electron configuration question?

Question

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10

This is the fullest electron configuration. Could someone please explain me why the orbitals are 4s, 3d, 4p not 4s,4p,4d? why the orbitals are 5s,4d,5p, not 5s,5p,5d? why 6s, 4f, 5d, 6p not 6s,6p,6d,6f? i am extremely confused. i think im oblivious to some fundamentals about the structure of the table, but i cant grasp a thing at the moment. Can someone please help me? :) Thanks.

Answer 1

To put it short and sweet....



The electrons take the lowest energy configurations within the nucleus. It just happens that the 3d is (very very slighly) higher in energy than the 4s so electrons fill the 4s shell first.

Answer 2

This atom has 7 shells around the nucleus, i.e. 1s - 7s. In this atomic configuration all 7 shells have 2 electrons in them, the maximum amount.

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10

There are sub-shells, atomic orbitals.

"The orbital names (s, p, d, f) are derived from the characteristics of their spectroscopic lines: sharp, principal, diffuse, and fundamental."

Look at the 6th shell: 6s2 4f14 5d10 6p6

There is a formula that controls this because you are considering the orbital as a wave and not a particle.

Think of it as the color blue. All particles with the spin the same as the color blue will be attracted to the color blue.

Answer 3

digital configuration of Carbon (C), Nitrogen(N), Oxygen(O) are as follows: C(6)- 1s2 2s2 2Px1 2Py1 N(7)- 1s2 2s2 2Px1 2Py1 2Pz1 O(8)- 1s2 2s2 2Px2 2Py1 2Pz1 right here, Carbon has 2 electrons in its 2p subshell. So, there is an probability to make solid 0.5-crammed P orbital interior the final shell of Carbon by gaining a million electron. So, to earnings stability Carbon can quite earnings a million electron. on the otherhand, Nitrogen has already 0.5-crammed P orbital interior the final shell. So, it won't settle for added electrons while you evaluate that's solid sufficient. to earnings greater electrons it has to break its stability. because of the fact of this, Nitrogen is way less constructive(greater constructive) than that of Carbon In case of Oxygen, there r 4 electrons in P orbital. that's neither 0.5-crammed nor finished-crammed. So it quite is not solid sufficient to earnings greater electrons. So, oxygen additionally can earnings greater electrons in its outermost shell ensuing in a greater constructive eelectronic affinity than Nitrogen. consequently, digital affinity of Nitrogen> Carbon & Oxygen

Answer 4

I think they are listed in the order they are out from the nucleus.

I know 3d is further out 4s.

Answer 5

Perhaps some of the information in the links below can help:



http://library.thinkquest.org/10429/low/eleconfig/electron.htm



http://antoine.frostburg.edu/chem/senese/101/electrons/index.shtml



http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson36.htm



http://education.jlab.org/qa/electron_config.html



http://creationwiki.org/Electron_configuration



http://www.sparknotes.com/testprep/books/sat2/chemistry/chapter4section5.rhtml



http://kaffee.netfirms.com/Science/activities/Chem/Activity.Electron.Configuration.html



http://www.sciencegeek.net/Chemistry/taters/Unit2ElectronNotations.htm