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Edexcel Unit 5:Transition Metals and Organic Chemistry 6CH05 (15th June 2015)

This poll is closed

What do you think the grade boundary for an A will be?

78+ 8%
75-77 10%
72-74 25%
69-71 31%
66-68 13%
63-65 4%
60-62 3%
57-59 3%
Less than 563%
Total votes: 112
(edited 8 years ago)

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Reply 1
I've got some good resources for the reactions of copper and chromium if anybody wants them? Its something they like to ask tbh its probably better if you learn the reactions! :/ that's what I've done!!
Reply 2
Original post by bladex
I've got some good resources for the reactions of copper and chromium if anybody wants them? Its something they like to ask tbh its probably better if you learn the reactions! :/ that's what I've done!!

Yeah thanks that would be amazing! Could you post them here?
Also could you link this thread in the unit 4 thread? :3
Reply 3
ImageUploadedByStudent Room1428798930.519853.jpg
So, I have been trying to solve this problem using molar volumes and can't think of another way of doing it. I would really appreciate it if someone could explain why the answer is D.


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Original post by bladex
I've got some good resources for the reactions of copper and chromium if anybody wants them? Its something they like to ask tbh its probably better if you learn the reactions! :/ that's what I've done!!

Oh, yeah. I definitely need some extra memorising on that part. You're kind to share them :biggrin:
Reply 7
Original post by Wannabe007
Thank you SO much ! :ahee:


No problem some of it might be too much for what we need to know but some of it is really useful :smile:
watching.
A2 is very frustrating, oh my.
Thank you though
Reply 9
hey guys...i've been trying to figure out how to solve question 23c on the june 2014 IAL unit 5 paper but I can't seem to figure it out. I tried checking the marking scheme but couldn't seem to figure out how they did it. If any of you have tried it could you please explain it to me? Thanks
Reply 10
Original post by jshah1997
hey guys...i've been trying to figure out how to solve question 23c on the june 2014 IAL unit 5 paper but I can't seem to figure it out. I tried checking the marking scheme but couldn't seem to figure out how they did it. If any of you have tried it could you please explain it to me? Thanks


I can remember not being able to do that question as well with the weird CxHy thing :/
Reply 12
Original post by daisychain_
These are brilliant, thank you! Do we have to learn everything on the sheets or the reactions only?


I'm only learning the reaction pathways because that's the thing they usually ask but I've took a quick look over the other stuff but haven't properly learnt it :smile:
May anyone give me an example of a deprotonation reaction in the transition topic please?
Original post by frozo123
May anyone give me an example of a deprotonation reaction in the transition topic please?


Deprotonation reactions involve water ligands losing hydrogen ions (proton) to a proton acceptor such as an hydroxide ion.
[Cu(H2O)6]2+ + OH- [Cu(OH)(H2O)5]+ + H2O
Original post by Ripper Phoenix
Deprotonation reactions involve water ligands losing hydrogen ions (proton) to a proton acceptor such as an hydroxide ion.
[Cu(H2O)6]2+ + OH- [Cu(OH)(H2O)5]+ + H2O


So that is never ligand exchange? correct?
Original post by frozo123
So that is never ligand exchange? correct?


its partial ligand exchange not full ligand exchange
Original post by Ripper Phoenix
its partial ligand exchange not full ligand exchange


Why is it partial ligand exchange if it's technically not replacing the ligand but actually deprotonating it but looking like it has exchanged? or have I just answered my own question?
Original post by frozo123
Why is it partial ligand exchange if it's technically not replacing the ligand but actually deprotonating it but looking like it has exchanged? or have I just answered my own question?


good!! you got it..thats deprotonation haha not ligand exchange
Reply 19
Guys I have a question,

When you react [Cu(H20)6]2+ WITH NH40H , what is the product?
My notes say Cu(OH)2, and in excess NH4OH you get [Cu(NH3)4(H2O)2]2+.
But why is it when you add NH3 you get [cu(OH)2(H20)4].

Does anyone get what I am trying to ask? And understand why this occurs, and whether this is right.

Thanks

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