nodes of the 4s mean that the electrons experience less shielding than the 3d's which have no nodes. In the neutral atoms of the 1st transition series the 4s is lower in energy because of this weaker shielding. We say the 4s is more penetrating.
When you ionize the atom though the electron comes from the 4s. This is because you are reducing the shielding and because the 4s is less affected by the shielding in the 1st place, the 3d drops in energy more than the 4s.

Pretty much sums it up i think - not rigorous QM but the reasoning I was taught at uni.
1
reply
Pigster
Badges: 20
Rep:
?
#16
Report 6 years ago
#16
Can you, in a similar nutshell, explain Cr?

I never did chem. beyond 1st year, but know about nodes.
0
reply
KombatWombat
Badges: 7
Rep:
?
#17
Report 6 years ago
#17
(Original post by Pigster)
Can you, in a similar nutshell, explain Cr?

I never did chem. beyond 1st year, but know about nodes.
It's a result of exchange energy. Basically for each pair of spin paired electrons you get '-K' exchange energy (K is positive, so -K is stabilising). So for 3d^5 you have -10 K exchange energy and it turns out this is more favourable than putting an electron into the lower energy 4s orbital.

This can be shown by proper quantum mechanics (it's something we did in our second year), but spin pairing is more stable because the pair of electrons can't be found in the same place, so there's lower electron-electron repulsion. (Why they can't be found in the same place takes a healthy dose of quantum mechanics).
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Feeling behind at school/college? What is the best thing your teachers could to help you catch up?

Extra compulsory independent learning activities (eg, homework tasks) (6)
4.41%
Run extra compulsory lessons or workshops (23)
16.91%
Focus on making the normal lesson time with them as high quality as possible (24)
17.65%
Focus on making the normal learning resources as high quality/accessible as possible (19)
13.97%
Provide extra optional activities, lessons and/or workshops (40)
29.41%
Assess students, decide who needs extra support and focus on these students (24)
17.65%

Watched Threads

View All