Cu has an unpaired electron in 4s, but it is diamagnetic. I thought that it has to be paramagnetic. What am I missing?
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4 Answers
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What is missing here ist the relation to physical/chemical status of that "copper". What are You talking about? Copper metal? Copper atoms as vapour? Copper ions I or II in aqueous solution? As a matter of fact, copper atoms are paramagnetic (one unpaired electron is enough despite the number of the paired ones!) BTW, the Herren Stern and Gerlach knew that silver atoms are paramagnetic, guess why!. In bulk copper metal the odd electron is sent into the pool of electrons making the metallic bond, thus the metal is diamagnetic, the same is for Cu+ salts, whreas Cu++ salts are paramagnetic.
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1$\begingroup$ I was talking about metal copper. I asked because if you take, for example, Na, which also has an unpaired electron, or Ag, they are paramagnetic. But Cu (metal Cu) is diamagnetic. That's the reason for my question. Thank you very much for your response. $\endgroup$– neutrinoCommented Mar 28, 2013 at 23:32
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3$\begingroup$ Going by this argument, shouldn't sodium metal be diamagnetic as it exists as $Na^+$ in its metallic structure? $\endgroup$– UserCommented Feb 24, 2018 at 15:25
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2$\begingroup$ II share neutrino's question and still don't see a satisfactory answer. I arrived here looking for an explanation of which elements with one electron beyond closed shells are paramagnetic in solid form, and why. I still don't know... It seems the alkalai metals are paramagnetic, while copper, silver and gold for example are diamagnetic... $\endgroup$ Commented Sep 16, 2019 at 3:47
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$\begingroup$ @neutrino silver is not paramagnetic davidfiedler.com/levitation.htm $\endgroup$ Commented Aug 15 at 10:54
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You can't derive the existence or magnitude of diamagnetic properties just from unpaired electrons. On the contrary, diamagnetism mostly comes from the complete shells. They behave as electric current loops that orient themselves in a certain way in the external magnetic field.
Copper has lots of these complete shells, so the diamagnetic contribution is large. There's also the opposite contribution from the unpaired electron but it's just one electron and the paramagnetic contributions "scale" with the number of these electrons and one is too few. So the diamagnetic terms win.
answered Mar 28, 2013 at 6:20
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4$\begingroup$ If this reasoning were right then all heavy elements would be diamagnetic, which isn't the case. $\endgroup$ Commented Sep 16, 2019 at 3:28
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$\begingroup$ This statement isn't implied by my statement. $\endgroup$ Commented Sep 16, 2019 at 14:47
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2$\begingroup$ ok, let's narrow it down to all heavy alkalai elements. For example, apparently caesium and francium are paramagnetic, but they have just one electron outside lots of closed shells. Your statement seems to imply that these elements should be diamagnetic. (I don't understand why they are paramagnetic while, say, gold is diamagnetic.) $\endgroup$ Commented Sep 17, 2019 at 15:08
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5$\begingroup$ "And indeed it wins". What determines when the paramagnetism wins? Rubidium, caesium, francium, copper and silver and gold all have one unpaired electron. The first three are paramagnetic, the last three are diamagnetic. Why? Your statement was that copper is diamagnetic since it has only one unpaired electron, so its diamagnetic effects win out. By this logic, rubidium, caesium, and francium would all be diamagnetic, but they're not. [Note, by the way, that paramagnetism can also receive contributions from orbital angular momentum.[ $\endgroup$ Commented Sep 19, 2019 at 21:08
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5$\begingroup$ I looked into it some more. The metals I mentioned above all have both diamagnetic and paramagnetic contributions. Their paramagnetism comes from the spin of conduction electrons, so is strongly limited by the Pauli exclusion principle ("Pauli paramagnetism"). This is relatively weaker than paramagnetism of materials with unpaired magnetic moments anchored on fixed atoms. It's also relatively temperature independent, as is diamagnetism, and similar in magnitude to the diamagnetic susceptibility. So indeed for these metals it comes down to a balance that could go either way. $\endgroup$ Commented Nov 3, 2019 at 6:12
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As Lubos has already told, the magnetic property originates from the extent to which electrons are filled in different orbitals.
If you have a look at the periodic table arrangement of d-block elements, you can see that Copper, Silver and Gold are on the same group due to their similar configuration $(n-1)d^{10}\ ns^1$. I don't know whether gold is diamagnetic (My periodic table shows: no data for Gold's magnetic properties). But from the look of copper and silver, gold should be diamagnetic too. These elements have still got an unpaired electron. So, they are paramagnetic. But it's so negligible that their diamagnetic property is enhanced.
It's probably determined by the magnetic moment $\mu=\sqrt{n(n+1)}\ \text{BM}$ (Bohr Magneton) where $n$ is the number of unpaired electrons. Hence, the more number of unpaired electrons an element has, the more paramagnetic character it has.
answered Mar 28, 2013 at 7:00
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Copper is diamagnetic because the upaired electrons in 4s orbitals is delocalized to make metallic bonds.
