Orbital Electrons.. energies and what comes off first

[SuperSaiyan3]

So say you are comparing the electron structure of CHROMIUM (Cr).

1. would 3d orbital electrons be of higher energy than its 4s electrons? Technically, the 4 (principal number = n) would be higher, would it not?
2. 4s orbital fills up before 3d orbital electrons.
3. then why would an electron from a 4s orbital electron get pulled off FIRST before a 3d electron orbital??

dazed & confused.

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[thebillsfan]

yeah I was JUST going to post this ?. haha

Well I know that 4s is farther away from the nucleus because its in the n=4 subshell. 3d is higher in energy since it fills up after 4s. well i guess i just answered my own question!

 

[thebillsfan]

but I just did a TPR problem that says: "the electron furthest from nucleus is the one w/ highest energy. since the electron configuration of Ni is (Ar)4s^23d^8, the electron in question is in the d sublevel.

WHAT?

There's an inconsistency...in this case, the electrfon furthest fromthe nucleus is NOT the one with the highest energy, correct? Is TPR wrong on this one?

 

[SuperSaiyan3]

yeah I was JUST going to post this ?. haha

Well I know that 4s is farther away from the nucleus because its in the n=4 subshell. 3d is higher in energy since it fills up after 4s. well i guess i just answered my own question!

so you conclude that the 4s electron will be pulled off first eh?

 

[Omni]

so you conclude that the 4s electron will be pulled off first eh?

Here's how the electrons fill.

1 s
2 s p
3 s p d
4 s p d f
5 s p d f g

Take your pencil and put it on the first s. Drag a line and take it through any letter that appears 1 down + 1 left. Once you reach the number, pick up your pencil and start on the next letter and draw the line in the similar way. Keep doing that until you've gotten down everything. The order that the orbital letters get crossed through the line is the order they fill up.
When you draw the lines, you should get this order:

1s 2sp 3sp 4s 3d 4p 5s etc. etc. until you get to the end.
Another way to see which order it goes in is by using the following equation: n+l
where n is the principle quantum number and l is the azimuthal quantum number (s=0, p=1, d=2, etc. etc.)

so, lets saw we want to know which orbital fills up first, 5s or 4d. Using the n+l equation, we get that 5s=5+0=5 and 4d=4+2=6. The orbital with the lowest sum gets filled first, so 5s gets filled first since 5<6.
But in another situation, lets saw we are comparing 4s and 3p. Both of them equal 4, so which one gets filled first? In that case, the one with the lowest quantum number (the lowest n) gets filled first. In this case, 3p will fill first.
In the end, drawing the chart is easier.

So now lets look at this configuration: 1s 2sp 3sp 4s 3d 4p 5s
In which order are the electrons stripped? After you get your configuration, put everything in order. so 1s 2sp 3sp 4s 3d 4p 5s will be
1s 2sp 3spd 4sp 5s. Electrons will be stripped in this order, first from 5s, then from 4p, then 4s, all the way to the end.

Hope this helps!

 

[wanderer]

So say you are comparing the electron structure of CHROMIUM (Cr).

1. would 3d orbital electrons be of higher energy than its 4s electrons? Technically, the 4 (principal number = n) would be higher, would it not?
2. 4s orbital fills up before 3d orbital electrons.
3. then why would an electron from a 4s orbital electron get pulled off FIRST before a 3d electron orbital??

dazed & confused.

Moving up in atomic number is not simply reverse of ionizing metals of higher atomic number.
For example when you go from Sc to V you are adding 2 protons and 2 electrons, but when you take away the two electrons from V, you still have those two extra protons, so it doesn't necessarily follow that because you added 3d after 4s were added that you'd take away the 3d first when you're ionizing. I'm guessing the real explanation is more advanced than anything you'd learn in Chem 1.

 

[Schemp]

I think it depends on what format the thing you're looking at chooses to use--there doesn't seem to be a consensus. Sometimes they write 3d before 4s (but after 3p), sometimes 3d comes after 4s.

From what I understand, the 3d and 4s orbitals are degenerate, or near enough as makes little difference. In other words, they're the same energy. That's why when you fill up the orbitals for Chromium, which has 6 valence electrons, you get 4s^1 3d^5. It puts one in the 4s, then one in each empty 3d orbital, then it will start pairing them.

Long story short, they're basically the same energy and a test question should only give you a correct answer that has one form or the other, not both.

Hope that helps.

 

[thebillsfan]

I think it depends on what format the thing you're looking at chooses to use--there doesn't seem to be a consensus. Sometimes they write 3d before 4s (but after 3p), sometimes 3d comes after 4s.

From what I understand, the 3d and 4s orbitals are degenerate, or near enough as makes little difference. In other words, they're the same energy. That's why when you fill up the orbitals for Chromium, which has 6 valence electrons, you get 4s^1 3d^5. It puts one in the 4s, then one in each empty 3d orbital, then it will start pairing them.

Long story short, they're basically the same energy and a test question should only give you a correct answer that has one form or the other, not both.

Hope that helps.

So, to confirm, then, elemental chromium and Cu^2+ both DO have a 4s13d5 electron config? In the case of copper, the two 4s electrons come off and then one of the 5d electrons goes into the s shell to fill it?

 

[SuperSaiyan3]

Here's how the electrons fill.

1 s
2 s p
3 s p d
4 s p d f
5 s p d f g

Take your pencil and put it on the first s. Drag a line and take it through any letter that appears 1 down + 1 left. Once you reach the number, pick up your pencil and start on the next letter and draw the line in the similar way. Keep doing that until you've gotten down everything. The order that the orbital letters get crossed through the line is the order they fill up.
When you draw the lines, you should get this order:

1s 2sp 3sp 4s 3d 4p 5s etc. etc. until you get to the end.
Another way to see which order it goes in is by using the following equation: n+l
where n is the principle quantum number and l is the azimuthal quantum number (s=0, p=1, d=2, etc. etc.)

so, lets saw we want to know which orbital fills up first, 5s or 4d. Using the n+l equation, we get that 5s=5+0=5 and 4d=4+2=6. The orbital with the lowest sum gets filled first, so 5s gets filled first since 5<6.
But in another situation, lets saw we are comparing 4s and 3p. Both of them equal 4, so which one gets filled first? In that case, the one with the lowest quantum number (the lowest n) gets filled first. In this case, 3p will fill first.
In the end, drawing the chart is easier.

So now lets look at this configuration: 1s 2sp 3sp 4s 3d 4p 5s
In which order are the electrons stripped? After you get your configuration, put everything in order. so 1s 2sp 3sp 4s 3d 4p 5s will be
1s 2sp 3spd 4sp 5s. Electrons will be stripped in this order, first from 5s, then from 4p, then 4s, all the way to the end.

Hope this helps!

It did help. Thank you so much. Now I do not have to memorize that mess of a zig zag line that kaplan gave me.

Great explanation.

 

[Schemp]

So, to confirm, then, elemental chromium and Cu^2+ both DO have a 4s13d5 electron config? In the case of copper, the two 4s electrons come off and then one of the 5d electrons goes into the s shell to fill it?

Do you mean Fe^2+, and not Cu^2+? Elemental Cr and Fe^2+ would have the same electron configuration, and yes, I believe it would be 4s1 3d5.

I'm not sure what you mean by the copper electrons coming off. There shouldn't be anything in the 5d orbitals. Maybe you meant 3d. If so, you're a little off I think. Two electrons wouldn't come out of one orbital. If you had at least 6 electrons, each of the 5 3d orbitals and the 4s orbital will each only have one electron. If you have 7 electrons, however, one of these 6 orbitals will have to have two electrons, each with opposite spin. In that case one of those paired electrons would be removed first.

I know it's kind of confusing cause there seems to be no consensus on whether 4s or 3d comes first, but again, only one form should appear with the correct answer. They wouldn't ask you the electron configuration of Cr and give both 3d5-4s1 and 4s1-3d5 as possible answers.

I guess the takeaway point is always remove from paired orbitals first, then lone electron orbitals when no paired ones are left.