[Bob Sacamano]

3M hydrochloric acid was added to Sn(s) and KI(s) and heated under reflux in an inert atmosphere. Why must the reaction be carried out under an inert atmosphere?

Initially I had thought it was to prevent the moisture in the air from entering the system. But there is already plenty of water in the solution because aq HCl was added. Then I thought that the the following reaction would take place:

Sn (s) + 2 HCl (aq) → SnCl2 (aq) + H2 (g)

Then I have the Sn(II) required to form the stannous iodide but its bonded to Cl... And besides that still doesn't explain why the reaction must be carried out under an inert atmosphere.

[Borek]

What is atmosphere composition?

[Bob Sacamano]

N₂ - inert
H₂O - already present in the system
Ar - inert

Leaving oxygen as the most likely suspect.

So assuming the inert atmosphere is to prevent the oxidation of the SnCl₂, the SnCl₂ must be somehow converted to SnI₂ making SnCl₂ an intermediate.

But I don't get why a chlorinated intermediate is formed first? why isn't the SnI₂ formed immediately?

[Borek]

Perhaps because HI is much more expensive and difficult to obtain?

[Bob Sacamano]

Yes, but the potassium iodide is also in solution. Why do we need the HCl to form stannous chloride before it is converted to stannous iodide?

Previously we synthesized tin(iv) iodide but did not use HCl as a solvent.

Is the stannous chloride complex somehow required to form as an intermediate in the process of synthesizing stannous iodide?

[Borek]

You need to oxidize tin - what will be oxidizing it in just a KI solution?

How did you made SnCl₄?

[Bob Sacamano]

We didn't make SnCl₄ to get SnI₄. We just heated Sn and KI in a glacial acetic acid / acetic anhydride solution at very high temperature and then crystallized the SnI₄ out of the solution.

I don't see why the Cl was needed for the one reaction but not the other.

[Borek]

Was the solution in contact with the air (oxygen)?

[Bob Sacamano]

Ahhhhh. I see. In that reaction the Sn is being oxidized by atmospheric oxygen to yield tin dioxide. And in the other reaction the tin is being oxidized by the Cl.

But why does the Cl induce the 2+ oxidation state and oxygen induce the +4 oxidation state?

[Borek]

It is not Cl^- that does the oxidation.

And atmospheric oxygen was not to yield tin dioxide. I think it is a clever combination of reagents - including acetic anhydride - that removes any traces of water that can be produced in the oxidation process, thus leaving just Sn(IV) that precipitates with iodide.

[Bob Sacamano]

Hmmm.

SnI₂ case:

Sn_(s) + 2H⁺ Sn²⁺ + H₂

Now, we still have the Cl^- and I^- in solution and have oxidized the Sn to a 2⁺ state. Why does stannous iodide now crystalize instead of stannous chloride? Since chloride is a better nucleophile won't it bond preferentially?

SnI₄ case:

The acetic anhydride will react with any water in the system to produce more acetic acid. Is H⁺ again acting as an oxidizing agent in this reaction? If so, why is this Sn complex oxidized to a 4+ state? The higher heat? and why is this system allowed to interact with atmospheric oxygen?

[Borek]

I guess the solubility of the iodides is much smaller.

You are right about H⁺ being an oxidizing agent.

Oxygen is much stronger oxidizing agent than H⁺.

[Bob Sacamano]

These are my proposed mechanisms:

SnI₄ case:

Sn_(s) + O₂ SnO₂

SnO₂ + 4I^- + 4H⁺ SnI₄ + 2H₂O

SnI₂ case:

Sn_(s) + 2H⁺ Sn²⁺ + H₂

Sn²⁺ + 2I^- SnI₂

Am I on the right track without these?

[Borek]

I told you - I doubt SnO₂ was present in the meantime.

Seems to me like you learned organic chemistry before inorganic, people that go that way try to find some strange mechanisms for inorganic reactions. That's not the way inorganic chemistry is done

You have to write probable half reaction for oxidizing agent - how it gets reduced. How the tin is oxidized is quite obvious, it just loses 4 electrons.

[Bob Sacamano]

Ok I'll take another stab at it. Although I must admit that my redox chemistry is lacking.

Sn_(s) Sn⁴⁺ + 4e^-

O₂ + 4e^- 2O₂^2-

2O₂^2- + 4H⁺ 4H₂O

Sn⁴⁺ + 4I^- SnI₄

That's my best guess. Although I still don't see why stable O₂ would take up the electrons to form O^2-.