Why are noble gases really hard to ionize, and why do they have positive electron affinities?

Because ionization of noble gases requires either violating the Pauli Exclusion Principle or taking away what makes the noble gas stable.

Read below for a more elaborated answer.

DISCLAIMER: LONG ANSWER!

WHY AN OCTET RULE?

First off, it would be helpful to you if you read this formal explanation on why the octet rule is a thing.

It requires that you have seen quantum numbers before (#n, l, m_l, m_s#) and have some idea of what the Pauli Exclusion Principle is.

You should read the whole explanation, but the gist of it is that no two electrons can occupy the same quantum state (Pauli Exclusion Principle).

THIS IS A VERY STRICT PRINCIPLE. IF TWO ELECTRONS TRY TO OCCUPY THE SAME QUANTUM STATE, THEY BOTH DISAPPEAR BECAUSE IT IS IMPOSSIBLE.

For the second period on the periodic table, if you go through the derivation, it works out that a maximum of #8# unique quantum states are available for the valence electrons, giving the "octet rule" we speak of.

It just means that a maximum of #8# unique valence electrons can exist within the #2s# and #2p# orbitals in total at any given time. That would apply to neon, specifically (electron configuration: #1s^2 2s^2 2p^6#), so take that as the first example.

WHY DOES THE OCTET RULE HAVE AN EXCEPTION FOR THE THIRD ROW AND ON?

I will go through a similar derivation, but for the fourth period/row in the periodic table, where we see krypton (#Z = 36#).

Krypton's electron configuration is:

#color(blue)(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6)#

Krypton does NOT have #18# valence electrons. It has #8#. For the fourth row (not quantum level) in the periodic table, we have the following quantum numbers, representing the #3d#, #4s#, and #4p# orbitals.

THE 3D ORBITALS

#n = 3#:

#l = 2#
#m_l = 0, pm1, pm2#
#m_s = pm"1/2"#

With #l = 2#, we have five values of #m_l#, giving five different #3d# orbitals (#d_(xy), d_(xz), d_(yz), d_(x^2 - y^2), d_(z^2)#), and thus five different quantum states for the electron.

Additionally, with #m_s = pm"1/2"#, the electron can be spin-up or spin-down, for each of those five different quantum states.

That means we can have a total of #5 xx 2 = \mathbf(10)# 3d electrons that can exist within the five #3d# orbitals, which we do. However, they are NOT valence.

THE 4S ORBITAL

#n = 4#:

#l = 0#
#m_l = 0#
#m_s = pm"1/2"#

With #l = 0#, we have one value of #m_l#, giving one unique #4s# orbital, and thus one quantum state for the electron so far.

Additionally, with #m_s = pm"1/2"#, the electron can be spin-up or spin-down, for that quantum state.

That means krypton has a total of #1 xx 2 = \mathbf(2)# valence electrons that can exist within the one #4s# orbital, and it does already.

THE 4P ORBITALS

#n = 4#:

#l = 1#
#m_l = 0, pm1#
#m_s = pm"1/2"#

With #l = 1#, we have three different values of #m_l#, giving three unique #4p# orbitals (#4p_x,4p_y,4p_z#), and thus three different quantum states for the electron so far.

Additionally, with #m_s = pm"1/2"#, the electron can be spin-up or spin-down, for each of those three unique quantum states.

That means krypton have a total of #3 xx 2 = \mathbf(6)# valence electrons that can exist within the three #4p# orbitals, and it has all #6# already.

WHAT DOES ALL THAT MUMBO JUMBO MEAN?

If you at home have been keeping track, that's #10 + 2 + 6 = 18# outermost electrons that Krypton can have: #10# from the #3d#, #2# from the #4s#, and #6# from the #4p# orbitals.

However, the #3d# electrons are NOT valence because they are energetically much lower than the #4s#. So, there are ONLY #8# valence electrons.

If there's anything you should remember from all this, it's that:

The Pauli Exclusion Principle is a VERY strict rule that says that if you try to go beyond the number of quantum states that it allows for the particular atom, it won't work.

You can take away electrons, to be sure, though it's still hard. But you would have a very hard time adding more to krypton. It would be a destabilizing action.

IONIZATION ENERGY AND ELECTRON AFFINITY OF KRYPTON

What does that mean? Let's examine the first ionization energy and the first electron affinity.

#"IE1"_"Kr" ~~ color(blue)("1350.7 kJ/mol")#

#"EA1"_"Kr" > color(blue)("0 kJ/mol")#

You can see a large value for the ionization energy, which makes sense, but something positive for the electron affinity. It couldn't be measured, so it was never reported. That just means krypton doesn't want any more electrons. in other words:

If all possible quantum states are occupied, like with krypton (and other noble gases), the element is stable.

Stable is good! It's like giving someone every material and immaterial possession they could ever want in their entire life and filling all their closet space. If you took away one of those things (like a giant flat-screen TV or their job), they would be furious, and perhaps even wrestle you for it back.

That's probably how krypton feels when you try to take away an electron. It's like losing some of the things you love in life. :)

If you gave them another job or another possession, they wouldn't want another job to busy themselves too much, nor another possession because everything is already in its place and there is literally no room for anything else.

That's probably how krypton feels when you try to give it an extra electron. It doesn't have room for more.