#1
Okay, this is the pit, so one has come here in search of the resident scientists

Basically one aspect of ionisation energies in chemistry that i've never got is deducing what period an element is in based off a graph such as this:



And also, a table:

Element x: IE1 736 /IE2 1450/ IE3 7740/IE4 10500/IE5 13600/IE6 18000/ IE7 21700/IE8 25600

So whats the method of counting? I've seen it briefly but not long enough to know how to apply it

#3
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#4
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i see your plan of not being helpful is proving just as apparant

And in what way did you help the thread starter?

#5
right, the electrons are in shells of 8, as you get to the next shell of 8, theyre significantly closer to the nucleus, so it has a greater effect on them, and its harder to get rid of electrons.

on the graph, you have to take 7 electrons off before theres a significant jump, so you have to take 7 electrons off before you get to the next shell so the outer shell has 7 electrons so its in period 7

hope this helps, just ask if it didnt



EDIT: for the table, the jump is after 2 ionisation energies, so its group 2.
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Last edited by stevo_epi_SG_wo at Nov 5, 2007,
#6
well at the far right you can see there are 2 results then it jumps down a bit.

Those 2 are group 1 and 2 (working right to left) filling the S Sub shell (alkaline and alkali earth metals)

It then jumps down and there are 8 here, this is the p sub shell being filled (actinoids) after this is another jump, there are 6 here, the d sub shell being filled and now your back into your groups, going right to left again 3 4 5 6 7 8, the last dot there is going back to the s shell, group 1
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#7
Quote by Storm_Bringer_
well at the far right you can see there are 2 results then it jumps down a bit.

Those 2 are group 1 and 2 (working right to left) filling the S Sub shell (alkaline and alkali earth metals)

It then jumps down and there are 8 here, this is the p sub shell being filled (actinoids) after this is another jump, there are 6 here, the d sub shell being filled and now your back into your groups, going right to left again 3 4 5 6 7 8, the last dot there is going back to the s shell, group 1


um, how do you know the graph doesnt continue after 17?

the logic is there (except 'd' sub-shells, where did that come from, typo maybe?) anyway, i think you did it a bit backwards mate
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#8
Quote by stevo_epi_SG_wo
um, how do you know the graph doesnt continue after 17?

the logic is there (except 'd' sub-shells, where did that come from, typo maybe?) anyway, i think you did it a bit backwards mate


Well, chlorine only has 17 electrons... I think
#9

Thanks!

And with to Stevo, if we're only given say 8 sets of IE's, we just state what period it'll be in
#10
Quote by Deliriumbassist
Well, chlorine only has 17 electrons... I think


....and mercury has 80, you cant assume it stops there. if i was in a flaming mood i could rip it to shre.......

lets not go there
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#11
Quote by philipisabeast

Thanks!

And with to Stevo, if we're only given say 8 sets of IE's, we just state what period it'll be in



yea, basically count (in ascending order) till theres a jump, et voila!
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#12
Quote by stevo_epi_SG_wo
....and mercury has 80, you cant assume it stops there. if i was in a flaming mood i could rip it to shre.......

lets not go there


Well, as it says Chlorine at the top of the graph, you can assume it stops there

If chlorine has 17 electrons, that is. I haven't done chemistry for a couple of years.
#13
Quote by Deliriumbassist
Well, as it says Chlorine at the top of the graph, you can assume it stops there

If chlorine has 17 electrons, that is. I haven't done chemistry for a couple of years.


touche.....lol, but they wont tell you that if youre supposed to sus what group its in

but going right to left isnt wise
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#14
Quote by stevo_epi_SG_wo
touche.....lol, but they wont tell you that if youre supposed to sus what group its in

but going right to left isnt wise


The graph should be going left to right, surely? Because the first ionisation energy refers to the removal of an electron from the outer shell?
#15
Quote by Deliriumbassist
The graph should be going left to right, surely? Because the first ionisation energy refers to the removal of an electron from the outer shell?


yea, start with the first electron, then keep going in till it jumps, not start at the 1s level and work out, cuz it might not be that, just the last two in whatever other shell
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#16
Quote by stevo_epi_SG_wo
yea, start with the first electron, then keep going in till it jumps, not start at the 1s level and work out, cuz it might not be that, just the last two in whatever other shell


Well, the 1s electrons can't leave due to shielding and nuclear attraction... so if the TS made that graph himself, he really should flip the line over.

EDIT: It's all coming back now... this was the stuff I was good at in chemistry.
#17
That graph goes 7 in the outer shell, 8 in the middle, 2 in the final (inner) shell.

It takes more and more energy to remove the electrons from the atom/ion because they're closer to the nucleus and more tightly held by it.

It's a graph of all the IEs of Cholrine, as it says at the top of the picture.

Quote by Storm_Bringer_
well at the far right you can see there are 2 results then it jumps down a bit.

Those 2 are group 1 and 2 (working right to left) filling the S Sub shell (alkaline and alkali earth metals)

It then jumps down and there are 8 here, this is the p sub shell being filled (actinoids) after this is another jump, there are 6 here, the d sub shell being filled and now your back into your groups, going right to left again 3 4 5 6 7 8, the last dot there is going back to the s shell, group 1


Chlorine has no electrons in d shells.
#18
^ To delirium: Nope got the graph off the net

But it shows that as an electron gets removed from each sub shell (S then P) the changing IE for each electron to be removed

So for the furthest ones out, there is the lowest intermolecular charge holding them so they can be removed easier and so on and so forth
#19
I've just realised all my info is wrong in this thread, lol... and I know why, Meths is my witness in chat

The electron configuration is 1s2 2s2 2p6 3s2 3p5, methinks

so no electrons in the d shell. The big jumps go to the next shell, and reading right to left gives you 2 electrons, 8 electrons, 7 electrons.

So no electrons in a d shell.
Last edited by Deliriumbassist at Nov 5, 2007,
#20
Quote by philipisabeast
^ To delirium: Nope got the graph off the net

But it shows that as an electron gets removed from each sub shell (S then P) the changing IE for each electron to be removed

So for the furthest ones out, there is the lowest intermolecular charge holding them so they can be removed easier and so on and so forth


Intra-atomic*

The furthest electrons out are easiest to remove. You can see in the graph that it takes successively more energy to remove each electron. There are also larger jumps. This is when you need to 'break' into the next (inner) shell. From the graph you can see there are 7 in the outer shell (all relatively easy to remove) and then 8 which take slightly more energy. These are in the middle shell. You also get 2 at the end which take the most energy.

Electorn configuration is 2.8.7.
#21
Basically as you go along a period in the periodic table the ionisation energy increases as they is a greater force of attraction between the positive nucleus and the negative electrons, making the energies higher

However as you go down a group in the periodic table the ionisation energy decreases as there are more electron shells to shield the outermost ring of electrons. This means the pull on the electrons is weaker as they are further away from the positive nucleus.


EDIT it has not been known for more than 4 or 5 electrons to be removed from an element
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Last edited by fire within at Nov 5, 2007,