# Why is disilyne bent?

Why is disilyne $\ce{Si2H2}$ bent when the steric number of silicon is 2?

• Short answer is: because VSEPR theory is oversimplified model of reality. Sep 21 '14 at 10:25
• "The equilibrium geometry of disilyne is not linear, but is twisted. The potential surfaces of acetylene and disilyne have a critical internuclear distance between the central atoms, where the stable geometry changes from linear to twisted forms. The R-dependence of the valence-shell electron energy causes the difference in the structure of the molecules." dx.doi.org/10.1016/0009-2614(82)87044-9 Sep 21 '14 at 10:30
• Could you elaborate that in an answer? Why is that particular form more abundant than linear?
– EJC
Sep 21 '14 at 10:32
• So electronic structure calculations cited above predict twisted equilibrium geometry for $\ce{H2Si2}$ with the only issue that the calculations are rather crude (HF/STO-3G). :D Sep 21 '14 at 10:33
• The twisted form of $\ce{H2Si2}$ is more energetically stable, and thus, it is more likely to find $\ce{H2Si2}$ in this form. The question why the twisted form is more stable does not have an answer you are looking for, except for that this is just the way nature works. :D It can be explained in terms of electronic structure theory as in the paper linked in one of my previous comments. Sep 21 '14 at 10:38

The full text of the article describing the preparation, characterization and reactivity of the compound can be found here. See figure 4 in the paper for a MO explanation as to why the trans-bent geometry occurs. Basically, the authors suggest that, "the bending is thought to be the result of the mixing of an in-plane $\ce{\pi}$-orbital with a $\ce{\sigma^{\ast}}$ orbital whose energies are close enough to cause the interaction."
• I was impressed by the fact (above link, Section 4, paragraph 2) that the bond length contraction going from $\ce{Si=Si}$ to $\ce{Si#Si}$ is half that for $\ce{C=C}$ to $\ce{C#C}$. That seems significant to me.