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Chemically Speaking, What's Bad About Inhaling Plastics?
I hear about things like dioxin, but I really don't see what's dangerous about plastics-shouldn't you be ble to just breathe it out, or pass it out of your system? How would it cause cancer?

Check out this information

The primary dangers from plastics themselves come when the monomer, and any compound put in to make the polymer matrix flexible, separate from the polymer and are released (leached) into whatever the plastic is containing. There are several such compounds which are known to be problematic to human biochemistry:

Phthalates - These are aromatic diesters, with a structure incorporating a phenyl ring with two ester linkages, typically to alkanes like ethyl or butyl groups. Phthalates are used in certain plastics to link the monomers, such as in poly-ethyl terephthalate (PET aka polyester), and in others as a plasticizer to make the normally rigid plastic flexible (PVC). Phthalates mimic certain hormones in the human body and so can cause issues with growth and maturation, as well as feminizing characteristics in males (reduced body hair, weight gain, moobs). PET is OK for packaging and storage of cold foods (it is, for instance, the plastic used for soda bottles), but you don't want to heat it because that will degrade the plastic to release the phthalate.
The ones used in PVC are especially problematic because they're not well-bonded; instead of being an integral piece of the polymer chain, they form weak hydrogen bonds between the polymer chains (thus the flexibility, even in relatively thick layers of the plastic), and so these phthalates easily leach out into water or "off-gas" into the air, even without heat. These phthalates are a big part of the "plastic smell" of new plastic products, and of water stored in them; the more flexible, and the longer the plastic has been kept enclosed, the more intense the smell. There was a big stink (NPI) in the early 90s when PVC was being used for water supply lines in home construction, and then they found all these problems with the stuff and a lot of homebuilders and chemical companies went bust paying for all this piping to be replaced with copper.

Bisphenols - These are variations of 2,2-bis-(dihydroxyphenyl)propane (which is the infamous Bisphenol A or BPA). Bisphenol A is the monomer for most of the original formulations of high-impact polycarbonate, such as Lexan. Again, when heated, the polymer breaks down and leaches the water-soluble monomer, which is structurally similar in many ways to phthalate and again has hormone-mimicking, endocrine-disrupting properties, which is why newer polycarbonates intended for food storage are made with other bisphenol monomers (most of which have similar endocrine properties, but because they aren't Bisphenol A they're still legal for now and allow the manufacturer to label the product "BPA-free").

Vinyl Chloride - The monomer of poly-vinyl chloride or PVC, this is typically a hazard in production facilities for PVC and when old, aged PVC begins to break down. The vinyl chloride monomer is a known carcinogen, which is the other reason PVC is no longer used for food packaging or water supply, and is being phased out from wastewater lines in favor of nylon.

As far as answering the question "shouldn't we just be able to pass it out of our system", the same could naively be asked of a whole host of chemicals that are bad for us. The biggest problem with that statement is the First Law of Toxicology; anything, in sufficient amounts, is harmful, even if it's absolutely essential for life in the right amounts. Selenium is an essential trace mineral in human biochemistry, required in certain protein synthesis mechanisms, but the LD50 blood concentration is about 3 mg/kg, and harmful effects such as stunted growth begin to be observed at concentrations within foodstuffs of as low as 4 to 5 ppm. Heck, you can die from drinking too much water, faster than you could from lack of it (it causes a condition called hyponatremia where the added water dilutes electrolytes to the point the body can no longer function).
What causes many chemicals to be harmful is in fact our body's own affinity for them as a beneficial chemical. Either the chemical is beneficial to us, or it looks like one that is, and the chemical is normally so rare in our diet and environment that our bodies grab all of it that they find, without an "off switch" to regulate intake.
Which leads to the second problem with that statement; sometimes our bodies don't efficiently metabolize and dispose of chemicals. Vitamin A, for instance, is an essential nutrient, and our body absorbs all it can, and gets rid of it only very slowly. As a result, it "bioaccumulates", and eventually becomes toxic. That's why beta carotene is the preferred "form" of Vitamin A in supplements; it's in fact an intermediate that our bodies can use to synthesize the vitamin as needed, but being water-soluble unlike actual Vitamin A, any unused excess of beta carotene is simply flushed out via the kidneys.
That process in itself can cause problems; the latest round of drug recalls the lawyers are running commercials on are for drugs that did exactly what they were supposed to, but were then metabolized into carcinogenic compounds that concentrated in the urine and caused various cancers of the excretory system.
And the last problem, tying back into the case in point, is that even if the body selectively absorbs it, efficiently processes it, and it isn't acutely toxic in any form or intermediate along the way, as long as there's an input of the chemical into your system, you'll always have some of it. Plastics are ubiquitous in consumer products, and especially food prep and packaging. As a result, virtually every resident of the industrialized world has a detectable amount of artificial phthalates in their blood, because it is simply impossible not to come into contact with them. Even USDA-certified organic foods have phthalates in them, because the cows eat food that's been stored or served in plastic bins and it shows up in their milk and meat, or the vegetable was fertilized with compost stored in a plastic bin or watered from a PVC irrigation system (even the farmer's market items from someone's backyard were likely grown using water collected in a plastic rain catchment and/or dispensed from a hose or a drip irrigator system that leaches plasticizer). What you store it in or prepare it with once it gets to you is immaterial; the damage is done.


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Geometrical isomerism in cycloalkenes
Why doesn't cyclohexene have a cis-trans isomer whereas cyclodecene has one?
Is strain a factor?

You simply can't make a transoid molecule of cyclohexene because there is not enough "length" in the carbon chain.
I don't know if it is "strain" per se--but a putative transoid cyclohexene would have extraordinarily exotic bonding arrangement--thus being very high in energy.

Yes, in cycloalkenes with of 7 or less carbon atoms, the high ring tension is a prohibitive factor, but in higher cycloalkenes it's not that great so they can show the geometrical isomerism.

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What is the Molecular Formula for CH3-C≡C-CH2-CH3
I am taking an Organic Chemistry class in my college and I was wondering what the Molecular Formula for CH3-C≡C-CH2-CH3 was.
It would also be nice to have the class of compound.

Seems like 2-Pentene http://en.wikipedia.org/wiki/Pentene

@RyanArmstrong, 2-pentene is not the correct name for this compound. The triple-bond makes it an alkyne, not an alkene. Jerry has correctly identified the compound in his answer below as 2-pentyne/pent-2-yne.

Are you sure you didn't mean the name of this molecule? Because strictly speaking, the molecular formula of the molecule you posted would be $\ce{C5H8}$. Pent-2-ene has molecular formula $\ce{C5H10}$.
The molecule you asked about has a triple bond, meaning that it's an alkyne and in this case, pent-2-yne. If you want the wiki page, it's here.
Count the number of carbon atoms and the number of hydrogen atoms to be sure.

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Racemization of biphenyl compounds
How does the effect of heat and effect of substituents affect racemization of biphenyl compounds?

Preparation: Take the parent compound per se, biphenyl.  The bond keeping the two phenyl rings together leads to one carbon atom we call C1, the other C1'.  Now if you label the other carbon atoms in each  phenyl ring seperately in clockwise direction, you'll get C2, C3, C4, C5, C6 in one cycle; and C2', C3', C4', C5', C6' in the second.  (I haven't a credit to upload a picture yet.)
Now: The larger, the more sterically demanding substitutents on C2, C6, C2' and C6' are (condition), the higher the energetical barrier is to interconvert the conformers into each other.  Eventually, this may lead to atropisomerism, and the isomers are atropisomers.
Note, please, sometimes there may be already a preference for one or the other conformation.  If, for example, you substitute C2 and C2' of biphenyl by nitrogen, free molecules of this compound 2,2'-bipyridine energetically favour a conformer with the nitrogens as far as possible separated (similar to a s-trans conformation).  Offering a cation to bind, to complex to, however may lead to a preference towards the s-cis-like conformation.

Well, what if there are sone substituents likr the nitro or methoxy group present on diffrent positions? Also how does heat have an effect on racemization?

The first part of my answer was based on sterics.  In other words, the bigger the atoms / groups facing each other across space of different phenyls (i.e. not bound on adjacent/vicinal carbon atoms of one and the same phenyl ring), the more likely the two rings are even more out of a plane in common.  You are right, a second potential contribution is electrostatics. E.g., an H-donor on phenyl #1 on C2, an H-acceptor on the other ring, say on C2' may actually lower the torsion angle between the phenyls below the one only based on occupied volume by the substitutents. This later as an example.


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How do the stored chemicals in batteries drive electronic current from their terminal to outside devices?
I have got around to at least read on Wikipedia, and it does not clarify how this works in any sensible manner.
I wondered how the chemicals make electrons "move". How is the basic understanding of this?

A battery, at its heart, is what we call a redox cell: some compounds are stripped of their electrons, and others gain those electrons. The movement of electrons between these two compounds can be harnessed to provide electricity, but that's taking us a step too far. What we want to understand here is why the electrons move. To do this, we'll have to understand the redox cell. As much as I hate to assume things, you're going to have to understand a bit about ions and equilibrium to understand this explanation.
(Note: the explanation I'm about to use is heavily taken from the excellent Chemguide site, run by Jim Clark. I'll include a link at the end.)
Let's say I place a bar of magnesium in a water bath. I'll take a look at the magnesium bar a short while after I place it in the water. If I could visualize atoms and electrons, this is what I'd see:

Some of the atoms of magnesium leave the bar and become ions in the solution, solvated by water (and other things, but that's not crucial to this explanation). Since they become ions, they leave electrons on the metal bar. Because the ions are positively charged, they are attracted to the negative charge on the metal bar, and tend to hang around there.
If I wait a little longer, I'll see more and more ions coming off the bar. As they start to build up in the water around the metal, every once in a while, one of the ions will hit the bar and reclaim its electrons. This is what we'll see then:

If I wait long enough, eventually, the number of atoms coming off the metal will be equal to the number of ions reabsorbing onto the metal. If I take out the "bites" in the metal bar to simplify the situation, this is what I see:

Notice how there's quite a buildup of electrons on the magnesium. This is because magnesium is a very active metal, like lithium. What if we used a much less active metal, one that doesn't have a large propensity to lose atoms into solution? An example of such a metal would be copper. Here's a side-by-side of the two:

Because the magnesium is a lot more active, it tends to lose ions more, and so it has more electron buildup. If we connect the two with a wire, we give the electrons a way to flow from the high concentration to the low concentration. We can then place something along the wire in the middle (say, a lightbulb) and harvest some energy!

This, of course, is a hilariously oversimplified scheme. For one, as the electrons go to the copper, more and more magnesium ions break off to replace them. At some point, there are so many positive ions in the solution that they draw the electrons back toward them, through electrostatic attractions (which is a fancy way of saying that positive and negative charges attract each other). To solve this, we'd put a salt bridge between the solutions that lets ions migrate back and forth.
There are other problems too: for example, this setup wouldn't work in real life because there are no ions in solution: real cells use salt solutions of the metal e.g. $\ce{MgCl2}$ or $\ce{CuSO4}$. There's also the issue that this battery is liquid and wouldn't do so well if you tipped it over, the fact that the voltage (which is really the difference between the charge on the metal bars) is only about 1.1 volts, yada yada. That's not central to the explanation.
The fact that certain compounds are more likely to gain electrons and some are more likely to lose them is the central reason why things like batteries can exist. If you'd like to read more about it, I pulled this explanation from chemguide.co.uk, run by Jim Clark. I am currently ready to graduate with a B.Sc in chemistry, and his website is still one of the best resources for conceptual learning that I've come across---as a matter of fact, I learned about things on that website (eutetic mixtures and azeotropes) that I still have yet to come across in four years of university-level study. If you'd like to read more into redox reactions, the redox equilibria page is at this link right here.


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Does variations in air pressure affect the melting point of PLA and ABS while 3D Printing?
Tight temperature controls are needed to ensure that the filament used in fabrication does not begin to "ooze". In places where the air pressure rises and drops during the course of the day there may be more or less ooze taking place. Controlling ooze is paramount to obtain better print quality.
Question: Given Boyle's law, shouldn't the melting point of PLA and other plastics used in the Fused Filament Fabrication 3D printing methods change based on the ambient air pressure of the machine's location? Should air pressure be taken into account when calculating how much heat to apply to the print nozzles? Does humidity affect PLA's melting point?

Boyle's law refers to properties of an idealized gas, where the product of pressure and volume remains constant.  Taking into account your description, as well from filament-based 3D printers seen so far, my answer is no, Boyle's law does not apply here.
If running by extrusion, polylactic acid (PLA) is heated to overpass melt, however this liquid is not heated to become a gas. Temperature control clearly is of importance, because the viscosity of the polymer depends on it.  And viscosity co-influences the precision of details you may print, actually down to the scale of less as a tenth of a millimetre.  Without experience to run one 3D print in at sea level, and the same again during an intercontinental flight, I assume it is more likely, machine control takes into account the pressure difference between ambient air and polymer melt, and not (yet) the (in comparison) lesser variations of air pressure.
The chemical structure of PLA shown in Wikipedia's article shows a relatively low number of functional groups susceptible to water, relatively low to the inner backbone of the polymer (in the structural scheme, this is written in the square brackets).  The article refers to typical molecular weights of 130 kDa when PLA is produced.  These two let me say, humidity does not play an important role.


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Confusion in unit cells of crystal system
Unit cells are divided into two main types

Primitive
Non-primitive

Primitive includes simple cubic lattice whereas non-primitive includes fcc bcc end centered..
Among the seven types of crystal systems orthorhomic crystal system posses both primitive as well as non-primitive unit cells i.e. face centred body centred rnd centred and simple cubic but the cubic system posses primitive , body centred and face centred arrangements only.
My question is if face centred is possible in cubic system why is not end centred type of unit cell arrangement possible in cubic system if there is no sort of hindrance in vacant space as in orthorhombic which posses all types of unit cell arrangements.

I'm a bit confused. What's going on here? Essence of the cubic system is equivalence of a, b and c axes, namely, presence of 3-fold axis in the body-diagonal direction of the cube. There's no chance of "base-" centred cubic.

Actually there is. You should do some research before answer questions since most of the time a reference is needed. http://www.ndt-ed.org/EducationResources/CommunityCollege/Materials/Structure/metallic_structures.htm and http://en.wikipedia.org/wiki/Cubic_crystal_system

First, a note: the classification of crystal systems, or the reduction of possible lattice types into primitive Bravais lattices, has nothing to do with “space available” or “hindrance”. It is purely a mathematical property having to do with the symmetry of the lattices.
To answer your question, the base-centered cubic lattice is not a Bravais lattice, because it is equivalent to a simple tetragonal lattice:

i got what you had to say about the space thing but i am still confused as. to why cubic systems do not possess end centred arrangement if face centred is possible?

@AanalDesai it is possible to have a base centered cubic system, but it is not a minimal representation of the system. Just like the unit cell must be the smallest cell that can reproduce the full periodic system by a set of translation, the Bravais lattices are by definition the smallest lattices. Imagine a base-centered cubic system (drawing on the left): you can represent this particular lattice as “base-centered cubic”, but you can also view it as a “simple tetragonal” lattice… and the second representation features a smaller unit cell, hence it is the irreducible one.

I had one more query, Wouldn't a face centred simple cubic be an in-centred tetragonal then..?


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At what temperature will proteins and fats boil in a vacuum?
A question was asked on another stackexchange site:

This is not nice perspective, but eventually it will happen. An astronaut falls out of spaceship because of damage caused by collision with other object, or because of suit decontamination. The fluids from the body would evaporate, and if any bacteria would survive, than only as spores. Does it mean the perfect mummification of the body? Or there will be some decay, caused by enzymes from damaged cells, for example?

Given that the water in the body will outgass taking with it much of the rest of the body, are there any cited sources that specify the pressure at which proteins, fats and bones will break down and become vapour?

You're talking about the human body being exposed to outer space and asking about what would happen to it's internal structure. Bones, fats and proteins are within the enclosed system that is the human body, they are not subjected to the low pressures of space. The answer to the pressure at which proteins, fats and bones will break down and become vapour is not related to the answer to what would happen to them in the human body if the body is exposed to the vacuum of space. Nevertheless, I'll try to provide both answers.
The answer to the second question (short-term) can be found on NASA's website at "ask an astrophysicist". It's actually quite surprising for someone used to sci-fi movies:

You do not explode. Your blood does not boil. You do not freeze. You do not instantly lose consciousness.

The question you linked talks about "decay", or the effects of long time exposure. That's hard to predict, but at some point the body would be attracted by a planet and collide and you wouldn't have anything to boil anymore. Now to the chemistry part of it:
There isn't much (if any) data on the boiling point or vapor pressure of the compounds you're asking - it's hard to reproduce and it's more a curiosity than anything else. So let's talk about what would happen if you have fat, bones and proteins directly exposed to outer space pressures and temperatures, which seems to be your point anyway. Temperatures in space go from -270°C (in outer space, unexposed to sunlight) to +120°C (near the earth, exposed to sunlight). Since you're asking about boiling point let's talk about what will happen at the highest temperature.
Let's start with the easier part - the bones. It is safe to assume that the inorganic components will be tougher than the organic ones. The inorganic part of bones is mostly a specific type of apatite, which is a rock. It can be found in moon rocks. It will not boil. So at least the inorganic part of the bone would remain somewhat intact.
Fat is made of triglycerides. It'll probably break down to glycerol and the correspondent acid. Then it would decompose and boil.
Proteins denature at harsh conditions, you'll be left with amino acids at some point. It's hard to tell what would happen since there isn't much data on it, but given the harsh conditions, which include exposure to all sorts of radiations, the amino acids certainly wouldn't last long and at some point you'd be left with more basic molecules that don't resemble a protein anymore. So what can we conclude? Exposed as a whole, the internal structure of the human body would remain somewhat intact for a while (physically speaking). It's hard to predict what happens then, but given that the outer space is such a hostile environment, it's unlikely that the body would last long. Exposed as individual components they would last much, much less. From bones, fat and proteins, it's likely that only the bones stand a chance.


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Reaction with carbon
Usually maximum time we see then after cooking by gas there is no black spot on the pot. But continually burning can affect the pot surface and makes black powder.
My question is:

What is the most acceptable reason that black powder(carbon) take places on the pot surface. Is it related to the burning temperature? or fuel reacts with the pot surface?

Please clarify your question with an example.  Is the pot empty?

Will the answer differ for empty and filled pot? how?

Your question is not clear - are you burning something that is in the pot, or are you concerned with the flame from the stove forming soot on the bottom of the pot?

I'm concerning with that, how black spot is on the pot. What reaction behind this.

See this question

For combustion to occur you need 3 things at the same place: fuel, oxygen and heat.
If the air intake is dirty or not big enough, you get a flame that has an orange color, and on a cooler surface deposits C before it has a chance to burn.
The reaction with the pot is only a physical one of heat transfer and maybe C deposition.

Can you write some carbon depositions reactions which incorporated with the pot?

Temperature is not a thing in the sense that it is a substance like fuel or oxygen; it is a measure of thermal energy.  It is more syntactically correct to say fuel, oxygen and heat.

One possibility for propane: $\ce{C3H8 + 4O2⟶ 4H2O + 2CO2 + C}$ but there are different mixes for the gas used for cooking. @bobthechemist: you're right.


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bond energy calories
I am attempting to optimize interactions for a series of random and test molecules to a protein using docking software. The software uses QM/MMGBSA to calculate free energy in joules (yes it is not a measurement but it is very corollary). Larger molecules automatically have more free energy because they have greater potential energy. I would like to divide the free energy by the amount of atoms in the molecule or the molecules total Dalton. The idea would be to normalize the energy and provide the amount of calories associated with the bond. I asked a professor at a university about measuring the efficiency of a bond in this manner and he said he was unsure what joule/mass would provide. It does appear that the amount of bond calories is able to properly classify with correct magnitudes of viability 2 positive controls and 1 negative but I need this idea to pass peer review. Is my rationalization of the joule/Dalton reasonable?
I am measuring the binding energy of azttp (AZT triphosphate), gtp (Guanine triphosphate) and AZTppppA (AZTP+AMP) on HIV Reverse Transcriptase wild type (8ICO). I am used 8 high quality conformers and independent docks with QM/MMGBSA for my negative sample (AZT+AMP), 5 conformers for my major positive sample (AZTTP) and 1 conformer for the minor positive sample (gtp). I have two or three high quality conformers through the same processes for each of my test samples.
HIV RT has a tunnel where nucleotide analogs pass through with a pocket where if a molecule binds effectively it is accepted but if it fails it is phosphorylated and excreted. I am attempting to create a metric which can be used to measure the effectiveness of a binding and allowing me to infer (only) probability of rejection. Since the window/box for the docking is small and does not change for each test subject I would like to consider HIV-RT a constant.

Could you be a bit more specific about what you are measuring and what you are trying to measure? Typically docking calculations yield the CHANGE in free energy between the bound and unbound state of a ligand-protein complex.
As you pointed out this can be different for larger/smaller molecules but is more highly influenced by the types of groups interacting, conformation of the ligand in the pocket etc.

@CTKlein i updated the question with more information as per your request

Is my rationalization of the joule/Dalton reasonable? || No. Standard scheme for measuring bond energies is to calculate free energies for complex and subtract energies of two separate molecules. In your case you probably have to calculate free energies for biological conditions and then derive constant of reaction. After that you have to solve classical problem of amount of bound/unbound molecules using some known (medically meaningful) starting concentrations

@permeakra thank you for your response but larger molecules have a greater probability of forming larger complexes as CTKlein points out

@caseyr547 first, masses are irrelevant in the case then =) for obvious reason. Second, chelate effect does exist. And third, I don't think 'mass' or 'size' has any meaning. AFAIK, in medicine new substance has to be 1) efficient ligand 2) specific ligand 3) not unreasonably costly per use case. 4)not extremely toxic. Joule/Dalton does not address any of this issues.

@permeakra I'm sure your great with chemistry but as CTKlein pointed out and i observe mass effects the efficiency to some measurable degree. While I'm sure your comments about the constant of reaction rate would work well in non static situations that is not the case here. Also calculating the free energy in a biological condition is impossible given our current technological advancement.

That's not really the point I was making. I was pointing out that molecules of different size will have to adjust accordingly in a binding pocket and/or the binding pocket may change its conformation. Inherently, this has nothing to do with the mass of the ligand.
Also I'm still not clear what exactly you are measuring when you say "free energy". As mentioned, a docking calculation will give you a the CHANGE in free energy. In your case that would be G(complex)-G(ligand)-G(protein). This value is a metric for how tightly the ligand binds which may or may not be what you're interested in.

@CTKlein The formula for deta G which I am using is dG = E_complex-minimized- (E_ligand_minimized + E_receptor_minimized). My hypothesis which I observe to be correct in 3 cases is: The probability of forming more complexes and generating a lower dG is higher when there are more atoms to form complexes in the ligand molecule.

@caseyr547 Once more. ligands with several points of binding binds more efficiently if the centers can assume fitting positions. It is not an artificial effect and is quite well-known in coordination chemistry. If you want metric measuring effectiveness of binding it is exactly free energy of binding. There is completely different matter, that you have to compare it with binding energy of water at same sites and to measure actual preference, but that's different matter. So you probably have either to live with what you have or reformulate the question

@permeakra summed it up nicely. If the pocket is big enough to accommodate the larger ligands you are testing then they are likely binding in more locations than the smaller ligands, hence lower free energy of binding. There is a multitude of interactions that need to be considered when measuring binding free energies (water in the pocket, types of chemical groups on the ligands/pocket etc.). While it would be wonderful if a metric as simple as mass were a good indicator for binding energy, it simply isn't the case.

@CTKlein ok the pocket is big enough to accommodate massive ligands and they are binding in more locations. Can I divide the energy by the amount of binding locations?

@CTKlein the goal would be to average the energy contribution per bond.

@permeakra yes the energy alone gives the metric of quantity but it does not give the metric of quality of the bond. 6 random bonds with 10 atoms total is less than 6 specific bonds with 10 atoms and is less than 3 specific bonds with 4 atoms total. The energy for the 6 random bonds would be greater than the energy of the 3 specific bonds but on average the quality would not be as good.

@caseyr547 I see. In this case I'd considered two options. Option one is to dock as many molecules of water simultaneously as I can and then check how many water molecules cover binding site of the ligand in question. Another option is to check the surface of contact between the binding ligand and the host protein. Both metrics are hard to measure, but they have some meaning at the very least. The last option is to divide by number of atomic contacts between the ligand and the protein.]

@permeakra yes ok dividing the energy by the number of contacts is possible but it has to be done by hand. i have a program which will count the number of contacts for individual ligands. if i did it right, gtp has 178 contacts and aztamp has 400. gtp has a complex dg of -36 and aztamp likewise is -79. From this, Can I safely conclude that aztamp contacts are individually no better than gtp?


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How to find the number of valence electrons?
I need to understand the following:
Considering an element Sulfur - S which has 16 electrons.
How do we calculate the number of valence electrons of S?
Please correct me if I am wrong:2+2+6+2+4=16. So, the valence electron = 4+2=6
Is it that always valence electron is the addition of the last two shells? In that case for Na, sodium, if we take 11 and arrange it as 2+2+6+1, then the valence electron is 1 or 6+1?
Now, considering the fact valence electron is 1, if we go by, up down up down rule, then as it is 1, hence the spin is up?
For 6: is it up(1) down(2) up(3) down(4) up(5) down(6) as it is both up/down hence it would be +-1/2 spin?
For Phosphorous 2,8,5, what would be the valence electron? 5 or 13?
If 5 then up down up down up, will it have an up spin?
Is it that for even numbers like 4,6 the spin will be both up/down i.e. +-1/2?

On the matter of topicality, the underling QM is certainly on-topic on physics.se (and especially the difficult calculation needed to determine the structure of outer shells in high Z atoms), but the notion of "valence" is something that I usually associate with chemistry. Thoughts from the crowd?

'p' and 'd'.
There is one type of 's' orbital. It is spherical. So an s orbital can hold 2 electrons only.
There are three types of 'p' orbitals. They are the same shape (like a dumbell) but are oriented in different directions - along the x, y and z axes.  So a 'p' orbital can hold $2$x$3$ = 6 electrons.
There are five types of 'd' orbitals (10 electrons) and seven types of 'f' orbitals (14 electrons).
Fill up the orbitals in the following order (following the red arrows), which is increasing order of energy. Lowest energy fills up first.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

So let's take the example of sulfur.
The total number of electrons is 16. We'll distribute them in the following manner.
$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons in the last (3rd) orbital. Therefore the valence number is 6.
It is not simply the last 2 numbers while writing the electronic configuration.
For Phosphorous, we have:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^3$, which gives us 5 electrons in outermost (3rd) orbit.
For Lithium (3 electrons), we have:
$1s^2, 2s^1$, which gives us a total of 3. Outermost orbit is the 2nd orbit and the total number of electrons in it is just one. Therefore valency of Lithium is one.
However, it is much simpler to do it the way I've shown above as with Sodium. Don't need to bring in all the orbitals
For the spin, just remember that you have to fill up the entire orbital with up spin electrons before you start putting the down spin electrons. Don't just alternate up and down.
Let's talk about the 'p' orbital. Each 'type' of the p-orbital must have up-spin electrons before adding the down-spin. So there will be 3 up spin electrons before you start adding the down-spin ones. Also, you can not move on to the next orbital without completely filling the previous one with an EQUAL number of up and down spins.
For Phosphorous; let's say we've filled the 1s, 2s, and 2p orbitals. Now we add one up spin to the 3s, followed by a down-spin. Which leaves us with 3 more electrons to be added to the valence orbit (3rd). It is a 'p'-orbital, which needs to have 3 up spin electrons before any downspin ones can be accommodated. But we have only 3 remaining! So all will be up-spin electrons in the valence orbit. So we'll have a total spin of +3/2.
Hope I didn't complicate too much, but I've actually tried to make it as short as possible. I've still left out quite a bit.

Out of curiosity, the "electronic configuration", like $(1s)^2 (2s)^1$ for Li or any more complicated case, is a result of Hartree-Fock calculation or from some model parametrized from experimental atomic spectroscopy? If it is the later one, how it is done in detail? I usually find book just gives result of electronic configuration with some explaination why it is like that, does not say how do we know that

I actually don't know, haha. We haven't studied the reasoning behind this stuff yet. I agree, no where does it give the reason. Have you tried Wikipedia?

wiki  "It is through the analysis of atomic spectra that the ground-state electron configurations of the elements were experimentally determined."  No details about how the electronic configuration is determined. My question is about, the determination may be ambiguous, both $2s^2$ and $2p^2$ can form $^1S_0$ state.

@user26143 Well, the wave-mechanical model (which includes all this) is just a theory. As was Neil Bohr's theory, and Thomson's theory, and Rutherford's theory. They all were proved to be wrong - or incomplete. For now, this model explains our observations to a satisfactory extent. If another theory/model comes up which can explain them better, or can improve on the drawbacks (if any) of this model, then that'll be accepted as the new one. I don't think there's any proof to it.

Well, Bohr's theory, quantum mechanics, and quantum field theory all have mathematical formulation with falsifiable predictions. I am not even sure what is the underlying mathematics about the "electronic configuration" and what is the falsifiable prediction it has... Besides, I don't know how do people write the electronic configuration like copper is as wiki and/or other sources, but in not other form. As wiki said, "... were experimentally determined" there should be a definite procedure for such determination, rather than proof (We don't need to prove an "experimental determination" ;))

It is worth mentioning the case of the rare earths, in which the nominal valence band actually lies closer to the nucleus (spacially) than one of the already filled shells resulting in a group of materials with low reactivity and very little chemical variation from element to element.

I think a more 'robust' definition for valence electrons are 'those that can form bonds'.  'Outermost electrons' can be slightly ambiguous in some cases I think.

It is not so difficult I could not understand your method of finding valence electrons, easiest way I could explain is try to find the atomic number of the element. For example chlorine it's atomic number is 17 now find it's electronic configuration which is 2,8,7. Now we know that there are 3 shells in it hence third shell would be it's last shell and valence electrons are the electrons of valence shell therefore valence electrons would be 7.

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text:
Do ions also form special electron configurations? (just as Copper and Chromium do)
Do ions also form special electron configurations? (just as Copper and Chromium do)
I know that because copper has 1 electron in its s-orbital (highest) because it wants to have a full d-orbital (d^10)
But do ions also act this way?

What do you mean by special?

that it will remove an electron from the s orbital to the d-orbital to have a stable half filled s-orbital and stable fully filled d-orbital

Check out the NIST database for electronic configurations for atoms and ions in the gas phase. If you play around a bit, you will find that yes, some ionizations will not follow the expected trends from the aufbau principle. For example, search for rhenium and compare the configurations between $\ce{Re^{11+}}$ and $\ce{Re^{17+}}$, and you'll see some of them are unexpected (and rather hard to explain). Osmium shows an anomaly going from $\ce{Os^{+}}$ to $\ce{Os^{2+}}$. There are other elements with ions containing aufbau anomalies, but I can't remember many from the top of my head.

In short, yes, the electrons will change orbitals to fill shells (even if it is an ion). This is why the transition elements can have more than one ion.


text:
Finding electron configuration of lanthanide ions
When I have a gadolinium ion ($\ce{Gd^3+}$), how can I calculate its electron configuration?
$$\ce{Gd} : \mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$
Do I need to first subtract the 2 electrons in the $\mathrm{s}$-orbital and then the $\mathrm{d}$-/$\mathrm{f}$-orbital?

When determining electron configurations of cations, one takes the electron configuration of the neutral atom, and then starts removing electrons from the orbital with highest n value.  If there is a tie, you then choose the orbital with highest l value.

I agree with bobthechemist's comment.
When you are determining the electronic configuration of anions/cations, you will first need to determine electronic configuration of neutral atom. Like in the case above:
$$\ce{Gd} : \mathrm{[Xe] (6s)^2 (4f)^7 (5d)^1}$$
Now arrange this configuration according to ascending value of $n$:
$$\mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$
Now, according to your need start subtracting electrons from highest value of $n$, like for $\ce{Gd^3+}$, the electronic configuration will be:
$$\ce{Gd^3+} : \mathrm{[Xe] (4f)^7}$$
See this to know more on calculating electronic configuration of cations and anions.


text:
Special electron configurations in p block on periodic table?
Gold and Chromium have weird electron configurations, because electrons are differently divided : eg an electron jumps from a s-orbital to a d-orbital to have a half-filled s orbital and a fully/half filled d-orbital, which is much more stable.
But does this also occur with the p-orbitals? eg why does Fluorine not transfer an electron from its s orbital to its p orbital to have a  half-filled s orbital and a fully filled p-orbital, that would be more stable?

The simplest answer to this problem is related to relative energy differences between the Nd and (N+1)s orbitals in transition metals and Ns and Np orbitals. The Nd orbitals in transition metals are at a slightly lower energy than the (N+1)s orbitals, so for certain atoms there are some unusual changes in the electron configurations, e.g. Cr and Cu. However, the energy differences between the Ns and Np orbitals are quite large, with Np being higher in energy. In these cases, there is a substantial energetic penalty in moving an electron from the Ns to Np orbital.
The electron configuration problem for transition metals is complex, but this J Chem Ed article has a nice description of the theory, although it is somewhat technical.


text:
What is the structure of N₅P₃?
What is the structure of the molecule $\ce{N5P3}$? $\ce{N5P3}$ is not documented online, so could anyone please comment on the structure of the molecule:

How many ligands does each phosphorous atom have?
How many ligands does each nitrogen atom have?

I'm not sure why previous answer cited only the first paper on phosphorus(V) nitride $\ce{P3N5}$. A couple of years later the same team reported three polymorphs $\ce{α,β,γ-P3N5}$ with three-dimensional network. For $\ce{α-P3N5}$ and $\ce{γ-P3N5}$ crystal structures have been refined. Both modifications demonstrate vastly different structural types and coordination of pnictogens.
Low-pressure phase $\ce{α-P3N5}$
Data obtained via powder neutron diffraction with higher resolution by using synchrotron radiation. The crystal structure was determined by direct methods and refined by the Rietveld method. [1, 2]
Structure description from [1]:

In $\ce{α-P3N5}$ zweier single chains occur parallel to $[110]$ and
  $[1\bar{1}0]$, in which the $\ce{PN4}$ tetrahedra are connected alternately through edge- and corner-sharing. These zweier single chains are linked together through additional vertex-sharing $\ce{PN4}$ tetrahedra. [...]  The network structure of connected $\ce{PN4}$ tetrahedra in $\ce{α-P3N5}$ could be characterized by analogy to condensed silicates by the specific distribution of the occurring $\ce{P_nN_n}$ ring sizes.

Liebau formula: $_\infty^3\ce{[P_3^{[4]} N_3^{[2]} N_2^{[3]}]}$ *

Figure 1. Crystal structure of $\ce{α-P3N5}$, view along $[010]$. Purple coordination tetrahedra correspond to $\ce{[PN4]}$.
Low-pressure phase $\ce{β-P3N5}$
$\ce{β-P3N5}$ was reported in a polymorphic mixture with $\ce{α-P3N5}$ and is it's stacking modification (no further crystallographic investigation has been published) [1]:

On the basis of the solved crystal structure of $\ce{α-P3N5}$, we are now working on the determination of structural models for $\ce{β-P3N5}$, and disordered $\ce{P3N5}$. The HRTEM results show that similar to the situation in the polytypes of the different
  stacking variants are based on the same structural modules, which are linked in different ways.

High-pressure phase $\ce{γ-P3N5}$
Data obtained via powder X-ray diffraction. The crystal structure of $\ce{γ-P3N5}$ was determined by direct methods and refined using the Rietveld method. [3]
Structure description from [3]:

The novel high-pressure modification consists of a polymeric three-
  dimensional network structure of linked $\ce{PN4}$ tetrahedra and
  tetragonal-pyramidal $\ce{PN5}$ units. [...] Rods from trans-edge-sharing $\ce{PN5}$ units running along $[010]$ are linked by vertices, leading to the formation of layers perpendicular to $[100]$. [...]  In accordance with the pressure–coordination rule a partial increase of the coordination number of the phosphorus atoms from four in $\ce{α-P3N5}$ to five in $\ce{γ-P3N5}$ is observed. [...] The density of $\ce{γ-P3N5}$ is $32\%$ higher than the density of $\ce{α-P3N5}$. [...] Therefore the high-pressure phase could show interesting materials properties (e.g. great hardness or low compressibility).

Liebau formula: $\ce{P_5^{[5]}[ P^{[4]} N_3]N_2}$ *

Figure 2. Crystal structure of $\ce{γ-P3N5}$, view along $[010]$. Purple coordination tetrahedra correspond to $\ce{[PN4]}$; yellow coordination pyramids — to $\ce{[PN5]}$.
* using Liebau's nomenclature for silicates, where the superscripted number in square brackets after the element symbol denotes the coordination number.
References

Horstmann, S.; Irran, E.; Schnick, W. Synthesis and Crystal Structure of Phosphorus(V) Nitride α-P3N5. Angewandte Chemie International Edition in English 1997, 36 (17), 1873–1875. https://doi.org/10.1002/anie.199718731.
Horstmann, S.; Irran, E.; Schnick, W. Phosphor(V)-nitrid $\ce{α-P3N5}$: Synthese ausgehend von Tetraaminophosphoniumiodid und Kristallstrukturaufklärung mittels Synchrotron-Pulver-Röntgenbeugung. Zeitschrift für anorganische und allgemeine Chemie 1998, 624 (4), 620–628. https://doi.org/10.1002/(SICI)1521-3749(199804)624:4<620::AID-ZAAC620>3.0.CO;2-K.
Landskron, K.; Huppertz, H.; Senker, J.; Schnick, W. High-Pressure Synthesis of $\ce{γ-P3N5}$ at 11 GPa and 1500 °C in a Multianvil Assembly: A Binary Phosphorus(V) Nitride with a Three-Dimensional Network Structure from $\ce{PN4}$ Tetrahedra and Tetragonal $\ce{PN5}$ Pyramids. Angewandte Chemie International Edition 2001, 40 (14), 2643–2645. https://doi.org/10.1002/1521-3773(20010716)40:14<2643::AID-ANIE2643>3.0.CO;2-T.

$\ce{N5P3}$ is more commonly written as $\ce{P3N5}$, and known as triphosphorus pentanitride. It's a crystalline solid at ambient conditions and not a molecular compound. From the first publication that reported the pure compound and its structure [1]:

In the solid a three-dimensional cross-linked network structure of corner sharing $\ce{PN4}$ tetrahedra has been identified with 2/5 of the nitrogen atoms bonded to three P atoms and 3/5 of the nitrogen atoms bonded to two P atoms.

Only the combination of spectroscopical and diffraction methods, especially electron microscopy for structural analysis, enables a detailed structural and crystallographic characterization of $\ce{P3N5}$ as no conventional single-crystal data for structure determination have been available for this compound.
A representative for an unique binary structure type was found for the first time which combines a tetrahedral structure with the very rare molar ratio of 3:5 for the constituent elements.

References

Schnick, W.; Lücke, J.; Krumeich, F. Chem. Mater. 1996, 8 (1), 281–286 DOI: 10.1021/cm950385y.


text:
What is the oxidation state of aluminium in NaAlO₂?
I guess that sodium loses an electron so has an oxidation state of +1 and that oxygen has an oxidation state of −2. But what oxidation state does the aluminium atom have in sodium aluminate ($\ce{NaAlO2}$)?

For every compound the sum of the oxidation states of the constituent atoms must be equal to the charge of the compound. So, since your compound is neutral the individual oxidation states have to add up to $0$. And as you have already determined the oxidation states of 2 atoms in the system the last one must be $+3$, since $1 + 3 - 2 \cdot 2 = 0$.

Indeed… also, you can confirm this by noting that +3 is the most common oxidation state for aluminium (look at its position in the periodic table)

But how does it then work with the electrons : does aluminium get electrons from sodium or ..?

@LievenB Aluminum is situated in the 3rd main group. So it has 3 valence electrons. Now in $\ce{NaAlO2}$ it has an oxidation state of $+3$ meaning (loosely speaking) that it gives away it's three electrons. Sodium is has an oxidation state of $+1$, so it gives away one electron. The two oxygen atoms, being much more electronegative than the metals take these four electrons and thus have an oxidation state of $-2$ each.

so if were to visualise the molecule, the oxygen atoms would come in the middle, both snatching the electrons from the sodium and aluminium?

No, $\ce{NaAlO2}$ consists of the anion $\ce{AlO_{2}^{-}}$ and the cation $\ce{Na+}$. Aluminum will be in the center between the two oxygens. Sodium is only some kind of counterion and will have the highest probability density at one of the two oxygens.

Even found a picture of $\ce {NaAlO2}$ here: http://img1.guidechem.com/chem/e/dict/37/7758-17-0.jpg

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authors: Рейимберды Курбанов


text:
Oxidation state of metal in metal oxo complexes always positive?
Do all metals in metal oxo complexes (eg $\ce{Fe}$ in Ferrate, $\ce{Mo}$ in $\ce{MoO4}$,...) have positive oxidation states?

Yes. Oxygen is much more electronegative than all of the transition metals, so it is given a formal -2 charge. Keep in mind, though, that these values are not intrinsic properties of these elements/molecules. Their use is intended to help us rationalize some of the chemistry and properties associated with these materials. As an interesting example, consider oxygen difluoride, $\ce{OF_2}$. Fluorine is the most electronegative element (that forms compounds with other elements) and is given a formal charge of -1. To balance the formal charges then, oxygen must have a +2 charge!

And as a result, oxygen difluoride is seriously nasty stuff. Oxygen yearns to achieve that "natural" -2 ox state, and it will do so by oxidizing anything even slightly more willing to donate an electron. Transition metals are a gimme for a violent oxidation/halogenation reaction, and the substance is one of few strong enough to attach fluorine to xenon.


text:
Does the hydrolysis speed increase when more H+ ions are in the solution?
In hydrolysis of polysaccharides, does the hydrolysis speed increase when more $\ce{H+}$ ions are present in the solution?
So if I were to use $\ce{H2SO4}$ instead of $\ce{HCl}$ (same volumes) then $\ce{H2SO4}$ will be the most effective?

The glycosidic bond of a polysaccharide chain is essentially an acetal moiety, and hydrolysis of acetals proceeds under acid-catalyzed conditions. Hence, the addition of acid, within some ambit of concentration, will accelerate the hydrolysis reaction. The optimum proportion of acid, however, is probably something that needs to be determined experimentally on a case-by-case basis. It isn't necessarily true that the rate of the reaction will continuously increase, without bound, as the concentration of catalyst increases.
The mechanism proceeds through a series of proton transfers and SN1 reactions. As far as I'm aware, the rate-determining step is typically the actual cleavage of the glycosidic bond after an initial protonation of the oxygen of the leaving group. Given that, I would expect increases in acid concentration to have marginal effect beyond some threshold value. That's just a surmise on my part, however.
I did find one study that examined the kinetics of starch hydrolysis, which found the optimum pH to be 3 (with values of 2-5 being tested).

so lower pH values than 3 would have a negative effect on the reaction according to that study?

@LievenB, all I can say is that a pH of 3 was better in that study than 2-5. It may be that pH 3 is a globally optimum value under the specific conditions of that experiment, or it may be that it was simply the best out of the set of values tested, and that values lower than 2 might be even better. Personally, the latter case would seem like fairly aberrant behavior to me, but I don't have enough information to rule it out, and I would be cautious in extrapolating too much. It does, however, refute the notion that lowering pH by some arbitrary amount always accelerates the reaction.

There are acid promoted hydrolysis and base promoted hydrolysis depending on the two different mechanisms. More $\ce{H+}$ do increase the reaction rate of the former but will decrease the latter.
Give me your specific reaction so I can figure out what kind of mechanism it undergo.
In the comparison of sulfuric acid vs hydrochloric acid. Sulfuric acid is available in much higher concentration compare to hydrochloric acid (around 12M maximum). It is preferred when you need very concentrate acid. When using at same concentration, sulfuric acid may be little more better than hydrochloric acid in terms of $\ce{H+}$ amount but will not be big difference because the second deprotonation is not significant under highly acidic condition.

it is the catalysed reaction between H2SO4 and Cellulose decomposition into glucose (or other monosaccharides)

Cellulose hydrolysis can be done in both acidic or basic condition (very strong base in non-aqueous solvent). Acidic condition is easier and preferred, and it is acid catalyzed process.


text:
Is the current situation possible for recreaction in a lab?
I want to convert cellulose to another form (crystalline => amorphous). Cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous in water.(Is the pressure achievable in the lab?)
Is this doable in the lab and how should I begin on reproducing that situation (which tools would I need) ?
I'm guessing that a bunsen burner would produce high enough temperatures (but I'm not quite sure about the pressure)
EDIT to my question, following up the advice of bobthechemist...
Crystalline substance to amorphous - Other ways?
You can convert crystalline cellulose to the amorphous, more reactive form by heating it to a temperature of around 320°C under a pressure of 25MPa.
But are there other, less extreme ways, to achieve that amorphous form of the compound?
So because I can't achieve these high temperatures, how can I achieve it in another way?

That kind of pressure can be achieved in commercial high pressure reactors as far as I know. So if you have access to such equipment and know how to handle it safely, this should be possible.

and without being able to use high pressure reactors? would I stand a chance?

A pretty good chance at injuring yourself, I'd say. I wouldn't dare to try such pressures with any improvised equipment, you're very likely to just blow the whole thing up.

ok - so there is no possibility what so ever to do that?

That's about 250 times atmospheric pressure.  A quick back of the envelope calculation suggests that a 160 pound person standing on a block that is 0.5x0.5 cm could generate 25 MPa of pressure.  You'd need to be a very good acrobat to do that over a bunsen burner.

25MPa is just north of 3600psi for those of us that haven't converted to SI units; 320*C ~= 608*F. Your average lecture bottle is rated to about 3000psi, while compressed gas cylinders can go as high as 6000psi. Trouble is, most of them don't stand up well to being heated by an open flame. Temperature is another beast; your Bunsen burner can easily achieve a flame temperature well in excess of 600*F (and a simple variation, the Meker burner, can hit more than twice that), but your Pyrex glassware is only rated to 500*F, and that for short periods. Bunsen burners are often used for glassworking on the lab bench, creating custom tubing rigs by softening straight tubing and then bending or stretching to create bends and bottlenecks. If the glass can be molded over a Bunsen burner, it's not going to be able to sustain the temperatures you're talking about.
The short of it is that both the temperature and pressure are possible, but neither are achievable with a benchtop apparatus in a fume hood; you'll need an HTHP reactor vessel rated for your experimental conditions. They exist, but whether you can get access to one is another question only your superior can answer. If you were planning on visually studying the change in the cellulose, you'll need a sapphire viewport, and these get very expensive very quickly.
Once you have the right tool for the job, it becomes a rather academic task of calculating how much volume of reactants are required to achieve the desired pressure in the vessel when heated to temperature, then filling the vessel and evenly heating it. Only two of the vessels in the linked column hold any appreciable amount of material at the proper temp and pressure, and one of them will just barely meet your needs; if you unevenly heat any of these vessels, by the time the far end reaches the proper temperature the spot being heated will have failed.


text:
Transition state and free energy
We have products $\ce{A + B}$ combining to form $\ce{C + D}$ through the transition state $\ce{X}$.

Are all reactions at least virtually reversible? Is the difference between a reversible reaction and an irreversible reaction simply that for the former, $\Delta G$ is small, where as the irreversible reaction will have a difference in free energy in products vs. reactants, that the reverse reaction simply doesn't occur, but theoretically could?
Why do concentrations affect the free energy? Is it because you sum the free energy for all reactant molecules, as well as for all the product molecules?
Following is from my textbook:

[...] the transition state is too unstable to exist for long. It collapses to either substrate or product, but which of the two accumulates is determined only by the energy difference between the substrate and the product -- the $\Delta G$ of the reaction.

I would appreciate if someone could clarify this for me. Feel free to be very nitpicky, I don't want an understanding that is "approximately" correct or "functionally" good.

What I mean is that if you have all products and reactants in solution, could reactions theoretically proceed in both directions, despite the sign of ∆G? And the energetically unfavourable reaction just happens a lot less, and depending on the magnitude of ∆G, virtually not at all. Since ∆G=0 at equilibrium. And I guess at equilibrium all reactions still proceed in both directions, but at the same rate?

That's exactly right.

According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible.
And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your situation.

Each reaction is microscopically reversible. So either your reactants $\ce{A}$ and $\ce{B}$ or your products $\ce{C}$ and $\ce{D}$ can combine forming the transition state ($\ce{TS}$). As your text book says, the $\ce{TS}$, being an energy maximum, is very unstable and collapses very fast. Since the transition state represents a maximum of the energy curve the collapse may go in either the direction of the reactants or the direction of the products with equal probability, pretty much in the same way a ball placed at the top of hill may roll down on either side of it. So, once you got to the transition state you have a 50:50 situation.
The thing that determines which way your reaction goes is how easy it is to make the transition state or to "get the ball to the top". In order reach the transition state you have to overcome a thermodynamic barrier, which is the difference of the free energies of products and reactants $\Delta G$, and a kinetic barrier, which is the activation energy $E_{\mathrm{a}}$. If $\Delta G$ is a very large and negative then the reaction will proceed in the direction from $\ce{A}+\ce{B}$ to $\ce{C}+\ce{D}$ and will be irreversible because then it will take considerably more effort to "climb" the energy hill coming from the $\ce{C}+\ce{D}$ side than coming from the $\ce{A}+\ce{B}$ side. This means that in a given time a lot more $\ce{A}$ and $\ce{B}$ molecules are "elevated" to the $\ce{TS}$ than $\ce{C}$ and $\ce{D}$ molecules. So, even if $\ce{TS}$ will revert to $\ce{A} + \ce{B}$ 50 % of the time the reverted molecules will quickly be back "on top of the hill" in order to have another try while for $\ce{C}$ and $\ce{D}$ this takes much longer - the reaction rate for the forward reaction is much higher than the reaction rate for the backward reaction. So over time more and more $\ce{A} + \ce{B}$ is converted to $\ce{C}+\ce{D}$ until equilibrium is estabilished and the ratio of reactants/products is equal to the ratio of the reaction rates.
When you have a reversible reaction and $\Delta G$ is very small the "hill to climb to the transition state" is nearly the same height irrespective of whether you come from the product or from the reactant side. So, forward and backward reaction proceed nearly at the same rate.
Concerning your second question: I'm sorry, but I couldn't find a non-mathematical explanation as to why $G$ depends on the concentration. The Gibbs free energy is the thermodynamic potential of an ensemble at constant pressure and temperature. For variable particle numbers $N_{i}$ its fundamental relation is:
\begin{equation}
  dG = -S dT + pdV + \sum_{i} \mu_{i} dN_{i} \ .
\end{equation}
So, it is connected to the particle numbers (and thus to the concentrations) of the species in the system.

I'd add a tiny comment: most reactions are equilibria, so they end up with both forward and backward reaction rates equal. I write most, and not all, because there is one exception: reaction between different solid phases. There, the reaction quotient is 1 (the different solid phases all have activity equal to one), and thus the reaction proceeds in the direction of negative ∆G until all reactants are consumed.

@F'x Interesting. I've never thought of that. If you would add an answer featuring this exceptional situation I'd certainly upvote it.

@Phillip. You state that there is a thermodynamic barrier and a kinetic one, which is the activation energy. There is no barrier in thermodynamic terms as these only tell us about starting and ending states.  Any reactions barriers are that due to potential energy because along the reaction coordinate bond lengths change and so potential energy curves have  minima at different values.

@Philipp How we calculate the Gibbs free energy at the initial and final state in order to make the diagram? Do we calculate it with a specific amount of reactants and products? I mean we calculate first the Gibbs enegy for a container only with reactants (at constant $T$ and $P$) and then we do the same thing for products?

First you are correct, there is no fundamental difference in reactions being described as reversible or irreversible, unlike in thermodynamics. A reaction will be called irreversible (a)if the product is removed from reaction i.e. by precipitation or physical removal and (b) if the rate of returning from product to reactants is so slow that it cannot be measured. This is somewhat subjective, but the lifetime of the back reaction can be years, so in practice its clear cut.
Secondly, the equilibrium constant for a reaction $\ce{aA + bB <=>cC + dD} $ is defined as $$K_p=\frac{P_C^cP_D^d}{P_A^aP_B^b}$$ where $P_A^b$ etc. represents the pressure at equilibrium. As the equilibrium constant must be dimensionless so pressure is understood to be divided by unit pressure of 1 bar. If the reaction is in solution then concentrations are used instead of pressure, then these are understood to be divided by $\ce{1 mol~dm^{-3}}$ (Technically instead of pressure or concentrations activities are used but for most uses pressure and concentration are sufficient). The standard free energy is defined as$$ \Delta G^0 = -RTln(K_p)$$ and this is how the pressures/concentrations come into the equilibrium constant.
The transition state is reached only when the reactants have sufficient energy to do so. By the Boltzmann distribution the number of energetic molecules rapidly diminishes as the energy is increased. Thus most collisions do not result in reaction, unless the activation barrier is very small compared to average (thermal ) energy. The reverse reaction has a larger barrier to surmount if the reaction is exothermic, thus fewer collisions result in reaction back to products than the other way round.
The transition state is at the top of the barrier between reactants and products, in solution it may be crossed and recrossed many times before reaction is complete, it all depends upon how quickly the energy the 'molecule' has at the transition state can be removed by collisions or by energy being redistributed internally into the products as vibrations and rotations or as kinetic energy of separation.
The activation energy barrier exists because the product will have different arrangement of bonds and atoms from the reactant and so different average internuclear separation in the reaction coordinate. The  activation energy can be imagined as the point where two potential energy parabolas cross, one being slightly displaced in position and energy vs. the other. This is what the normal reaction profile drawn in text books and elsewhere represents. It is usually shown as one line(the bottom of the first parabola), a curved line to a maximum (crossing point, transition state) then onto another lower line to represent product energy at the minima of the second parabola.

First question:
The answer is true from a reaction view. You are right. No matter how the reaction is favored in terms of Gibbs free energy change. The reverse process will happen in small or tiny amount.
From a thermodynamics view, they are different. Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but irreversible reaction has energy loss.
Second question:
Gibbs free energy change is a function depending on concentration for solution and partial pressure for gas. So concentration will obviously affect it.
Do not confuse it with the standard Gibbs free energy change, which assume that you have a standard condition (1 Pa pressure or 1 mol/L concentration). Since the concentration or partial pressure is a constant (precondition), the standard Gibbs free energy change is independent to concentration or partial pressure.
Both of your question is not related to transition state. Transition state is part of the chemical kinetics, not thermodynamics.

I don't understand what you mean by “Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but irreversible reaction has energy loss” — or rather: I think I understand it, and I disagree

If you understand reversible process in physics and treat reaction as a process. A restrict defined reversible reaction is an ideal process that goes so slow that at the reaction is kept very close to equilibrium until finish. In that way, all energy (free energy change) is transformed to work or other kind of energy without loss (heat). That is how the textbook "Chemistry A molecular approach" (Nivaldo J. Tro) defined "reversible" and "irreversible" reaction. On page 798 in second edition.

@ Ian Fang There is no fundamental difference in reactions being described as reversible or irreversible, unlike in thermodynamics. A reaction will be called irreversible (a)if  the product is removed from reaction i.e. by precipitation or physical removal and (b) if the rate of returning from product to reactants is so slow that it cannot be measured. This is somewhat subjective, but the lifetime of the back reaction can be years, so in practice its clear cut.


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Predict photochemical decomposition of organic compound?
Is it possible to predict wether an organic compound will photochemically decompose?
Also, is it possible to predict the products which it will decompose into?

Just as an experienced synthetic chemist can make an educated guess about how compound A and compound B will react, an experienced photochemist can usually do the same.  There are many, many well-known photochemical reactions that are mechanistically understood and sometimes even used in synthetic procedures, some commercially.  A few such reactions are the
Norrish Type I reaction

Norrish Type II reaction

Paterno-Buchi reaction

di-pi-methane reaction

photochemical induced halogenation, olefin dimerization and so many more.  So yes, an experienced photochemist can make reasonable predictions about which chemicals will react and what the possible products might be.


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Transesterification of propylene carbonate and methanol
I'm going to try out the transesterification of propylene carbonate and methanol. I know that the mechanism is as follows :

-What are the products yielded after the reaction?
-And which catalyst should I use? (reaction is happening at normal air pressure and at room temperature)

What you have listed is not a mechanism. Also, those structures are not propylene carbonate nor methanol.  Maybe you should look at the actual structures and try to figure out yourself.

the mechanism is an example mechanism (See the wikipedia page about transeserification)

I know it is used to obtain  dimethyl carbonate (DMC) I think the others products should vary from which catalyst you choose. Have a look to Transesterification of propylene carbonate with methanol using $\ce{MgAlCO3}$
hydrotalcite as solid base catalyst: Indian Journal of Chemistry
Vol. 49A, September 2010, pp. 1182-1188 to choose the catalyst. I think there are two little typos in Scheme 2 where the mechanism is shown: miss $\ce{^-}$ from $\ce{CH_3O^-}$ (first step) and miss $\ce{C}$ from one part ($\ce{CH_3CO}$) of DMC formula (second step) but is a good article.


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Organic compounds that react with carbonate ions
I want to remove carbonate ions from dissolved potassium carbonate (potash) by letting an organic compound react with the carbonate ions. I don't have a very good knowledge of organic chemistry, but are there certain types of organic molecules that can react with these carbonate ions to form new compounds?
Additionally, are there reactions in which the carbon atom from the carbonate ion can be inserted into the main carbon chain of another organic compound (making, for example, propane from ethane)?

I just rediscovered this question, are you still looking for answers or was the one provided by Al enough? Or do you want to bind carbonates in this form?

Cook it up with ethylenediamine or diethylenetriamine- but don't bet on it.  What do you plan to have as the counterion for $\ce{K^+}$?  Check your carbon oxidation states for the second half of your question.
No, wait!  Add citric acid.  Yeah, that's the ticket.  Plop, plop, fizz, fizz.


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Free automatic organic synthesis Linux command line program
I'm trying to find a free automatic organic synthesis Linux command line program which take mol or smile files as input. I have random molecules and I'd like to know how difficult above a baseline they will be to synthesize. An implementation of synthetic accessibility and not a full step wise synthesis program would also be acceptable. Thanks for the help.

I would be very surprised if such a program exists for anything but a small subset of similarly structured molecules.  Additionally, refer to the help section with respect to questions that are purely of a simulation nature.

A general solution to this problem, by the way, would be literally worth zillions of dollars.

And probably a Nobel Prize (with its additional 8 million Swedish Kronor).

@bobthechemist awh well then you can successfully be surprised as I found a solution.

The SAscore is now implemented in RDkit and available for download.

If you're wondering what "SAscore" is, see the following open-access article: "Estimation of synthetic accessibility score of drug-like molecules based on molecular complexity and fragment contributions" http://www.jcheminf.com/content/1/1/8


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Absolute limits on solubility of immiscible solutes?
I know that for a particular, immiscible solute, there is a limit on how much of it can be dissolved in a solvent at a given temperature. However, I was wondering if there is only so much "total solute" that a given solvent can dissolve. As a simple example, suppose you saturate water with salt, then start dissolving sugar into it. Does the fact that it is saturated with salt limit how much sugar it can dissolve? My mental image is one of each water molecule being able to suport only so many solvation structures, regardless of the chemicals participating in the structure.

If you speak about solubility, you are out of domain of simple approximations and the only correct answer is the experiment. For the example given, see http://dx.doi.org/10.1111/j.1365-2621.1992.tb01192.x
The problem is the difficulty to quantify the effect of temperature, which is very important here.

Wow, thanks for the uber-specific, relevant link :) So I guess the answer is YES...the solutes in multi-solute solutions do not behave linearly or independently with reagrd to their phase partitioning. Thanks!!


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Is it customary to express heat loss as a negative number?
In my chemistry class, we had a problem akin to the following:

If an object with mass bla and specific heat bla is placed into a bla g calorimeter with an initial temperature of bla and a specific heat of bla, and the temperature of the calorimeter rises to bla at the end of the experiment, how much heat was lost by the object?

I performed the calculations and got an answer of, let's say, $8 J$. When I submitted this answer, I was told that it was wrong, and the correct answer would have been $-8 J$. Wouldn't this response to the above question mean that the object gained heat?

Sign conventions have always been a bit of an issue in chemistry, especially in introductory courses. I agree with your interpretation in this case, but what might have been meant by the question was to state the heat change of the object, in which case -8J follows standard convention.

It is better to say it as a convention, than saying it as a customary. It is convention that, one hand we consider as right, and the other as left. One may assume the vice versa case. It would not be wrong.

I guess what I'm asking is whether I'm right or wrong.

Note that physicists and chemists use different sign conventions for work, and I've seen some variation in the convention for heat, too.

In cases such as these, I think it can be useful to explicitly write out that "$8 J$ of heat is released", or "considering the convention that negative energies correspond to heat release, the energy involved in the process is $-8 J$" to cut through any ambiguity.

Your assumptions are correct, and from the wording of the question, the "correct" answer doesn't make sense, unless like you said, the object gained heat, which is possible; people have negative profits all the time. Sign conventions are not flawed in chemistry, but you just have to be careful when wording a question. To answer the title of your question, yes it is not only customary, it is necessary to express heat loss as a negative number, but the question still has to be phrased correctly--using the term "lost" serves as a negation in this use of language, the same way the unary operator in -1 serves as negation for a number. If heat loss were not expressed negatively, it wouldn't make sense because the system of an exothermic reaction must lose energy for its surroundings to gain it.


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How to interpret a MALDI mass spectrum?
I have some questions regarding how to read a MALDI mass spectrum in a paper. I am a biologist by training, so I would like some analytical chemists to help me out.
The researchers ran a western blot, and immunoblotted for a protein of interest. They cut out this band from the western blot, subjected the protein to trypsin digest, and used these peptide fragments for MALDI-TOF analysis. I know for a fact there will be other proteins at the same apparent molecular mass which would have also been included in the analysis, which should be taken as contaminants in the resulting spectrum.
They produced a graph with the absorbance intensity x 1,000 on the dependent axis, and mass:charge (m/z) ratio on the independent axis.
The authors point to two peaks, which they say correspond to two peptide fragments from the tryptic digest of their protein of interest. They provide the supposed peptide fragment sequences and m/z ratio data for these two peaks, and I checked to make sure they are the same as reported. The authors expect to see two peaks: protein of interest is digested to two fragments, A + B.
I have two questions:

Peak 'A' and peak 'B' are of different intensities. Peak 'A' is 20-fold greater than peak 'B'. However, the tryptic digest must yield equimolar amounts of peptide fragments A and B. Is it possible that these intensity peaks can be 20-fold different? Wouldn't the absorbance intensity correlate to total analyzed peptide mass?
Is it normal to multiple absorbance intensity by a (seemingly arbitrary) factor, in this case, 1000? Since the absorbance intensity of peak 'B' is < 0.5 (after scaling up), could this represent noise rather than signal in the analysis?

Could you provide a link or a reference to the article in question? I am a bit puzzled: usually identification of a protein based on only two peptides should be considered very carefully in a peptide mass-fingerprint (PMF) analysis, and is probably not valid if many other proteins are present (as peptides with similar masses could originate from other proteins). Did the authors perform any MS/MS experiment to confirm the identification step?

Please see the wiki articles for the general method:

http://en.wikipedia.org/wiki/Mass_spectrum
http://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization
http://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry

Then let's start with question 2: yes, you can and you should. The spectrometer measures in arbitrary units, therefore you can speak only about relative abundances.
And question 1: it depends how easily is the given molecule ionized. So if the fragment A is rich in polar aminoacids, whereas fragment B is mostly hydrophobic and difficult to ionize, than you should expect this big difference in signal intensities.

Thank you. So if we assume that A is twice as massive as B, and are equally ionizable, then there should be a discrepancy in their relative signal intensities. Shouldn't there be some form of internal control, or is that only if one wishes to quantify absolute peptide masses from the spectrum?

If the fragments are equally ionziable, then the intensity should be similar. Of course, the m/z will be twice as large for the bigger one. Maybe I do not fully understand your problem. It's m/z which is the important characteristic, whereas intensity tells you little bit about ionizability, ion stability and quite indirectly about abundance.

Actually you explained the answers to my questions well. I do see the m/z are different by a factor of two. I would assume that ionizability and ion stability would be the same for small peptides, and so I should expect similar band intensities. I think that is enough to be critical of the results I have found in paper.

@leonardo: There are no real reasons for the ionization efficiencies to be similar for all small peptides. Physico-chemical properties of small peptides can cover a wide range, and you cannot assume that "small peptides" is enough to allow for similar ionization efficiencies.

In addition, I would like to point out that the intensities in the spectrum are not only related to ionization efficiencies: the entire workflow has to be analyzed, from the initial peptide extraction from the digested band to the mass analysis. For instance, peptide extraction from a gel can lead to strongly different peptide abundances due to pore sizes in the gel if the peptides have very different sizes. (BTW, did they perform the digest directly on the membrane blotted protein or in gel?) A TOF analyzer can also introduce strong variations in intensity when the m/z values are far apart...

I know this is an extremely late response, but perhaps this will help another person who is looking at this.
for your first question, the degree of ionization in Maldi depends mostly on the acid-base equilibrium between the matrix substance and the analyte, that is, it's Ka value. And as the above answer has already mentioned, a good way to predict this would be to see how many polar amino acids residues there are. So it is entirely possible that even if the theoretical ratio of two peptides are 1:1, you might obtain a different ratio in the spectrum.
Someone in the comments section has asked whether the authors of the paper in question has done any MS/MS experiments. I happen to have the same question. So far, I have heard of anyone doing peptide sequencing using just MS:it is either done by Edman degradation or by MS/MS experiments (through de novo peptide sequencing). Since you mentioned that they did do sequencing, I'm starting to wonder whether you actually meant a MALDI-TOF/TOF. In that case, fragmentation of individual peptides chosen for the second TOF experiment would depend on the collision induced dissociation parameters. Here is a website detailing the fragment types: https://en.m.wikipedia.org/wiki/De_novo_peptide_sequencing


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What is the number of unpaired electrons in cobalt(II) tetrathiocyanate?

Find the number of unpaired electrons present in the d orbital (whose lobes are present along the axis) for the complex $\ce{[Co(SCN)4]^{2-}}$.

Since $\ce{{SCN}^{-}}$ is a weak ligand I did not pair the electrons, but still I got the wrong answer. The correct answer is 0. But how?

"whose lobes are present along the axis" seems very ambiguous to me. Could you clarify what it means?

@ssavec i applied crystal field theory ...... i filled the d orbitals according to hunds rule..... since Cobalt is in +2 oxidation state hence $Co=[Ar]4S^2 3d^7$ hence $Co^{2+}=[Ar]3d^7$ Now i filled up the orbitals.

@chipbuster i think this refers to the $e_g$ set of orbitals because $e_g$ set contains $d_{x^2-y^2}$ and $d_{z^2}$ whose lobes are directed along the axis as the lobes of $d_{xy}$ $d_{yz}$ and $d_{xz}$ are present between the axis.

@Hardik which geometry you assumed, having $\ce{M L_4}$ complex?

@ssavec i got my mistake...... i filled up the orbitals according to octahedral splitting...... instead i had to do it with tetrahedral. Thankyou.

According to the tetrahedral splitting in crystal field theory, the filling of the orbitals is just opposite to the filling in case of octahedral splitting. Thus filling up the elecrons according to it and obeying the hunds rule (since it is a weak ligand) the number of unpaired electrons are 0 in $d_{x^2−y^2}$ and $d_{z^2}$ whose lobes are directed along the axis.


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Interpretation of the following mass spectrum

I'm given the above spectrum. $M^+ = 102$, probably $\ce{C_6H_14O}$ or $\ce{C_5H_10O_2}$.
I'm thinking of ethyl propionate. (that picture also shows the fragments I expect). I can account for the following peaks:
m/z = 73: Loss of ethyl from acyl side (expected lowish intensity)
m/z = 57: Loss of alkoxy to form $\ce{CH3CH2CO+}$ (moderate intensity)
m/z = 45: Alkoxy ion (weak)
m/z = 29: ethyl carbocation, moderate intensity.
I can't account for 74, since according to my book the alkoxy side would not proceed through normal Mclafferty, but the Mclafferty + 1 rearrangement... That should yield m/z=75.
I proposed methyl butyrate, and of course that's not right. My instructor wrote to pay attention to the m/z = 57 (M-45) and m/z = 85 (M-17). This fits well with carboxylic acids, but again cannot account for the m/z = 74. McLafferty of carboxylic acids would result in a fragment of m/z=60...
Can someone help me out, I've come to a dead end for hours now.

I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.
But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!
The spectre is of: 2-methyl-butanoic acid
M-17 and M-45:
Alpha cleavage of carbonyl => loss of hydroxyl radical or carboxyl radical

m/z = 74:
Loss of ethene by McLafferty rearrangement

m/z = 45:
$\ce{COOH^+}$ fragment

M-15 and M-29:
Loss of methyl radical or ethyl radical

m/z = 29:
Ethyl carbocation

For m/z 74 it could be methyl ester:  in the loss of a $\ce{CH2=CH2}$ by McLafferty rearrangement.
$\ce{CH2=C(OH)OCH3+}$

This doesn't seem likely to me, especially considering the intensity. Do you suggest a mechanism :-)?

May be I'm wrong, but I've thought in the lost of a CH2=CH2 by a McLafferty rearrangement.

Okay I just read it again. It is the ethyl ester that fits with many of the peaks, but I was troubled because I thought the book said it would only undergo McLafferty+1 rearrangement (with double proton transfer, M/Z=75), but apparently that's just an additional mode of fragmentation to the normal McLafferty arrangement. This gives [CH3CH2C(OH)O]+ m/z=74. I would like to explain the m/z=85 though, now that my instructor pointed my attention to it. hmm

Possibilty: Loss of a hydroxyl group (m = 17).


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Most economical method to convert potassium oxide to potassium nitride.
I have a compound potassium oxide($\ce{K2O}$) and I am trying to convert it into potassium nitride ($\ce{K3N}$).
Here are a few possible reaction methods :

Split potassium oxide into the constituent elements potassium and oxygen  (Not very good  as a lot of energy is required)
$\ce{K2O + N2 -> K3N + O2}$ (however this won't happen(?) because nitrogen has a lower electronegativity value than oxygen does)

and so on.
Does anyone know how can I convert the oxide to the nitride, releasing oxygen in the process, by not having to use a lot of energy-requiring processes?

The second reaction can happen- Thermodynamics are altered by concentration differential. Lots of Nitrogen and little K2O. It will need catalysing to break the nitrogen to atoms which is a kinetics problem. Potassium Nitride is not all that stable. Just look at Sodium Nitride in wikipedia.

@user2617804 How would I be best to begin on doing more research for the process that you mentioned above?

@user2617804 - Potassium nitride $\ce{K3N} \ne$ potassium azide $\ce{KN3}$

Do you really want to go there?
Based on this entry from Wikipedia the compound is synthesized and stable only below room temperature, and on top of this requires elemental potassium:

This compound was produced by the reaction of potassium metal and liquid nitrogen at 77 K (−196.2 °C; −321.1 °F) under vacuum:[1]
$\ce{6K + N2 → 2K3N}$
This compound decomposes back into potassium and nitrogen at room temperature.

You may want to check that the target is actually $\ce{K3N}$ versus the more plausible $\ce{KN3}$.
Cited Reference

Fischer, D.; Cancarevic, Z.; Schön, J. C.; Jansen, M. (January 2004). "Zur Synthese und Struktur von K3N". Zeitschrift für anorganische und allgemeine Chemie (in German). 630 (1): 156–160. https://doi.org/10.1002/zaac.200300280


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Evaporating the water of a solution of only carbonate ions and water gives?
I have a solution with only $\ce{CO_3^{2-}}$ ions and water.
What would happen with the carbonate ions if I were to evaporate the water of that solution (in a vacuum)?

I'm pretty sure that you also have some counter ion in there

but what if the ion was selectively let through by a semi penetrable membrane into this solution?

You can't selectively remove one type of charge and sustain the difference indefinitely. Because the electrostatic  force constant is so huge, without work being done in the form of an opposing electric field, significant net charges would quickly be neutralized by tearing out the opposite charge from whatever they can, far before getting to macroscopically measurable amount of excess ions. Leaving the system in a vacuum is not enough as the charge would jump through empty space.

so would it be possible to do or don't the laws of physics/chemistry allow it to happen?

There is nothing that says solutions must be exactly electrically neutral, just that they can't have too much excess charge. It's a problem of quantifying the limit. After some research, it seems that you theoretically could prepare a solution with excess $\ce{CO^{2-}_3}$, though the excess would have to be quite small (apparently well under $1\ \mu M$). I'm not exactly sure what would happen if you removed the water entirely in a vacuum. Maybe the ions would become gaseous and float around until they hit a surface.

Here are a few sources: 1 2 3

I would love to add some quantum chemical bone to the stew we are trying to cook here.
Let us assume that it is possible to have a solution of carbonate ions, like already discussed by Nicolau in the comments. Then Rauru is also right that you will have to consider several equilibria:
\begin{aligned}\ce{
(1) && 2H2O &<=> H3+O + {}^{-}OH \\
(2) && CO3^{2-} + H2O &<=> HCO3- + {}^{-}OH\\
(3) && HCO3- + H2O &<=> H2CO3 + {}^{-}OH
}\end{aligned}
Now $(3)$ seems a little far fetched, so we will ignore it.
When you do now remove water from the equilibrium through (gentle) evaporation, you will have a mixture of only $\ce{CO3^{2-} + HCO3- + {}^{-}OH}$. Further assuming this is possible you will always have some collisions within this ensemble. Statistical it is therefore always possible to also remove $\ce{HCO3- + {}^{-}OH}$ via $(2)$ from the system. Also single entities of carbonate ions will repel each other as states by G M. So basically it all boils down to on question: Is carbonate stable in vacuum? What happens to carbonate if you put it into vacuum?
Now we have some quite good approximations to these particular conditions (gas phase and 0K), but I do not want to go into detail. I did a calculation on the carbonate with Gaussian09 and as a cheap shot BP86/cc-pVTZ level of theory. The only thing I want to show is the highest occupied molecular orbital.

What can clearly see is that this is no bonding orbital. There are lone pair contributions from all Oxygen atoms (due to imposed symmetry constraints). In addition to the high negative charge, electrons will not be stabilized by the field of the nuclei. Also, Orbitals only vanish at infinity, giving electrons a non-zero probability to actually leave the compound. Furthermore, there are two electrons in this orbital, so the spins of these are paired, which also costs energy.
My suspicion therefore is that the carbonate at least loses one electron. To where? To anywhere. (Also mentioned by Nicolau in the comments.) I do not have any actual proof for this, as it quite hard to find people actually doing experiments and/ or calculations on that topic. I remember that one question of a Professor of mine in the first semester was: "What happens to sulphate if you shoot it into space?" The answer was given without any further explanation and I never bothered to ask: "It looses an electron."

Note:
Another point that should always be considered when dealing with gas phase reactions is, that when ions have enough kinetic energy they might react in the strangest ways. One of them might be
$$\ce{2CO3^{2-} -> 2CO2- + O2^{2-}}.$$
Also collision with any particle may cause electrons to be transferred.

Theoretically you should have an explosion. Because every $\ce{CO_3^{2-}}$ ions exerts a force against the other. You should be able to calculate it with coulomb equation:
$$|F|=K_e\frac{q_{\ce{CO_3^{2-}}}^2}{r^2}$$
Of course you have to consider the force exerts from every particle to each others, summing every force field assuming a certain          distance: a very hard calculation.
By the way your assumption are quite hard to reach in great scale as say Raur Ferro. But a practical application of electrostatic forces that could be enlightening is Electrospray ionization (ESI), a technique used in mass spectrometry to produce ions.
 In this technique the solvent from an areosols that hold the analyte is forced to evaporate (with various expedients) from a charged droplet until the drop becomes unstable due the charges. At a certain point the droplet undergoes to what is called Coulomb fission, so the original droplet explodes creating many smaller droplets. I know this is not the same thing but this can show you how electrostatic forces can act. I think however that this question is more related to physic than chemistry.

As a theoretical experiment, you must think on different aquieus equilibria:
$\ce{H2O <=> H+ + OH-}$
$\ce{CO3^{2-} <=> HCO3- <=> H2CO3}$
$\ce{H2CO3}$ will be formed with water's $\ce{H+}$, and in vacuum, possibly will decompose to water and $\ce{CO2}$.
At the end, may be a mixture of $\ce{OH-}$ and $\ce{CO3^{2-}}$ will remain.


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Split carbonate ions?
Can you split carbonate ions into their constituent elements directly in the carbonate form? (so not first producing $\ce{CO2}$ and then decomposing the carbon dioxide)

(edit:)
I know that when running electricity through an aqueous solution containing carbonate ions and other element ions, the carbonate ions will act as individual groups and will not split (only after chemical reactions (for example the thermolysis of metal carbonates will produce carbon dioxide)) So this was just out of interest - how can the carbonate ion itself be split into its constituent elements if it can't be split with electricity (or at high voltage, could it then be split?)?

This seems to be a homework question.
Can you please add what you have tried yet to solve your problem?

I'm not sure if it's homework, but questions of the type “why is X stable?” are not very suitable for our site… more interesting are questions of the type “I would have thought X decomposes into Y and Z, with this equation, because of S and T. Why doesn't in happen in these conditions?”

@F'x It isn't a homework question, it was just out of interest. I've edited my question, (done some more effort)

I tend to agree with @F'x.  The present question boils down to "once $CO_2$ is produced in the decomposition of carbonate, how can the carbon dioxide be further decomposed into its constituent elements."

I understand what you mean - but can the CO32- ion immediately (without producing CO2) be split into its constituent elements?

Plasma torch will disintegrate in just fine, only it's not a typical chemistry modus operandi.


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Explaining redox reactions to a 6 year old?
In a similar vein and for pretty much the same reason as I post this question on the Physics.SE site, what would be a an effective means to promote understanding about redox reactions to a 6 year old?  In particular, what (if any) simulations are available?

Why specifically redox reactions? Seems quite a challenge for a six year old to understand even what a chemical reaction is in general.

Long story behind this, but after seeing a 'redox front' in a road cross section, this child got interested in what caused it.

I might suggest you take a look at What's Chemistry All About by Frith, Gillespie and Larkum.  (ISBN: 1409547078)  It is a very approachable book that in my mind pays close attention to the science jargon introduced to its audience.  Redox reactions are described in the section "Let's make things happen" under the types of reactions.  A bit of history about where the name oxidation comes from and the safety symbol that you will see on an oxidizing agent are both included in the brief discussion.
The example reactions given are combustion and firecrackers, which I think are appropriate and accessible to a child.
As for experiments to do, I'll suggest another book, the Illustrated Guide to Home Chemistry Experiments by Thompson. (ISBN: 0596514921) There is a chapter on electrochemistry with two experiments that might be of interest to a young learner: the electrolysis of water to make oxygen and hydrogen gas, and the oxidation of iron.  Of course, there's also the "make your battery" type experiments which are always fun.
As for simulations, I would suggest heading on over to the PHet website and have a stroll through all the activities there.  I don't recall off hand what they have for electrochemistry, but there's plenty to keep a budding scientist interested.

I was taught using the OIL RIG acronym, where oxidation is loss, and reduction is gain. I would also utilise a picture of some sort showing how a oxidant is reduced, and a reducer is oxidised. I would also mention that most oxidising agents have oxygen in them, whereas reducing agents can have hydrogen. Good luck.

Let's speak the Redox with an exempla : Corrosion off a piece of Iron.And Oxygene around you and me. As child of 6 years old had probably seen this, it will directly rely to everyday 's live.
Say this to him : You cannot see the "messy movements" in this Iron because your eyes are too big.......and because you can see the moon and stars.......but not all stars in the universe because we are too small.
The idea off beeing somewhere in the middle off "The Scale" is an important step to start imaging the all around and inside off you. Back the Redox.
The iron, as all the rest is made of 3 sweeties, called Neutron...not an easy name but this is exiting for a child the knonw some specials words. As 3 new words is fine at C yers old,continue with Electron, an other one to love and Proton. Those 3 sweeties are bricks to create all you can see, fill and  smell.
Redox is when sweeties are given and received.I can not receive if nothing is given. Both are related and if one does not happens, the other one too.
A child should ask how it is possible to have only 3 sweeties too have all around with différences. More or less you have always those
3 sweeties together exept when speaking e.g. nuclear explosion. Just catch his meaning with powerfull stuffs. Back on redox,now to have Iron you need those 3 sweeties with 25 other teams of those 3 members: E-, P+ and N°. This make 26 teams to make Iron, the very kind Iron can give some Electrons too Oxygene we need to be alive. And the volume off Iron increase as the oxygene who received some electrons must stay around the iron or leave. As they are now togheter, they will stay togheter as long as you do not found and other team off sweeties able to give back electrons too Iron or to stoll the electron owned by oxygene. Use off this ugly stuff called chlorine can make it true.
Next : Explain the exeption off nuclear phenomena ?
Regards.