Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Revned on May 03, 2005, 01:24:35 AM
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There are 2 possible oxidation states of copper, those are Cu1+ and Cu2+.
The electronic configuration of Cu1+ is : 1s2 2s2 2p6 3s2 3p6 3d10
The electronic configuration of Cu2+ is : 1s2 2s2 2p6 3s2 3p6 3d9
As you should know Copper II is more stable than Copper I. But the thing i find weird is that Cu1+ has a full outer d sub-shell which is in theory highly stable, and the Cu2+ has a partially filled d sub-shell which in theory is very unstable. So what is going on here?
The only answer i could find to that is that Copper I is disproportiaonate, which means that if put in an aquous solution it will auto oxidate and reduce to a solution of Copper II and Copper. I beleive that is not a proper answer to my question, so maybe someone here could help me understand this problem better.
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You are quite correct in that it appears at first sight that Cu+ should be more stable than Cu2+ but IN AQUEOUS CONDITIONS it isn't.
Why?
Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The Cu2+ ion has a greater charge density than the Cu+ ion and so forms much stronger bonds releasing more energy.
The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the Cu+ ion loses an electron to become Cu2+ which can then release this hydration energy.
Most processes are a balance between their relative energy gain and loss.
A similar argument can be applied to the solid forms. only to find the relative stability a Born Haber cycle would have to be constructed. In this case it comes down to a balance between the greater lattice enthalpy of the Cu2+ compound (compared to the Cu+ compound) and the second ionisation energy.รง
I haven't done the sums so I don't know the result!
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There are 2 possible oxidation states of copper, those are Cu1+ and Cu2+.
The electronic configuration of Cu1+ is : 1s2 2s2 2p6 3s2 3p6 3d10
The electronic configuration of Cu2+ is : 1s2 2s2 2p6 3s2 3p6 3d9
As you should know Copper II is more stable than Copper I. But the thing i find weird is that Cu1+ has a full outer d sub-shell which is in theory highly stable, and the Cu2+ has a partially filled d sub-shell which in theory is very unstable. So what is going on here?
The only answer i could find to that is that Copper I is disproportiaonate, which means that if put in an aquous solution it will auto oxidate and reduce to a solution of Copper II and Copper. I beleive that is not a proper answer to my question, so maybe someone here could help me understand this problem better.
are you sure that's the correct configuration for copper I and II? Where are the missing electrons for copper I?
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I don't think that is the correct chemical notation.
CUI AR[18]4s2 3d10
CUII AR[18]4s2 3d9
You should get used to writeing chemical notation in shorthand because it's easier to read.
p.s. I haven't taken chemistry since last yeare, so I'm a bit rusty and am not sure that what I put is correct.
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I don't think that is the correct chemical notation.
CUI AR[18]4s2 3d10
CUII AR[18]4s2 3d9
You should get used to writeing chemical notation in shorthand because it's easier to read.
p.s. I haven't taken chemistry since last yeare, so I'm a bit rusty and am not sure that what I put is correct.
The electronic configuration of Copper is 1s2 2s2 2p6 3s2 3p6 3d10 4s1 OR [Ar] 3d10 4s1
Electrons in Copper fill the 3d sub-shell before the 4s sub-shell as an exeption to the other elements because a full 3d sub-shell and a half filled 4s sub-shell is more stable than having a partially filled 3d sub-shell and a full 4s sub-shell.
So as you should know copper loses the electrons in the 4s sub-shell before the 3d sub-shell when it forms ions. Leaving the configuration of its ions to this:
Cu+ is : 1s2 2s2 2p6 3s2 3p6 3d10
Cu2+ is : 1s2 2s2 2p6 3s2 3p6 3d9
which is the same as the shorthand notation of:
Cu+ is : [Ar] 3d10
Cu2+ is : [Ar] 3d9
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The only answer i could find to that is that Copper I is disproportiaonate, which means that if put in an aquous solution it will auto oxidate and reduce to a solution of Copper II and Copper. I beleive that is not a proper answer to my question, so maybe someone here could help me understand this problem better.
Off course, disproportionation can take place, but you should take into account more interactions in the solution - copper forms different complexes and aqua ions and stability of Cu(I) or Cu(II) depends on environment in solution.
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...am not sure that what I put is correct.
Then don't post in threads, when you are not confident of the answer. ::)