**10 points for you if you help me!** Question about electron arrangement in the quantum mechanical model!?

Question

I have a question about the Aufbau principle. I was thinking first you fill up all the energy level 1 orbitals, then the energy level 2, then the energy level 3, etc., but that wasn't working out too well for me. Is THIS the order in which they will be filled?

1s
2s
2p
3s
3p
4s
3d
4p
5s
4d
5p
6s
4f
5d
6p
7s
5f
6d
7p

If so, how do you write the electron configuration? For example, I know that for nickel, it's 1s(^2)2s(^2)2p(^6)3s(^2)3p(^6)3d(^8)4s(^2)... but why wouldn't the 4s(^2) be written before the 3d(^8) since that orbital was "filled" first?

Lastly, I thought that all orbitals in a principal energy level must be filled with 1 electron, to give it a parallel spin, before you pair the electrons. If that is so, than how come, in phosphorus for example, 3s gets a pair of electrons, whereas 3d doesn't even have one? Am I getting mixed up and it's that all "electron slots" in an atomic orbit must have 1 electron before pairing them?
Please help and please explain well! THANKS

Answer 1

http://www.thecenterforadvancedlearning.org/chemistry/EConfigs.pdf

http://www.kmhsmagnet.com/staff/tbrown/ppts/Chemistry%20Ch.%2013%20Powerpoint%20-%20Electrons%20in%20Atoms.ppt#1

http://users.stlcc.edu/gkrishnan/electron3.html

Answer 2

First answer: yes that is how they will fill, the subshells are slightly less stable than the key shell and the 4s filling before 3d is the first example.

Second answer: It would actually be the most correct to write out all the filled subshells first, then the 4s2 since it is filled then 3d2, 3d2, 3d2, 3d1, 3d1 as each portion of the d subshell can only hold 2 e's as well (neg and pos 1/2 spin) and they fill the unoccupied before pairing up.

Third answer: you can induce the electrons to jump into subshells via energy application, but in its ground state the 3-3p subshells are much more stable than any of the 5-3d subshells therefore you start filling the p's before the d's and then come back and fill them up later.

Answer 3

Even though the order at which the shells are to be filled 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p etc.


the 4s cannot be filled before the 3d because the 4s has a lower energy than the 3d orbitals. so the law also says that the orbital with the lower energy gets filled first.( remember).
the issue of electrons filling the 3p orbitals before the 3d orbitals still has to do with the energy levels of the orbitals. it will cost the lone electron less energy to get itself into the 3p orbital than the 3d orbital that has a higher energy. so it'll simply chose to enter the 3p orbital.

Answer 4

First answer: particular it rather is how they're going to fill, the subshells are slightly lots much less solid than the main crucial shell and the 4s filling previously 3-D is the 1st get on a similar time. 2d answer: it is going to rather be the main ideal to place in writing down out each and all of the crammed subshells first, then the 4s2 via fact it rather is crammed then 3d2, 3d2, 3d2, 3d1, 3d1 as each and each part of the d subshell can only carry 2 e's as properly (neg and pos a million/2 spin) and that they fill the unoccupied previously pairing up. 0.33 answer: you're able to finally finally end up interior the electrons to bounce into subshells by means utility, yet in its floor state the three-3p subshells are so lots extra ideal solid than any of the 5-3-D subshells subsequently you initiate filling the p's previously the d's and then come back and fill them up later.