[wakka]

I'm just confused a bit about how to visualize subshell filling. Now i know 1s and 2s hold 2 electrons which travel around in a spere shape, 3s travels over 2s i assume. then when the p shell comes in, you have 3 sets of two electrons travelling in figure 8's on the x, y and z axis' around the 1s and 2s spheres. what i don't get is when you get the D subshells coming in, how they travel... for example when you have 4s and then 3d... if the 4s is a sphere overlapping the 3s sphere, then why would electrons go underneath the 4s sphere and travel around the 3s?? Same thing for the f subshells. this may be a newb question but hey, i am a newb 

[Morphic flip]

D subshells which are full or half full are more stable than that of the next s subshell.
The lowest energy state is always filled first.
Have you read the Pauli exclusion principle and electron interactions?

[xstrae]

As a general rule, a new electron enters an empty orbital for which the value of 'n + l" is minimum. Here, 'n' is the principal quantum number and 'l' the azimuthal quantum number. This happens because the D subshell happens to be more stable than the s subshell. Try reading about the 'aufbau principle' and 'paul's exclusion principle'

http://en.wikipedia.org/wiki/Aufbau_principle

[Stewed_ant]

i had the same problem and still do, trying to vizulalize it all, the only thing i can offer is i try to think of it as not a shaped orbital but just as an area where there is an increase in probability of finding an electron... it helps me sometimes. The shells overlap each other, or the probabilities over lap each other D sub shells look like strange donut things.

This may help
http://en.wikipedia.org/wiki/Electron_configuration

[cuongt]

one thing i dont get about subshell notation is something like this

1s2,2s2,2p6,3s2,3p6,  then 4s2 or 3d10?  my teacher was confusing me so much. then wen it loses it loses 4s2 first or something.

[tamim83]

then 4s2 or 3d10?  my teacher was confusing me so much. then wen it loses it loses 4s2 first or something.

The 4s orbital is filled before the 3d orbitals.  In short, energy is the reason for this.  In transition metals the d and s orbitals are quite close in energy (the d orbitals are only slightly lower in energy than the s orbital of the next principle quantum number).  As a result, electrons will fill the s orbitals first because it is more favorable to have a half or completely filled s orbital than one (or two) electrons in a d orbital (with a great deal of empty orbitals left over).  Only after the s orbital fills up does electrons start to fill the d orbitals. 

 And yes, transition metals lose the s electrons before losing d electrons since they are slightly higher in energy than the d electrons.  Or in other words, they are in the "outer shell" so they are lost first. 

[cuongt]

thanks a lot legend 

[Mitch]

The reason why electrons fill the 4s subshell before the 3d is because the 3d electrons are closer to the nucleus and will repel each other in the 3d more strongly then if they were in the higher energy and larger orbit of the 4s subshell. So yes, the 3d is lower in energy, but the energy of configuration would be lower if the electrons occupied the 4s.

[tamim83]

Hmm, I never heard it explained that way before Mitch.  That actually makes a lot of sense to me.  I just always thought it had something to do with not wanting a lot of partially filled orbitals (i.e.-far too much space) as opposed to having less space.  Interesting. 

[cuongt]

btw mitch is that is a highschool explanation rite? not a university