Question

As each electron is removed, one at a time, from an atom of aluminum, more ionization energy is required. Between which two successive ionization energies would you expect the greatest difference (largest change in value)? Explain your answer in terms of attraction, repulsion, and electron arrangement.

Answer 1

The electronic configuration of Al is:

1s2 2s2 2p6 3s2 3p1

Whilst I don't have the exact figures in front of me @sb1028 , my instinct would be that the greatest difference would be between the third and fourth ionisation energy.

The first ionisation energy is the minimum energy required to remove the most loosely bound electron from the valence shell of a neutral gaseous atom. The valence shell for Al is n = 3 (i.e. 3s and 3p electrons). In this case, the higher energy (more shielded from positive charge of the nucleus) 3p electron will be removed, to form the Al+ ion.

The second ionisation energy for is the minimum energy required to remove the most lossely bound electron from the Al+ gaseous ion. So, one of the 3s electrons will be removed to form the Al2+ ion.

The third ionisation energy will be the energy required to remove the final 3s electron, thus converting Al2+ to Al3+. This will require a relatively low energy input, as the 3s electron is sitting in the valence shell on its own. In removing it, we are creating a new full outer shell of electrons, which is favorable for any atom to have due to it being low in energy (n = 2 level, with the 2s and 2p orbitals). The electronic configuration of Al3+ is:

1s2 2s2 2p6

which is the same as the Noble gas of neon. The full outer shell induces a special stability and makes this particular configuration relatively unreactive. Again, the p-electrons are more shielded from the positive charge of the nucleus than the s-electrons and are therefore higher in energy. Thus, in order to convert from Al3+ to Al4+ and remove a 2p-electron electron from this ion, we need a large fourth ionisation energy.

So, based on this I would say the difference between the third and fourth ionisation energies for Al would possibly be the greatest. Hope that helps you out a bit! :)

Answer 2

11th and 12th ionization energy will have the greatest difference because of the attraction on the closest electron shell.

Answer 3

I think you may actually be right @Sachintha , it would be for the same reason as the difference between the third and the fourth IE but the shells/orbitals we are now taking electrons from will lie closer to the positively charged nucleus. I don't have the exact figures though so I'm not fully sure!

Answer 4

https://www.webelements.com/aluminium/atoms.html
Those data might be accurate.