# How does VB theory explain the bond angles in SiO2?

Silicon dioxide has a huge variety of structures. Most of them have a tetrahedral $\ce{SiO2}$ unit cell — the $\ce{O–Si–O}$ angle is $109.5°$, accordingly. The VB/hybridization approach to this tetrahedron would consequently assign $4\text{ }\mathrm{sp^3}$ orbitals to the silicon atom.

Now, the interesting part about silicon dioxide is the flexibility of its $\ce{Si–O–Si}$ angles; they vary from $100$ to $170$ degrees, depending on the polymorph. How is this possible? Since the oxygen bonds to two silicon atoms, one would expect a $\mathrm{sp^3}$ hybridization, with two orbitals filled with lone pairs — hence, a $104°$ angle like in water, or a slightly bigger one (since silicon is less electronegative than hydrogen — thus, the $\ce{Si–O}$ bond is more polar than the H–O bond and the angle bigger). But never an angle as large as $170°$: that would almost point at an $\mathrm{sp}$ hybridization of the oxygen atoms. How is such an angle energetically more convenient than the typical bent one?

• Most of the silicon dioxide polymorphs show deviation from the ideal (predicted) angles, but they're metastable and will eventually revert to $\alpha$-quartz. See The Quartz Page for a more thorough discussion. Also note that the angles are for Si-O-Si and not O-Si-O. – Todd Minehardt Dec 27 '15 at 19:07
• @ToddMinehardt If the varying angles are $\ce{Si-O-Si}$ rather than $\ce{O-Si-O}$, does that mean $\angle (\ce{O-Si-O}) = 109^\circ$ always? If so, OP’s question is moot … – Jan Dec 27 '15 at 20:20
• @Jan - Yes, the question is more or less moot. It is the network arrangement of $\ce{SiO4}$ tetrahedra in various polymorphs from which the variation in $\ce{Si-O-Si}$ angles arises. Each of the tetrahedra are bonded to the surrounding ones ${\it via}$ an oxygen atom - see figures 3.04, 3.05, and 3.06 on this page for a clear pictorial. The $\ce{O-Si-O}$ angles are as expected for a tetrahedral geometry and do not deviate from predicted by much at all. – Todd Minehardt Dec 27 '15 at 21:48
• That the Si–O–Si bond angles arise from the arrangement of the SiO4 tetrahedrons is obvious, as well as the fact that the O–Si–O angles are always 109.5° in stable polymorphs (since we're talking about tetrahedrons). The question is: how does VB theory represent the bonding between oxygen and silicon with these extreme variety of angles? How does it justify the existence of non-109° bonds around an oxygen molecule? Why do these bond angles differ so drastically from other X–O–X bonds such as those in water, according to VB theory? – Corundum Dec 27 '15 at 23:34
• There is also the possibility that valence bond theory fails for a silicon dioxide. There are many compounds for which VB can't explain the structure. – Tyberius Aug 3 '17 at 2:51