Cu has an anomalous electron configuration. $\ce{Cu ~=~ 1s^2~2s^2~2p^6~3s^2~3p^6~4s^1~3d^{10}}$, it does not follow the usual pattern. In this case, the 3d subshell is filled before the 4s, which usually happens in the reverse order ($\ce{1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^{9}})$.
My question is, can you tell by an element's position on the periodic table (Group #, Row #, Block, etc.) that it will have an anomalous electron configuration? Do I have to memorize which elements have this property?
I am noting that Cu is in the 1B Group of Transition metals which implies that it has one valence electron, this is consistent with its configuration. will this be true for all cases?
Also: $\ce{Fe ~=~ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$ and is a transition metal as well (Group 8). This implies it must have 8 valence electrons. But following the definition of valence electrons (The number of electrons in outermost shell), $n = 4$ is the outermost shell and I get a conundrum. Obviously if I add $\ce{4s + 3d}$ electrons everything is good. Why is this?
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