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Writing the ground state electron configuration for the given atoms

Bive the expected ground state electron configurations for atoms with the following atomic number: A. Z=55 B. Z=40 Can someone please work one of the two out and explain how to do this type of problem thank you

kingchemist Tue, 10/11/2011 - 05:24

The blocks, s, p, d and f refer to the orbitals that are currently being filled.

For groups 1 and 2, this is the s orbital, so these groups are the s block elements.

For groups 3 (Boron), 4, 5, 6, 7 and 8 (noble gases), it is the p orbitals that are being filled so these are the p block elements. Some courses call these groups 13-18)

In the first transition series (Sc to Zn) the d orbitals are being filled, so these are the d block elements.

Elements making the f block are filling their f orbitals.

http://www.gordonengland.co.uk/elements/blocks.htm

is useful. If you click on the element, it shows you data including the electron arrangement.

Also the periodic table on

http://www.webelements.com/

is also excellent.

 

The order of filling of orbitals is shown on

http://www.chemguide.co.uk/atoms/properties/atomorbs.html


I would start with simple electron arrangements and develop the notation from that, using some simple rules

The first energy level can hold up to 2 electrons
The second energy level can hold up to 8 electrons
The 3rd energy level can hold 8 electrons but in some cases this can be up to 18 electrons
The 4th energy level can hold up to 32 electrons.

Energy levels have orbitals. Every energy level has an orbital called the s orbital. eg. in the 1st energy level this is the 1s orbital, 2s in the 2nd energy  level etc. All s orbitals can hold up to 2 electrons
Energy levels from the 2nd upwards also have p orbitals of which there can be up to three, each capable of holding up to 2 electrons. So the 2p orbitals can hold up to 6 electrons, the 3p orbitals can hold up to 6 electrons.
Energy levels from 3rd upwards have an s orbital (3s, 4s etc), up to three p orbitals (eg 3p, 3p, 3p) and possibly up to five d orbitals making the total electrons possible in the 3rd energy level to 18 (2 in the 3s, 6 in the 3p orbitals, and 10 in the 3d orbitals)
Energy levels from the 4th energy level upwards are like the 3rd energy level with 4s, 4p and 4d orbitals. However, they can accommodate up to an extra 14 electrons in seven 4f orbitals.

The order in which these are filled is important and there are diagrams to help the filling. However s are filled before p orbitals in the same level.

Let us consider a simple electron arrangement that you will have access to
Copper = 2, 8, 18, 1
1st level = 2 electrons. The only orbital here is the 1s orbital, so there are 2 electrons in this = 1s2. First level is now full. Next electrons go into the 2nd level.
There are 8 electrons here (2nd number in the simple arrangement) so the first two of these fill the 2s orbital (= 2s2). This leaves 6 other electrons to accommodate in the 2nd energy level. This go into the 2p orbitals, one a time at first until the three are half filled, and then the remaining 3 electrons share the p orbitals (2p2 2p2 2p2)
There are 18 electrons in the 3rd energy level - 2 in the 3s orbital, 6 in the 3p orbitals which leaves 10 more to be accommodated in the 3rd energy level - so these fill the d orbitals (3d2 3d2 3d2 3d2 3d2) and now the 3rd level is totally full
The simple arrangement says that there is one electron in the 4th energy level and the lowest energy orbitals available here is the 4s orbital (= 4s1)

So copper = 1s2 2s2 2p2 2p2 2p2 3s2 3p2 3p2 3p2 3d2 3d2 3d2 3d2 3d2 4s1

Now the nearest noble gas, before Cu is Ar 2,8,8 or 1s2 2s2 2p2 2p2 2p2 3s2 3p2 3p2 3p2 and so we can simplify copper to
[Ar] 3d2 3d2 3d2 3d2 3d2 4s1

An easier one is potassium, 2,8,8,1 which by the same reasoning is 1s2 2s2 2p2 2p2 2p2 3s2 3p2 3p2 3p2 4s1  (no 3d orbitals involved as 3d orbitals are not filled until after the 4s orbital is filled. So this can be represented as [Ar] 4s1

A more complex one is Hg = 2,8,18,32,18,2
If there are only 2 electrons in an energy level, only the s orbital is filled = 1s2 and 6s2
If there are 8 electrons in an energy level, the s orbital and all three p orbitals are filled in that level (= 2s2 2p2 2p2 2p2; 3s2 3p2 3p2 3p2; 4s2 4p2 4p2 4p2; and 5s2 5p2 5p2 5p2;

If there are more than 8 electrons in an energy level and not more than 18 electrons, then as well as s and p, d orbitals are involved (= 3d2 3d2 3d2 3d2 3d2)

If there are more than 18 electrons in an energy level and not more than 32 electrons, then as well as s, p and d, f orbitals are involved (= 4f2 4f2 4f2 4f2 4f2 4f2 4f2)

Putting all this together gives us
1s2 2s2 2p2 2p2 2p2 3s2 3p2 3p2 3p2 3d2 3d2 3d2 3d2 3d2 4s2 4p2 4p2 4p2 4d2 4d2 4d2 4d2 4d2 4f2 4f2 4f2 4f2 4f2 4f2 4f2 5s2 5p2 5p2 5p2 5d2 5d2 5d2 5d2 5d2 6s2

This can be shortened by grouping p orbitals and d orbitals
1s2 2s2 2p62 3s2 3p2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 or
[Xe] 4f14 5d10 6s2