quantime
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studying the borohydrate anion
I am trying to understand the borohydrate ion, BH4(-)
Boron, the atomic number = 5, so elemental electronic configuration is 1s2 2s2 2p1
molecular sp hybridization can allow averaging the 2s2 + 2p1 = 2sp3 and so there are 3 unfilled MO's that can act as bonds
How does the fourth hydrogen possibly get in there?
I can imagine if you bring an H with an attached electron and
cram it into the hybridized valence (like a Lewis diagram because it has four spots) you get BH4(-). This is an argument for the correct charge on the
ion, but not for how or why the boron accepted the 4th hydrogen and electron.
I can't get my head around how the sp3 hybrid subshell gets extended from the three MO bonds to four. Why does this happen?
It seems the energy state would strongly lean to BH3 and it would be a difficult, if impossible to have BH4(-).
How can BH4(-) even exist?
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forgottenpassword
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Maybe this helps?
http://books.google.co.uk/books?id=qUucAQAAQBAJ&pg=PA98&...
And this:
http://books.google.co.uk/books?id=2RgbAgAAQBAJ&pg=PA357...
[Edited on 2-7-2014 by forgottenpassword]
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NexusDNA
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Please correct me if I'm wrong, but borohydride works by accepting the 2 electrons of a H- into the unoccupied 2pz orbital of borane.
Bromine, definitely bromine.
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quantime
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very helpful - yes borohydride
Thanks forgotten password! I hope you remember your password.
[Edited on 2-7-2014 by quantime]
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NexusDNA
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Oh, and btw, welcome to Science Madness! 
Bromine, definitely bromine.
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quantime
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still not right yet
The question is why in hell does an H(-) anion stick into a borane BH3 to form borohydride? It seems the molecular orbitals of borane somehow attracts
a proton and two electrons which is neither electrophilic nor nucleophilic so that makes no sense, and flips from trigonal planar into trigonal
tetrahedral configuration. What makes that happen? Borohydride seems like the higher energy configuration. It seems like borohydride would not happen.
But obviously it does, so
I don't understand something basic about molecular orbitals, and I can't make sense of those weird energy diagrams after staring at them for hours
over several days.
Please I just want to understand what is going on here. This is a beginnings chemistry question that I can't find any answer to.
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forgottenpassword
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Read the commentary under figure 7.6 in the first book I linked to.
Now look at figure 7.5. The HOMO of LiH and the LUMO of BH3 can combine to form bonding and anti-bonding orbitals, shown by the two energy levels.
Since the bonding orbital is of lower energy than the starting materials, the product is more stable than the reactants.
[Edited on 4-7-2014 by forgottenpassword]
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DraconicAcetate
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Look at it this way. BH3 has a boron which is sp2 hybridized and has an empty p orbital. It is electron-deficient, since it
only has six shared valence electrons instead of a full octet. If it reacts with a hydride ion (which doesn't have a strong hold on its electrons
anyway), it can have a full octet with an sp3 hybridization.
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quantime
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very interesting
DraconicAcetate - Please, yes, BH3 has 2spx 2spy 2spz hybridized valence orbitals. All three valence orbitals are full. No orbitals are available for
bonding. That is where I break down.
In general is there some property where molecules want four valence orbitals? I thought it was the s + 3p that makes atoms want the four orbitals.
A full sp3 hybridization would have four hybridized degenerate orbitals. What could possibly be the four directions of a full sp3 hybridization?
sp(x) sp(y) sp(z) sp(*)
* - points into the 4th dimension? 
Does this story sound accurate? A BH4(-1) anion does not form by itself in isolation. The formation of this ion requires a mechanism. A typical
mechanism is LiH + BH3 whereby the LUMO of LiH and the HOMO of BH3 crash together to create a new covalent bond. The crash results in the positive
ionization of Li, the negative ionization of Li's H, the creation of a bonding MO, and the transfer of the hydronium into the new bond. The result is
an ionic compound with Li(+) on one side and a BH4(-) on the other side of the dipole. All of this crazy scary not intuitive stuff has to happen in
order to make a BH4(-) ion. Chemistry is absurdly complicated.
[Edited on 6-7-2014 by quantime]
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DraconicAcid
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| Quote: | | DraconicAcetate - Please, yes, BH3 has 2spx 2spy 2spz hybridized valence orbitals. All three valence orbitals are full. No orbitals are available for
bonding. That is where I break down. |
No, it has three hybrid orbitals (three sp2) and an empty p orbital. That's four valence orbitals.
| Quote: | A full sp3 hybridization would have four hybridized degenerate orbitals. What could possibly be the four directions of a full sp3 hybridization?
sp(x) sp(y) sp(z) sp(*)
* - points into the 4th dimention? |
The four sp3 orbitals are 109.5 degrees apart, not 90. Molecules are three-dimensional objects, not four.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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quantime
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does an orbital exist if there are no eletrons in it?
No, it has three hybrid orbitals (three sp2) and an empty p orbital. That's four valence orbitals
I am amazed. How can an orbital exist that has no electrons in it? Do all possible orbitals exist then? All possible orbitals are also empty.
Please, I am missing something important.
[Edited on 6-7-2014 by quantime]
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Brain&Force
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It's a tetrahedral shape, isoelectronic with methane.
Good to see you've got your account back, Draconic...um...(insert creative ending here)! It's great to be an international hazard again 
[Edited on 6.7.2014 by Brain&Force]
At the end of the day, simulating atoms doesn't beat working with the real things...
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quantime
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sz2 + px + py = sp3?
sz2 what is that?, two s orbitals along the z-axis?
is sz2 one orbital or two?
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kmno4
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First thing: "hybridization" is only mathematical trick, most chemists use this therm so often, that they soon start to belive that it is reality
Second thing: "BH3" is very unstable molecule, far less stable than BH4(-) ion. There is "dimeric" molecule, well described, namely B2H6.
But there is no any equilibrium BH3↔B2H6 in gas phase (for example).
| Quote: | | How can an orbital exist that has no electrons in it? |
And what is the orbital ? Mathematical calculation.
[Edited on 6-7-2014 by kmno4]
Слава Україні !
Героям слава !
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blogfast25
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| Quote: | Quote: Originally posted by kmno4  | First thing: "hybridization" is only mathematical trick, most chemists use this therm so often, that they soon start to belive that it is reality
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[Edited on 6-7-2014 by kmno4] |
You're oversimplifying, to the extent of coming across as simplistic as the people you accuse.
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Nicodem
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Actually, he is not oversimplifying anything. The hybrid orbitals are a pretty much mathematical simplifications (just models, they don't represent
reality, but can be useful to some limited extent to predict empirical observations). For some reason they are still commonly wrongly presented to
students as some definitive model instead of a mathematical approximation. In scientific articles it is still common to describe the bonding geometry
using the spx and related terms, but this is only due to lack of better terminology. The model can be used to predict certain spectroscopic
measurements to some extent, but it is way less useful than students are made to believe. It is best avoided to invoke such models without a deeper
understanding of the mathematics behind it. Unfortunately, the ones who know least about it, tend to invoke it the most.
For a better explanation, see DOI: 10.1021/ed100155c and the follow-ups DOI: 10.1021/ed200615j and DOI: 10.1021/ed200746n.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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blogfast25
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Hmm... what I see in these sources is a debate. Not quite cut and dried.
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quantime
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hybridized orbitals math tricks
I read chapter 6 and 7 of the first link again last night, and I came to a similar conclusion. The hybrid orbitals are a linear combination of
orbitals, like vectors in a vector space, that point towards the tetrahedral angles. The hybrid orbitals don't really exist. By employing hybrid
orbitals one is making the assumption of tetrahedral angles and thus the assumption of four orbitals.
Working theory for the moment:
Hybrid orbitals help when you already know the configuration, but don't help describe the dynamics of why a molecule gets into a configuration.
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