The Student Room Group
A normal Fe atom has an electron arrangement of:

1s2 2s2 2p6 3s2 3p6 4s2 3d6

You're removing 2 electrons from it to generate the Fe2+ ion, which are removed from the 4s orbital first (this is always the case in transition chemistry - as far as A Level is concerned)

So the correct arrangement is

1s2 2s2 2p6 3s2 3p6 3d6.
TwilightKnight
A normal Fe atom has an electron arrangement of:

1s2 2s2 2p6 3s2 3p6 4s2 3d6

You're removing 2 electrons from it to generate the Fe2+ ion, which are removed from the 4s orbital first (this is always the case in transition chemistry - as far as A Level is concerned)

So the correct arrangement is

1s2 2s2 2p6 3s2 3p6 3d6.


ah okay, thanks :smile:

I simply realised that it lost two electrons, so did the configuration of Chromium.
prince_of_paupers
ah okay, thanks :smile:

I simply realised that it lost two electrons, so did the configuration of Chromium.


The reason Chromium and Copper have that weird half filled s orbital is because it is the most stable arrangement for the number of protons located in the nucleus.

Always take the electrons from the S orbital first.
Your booklet say right for fe² E.C. is correct acc. To your bookletBecause first e- removed from 3d6 but after removing 1 e- it is half filled
Reply 5
I thinks it 1s2 2s2 2p6 3s2 3p6 3d4 4s2 am not sure in this answer and please respond
Original post by TwilightKnight
The reason Chromium and Copper have that weird half filled s orbital is because it is the most stable arrangement for the number of protons located in the nucleus.

A half-filled or completely filled subshell results in a stability benefit for an atom.

Cr has [Ar] 4s13d5 meaning both 4s and 3d subshells are half-full.

Cu has [Ar] 4s13d10 meaning 4s subshell is half-full and 3d subshell is completely full.

Both of these electron configurations maximise stability for the atoms. It's not related to number of protons.
Reply 7
Original post by ChemistryWebsite
A half-filled or completely filled subshell results in a stability benefit for an atom.

Cr has [Ar] 4s13d5 meaning both 4s and 3d subshells are half-full.

Cu has [Ar] 4s13d10 meaning 4s subshell is half-full and 3d subshell is completely full.

Both of these electron configurations maximise stability for the atoms. It's not related to number of protons.

Apparently not: http://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html

And anyway, this is a nine year-old thread. Why hasn't anyone point that out? Let the dead lie.
Original post by Pigster
Apparently not: http://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html

And anyway, this is a nine year-old thread. Why hasn't anyone point that out? Let the dead lie.

Out of curiosity, are you really teaching that blog to your students?

The only specific mention of the reason for the exceptional electron configurations in a specification is here in the SQA Advanced Highers spec.https://www.sqa.org.uk/files_ccc/AHChemistryCourseSpec.pdf
This includes on page 10 "There is a special stability associated with half-filled and full subshells."

Other specs seem to shy away from being specific.
(I deal with Highers / Advanced Highers students as well as A level - I have no idea what level the OP or the recent respondent are studying, of course. None of us has.)




Original post by Manooha
I thinks it 1s2 2s2 2p6 3s2 3p6 3d4 4s2 am not sure in this answer and please respond

No, that's incorrect. The 4s electrons are the first to be lost from transition metals when oxidised.
There aren't 4s electrons in the Fe2+ ion, but there are six 3d electrons.
Reply 9
use periodic table

1. drop both 4s elestrons on the left side
2. count the number of 3d electrons til you reach Fe

did you study at all?
Original post by gaia0619
use periodic table

1. drop both 4s elestrons on the left side
2. count the number of 3d electrons til you reach Fe

did you study at all?

Does your method always work? i.e how do you explain Cr and Cu?
for Cr and Cu, you need to start from 3d5 and 4s1 / 3d10 and 4s1


you know why
Original post by gaia0619
did you study at all?

You've got to stop being so condescending. Not everyone has the same ability as you do.

If someone posts a question, they should show that they've had a go at it at first - but shaming people isn't helpful to anyone.
Original post by gaia0619
for Cr and Cu, you need to start from 3d5 and 4s1 / 3d10 and 4s1


you know why

Humour me... Why?

And not "because it works."

And not "because a half-filled or fully-filled sub-shell is more stable." Because why isn't V [Ar] 3d5 rather than [Ar] 3d3 4s2? Or why does Cu more happily form Cu2+ ([Ar] 3d9) over Cu+ ([Ar] 3d10) if a fully-filled sub-shell is so stable?

Why is your rule fill the 4s sub-shell first and then the 3d subshell, except when you need to fill the 3d first and then the 4s. Why?
dun you have something more interesting to work on?

whem empty, energy level: 4s lower than 3d

when occupied, energy level: 3d lpwer than 4s
Original post by Pigster
Humour me... Why?

And not "because it works."

And not "because a half-filled or fully-filled sub-shell is more stable." Because why isn't V [Ar] 3d5 rather than [Ar] 3d3 4s2? Or why does Cu more happily form Cu2+ ([Ar] 3d9) over Cu+ ([Ar] 3d10) if a fully-filled sub-shell is so stable?

Why is your rule fill the 4s sub-shell first and then the 3d subshell, except when you need to fill the 3d first and then the 4s. Why?

dun you have something more interesting to work on?

whem empty, energy level: 4s lower than 3d

when occupied, energy level: 3d lpwer than 4s
Original post by gaia0619
use periodic table

1. drop both 4s elestrons on the left side
2. count the number of 3d electrons til you reach Fe

did you study at all?


Original post by gaia0619
dun you have something more interesting to work on?

whem empty, energy level: 4s lower than 3d

when occupied, energy level: 3d lpwer than 4s

I've been on holiday, hence my week delay replying.

Don't your two pieces of advice contradict each other?

Aren't you advocating 4s first in your first post and 3d first in your second?

And if, in your second, you state that 3d are lower (my addition: and hence fill first), why is (for example) V: [Ar] 3d3 4s2? and not [Ar] 3d5? Do you have a rule?