Edexcel A2 Chemistry Unit 5 ~ 22nd January 2013

I never mark my papers/count the marks.


no idea, perhaps a bit lower than the past years?

lol, quick response Im pretty sure it will be lower. by how much, not sure.

lol sorry xD
yeah it should be lower, seems like it wasn't 'that easy'
I A* is always 72-73/90 , only in january 2012 it was 80/90.
I don't know if it can fall below 70, we will see

i think an A could be around 60-63. how did you find it?

Strange paper but not difficult as hell I would say xD

I did not know 3 of the copper compounds in the first question, never seen them in the book, depending on the mark scheme, I will have a chat with Edexcel later :J
I could do the rest in the first question as James A, till that 4 marker, lol

Also I realize that I did not draw the additional carbon chain in that polyamide in last question. I probably drew Kevlar by forgetting that CH2 group xD /facepalm

Strange paper but not difficult as hell I would say xD

I did not know 3 of the copper compounds in the first question, never seen them in the book, depending on the mark scheme, I will have a chat with Edexcel later :J
I could do the rest in the first question as James A, till that 4 marker, lol

Also I realize that I did not draw the additional carbon chain in that polyamide in last question. I probably drew Kevlar by forgetting that CH2 group xD /facepalm

Same, I may contact edexcel for the copper names, that was a joke. The drawing a polymer q? Lol, I dont even remember. Wasnt kevlar lol, but it did have conh link I think.

I must admit though, how the **** are we supposed to have known what the black solid was?? arghhhhhh so annoyed man.

hopefully low grade boundaries, they will have to!

I put Copper oxide, no clue if thats right or whatever. I just made a guess, using some stupid logic like the heat removed the water and oh ligands, had no freaking clue. Was the q on why copper iodide was colourless to do with the splitting of d orbitals?

you talk about no d-d transitions taking place from lower energy ligands to higher energy ligands, hence colourless. Then you say the opposite for the other compound (it was a comparison question) which was coloured.

What did you put for the absorbic acid q about why four tablets are used rather than two? I put to increase the reliability of the experiment.

you talk about no d-d transitions taking place from lower energy ligands to higher energy ligands, hence colourless. Then you say the opposite for the other compound (it was a comparison question) which was coloured.

What did you put for the absorbic acid q about why four tablets are used rather than two? I put to increase the reliability of the experiment.

Same, I may contact edexcel for the copper names, that was a joke. The drawing a polymer q? Lol, I dont even remember. Wasnt kevlar lol, but it did have conh link I think.

I must admit though, how the **** are we supposed to have known what the black solid was?? arghhhhhh so annoyed man.

hopefully low grade boundaries, they will have to!

you talk about no d-d transitions taking place from lower energy ligands to higher energy ligands, hence colourless. Then you say the opposite for the other compound (it was a comparison question) which was coloured.

What did you put for the absorbic acid q about why four tablets are used rather than two? I put to increase the reliability of the experiment.

oh, and the 4 marker on why phenol was more reactive then the other compound, benzene methanol or something like that. 4 marks? Wth was up with that q...

lul yes I know it wasn't kevlar ~_~
I am just saying that I forgot this red circled part in my chain on both sides :/
(it was something like this)



I still don't know what that pink solid + blue solution is xD



You had to give the electronic configuration, showing that one is 3d10 the other is not (3d9 I think) then start writing about how igands split d orbitals ...and d-d transitions taking place/not taking place etc.

Using four tablets gives you that volume of titre. using less would lower the titre (and the mean value) so higher %error (or %uncertainty) therefore not reliable bla bla xD

The lone pair of electrons on the oxygen atom in phenol is drawn into the benzene ring, which increases the electron density in the delocalised ring, hence phenol is more susceptible to electrophillic attack, hence faster reactions. I put for the other molecule that methoxy group (i think that was it, forgot) draws electron density away from the benzene ring, hence making it less susceptible to electrophillic attack.

The lone pair of electrons on the oxygen atom in phenol is drawn into the benzene ring, which increases the electron density in the delocalised ring, hence phenol is more susceptible to electrophillic attack, hence faster reactions. I put for the other molecule that methoxy group (i think that was it, forgot) draws electron density away from the benzene ring, hence making it less susceptible to electrophillic attack.

Yeyy, I think I got all the above qs right, so thats like another 6 marks lol
Oh, the pink solid is pure copper I believe. I didnt get this either, but I did remember that copper has a salmon pink colour in some form? I think its just copper. Had no clue about that or the blue solution in the exam.

copper and pink?
is that even possible? xD

I wrote it doesn't have lone pair of electrons (the carbon atom attached to the ring), so can't donate and is therefore less reactive..

I wrote it doesn't have lone pair of electrons (the carbon atom attached to the ring), so can't donate and is therefore less reactive..



copper and pink?
is that even possible? xD
(edit/ seems like it is? lol)


what about the hydrogen environments and that IR data?

LOOOOL told you it was pink The 4 marker on the h environments? I remember circling two diff environments and grouping them together as one letter, dont rememeber lool. Tell me the answers, maybe then it will come to mind. I seem to forget everything