|
Simple question, the average neutral copper atom has 2 valance electrons, and Sulfide 6. Wouldn't the two electrons jump from copper and bind to sulfide? The empirical formula is : $\ce{Cu + S -> Cu2S}$. So apparently two copper atoms bind with every one sulfide atom. Does that mean that the copper atoms each gives one electron away and leaves them with 1 valance electron each? |
||||
|
|
Nope. Only one:
In the transition metals, the previous shell is being filled. Generally, the valence shell can't have more than 8 electrons, but lower shells can. So, you have the two outer electrons filled in in Group 1 and 2, and the rest are filled in in the inner shell. With copper, we have 17 electrons in the inner shell, and 2 on the outer shell by this rule. But--it is sometimes more favorable for the inner shell to be "complete" with 18 electrons, leaving the outer shell with only one. This is much better explained if you understand what atomic orbitals are. Think of these as as "subshells". In copper, we have two possible shell configurations: 10 electrons in $3d$ and one in $4s$, or 9 in $3d$, and 2 in $4s$. The numbers signify the shell, and the letters signify the orbital type. There are $5$ d orbitals per shell (not all shells have $d$ orbitals, though), and 1 $s$ orbital. Each orbital holds two electrons. Here, we have the trade-off between having a fully filled $3d$ orbital set and having a fully filled $4s$ orbital. Fully filled orbital sets lead to stability. Anyway, since both choices lead to a fully filled orbital set, a lot of other external factors decide which electronic configuration copper uses. So you have both $\ce{Cu^+}$ and $\ce{Cu^{2+}}$. As a side note, $\ce{CuS}$ exists as well. |
|||||||
|
