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Alevel chemistry

can someone tell which reaction undergo complete substitution and which reaction does not under go complete substation for transitions metals very Confusing
Original post by angelina avalina
can someone tell which reaction undergo complete substitution and which reaction does not under go complete substation for transitions metals very Confusing


it depends, eg using excess or dilute
Reply 2
for AQA u need to know about fe2+, cu2+, fe3+ and al3+. with excess ammonia cu2+ undergoes incomplete ligand substitution whereas cobalt undergoes complete substitution. apparently the reactions of cobalt aren’t required knowledge tho. hope that helps a bit !
thank you
Original post by nnnn6666
for AQA u need to know about fe2+, cu2+, fe3+ and al3+. with excess ammonia cu2+ undergoes incomplete ligand substitution whereas cobalt undergoes complete substitution. apparently the reactions of cobalt aren’t required knowledge tho. hope that helps a bit !
Original post by angelina avalina
can someone tell which reaction undergo complete substitution and which reaction does not under go complete substation for transitions metals very Confusing


Note that when I refer to an ion like Cu^2+, I am referring to the corresponding hexaaqua complex. I’m just too lazy to type out the full formula.

You only really need to consider the chemistry of Cu^2+, since the other ions in solution you need to be aware of (Fe^2+, Fe^3+, Al^3+) don’t tend to undergo ligand exchanges with the compounds you are expected to know the results for (NH3, NaOH, Na2CO3).

Cu^2+ does not undergo full ligand exchange with excess ammonia for complicated reasons that you don’t need to know, but in excess ammonia, it does undergo partial ligand exchange to form [Cu(NH3)4(H2O)2]^2+ (when there is not an excess of ammonia, the Cu^2+ is deprotonated only to Cu(OH)2(H2O)4). The Al^3+, Fe^2+ and Fe^3+ ions are deprotonated by NH3 to form their respective hydroxides and do not react further in excess, so they do not undergo ligand exchange full stop.

All of the ions you need to be aware of are deprotonated by NaOH, but aluminium reacts further, forming [Al(OH)4]^- ions, so the precipitate of aluminium hydroxide redissolves.

Cu^2+ and Fe^2+ react with the carbonate ion in Na2CO3 to form precipitates of copper(II) carbonate (this isn’t quite the case, but A level doesn’t expect you to know this) and iron(II) hydroxide, respectively. Al^3+ and Fe^3+ react with the carbonate ion by an acid-base reaction, as both ions are small and highly charged, so they polarise any surrounding water ligands to release hydrogen ions - this leaves hydroxide ions, so you get precipitates of aluminium hydroxide and iron(III) hydroxide, respectively.

As such, the only example of ligand exchange you actually need to memorise of is the reaction of Cu^2+ with excess ammonia- any other examples of ligand exchange you may be assessed on will generally be given in the question.
(edited 9 months ago)
thank you
Original post by TypicalNerd
Note that when I refer to an ion like Cu^2+, I am referring to the corresponding hexaaqua complex. I’m just too lazy to type out the full formula.

You only really need to consider the chemistry of Cu^2+, since the other ions in solution you need to be aware of (Fe^2+, Fe^3+, Al^3+) don’t tend to undergo ligand exchanges with the compounds you are expected to know the results for (NH3, NaOH, Na2CO3).

Cu^2+ does not undergo full ligand exchange with excess ammonia for complicated reasons that you don’t need to know, but in excess ammonia, it does undergo partial ligand exchange to form [Cu(NH3)4(H2O)2]^2+ (when there is not an excess of ammonia, the Cu^2+ is deprotonated only to Cu(OH)2(H2O)4). The Al^3+, Fe^2+ and Fe^3+ ions are deprotonated by NH3 to form their respective hydroxides and do not react further in excess, so they do not undergo ligand exchange full stop.

All of the ions you need to be aware of are deprotonated by NaOH, but aluminium reacts further, forming [Al(OH)4]^- ions, so the precipitate of aluminium hydroxide redissolves.

Cu^2+ and Fe^2+ react with the carbonate ion in Na2CO3 to form precipitates of copper(II) carbonate (this isn’t quite the case, but A level doesn’t expect you to know this) and iron(II) hydroxide, respectively. Al^3+ and Fe^3+ react with the carbonate ion by an acid-base reaction, as both ions are small and highly charged, so they polarise any surrounding water ligands to release hydrogen ions - this leaves hydroxide ions, so you get precipitates of aluminium hydroxide and iron(III) hydroxide, respectively.

As such, the only example of ligand exchange you actually need to memorise of is the reaction of Cu^2+ with excess ammonia- any other examples of ligand exchange you may be assessed on will generally be given in the question.
also this may be stupid questions but I've memorised all the equations for example like this
2[FE[h20]6]3+ +3c032= 2[fe[H20]3[oh]3]+3c02
but some markschme it requires you to put oh first then h20 how would I know what to put first

the markschem said it should be 2[FE[oh]3[h20]3] + 3c02 I don't really understand why also with ligand substitution reaction I've memorise the h20 going first then the 0h
Original post by angelina avalina
also this may be stupid questions but I've memorised all the equations for example like this
2[FE[h20]6]3+ +3c032= 2[fe[H20]3[oh]3]+3c02
but some markschme it requires you to put oh first then h20 how would I know what to put first

the markschem said it should be 2[FE[oh]3[h20]3] + 3c02 I don't really understand why also with ligand substitution reaction I've memorise the h20 going first then the 0h

It’s not actually that important whether you list (OH) before (H2O) in the formulae of the hydroxides.

It’s more common to list (OH) first, so mark schemes generally give the formulae of hydroxides in the form M(OH)n(H2O)6-n (where n is the charge on the metal cation, M^n+). Even M(H2O)6-n(OH)n is a correct formula, so I reckon in an actual exam, they wouldn’t penalise it.
(edited 9 months ago)
thank you
Original post by TypicalNerd
It’s not actually that important whether you list (OH) before (H2O) in the formulae of the hydroxides.

It’s more common to list (OH) first, so mark schemes generally give the formulae of hydroxides in the form M(OH)n(H2O)6-n (where n is the charge on the metal cation, M^n+). Even M(H2O)6-n(OH)n is a correct formula, so I reckon in an actual exam, they wouldn’t penalise it.
Hi guys

For the reaction of aqueous ions in solution do I only need to know their reaction with Naoh, nh3 and na2co3? Or do I need to know the reactions in water and how they react with each other

And I don’t get acidity reactions
Original post by B7861
Hi guys

For the reaction of aqueous ions in solution do I only need to know their reaction with Naoh, nh3 and na2co3? Or do I need to know the reactions in water and how they react with each other

And I don’t get acidity reactions


You need to know the reactions of hexaaqua ions of Al^3+, Cu^2+, Fe^2+ and Fe^3+ with NaOH, NH3 and Na2CO3 only.

You should be aware that aqueous salts of these metals tend to exist as hexaaqua complexes.

Acidity arises from the central metal ion in the hexaaqua complex being relatively small and charged, leading to the polarisation of O-H bonds in the water ligands. As such, H^+ ions are released. The smaller and more charged the central ions are, the more polarising the hexaaqua complexes are and the more acidic they are.
Reply 11
Original post by B7861
Hi guys

For the reaction of aqueous ions in solution do I only need to know their reaction with Naoh, nh3 and na2co3? Or do I need to know the reactions in water and how they react with each other

And I don’t get acidity reactions

u also need to know about their reaction with hcl. when u say acidity reactions do u mean how aqua 3+ ions are more acidic than 2+ ions? in that case, it’s bc the 3+ ions have a higher charge density so it polarises the oxygen of the o-h bond. this weakens the o-h bond so more h+ are released.
[fe(h2o)6]3+ -> [fe(h2o)5(oh)]2+ + h+
(edited 9 months ago)
Original post by TypicalNerd
You need to know the reactions of hexaaqua ions of Al^3+, Cu^2+, Fe^2+ and Fe^3+ with NaOH, NH3 and Na2CO3 only.

You should be aware that aqueous salts of these metals tend to exist as hexaaqua complexes.

Acidity arises from the central metal ion in the hexaaqua complex being relatively small and charged, leading to the polarisation of O-H bonds in the water ligands. As such, H^+ ions are released. The smaller and more charged the central ions are, the more polarising the hexaaqua complexes are and the more acidic they are.


So do I need to know the acidity reactions? I do t even know what they look like
Original post by B7861
So do I need to know the acidity reactions? I do t even know what they look like

You need to be aware that any hexaaqua ion can act as an acid, so yes.

A general equation is as shown:

[M(H2O)6]^n+ (aq) <=> [M(H2O)5(OH)]^(n-1)+ (aq) + H^+ (aq)

Where n is the charge on the metal cation M^n+.
Guys what’s the fastest way to revise for chemistry practicals, not revised for it at all. And what type of practical stuff will come up in p1?
Original post by B7861
Guys what’s the fastest way to revise for chemistry practicals, not revised for it at all. And what type of practical stuff will come up in p1?


Generally it’s interpreting results of experiments (i.e test tube reactions), plotting graphs or calculations based on experimental results. You may as well have a look at that booklet by chemrevise on the practicals https://chemrevise.files.wordpress.com/2018/04/practical-guide-aqa1.pdf
Original post by TypicalNerd
Generally it’s interpreting results of experiments (i.e test tube reactions), plotting graphs or calculations based on experimental results. You may as well have a look at that booklet by chemrevise on the practicals https://chemrevise.files.wordpress.com/2018/04/practical-guide-aqa1.pdf

I’m good with test tube reactions, graphs and calculations but not with actual experiments. So remembering methods for 6 markers and some questions ask why a so and so substance was added

I’ll probably have a look at the chemrevise doc thanks

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