All roads lead to S4N4

Question

Why does thionyl chloride $\ce{(SOCl2)}$ when it reacts with liquid ammonia form a trace of $\ce{S4N4}$. What is the mechanism of the reaction forming $\ce{S4N4}$.

See discussion of sulfur chemistry for details.

Answer 1

Yes, you are indeed right in the comment section that the reaction between thionyl chloride and ammonia will give trace amount of tetrasulfur tetranitride(1) and the intended product is thionyl imide ($\ce{HNSO}$)(2).

1. From the 'thionyl chloride' entry from the same site you used in the question:

   Thionyl chloride reacts with ammonia giving rise to various complex products, the nature of which varies with the conditions.

2. As said, the main product from the reaction in gas phase is thionyl imide and trace amount of thiazyl chloride($\ce{NSCl}$) and sulfur dioxide. The reaction are as follows:

$$\ce{SOCl2(g) + 3NH3(g) -> HNSO(g) + 2NH4Cl(s)}$$ $$\ce{HNSO + H2O -> SO2 + NH3}$$

Ref. : Inorg. Chem., 1977, 16 (1), pp 216–217 DOI: 10.1021/ic50167a050

Answer 2

If we work out oxidation states, we find $\ce{S^{IV}}$ in thionyl chloride, and of course$\ce{N^{-III}}$ in ammonia. Thus formation of tetrasulfur tetranitride would involve the combination reaction

$\ce{4 S^{IV} +4 N^{-III} -> S4N4}$

But we see that to balance charges four electrons must be added, so sulfur tetranitride is actually a reduction:

$\ce{4 S^{IV} +4 N^{-III} + 4e^- -> S4N4}$

Hence a redox reaction must be occurring. Assuming the quite plausible notion that the oxidation half-reaction is additional sulfur being oxidized to $\ce{S^{VI}}$ ($\ce{S^{IV} -> S^{VI} +2e^-}$)*, we would form two moles of $\ce{S^{VI}}$ per mole of sulfur tetranitride. Then we could have something like this:

$\ce{20 NH3 + 6 SOCl2 -> S4N4 + 14 NH4^+ + 2 SO3NH2^- + 12 Cl^-}$

*This reference reports the formation of sulfate as well as sulfamate from thionyl chloride and ammonia, both of these species having $\ce{S^{VI}}$.