Confusion about gauche effect

This recent [question](https://chemistry.stackexchange.com/questions/56972/why-is-gauche-conformer-of-2-3-dimethyl-butane-more-stable-than-its-anti-conform) about the gauche effect led me to read this rather older [answer](https://chemistry.stackexchange.com/a/39204/30803). The problem is that [I notice I am still confused](http://lesswrong.com/lw/if/your_strength_as_a_rationalist/).


To take the case of **1,2 difluoroethane**, I drew these orbitals:
[![orbitals of C2H4F2 in the gauche (1,2) and anti (3,4) confromations](https://i.stack.imgur.com/LQiRj.png)](https://i.stack.imgur.com/LQiRj.png)
**Figure1:** orbitals of $\ce{C2H4F2}$ in the gauche (1,2) and anti (3,4) confromations


Now I understand that **1** (gauche) leads to *more* stabilization than **3** (anti). But **2** (gauche) leads to *less* stabilization than **4** (anti). I don't know which effect is strongest: the destabilization in **2** or the stabilisation in **1**. So I don't know why the gauche conformation (**1**,**2**) leads to more stabilization than the anti (**3**,**4**).


Did I draw something wong? Did I make a mistake in reasoning? Did I miss something else? Is this one too many error-acknowledgement questions?




In figures 2, it looks like you're saying the sigma of the C-F will be donating into the sigma\* of the C-H, and that causes stabilization. However, a C-F sigma bond is a much poorer donor than a C-H due to the high electronegativity.


Since there is little to no donation actually occurring in 2 (or 4, for the matter) those interactions are much less relevant than depicted in 1/3 where there is significant donation for C-H sigma into C-F sigma\*. Frankly I would never have even thought to draw 2/4 since there is so little electron density in the H lobe of the C-F sigma, but cool!


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