OmarEdExcel
Badges: 9
Rep:
?
#1
Report Thread starter 1 year ago
#1
Hello TSR,

I hope you are all alright. I have recently encountered a problem with the ionisation energies of Potassium.

Here is the question in the paper January 2017

http://prntscr.com/lww2x8

This is the mark schemes http://prntscr.com/lwvzqv

1 - What I don't understand is, why an electron is removed from 3s rather than 3p?

2. Why from 2p rather than 3s?

3. Why from 1s rather than 2p?

Isn't the electronic configuration of Potassium using subshell notation as follows:

1s2, 2s2, 2p6, 3s2, 3p6, 4s1

And since it is, isn't it supposed that the electrons are knocked out horizontally, from right to left? Meaning that 4s will be knocked out before 3p, 3p before 3s, 3s before 2p and 2s before 1s?

And by the way, if the subshells are arranged in increasing order of ionisation energy ( downwards ) as follows :

1s,
2s, 2p
3s, 3p, 3d
4s, 4p, 4d , 4f
5s, 5p, 5d, 5f
6s, 6p, 6d,
7s

Why the electronic configuration only includes s and p notation without any of the d and f ? For example, why potassium isn't :

1s2, 2s2, 2p6, 3s2, 3p6, 3d1?

Thank you in advance, your cooperation will be highly appreciated.

Best Regards,
OmarEdExcel.
Last edited by OmarEdExcel; 1 year ago
0
reply
BobbJo
Badges: 12
Rep:
?
#2
Report 1 year ago
#2
(Original post by OmarEdExcel)
Hello TSR,

I hope you are all alright. I have recently encountered a problem with the ionisation energies of Potassium.

Here is the question in the paper January 2017

http://prntscr.com/lww2x8

This is the mark schemes http://prntscr.com/lwvzqv

1 - What I don't understand is, why an electron is removed from 3s rather than 3p? 2. Why from 2p rather than 3s? 3. Why from 1s rather than 2p?
Questions are related and are answered together:

It says first the 4s electron is removed, then the 3p electrons are removed, then 3s, then 2p, then 2s, then 1s.
Once all 4s electrons are removed, 3p electrons are removed. 3p is found in inner shell and is closer to the nucleus. So much more energy is needed. This explains the rise.

Once all 3p electrons are removed, 3s electrons are removed. This is still in the same shell so not a big rise. When all 3s electrons are removed, 2p electrons are removed. This is found in an inner shell. Same logic for last.


Isn't the electronic configuration of Potassium using subshell notation as follows:

1s2, 2s2, 2p6, 3s2, 3p6, 4s1
Yes

And since it is, isn't it supposed that the electrons are knocked out horizontally, from right to left? Meaning that 4s will be knocked out before 3p, 3p before 3s, 3s before 2p and 2s before 1s?
Highest energy electron is removed first so yes order of removal is 4s -> 3p -> 3s -> 2p -> 2s -> 1s

And by the way, if the subshells are arranged in increasing order of ionisation energy ( downwards ) as follows :

1s,
2s, 2p
3s, 3p, 3d
4s, 4p, 4d , 4f
5s, 5p, 5d, 5f
6s, 6p, 6d,
7s

Why the electronic configuration only includes s and p notation without any of the d and f ? For example, why potassium isn't :
There is a principle called "Aufbau rule" which helps you write electronic configuration. It involves drawing diagonal lines. e.g diagonal 1 : 1s, diagonal 2: 2s, diagonal 3: 2p, 3s, diagonal 4: 3p 4s, diagonal 5: 3d, 4p, 5s
so order of filling is 1s 2s 2p 3s 3p 4s then 3d 4p 5s...

The lowest energy orbitals are always filled first. So 3d are filled after 4s.
0
reply
OmarEdExcel
Badges: 9
Rep:
?
#3
Report Thread starter 1 year ago
#3
Okay, step by step.

In the mark scheme, in the 2nd bullet point, it says that 3p electron is removed rather than 4s in the first sharp increase. Why?

Considering the electronic configuration :

1s2, 2s2, 2p6, 3s2, 3p6, 4s1

and that electrons are removed horizontally from right to left :

<------------------------------

Why it says by word in the mark scheme " the first sharp increase is when a 3p electron is removed for the first time / rather thna a 4s electron " ?

Thank you for your commitment to replies.
0
reply
BobbJo
Badges: 12
Rep:
?
#4
Report 1 year ago
#4
(Original post by OmarEdExcel)
Okay, step by step.

In the mark scheme, in the 2nd bullet point, it says that 3p electron is removed rather than 4s in the first sharp increase. Why?

Considering the electronic configuration :

1s2, 2s2, 2p6, 3s2, 3p6, 4s1

and that electrons are removed horizontally from right to left :

<------------------------------

Why it says by word in the mark scheme " the first sharp increase is when a 3p electron is removed for the first time / rather thna a 4s electron " ?

Thank you for your commitment to replies.
The sequence of removal is 4s first then 3p as you said

When 4s electron is removed, the next electron to be removed is a 3p electron. This 3p electron, rather than 4s, is found in an inner shell and there is a sharp rise in I.E
0
reply
OmarEdExcel
Badges: 9
Rep:
?
#5
Report Thread starter 1 year ago
#5
With that being said, that means that it is an inappropriate statement done by the mark schemes. If 4s is removed before 3p, then they shouldn't have said 4s is removed rather than 3p. it is this word rather that is causing a problem to me. But, they did so to explain the sharp increase and not the order, I get it now. I also find that a decent explanation for the sharp increases is that each an every time there is a sharp increase we notice that we are entering another subshell, and the ionisation energies increase as the subshells are gradually increasing in terms of how close they are to the nucleus.

Nevertheless, I am very thankful overall once again BobbJo. By the way, is there any way by any means of possibility I can contact you outside TSR for help? Thank you in advance.
Last edited by OmarEdExcel; 1 year ago
0
reply
BobbJo
Badges: 12
Rep:
?
#6
Report 1 year ago
#6
(Original post by OmarEdExcel)
With that being said, that means that it is an inappropriate statement done by the mark schemes. If 4s is removed before 3p, then they shouldn't have said 4s is removed rather than 3p. But, they did so to explain the sharp increase and not the order, I get it now. I also find that a decent explanation for the sharp increases is that each an every time there is a sharp increase we notice that we are entering another subshell, and the ionisation energies increase as the subshells are gradually increasing in terms of how close they are to the nucleus.

Nevertheless, I am very thankful overall once again BobbJo.
It is not inappropriate. You are having a misunderstanding

"The first sharp increase is when a 3p electron is removed rather than a 4s electron" is what ms says

You are reading "4s removed rather than 3p"

Potassium is 1s2 2s2 2p6 3s2 3p6 4s1
They are removing the 4s electron first

we are left with
1s2 2s2 2p6 3s2 3p6

a 3p electron is removed now rather than 4s
1
reply
OmarEdExcel
Badges: 9
Rep:
?
#7
Report Thread starter 1 year ago
#7
Exactly, that sums it up, and you were right about the way I read it. It is the removal of 3p that is causing the sharp increase. Thank you once again.
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

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (173)
14.56%
I'm not sure (54)
4.55%
No, I'm going to stick it out for now (347)
29.21%
I have already dropped out (35)
2.95%
I'm not a current university student (579)
48.74%

Watched Threads

View All