# Can linear SOCl2 exist?

Specifically, instead of its normal structure, this Lewis structure seems to be valid as a constitutional isomer:

$\ce{Cl-O-S-Cl}$

So what prevents this from existing, or if it already exists how is it made?

• Anything with Cl-O bond is a strong oxidizer, sulfur in the oxidation state +2 is a reducing agent, so the thing will pretty quickly transform into common $\ce{SOCl2}$. – Ivan Neretin Sep 18 '15 at 13:26
• Just for clarification: You do mean ‘linear’ $\ce{Cl-O-S-Cl}$ as opposed to trigonal pyramidal $\ce{O-SCl2}$, correct? – Jan Oct 1 '15 at 12:20

Not for long. Sulfur is in this case a reducing agent (love to give electrons away). And Oxygen is an odixidizing agent. It's just not stable or favorable. But sure you can make it.

• Ivan's comment is better than this answer - you need to back up claims about possibility of creation in the answer – Mithoron Sep 18 '15 at 14:12
• Not quite. I failed to add a discrediting note like "...should this thing exist, it will..." – Ivan Neretin Sep 18 '15 at 15:22
• @Ivan Neretin Yes, it looks like it would be better, but still... – Mithoron Sep 19 '15 at 0:02

I highly doubt that it could exist. The co-linear arrangement, $C_{\infty\mathrm{v}}$, of $\ce{Cl-O-S-Cl}$ is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the $S=0$ potential energy surface. It has two imaginary bending modes, $\nu_1=371.39\mathrm{i}$, $\nu_2=108.23\mathrm{i}$. (Please note that they are actually perpendicular to each other.)

That basically means, that distortion along these modes will lead to an arrangement of the nuclei which is lower in energy. Since this is a second order saddle point, this arrangement can itself only be a transition state (first order saddle point), which also cannot be considered stable. (Compare also here.)
Therefore distortion among $\nu_1$ would probably lead to $\ce{Cl-O}$ and $\ce{S-Cl}$ fragmentation. While distortion along $\nu_2$ might preserve the $\ce{S-O}$ bond, fragmenting the chlorine atoms/ chloride ions - but that is pure speculation of a hypothetical case.

Let's try a very tempting ad-hoc explanation of why this is the case. In that case let us assume all atoms are $\mathrm{sp}$ hybridised and the bonding axis is $z$. The Lewis structure that seems to be valid has therefore three lone pairs ($\mathrm{sp}$, $\mathrm{p}_x$, $\mathrm{p}_y$) at each chlorine and two lone pairs ($\mathrm{p}_x$, $\mathrm{p}_y$) at the oxygen and sulfur. The three σ bonds are formed by the $\mathrm{sp}$ orbitals.
That would result in four lone pairs that are co-aligned in the same plane. Overlap is inevitable, and in the most simplest terms of VSEPR theory, we would call that lone pair repulsion. (Please note that for strictly VSEPR theory we would have to assume tetrahedral hybridisation.)
In other words, same spin electrons have to avoid occupying the same space and since the orbitals are so close together this will rise the energy a lot. You can probably call it Pauli repulsion.

The MO picture is too messy, so that I won't show it here. Basically most of the occupied orbitals are anti-bonding with respect to one of the bond. The HOMO is of π symmetry and anti-bonding with respect to every bond.

• This is a good answer, but unfortunately I wasn't really interested in sp hybridized linear, I meant it in the "straight chain alkane" linear sense where the oxygen and sulfur are likely sp3 hybridized. – user19026 Oct 2 '15 at 0:24
• @Kainui you mean in a zig-zag fashion? Well then it is not linear, is it? – Martin - マーチン Oct 2 '15 at 3:18
• It's not linear in that symmetry-phys chem sense, but in sense of org chem, i.e. 'linear alkane chain', it is somehow "linear". – pH13 - Yet another Philipp Oct 2 '15 at 6:38
• Yeah exactly @ph13 , I mean it seems kind of funny to imagine an oxygen in a truly linear state, I can't think of a single case where oxygen is truly 'linear' in that sense lol. – user19026 Oct 2 '15 at 12:54