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ocr a f325 revision thread

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Original post by Gulzar
Has anyone got the f325 jan 2013 paper? It would be such a great help for anyone to send me it :biggrin:


hey buddy want to revise
Reply 381
Original post by otrivine
hey how are you :smile:

up for some revision


Do you know the answer to the question? Balancing redox equations?
Original post by Better
Do you know the answer to the question? Balancing redox equations?


yep I do :smile:
Reply 383
Original post by otrivine
yep I do :smile:


Right cool. If you're not going to tell me I don't see why you replied.

Anyone else ? Please look above and provide any help if you can.

Would really appreciate it. Balancing REDOX EQUATIONS not too sure about when to use H20 or H+/OH- the book doesn't clarify.
Original post by Better
Right cool. If you're not going to tell me I don't see why you replied.

Anyone else ? Please look above and provide any help if you can.

Would really appreciate it. Balancing REDOX EQUATIONS not too sure about when to use H20 or H+/OH- the book doesn't clarify.


oh you want me to explain, I thought u just said if you know redox, which part do u need help in
Reply 385
Original post by Better
Right cool. If you're not going to tell me I don't see why you replied.

Anyone else ? Please look above and provide any help if you can.

Would really appreciate it. Balancing REDOX EQUATIONS not too sure about when to use H20 or H+/OH- the book doesn't clarify.


Sorry cant revise otrivine, but was just skimming through this thread

Basically if you have more oxygens on the left of the equation then on the left you pretty much always add a H2O to the right hand side, so that you'be balance the number of O. Then to balance Hydrogens you add H+ to the left hand side.
Reply 386
Original post by Dr00n
Sorry cant revise otrivine, but was just skimming through this thread

Basically if you have more oxygens on the left of the equation then on the left you pretty much always add a H2O to the right hand side, so that you'be balance the number of O. Then to balance Hydrogens you add H+ to the left hand side.


Ah kk thanks so much mate, appreciate it, I can get on with the rest of the Exam Style Q's now!

:smile:
I don't suppose anyone would be able to explain how you're able to work out how many ligands a complex ion will have in it and what all the different colour changes will be as we basically got told in lesson that if in doubt just put 6 around and go from there but that's hardly accurate enough for me to accept.


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Original post by MathsNerd1
I don't suppose anyone would be able to explain how you're able to work out how many ligands a complex ion will have in it and what all the different colour changes will be as we basically got told in lesson that if in doubt just put 6 around and go from there but that's hardly accurate enough for me to accept.


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Hi BUDDY how are you, want to revise
Original post by otrivine
Hi BUDDY how are you, want to revise


Hi, could you help me with my problem? I'm just about to start on some Unit 1, just got a little bit left to learn for it then I'll be doing abit of this unit.


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Original post by MathsNerd1
I don't suppose anyone would be able to explain how you're able to work out how many ligands a complex ion will have in it and what all the different colour changes will be as we basically got told in lesson that if in doubt just put 6 around and go from there but that's hardly accurate enough for me to accept.


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Regarding the colours, you need to learn all of them by heart.
Secondly, the number of ligands, you just count them in the structure to see how many there are, for example in [Cu(h2o)6]2+ there are 6 ligands and hence, 6 coordinate bonds(dative covalent)
Original post by otrivine
Regarding the colours, you need to learn all of them by heart.
Secondly, the number of ligands, you just count them in the structure to see how many there are, for example in [Cu(h2o)6]2+ there are 6 ligands and hence, 6 coordinate bonds(dative covalent)


I understand all that, it's just the fact when you react them with something else and the ligand changes too, would they provide us with the equation so we would know how many coordinate bonds there should be, or are we just meant to know those aswell?


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Original post by MathsNerd1
I understand all that, it's just the fact when you react them with something else and the ligand changes too, would they provide us with the equation so we would know how many coordinate bonds there should be, or are we just meant to know those aswell?


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do you have an example, of this case, it would be easier to explain to you
Original post by otrivine
do you have an example, of this case, it would be easier to explain to you


Erm it's hard to think of an example as the teacher was just rambling on about it but we reacted copper (II) with ammonia and cobalt (II) with HCL although I'm not sure those are cases where the number of coordinate bonds changes but he did say, if in doubt just go with 6 which won't always be correct so I just want to understand an actual process of working it out in each case.


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Original post by MathsNerd1
Erm it's hard to think of an example as the teacher was just rambling on about it but we reacted copper (II) with ammonia and cobalt (II) with HCL although I'm not sure those are cases where the number of coordinate bonds changes but he did say, if in doubt just go with 6 which won't always be correct so I just want to understand an actual process of working it out in each case.


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Ok, so, I presume you know the equations , yes your teacher is correct, because it is 6 coordinate bonds!
Original post by otrivine
Ok, so, I presume you know the equations , yes your teacher is correct, because it is 6 coordinate bonds!


But will there ever be a situation where you start with 6 coordinate bonds and after reacting with a solution will change the number of coordinate bonds, maybe reduce it to 4?


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Original post by MathsNerd1
But will there ever be a situation where you start with 6 coordinate bonds and after reacting with a solution will change the number of coordinate bonds, maybe reduce it to 4?


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yes this is the case when copper(II) reacts with the concentrated HCL

so when [Cu(h2o)6] gives [Cucl4]2-

coordinate number changes from 6 to 4
Original post by otrivine
yes this is the case when copper(II) reacts with the concentrated HCL

so when [Cu(h2o)6] gives [Cucl4]2-

coordinate number changes from 6 to 4


Okay but would they just expect you to know that or would they provide you with an equation so you're able to work it out for yourself?


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Original post by MathsNerd1
Okay but would they just expect you to know that or would they provide you with an equation so you're able to work it out for yourself?


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No, you may be asked to write the equation and to get the number of coordinate bonds
Reply 399
Hey guys, I'm aiming for an A* like many people here does anybody have stretch and challenge questions for this unit (F325) and the last unit (F324). Would really appreciate it. Also, does anybody have the past papers and markschemes for these two topics! Thanks!!!

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