ligand substitutions
[Cu(H2O)6]2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]2+ + 4H2O
'When a small amount of ammonia is added a pale blue precipitate of [Cu(NH3)4(H2O)2]2+ forms.' I did a bit of research into how this occurs and found that these reactions occur:
NH3 + H2O <=> NH4+ + OH-
Cu2+ + 2OH- => Cu(OH)2
But then my book says 'on addition of an excess of aqueous ammonia, the pale blue precipitate dissolves and a deep blue solution is formed containing [Cu(NH3)4(H2O)2]2+'. What exactly is occuring in this step? How does the Cu(OH)2 form [Cu(NH3)4(H2O)2]2+?
'When a small amount of ammonia is added a pale blue precipitate of [Cu(NH3)4(H2O)2]2+ forms.' I did a bit of research into how this occurs and found that these reactions occur:
NH3 + H2O <=> NH4+ + OH-
Cu2+ + 2OH- => Cu(OH)2
But then my book says 'on addition of an excess of aqueous ammonia, the pale blue precipitate dissolves and a deep blue solution is formed containing [Cu(NH3)4(H2O)2]2+'. What exactly is occuring in this step? How does the Cu(OH)2 form [Cu(NH3)4(H2O)2]2+?
(edited 10 years ago)
Original post by tazmaniac97
anyone?
I'm not really sure on this, if you're doing OCR chemistry then you really do not need to know why. You just need to know that it happens.
Original post by fletchdd02
I'm not really sure on this, if you're doing OCR chemistry then you really do not need to know why. You just need to know that it happens.
I know I don't need to know, but I want to.
[Cu(H20)6]2+ - Pale blue solution
When excess OH- is added you form Cu(H20)4(OH2)- Pale blue ppt.
When X.S. Ammonia is added a deep blue solution of [Cu(NH3)4(H2O2)]2+ forms...
I don't really understand your question?
'Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand.
If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base. In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred.'
When excess OH- is added you form Cu(H20)4(OH2)- Pale blue ppt.
When X.S. Ammonia is added a deep blue solution of [Cu(NH3)4(H2O2)]2+ forms...
I don't really understand your question?
'Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand.
If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base. In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred.'
(edited 10 years ago)
On addition of a small amount of OH- or NH3 you get Cu(OH)2(H2O)4
In excess ammonia you get [Cu(NH3)4(H2O)2]2+
In excess ammonia you get [Cu(NH3)4(H2O)2]2+
Original post by tazmaniac97
[Cu(H2O)6]2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]2+ + 4H2O
'When a small amount of ammonia is added a pale blue precipitate of [Cu(NH3)4(H2O)2]2+ forms.' I did a bit of research into how this occurs and found that these reactions occur:
NH3 + H2O <=> NH4+ + OH-
Cu2+ + 2OH- => Cu(OH)2
But then my book says 'on addition of an excess of aqueous ammonia, the pale blue precipitate dissolves and a deep blue solution is formed containing [Cu(NH3)4(H2O)2]2+'. What exactly is occuring in this step? How does the Cu(OH)2 form [Cu(NH3)4(H2O)2]2+?
'When a small amount of ammonia is added a pale blue precipitate of [Cu(NH3)4(H2O)2]2+ forms.' I did a bit of research into how this occurs and found that these reactions occur:
NH3 + H2O <=> NH4+ + OH-
Cu2+ + 2OH- => Cu(OH)2
But then my book says 'on addition of an excess of aqueous ammonia, the pale blue precipitate dissolves and a deep blue solution is formed containing [Cu(NH3)4(H2O)2]2+'. What exactly is occuring in this step? How does the Cu(OH)2 form [Cu(NH3)4(H2O)2]2+?
Ahh. I understand you question and was struggling with this myself.
Here is how..
When you add small amount of NH3 sol to the Copper hydroxide you are actually adding it to [Cu(OH)2(H2O)4](s) people tend to leave the H2O part out. Check out this. it may help!
http://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top
Original post by Future_Dr
Ahh. I understand you question and was struggling with this myself.
Here is how..
When you add small amount of NH3 sol to the Copper hydroxide you are actually adding it to [Cu(OH)2(H2O)4](s) people tend to leave the H2O part out. Check out this. it may help!
http://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top
Here is how..
When you add small amount of NH3 sol to the Copper hydroxide you are actually adding it to [Cu(OH)2(H2O)4](s) people tend to leave the H2O part out. Check out this. it may help!
http://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top
Oh no wonder! In the book it said that the precipitate reactions on spread 2.3.3 occur, and when I looked back at that spread that's the reaction I saw (the one I wrote in the OP). Thank you
Original post by yarshad
[Cu(H20)6]2+ - Pale blue solution
When excess OH- is added you form Cu(H20)4(OH2)- Pale blue ppt.
When X.S. Ammonia is added a deep blue solution of [Cu(NH3)4(H2O2)]2+ forms...
I don't really understand your question?
'Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand.
If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base. In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred.'
When excess OH- is added you form Cu(H20)4(OH2)- Pale blue ppt.
When X.S. Ammonia is added a deep blue solution of [Cu(NH3)4(H2O2)]2+ forms...
I don't really understand your question?
'Ammonia solution can react with hexaaqua metal ions in two quite distinct ways, because it can act as a base as well as a ligand.
If you add a small amount of ammonia solution you get precipitates of the metal hydroxide - the ammonia is acting as a base. In some cases, these precipitates redissolve when you add more ammonia to give solutions in which a ligand exchange reaction has occurred.'
Original post by bananarama2
On addition of a small amount of OH- or NH3 you get Cu(OH)2(H2O)4
In excess ammonia you get [Cu(NH3)4(H2O)2]2+
In excess ammonia you get [Cu(NH3)4(H2O)2]2+
Original post by Future_Dr
Ahh. I understand you question and was struggling with this myself.
Here is how..
When you add small amount of NH3 sol to the Copper hydroxide you are actually adding it to [Cu(OH)2(H2O)4](s) people tend to leave the H2O part out. Check out this. it may help!
http://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top
Here is how..
When you add small amount of NH3 sol to the Copper hydroxide you are actually adding it to [Cu(OH)2(H2O)4](s) people tend to leave the H2O part out. Check out this. it may help!
http://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top
Can you guys answer one more question please? In this reaction:
[Cu(H2O)6]2+ + 4Cl- <=> [CuCl4]2- + 6H2O
Why don't the Cl- ions react with water and form OH- ions and create a precipitate, like with ammonia?
(edited 10 years ago)
Original post by tazmaniac97
Can you guys answer one more question please? In this reaction:
[Cu(H2O)6]2+ + 4Cl- <=> [CuCl4]2- + 6H2O
Why don't the Cl- ions react with water and form OH- ions and create a precipitate, like with ammonia?
[Cu(H2O)6]2+ + 4Cl- <=> [CuCl4]2- + 6H2O
Why don't the Cl- ions react with water and form OH- ions and create a precipitate, like with ammonia?
Cl- ions do not have properties of acting as a base , only a ligand, so full substitution occurs i'm guessing
Original post by tazmaniac97
Can you guys answer one more question please? In this reaction:
[Cu(H2O)6]2+ + 4Cl- <=> [CuCl4]2- + 6H2O
Why don't the Cl- ions react with water and form OH- ions and create a precipitate, like with ammonia?
[Cu(H2O)6]2+ + 4Cl- <=> [CuCl4]2- + 6H2O
Why don't the Cl- ions react with water and form OH- ions and create a precipitate, like with ammonia?
It's a different story with Cl- since they are very big ions or something. Try looking at the website I posted and go back to other menus.. look around and there is lots of info that should really help you. Nevertheless. It's not really required in most specs and I don't do OCR but don't think it's required there too.
Go with the above post.... because they are not bases.
Original post by yarshad
Cl- ions do not have properties of acting as a base , only a ligand, so full substitution occurs i'm guessing
Why can't it act as a base? It's negatively charged so I thought it would be able to accept protons?
Original post by tazmaniac97
Why can't it act as a base? It's negatively charged so I thought it would be able to accept protons?
Chloride are better ligands- because they're bigger, only 4 can fit aroind the central metal ion, meaning 4 chlorides added and 6H2O taken away (this is change in co-ordination number) and this results in a positive entropy change because more species = more disorder so this is favoured, and Cl- substitutes.
I think :P
Posted from TSR Mobile
A base is an electron pair donor, cl-is more stable with an extra election so it won't give it away.
A base is an electron pair donor, cl-is more stable with an extra election so it won't give it away.
Original post by TaraBelle
Chloride are better ligands- because they're bigger, only 4 can fit aroind the central metal ion, meaning 4 chlorides added and 6H2O taken away (this is change in co-ordination number) and this results in a positive entropy change because more species = more disorder so this is favoured, and Cl- substitutes.
I think :P
I think :P
Original post by yarshad
Posted from TSR Mobile
A base is an electron pair donor, cl-is more stable with an extra election so it won't give it away.
A base is an electron pair donor, cl-is more stable with an extra election so it won't give it away.
Both these answers seem reasonable, but I don't know which one's right
... or maybe they both are.But it is possible right for Cl- to react with H2O right , even though it's not favorable (bold bit)? Or is it completely impossible?
Original post by tazmaniac97
Both these answers seem reasonable, but I don't know which one's right ... or maybe they both are.
But it is possible right for Cl- to react with H2O right , even though it's not favorable (bold bit)? Or is it completely impossible?
... or maybe they both are.But it is possible right for Cl- to react with H2O right , even though it's not favorable (bold bit)? Or is it completely impossible?
You can think about why cl- is a poor base in the same way as HCl is a strong acid. If Cl- deprotonated something it would form HCl which is very acidic.... It's all about the stabilities of the conjugate acid-base pairs. Equally NH3 is a strong base as it's conjugate acid NH4+ is a weak acid.
It can be rationalised differently in terms of availability of lone pairs and stabilisation of charge, but at A-level you usually focus on the conjugate pairs.
CuCl4 forms a tetrahedral shape due to the steric bulk of the ligand. Also a factor is complex electronic effects that you will not cover unless you study chem at uni. The important result is that Cl- gives less of a reason for the octahedral geometry about the metal compared to NH3 so it relaxes to the less sterically crowded tetrahedral shape. The substitution of the (I think 3rd, might be 4th) NH3 ligand is driven by the favourable entropy as two ligands are expelled. This is because this is the point where the complex changes from octahedral to tetrahedral geometry.
As to your original post... this is called ligand exchange, it's not a complex process. It operates by two main mechanisms for complexes of this sort. These are the associative and dissociative mechanism. In the A mechanism the incoming ligand coordinates to form a 7 coordinates species and then a ligand is lost to form an octahedral complex. In the D mechanism, a ligand is lost to form a 5 coordinate species, then a ligand attacks to reform an octahedral complex. It is possible to predict which will happen, but the theory is 3rd year undergrad level so I won't explain it. This reaction will happen by the D mechanism.
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