do half-filled s orbitals offer stability?

[aviary17]

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I thought Fe3+ would take on the configuartion [Ar]4s1 3d5 because then each subshell would be half-full. However, the answer was that the electrons wold be taken off in order of highest E to lowest E, and thus it would be [Ar]3d6. I have often wondered about this topic - does a half-filled s orbital offer stability?

 

[vapremed]

Firstly, Fe3+ has 3 electron removed, obviously, so you get [Ar]3d5.

I think you may be referring to Fe2+. With that ion it would be [Ar]3d6. The best way to picture it is to just notice that n, the principle quantum number (4 in the case of 4s) is the first determinant of energy. So any electron with a N of 4 is going to have more energy than one in a 3d orbital, so that will be removed first.

 

[rocuronium]

When transition metals form ions, they lose electrons from the subshell with the highest principle quantum number first. That's why the 4s electrons are lost before the 3d electrons. I'm not sure of the exact reasoning behind this, but it seems to be the general rule.

 

[Vihsadas]

The real rule of why the lowest energy orbitals fill first is related to the number of nodes in each orbital. The order of orbital energies is given by summing the quantum numbers n and l.
The orbitals with the lower (n + l) value will be the lower energy orbital. If two orbitals have the same (n + l) value, then the orbital with the lower n value fills first.

To help remember this without having to calculate (n + l), just use the following diagram:
http://www.steve.gb.com/images/science/orbital_filling.png

Just draw diagonal lines from top right to bottom left to determine the filling order.

 

[rocuronium]

To help remember this without having to calculate (n + l), just use the following diagram:
http://www.steve.gb.com/images/science/orbital_filling.png

Just draw diagonal lines from top right to bottom left to determine the filling order.

That link doesn't seem to work.

I do know what diagram you are talking about. But according to that, wouldn't you expect to lose the 3d subshell electrons before the 4s electrons?

 

[spo01]

That link doesn't seem to work.

I do know what diagram you are talking about. But according to that, wouldn't you expect to lose the 3d subshell electrons before the 4s electrons?

When transition metals form ions, they lose electrons from the subshell with the highest principle quantum number first. That's why the 4s electrons are lost before the 3d electrons. I'm not sure of the exact reasoning behind this, but it seems to be the general rule.

Yeah it was considered one of those exceptions when I read about it. This is specifically mentioned in Examkrackers and they even wrote that it 'is probably bordering on a little too much knowledge to actually be tested'. Sorry I don't have my books with me at the moment or else I'd type up exactly what it said but it was basically what you said.

 

[rocuronium]

Yeah it was considered one of those exceptions when I read about it. This is specifically mentioned in Examkrackers and they even wrote that it 'is probably bordering on a little too much knowledge to actually be tested'. Sorry I don't have my books with me at the moment or else I'd type up exactly what it said but it was basically what you said.

Yeah I read that in EK as well. I'm curious about the reason behind it though. I find it a lot easier to remember if I know why!

 

[Vihsadas]

That link doesn't seem to work.

I do know what diagram you are talking about. But according to that, wouldn't you expect to lose the 3d subshell electrons before the 4s electrons?

Yeah you would, I didn't even think to check for that exception because I really don't think it's MCAT relevant. I'm surprised that the OP found that in a prep book.

EDIT: NVM I just saw the note from the EK book.