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Original post by Gnome :)
The electron configuration is 1s2 2s2 2p6 3s2 3p4
So from 1s2 2s2 2p6 and 3s2 you get 6 full orbitals
In 3p4 it's slightly different. Electrons "prefer" being on their own in this context due to electrostatic repulsion. The 3p subshell can have 3 pairs of electrons, or 3 orbits, so 3 electrons will go into separate orbitals. The remaining electron then has to join one of these electrons. So the 3p subshell is made of 3 orbitals; 2 with a single electron, one with a full orbital.
Therefore you have 7 full orbitals, and 2 half filled orbitals


Thank you so much for reminding me that they fill up one at a time! I completely forgot :smile:
Not long now, how is everyone coping? I'm just about seeing the light at the end of the tunnel.
Original post by HarryMWilliams
Not long now, how is everyone coping? I'm just about seeing the light at the end of the tunnel.



hopefully i'll be seeing that soon too :smile: I really want an A* but I can't seem to even get an A grade in unit 5. It's quite off putting - I think i'm trying hard, i guess i just need to change my strategy. Have you started papers yet? :smile:
Original post by jojo1995
hopefully i'll be seeing that soon too :smile: I really want an A* but I can't seem to even get an A grade in unit 5. It's quite off putting - I think i'm trying hard, i guess i just need to change my strategy. Have you started papers yet? :smile:


Yah - I've done one or two, I'm focussing on the organic stuff at the moment. I quite like Unit 5 over the others, it's not 'easy' per se, but just nice. :biggrin:
Original post by HarryMWilliams
Yah - I've done one or two, I'm focussing on the organic stuff at the moment. I quite like Unit 5 over the others, it's not 'easy' per se, but just nice. :biggrin:


yeah... you need to know the organic inside out - the jan grade boundaires are so high... for the one i just did, it was 80 FOR AN A* ??? WHY lol - I prefer unit 4 by 100% :tongue:
Hi, does anyone know how to calculate the percentage error of a beaker, pipette, thermometer and anything else please ? :smile: thank you in advance
Reply 1106
A hydrated transition metal ion is colourless. Which of the following could be the
electronic configuration of this ion?
A [Ar] 3d54s2
B [Ar] 3d8
C [Ar] 3d104s2
D [Ar] 3d10

why D and not C?
Original post by jojo1995
yeah... you need to know the organic inside out - the jan grade boundaires are so high... for the one i just did, it was 80 FOR AN A* ??? WHY lol - I prefer unit 4 by 100% :tongue:



I quite like Unit 4 too actually - A2 is far much more interesting and enjoyable than AS.
Original post by Ryejd
A hydrated transition metal ion is colourless. Which of the following could be the
electronic configuration of this ion?
A [Ar] 3d54s2
B [Ar] 3d8
C [Ar] 3d104s2
D [Ar] 3d10

why D and not C?



ion means it has lost electrons, and transition metals have thier outer electrons in 3d and 4s sub shells. so in order for it to be an ion it must have lost some electrons. electrons are lost from the 4s first :smile:

hopefully i've answered your question, i am quite bad at explaining.
Original post by Ryejd
A hydrated transition metal ion is colourless. Which of the following could be the
electronic configuration of this ion?
A [Ar] 3d54s2
B [Ar] 3d8
C [Ar] 3d104s2
D [Ar] 3d10

why D and not C?

Hey, :smile:
C displays the electron configuration of an atom of zinc as the s-subshell is full.
D on the other hand does not have any electrons in the s-subshell which indicates that 2 electrons had been lost. This shows that it is the transition metal ion.
(edited 10 years ago)
Reply 1110
Guys how are unit 5 papers? I done most of my revision,but haven't done a single paper yet. It looks scary lol :no:

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Original post by Tackla
Guys how are unit 5 papers? I done most of my revision,but haven't done a single paper yet. It looks scary lol :no:

Posted from TSR Mobile


The January 2013 (I think) paper with the copper question... good luck! :biggrin:

Otherwise, they aren't too bad. Some of the organic questions are just repeated (effectively) in each paper so I would suggest you learn those verbatim.
Hey, could someone please just check my thought process is correct.

When a redox reaction produces a solid, the equilibrium moves to the right (so the constant increases) so the value for the cell's emf will increase? Which explains why, despite a negative Ecell, it could still occur?
Reply 1113
Original post by jojo1995
Hi, does anyone know how to calculate the percentage error of a beaker, pipette, thermometer and anything else please ? :smile: thank you in advance


Hi to calculate the percentage error for the apparatus
%error
= the uncertainty in the equipment divided by the reading an then you times 100%

This kind of questions they will sometimes give you the titre reading so u use that and divide with the reading and times with 100%.
Hope you understand what I meant. If you don't then let me know yeah.
Cheers. :smile:😊
If they don't give you the titre reading then u have to calculate it by yourself where you 2 times with the uncertainty of the equipment and then divide with the reading and times with 100%.



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Original post by ayeswary
Hi to calculate the percentage error for the apparatus
%error
= the uncertainty in the equipment divided by the reading an then you times 100%

This kind of questions they will sometimes give you the titre reading so u use that and divide with the reading and times with 100%.
Hope you understand what I meant. If you don't then let me know yeah.
Cheers. :smile:😊
If they don't give you the titre reading then u have to calculate it by yourself where you 2 times with the uncertainty of the equipment and then divide with the reading and times with 100%.



Posted from TSR Mobile



Thank you so much :smile: do you know how I'd calculate the uncertainty ?
Original post by jojo1995
Thank you so much :smile: do you know how I'd calculate the uncertainty ?

So if you have a burette that measures to 0.1cm3, you uncertainty is +/- 0.05, so 0.05*2=0.1 :smile:
Reply 1116
Original post by Gnome :)
So if you have a burette that measures to 0.1cm3, you uncertainty is +/- 0.05, so 0.05*2=0.1 :smile:


What? In physics the uncertainty is = the smallest scale division that your equipment measures why is it now the uncertainty is that/2? (shouldnt the uncertainty of the burette be 0.1cm3?
Reply 1117
Original post by JRP95
What? In physics the uncertainty is = the smallest scale division that your equipment measures why is it now the uncertainty is that/2? (shouldnt the uncertainty of the burette be 0.1cm3?


Yes the smallest division for the burettte is 0.1 cm3. But to calculate the percentage error we use the uncertainty which is the smallest division divided by 2. In other words 0.1/2 = 0.05 cm3. Uncertainty is an estimate attached to a measurement which characterizes the range of values within which the true value is said to lie. That's why we use 0.05 instead of 0.1. Well in physics is a different thing. I am not too sure about physics cause have dropped the subject and only focussing on the other two sciences. So hope u understand yeah. :smile:
Cheers :smile:


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Original post by JRP95
What? In physics the uncertainty is = the smallest scale division that your equipment measures why is it now the uncertainty is that/2? (shouldnt the uncertainty of the burette be 0.1cm3?

The calculation you need is:
(Uncertainty in equipment / Total amount measured) x 100 = % uncertainty

Remember that when reading a burette you take a reading twice (Start and end of titration) so you need to multiply the uncertainty by 2. It's also worth just remembering that the standard uncertainty for a burette is ±0.05cm3

So for example if you did a titration and got a titre value of 10cm3, the uncertainty would be:
2 x (0.05/10) x 100 = 1%
(edited 10 years ago)
what exactly do you mean by synoptic by the way? :P I am new to A2 :/
:colondollar:

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