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What makes some metals melt at higher temperature? |
Many diagrams of orbitals I've seen involve electrons moving through a center point—where the nucleus is. How can this be? Clearly they don't actually pass through the nucleus, so what does happen? |
How do orbitals coexist with a nucleus? |
The $p$ orbitals, for example, have a nodal plane where the nucleus is, which means the electron density is zero there.
An orbital doesn't represent a path the electrons take when moving. An orbital is a region of probability. To make things clear and definite, when we draw an orbital we only draw the region where 95% (for example) of the probability lies. The fact that the $p$ orbitals have a nodal plane simply means that the probability of finding an electron on that plane vanishes.
A [positivist](http://en.wikipedia.org/wiki/Positivism) would thus consider an electron's trajectory around a nucleus nonsensical, since by the uncertainty principle, we can never measure it.
The use of orbitals is for visualizing electron density - where _can_ the electron be, most of the time? It's a very useful way of interpreting phenomena such as chemical reactivity (think of SN<sub>2</sub>) or stability (think of benzene's $p$ orbitals). |
*Although I can't think of any drug examples other than thalidomide, here's information on thalidomide's mechanism:*
The chiral carbon of thalidomide can tautomerize in basic conditions into an enol, which is achiral. A reversal back to the ketone results in a mix of *(R)* and *(S)* enantiomers.
![Chemical diagram for s-thalidomide][1]
In the body, this tautomerization is generally catalyzed by basic amino acids. Specifically, albumin is the main catalyst in humans.
While this is beyond the scope of your question, the reason that only *(S)*-thalidomide causes birth defects is that it can insert itself into DNA and suppresses certain genes necessary for embryonic development.
<hr>
1. [Reddy - *Chirality in Drug Design and Development*][3]
2. http://www.ncbi.nlm.nih.gov/pubmed/9860497
3. http://www.ncbi.nlm.nih.gov/pubmed/9499573
[1]: https://i.stack.imgur.com/w1k3n.png
[2]: http://www.ncbi.nlm.nih.gov/pubmed/9860497
[3]: http://books.google.com/books?id=HibwXSgofi0C&pg=PA330&lpg=PA330&dq=albumin+thalidomide+racemization&source=bl&ots=8aLGIAbZa9&sig=uaQhWP7tAzUScQiQ0AuRom-v09k&hl=en&sa=X&ei=mHubT43TOeGD6AGG1oXuDg&ved=0CB8Q6AEwAA#v=onepage&q&f=false |
I think there are two properties at play here: atomic velocity and stability of the lattice structure.
Recall that temperature is a measure of the average kinetic energy of the molecules, so $v \propto \sqrt{\frac{T}{m}}$, or to achieve the same velocity, $T \propto m$. So at the same temperature, two heavier atoms will move more slowly past each other than two lighter molecules, giving them more time to interact.
Now recall that in a solid molecules are held in a lattice by intermolecular forces, and in a liquid atoms have enough energy that the forces between them are no longer strong enough to hold the atoms in a lattice. So the more stability a substance's atoms gain from being in their solid, crystalline arrangement, the higher its melting point will be. Now why would the metals in that area gain more stability in their solid forms? I'm fairly sure the answer has to do with the completeness of orbitals and half-orbitals. I'm not sure of the particulars of the delocalized electron cloud in metals, but I think it's likely that it allows these metals to in some manner fill or empty their incomplete orbitals.
|
>*"Both Vanadium and its 3+ ion are pramagnetic. Use electron configuration to explain why this is so"*
The electron configuration of Vanadium is :
#$[Ar]4s^23d^3$
If drawn in a box diagram, it would be shown that the electron in the $d$ orbitals aren't paired. This explains how an uncharged Vanadium is paramagnetic, but if Vandium loses 3 electrons, the only shells left are ones with paired electrons.
How can a Vanadium 3+ ion be paramagnetic if it loses all its unpaired electrons?
Much Thanks |
How is Vanadium's 3+ ion paramagnetic? |
When [**N,N-dimethylaniline**][1] is reacted with $\ce{H_2SO_4}$ and $\ce{HNO_3}$ it gives mainly the *meta* product, even though $\ce{NMe_2}$ is an *ortho* / *para* directing group. Why is this?
[1]: http://en.wikipedia.org/wiki/Dimethylaniline |
>*"Both Vanadium and its 3+ ion are pramagnetic. Use electron configuration to explain why this is so"*
The electron configuration of Vanadium is [Ar] 4s<sup>2</sup> 3d<sup>3</sup>.
If drawn in a box diagram, it would be shown that the electron in the $d$ orbitals aren't paired. This explains how an uncharged Vanadium is paramagnetic, but if Vandium loses 3 electrons, the only shells left are ones with paired electrons.
How can a Vanadium 3+ ion be paramagnetic if it loses all its unpaired electrons? |
When I look around for why Copper and Chromium only have 1 electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.
What I'm really looking for is **why** the d orbital more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each subshell of the d orbital is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.
**Why does this not happen with in other columns as well?** Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like the Flourine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.
[1]: http://en.wikipedia.org/wiki/Electron_configuration |
Why are do elements in columns 6 and 11 assume 'abnormal' electron configurations? |
![Diagram of P orbital vs. S orbital distance from nucleus][1]
My chemistry book explains that even though electrons in the 2p orbital are closer to the nucleus on average, electrons from the 2s orbital spend a very short time very close to the nucleus (penetration), so it has a lower energy. Why does this tiny amount of time spent close to the nucleus make such a big difference? It seems like it should be the average distance that matters, not the smallest distance achieved at any one point, in determining stability. What makes that momentary drop in energy so important that it is outweighs all the time spent farther away from the nucleus with a higher energy?
[1]: https://i.stack.imgur.com/Tnn32.gif |
Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus? |
There are some stereochemical reactions that result in the presence of enantiomers. When moving forward with a practical organic synthesis, how does one usually separate them in order to continue with one of the enantiomers? |
How do you separate enantiomers? |
The most stable cycloheanxe form is in chair conformation but on the other hand, the bigger the side-chain of the cyclohexane is, its hydrogens become more equatorial rather than axial, which brings them closer to the cyclohexane's hydrogens.
Shouldn't they arrange in a less energetic, but a more stable conformation? |
Why bigger the side-chain is, its hydrogens tend to be more equatorial? |
The most stable cyclohexane form is the chair conformation but on the other hand, the bigger the side-chain of the cyclohexane is, its hydrogens become more equatorial rather than axial, which brings them closer to the cyclohexane's hydrogens.
Shouldn't they arrange in a less energetic, but a more stable conformation? |
>*"Both Vanadium and its 3+ ion are paramagnetic. Use electron configuration to explain why this is so"*
The electron configuration of Vanadium is [Ar] 4s<sup>2</sup> 3d<sup>3</sup>.
If drawn in a box diagram, it would be shown that the electron in the $d$ orbitals aren't paired. This explains how an uncharged Vanadium is paramagnetic, but if Vandium loses 3 electrons, the only shells left are ones with paired electrons.
How can a Vanadium 3+ ion be paramagnetic if it loses all its unpaired electrons? |
I have a solution of copper acetate and I would like to play around with the ligands to get different colors.
Background: The copper acetate was made through mixing vinegar (5% acetic acid), NaCl, and C$_{\textrm{(s)}}$. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room.
Question(s):
What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color?
Will heating or cooling the solution change the color and/or ligands? |
How may copper acetate ligands be manipulated to change colors? |
When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.
What I'm really looking for is **why** the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each subshell of the d orbital is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.
**Why does this not happen with the other columns as well?** Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like flourine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.
[1]: http://en.wikipedia.org/wiki/Electron_configuration |
Why do elements in columns 6 and 11 assume 'abnormal' electron configurations? |
Phosphate buffer is a common buffer in biological applications, it's especially popular for NMR. Magnesium ions are necessary for many biological systems to work, though the amount of magnesium ions you can add in a phosphate buffer is limited by the relatively low solubility of magnesium phosphate.
What I'm curious about is whether the phosphate buffer can affect the availability of magnesium ions even in concentrations where no visible precipitation occurs? Common phosphate buffer concentrations are around 10-100 mM, and magnesium ions are usually 5 mM or below. At those concentrations I see no visible precipitation, but is the magnesium actually fully available or is the phosphate sequestering it and thereby lowering the effective magnesium ion concentration in the solution?
Even just calculating the solubility of magnesium salts in phosphate buffer is not straightforward, as there are multiple species of magnesium phosphate and possibly even other salts like $\ce{MgKPO4}$.
Does phosphate buffer affect the availability of magnesium ions in solution compared to a non-interacting buffer? If that is so, is there a way to calculate the size of that effect? |
I was tasked with figuring out whether Carbon or Nitrogen has a more negative electron affinity value. I initially picked Nitrogen, just because Nitrogen has a higher Z$_{eff}$, creating a larger attraction between electrons and protons, decreasing the radius, causing a higher Ionization Energy, and therefore decreasing the electron affinity value, but I was actually wrong, and the solutions manual explains it as this:
>"As you trace from C to N across the periodic table you would normally expect N to have the more negative electron affinity. However, N has a half-filled p sublevel, which lends it extra stability, therefore it is harder to add an electron."
Are there any major exception rules when comparing electron affinity? I'm hesitant to use Nitrogen as an exception, because I don't know how far it extends. If Nitrogen has a more positive EA than Carbon, does that also extend to Boron, or Aluminium, or Phosphorus?
I later found that this also applies when comparing Silicone and Phosphorus. The explanation was the same.
What exception rules should be noted when comparing electron affinity? Are there any at all? And how far does the exception with atoms with half-filled p sublevels extend?
*Note: I'm not entirely sure which tags fit this question, so I encourage anyone who knows to edit it in.*
|
Are there any major exceptions when comparing electron affinity? |
Most periodic tables only feature one Hydrogen atom, on the top of the first group. But some, like the one I was given, also show Hydrogen in the 7th group, to left of Helium.
Why are there two Hydrogen atoms? What's the difference between the two, and why do we work with the left one more often than the right? |
Why are there two Hydrogen atoms on some periodic tables? |
I have a solution of [copper acetate][1] and I would like to play around with the ligands to get different colors.
Background: The copper acetate was made through mixing vinegar (5% acetic acid), NaCl, and C$_{\textrm{(s)}}$. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room.
[edit] I neglected to mention that the experiment simultaneously produced a 0.5 inch deposit of what appears to be [Copper Carbonate][2] on the bottom of the 1 Liter beaker. Also, I have let the copper acetate solution evaporate for several years now. The former 1L is now 1/2 liter and has begun precipitating crystals (like the ones on the wiki page). [/edit]
Question(s):
What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color?
Will heating or cooling the solution change the color and/or ligands?
[1]: http://en.wikipedia.org/wiki/Copper_acetate
[2]: http://en.wikipedia.org/wiki/Copper_carbonate |
I have a solution of [copper acetate][1] and I would like to play around with the ligands to get different colors.
Background: The copper acetate was made through mixing vinegar (5% acetic acid), NaCl, and C$_{\textrm{(s)}}$. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room.
[edit] I neglected to mention that the experiment simultaneously produced a 0.5 inch deposit of what appears to be [Copper Carbonate][2] or [Verdigris][3] on the bottom of the 1 Liter beaker. Also, I have let the copper acetate solution evaporate for several years now. The former 1L is now 1/2 liter and has begun precipitating crystals (like the ones on the wiki page). [/edit]
Question(s):
What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color?
Will heating or cooling the solution change the color and/or ligands?
[1]: http://en.wikipedia.org/wiki/Copper_acetate
[2]: http://en.wikipedia.org/wiki/Copper_carbonate
[3]: http://en.wikipedia.org/wiki/Verdigris |
Overall ring strain seems to be a big issue when it comes to organic chemistry. That is why cyclopentane may be in an "envelope form" or why cyclobutane may be in a kinked, kite form. Both of these example molecules are not planar. So, why is epoxide (oxacyclopropane) a stable molecule, if at all? |
What makes an epoxide stable? |
I have a solution of [copper acetate][1] and I would like to play around with the ligands to get different colors.
Background: The copper acetate was made through mixing vinegar (5% acetic acid), NaCl, and C$_{\textrm{(s)}}$. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room.
[edit] I neglected to mention that the experiment simultaneously produced a 0.5 inch deposit of what appears to be [Copper Carbonate][2] or [Verdigris][3] on the bottom of the 1 Liter beaker. Also, I have let the copper acetate solution evaporate for several years now. The former 1L is now 1/2 liter and has begun precipitating crystals (like the ones on the wiki page). Fun fact: During the Renaissance, glacial [acetic acid][4] was made by [dry distilling metal acetates][5] and primarily copper(II) acetate.[/edit]
Question(s):
What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color?
Will heating or cooling the solution change the color and/or ligands?
[1]: http://en.wikipedia.org/wiki/Copper_acetate
[2]: http://en.wikipedia.org/wiki/Copper_carbonate
[3]: http://en.wikipedia.org/wiki/Verdigris
[4]: http://en.wikipedia.org/wiki/Acetic_acid
[5]: http://alchymie.ca/radicalvinegar.htm |
Well, you have to ask: *how* stable? and *compared to what*?
Peroxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does [cyclopropane](http://en.wikipedia.org/wiki/Cyclopropane), which also contains a three-membered ring.
So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from [these lecture notes](http://chemistry.illinoisstate.edu/standard/che38037/handouts/380.37thermo.pdf), which indicate that ***“ethylene oxide has less ring strain than cyclopropane”***. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°. |
According to [wikipedia and the references given therein][1], $\pi\cdots\pi$ stacking interactions are the result of interaction between the quadrupole moments of two aromatic rings, rationalising the stabilisation of perpendicular and offset-parallel association modes of the benzene dimer. Conventional wisdom is that the supramolecular complexes of buckminsterfullerene and functionalised corannulenes (the so-called buckycatchers) are stabilised by $\pi\cdots\pi$ stacking. How can the quadrupole interaction explanation be reconciled with the pseudo-spherical symmetry of buckminsterfullerene? Surely the interaction must be entirely Van der Waals in nature?
[1]: http://en.wikipedia.org/wiki/Pi-pi_interaction |
Are buckminsterfullerene-corannulene complexes actually stabilised by π stacking? |
This question raises more important issues than just the technical "why methyl ester," so I'll address those too.
The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. [This report][1] says it is approximately 10x less sweet (see Table VI on page 2689 and the entry 'Asp-Phe-OEt' with "++" being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.
A toxicology class I once took can mostly be summed up by a cliche phrase, "the dose makes the poison." Even too much of a **good** thing is by definition a **bad** thing, for example too much [oxygen][2], [water][3] or even the [sun][4] can lead to death. We also say that too little of any compound won't cause *measurable* harm.
It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a [CNS depressant][5]. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for enzymes which metabolize ethanol, called [alcohol dehydrogenase][6]. ADH lead to biochemical pathways that allow the safe removal of ethanol, but methanol is a different case. ADH instead make formaldehyde, which is then converted to [formic acid][7] by [ALDH][8]. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a [fire ant][9]. In low quantities it is fairly benign. [Formaldehyde][10] is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind.
Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly 1.8 g of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, 8 hours after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~ 200 lbs person ingests 1.8 g of aspartame. Let's stop for a second and put this in terms of something we can imagine.
I don't have any reliable numbers, but let's just say that for a 12 oz diet pop there are 200 mg of aspartame. That means for every liter of pop, there are approximately 564 mg of aspartame. That means 3 liters of pop would get me to 1.8 g of aspartame. That means I would need to drink more than 3 L in one sitting to overwhelm the excretion systems. This suggests an [acute poisoning][11] isn't very plausible.
There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. "Sure, there is no acute toxicity, but what happens in the long-term with a low dose?" For the answer to this question we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by [uncertainties][12] in their measurements and this is just the general nature of science.
Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledgebase. Once in a while the general community is [proven wrong about something long thought to be true][13] and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar in a non-toxic dose is probably killing hundreds of thousands of people due to long-term exposure.
[1]: http://pubs.acs.org/doi/pdf/10.1021/ja01038a046
[2]: http://en.wikipedia.org/wiki/Oxygen_toxicity
[3]: http://en.wikipedia.org/wiki/Water_intoxication
[4]: http://www.epa.gov/sunwise/uvandhealth.html
[5]: http://en.wikipedia.org/wiki/CNS_depressant
[6]: http://en.wikipedia.org/wiki/Alcohol_dehydrogenase
[7]: http://en.wikipedia.org/wiki/Formic_acid
[8]: http://en.wikipedia.org/wiki/Aldehyde_dehydrogenase
[9]: http://en.wikipedia.org/wiki/Fire_ant
[10]: http://en.wikipedia.org/wiki/Formaldehyde
[11]: http://en.wikipedia.org/wiki/Acute_toxicity
[12]: http://en.wikipedia.org/wiki/Uncertainty
[13]: http://en.wikipedia.org/wiki/Missoula_Floods#Flood_hypothesis_proposed |
Since the steam pressure of ammonia is higher than that of water, I would expect distillation to be a reasonable way of seperating a mixture of both.
However, in industrial applications known to me there's always a step of stripping the ammonia chemically from the vapor. Why is this necessary? |
Why is distillation no viable way to seperate ammonia from water? |
Why is distillation not a viable way to seperate ammonia from water? |
Why does for example oxygen turn into gas at a much lower temperature than water?
(Does it have anything to do with the molecular structure? A water molecule does have a more complex structure than oxygen, though the R-410A (a mixture of two gases commonly used in heating pumps) is much more complex than water, and it boils at -48.5 degrees Celsius.) |
Why does different substances have different boiling points? |
For example, why does for example oxygen turn into gas at a much lower temperature than water?
Does it have anything to do with the molecular structure? A water molecule does have a more complex structure than oxygen, though the R-410A (a mixture of two gases commonly used in heating pumps) is much more complex than water, and it boils at -48.5 degrees Celsius. |
Why do different substances have different boiling points? |
This question raises more important issues than just the technical "why methyl ester," so I'll address those too.
The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. [This report][1] says it is approximately 10x less sweet (see Table VI on page 2689 and the entry 'Asp-Phe-OEt' with "++" being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.
A toxicology class I once took can mostly be summed up by a cliche phrase, "the dose makes the poison." Even too much of a **good** thing is by definition a **bad** thing, for example too much [oxygen][2], [water][3] or even the [sun][4] can lead to death. We also say that too little of any compound won't cause *measurable* harm.
It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a [CNS depressant][5]. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for enzymes which metabolize ethanol, called [alcohol dehydrogenase][6]. ADH lead to biochemical pathways that allow the safe removal of ethanol, but methanol is a different case. ADH instead make formaldehyde, which is then converted to [formic acid][7] by [ALDH][8]. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a [fire ant][9]. In low quantities it is fairly benign. [Formaldehyde][10] is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind.
Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly 1.8 g of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, 8 hours after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~ 200 lbs person ingests 1.8 g of aspartame. Let's stop for a second and put this in terms of something we can imagine.
I don't have any reliable numbers, but let's just say that for a 12 oz diet pop there are 200 mg of aspartame. That means for every liter of pop, there are approximately 564 mg of aspartame. That means 3 liters of pop would get me to 1.8 g of aspartame. That means I would need to drink more than 3 L in one sitting to overwhelm the excretion systems. This suggests an [acute poisoning][11] isn't very plausible.
There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. "Sure, there is no acute toxicity, but what happens in the long-term with a low dose?" For the answer to this question we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by [uncertainties][12] in their measurements and this is just the general nature of science.
Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledgebase. Once in a while the general community is [proven wrong about something long thought to be true][13] and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar, with a dose too small to cause acute poisoning, is probably killing hundreds of thousands of people due to long-term exposure.
[1]: http://pubs.acs.org/doi/pdf/10.1021/ja01038a046
[2]: http://en.wikipedia.org/wiki/Oxygen_toxicity
[3]: http://en.wikipedia.org/wiki/Water_intoxication
[4]: http://www.epa.gov/sunwise/uvandhealth.html
[5]: http://en.wikipedia.org/wiki/CNS_depressant
[6]: http://en.wikipedia.org/wiki/Alcohol_dehydrogenase
[7]: http://en.wikipedia.org/wiki/Formic_acid
[8]: http://en.wikipedia.org/wiki/Aldehyde_dehydrogenase
[9]: http://en.wikipedia.org/wiki/Fire_ant
[10]: http://en.wikipedia.org/wiki/Formaldehyde
[11]: http://en.wikipedia.org/wiki/Acute_toxicity
[12]: http://en.wikipedia.org/wiki/Uncertainty
[13]: http://en.wikipedia.org/wiki/Missoula_Floods#Flood_hypothesis_proposed |
If dry ammonia gas is passed through anhydrous copper sulfate, will it turn blue (due to the formation of tetraamminecopper(II) complex)? Or will silver chloride form diamminesilver(I) complex in liquid ammonia?
If the above complexes do not get formed in non-aqueous mediums, why is it so? |
Is aqueous medium necessary for complex formation? if so, why? |
Over the course of my studies, I have switched largely from using Z-matrix representations of molecular geometries in calculations to Cartesian representations.
The software that I use now makes it easy to add the sorts of constraints/restraints/transits that I would have previously used Z-matrices for, and I know that Z-matrix geometries can be problematic in large molecules<sup>*</sup> where minute changes in a bond angle or dihedral (due, for instance, to rounding errors/low-quality gradients) can result in large movements in peripheral atoms.
<b>What pros or cons exist for either geometry definition that I don't know about? What circumstances recommend one representation over another?</b>
*Or small molecules with silly Z-matrices. |
Pros and cons of Cartesian vs. Z-matrix representations of molecules? |
Well, you have to ask: *how* stable? and *compared to what*?
Epoxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does [cyclopropane](http://en.wikipedia.org/wiki/Cyclopropane), which also contains a three-membered ring.
So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from [these lecture notes](http://chemistry.illinoisstate.edu/standard/che38037/handouts/380.37thermo.pdf), which indicate that ***“ethylene oxide has less ring strain than cyclopropane”***. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°. |
Pennsylvania State University provides irradiated NaCl (180,000 rads of gamma radiation) to teachers for a very entertaining demonstration. The salt is orange after irradiation. When it is placed on a hot plate, you see a flash of visible light and the salt returns to it normal color. The explanation provided with the salt is that "The heat allows the electrons that were trapped in the crystal structure (when the energy was deposited) to return to their original position." This language evokes a picture of a particle electron in a crystal jail for me. The explanation goes on to say that the energy is released as a photon of visible light. (I understand this part of the explanation.) So, my question is, how are electrons trapped in a crystal structure? What keeps them in an excited state? |
How is the electron "trapped in the crystal" in irradiated NaCl? |
**Cartesian Space**
In Cartesian space, three variables (XYZ) are used to describe the position of a point in space, typically an atomic nucleus or a basis function. To describe the locations of two atomic nuclei, a total of 6 variables must be written down and kept track of. The general ruling is that for Cartesian space, 3N variables must be accounted for (where N is the number of points in space you wish to index).
**Internal Coordinates**
Z-matrices use a different approach. When dealing with Z-matrices, we keep track of the relative positions of points in space. Cartesian space is 'absolute' so to speak. A point located at (0,0,1) is an absolute location for a coordinate space that extends to infinity. However, consider a two atom system. The translation of the molecule through space (assuming a vacuum) will have no affect on the properties of the molecule. An H2 molecule centered around the origin (0,0,0) is no different from the same H2 molecule being centered around (1,1,1). However, say we increase the distance between the hydrogen atoms. We now have altered the molecule in such a way that the properties of that molecule has changed. What did we change? We simply changed the bond length, one variable. We increased the distance between the two atoms by some length R. With Z-matrices, we keep tabs on internal coordinates: bond length (R), bond angle (A), and torsional/dihedral angle (T/D). Using internal coordinates reduces our 3N requirement set by the Cartesian space down to a 3N-6 requirement (for non-linear molecules). For linear molecules we keep tabs on 3N-5 coordinates. When performing complex computations, the less you have to keep track of, the less expensive the computation.
**Symmetry**
Consider the following molecule, H2O. We know from experience that this molecule has C2V symmetry. The OH bond lengths should be equivalent. When using some sort of optimizing routine, you may want to specify symmetry in your system. With a Z-matrix, the process is very straightforward. You would construct your Z-matrix to define the OH(1) bond as being equivalent to the OH(2) bond. Whatever program you use should automatically recognize the constraint and will optimize your molecule accordingly giving you an answer based off a structure that is constrained to C2v symmetry. With Cartesian space this is not guaranteed. Rounding errors can cause your program to break symmetry, or your program may not be very good at guessing the point group of your molecule based on the Cartesian coordinates alone.
**Picking the Right One**
As a preface, programs like Gaussian convert your Cartesian coordinate space (or your pre-defined Z-matrix) into redundant internal coordinates before proceeding with an optimization routine unless you specify it to stick with Cartesians or your Z-matrix. I warn you that this makes your calculation much more expensive. I find that I will explicitly specify 'Z-matrix' when I know I'm dealing with high symmetry and when I know my Z-matrix is perfect.
You will want to use Z-matrices on systems that are rather small. If dealing with systems with high symmetry, Z-matrices are almost essential. They can be rather tricky to implement and you will likely spend some time figuring out the proper form of your Z-matrix through trial-and-error. If you wish to scan a particular coordinate, Z-matrices are also very helpful as you can tell a program to scan across a bond length, angle or torsion with ease (as long as you've properly defined that coordinate in your Z-matrix).
I use Cartesian coordinates for large systems, systems with very little or no symmetry, or when I'm in a hurry.
|
What you're describing is thermoluminescence, which is used as a paleochronology method (aka Trapped Electron (TE) dating, afaik). The [wikipedia article][1] has a good synopsis. Simply, electrons get bound to potential wells generated in the radiation-defected structure. When you heat the crystal up you allow it to anneal and return to a more regular structure. The defected crystal is stuck in an energetic local minimum.
You can also do this trick with X-rays. Calcite (or was it aragonite?) exposed to X-rays will turn light blue for a while.
[1]: http://en.wikipedia.org/wiki/Thermoluminescence_dating |
To my understanding, [cationic polymerization](http://en.wikipedia.org/wiki/Cationic_polymerization) is a type of [living polymerization](http://en.wikipedia.org/wiki/Living_polymerization) but my teacher's notes and Wikipedia seem to suggest they're [different](http://en.wikipedia.org/wiki/Living_cationic_polymerization). I can't see the difference.
The same goes for anionic polymerization. From what I've studied so far, the difference must be subtle, if there is one.
Is there such a difference? |
I've always seen benzyne (benzene with a triple bond) classified as an 'intermediate'. I really don't see why it needs to be an intermediate, though. The possuble reasons I can come up with are flimsy:
- It has an unstable sp2-sp2 $\pi$ bent bond: We have bent bonds with more angle strain in cyclopropane, which is not an intermediate. Though those bonds are $\sigma$ bonds.
- It immediately reacts with itself to form a dimer(can't recall the name) in the absencr of other reagents : There are many other molecules which spontaneously dimerize (eg $\ce{NO2}$ at certain temperatures), yet they are not classified as 'intermediates'.
So, what makes benzyne an 'intermediate'?
More generally (but possibly too broad), what makes a molecule an 'intermediate'? |
Why is benzyne an intermediate? |
You can buy NMR tubes in a huge variety of qualities, with an equally huge difference in price between the cheapest and the most expensive NMR tubes. They are usually rated for a specific spectrometer frequency, e.g. 300 MHz+ or 600 MHz+.
What is the difference between these different grades of NMR tubes? And how large is the effect on the quality of the spectra if a lower-grade tube is used?
|
What benefits do higher-rated NMR tubes provide? |
There are two kinds of bonds **between** polymers in solid material: primary and secondary. Primary bonds can be also called cross-linked bonds. What is the difference between the two (strength, type, response to stress/strain, etc)? How do they compare to bonds inside a polymer? |
What is the difference between cross-linked bond and secondary bond in polymers? |
Intermediate is a middle step in the chemical process: reactants -> intermediate(s) -> products. Benzyne is formed from benzene and is highly reactive. Thus, it is never the first or last step of the process. To my knowledge benzyne is rarely observed as a product, because its reacting fast with other molecules because its triple bond.
What makes a molecule an intermediate? It is a highly reactive "middle product" of a reaction.
http://academic.scranton.edu/faculty/cannm1/advancedorganic/advancedorganicmodule.html
http://en.wikipedia.org/wiki/Aryne
|
A general rule is: crap goes in, crap comes out. A large-sample low-field 1D NMR at room temperature is usually only minimally affected by using a cheap NMR tube. There are important differences though and I'll highlight a few.
The first distinction between prices is what the tube is constructed from: quartz obviously costs more than borosilicate. Why would a chemist ever use the more expensive quartz? You can [heat/cool quartz faster][1] (nice for thermal studies), the UV cutoff is lower (think 190 nm opposed to 320nm) which is important for photolysis, you can work with quartz at higher temperatures (around 1300 deg C instead of 250 C), and the purity of quartz is better controlled than your typical pyrex. There are different grades of quartz, fused and synthetic, and there are different grades of borosilicate, such as the high-quality pyrex or the lower-quality Class B, each comes with its own limitations as far as purities are concerned and so forth.
Three more important parameters have to do with the manufacturing of your tube are: concentricity, camber and wall thickness. Lower quality tubes will tend to have less precision & accuracy over each of these parameters and as a result your sample may wobble while spinning (introducing problems such as modulation sidebands). A particularly bad tube can hit your RF coils and cause damage to your probe over time slowly or quickly if it is ignoring any reasonable standard -- even more apparent for a tube at this level of "quality" is that it may be easier to break while acquiring your sample and we all should be aware of how much fun that is for everybody involved.
Shimming can deal with impurities present in the glass (such as ferric oxide) and increased impurities in the glass/inhomogeneities will result in taking longer to get a good shim. Time is money.
A lot of these things have lower tolerances in more complex experiments and at higher fields. It really does depend on your particular experiment and what you're hoping to get out of it.
[1]: http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion |
I've always seen benzyne (benzene with a triple bond) classified as an 'intermediate'. I really don't see why it needs to be an intermediate, though. The possible reasons I can come up with are flimsy:
- It has an unstable <em>sp</em><sup>2</sup>–<em>sp</em><sup>2</sup> $\pi$ bent bond: we have bent bonds with more angle strain in cyclopropane, which is not an intermediate. Though those bonds are $\sigma$ bonds.
- It immediately reacts with itself to form a dimer(can't recall the name) in the absencr of other reagents : There are many other molecules which spontaneously dimerize (eg $\ce{NO2}$ at certain temperatures), yet they are not classified as 'intermediates'.
So, what makes benzyne an 'intermediate'?
More generally (but possibly too broad), what makes a molecule an 'intermediate'? |
Intermediate is a middle step in the chemical process: reactants -> intermediate(s) -> products. Benzyne is formed from benzene and is highly reactive. Thus, it is never the first or last step of the process. To my knowledge benzyne is rarely observed as a product, because its reacting fast with other molecules because its triple bond.
What makes a molecule an intermediate? It is a highly reactive "middle product" of a reaction.
See [here](http://academic.scranton.edu/faculty/cannm1/advancedorganic/advancedorganicmodule.html) and [there](http://en.wikipedia.org/wiki/Aryne). |
**Cartesian Space**
In Cartesian space, three variables (XYZ) are used to describe the position of a point in space, typically an atomic nucleus or a basis function. To describe the locations of two atomic nuclei, a total of 6 variables must be written down and kept track of. The general ruling is that for Cartesian space, 3N variables must be accounted for (where N is the number of points in space you wish to index).
**Internal Coordinates**
Z-matrices use a different approach. When dealing with Z-matrices, we keep track of the relative positions of points in space. Cartesian space is 'absolute' so to speak. A point located at (0,0,1) is an absolute location for a coordinate space that extends to infinity. However, consider a two atom system. The translation of the molecule through space (assuming a vacuum) will have no affect on the properties of the molecule. An H2 molecule centered around the origin (0,0,0) is no different from the same H2 molecule being centered around (1,1,1). However, say we increase the distance between the hydrogen atoms. We now have altered the molecule in such a way that the properties of that molecule has changed. What did we change? We simply changed the bond length, one variable. We increased the distance between the two atoms by some length R. With Z-matrices, we keep tabs on internal coordinates: bond length (R), bond angle (A), and torsional/dihedral angle (T/D). Using internal coordinates reduces our 3N requirement set by the Cartesian space down to a 3N-6 requirement (for non-linear molecules). For linear molecules we keep tabs on 3N-5 coordinates. When performing complex computations, the less you have to keep track of, the less expensive the computation.
**Symmetry**
Consider the following molecule, H2O. We know from experience that this molecule has C2V symmetry. The OH bond lengths should be equivalent. When using some sort of optimizing routine, you may want to specify symmetry in your system. With a Z-matrix, the process is very straightforward. You would construct your Z-matrix to define the OH(1) bond as being equivalent to the OH(2) bond. Whatever program you use should automatically recognize the constraint and will optimize your molecule accordingly giving you an answer based off a structure that is constrained to C2v symmetry. With Cartesian space this is not guaranteed. Rounding errors can cause your program to break symmetry, or your program may not be very good at guessing the point group of your molecule based on the Cartesian coordinates alone.
**Picking the Right One**
As a preface, programs like Gaussian convert your Cartesian coordinate space (or your pre-defined Z-matrix) into redundant internal coordinates before proceeding with an optimization routine unless you specify it to stick with Cartesians or your Z-matrix. I warn you that specifying your program to optimize using Cartesian coordinates makes your calculation much more expensive. I find that I will explicitly specify 'Z-matrix' when I know I'm dealing with high symmetry and when I know my Z-matrix is perfect.
You will want to use Z-matrices on systems that are rather small. If dealing with systems with high symmetry, Z-matrices are almost essential. They can be rather tricky to implement and you will likely spend some time figuring out the proper form of your Z-matrix through trial-and-error. If you wish to scan a particular coordinate, Z-matrices are also very helpful as you can tell a program to scan across a bond length, angle or torsion with ease (as long as you've properly defined that coordinate in your Z-matrix).
I use Cartesian coordinates for large systems, systems with very little or no symmetry, or when I'm in a hurry.
|
Given an organic compound, is there any way to decide if it is volatile (or compare volatility--everything is volatile in the end)?
Volatility is due to the tendency to evaporate. The conclusion I can draw from this is that a more volatile compound will be lighter and/or will have less intermolecular forces. But this doesn't seem to work--I recall that methanol is _less_ volatile than ethanol ([citation needed]).
So are there any general guidelines for this or is it another experimental property? |
The main chemistry prep room at my institution (and also at the ambulance station) is labelled with the standard green compressed gas sign. However, there is a small '2' included within the diamond. None of the chemistry technicians, tutors or ambulance staff seem to know what it is referring to and I can't find any reference of it on the internet. Would anyone here happen to know?
![Example of Compressed Gas (2) Hazard Sign][1]
The related products of this [online provider][2] show that other hazards have different numbers but some also have two (e.g. flammable). Is there a standard that is being referred to here?
[1]: https://i.stack.imgur.com/9q3n8.jpg
[2]: http://www.directa.co.uk/site/scripts/product_browse.php?product_id=8174&category_id=481 |
What is the meaning of the '2' on some Compressed Gas Hazard signs in the lab? |
To my understanding, [cationic polymerization](http://en.wikipedia.org/wiki/Cationic_polymerization) is a type of [living polymerization](http://en.wikipedia.org/wiki/Living_polymerization) but my teacher's notes and Wikipedia seem to suggest they're [different](http://en.wikipedia.org/wiki/Living_cationic_polymerization). I can't see the difference.
The same goes for anionic polymerization. From what I've studied so far, the difference must be subtle, if there is one.
Is there such a difference?
_For some unknown reason, my question doesn't appear in the question list. At least I don't see it..._ |
Diborane has the interesting property of having two 3-centered bonds that are each held together by only 2 electrons (see the diagram below, from [Wikipedia](http://en.wikipedia.org/wiki/Diborane)). These are known as "banana bonds."
I'm assuming there is some sort of bond hybridization transpiring, but the geometry doesn't seem like it is similar to anything I'm familiar with Carbon doing. What sort of hybridization is it, and why don't we see many (any?) other molecules with this bond structure?
![enter image description here][1]
[1]: https://i.stack.imgur.com/9L9j2.png |
What makes diborane an ideal situation for banana bonds? |
Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.
$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.
**So the basic reason for this happening is due to an incomplete octet wanting to fill itself.**
Note that "banana" is not necessarily only for [3c2e][1] bonds. Any [bent bond][2] is called a "banana" bond.
Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.
_Sneaks off and checks Wikipedia._ Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.
I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.
<sup>*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.</sup>
[1]: http://en.wikipedia.org/wiki/Three-center_two-electron_bond
[2]: http://en.wikipedia.org/wiki/Bent_bond |
To my understanding, [cationic polymerization](http://en.wikipedia.org/wiki/Cationic_polymerization) is a type of [living polymerization](http://en.wikipedia.org/wiki/Living_polymerization) but my teacher's notes and Wikipedia seem to suggest they're [different](http://en.wikipedia.org/wiki/Living_cationic_polymerization). I can't see the difference.
The same goes for anionic polymerization. From what I've studied so far, the difference must be subtle, if there is one.
Is there such a difference? |
What makes banana bonds possible in diborane? |
Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.
![enter link description here][1]
As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.
$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.
**So the basic reason for this happening is due to an incomplete octet wanting to fill itself.**
Note that "banana" is not necessarily only for [3c2e][2] bonds. Any [bent bond][3] is called a "banana" bond.
Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.
_Sneaks off and checks Wikipedia._ Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.
I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.
<sup>*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.</sup>
[1]: http://upload.wikimedia.org/wikipedia/commons/thumb/6/63/Diborane-2D.png/320px-Diborane-2D.png
[2]: http://en.wikipedia.org/wiki/Three-center_two-electron_bond
[3]: http://en.wikipedia.org/wiki/Bent_bond |
If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds chemical **and** physical?
From the freedictionary.com, chemical bond:
Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.
What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these "several forces" excactly?
In the answer I would like to see the lists of chemical and physical bonds. |
What is the difference between physical and chemical bonds? |
Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.
![enter link description here][1]
As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.
$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.
[It seems that there are a lot more compounds with 3c2e geometry](http://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane/229#comment250_229). I'd completely forgotten that there were entire homologous series' under 'boranes' which all have 3c2e bonds (though not the same structure)
And there _are_ Indium and Gallium compounds as well. Still group IIIA, though these are metals. I guess they, like $\ce{Al}$, still form covalent bonds.
**So the basic reason for this happening is due to an incomplete octet wanting to fill itself.**
Note that "banana" is not necessarily only for [3c2e][2] bonds. Any [bent bond][3] is called a "banana" bond.
Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.
_Sneaks off and checks Wikipedia._ Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.
I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.
<sup>*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.</sup>
[1]: http://upload.wikimedia.org/wikipedia/commons/thumb/6/63/Diborane-2D.png/320px-Diborane-2D.png
[2]: http://en.wikipedia.org/wiki/Three-center_two-electron_bond
[3]: http://en.wikipedia.org/wiki/Bent_bond |
In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is **the sharing of electrons between two or more nuclei**. In contrast, other interactions are often said to be *intermolecular* (which is somewhat more specific than the term “physical”.
---
In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).
1. Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In [*The Nature of the Chemical Bond*][1] (1960), he gives the following definition:
![enter image description here][2]
**A bond is what links atoms into molecules**, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…
2. The complete opposite: **consider all interactions as chemical bonds**, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like [this one](http://books.google.com/books?id=bLgkS8pE1N0C&pg=PT175):
![enter image description here][3]
This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).
3. You can classify interactions by energy: decide that **chemical bonds are those interactions that have an energy higher than a certain threshold**, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.
4. Finally, what I believe is the most common description is to look at **the nature of the interaction**, and classify it following a certain **convention**. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.
5. I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the *New Oxford American Dictionary* as well, which I quite like:
> **chemical bond**<br> a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.
[1]: http://books.google.com/books?id=L-1K9HmKmUUC
[2]: https://i.stack.imgur.com/l6K65.png
[3]: https://i.stack.imgur.com/RtWts.png |
In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is **the sharing of electrons between two or more nuclei**. In contrast, other interactions are often said to be *intermolecular* (which is somewhat more specific than the term “physical”.
---
In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).
1. Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In [*The Nature of the Chemical Bond*][1] (1960), he gives the following definition:
![enter image description here][2]
**A bond is what links atoms into molecules**, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…
2. The complete opposite: **consider all interactions as chemical bonds**, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like [this one](http://books.google.com/books?id=bLgkS8pE1N0C&pg=PT175):
![enter image description here][3]
This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).
3. You can classify interactions by energy: decide that **chemical bonds are those interactions that have an energy higher than a certain threshold**, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.
4. Finally, what I believe is the most common description is to look at **the nature of the interaction**, and classify it following a certain **convention**. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.
5. I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the *New Oxford American Dictionary* as well, which I quite like:
> **chemical bond**<br> a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.
Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is **sharing (or transfer) of electrons**. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).
Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…
[1]: http://books.google.com/books?id=L-1K9HmKmUUC
[2]: https://i.stack.imgur.com/78asf.png
[3]: https://i.stack.imgur.com/RtWts.png |
If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds chemical **and** physical?
From the freedictionary.com, chemical bond:
> Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.
What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these "several forces" excactly?
In the answer I would like to see the lists of chemical and physical bonds. |
[Neil Bartlett](http://en.wikipedia.org/wiki/Neil_Bartlett_(chemist))(1932-2008) first synthesized $XeF{_4}$ (and $XeF{_6}$) in 1962. In the [synthesis](http://en.wikipedia.org/wiki/Xenon_tetrafluoride#Synthesis), a nickel chamber is used, and heated to 400°C, causing the formation of $NiF{_4}$.
The $XeF{_4}$ molecule is planar. Normally, when I think of a metal catalyst, the reactant stands up off of the surface to be "harvested" by the other reactant (like $C{_2}H{_4}$, for example), and the attack is on one side of the plane.
It would seem as though the fluorines would need to configure themselves in such a way that all could be "forced" on the $Xe$ at the same time. What does this configuration look like, and why aren't two $F$s pulled off instead of 4? |
What is the configuration of nickel tetrafluoride during the synthesis of xenon tetrafluoride? |
I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, [$\ce{PH3}$][1] has _nearly_ pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a _nearly_ pure $s$ orbital.
Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.
While answering [this question][2], I realised that it's not that easy predicting the numbers.
Take the same image of $\ce{B2H6}$:
![enter link description here][3]
At first, seeing the 97°, I thought "well, the inner $\ce{B-H}$ bonds will be almost pure $p$". But, with that, I couldn't figure out where the 120° came from, because that is for perfect $sp^2$.
I then realised that I was being stupid and it wasn't that simple--just because you have a 120° doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.
How does one generally go about predicting such "nonuniform hybridizations" if one knows the bond angles?
Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.
[1]: http://en.wikipedia.org/wiki/Phosphine
[2]: http://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane
[3]: http://upload.wikimedia.org/wikipedia/commons/thumb/6/63/Diborane-2D.png/320px-Diborane-2D.png |
[Neil Bartlett](http://en.wikipedia.org/wiki/Neil_Bartlett_(chemist))(1932-2008) first synthesized $XeF{_4}$ (and $XeF{_6}$) in 1962. In the [synthesis](http://en.wikipedia.org/wiki/Xenon_tetrafluoride#Synthesis), a nickel chamber is used, and heated to 400°C, causing the formation of $NiF{_4}$, but it not part of the reaction.
It would seem as though the fluorines would need to configure themselves in such a way that all could be "forced" on the $Xe$ at the same time. What does this configuration look like, and why aren't two $F$s pulled off instead of 4? |
What is the configuration of fluorine during the synthesis of xenon tetrafluoride? |
[Neil Bartlett](http://en.wikipedia.org/wiki/Neil_Bartlett_(chemist)) (1932–2008) first synthesized $\ce{XeF{_4}}$ (and $\ce{XeF{_6}}$) in 1962. In the [synthesis](http://en.wikipedia.org/wiki/Xenon_tetrafluoride#Synthesis), a nickel chamber is used, and heated to 400°C, causing the formation of $\ce{NiF{_4}}$, but it not part of the reaction.
It would seem as though the fluorines would need to configure themselves in such a way that all could be "forced" on the $\ce{Xe}$ at the same time. What does this configuration look like, and why aren't two $\ce F$'s pulled off instead of 4? |
[Neil Bartlett](http://en.wikipedia.org/wiki/Neil_Bartlett_(chemist)) (1932–2008) first synthesized $\ce{XeF{_4}}$ (and $\ce{XeF{_6}}$) in 1962. In the [synthesis](http://en.wikipedia.org/wiki/Xenon_tetrafluoride#Synthesis), a nickel chamber is used, and heated to 400°C, causing the formation of $\ce{NiF{_4}}$, but it not part of the reaction.
It would seem as though the fluorines would need to geometrically configure themselves in such a way that all could be "forced" on the $\ce{Xe}$ at the same time.
What does this configuration look like, and why aren't two $\ce F$'s pulled off at a time instead of 4? |
What is the geometric configuration of the four fluorine atoms during the synthesis of xenon tetrafluoride? |
I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.
I'll explain via an example:
Let's take an $\ce{N3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).
Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry--we have a distinct triplet of orthogonal directions separate from the others.
This leads to these questions:
How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they _thenselves_ wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)
So, I'd like a clear explanation of how/why the symmetry breaks. |
Symmetry lost in orbitals? |
I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.
I'll explain via an example:
Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).
Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.
This leads to these questions:
How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they _themselves_ wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)
So, I'd like a clear explanation of how/why the symmetry breaks. |
Let's say I synthesized or isolated a chiral molecule and want to know the absolute configuration (R or S) of that molecule. I could obviously solve the structury by X-Ray crystallography, but that's a lot of work and I might not want to do that or might not have access to the necessary equipment.
What other ways exist to determine the absolute configuration of a molecule experimentally? |
How do I determine the absolute configuration experimentally? |
I remember seeing a periodic table that had the top-left corner of the d-block shaded and marked as "brittle". If I recall correctly, the elements were $\ce{Sc,Ti,V,Cr,Mn,Y,Zr,Nb,La}$. I think (Sorry, I can't find much on it on the Net)
The triangular shaded area is a clear indicator of a groupwise and period wise trend (similar to how the nonmetals form a triangle due to the electron affinity/Ionization potential trends).
So, I ask, what are the trends leading to the brittleness of a metal, and, more importantly, _why_ are these trends there? (usually trends can be explained in terms of simpler things like atomic size/sheilding/etc) |
A Chemdrawing software? |
Besides elucidating or verifying a chemical structure, NMR can also be used e.g. for quantifying a mixture of different chemicals.
Depending on the spectrum and specific substance, integrating the NMR signals for the same molecule can result in significant variations from the actual ratio of nuclei, which is known from the chemical structure.
What factors do affect the quantification of NMR signals? What precautions should one take during acquisition and processing of NMR spectra to make sure that the spectra can be analyzed quantitatively? |
What factors are important for quantitative analysis of an NMR spectrum? |
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