Transition metals

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Kaneki Ken
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#1
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#1
the colour changing equations,
we werent taught it properly and im stuggling to understand how it works and the equations needed to be known
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Rexx18
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#2
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(Original post by Kaneki Ken)
the colour changing equations,
we werent taught it properly and im stuggling to understand how it works and the equations needed to be known
Which exam board?
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Kaneki Ken
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#3
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#3
(Original post by Rexx18)
Which exam board?
OCR
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Rexx18
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#4
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Can't really help with OCR since I do AQA, but I guess I could give you the ones from AQA, a lot of them are bound to be similar to OCR.

We're required to know the reactions of copper(II), iron(II), aluminium(III) and iron(III) with hydroxides, ammonia, excess hydroxides, excess ammonia and carbonates.

In general, for these 4 metals, all of them will form a precipitate with hydroxide ions and with ammonia (as long as it's not in excess). It's important to note that, as a general rule, something that precipitates cannot have an overall charge. This fact will help you set up the equations. The reactions are as follows:

[Cu(H2O)6]2+ + 2OH- --> Cu(OH)2(H2O)4 + 2H2O (pale blue solution changes to a blue precipitate)
[Cu(H2O)6]2+ + 2NH3 --> Cu(OH)2(H2O)4 + 2NH4+ (pale blue solution changes to a blue precipitate)

With the first reaction, each hydroxide ion pulls off a H+ from one of the 6 water ligands, which is why you end up with the precipitate that you do. The 2 hydroxide ions turn into 2 waters and the 2 water ligands turn into 2 OH- ligands. With the second reaction, each ammonia again pulls of a H+ ion from one of the 6 water ligands. This forms 2 ammonium ions.

The reactions for iron(II) are the same, except the colour difference.

[Fe(H2O)6]2+ + 2OH- --> Fe(OH)2(H2O)4 + 2H2O (pale green solution changes to a green precipitate)
[Fe(H2O)6]2+ + 2NH3 --> Fe(OH)2(H2O)4 + 2NH4+ (pale green solution changes to a green precipiate)

The reactions for aluminium(III) and iron(III) are again the same, just with one additional H+ ion being removed from the water ligands (remember, we need an overall charge of 0 for something to precipitate).

[Al(H2O)6]3+ + 3OH- --> Al(OH)3(H2O)3 + 3H2O (colourless solution changes to a white precipitate)
[Al(H2O)6]3+ + 3NH3 --> Al(OH)3(H2O)3 + 3NH4+ (colourless solution changes to a white precipitate)

[Fe(H2O)6]3+ + 3OH- --> Fe(OH)3(H2O)3 + 3H2O (yellow solution changes to a brown precipitate)
[Fe(H2O)6]3+ + 3NH3 --> Fe(OH)3(H2O)3 + 3NH4+ (yellow solution changes to a brown precipitate)

As you can see, the equations for all 4 of these metals are relatively straightforward. You pull off the correct number of protons from the water ligands until the overall charge is 0, at which point it will precipitate out. As for the colour changes, remember, one of the things which affects deltaE is the type of ligand. We're changing from 6 water ligands to either 2 OH- and 4 H2O or to 3 OH- and 3 H2O. This brings about a change in deltaE and a different colour forms.

You might need to know that copper reacts with excess ammonia. The reaction for this is shown below:

[Cu(H2O)6]2+ + 4NH3 --> [Cu(NH3)4(H2O)2]2+ + 4H2O (pale blue solution changes to a deep blue/royal blue solution)

As you can see, this reaction is slightly different. When ammonia is in excess, it no longer pulls off protons from the water ligands, but rather, it substitutes for them. It's important to note that copper only does a partial substitution. Only 4 water ligands will be replaced for ammonia.

You might also need to know that aluminium reacts with excess hydroxide ions. The reaction for this is shown below:

[Al(H2O)6]3+ + 4OH- --> [Al(OH)4(H2O)2]- + 4H2O (colourless solution remains colourless)

For the aluminium reaction, you can see it's very similar to the one where the hydroxide ions aren't in excess. The hydroxide ions still pull off a proton from the water ligands, the only difference is that it's 4 hydroxide ions this time, not 3. Remember, since the overall charge is now -1, it can not be a precipitate.

Finally, you might need to know the reactions of these 4 metals with carbonates. Just like with hydroxides, it's easiest to write these as ionic equations. They are outlined below:

[Cu(H2O)6]2+ + CO32- --> CuCO3 + 6H2O (pale blue solution changes to a green-blue precipitate)
[Fe(H2O)6]2+ + CO32- --> FeCO3 + 6H2O (pale green solution changes to a green precipitate)

2[Al(H2O)6]3+ + 3CO32- --> 2Al(OH)3(H2O)3 + 3CO2 + 3H2O (colourless solution changes to a white precipitate and effervescence of carbon dioxide)
2[Fe(H2O)6]3+ + 3CO32- --> 2Fe(OH)3(H2O)3 + 3CO2 + 3H2O (yellow solution changes to a brown precipitate and effervescence of carbon dioxide)

As you see for all these equations, the reactions of the 2+ metals are very closely related. The same goes for the 3+ metals. Key points to remember are that things will only precipitate if they lose the overall charge and that 3+ metals will form carbon dioxide gas when reacting with carbonates. Writing out the reactions with hydroxide ions and carbonates is also much easier if you do the ionic equations.
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Rexx18
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The last equation I can provide you with (which you may or may not need to know) is the reaction between copper(II) and chloride ions. The reaction is as follows:

[Cu(H2O)6]2+ + 4Cl- --> [CuCl4]2- + 6H2O (pale blue solution changes to a blue-green or green-yellow solution)

The reason that only 4 chloride ligands substitute for the 6 water ligands is due to their size. Chloride ligands are larger, therefore there is only enough space to fit 4 of them onto the copper ion. If you tried to fit more than 4 on, you'd get repulsion and it would quickly become unstable.
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Kaneki Ken
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#6
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(Original post by Rexx18)
The last equation I can provide you with (which you may or may not need to know) is the reaction between copper(II) and chloride ions. The reaction is as follows:

[Cu(H2O)6]2+ + 4Cl- --> [CuCl4]2- + 6H2O (pale blue solution changes to a blue-green or green-yellow solution)

The reason that only 4 chloride ligands substitute for the 6 water ligands is due to their size. Chloride ligands are larger, therefore there is only enough space to fit 4 of them onto the copper ion. If you tried to fit more than 4 on, you'd get repulsion and it would quickly become unstable.


wow!
thanks so much i really appreciate ur help!!!!!!!!!!!!
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FusionNetworks
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Check your specification. Each exam board is different. For OCR, you need to learn the colour changes and equations associated with NaOH and Ammonia in transitional elements: Cu2+, Fe2+/3+, Cr3+ And Mn 2+. Also, the CuCl4 2- posted above. (Off the top of my head, so double check spec!)
There is no way about it, you just have to memorise all the colour changes and the equations, sorry
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Kaneki Ken
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#8
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(Original post by FusionNetworks)
Check your specification. Each exam board is different. For OCR, you need to learn the colour changes and equations associated with NaOH and Ammonia in transitional elements: Cu2+, Fe2+/3+, Cr3+ And Mn 2+. Also, the CuCl4 2- posted above. (Off the top of my head, so double check spec!)
There is no way about it, you just have to memorise all the colour changes and the equations, sorry
deep lol
thanks alot for informing me appreciate it
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