A Diabolically Difficult Chemistry Puzzle

2022-07-24

Tagged: chemistry

In 2012, I was a mentor at the USNCO Study Camp, which prepares and selects the top students in the U.S. to send to the International Chemistry Olympiad. I wrote a question for the final exam which only 2 of the 20 students completed. Both students admitted to brute-forcing the system of equation in 5 unknowns. I found it hilarious at the time that I’d stumped the entire camp with a seemingly trivial problem. The students were not as amused! Without further ado:

Balance the following equation.

BHX4X−+BFX3→BX11HX14X−+BFX4X−+HX2(g)\ce{BH4- + BF3 -> B11H14- + BF4- + H2(g)}

Source: https://doi.org/10.1016/S0022-328X(03)00460-1

This 11-boron, 14-hydrogen species has C5vC_{5v} symmetry (dodecahedron with missing vertex), and its parent species BX11HX11X2−\ce{B11H11^2-} is apparently fluxional, rapidly interconverting with no one boron identifiable. If you’re wondering how I stumbled across such an obscure reaction, I decided at some point that boron was probably a simple enough element that I could learn everything there was to know about boron if I’d really tried. Boy, was I wrong! It turns out a lot of research went into obscure borohydride species as potential energy-dense rocket fuels during the Cold War.

A hint and solution for this equation comes after the spoiler break.

v v v v v v















Hint

This is a redox equation. Find the two half-reactions.

v v v v v v















Solution

The borohydride and fluoroborate species all have boron in their (+3) oxidation state, while the B11 species has a 15/11 oxidation state, which is about (+1.36) on average. The hydrogen, on the other hand, goes from hydride at the (-1) oxidation state to hydrogen gas at the (0) oxidation state. We’ll have to find the right ratios that lead to net electron transfer balance.

In a traditional redox reaction, typically what happens is that you first insert electrons to form the correct half-reaction, and then use HX2O\ce{H2O}, HX+\ce{H+} or OHX−\ce{OH-} species to balance oxygens, hydrogens, and charges. As a simple example, for the reduction of Cr(+6) to Cr(+3):

CrOX4X2−→CrX3+\ce{CrO4^2- -> Cr^3+}

CrOX4X2−+3 eX−→CrX3+\ce{CrO4^2- + 3e- -> Cr^3+}

CrOX4X2−+3 eX−→CrX3++4 HX2O\ce{CrO4^2- + 3e- -> Cr^3+ + 4H2O}

CrOX4X2−+3 eX−+8 HX+→CrX3++4 HX2O\ce{CrO4^2- + 3e- + 8H+ -> Cr^3+ + 4H2O}

How do these steps apply to this problem? This isn’t an aqueous reaction - borohydrides are pyrophoric in air! Instead of using HX+\ce{H+} to balance our charges, we’ll use HX−\ce{H-}.

11 BHX4X−+20 eX−→BX11HX14X−+30 HX−\ce{11BH4- + 20e- -> B11H14- + 30H-}

2 HX−→2 eX−+HX2\ce{2H- -> 2e- + H2}

Balanced reaction:

11 BHX4X−→BX11HX14X−+10 HX−+10 HX2\ce{11BH4- -> B11H14- + 10H- + 10H2}

Free hydride is not in the final equation - this is where the fluoroborate species come in. You could see the role of BFX3\ce{BF3} as a hydride sponge.

4 BFX3+4 HX−→3 BFX4X−+BHX4X−\ce{4BF3 + 4H- -> 3BF4- + BH4-}

Balancing and cancelling the hydrides, the final balanced equation is

17 BHX4X−+20 BFX3→2 BX11HX14X−+15 BFX4X−+20 HX2\ce{17BH4- + 20BF3 -> 2B11H14- + 15BF4- + 20H2}