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Sciencemadness Discussion Board » Fundamentals » Chemistry in General » Short report on making K2[MoOCl5] and K2[MoOBr5]

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Bedlasky

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posted on 22-12-2021 at 17:52

Short report on making K2[MoOCl5] and K2[MoOBr5]

Hi,

I made K2[MoOCl5] and K2[MoOBr5]. The later I finished today. Synthesis alone doesn't take long, but drying take literally weeks. Both are usefull reagents for making other Mo(V) complexes (but I personally will use them for some test tube reactions and analytical chemistry). I followed this procedure (with some minor changes):

Preparation of [MoOX5]2- (2).png - 119kB

I did all calculations on 5 g of MoO3, I just didn't write them somewhere. So KI, KOH, KBr - all in stoichiometric amounts. HCl/HBr in excess of course. Yield of K2[MoOBr5] is lower because I overheat ammonium heptamolybdate and lots of it turned in to mixed Mo(VI)/Mo(V) oxide which partially didn't react with HBr.

Preparation of K2[MoOCl5]

Starting materials:

36% HCl

(NH4)6Mo7O24.4H2O

KOH

KI

Procedure:

Heat ammonium heptamolybdate until it decompose to MoO3. Overheating is really easy, mix it well during heating. Your MoO3 probably will contain some Mo(VI)/Mo(V) oxide anyway, but small amount doesn't matter. Let the oxide cool down to RT. If some blue oxide doesn't dissolve in KOH in later step, add few drops of hydrogen peroxide and then bring it to boil to destroy excess of hydrogen peroxide. If you have MoO3 on hand, skip this step.

Then prepare 40% KOH and add MoO3 in to it. Solution will be hot, so cool it in water bath. After cooling add 25 ml of 36% HCl. Most of KOH is neutralized by MoO3, so don't worry. Solution will heat up, but not to boil. Finally add KI and heat up the solution. Following reaction occurs:

2K6Mo7O24 + 14KI + 2KCl + 68HCl = 14K2[MoOCl5] + 7I2 + 34H2O

Boil it until iodine stop escaping from the solution. HCl will quickly boil away during reaction, so add HCl as you need. When reaction is complete, boil dark brown solution down until solid starts to crystallize, place it on steam bath and let it evaporate (solid doesn't must be dry, it can be wet). You get dark brown solid. Treat this solid with 1-2 ml of 36% HCl. Solid turns from dark brown to green. Discard green solution and put solid in to the dessicator with NaOH. KOH works better from my experience, but this is probably caused by particle size (my KOH is big flakes, while NaOH small beads). Drying process will take cca 4 weeks. Change periodically NaOH.

Preparation of K2[MoOBr5]

Starting materials:

48% HBr

(NH4)6Mo7O24.4H2O

KBr

Procedure:

Make MoO3 as in previous synthesis. Dissolve it in HBr and add KBr. Bring solution to boil. Following reaction occurs:

2MoO3 + 4KBr + 8HBr = 2K2[MoOBr5] + Br2 + 4H2O

When bromine stop escaping from the solution, boil it down until solid starts to crystallize. Then let it evaporate on steam bath. Treat brown solid with 1-2 ml of 48% HBr, discard the solution and put solid in to the dessicator with NaOH. Drying process will take again cca 4 weeks. Change periodically NaOH. K2[MoOBr5] is dark brown solid.

Both complexes are soluble in conc. HCl/HBr. In water they hydrolyse, chloride/bromide ligands are replaced by water:

[MoOX₅]^2- + H₂O <--> [MoOX_(5-x)(H₂O)_x]^(x-2)+ + X^-

These complexes tend to dimerize to [Mo₂O₃X_(8-x)(H₂O)_x]^(x-4)+ and [Mo₂O₄X_(6-x)(H₂O)_x]^(x-4). In these complexes, two MoO units are linked through one or two oxygen bridges, every MoO unit have 4 or 3 other ligands (halide, water etc.) - [4LOMo-O-MoOL4] and [3LOMo=(O)2=MoOL3], where L is ligand.

*Side note: After dissolving MoO3 in HCl/HBr, you can observe formation of light yellow/yellow solution. This is caused by [MoO2Cl4]2-/[MoO2Br4]2- ions.

[Edited on 23-12-2021 by Bedlasky]

woelen

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posted on 24-12-2021 at 00:08

Very interesting syntheses. I hope to find some time for these during the christmas holidays.

There is one question I have. For making the K2MoOCl5 you used KI, stoichiometric amount. It is not easy to get a stoichiometric amount of that, because it is somewhat hygroscopic and the other starting materials also may have impurities. What would be better, having a little excess MoO3, or having a little excess KI. I can imagine that in the latter case, you might have contamination with iodide ligands in your end-product.

Isn't the use of K2S2O5 as reductor an alternative. You can boil off excess SO2 easily, and I can imagine that the presense of sulfate ions, which will be formed from this is not that severe. It may lead to a little K2SO4 impurity, but that would be less severe than having iodide contamination in your product.

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Bedlasky

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posted on 24-12-2021 at 00:22

Sulfites and bisulfites don't react with molybdate. Thiosulfate does, it can reduce it to V oxidation state. But your product will be contaminated with tetrathionite and other stuff. Another way to go is dissolve some Ti powder in HCl and use it as reductor. TiO2 can be filtered off. But be careful because TiCl3 is capable reduce molybdate to Mo(IV).

You can dry KI for better accurancy. If you add exces MoO3, your product will be contaminated with molybdenum blue. I personaly added very small excess of KI, I don't think that some minor KI contamination will cause problems in other reactions. And most of KI contamination was removed during HCl washing.

Sciencemadness Discussion Board » Fundamentals » Chemistry in General » Short report on making K2[MoOCl5] and K2[MoOBr5]