Ionisation enrgy Watch

nisha.sri
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I was doing a paper and it said 'state all the ionisation numbers that are involved in a removal of an electron from an S subshell' and they gave you loads of figures in table. How would you work it out ?
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jazz_xox_
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What exam board is this?
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nisha.sri
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(Original post by jazz_xox_)
What exam board is this?
OCR
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jazz_xox_
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(Original post by nisha.sri)
OCR
Oh I'm on AQA but never seen a question like this!
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nisha.sri
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(Original post by jazz_xox_)
Oh I'm on AQA but never seen a question like this!
Yh i really dont know what to do
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jazz_xox_
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(Original post by nisha.sri)
Yh i really dont know what to do
Could you post a pic of the question?
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Biomed1018
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(Original post by jazz_xox_)
Could you post a pic of the question?
^^
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nisha.sri
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nisha.sri
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(Original post by jazz_xox_)
Could you post a pic of the question?
(Original post by Biomed1018)
^^
Here you go
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Tasnim:)
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I would say write out the electronic configuration of aluminium and see if you can use it with the ionisation energy table given if that makes sense?
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RickHendricks
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(Original post by nisha.sri)
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u gonna need periodic table for this:

look at where aluminium is in the periodic table, and it's gonna tell u its electronic configuration, which is

1s2, 2s2, 2p6, 3s2 3p1

during ionisation electrons are removed from the outermost shell, meaning that from this information, the first electron is going to be removed from a P orbital, then the next two are going to be removed from s orbita (so 2 so far) and the next 6 from a p orbital, and the next 2 from an S orbital.

Since the table cares about the first 11 electrons, go down in the configuration:

first ionisation: P orbital
Second ionisation: S orbital
Third ionisation: S orbital
Fourth ionisation: P orbital
5th ionisation: P orbital
6th ionisation P orbital
7th ionisation: P orbital
8th ionisation: P orbital
9th ionisation: P orbital
10th ionisation: S orbital
11th ionisation: S orbital

from this list, there are exactly 4 electrons removed from an S orbital
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nisha.sri
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(Original post by RickHendricks)
u gonna need periodic table for this:

look at where aluminium is in the periodic table, and it's gonna tell u its electronic configuration, which is

1s2, 2s2, 2p6, 3s2 3p1

during ionisation electrons are removed from the outermost shell, meaning that from this information, the first electron is going to be removed from a P orbital, then the next two are going to be removed from s orbita (so 2 so far) and the next 6 from a p orbital, and the next 2 from an S orbital.

Since the table cares about the first 11 electrons, go down in the configuration:

first ionisation: P orbital
Second ionisation: S orbital
Third ionisation: S orbital
Fourth ionisation: P orbital
5th ionisation: P orbital
6th ionisation P orbital
7th ionisation: P orbital
8th ionisation: P orbital
9th ionisation: P orbital
10th ionisation: S orbital
11th ionisation: S orbital

from this list, there are exactly 4 electrons removed from an S orbital
OMG thank you so much ! Stupid me didnt even read the question properly i didnt even see the word aluminium
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RickHendricks
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(Original post by nisha.sri)
OMG thank you so much ! Stupid me didnt even read the question properly i didnt even see the word aluminium
*stares into the wall*
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Rohit_Rocks10
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(Original post by nisha.sri)
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The electron distribution for aluminium is 1S2 2S2 2P6 3S2 3P1

From the above electronic configuration you can see that

1st ionisation number involves removal of P electron.


2nd ionisation number involves removal of S electron.


3rd ionisation number involves removal of S electron.


4th ionisation number involves removal of P electron.


5th ionisation number involves removal of P electron.


6th ionisation number involves removal of P electron.


7th ionisation number involves removal of P electron.


8th ionisation number involves removal of P electron.


9th ionisation number involves removal of P electron.


10th ionisation number involves removal of S electron.


11th ionisation number involves removal of S electron.
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nisha.sri
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(Original post by Rohit_Rocks10)
The electron distribution for aluminium is 1S2 2S2 2P6 3S2 3P1

From the above electronic configuration you can see that

1st ionisation number involves removal of P electron.


2nd ionisation number involves removal of S electron.


3rd ionisation number involves removal of S electron.


4th ionisation number involves removal of P electron.


5th ionisation number involves removal of P electron.


6th ionisation number involves removal of P electron.


7th ionisation number involves removal of P electron.


8th ionisation number involves removal of P electron.


9th ionisation number involves removal of P electron.


10th ionisation number involves removal of S electron.


11th ionisation number involves removal of S electron.
Thankss yh i didnt read the question properly i didnt even see aluminium but thank you anyways
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Rohit_Rocks10
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(Original post by nisha.sri)
Thankss yh i didnt read the question properly i didnt even see aluminium but thank you anyways
Ohh, ahah! Anytime!
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Anonymouspsych
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(Original post by nisha.sri)
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Oh that's quite easy and I do AQA too. When you posted the question initially without the picture I didn't know what the question was on about xD. So think about the electronic configuration of aluminium so it would be 1s2 2s2 2p6 3s2 3p1
You know the 1st ionisation energy number will refer to the outermost electron being removed so in this case the 3p1 electron.
You just work downwards until you get the the s subshell so the answer to di) would be 2nd,3rd,10th and 11th
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nisha.sri
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(Original post by Anonymouspsych)
Oh that's quite easy and I do AQA too. When you posted the question initially without the picture I didn't know what the question was on about xD. So think about the electronic configuration of aluminium so it would be 1s2 2s2 2p6 3s2 3p1
You know the 1st ionisation energy number will refer to the outermost electron being removed so in this case the 3p1 electron.
You just work downwards until you get the the s subshell so the answer to di) would be 2nd,3rd,10th and 11th
Yh i didnt read the question properly i didnt even see the al lool
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