Stereochemistry in Sn2 reaction

I'm confused why the absolute configuration always has to change after an $S_N2$ reaction:

For example, in primary alkyl halide above there are 4 planes Nu can attack; and absolute configuration is changed after an $S_N2$ reaction because the nucleophile (Nu) only attacks the 'black plane'.
I can understand that the nucleophile wouldn't prefer red or green plane because of steric hindrance, but here's my question:

Why doesn't the nucleophile attack the 'blue plane'?

The attacking nucleophile does not just shove its electron pair to the carbon from some random direction. No, it must donate that pair to the antibonding orbital ($\sigma^*$) of the $\ce{C-X}$ bond, thus weakening and eventually breaking that bond. And it just so happens that the antibonding orbital is mostly located on the side opposite to X.

Upd. Thought I'd better look up a nice picture. Here's one: (from this page)

• Ug. F as a leaving group. Ug. – Lighthart Mar 16 '16 at 21:03
• Yeah, yeah, I know, but it would suffice to illustrate the basic idea. – Ivan Neretin Mar 16 '16 at 21:23

The reason behind which 'plane' the nucleophile attacks has to do with Molecular Orbital Theory.

But here's a short explanation for this: the C-X bond appears because two atomic orbitals interact to form two molecular orbitals: a filled bonding orbital ($\sigma$) and an unfilled anti-bonding orbital ($\sigma^*$).

For the C-X bond to break, the nucleophile has to attack $\sigma^*$, which means that the nucleophile has to get a good overlapping of its filled orbital with the unfilled anti-bonding orbital of C-X. Because of this attack, the C-X bond breaks and the C-Nu bond forms.

To give you an image of where the anti-bonding orbital is located, think of the C-X $\sigma$-bond; behind that bond on the carbon side, there lies the biggest portion of $\sigma^*$ (see Ivan's image). If we consider that, this orbital intersects the 'black plane' that you were talking about.

Therefore, the nucleophile can't attack the other 'planes' besides the 'black plane' because that doesn't assure a good overlapping between orbitals.

As for the change in configuration, in primary alkyl halides (the example you gave) we don't have a change of configuration because the carbon atom that's bonded to the halogen is not chiral.

But with a chiral compound, (i.e. (R)-2-chlorobutane), we get a change in configuration because the nucleophile attacks 'from the back', reversing it (i.e. (S)-butan-2-ol.)