janelee Posted July 16, 2008 Share Posted July 16, 2008 (edited) How does the organic matter Rubidium tetrabromoaurate synthesize? Can any expert give some suggestion to me. Thank you in advance! Edited July 17, 2008 by Dak removed link Link to comment Share on other sites More sharing options...
Gilded Posted July 16, 2008 Share Posted July 16, 2008 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. Link to comment Share on other sites More sharing options...
hermanntrude Posted July 16, 2008 Share Posted July 16, 2008 moved to inorganic. I havent a clue how you'd make this stuff to be honest Link to comment Share on other sites More sharing options...
Gilded Posted July 16, 2008 Share Posted July 16, 2008 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. Link to comment Share on other sites More sharing options...
hermanntrude Posted July 16, 2008 Share Posted July 16, 2008 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 Link to comment Share on other sites More sharing options...
Gilded Posted July 16, 2008 Share Posted July 16, 2008 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. Link to comment Share on other sites More sharing options...
YT2095 Posted July 17, 2008 Share Posted July 17, 2008 RbBr + AuBr3 = RbAuBr4 all as Aqueous. HBr + AuBr3(aq) ---> H+[AuBr4]− but other Halide sources can be used here also, RbBr for instance Link to comment Share on other sites More sharing options...
Tartaglia Posted July 24, 2008 Share Posted July 24, 2008 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 Link to comment Share on other sites More sharing options...
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