Garfield Posted September 21, 2005 Posted September 21, 2005 Orbitals get "filled" with electrons starting from the orbital that has the lowest energy- 1s, than 2s<2p<3s<3p<4s<3d<4p<4d(?)... How does the "sequnce" qontinue?
LucidDreamer Posted September 21, 2005 Posted September 21, 2005 4d<5p<6s<4f<5d<6p<7s<5f <6d<7p <8s <5g <6f <7d <8p
timo Posted September 21, 2005 Posted September 21, 2005 I´m no chemicist but this "the orbitals get filled one after the other" is a rule of thumb, at most. E.g. the Lanthanides have one 5d electron. After this, the 4f orbital is filled first before the elements continue on with filling the 5d. You should be able to find the electron configuration in any semi-decent periodical table.
jdurg Posted September 21, 2005 Posted September 21, 2005 Yes, it is only a 'rule of thumb'. Take gold, for example. Gold has only 1 6s electron while osmium, which is three spots to gold's left, has 2 6s electrons.
Garfield Posted September 22, 2005 Author Posted September 22, 2005 4d<5p<6s<4f<5d<6p<7s<5f <6d<7p<8s<5g<6f<7d<8p No 5s orbital? And how does the g orbital look like? I know, I'm a new-be. We're only learning "basics" at school.
DQW Posted September 23, 2005 Posted September 23, 2005 4p < 5s < 4d got missed. Also, note that these are the order of filling orbitals in isolated gaseous atoms (not, for instance, in a crystalline solid, where near-neighbor interactions slightly alter the energy levels). The "order" can be generated from the following two rules : 1. An orbital with a lower value of n+l fills first, 2. In case of a tie, the orbital with the lower n fills first.
rthmjohn Posted September 23, 2005 Posted September 23, 2005 g orbitals don't exist in ground-state atoms, only in excited ones. As far as their structure... I wouldn't bother until you get into advanced calculus... or you could look online for some pictures, but the g orbitals themselves (g has NINE suborbitals) have an EXTREMELY COMPLEX three-dimensional arrangement
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