Why does boron banana bond here?

[Vriska]

BH4- + H^+ - > H2 + H2B6

Isn't hydride just a H -, simply putting that with an H+ should give a stable BH3 + H2. Why's the banana bonding happening?

Also in NaBH4 + I2 - > B2H6 + 2NaI + H2, can i imagine a oxidation like 2BH4- -> BH3 + H2 + 2e? So I can break I2 as 2I -?

 

[Borek]

Why's the banana bonding happening?

Beware: it is a question that has no answer. Science doesn't bother with "why", it is about "how". We do know the most stable system is the one with the lowest energy. When we calculate which system has the lowest energy in this case, turns out it is the one with banana bonds. Why? Because that's the way it is.

 

[HAYAO]

Beware: it is a question that has no answer. Science doesn't bother with "why", it is about "how". We do know the most stable system is the one with the lowest energy. When we calculate which system has the lowest energy in this case, turns out it is the one with banana bonds. Why? Because that's the way it is.

Well then, I think he's asking how the one with banana bonds become the lowest energy, not that I can give any answers here.

If I were more knowledgeable with computational chemistry, then I might have been able to give an answer. For example, how the angle of the bond affects the system energy compared to the isolated BH₃, and stuff like that. That is, of course, if we can limit the main contributing factor to one or two so that it is easier to compare. With more than three factors it becomes too complicated for us to comprehend, which I believe would be the actual case for the question in the OP, hence the answer "because that's the way it is".

 

[Borek]

What you are suggesting is a post factum rationalization that has no predictive power for other cases - so it is useless as a tool.

 

[HAYAO]

I don't understand what you mean.

 

[Vriska]

Beware: it is a question that has no answer. Science doesn't bother with "why", it is about "how". We do know the most stable system is the one with the lowest energy. When we calculate which system has the lowest energy in this case, turns out it is the one with banana bonds. Why? Because that's the way it is.

Well, alright then - Would BH4 - + H+ -> BH3 + H2 be a chemist's educated guess at predicting the reaction and the occurrence of B2H6 a small surprise?

 

[Borek]

Would BH4 - + H+ -> BH3 + H2 be a chemist's educated guess at predicting the reaction and the occurrence of B2H6 a small surprise?

Yes.

 

[Borek]

I don't understand what you mean.

What I mean is that you are trying to find a way of "explaining" something that works only for this single case, and is already based on what you know about the geometry. Such an explanation won't work for other molecules, thus it doesn't add anything new to what we already know.

 

[DrDu]

Boron is a curious case, as it is in between being a metal and a non-metal. With metals, these multi-center bonds are normality, also in compounds involving hydrogen, it is unusual for non-metals.
Here a funny example of four hydrogens bridging two niobium atoms:
http://pubs.rsc.org/en/content/articlehtml/2013/DT/C2DT32798H

 

[HAYAO]

What I mean is that you are trying to find a way of "explaining" something that works only for this single case, and is already based on what you know about the geometry. Such an explanation won't work for other molecules, thus it doesn't add anything new to what we already know.

Oh, okay. Well that's not what I meant. It must be my bad phrasing.

I meant to say that any outcome of a chemical reaction is a combination of several factors. If we go further into the quantum level, we have to consider even more factors. In some cases where we compare two possible products, the difference between the two is simple because out of all the factors that affect the reaction process to yield one of these two, only very few of them actually matters. However, some cases are extremely complicated where too many factors affect the reaction process, and we cannot comprehend these factors at once. I believe the latter is what we call "it's just the way it is", because of the blackbox that entail when we just consider the energy of the products by quantum chemical calculation. But it's "not" just the way it is. Our brain just don't have the computing power to comprehend the factors that affect the complicated reaction.

For example, some reaction is as simple as steric hindrance and the outcome of the reaction is intuitively easy to predict, while some reaction can be much more complex and we must consider not only steric hindrance, but temperature, pressure, electronegativity, strength of a bond, etc. If all these factors contributes to the outcome, then it cannot be easily determined "how" that outcome came to be.I mentioned this aspect of reactions because from the context of OP's post, I thought that OP's question of "why" is just a bad phrasing and he actually meant "how", like you said would be the relevant question here. I thought that therefore, one should answer to him "how" the outcome happened, but I also mentioned that answering the question "how" is not always easy.

 

[TeethWhitener]

Why? Because that's the way it is.

NO NO NO! This is how religions start. "We observe water to boil at 100°C. Why? Because that's the way it is." Do you notice how it just shuts down discussion?

To answer the question, the standard high school chem line here is that BH₃ violates the octet rule. (In fact, at higher temperatures--where entropy is more important in the free energy balance--free BH₃ monomer is observed.) The octet rule violation is not too far off the mark, as it turns out. Boron is more stable with a filled octet (because of the standard filled shell arguments), and to get there, the empty p_z orbital on one BH₃ molecule's boron center interacts with the filled σ orbital of a B-H bond on another BH₃. Rehybridization of the boron centers from sp² to sp³ lowers the energy of the system further (this happens ostensibly because the D_2h symmetry of the observed diborane molecule keeps us from shoving certain electrons into higher energy levels that would be required in a lower symmetry system where rehybridization did not occur).

 

[Borek]

NO NO NO! This is how religions start.

That's because "why" is the wrong question. I can answer all your explanations with a single "why?" and you will be back at where you started. And if not, I will ask "why?" till you start to bleed and give up, kids do it routinely with their parents However, if you answer question "how do we find state of the system?" with single statement "system will get to the lowest energy state" you give a universal recipe - and you don't need to look for rationalizations and explain "why's".

To use another example of what I mean: question "why is π 3.141592...?" doesn't make much sense. "How is π defined?, "What is value of π?", "How to calculate value of π?" are questions that we can answer, but trying to answer "why is π 3.141592...?" is a waste of time.

 

[Borek]

In some cases where we compare two possible products, the difference between the two is simple because out of all the factors that affect the reaction process to yield one of these two, only very few of them actually matters. However, some cases are extremely complicated where too many factors affect the reaction process, and we cannot comprehend these factors at once.

OK, I have no problem with this approach, as long as we clearly state from the very beginning that it is a partial answer that happens to work because we can ignore other factors. Unfortunately, way too often such answers are treated as the "real" answers, leaving students with a feeling every case has a single, simple "why" answer behind. Pedagogically dangerous IMHO.

 

[HAYAO]

OK, I have no problem with this approach, as long as we clearly state from the very beginning that it is a partial answer that happens to work because we can ignore other factors. Unfortunately, way too often such answers are treated as the "real" answers, leaving students with a feeling every case has a single, simple "why" answer behind. Pedagogically dangerous IMHO.

I completely agree. Chemistry is all about approximations IMO, and obviously, each type of approximations only work for particular cases where such approximation is valid. Most people seems to learn things top-down instead of bottom-up without being taught that they are an approximation, hence they take these approximations for granted and believe that they are the underlying universal truth or something, which is just wrong.

 

[Vriska]

NO NO NO! This is how religions start. "We observe water to boil at 100°C. Why? Because that's the way it is." Do you notice how it just shuts down discussion?

To answer the question, the standard high school chem line here is that BH₃ violates the octet rule. (In fact, at higher temperatures--where entropy is more important in the free energy balance--free BH₃ monomer is observed.) The octet rule violation is not too far off the mark, as it turns out. Boron is more stable with a filled octet (because of the standard filled shell arguments), and to get there, the empty p_z orbital on one BH₃ molecule's boron center interacts with the filled σ orbital of a B-H bond on another BH₃. Rehybridization of the boron centers from sp² to sp³ lowers the energy of the system further (this happens ostensibly because the D_2h symmetry of the observed diborane molecule keeps us from shoving certain electrons into higher energy levels that would be required in a lower symmetry system where rehybridization did not occur).

Oh, the monomer thing answered another question I had - was wondering how B2H6 breaks in water to give boric acid and hydrogen, makes sense that the reaction gives enough energy to break it into monomer and react in the usual way.

I read that 3 is the highest "group oxidation number" oh unless the two hydrogens that boron is connected to each have a .5 oxidation state? does it?

Also you said sp2 - > sp3 lowers energy. Really? how?

 

[TeethWhitener]

I can answer all your explanations with a single "why?" and you will be back at where you started.

This isn't true: You've now been informed of a whole host of factors that are relevant to how energy levels are determined.

And if not, I will ask "why?" till you start to bleed and give up

I honestly don't see anything wrong with scientific curiosity. I would contend that there's far too little of it nowadays.

"system will get to the lowest energy state"

In principle, yes. Functionally, no. You can store a gas-tight container of H₂ and O₂ essentially indefinitely even though H₂O is lower in energy. Same idea for the slogan, "A diamond is forever."

I read that 3 is the highest "group oxidation number" oh unless the two hydrogens that boron is connected to each have a .5 oxidation state? does it?

I don't quite understand you here. Can you clarify?

Also you said sp2 - > sp3 lowers energy. Really? how?

The symmetry of a system determines the structure of its energy levels. The most well-known symmetry effect that one encounters in chemistry is known as Jahn-Teller distortion, where a highly symmetric transition metal complex can lower its energy by breaking its symmetry (usually from O_h to D_4h). [A similar idea exists in solid-state physics: the Peierls distortion.] The most well-known one in physics is probably either Cooper pairing or the Higgs mechanism. Usually, the stabilization comes from lowering symmetry, but there can be cases where raising the symmetry provides the correct arrangement of energy levels to lower the system's energy.

 

[Borek]

 

[Vriska]

I don't quite understand you here. Can you clarify?

The symmetry of a system determines the structure of its energy levels. The most well-known symmetry effect that one encounters in chemistry is known as Jahn-Teller distortion, where a highly symmetric transition metal complex can lower its energy by breaking its symmetry (usually from O_h to D_4h). [A similar idea exists in solid-state physics: the Peierls distortion.] The most well-known one in physics is probably either Cooper pairing or the Higgs mechanism. Usually, the stabilization comes from lowering symmetry, but there can be cases where raising the symmetry provides the correct arrangement of energy levels to lower the system's energy.

that's neat I guess, but what does lowest energy state mean? is it referring to bond strength? I don't know because SF2 has a stronger bond compared to SF6 but SF2 is more symmetric.

What i meant by the B2H6 was, since boron has a maximum oxidation number of +3, the two hydrogens on it must have a oxidation number of .5. Is this the case? Does the fractional number have an impact on bond stability?

 

[TeethWhitener]

Yes, I’ve seen this. As much as I enjoy some of Feynman’s work, this strikes me as grammar nitpicking masquerading as bad philosophy. And I (perhaps unfairly to Feynman) have to admit that I find it pretty disingenuous.

what does lowest energy state mean?

The easiest example is the Jahn Teller effect:
https://en.m.wikipedia.org/wiki/Jahn–Teller_effect
I don’t know what your background is, but if you look at the energy level diagrams in that link, when the molecule changes symmetry, the arrangement of its energy levels changes as well.

I don't know because SF2 has a stronger bond compared to SF6 but SF2 is more symmetric.

SF₂ has lower symmetry than SF₆. More precisely, the symmetry group of SF₂ (C_2v) is a subgroup of the symmetry group of SF₆ (O_h).

What i meant by the B2H6 was, since boron has a maximum oxidation number of +3, the two hydrogens on it must have a oxidation number of .5. Is this the case? Does the fractional number have an impact on bond stability?

Why? Two borons, each with oxidation number of +3, are balanced out by 6 hydrogens of oxidation number -1 each.

 

[Vriska]

Yes, I’ve seen this. As much as I enjoy some of Feynman’s work, this strikes me as grammar nitpicking masquerading as bad philosophy. And I (perhaps unfairly to Feynman) have to admit that I find it pretty disingenuous.

The easiest example is the Jahn Teller effect:
https://en.m.wikipedia.org/wiki/Jahn–Teller_effect
I don’t know what your background is, but if you look at the energy level diagrams in that link, when the molecule changes symmetry, the arrangement of its energy levels changes as well.

SF₂ has lower symmetry than SF₆. More precisely, the symmetry group of SF₂ (C_2v) is a subgroup of the symmetry group of SF₆ (O_h).

Why? Two borons, each with oxidation number of +3, are balanced out by 6 hydrogens of oxidation number -1 each.

Wow, pure math in chemistry. That's pretty darn neat.

The boron thing was because two hydrogens in B2H6 are different... I don't know, you have different oxidation numbers in hydrogen peroxide don't you, due to the peroxy bond?

 

[TeethWhitener]

The boron thing was because two hydrogens in B2H6 are different... I don't know, you have different oxidation numbers in hydrogen peroxide don't you, due to the peroxy bond?

The hydrogens in hydrogen peroxide are equivalent, so that's not the best example. One could consider an example like ethanol, where the hydrogens on the C are in a different chemical environment (and have different observable properties) than the hydrogen on the O. But even in this case, if we're merely evaluating oxidation numbers, the H's are all +1. The only weird thing about borane with respect to oxidation number is that the H's are in the -1 oxidation state, but this is reflected in the fact that borane is a very good reducing agent (the H's want to go from -1 to 0 or +1 by giving their electrons to a suitable substrate).