How does the organic matter Rubidium tetrabromoaurate synthesize? Can any expert give some suggestion to me. Thank you in advance!

Edited July 17, 200817 yr by Dak
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The simplest synthesis I can think of is mixing gold with a solution of nitric acid and hydrobromic acid to form auric acid which can then react with rubidium to form rubidium tetrabromoaurate. Rubidium tetrabromoaurate isn't organic btw.

moved to inorganic.

 

I havent a clue how you'd make this stuff to be honest

Hmm actually the auric acid I mentioned isn't auric acid if it's made with HBr instead of HCl, as auric acid is HAuCl4. "Bromoauric acid" maybe? Perhaps one of the chemistry experts knows if the synthesis I mentioned will actually work.

Hmm actually the auric acid I mentioned isn't auric acid if it's made with HBr instead of HCl, as auric acid is HAuCl4. "Bromoauric acid" maybe? Perhaps one of the chemistry experts knows if the synthesis I mentioned will actually work.

 

i'm not sure if auric acid exists. [ce]HAuCl4[/ce] is hydrochloroauric acid, so i guess the bromo one is hydrobromoauric acid

i'm not sure if auric acid exists. [ce]HAuCl4[/ce] is hydrochloroauric acid, so i guess the bromo one is hydrobromoauric acid

 

Oh. I recall the product of the rather famous gold + aqua regia reaction ([ce]HAuCl4[/ce]) being called auric acid, but that's probably just a trivial name.

RbBr + AuBr3 = RbAuBr4

all as Aqueous.

 

 

HBr + AuBr3(aq) ---> H+[AuBr4]−

but other Halide sources can be used here also, RbBr for instance

AuBr4- ought to be fairly easily synthesised either as YT2095 says from RbBr and AuBr3.

 

There may also be other ways. This is because AuX4- is d8 square planar and these compounds are extremely labile as regards substitution. Thus displacement of Cl- ligands from AuCl4- would probably work in the presence of excess RbBr. The trans directing nature of Cl- and Br- should be similar. Some of the original work on substitution of d8 square planar complexes was carried out on such gold complexes, (1940's and 1950's) which seems a little surprising to me as d8 square planar Pd(II), Pt(II) Rh(I) and Ir(I) have been more thoroughly studied