[Catsceo]

Is H₃BiCl₆ possible to synthesize?  Would it be any stronger or similar to HSbF₆?

[TheUnfocusedOne]

http://books.google.com/books?id=YmBiXJ7qRtAC&pg=PA260&lpg=PA260&dq=H3BiCl6&source=bl&ots=3rnGQnNawG&sig=hT2lQtCY0fnq1mOx7y9M8KbR124&hl=en&ei=j6L-TNvND8P98AaRu4SFBw&sa=X&oi=book_result&ct=result&resnum=3&ved=0CB8Q6AEwAg#v=onepage&q=H3BiCl6&f=false

Its possible.

[AndersHoveland]

Sharing my quick thoughts about possible super acids, and stronger fluoride ion acceptors:

Can I suggest that
(CF3)3SbO , triflic anyhydride (FSO2)2O, and HOSO2CF3 will react to form the acid
H(+) Sb(CF3)3(OSO2CF3)3(-).

This acid would be a useful superacid, much safer because it will not hydrolyze into toxic hydrogen fluoride gas. I would like to point out that I can pour concentrated perchloric acid on my skin if I quickly wash it off because a protective layer of amine perchlorates (that has a low solubility) form. If the danger of HF can be eliminated from a certain type of superacid, it might be possible to do the same. I'm not suggesting playing around with the stuff, just that it could be less of a danger than HSbF6


Sb(CF3)3 has already been prepared.
"Bis(trifluoromethyl)cadmium complexes react with antimony trihalides in polar solvents to form Sb(CF3)3 in high yield." Dieter Naumann, Wieland Tyrra and Ferdinand Leifeld

I am currently aware of H+B(CF3)4-, and carborane but these super-acids are expensive or not readily obtainable. H(+) Sb(CF3)3(OSO2CF3)3(-) could more readily be prepared and useful in the lab.

I think that BF4- might react with SF4, although I think most chemists would think this is impossible. Although BF3 is a stronger Lewis acid, in this case SF4 may act like the Lewis acid.

BF4- and 4 SF4 --> B(SF5)4-

B(CF3)4 (-) anion has already been prepared and found to hold its F- more strongly than even SbF6-. B(SF5)4- should be even better, possibly more than carborane anions, allowing stabilization of new extreme cations not currently possible.

I would not agree that BF4- is inert, it likely is fairly reactive, but the equilibrium usually favors BF4-, thus it should be able to temporarily donate a F- to SF4. Then the SF5- can stick back on to the boron. B(SF5)4- would likely be very non-coordinating. The parent compound SF6 is highly inert.


Of course SClF5 is quite reactive and could be used to introduce the SF5 group to the Boron atom.

consider the B(CF3)4- ,which has been prepared, formed by B(CN)4- with ClF3.
The corresponding acid H+ B(CF3)4- is more acidic than HSbF6!


B(CF3)4- anion is known to hold on to its fluoride ion more than even SbF6- , that is to say that the parent compound B(CF3)3CF2 (I am not sure if the parent compound is real or hypothetical, but it was printed in some literature) is a stronger F- abductor than even SbF5. I believe it may have been your group that created N5+B(CF3)5- by precipitating out CsSbF6, in an attempt to create a more stabilized pentazolium salt.
The thought occurred to me, if B(CF3)4- is better than SbF6-, than would not Sb(CF3)6- be even less coordinating? Possibly this could be done by using (CH3)4N+, Sb(CN)6- and reacting with ClF3. As you already well know (CH3)4N+ is not attacked by ClF3. Then the tetramethyl ammonium cation would be exchanged for a hydrogen cation, leaving HSb(CF3)6, perhaps even more acidic than carborane acids.
And since Au2F10 is a stronger F- abductor (on wikipedia) than SbF5, perhaps Au(CF3)5(CF2), if it could be made, would be an extremely strong fluoride ion abductor, possibly enabling the stabilization of new extreme cations. Au(CF3)5(CF2) may be more likely to exist than Sb(CF3)5(CF2) because since AuF5 is electron deficient enough to dimerize, the gold could possibly share electrons with the carbon, which is only otherwise bound to the two fluorine atoms.

[Catsceo]

Sharing my quick thoughts about possible super acids, and stronger fluoride ion acceptors:

Can I suggest that
(CF3)3SbO , triflic anyhydride (FSO2)2O, and HOSO2CF3 will react to form the acid
H(+) Sb(CF3)3(OSO2CF3)3(-).

This acid would be a useful superacid, much safer because it will not hydrolyze into toxic hydrogen fluoride gas. I would like to point out that I can pour concentrated perchloric acid on my skin if I quickly wash it off because a protective layer of amine perchlorates (that has a low solubility) form. If the danger of HF can be eliminated from a certain type of superacid, it might be possible to do the same. I'm not suggesting playing around with the stuff, just that it could be less of a danger than HSbF6


Sb(CF3)3 has already been prepared.
"Bis(trifluoromethyl)cadmium complexes react with antimony trihalides in polar solvents to form Sb(CF3)3 in high yield." Dieter Naumann, Wieland Tyrra and Ferdinand Leifeld

I am currently aware of H+B(CF3)4-, and carborane but these super-acids are expensive or not readily obtainable. H(+) Sb(CF3)3(OSO2CF3)3(-) could more readily be prepared and useful in the lab.

I think that BF4- might react with SF4, although I think most chemists would think this is impossible. Although BF3 is a stronger Lewis acid, in this case SF4 may act like the Lewis acid.

BF4- and 4 SF4 --> B(SF5)4-

B(CF3)4 (-) anion has already been prepared and found to hold its F- more strongly than even SbF6-. B(SF5)4- should be even better, possibly more than carborane anions, allowing stabilization of new extreme cations not currently possible.

I would not agree that BF4- is inert, it likely is fairly reactive, but the equilibrium usually favors BF4-, thus it should be able to temporarily donate a F- to SF4. Then the SF5- can stick back on to the boron. B(SF5)4- would likely be very non-coordinating. The parent compound SF6 is highly inert.


Of course SClF5 is quite reactive and could be used to introduce the SF5 group to the Boron atom.

consider the B(CF3)4- ,which has been prepared, formed by B(CN)4- with ClF3.
The corresponding acid H+ B(CF3)4- is more acidic than HSbF6!


B(CF3)4- anion is known to hold on to its fluoride ion more than even SbF6- , that is to say that the parent compound B(CF3)3CF2 (I am not sure if the parent compound is real or hypothetical, but it was printed in some literature) is a stronger F- abductor than even SbF5. I believe it may have been your group that created N5+B(CF3)5- by precipitating out CsSbF6, in an attempt to create a more stabilized pentazolium salt.
The thought occurred to me, if B(CF3)4- is better than SbF6-, than would not Sb(CF3)6- be even less coordinating? Possibly this could be done by using (CH3)4N+, Sb(CN)6- and reacting with ClF3. As you already well know (CH3)4N+ is not attacked by ClF3. Then the tetramethyl ammonium cation would be exchanged for a hydrogen cation, leaving HSb(CF3)6, perhaps even more acidic than carborane acids.
And since Au2F10 is a stronger F- abductor (on wikipedia) than SbF5, perhaps Au(CF3)5(CF2), if it could be made, would be an extremely strong fluoride ion abductor, possibly enabling the stabilization of new extreme cations. Au(CF3)5(CF2) may be more likely to exist than Sb(CF3)5(CF2) because since AuF5 is electron deficient enough to dimerize, the gold could possibly share electrons with the carbon, which is only otherwise bound to the two fluorine atoms.

Do you mind fixing your chemical formulas?  They are very hard to understand.  Use the sub buttons to denote subscript.  Or better yet draw the compounds.

[AndersHoveland]

The N5(+) pentazolium cation has a structure: (-)N=N(+)--N(-)--N(+)=N(-)

The B(CF3)4 (-) anion is a boron atom connected to four --CF3 groups.
It is difficult to draw the corret structure with conventional techniques.
The extra electron resonates amongst all twelve fluorine atoms. The boron atom is electron deficient in its usual valence of +3, and so it can further share electrons.
I do not know whether there is any 3-center bonding on this anion, but it is quite possible that there is some aromaticity, as the nitroformate ion is known to have.
Thus the correct bonding structure may actually be quite complex- some configuration of delocalized 3-center bonding fields.

Sb(CF3)3(OSO2CF3)3(-) could also been drawn as

..O....O
..ll....ll
F3CSOSbOSCF3
..ll....ll
..O....O

with 3 (CF3) groups also attached to the antimony atom (Sb), which is not shown, and an additional third triflate ion group (also not shown) that is coordinated with the Sb atom. The triflate anion has the structure F3CSO2O(-), and the extra electron resonates around all three triflate groups, so that there is no distinguishing between triflate groups covalently bonded and triflate anions. To have a better conception, one must think in terms of fractional bonds and charges.
For example, with tetrafluoroborate ion BF4 (-) , it could be said that each bond has an order of 3/4 (three fourths), while each fluorine atom carries a charge of minus 1/4.

[Catsceo]

The N5(+) pentazolium cation has a structure: (-)N=N(+)--N(-)--N(+)=N(-)

The B(CF3)4 (-) anion is a boron atom connected to four --CF3 groups.
It is difficult to draw the corret structure with conventional techniques.
The extra electron resonates amongst all twelve fluorine atoms. The boron atom is electron deficient in its usual valence of +3, and so it can further share electrons.
I do not know whether there is any 3-center bonding on this anion, but it is quite possible that there is some aromaticity, as the nitroformate ion is known to have.
Thus the correct bonding structure may actually be quite complex- some configuration of delocalized 3-center bonding fields.

Sb(CF3)3(OSO2CF3)3(-) could also been drawn as

..O....O
..ll....ll
F3CSOSbOSCF3
..ll....ll
..O....O

with 3 (CF3) groups also attached to the antimony atom (Sb), which is not shown, and an additional third triflate ion group (also not shown) that is coordinated with the Sb atom. The triflate anion has the structure F3CSO2O(-), and the extra electron resonates around all three triflate groups, so that there is no distinguishing between triflate groups covalently bonded and triflate anions. To have a better conception, one must think in terms of fractional bonds and charges.
For example, with tetrafluoroborate ion BF4 (-) , it could be said that each bond has an order of 3/4 (three fourths), while each fluorine atom carries a charge of minus 1/4.

I took the liberty to draw these now that you layed it out a little better.

N₅⁺:



B(CF₃)₄^-:



B(SF₅)₄^-:



And last but not least, Sb(CF₃)₃(trifyl)_3-:



Are these correct structures to what you are getting at?

[AndersHoveland]

Yes, basically the pictures above are correct. Of course, the resonance of the extra electric charge is very complicated. The extra charge is distributed over the fluorine atoms- all of them in each molecule, including
all the oxygen atoms in the triflate groups. Again, the resonance states are quite complex.

The N5(+) cation is not completely linear, it is somewhat skewed.

If you have interest in a theoretic superbase
http://www.sciencemadness.org/talk/files.php?pid=180106&aid=10726