Question

Explain the following observation using principles of atomic structure and/or bonding. "The second ionization energy of sodium is about 3 times greater than the second ionization energy of magnesium."

Answer 1

Let's first define what ionisation energy is:

"The ionisation energy (or the first ionisation energy to be more precise) is the energy needed to completely remove the most loosely bound electron from a neutral gaseous atom."

In terms of the "most loosely bound electron", we associate this to an electron in the shell which is furthest away from the positive charge in the nucleus, which the electrons are attracted to. The further away we are from the nucleus, the weaker this electrostatic attraction, plus we have the added effect of 'shielding' from electrons between the nucleus and the most loosely bound electron, thus reducing this attraction further.

So, the second ionisation energy will be the energy required to remove the most loosely bound electron on a neutral gaseous atom which has already lost one electron, and thus is an ion.

The reason why the second ionisation energy (IE) for sodium is higher than magnesium lies in their electronic configurations. You kind of need to understand what these are in order to answer this question, and how we fill electrons into the different orbitals at the different energy levels, building from the lowest energy level upwards. So, the electronic configuration of a sodium atom, with 11 electrons, is:
\[1s ^{2}2s ^{2}2p ^{6}3s ^{1}\]

If we remove the most loosely bound electron by meeting the first ionisation energy, we are left with a sodium cation with an electronic configuration of:
\[1s ^{2}2s ^{2}2p ^{6}\]

As you can see, the electron in the s-orbital in the third energy level (the furthest out) has been removed. Similarly, the electronic configuration for neutral magnesium is (with 12 electrons):
\[1s ^{2}2s ^{2}2p ^{6}3s ^{2}\]

As with sodium, if we remove one electron, we do so from the 3s orbital, furthest away from the nucleus and of the highest energy. however, there will still be one electron left in the orbital, and so the electronic configuration now looks like:
\[1s ^{2}2s ^{2}2p ^{6}3s ^{1}\]

(Continued)

Answer 2

So, we now want to remove a second electron from these sodium and magnesium cations. Following the same rules as before of removing the most loosely bound electron in the orbital of highest energy, which is furthest away from the nucleus:

- For the sodium ion, with an electronic configuration of:
\[1s ^{2}2s ^{2}2p ^{6}\]


If we remove an electron from the 2p orbital, then we are removing it from what we call a 'full subshell'. There is a special stability for an atom which has an electronic configuration containing a half-full or a completely full outer subshell, so it will not want to lose an electron easily in this case. So, the second ionisation energy would be relatively high as it would take a lot of energy to disrupt this stability.

-For the magnesium ion, with an electronic configuration of:
\[1s ^{2}2s ^{2}2p ^{6}3s^{1}\]

the next electron to be removed will come from the same orbital as the first electron to be removed. Indeed, the ion will want to lose this electron more than sodium, as in doing so its outer subshell would now be a full subshell (with 2p), and it would gain extra stability from this. So, this is why the second ionisation energy for the magnesium ion would be relatively low compared to that of the sodium ion.

It takes a bit of explaining, but once you get your head round it the answer here should hopefully be fairly clear, just understanding the electronic configurations and making sure to remember our rule on full and half-filled subshells. Hope that helps you out! :)