Why Birch reduction of alkyne is E-selective?

From Organic Chemistry(Clayden, Greeves and Warren, 2nd edition), pp. 682:



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> ![Birch reduction of alkyne](https://i.stack.imgur.com/LDnR4.png)
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> The sodium donates an electron to the LUMO of the triple bond (one of the two orthogonal $\pi^\ast$ orbitals). The resulting radical anion can pick up a proton from the ammonia solution to give a vinyl radical. **A second electron, supplied again by the sodium, gives an anion that can adopt the more stable *trans* geometry.** A final proton quench by a second molecule of ammonia or by an added proton source (t-butanol is often used, as in the Birch reduction) forms the *E* alkene.
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I have some questions on this reaction:


(1) Double bond on the vinyl anion blocks rotation, but how can anion adopts *trans* geometry between *cis* and *trans* without rotation?


(2) Why anion with *trans* geometry is more stable? If it is because of steric repulsion between $R^1$ and $R^2$, why electronic repulsion between lone pair and $R^1$ can be ignored?




[![Credits to Kurti (2019)](https://i.stack.imgur.com/1mwSE.png)](https://i.stack.imgur.com/1mwSE.png)


It's not rotation but inversion (remember amine lone pair inversion) from a linear $sp$ form to trigonal planar $sp^2$ that occurs. Lone pair and radical electron are held close to atom and repulse each other a lot so stable to keep them in anti dihedral positions. If the lone pair next to a radical electron was less repulsive, then it would have a lower energy transition state and be more readily formed according to Hammond's Postulate. But experimental data indicates that that is not the case.


Picture Citation: Kurti (2019)


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