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Arrangement of Electrons

Energy Levels or Shells

The simplest model of electrons has them orbiting in shells around the nucleus. Each successive shell is further from the nucleus and has a greater energy.

Sub Shells and Orbitals

This model can be further refined by the concept of sub shells and orbitals.

Sub shells are known by letters s, p, d, and f. The s sub shell can contain 2 electrons, p 6, d 10 and f 14.

Electrons occupy negative charge clouds called orbitals, each orbital can hold only 2 electrons. Each type of shell has a different type of orbital.

How we write electron configurations

Electrons fill the lowest energy level first this means it is generally easy to predict how the electrons will fill the orbitals (it gets more complicated with the transition metals).

Let’s look at building up the electronic arrangement (electron configuration) from hydrogen (Z = 1) as far as krypton (Z = 36).

The order of filling of orbitals in atoms (it is different for positive ions) is:

1s 2s 2p 3s 3p 4s 3d 4p (notice how the 3d comes before 4p)

By going through the elements in sequence, we will see the various rules come into play.

Hydrogen (Z = 1) It has one electron. It must be placed in the orbital with the lowest energy.

1s1

Helium (Z = 2) It has two electrons. There is room in the 1s orbital for a maximum of two electrons but they must have opposite spins. Electron spin is a difficult concept but think of it as being no more than a label of "up-spin" or"down-spin".

1s2

Lithium (Z = 3) It has three electrons. Although two can go into the 1s orbital, the third one must be placed in the 2s.

1s22s1

Beryllium (Z = 4) It has four electrons. The fourth can also go in 2s (but remember about the opposing spins)

1s22s2

Boron (Z = 5) It has five electrons. The fifth one must go in 2p. There are three 2p orbitals available and these are identified by their direction (px, py, pz). It doesn’t matter which of these is chosen when there is only one electron in one of these orbitals.

1s22s22p1

Carbon (Z = 6) It has six electrons. The sixth one must go in a different 2p orbital. We now need to specify the particular 2p orbitals we use. It doesn’t matter which combination of x, y, or z but they must be different.

1s22s22px12py1

Nitrogen (Z = 7) It has seven electrons. We now need the remaining 2p orbital.

1s22s22px12py12pz1

Oxygen (Z = 8) It has eight electrons. We can now complete the filling of the 2p orbitals in turn. Again it doesn’t matter which of the orbitals is completed first.

1s22s22px22py12pz1

Fluorine (Z = 9) It has nine electrons.

1s22s22px22py22pz1

Neon (Z = 10) It has ten electrons.

1s22s22px22py22pz2

or since all of the 2p orbitals are completed we can simply write: 1s22s22p6

Sodium (Z = 11) We now have no space in the second energy level and so have to start on the third.

1s22s22p63s1

This can be simplified (unless you are told to write the complete configuration) to [Ne]3s1

Magnesium (Z = 12) to argon (Z = 18) We continue through the whole of the period using the same concepts as we did from lithium to neon.

Magnesium is [Ne]3s2Argon is [Ne]3s23p6

Potassium (Z = 19) As before, we start on the fourth energy level because that is the next in the order of filling.

[Ar]4s1

Calcium (Z = 20) This is where you stopped for GCSE

[Ar]4s2

Scandium (Z = 21) We start on the set of 3d orbitals...

[Ar]4s23d1

and continue as expected (without worrying about the opposing spins this time) until we come to…

Chromium (Z = 24)

We might expect [Ar]4s23d4 but instead we get [Ar]4s13d5

This is because of the special stability of the half-filled set of 3d orbitals. An electron promoted from the 4s to the 3d. Although this requires energy, this is paid back by the extra stability.

Manganese (Z = 25) The gap in 4s is filled again

[Ar]4s23d5

Iron (Z = 26), Cobalt (Z = 27) and Nickel (Z = 28) proceed onwards as expected

[Ar]4s23d6 , [Ar]4s23d7 , [Ar]4s23d8

Copper (Z = 29) adopts the unexpected (like chromium) because of the special stability of the full 3d.

[Ar]4s13d10

Zinc (Z = 30) The gap in 4s is filled again

[Ar]4s23d10

Continue by filling the 4p orbitals (opposing spin is important again) until we get to

Krypton (Z = 36)

[Ar]4s23d104p6