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Find the number of water molecules "embedded" in sucrose
Recently I got a question as follows:

How many molecules of water are joined with 12 atoms of carbon in a molecule of sucrose ?
(A) 12
(B) 11
(C) 6
(D) 10

I am unable to understand the question.
What actually is the question ?

Sucrose is a carbohydrate.  This means that stoichiometrically, it can be regarded as adding water ($\ce{H2O}$) to elemental carbon $\ce{C}$.
The formula of sucrose is $\ce{C12H22O11}$.
Another way to write this formula is $\ce{C12(H2O)11}$.
Thus, sucrose can be regarded (stoichiometrically, but not structurally) as eleven waters plus twelve elemental carbons.  Therefore the correct answer is (b).
The question is not well-worded, and would be better stated as "How many water molecules are required to join with the 12 carbon atoms in sucrose to give the correct empirical formula for sucrose?".
In the comments Ivan Neretin offers an even better variant, which I will re-word as "How many water molecules could extremely powerful dehydrating agents such as $\ce{H2SO4}$ or $\ce{SO3}$ extract from sucrose, if we assume the only non-water product is pure carbon?".

That would be not quite good either. Water molecules don't just up and join carbon atoms on their own. I don't know how to salvage this question. Maybe, "How many water molecules can be snatched from sucrose by $\ce{H2SO4}$, if the product is assumed to be pure carbon."

Thanks & good point.   I added your idea to the answer.  Perhaps a better word than "join" is "embed", which may be suitably ambiguous for this somewhat vague question.

@Curt F I was quite thinking about how many water molecules can be joined with sucrose molecule with the help of hydrogen bond. But because there was no mention of Hydrogen bond  so I didn't knew what actually it means.

It really is a misleading 'trick' question. I just does not help to understand the structure of sucrose, in fact quite the opposite.

It's not really a trick question.  It requires stoichiometric thinking (all they had in the old days) instead of structural thinking, though.


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Replacing gas in a chamber
Is there a way to replace a gas (air) for example with another gas (say, helium) in a gas chamber? It does not need to be a complete replacement, but there would need to be a significant amount of the new gas. Thanks!

Pump out air, then let in another gas.

Assuming there is an inlet and an outlet, then you could "purge" the chamber which is just blowing the second gas through the chamber for some time. The downside of this technique is that you use many times the volume of the chamber of the second gas. If the chamber is small that really isn't much of a problem.

@MaxW  He didn't say he needs complete replacement, so 2 or 3 volumes would probably be adequate, particularly if the chamber is not mixed.

Chemists who need to control which gases are in their vessels (e.g. those who work with air sensitive compounds) do this routinely.
There are two ways to achieve gas replacement. The more reliable way is to work with vacuum-lines that allow switching the input to a vessel between a vacuum and an inert gas. First the air is removed and then the inert gas introduced.
Often, though, this is overkill. In many cases where liquids are involved the liquid needs to be degassed (to remove air dissolved in the liquid). In many cases it is good enough to bubble the inert gas through the liquid and the vessel for long enough to sweep out the unwanted gases. Bubbling the inert gas through the liquid and allowing it to escape the vessel is good enough if done for long enough.
If it isn't important to remove all traces of the original gas, this is usually an effective technique and gets more effective the longer the inert gas is allowed to flow.

First you would need to know if the gas is denser than air or not. You would make two holes in the container, one in the top, the other in the bottom. For helium start pumping it in through the top. The air will be pushed out the bottom until the container contains only helium. The denser air will flow through the bottom and the helium remain at the top (to an extent). The container would then need to be resealed to ensure the gas stays. For a dense gas like C0$_2$, you would do the inverse, pumping through the bottom instead.
As a side note, it would take less gas to purge the container if you pumped it slowly, letting the gas settle in layers. Pumping it too fast would cause a turbulent flow, pushing out both gasses at once instead of just the one you wanted to get rid of.

This makes perfect sense to me. Does whoever down-voted want to explain why this is a bad idea?

The answer given by @matt_black is technically sound and is how chemists actually tackle this problem, but I wanted to add a clever answer (which I read somewhere and can't remember the source for unfortunately) that I saw which works for gloveboxes and other compartments where you can still manipulate things in the compartment. The idea is that you can fill balloons with your replacement gas, put them in the chamber, then purge only the space between balloons with the new gas. By decreasing the volume that must be purged, you use much less gas. Finally, pop the balloons.
I've wanted to do this for a while when rebuilding a glovebox, but haven't wanted to clean up all the popped balloons.

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Electron orbital diagram of vanadium
I am learning about electron configurations and I came across the "exceptions" of $\ce{Cr}$ and $\ce{Cu}$ where electrons actually half fill the $\mathrm d$ orbitals instead of the expected $\mathrm s$ orbitals. I thought maybe vanadium would do the same, but my textbook gives the answer of an electron configuration exactly following the Aufbau principle.
I've attached a picture of my drawing for the textbook answer and what I had thought might happen — could someone explain why vanadium is not an "exception" like chromium or copper?

Electrons always fill in the lowest energy configuration possible.
Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order.  In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion.  In their case, the energy to promote an s electron to the d orbitals is compensated for by the effects of exchange energy and no repulsion.  This effect is called "the stability of half filled subshells", or something to that effect, in most textbooks.
The energy cost to promote 2 electrons from the s subshell, as would be the case for Vanadium, is too much to be compensated for by the exchange energy gained and the repulsion lost, so the lower energy configuration is to have two electrons in the s subshell and 3 in the d.
Here's a nice website that shows the electron configuration for all elements: with http://www.ptable.com/#Orbital

This answer is not quite right because in vanadium (and indeed all of the first row of transition metals), the $\ce{4s}$ subshell is higher in energy than the $\ce{3d}$ subshell.

Its very simple. You cant skip orpitals. Since 4s is smaller than 3d you will have to fill 4s first and than 3d. But u cant excite both the electrons from 4s cause it would take a very high amount of energy and even if u did the 4s becomes vacant half filling the 3d orbital and thats would be highly unstable. So the only possible stable configuration would be 4s2 3d3.
And in Cr and Cu the excites 1 electrons because by doing that both 4s and 3d orbital gains half filled i.e the stable configuration. So in vanadium if u excite 1 electron the d orbital will still not be half filled. But 4s orbital will be half filled. Since fully filled orbital are more stable than half filled the vanadium will have 2 electrons in 4s orbital.

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Do bare protons exist, even transiently, in aqueous solution?
Generally $\ce{H^+}$ ions in water remain in the form of hydronium ions ($\ce{H3O+}$), and not as "free" aqueous protons, as far as I've been taught.
My question: is this always the case, even when considering dynamics at extremely short time scales? By this I am not meaning interconversion with the Zundel ($\ce{H5O2+}$) or other states. I mean to say, does bare $\ce{H+}(\mathrm{aq})$ ever exist  on, say, a femtosecond time scale?

I think this is one of those things where there probably is a single proton floating around somewhere in a solution, but the formation of this is so unlikely. It means that you need to have just a hydrogen atom somewhere in solution and some other molecule which will take that electron and form a radical, which is certainly not stabilizing. Then, the proton needs to somehow not react with anything for a femtosecond. My bet is that this just doesn't happen...

This is probably more of a philosophical question - when is a H bonded to something,  and when is it not?  That's quite tricky to answer,  particularly in the liquid state...

Relevant: https://chemistry.stackexchange.com/questions/32558/why-are-hydrogen-ions-always-associated-with-another-molecule/32559#32559

According to Myths about the proton. The nature of H+ in condensed media Accounts of Chemical Research, vol. 46, pp 2567–2575:

But to put this into
  perspective, an electron-free proton has an estimated acidity $10^{56}$(!) times greater than 100%
  H2SO4, the threshold of superacidity.$^2$ Adding a bare proton to a molecule is a real event in the
  gas phase but only a solvated hydrogen ion can be transferred in condensed phases.

No.
The proton is always within some electron density, even when transitioning between the hydronium/water and whatever anion it was bonded with.

To expand on this answer:  Say a proton moves from one water molecule to another.  It does not jump off and become a bare proton, even for a moment.  Rather the second water molecule begins to form a bond with the proton as the first water molecule begins to loosen, leading to a transition state where the proton is partially bonded to both water molecules.


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What is the correct order of subshell energies?
I was taught in high school that the order is given by the Aufbau Principle:
1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f>5d>6p...
Why is the 3d subshell higher in energy than the 4s subshell if it is in a lower shell? Why not call it the 4d shell?
My current understanding is that the shell number indicates distance from the nucleus, while energy can be affected by other factors, such as subshell type.
Using the element bromine (e- config [Ar]4s2 3d10 4p5) as an example, there are three reasons why I think this is the case:

The 4s subshell would fill with electrons before the 3d subshell since it is lower in energy for some reason.
The 4p subshell would fill with electrons after the 3d subshell since it is higher in energy for some reason.
The valence subshells for Br are the 4s and 4p despite 3d being higher in energy than 4s because 4s and 4p are farther from the nucleus distancewise.

The shell number comes from certain mathematical procedure; it may be somewhat related to the distance from the nucleus, but does not indicate the said distance, nor can it be ascribed arbitrarily. So the 3d subshell is indeed 3 and not 4.

Have a look here, especially at the part "Why do states with the same n
but lower ℓ
values have lower energy eigenvalues?".

Also http://chemistry.stackexchange.com/questions/33297/why-do-3d-orbitals-have-lesser-energy-than-4s-orbitals-in-transition-metals

Thanks for your help. I will research more thoroughly next time.


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Why is imidazole considered an electron deficient heterocycle?
A paper I'm looking at1 mentions a few electron deficient heterocycles:

electron deficient heterocyclic compounds such as di(azines), benzothiazole and imidazole...

What are the criteria to determine whether a heterocycle is electron deficient or not? (Why is imidazole considered electron deficient, for example?)

Reference
(1) Bureš, F. Fundamental aspects of property tuning in push–pull molecules. RSC Adv. 2014, 4 (102), 58826–58851. DOI: 10.1039/C4RA11264D.

I'd point out that these aren't just heterocycles but heteroaromatic systems. The conjugation is important to the discussion.

As indicated in the comment below, a rule of thumb for these particular compounds, is that aromatic nitrogens with double bonds (e.g., not pyrrole) are "electron deficient."
The term is a bit ill-defined. It does not mean the aromatic system (which usually still has 4n+2 electrons) actually has a different electron count.
More broadly, there can be several ways of declaring "electron rich" or "electron deficient," for organic compounds, but in the context of a "push-pull" conjugated or aromatic system (in the paper you link), these refer to rings or functional groups that are electron donors or electron acceptors.
So:

Electron deficient - tends to gain negative charge as an electron accepting group or ring
Electron rich - tends to gain positive charge as a donor group or ring

The charge transfer need not be a full electron - it is almost always much less (e.g., a small fraction).
The second part of your question is how you can determine this.
Conjugated systems are typically imagined as a particle-in-a-box.. that the electrons are highly delocalized. The "electron donor" is a region with higher potential energy and the electron acceptor is a region with lower potential energy. Thus, the electrons have slightly greater probability on the electron acceptor.
Therefore, you determine "electron rich" or "electron deficient" in comparison with the rest of the conjugated system.
Typically, electron deficient rings have more negative HOMO and LUMO energies - making them more difficult to oxidize (i.e., they are "deficient" in electrons and it's hard to give up another) and easier to reduce (i.e., adding an electron makes them less "deficient").
Electron rich rings have more positive HOMO and LUMO energies - making them easier to oxidize and give up an electron.
Again, this is a rule of thumb and it's almost always partial charge transfer.

This answer is quite correct but fairly generic. A rule of thumb is that heteroarmoatic compounds with double bonded nitrogens (pyridine for example) are generally electron-deficient.

Absolutely true for the specific case, but there are plenty of push-pull conjugated systems without double-bonded nitrogens. So I went on the second (broader) question - when is it an acceptor? Truthfully, it depends on the rest of the conjugated system (i.e., electron deficient compared to what?)

Surely the electron deficiency refers to neighbouring aromatic atoms as opposed to the whole ring I.e. It should be clarified that the aromatic rings still have the same number of electrons (2n+2).

@Beerhunter You mean 4n+2 electrons. Yes, the aromatic rings still have 4n+2 electrons, but it does not refer to any neighboring atoms. The "electron deficiency" refers to the tendency for the ring to pull charge density from other conjugated systems. The charge transfer is rarely a full electron.

thank you all very much for your answers and comments! do you know the rationale behind that rule of thumb ("aromatic nitrogens with double bonds")?

@Zalumon Nitrogens are more electronegative than carbon, hence tend to withdraw electron density via inductive effect. This is what a "pyridine-type nitrogen", i.e. a nitrogen with a double bond, does. However, a "pyrrole-type nitrogen" (while still electron-withdrawing via the inductive effect) can also donate electrons via resonance. So, pyrrole is actually considered an electron-rich heterocycle.


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Trying to understand the causes and implications of kinks in fatty acid chains
I watched this video (The Deal with Fat by SciShow):
https://www.youtube.com/watch?v=mvvx2yQRbzQ
To summarize the main points I want to discuss:

Saturated fats are chains without any double-bonds and tend to result in solid structures because they stack well.
Unsaturated fats have at least one double-bond between carbon atoms, resulting in fewer overall hydrogens. This either results in a kink (cis structure) or a straight lockout (trans) depending on whether the hydrogens around the double bond are on the same side or not.
Kinked fats are harder to stack into lattices so they tend to be liquid, so most unsaturated fats are liquid.

My questions:

I thought there were some trans fats that were healthy / naturally occurring, such as those from certain meat products. What's the difference between the healthy natural trans fats and the trans fats made from partially hydrogenated oils and such?
What exactly causes the kink? Hydrogen atoms repelling each other?
If the hydrogens are repelling each other, then how do the chains even "stack" well in the first place if they are straight? Why wouldn't they all repel each other and remain liquid?
Why do unsaturated fats, which have one double bond and tend to be kinked, straighten out and result in solids when chilled?

The second part of the primary answer to the second question is backwards. A visual representation for something is not the reason why that something happens. IOW, while the answer is correct about the kink being visually represented, that idea that you add or get (by drawing) one more "zig zag" is not the reason why a kink forms.

@BikerJ I read the explanation as meaning that the double bond redirects the chain, resulting in more "zigs" than "zags" and therefore interfering with the stacking of the chains.

No, trans fats rarely occur in nature, and are generally thought to be very unhealthy.
No, like all neutral atoms, hydrogen atoms attract each other, albeit weakly (by London Dispersion Forces). That's why neatly stacked saturated fats are solid. The kink is to do with the structure of a hydrocarbon chain; their structural formula is usually drawn as a straight line, but it's actually more of a zigzag, each carbon at an angle to the last. Loosely speaking, the double bond interrupts that, so you get one more zig than zag, resulting in a kink.
See 2.
They don't need to straighten out to become solid; they just need to attract each other enough to overcome the tendency of warm molecules to keep moving around.

Some trans fats do occur in nature, though, and though they are a small percentage, they do exist in many meats, apparently. I've seen sources that suggest they shown not to raise cholesterol as semi-synthetic trans fats do. I don't know why that is/would be, and the source I saw that said it provided no source (it was Smart Balance's page about hydrogenated oils, I will look into it further, though).

Here is a source showing butter naturally enriched in trans fat had no measured negative effect. More info here.

Interesting! Thanks @DGS, I was deliberately guarded in my answer because I wasn't sure how rarely trans fats occur in nature, or how clear it is that they're bad for you across the board.


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What causes straight fatty molecules to form a lattice (e.g. saturated fats)?
What causes multiple chains of saturated fats to pack together and form a solid? It is said that because the chains are straight, they form a lattice and line up more easily.
But if I had a bunch of straight sticks and drop them into a pile, they don't magically form a lattice where everything lines up. So is there something about the fatty chains that causes them to line up and pack closely together into a lattice?

Nature abhors a vacuum, and your sticks aren't sticky.
When you have a jumble of sticks, there's lots of voids and spaces between them - the pile of sticks takes up much more room than the bundle. For wooden sticks, this is no problem, as air can easily fill in the voids between them. With saturated fats, this isn't possible. There really isn't anything that can fill the voids. The only way to fill them is to line up the chains in an ordered lattice.
Also, when the chains are in an ordered lattice they're making a large number of van der Waals interactions between them. Because the chains are straight, the best way to maximize these favorable attractive interactions is to pack the chains together in an ordered array. The van der Waals forces are comparatively small, but over a long chain it adds up. (I haven't ever seen this done, but I'm guessing that if you coated your wooden sticks with something sticky like honey, and shook them around a bit, your wooden sticks would also want to line up with each other and stick together in an ordered array.)
Temperature, of course, can overcome both effects. The additional kinetic energy in the molecules allows them to overcome the attractive interactions between the molecules, and higher temperatures mean that the molecules are moving faster, pushing away other molecules and effectively filling in the voids to some extent. When you get to a high enough temperature, the fat melts.
In unsaturated fats, the chains are kinked. This means that it's much harder to line them up to form all those van der Waals interactions, and there are intrinsically voids in even the most ideal packing. Thus it's much easier to break those interactions and add the additional voids needed for a liquid state, so the solid to liquid transition happens at a lower temperature.
One thing to keep in mind is that things like solid/liquid transitions are not absolute things. Where the solid to liquid phase transition happens is based on the relative stability of the solid versus the liquid. It isn't that the saturated fat can form a solid and the unsaturated fat can't, it's just that the solid saturated fat is more stabilize relative to the liquid saturated fat, than the solid unsaturated fat is to the liquid unsaturated fat. As the change in stability is much greater for saturated fat, it takes a higher temperature to melt the saturated fat than it does the unsaturated one.


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Why are reactions of acids with metals exothermic?
I realised the other day that I'm not sure why the reactions of strong acids with metals are exothermic. Any reaction is exothermic if there's more energy released from bonds being made than is used in breaking bonds, of course (using the word 'bonds' in a sufficiently loose sense).
In these reactions, metallic bonds are broken; covalent bonds are formed in hydrogen, but the hydrogen is in a gaseous state, so its molecules must be less attracted to each other than the hydrogen ions were to each other in the liquid. I'd vaguely guessed that the latter fact would make this an endothermic reaction (it works for sodium carbonate, after all) but evidently not!
Is there a rigorous way of calculating the total enthalpy change, or an easy explanation that I'm missing?
I guess I can calculate the enthalpy change of every one of the state changes involved, but it's not obvious that that's always going to come out negative (which I gather it does) so I feel like there's probably a less mathematical way of explaining this observation.

Keep in mind, the fact that hydrogen is a gas at room temperature implies that its intermolecular forces are weak, but says nothing about its intramolecular forces. A H-H covalent bond is actually quite strong, with a bond enthalpy on par with C-H bonds.

Also, you can't really overlook entropic driving forces in reactions that produce gasses.

Lighthart, yes, but this question seems to be about exothermic vs. endothermic, not exergonic vs. endergonic.

Yes there is a rigorous way of calculating the enthalpy change.
First write down an example reaction.  It might not be as easy as you think.  What species are really formed when zinc chloride (for example) dissolves in water?
The simple version below...
$$\ce{Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(g)}$$
...won't do, solvated zinc ions in hydrochloric acid are present in a variety of tetrahedral chloro-aqua complexes.
So maybe something like
$$\ce{Zn(s) + 2 H+(aq) + 2 Cl-(aq) + 2H2O(aq) -> ZnCl2(H2O)2(aq) + H2(g)}$$
This at least shows which bonds are being broken and which are being formed:

Zn-Zn bonds (metallic) are broken
H-H bonds are being formed
Zn-Cl and Zn-aqua (actually zinc-oxygen) bonds are being formed

To figure out the enthalpy change, you can figure out the approximate enthalpy of formation of each bond type.  H-H bonds are pretty "strong", i.e. the bond enthalpy is relatively high at 436 kJ/mol.  So when those bonds are formed, that enthalpy is released.  Zinc-water (i.e. zinc-oxygen) and zinc-chlorine bonds are also pretty strong at about 284 and 229 kJ/mol, respectively.  Zinc-zinc bonds enthalpies are only ~29 kJ/mol.
Add it all up, and you get a lot of enthalpy being released!


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Why are trans fats worse than saturated fats?
Saturated fat molecules have no double-bonded carbons, so they are long and straight, which means they stack easier and tend to form solids at room temperature, and solids are better at forming plaques in your arteries and interacting with cholesterol in the bloodstream.
Moving onward to hydrogenation, to my current understanding, if you hydrogenate vegetable oil all the way, you get saturated fats (no double bonds anywhere because you've shoved in so much hydrogen to occupy all the bonds). But for the purposes of making trans fats, it's usually partial hydrogenation.
I assume that most vegetable oils that get partially hydrogenated must be polyunsaturated fats to begin with (meaning more than one double-bond) because partial means you aren't replacing all the double bonds, so there must be at least two double-bonds present.
The result of partial hydrogenation means you remove these double bonds and start straightening out the chains, which makes them easier to stack (which is why trans fats tend to be solid or semisolid), like saturated fats.
My question:
Why are trans fats worse than saturated fats? It looks as though partial hydrogenation is simply taking a polyunsaturated fat and bringing it closer and closer to the status of a saturated fat, but not all the way. If the goal is to get a fat that is solid and lasts longer, why not just use a saturated fat?
And for whatever reason saturated fats aren't the answer: Isn't the end result of partial hydrogenation technically just another unsaturated fat? I thought unsaturated fats were considered healthy? So how are trans fats worse than fully-stackable saturated fats?
Or upon reflection maybe my question is, why are trans fats considered so much worse than other unsaturated fats? If all hydrogenation does is break up double bonds and insert hydrogen, straightening out any cis-kinks in the chain, doesn't this just generate the same acid with fewer double bonds and straighter chains? How is this any different than going out in nature and finding the same unsaturated fat, rather than going to the trouble of converting?
What's causing the difference, here? Where is my understanding breaking down?

Someone may very well give a more thorough answer but from what I understand, most naturally occurring unsaturated fats have their double bonds in the cis conformation which is generally higher in energy than the trans conformation.
Thus, the idea is that when you ingest trans fats (i.e. fats with a trans double bond) your body has to somehow hydrogenate this double bond before it can break it down further. This is more difficult for your body to do because it must either isomerize to a cis bond and then hydrogenate, or just hydrogenate the trans bond which is not easy for your body to do effectively. In some cases hydrogenation of trans fats can't actually happen because nature basically makes cis double bonds in fats (which sounds weird but is true) so your body has ways of breaking these fats down but not trans fats. So, the fat just gets stored or excreted without doing anything to it.

Does industrial hydrogenation do more than just break up double bonds / append hydrogen? Does it also somehow change cis double-bonds to trans double-bonds (how?) without adding any extra hydrogen at those junctions?

"nature basically makes cis double bonds in fats (which sounds weird but is true)" what about the natural trans fats found in milk? http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2596709/

I'm not saying they don't exist, but simply that lipids primarily contain cis double bonds. Pretty much all fatty acids are cis double bonds: Palmitoleic acid, oleic acid, myristoleic acid, linoleic acid, archadinoic acid. Also that article says that 2.7% of the 25% unsaturated fats in bovine milk is trans. That's .675% of the total.

When you say hydrogenate in your answer, do you mean hydrolyze?

Does hydrogenation somehow change cis to trans bonds? And are those trans fats harder to process in the body because we lack the right kinds of lipase enzymes to hydrolyze them? So why don't they just pass through the system, unabsorbed, without negatively reacting with anything (similar to how a fat blocker might interfere with a triglyceride's ability to get pulled apart into fatty acids by lipase)?

I don't know. I'm sure there's a lot written about this. Trans fats have been shown to increase the likelihood of heart disease, so one more biologically minded than myself might be able to make something out of that fact.

But no I mean I hydrogenate. As I understand it the body first hydrogenates the double bonds and then breaks down these saturated chains into smaller parts (by some process unknown to me).

If the oil is hydrogenated, wouldn't it be like burning a water soaked log where it produces less heat?  Less heat means less calories?

In my chemistry lab, one of the teaching experiments at my university involved nickel catalysed isomerisation of an alkene, as nickel is also the catalyst for hydrogenation.
This could explain why you end up with trans- fats from a cis- starting material.

I’m not entirely sure how this answers the question …

I'm not sure that this is an answer to the question, but rather a comment. Please use the post answer button only for actual answers. This is really a comment, not an answer. With a bit more rep, you will be able to post comments.

First of all, the concept that saturated fats are the culprit in heart disease is hotly debated and the AMA has been removing, and pulling back on general guidelines about saturated fat.  The opposing model would be the inflammatory model, and the primary pro-inflammatory fatty acid is linoleic acid (omega-6 PUFA).  Furthermore PUFA's are much more capable of becoming oxidized species and carrying oxidizing groups to arterial linings resulting in oxidation damage.
Most of the correlation of heart disease to fatty acid intake is independently explained by the trans fats that tended to be present in hydrogenated fats that also contained saturated fatty acids.
Mixed fats that contain PUFAs (polyunsaturated fatty acids) and MUFAs (monounsaturated fatty acids) can be hydrogenated to turn some of the UFAs into saturated fatty acids.  This prevents rancidity and raises the smoke point of the fat because PUFAs are the easiest to go rancid, and to smoke under heat.
There are two general instances in which fats are hydrogenated.  Liquid oils which are primarily MUFA and PUFA mixtures are hydrogenated to saturate those fatty acids, prevent onset of rancidity, and raise the smoke point.  The main reason for the hydrogenation is not to make the fat solid, but to make it saturated so that those things happen.  The second instance would be with a fatty acid mixture such as lard that contains saturated fatty acids, but also fractions of oleic acid and linoleic acid.  Hydrogenation makes the lard more homogeneously solid and turns primarily the large 18 carbon oleic and linoleic fractions into 18 carbon saturated stearic acid.  (There is actual evidence now that the primary reason that saturated fats were implicated in heart disease was because of the trans fat content in hydrogenated oils and lard.  The evidence is the reduction in new cases of heart disease since the banning of trans fats).
The natural UFAs tend to have a cis structure, but the trans structure is more sterically stable because the large hydrocarbon tails on either side of the double bond will be further separated.  Some of the broken double bonds are broken by the hyrdogenation process enough to allow a rotation before reforming and the majority of the products of the natural rotation will be trans once the double bond forms and locks down the rotation.
So why are trans fats made from primarily 18 carbon chains BAD?  Most, if not all of the correlated risk to intake can be explained by an increase in LDL levels with no corresponding increase in HDL levels, so its not the trans fats forming the plaques but the fact that they tend to get repackaged as LDL and possibly only a small fraction of LDL molecules that account for most correlated heart disease risk.  These LDL molecules also increase with higher dietary intake of fructose and linoleic acid and the findings are consistent with an inflammatory model (not a saturated fat model) of heart disease.

Saturated fat molecules have no double-bonded carbons, so they are long and straight, which means they stack easier and tend to form solids at room temperature, and solids are better at forming plaques in your arteries and interacting with cholesterol in the bloodstream.

This in invalid reasoning, because your arteries and the bloodstream are not at room temperature, they are at around 37 degrees Celsius where saturated fats are liquid.
Trans fats are unhealthy because the enzymes responsible for metabolizing them are "expecting" cis fats, and trans fats disrupt the whole metabolic chain. One of the side effects is raise of the low density lipoprotein cholesterol (LDL), which increases inflammation and risk of heart disease.

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Is there a nomenclature for linear alkanes with n carbons?
Is there a common nomenclature (not necessarily IUPAC) for naming linear alkanes with $n$ carbons without resorting to Greek/Latin numerical prefixes (hex-, hept-, etc.)?
For instance, is there a name for the higher linear alkane of $\mathrm{50}$ carbon atoms other than n-pentacontane? ("Not necessarily IUPAC" is meant in the same sense that $\alpha$, $\beta$, and $\gamma$ nomenclature  don't follow IUPAC rules but are nevertheless often used to designate relative positions of carbon atoms)

Informally you can address them as C50, once you set up the stage. But you should feed the IUPAC first.

I suspect the most compact way (and what most chemists would use) is to specify the generic formula like CH3(CH2)n(CH3) which is unambiguous and saves having to know any latin or greek numeric prefixes for big numbers.

^^ matt_black's post definitely seems, to me, to describe a method more common than using the Latin/Greek method, even.

That condensed formula certainly satisfies what I'm looking for, though I was personally hoping for something that rolls off the tongue a little more concisely. (Though of course that is not a qualification for being a answer.) In the same sense that the point group associated with a tetrahedral molecule can be concisely referred to as $T_d$

Great idea, something like n-[50]carbane would be useful.

@mykhal This was exactly the sort of answer I was looking for, though I was hoping there was something already in the literature. I was mostly curious if there was a standard convention for identifying n-alkanes without memorizing all the numerical prefixes. In hindsight, though, "50-carbon n-alkane" is probably concise enough for everyday use and I should probably just translate to IUPAC when necessary for publication.

This does not answer you question, just mentions an alternative name that does have that count arabic numeral in the name, but is not eliminating the greek/latin prefixes. One could use IR-7.4 INORGANIC CHAINS AND RINGS nomenclature that would be impractical, e.g. the mentioned n-pentacontane would become dohectahydridopentacontacarby[50]catena or maybe even 1,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19,19,20,20,21,21,22,22,23,23,24,24,25,25,26,26,27,27,28,28,29,29,30,30,31,31,32,32,33,33,34,34,35,35,36,36,37,37,38,38,39,39,40,40,41,41,42,42,43,43,44,44,45,45,46,46,47,47,48,48,49,49,50,50,50-dohectahydridopentacontacarby[50]catena (not to confuse with catenanes).

Bliss you :-) $\text{ }$

That sounds even worse than using the simpler nomenclature that comes from greek numbers. It also seems way unnecessary.

Sorry for the late response. I think this is a little unwieldy, to be honest, and I'm not sure what benefits this would confer.


text:
How is the reaction between Aluminium and Mercury Chloride possible?
Aluminium is coated with a thin layer of Aluminium Oxide. Why then do we write the reaction as

$$\ce{2Al + 3HgCl2 → 2AlCl3 + 3Hg}$$

Shouldn't we first write an equation showing how $\ce{HgCl2}$ reacts with $\ce{Al2O3}$ ?

I think the reaction takes place in an inert atmosphere. The reaction is simply a reduction of $\ce{Hg^2+}$ to $\ce{Hg}$

The reaction that happens is the one described. But the reason it happens is because the reaction with mercury salts disrupts the surface layer of alumina that otherwise protects the aluminium from routine oxidation (as many other substances do). Once the reaction has started no further protective layer forms so even minor defects in the protective alumina layer lead to a runaway disruption that allows further reaction. Small physical effects in the alumina are usually fixed in air, but won't "heal" when other reactions are occurring.

Socratic Q&A

The equation is correct assuming the aluminium is pure. In reality, $\ce{HgCl2}$ won't readily react with aluminium sheet since both are solids and $\ce{HgCl2}$ is a weakly-dissociated salt meaning that even in aqueous environment there won't be much dissociation into $\ce{Hg^2+}$ and $\ce{Cl^-}$. The molecule would simply go into solution as is.
Maybe there will be reaction if you mix aluminium and mercury(II) chloride powders and heat the mixture - still there will be formation of mercury(II) oxide rather than metallic mercury:
$$\ce{Al2O3 + 3 HgCl2  -> 2 AlCl3 + 3 HgO}$$
I think the reaction really goes off in aqueous solution or at least moist environment, where some mercury(II) chloride gives acidic reaction and we get some hydrogen chloride that will readily dissolve the oxides, allowing for the displacement reaction.
Note that aluminium amalgam is commonly prepared by first scratching the metal surface under water or under mercury (as otherwise it would immediately re-oxidize to alumina), grinding the two metals together or first etching the metal with dilute $\ce{HCl}$ or $\ce{NaOH}$. Under water, the aluminium is protected from air and will form hydroxide rather than oxide, which reacts much faster with aqueous $\ce{HgCl2}$.
Indeed aluminium with thick oxide layer reacts very slowly with $\ce{HgCl2}$. It eats away the oxide layer but at the cost of some dissolved mercury(II) turning into insoluble $\ce{HgO}$. There is a neutralization going on rather than displacement plus other side reactions (like formation of insoluble calomel - $\ce{Hg2Cl2}$, formation of oxychlorides etc.).
The typical procedure for preparing aluminium amalgam is therefore properly degreasing and etching aluminium with solvents and dilute base first, then washing off excess solvent, hydroxides and base (the metal is still wet) and then getting it into contact with dilute mercury(II) solution (which can now be very dilute as we don't waste $\ce{HgCl2}$ for the etching).

Interesting.... I am Libor too, also from Brno.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

HgO is decomposed by mild heat. Al2O3 is one of the most exothermic compounds.

What is the meaning of the sentence : "I am Libor too" , sent by Poutnik 2 hours ago ?


text:
Ullmann's Reaction (similar to Wurtz) - aryl chloride with deactivating groups - stability of free radical
Ullmann's Reaction:

Mechanism:

Question:
My textbook says with $\ce{X} = \ce{Cl}$, the reaction gives less yield compared to using $\ce{X} = \ce{I}$ since $\ce{Cu}$ is not as reactive as $\ce{Na}$ ($\ce{C - I}$ bond is weak).
But if aryl halides with substituents such as $\ce{NO_2}$ are used, they can help in obtaining good yield.
What I observe from both the mechanisms is that adding deactivating groups in the ortho or para position relative to the halogen should actually give lesser yield since the deactivating group would cause a partial positive charge to exist on the $\alpha$-carbon hence destabilizing the free radical intermediate.
Can anyone clarify my doubt? I suspect my last argument to be wrong regarding the $\alpha$-carbon. The $\alpha$-carbon is already electron-deficient and the partial positive charge should make it worse?
Courtesy: http://www.name-reaction.com/ullmann-reaction

Your assumption that electron-withdrawing groups destabilize radicals is wrong. Radicals are in fact stabilized by both electron-withdrawing and electron-donating groups and even more so if both types are attached to the same carbon atom (captodative effect).
I modified an image from Ian Fleming's Molecular Orbitals where you have the proof. In the image you can see that the combination of the carbonyl group's (which is electron-withdrawing) pi bonding/antibonding pair of orbitals with the p orbital of the radical results in an overall decrease in energy:

What if the first step in the mechanism was the RDS? Then the deactivating group would improve the reaction rate since it enhances the ability to knock out the halide.

I guess so. ...


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Which has more ionic character, LiF or K2O?
After drawing the structures of both and considering dipole moment, doesn't it seems that all K2O should have more ionic character than LiF?
My textbook says its LiF.

Li and F are very small atoms hence they form strong covalent bond.

So why they have more ionic character?

I do not know the answer, looks like I did not frame my comment properly. It appears more like an answer.

I don't think one can compare the ionic character of compounds where both the anion and cation are different. I have no idea on what basis one would decide which is more ionic.

http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound


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Why do unsaturated hydrocarbons burn with a sooty flame while saturated hydrocarbons burn with a clear flame when burnt in the air?
Why do saturated hydrocarbons burn with a clean flame while unsaturated hydrocarbons burn with a sooty flame when burnt in air?
My chemistry book says that it happens due to the amount of carbon being more than the amount of hydrogen in unsaturated hydrocarbons. But isn't that also the case with saturated hydrocarbons?

Also, am i wrong in thinking that it happens due to the bonds of unsaturated compounds being stronger than the single bonds present in saturated hydrocarbons due to which incomplete combustion of unsaturated hydrocarbons takes place?

Your observation is not true in general. The flame character depends a lot of how the hydrocarbon is burned. If the amount of air or oxygen is right, all hydrocarbons will burn cleanly. Eg Acetylene (or ethyne or C2H2) will burn cleanly with one of the hottest flames known if the gas/oxygen mixture is well controlled as it is in oxygen-acetylene welding torches.

Your observation that unsaturated hydrocarbons burn with a sooty flame is not always true and depends on the circumstances. There are more factors at play than the nature of the hydrocarbon that is burning.
The most important reason why flames are sometimes smoky is that the combustion is incomplete. This means that not enough oxygen was present to fully oxidise all the carbon to carbon dioxide or carbon monoxide gases (and the amount of CO will vary with the ratio as well as, given enough oxygen, it should also be completely combusted to carbon dioxide). If there is not enough oxygen in the flame there are many competing side reactions that generate sooty resides (the reactions are pretty complex). Some pyrolysis reactions of hydrocarbons may occur before combustion itself (if the mixing in the flame is less than perfect) and these may contribute to the generation of sooty products which may be harder to burn than the original hydrocarbon.
Incomplete combustion is as much a function of the environment of burning as it is the thing that burns. Some environments don't enable enough oxygen to get into the gas mixture to drive complete combustion and the reactions can easily leave a lot of sooty emissions. A candle, for example, burns with a smoky flame despite being made from mostly saturated long chain hydrocarbons. It does so because a candle wick doesn't do a good job of mixing air with the vapours of the burning hydrocarbon (this is deliberate as the prime role of candles is to provide light and this comes from incandescent particles of soot in the flame: a pure flame burning efficiently would provide little light by itself).
Even well controlled environment such as engines can't completely avoid side reactions. But Diesel engines (which use mostly saturated hydrocarbons) produce more sooty particles than petrol (gasoline) engines though petrol engines have far more unsaturated fuels. The exact reasons are complex and depend on the precise environment inside the engine.
In well controlled environment where the oxygen-fuel ration is set carefully most hydrocarbons burn cleanly with little soot. Gas-based cookers (using mostly saturated methane, ethane, propane or butane) are very clean because they control the gas mix very precisely (use the wrong burner and they may well be sooty!). Welding and cutting torches can happily use the highly unsaturated acetylene with a clean, non-sooty, flame (the oxygen-acetylene flame is one of the hottest achievable) because they control the gas mix very carefully leaving little room for side reactions.
Uncontrolled flames might well show your general observation that unsaturated fuels burn more sootily, but this is as much a product of the environment as it is the product of the hydrocarbon. It is hard, without a carefully controlled experiment, to tell whether the general observation is true at all.

Thanks for the answer, the comparison between burning acetylene in welding and the candle example clearly shows that the circumstances of combustion play a more significant role in deciding the soot produced as compared to the type of hydrocarbon.

I think that your book is actually saying the percentage of carbon in unsaturated hydrocarbon is more than saturated hydrocarbon. Let's take the hydrocarbon with two carbon, which are saturated hydrocarbon(ethane), unsaturated hydrocarbons(ethene and ethyne).
The percentage of carbon in ethane($\ce{C_2H_6}$)
=$\ce{[2(12)/{2(12) + 6(1)}]*100}$
= 80 %
The percentage of carbon in ethene($\ce{C_2H_4}$)
=$\ce{[2(12)/{2(12) + 4(1)]*100}}$
= 86 %
The percentage of carbon in ethyne($\ce{C_2H_2}$)
=$\ce{[2(12)/{2(12) + 2(1)}]*100}$
= 92 %
According from above, we can see unsaturated hydrocarbon have a higher percentage of carbon than saturated hydrocarbon. So, there is a sooty flame when burning unsaturated hydrocarbons and a clean flame when burning saturated hydrocarbon

How exactly does the increase in carbon content translate to incomplete combustion?

@Aniansh Is this about complete or incomplete combustion?

you just mentioned in your answer that the increased carbon content of unsaturated hydrocarbons leads to them burning with a sooty flame and that only happens when incomplete combustion takes place. I am asking you how is increase in carbon content related to sooty flame? By this regard, ethane should burn less efficiently as compared to methane. Also, does the carbon to hydrogen ratio in a hydrocarbons affect its combustion properties?

What I can say is the soot is carbon, if a higher content of carbon will produce a higher chance of soot. Second,   the carbon to hydrogen ratio in a hydrocarbons affect the efficiency of combustion. Regarding to the question of methane vs ethane, I still haven't have much research on it but I will try find the answer

Thanks for the answer ! It was helpful! I am sure someone else will post a complete answer in the future.


text:
Partial Pressure and Partial Molar Volume
When we define partial pressure and partial molar volume of a gas in a mixture, suppose $P$ is partial pressure of a gas in a mixture and $V$ is partial molar volume
then why the following equation doesn't hold
$PV=nRT$, (where $n$ is number of moles of gas) ?

It doesn't work because partial molar volume of a species in a gas mixture is defined as the partial derivative of the total volume of the system with respect to the number of moles of that species at constant total pressure and temperature of the system.


text:
What happens to the lost mass in radioactive decay?
We were studying kinetics in class which led me to wonder what happens to the decayed matter. After all, there is a loss in mass and according to mass energy equivalence, there should be some release in energy. So I thought that heat may be released by radioactive decay, but I doubt it.
Can anyone tell me what happens to the lost mass?

On the nuclear physics side, it goes in to kinetic energy of the various particles as well as any gammas. As those particles interact with the rest of the world, they will be thermalized resulting in heat  to the surroundings. If this did not happen, fission reactors would have a hard time making steam to turn turbines.

Suggest @Jon Custer move Comment to Answer

On the nuclear physics side, it goes in to kinetic energy of the various particles as well as any gammas. If it is a simple reaction (D+D -> 3He+n or 3H+p) you can determine the energy of each particle in the center of mass by simple kinematics. In something resulting in more particles (most fission events result in 2 nuclei and multiple neutrons) you get a spread of kinetic energies of the various bits. And, if one or more of the nuclei are left in an excited state than you can get emission of gammas, energetic photons.
As these daughter particles interact with the rest of the world, they will be thermalized through interactions with electrons and nuclei surrounding them, transferring kinetic energy, resulting in heat to the surroundings. Any gammas will eventually be absorbed/scattered, also resulting in energy transfer to the surroundings. If this did not happen, fission reactors would have a hard time making steam to turn turbines.


text:
What happens in the NH4Cr2O7 with HgSCN reaction?
Apparently this video is a reaction of $\ce{NH4Cr2O7}$ with $\ce{HgSCN}$.
The title of the video suggests the reaction opens a portal to hell, but I am pretty sure that is not the case.
What exactly goes on in the reaction?

In short, this is a mash-up of two well-known experiments: Pharaoh's serpent and Vesuvian fire. The two compounds do not really react with each other (well, at least not much); instead, each reacts on its own.

I can't comment on the metaphysical aspects ;)
This experiment is known as Pharaoh's snakes, and it is indeed impressive.
Upon heating, mercury thiocyanate decomposes to yield mercury vapour, carbon sulfide, and dicyan (ethanedinitrile):
$$\ce{2Hg(SCN)2 ->[\Delta] 2 Hg + 2CS2 + (CN)2}$$
As a result, the solid starting turns to a foam-like mass.
The effect is amplified due to the presence of ammonium dichromate, which oxidizes the carbon disulfide to sulfur dioxide and carbon dioxide.
The experimenter is apperently aware that quite a lot of rather harmful components are released and therefore has decided to run the show in the fumehood and closes the window immediately after igniting the mixture.

None of those products looks nice, but the cyanogen may just open a one-way portal to hell for some.


text:
What type of acid is found in tea?
We were trying the litmus paper test and we found out that tea is acidic. So the question came in my mind and my science teacher could not give me any satisfactory answer: what acids are found in tea?

I was going to suggest "tannic acid" as one possibility, but apparently while tea contains tannins, it doesn't contain "tannic acid". (Though the tannins in tea may still be acidic, despite not being "tannic acid".)

This might be helpful although it is related to coffee-http://chemistry.stackexchange.com/questions/18513/if-caffeine-is-an-alkaloid-why-is-coffee-acidic

Some of the major components of tea are polyphenols. They are fairly acidic and may dominate the pH of tea.
For some examples of the types of polyphenol that are common see this site on tea chemistry:

The major flavanols [a specific type of polyphenolic compounds] in tea are: catechin (C), epicatechin (EC), epicatechin gallate (ECG), gallocatechin (GC), epigallocatechin (EGC), and epigallocatechin gallate (EGCG).

EGCC has this structure:

from which some degree of acidity is likely to result from all those phenolic OH groups.

According to this analysis from 1965, there are several different acids in tea leaves:

Succinic, oxalic, malic, citric, isocitric, and five unknown acids have been found at appreciable levels (>0·1 μequiv./g. fresh wt.) in tea shoot tips. The major acids quantitatively were oxalic, malic and citric acids. The acids studied were recovered quantitatively from plant extracts, but they accounted for less than one-fourth the total acidity present in tea shoot tips.

However, some of these acids degrade as the leaves go from the plant to your teapot, in particular during the withering process:

Changes in the levels of organic acids take place in tea shoot tips during the manufacture of tea which are typical of senescing plant tissues. Notably, the levels of succinic and malic acids are markedly reduced during withering. The level of oxalic acid, however, remains nearly constant throughout manufacture.

Unfortunately, the full article is behind a paywall and I don't have access to further details.

Tea has Tartaric acid as tea is a little bit bitter in taste. Also many other foods have tartaric acid such as grapes and lemon.


text:
Calculating the heat of reaction for Mg metal and HCl

Adding $\pu{2.00g}$ $\ce{Mg}$ metal to $\pu{95.0 mL}$ of $\mathrm{1.00 M}$ $\ce{HCl}$ in a coffee-cup calorimeter leads to an increase of $\pu{9.2 ^\circ C}$. If the molar heat capacity of $\pu{1.00 M }$ $\ce{HCl}$ is the same as that for water $(\pu{C_p = 75.3 J mol-1 K-1)}$, what is the heat of reaction?

Two relevant equations are: $${\Delta H_{(rxn)} = -q}\\
{q = n \times C_p \times \Delta \mathrm T}$$
The answer given is $\pu{-44.4 kJ mol-1}$. I wonder if I'm not way off the mark here by using the wrong equations, because the answer I got was $\pu{122.83 J mol-1 \; ( 0.1773 mol \times 75.3 J \times 9.2)}$.
note: The heat of reaction = $\Delta H_{(rxn)}$

The equations you listed can be used to partially solve the problem.  However, with the equations you used, you have made one error which was in your choice of "moles."  Your moles value must match up with your $\ce{C_{p}}$, so you must find the moles of Water in $\pu{95.00mL}$!  Alternatively, you can use $4.184 \pu{\frac{J}{g\ ^\circ C}}$ and $\pu{1 \frac{g}{mL}}$ and $\pu{95.00mL}$ of water along with the equation $\mathrm{Q = n \times C_p \times \Delta T}$.
On the other hand, the premise of your solution is close but wrong, which is obvious if you examine the units:

$\pu{0.1773 mol} \times 75.3 \frac{\pu J}{\pu{mol K}} \times \pu{9.2 K} = \pu{122.8 J}$

The units for Heat of Reaction are $\pu{\frac{J}{mol}}$, not $\pu{J}$!
To finish off your solution, you must divide by the moles of reaction that occurred (given by the limiting reactant).  $\pu{-44.4 kJ}$ is the value calculated based on the moles of Mg reacted.  However, I suspect that the given answer is wrong because the balanced equation does not yield a 1:1 reaction ratio between $\ce{Mg}$ and $\ce{HCl}$, which means that $\ce{HCl}$ is the limiting reactant, which would make the Heat of Reaction Larger.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What exactly happens when vegetable oil is partially hydrogenated?
My current assumption is that partial hydrogenation requires a polyunsaturated fat, because partial implies at least one but not all, which means there must be at least two double bonds in the fat, hence polyunsaturated, right?
Assuming that is correct:
My current understanding is that when you bubble hydrogen (using some heavy metal catalyst) through a vegetable oil, somehow this results in transforming polyunsaturated fats to unhealthy forms of unsaturated fats called trans fats.
What exactly is happening, here? I initially assumed that adding hydrogen to the fat would do little more than making double-bonded carbons into single bonds and then attaching two hydrogen atoms to the now-free valence electrons.
But apparently it also removes cis-kinks and makes them into straight trans-lines, which doesn't change the number of hydrogens but rather changes the shape.
So what's going on here exactly? What do we say partial hydrogenation is actually doing to the polyunsaturated fat?

You actually don't need a polyunsaturated fat to be partially hydrogenated. The "partial" refers to the macroscopic view of the fat, rather than a per-molecule basis. That is, if you have, say, one mole of monounsaturated fats and hydrogenate the double bond in half a mole of them, then you've partially hydrogenated the original mole of fat. Taken individually, each molecule has either been hydrogenated or it's still monounsaturated - no single molecule has been "partially hydrogenated". However, the mole taken as a whole has only been partially hydrogenated - 50% of the double bonds that were originally present have been hydrogenated.
Regarding trans fats, the important thing to realize is that (parts of) the hydrogenation reaction is reversible. The addition of molecular hydrogen is not concerted. Each hydrogen is added in a step-wise fashion (roughly). It's in that one-hydrogen added, one-to-go stage where the reaction can reverse course and go back to a double bond. But prior to reforming, the catalyst has broken the double bond and you have free rotation around it. Which way it's put back together is a matter of chance.
So you start with a fat which has all cis bonds. (Because the stereo and regio-selective enzymes that make them only make cis fats.) You then start the hydrogenation process. Each time you add hydrogens some goes to completion, and you end up with saturated bonds. But a fraction of the reactions get part way and then reverse. But as the hydrogenation catalyst is not stereo- and regio-selective, it's just as happy reversing to a trans bond as it is to a cis bond. So it's those bonds - the ones which go part way and then reverse - which end up as the trans fat in the final product.
The reason why fully hydrogenated fats don't have this issue is because they're produced with an excess of hydrogen. Yes, some of the double bonds react part way and then reverse to trans fats, but the catalyst is not stereo- and regio-selective, and it's able to convert the trans bonds to saturated bonds just as easily as the cis bonds. All the trans fats that get formed in the reaction are subsequently destroyed by the complete hydrogenation.

How does one ensure that only (for example) half of each molecule is hydrogenated? I thought it was all processed in giant vats where everything mixes together, rather than specifically bombarding certain sections of individual molecules with hydrogen.

@user27186 You don't hydrogenate half of each molecule, you hydrogenate half the quantity of molecules ... I'll rephrase.

Maybe I misread. So you're saying that partial hydrogenation means that if we have a vat of unsaturated fat in, say, liquid form, then we fully hydrogenate a fraction of the vat? As opposed to hydrogenating all of it and getting a vat of solid saturated fat?

"Taken individually, each molecule has either been hydrogenated or it's still monounsaturated" Do you mean just unsaturated (since a polyunsaturated fat can be used too?)

Also what you're saying seems to contradict this spot in this video here: https://www.youtube.com/watch?v=6zU0atqAL4s&t=2m6s which implies that partial hydrogenation does refer to individual molecules as being "partially hydrogenated" due to only some of the double bonds being broken and replaced with hydrogen.

What I also still don't understand: If you just kept adding more hydrogen, more hydrogen, more hydrogen, would you eventually get plain saturated fat (which isn't considered as bad as trans fat)?

Vegetable oils which have been partially hydrogenated, are now partially saturated so the melting point increases to the point where a solid is present at room temperature. The degree of hydrogenation of unsaturated oils controls the final consistency of the product.

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text:
Alkane deprotonation by organolithiums
Butyllithium isomers are all stored in alkane solvents. Theoretically, equilibrium should (given enough time) lead to formation of mostly n-hexyllithium in the case of  n-BuLi stored in (mostly) n-hexane, I believe. I understand that the kinetics of deprotonating an alkyl proton are extraordinarily slow, apparently even in the case of the much stronger t-BuLi in pentane (which isn't merely thermodynamically neutral; it should be thermodynamically favorable for the lithium to exchange with the proton of a less substituted carbon).
I was wondering, though, does anyone know of any instance where a significant portion (or any observable portion, really) of something like t-BuLi has been noted to have reacted with alkane solvents?
All searches I've attempted have given me completely unrelated results, and even attempting to sift through them, I have come away with no mention of the idea, even.
Also, would the addition of some kind of lithium-binding reagent such as TMEDA or 15-crown-5 increase or decrease the rate here (assuming no Brønsted reaction between BuLi and the ligand)?
I ask because BuLi will typically deprotonate a heteroatom before doing nucleophilic addition, but will do nucleophilic addition (to a carbonyl) before deprotonating most "softer" carbon acids, suggesting alkyllithiums are pretty hard bases. I'm assuming making a more "naked" carbanion would make a harder base (more charge-driven than "HOMO-LUMO overlap"-driven), and I'm wondering if that could make it slower to react with an alkyl proton (though, an unstabilized carbanion is a fairly hard nucleofuge, so I could see the reaction going faster, too).

Nice question.  I've got my expert considering it.

Seems it's last problem with these solutions https://en.wikipedia.org/wiki/N-Butyllithium but generally your points seem OK.

@Mithoron What do you mean "last problem"?

Seems it's insignificant in comparison with reactions with air and decomposition via elimination of LiH. Also hexyllithium is very similar.

I would bet activation energy for this sort of reaction is quite high, and since primary and tertiary carbons aren't hugely different in stability the sort of H for Li exchange reaction would be very slow. For nBuLi, the reaction would practically not happen, and for sBuLi and tBuLi intramolecular rearrangement (proton shift) is more likely than reaction with solvent, resulting in nBuLi and iBuLi respectively.
Agreed that elimination and precipitation of LiH will be faster than alkyl exchanges in these solution, though I've never checked an old bottle of tBuLi for any precipitate so this is entirely speculative.
Investigating this would be quite challenging, as loss of base power in practice usually happens from slow exposure to O2 and H2O,which would resolve Li2O and LiOH respectively, also basic precipitates.


text:
What does the circle inscribed in a benzene ring represent?
I have seen benzene being depicted in two different ways:

For the right-hand structure, does the inscribed circle represent the movement of double bonds or electrons?

The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true.

It does not switch; it just stays there, in some sense "halfway between" these structures. There is no movement of double bonds, and in fact there are no double bonds. (Also, there is no movement of electrons either. They just sit there on their molecular orbitals, which by itself is an approximation, though a pretty decent one). All $\ce{C-C}$ bonds in benzene are the same; they are neither double nor ordinary. Read the Wikipedia page on aromaticity, or just look at this picture from it:

That's what the circle stands for.

I found the teaching of this concept is tied far too tightly to the representational aspect.  It shouldn't be taught as a 'double bond' where everything exists in a solid state but an electron probability distribution, which, in fact, is what it is.

I read the aromaticity, and didn't anderstand something "instead of being tied to one atom of carbon, each electron is shared by all six in the ring" the free electron shared with all of them or just with the electrons on his side?

To begin with, don't call that electron free, because it is not. As for your question, the right answer is closer to the first option. Each electron is delocalized over both sides of the ring.

so the electrons just delocalized on both sides, and when all electrons doing so, it looks like the bonds is moving back and forth?

To me, it does not look like that at all. A matter of habit, I guess. Yes, the electrons are just delocalized on both sides, but nothing is moving.

Will this delocalixation give stability to the ring and make the structure that contains the ring withstand more load to break?

Well, yeah, sort of.

But you can make the electrons move around the ring by applying an appropriate magnetic field — think NMR ;)

This is a typical example of the downfall of the Lewis structures. In fact, $\ce{C6H6}$ can be represented by more than one Lewis structure:

Do any single one of them actually represent benzene? No! If it did, then the $\ce{C-C}$ bond lengths would differ from each other. In reality, it is observed that all the bond lengths are equal.
There are two approaches to solving this problem: valence bond theory (VBT) and molecular orbital theory (MOT).

Introduction
An electron is described by the wavefunction $\psi$ where $$\psi(x)\equiv \langle x|\psi\rangle= \textrm{probability amplitude of finding the electron at} \, x;$$
where the state $|\psi\rangle$ is defined as $$|\psi\rangle = \int_{\textrm{all}\, x} |x\rangle\langle x|\psi\rangle \, \mathrm dx.$$
The probability of finding the electron in the region $x \pm \mathrm dx$ is given by $$\textrm{prob}(x,\mathrm dx)= |\psi(x)|^2\;\mathrm dx$$
Also, from the First Principles of Quantum Mechanics:

When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately.$^\dagger$

Valence bond theory
Consider hydrochloric acid. How can it be represented by Lewis structure?
$$\ce{H-Cl}\,\,\,\textrm{or}\,\,\, \ce{H^+Cl^-}$$
The former is described by the wavefunction $$\psi_{\ce{H-Cl}}(1,2)= \psi_{\ce{H}}(1)\psi_{\ce{Cl}}(2) + \psi_{\ce{H}}(2)\psi_{\ce{Cl}}(1)$$ and the latter by $$\psi_{\ce{H^+Cl^-}}(1,2)= \psi_{\ce {Cl}}(1)\psi_{\ce {Cl}}(2)$$
However, we know that $\ce{HCl}$ is neither 100% ionic nor 100% covalent. To properly describe the wavefunction of $\ce{HCl}$, quantum superposition is applied viz.
$$\psi_{\ce{HCl}}= \psi_{\ce{H^+ Cl^-}}C_1 + \psi_{\ce{H-Cl}}C_2$$
where $C_1$ is the probability-amplitude to find the electron-pair at $|\psi_{\ce{H^+ Cl^-}}\rangle$ provided it was prepared in $|\psi_{\ce{HCl}}\rangle$ that is, $C_1 =\langle \psi_{\ce{H^+ Cl^-}}|\psi_{\ce{HCl}}\rangle;$ also $C_2 = \langle \psi_{\ce{H-Cl}}|\psi_{\ce{HCl}}\rangle$ and $C_1^2 + C_2^2= 1.$

A better description of the wavefunction for the molecule is a superposition of the covalent & ionic descriptions, & we write (with a slightly simplified notation) $$\psi= \psi_{\ce{H-Cl}} + \lambda\psi_{\ce{H^+ Cl^-}}$$ with $\lambda$ some numerical coefficient. In general, we write $$\psi = \psi_\text{covalent} + \lambda \psi_\text{ionic}$$ [...]According to the general rules of quantum mechanics, in which probabilities are related to squares of wavefunctions, we interpret the square of $\lambda$  as the relative proportion of the ionic contribution. If $\lambda^2$ is very small, the covalent description is dominant. If $\lambda^2$ is very large, the ionic description is dominant.$^{\ddagger}$

Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'
In the same way, we can describe benzene as the following (un-normalised) wavefunction :
$$\psi = c_1\psi_{\text{Kek}1} + c_2\psi_{\text{Kek}2} + c_3\psi_{\textrm{Dew}1} + c_4\psi_{\textrm{Dew}2} + c_5\psi_{\textrm{Dew}3}$$
[It has been estimated that each Kekule canonical form has a relative contribution of about $0.408$ while the Dewar structures each has a relative contribution of $0.177\;.$ Thus the weight of two Kekule structures each is $0.39$ ie. they contribute to about $39\%$ to the resonance while each Dewar structure has relative weight is $0.073$ ie. they contribute to $7.3\%$ to the resonance hybrid. And thus many times, Dewar structures aren't taken into consideration in creating the resonance hybrid of benzene.]
Remember, the wavefunction on the left is real and all in the right are fictitious.
Resonance thus explains the delocalisation of electrons in unhybridised $p$ orbitals that gundergo $\pi$ bonding throughout the whole molecule. Thus, the circle represents the delocalisation of $\pi$ bonds over the whole molecule.

Molecular orbital theory
Molecular orbitals are one-electron wavefunctions which are formed due to linear combination of atomic or hybrid orbitals. Unlike in VBT where bonding occurs due to the coupling between atomic orbitals of the participating atoms and thus remain localised between those two atoms, in MOT, all atomic orbitals apart from those two atoms participate in the combination and thus the MOs are spread over all the nuclei of the molecules.
Delocalisation is thus inherent in the approach of MOT. The bonding influence of a single electron pair is spread over the entire molecule and not just the adjacent participating atoms of the molecule.
Now, let's come to the case of benzene.
All the unhybridised p orbitals of $\ce{C}$ atoms in $\ce{C6H6}$ combine to form MO. Since there are six such orbitals, six MOs would be formed.
The (un-normalised) wavefunction thus can be written as:
$$\psi= c_1\psi_1 + c_2\psi_2 + c_3\psi_3 + c_4\psi_4 + c_5\psi_5 + c_6\psi_6$$
Three bonding MOs and three antibonding MOs are formed; the former being lower in energy:

There are two pairs of degenerate orbitals; one is lower in energy while the other is in higher level. The electrons occupy the MOs of lower energy level.
Each MO is spread either all around, or partially around, the $\ce{C6}$ ring  which corresponds to delocalised $\pi$ bonding. Thus, each electron pair helps in binding together several of the $\ce C$ atoms.
Conclusion:
Due to delocalisation, all $\ce{C-C}$ bonds get extra stability which is due to the lowering of energy levels of the electrons. While VB theory explains with imaginary structures, MO theory explains it with the inherent delocalisation in its definition.
But the crux of the point is, the bonded electron pairs never remain confined or localised between two participating atoms; it spreads over the whole molecule; it delocalises.
The circle in that structure you mentioned represents this delocalisation of $\pi$ bond over the entire ring.

References:
$^\dagger$ The Feynman Lectures of Physics Vol. III by Richard P. Feynman, Matthew Sands, Robert B. Leighton.
$^\ddagger$ Elements of Physical Chemistry by Peter Atkins and Julio de Paula.

Is Dewar benzene a resonance structure or different molecule than benzene?

Whilst the circle drawn on a benzene ring is done to represent the resonance stabilised delocalised pi orbital electrons as comprehensively explained by the previous author, there is another more practical reason why a circle is drawn in this manner to represent carbon bonding in benzene. If you consider the case of cyclohextriene [and consider the position double bonds in cyclohextriene occupy] apart from the major fact the 6 carbon atoms in these two compounds are very differently bonded, when they are written in reactions both would actually appear to be identical compounds, [which of course they definately are not]. So benzene's structure is also written this way to differentiate between these two compounds otherwise this situation would cause much confusion in writen chemical reactions separately involving both benzene and cyclohextriene!

Thanks for the explanation, one question, Why the electrons are stabilised in the ring?

@studentrr:  Electrons are stabilised because now they have more space to move about thus lowering the kinetic energy without actually increasing the potential energy . Electrons always want to lower their Coulomb potential energy without making high kinetic energy by confining in a smaller place which can be attributed to Uncertainty Principle.

Thanks a lot for explaining on energy level, So you mean that electron consume energy when it moves to each atom, and so lowers it's kinetic energy? but didn't understand a bout potential energy?

@studentrr: Okay! I don't encourage long comment session on this. You can come at Chem Chatroom: http://chat.stackexchange.com/rooms/3229/the-periodic-table where you can make any query and hopefully might get an intuitive chat from the users; so do come there; it is not the right place to make comments after comments.


text:
Polar rearrangements: What is this reaction called?
I have been trying to find more information about the given reaction but cannot seem to find it? I want to know the name so I can read up more information on it. Can anyone help?

It is a Wagner–Meerwein rearrangement. Your particular example has the following mechanism:

Edit: Regarding your comment, the best way to see what is happening is to draw the product so that it resembles the reactant (like in the image below) and then in another step draw it in a more reasonable conformation:

Thank you!, however I do have another question if you don't mind,  from steps 2 to 3 in your mechanism, I understand how the methyl group moves but it's very hard to visualise, could you be of any assistance, please?

Related to followup question in comment:
The molecule is redrawn from a different orientation.
This ring system is bicyclic and is basically a five-member ring stitched to a five-member ring stitched to a six-member ring.  The original six-member ring is shrinking to a five-member ring and one of the original five-member rings is expanding to the new six-member ring.
Track where the geminal dimethyl groups are, they do not shift.  This will help you sort it all out


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Calculate the number of Carbon atoms in .0400g C7H5NO3S
I understand how to get from grams to moles and from moles to molecules with Avagadro's number. That gives us 1.31x10^20 molecules of C7H5NO3S. Now we take that and multiply by 7 to get the individual Carbon atoms. This is where I get confused because my calculator says 9.17x10^20 but my book says 9.20x10^21. Please help explain!

Based on the mass and molecular formula you've given, I get $9.21\times 10^{20}~\mathrm{C~atoms}$, essentially in agreement with your number.

Yes, the book answer is wrong.

Are you sure that $\ce{C7H5NO3S}$ is the molecular formula? Could it instead be the empirical formula?

My book doesn't actually specify, I was just assuming that this was the molecular formula.

It was originally 40.0mg.

If they didn't specify, then it probably was meant to be the molecular formula, yeah. @orthocresol is right, the book is wrong.

.0400g is a really non-chemical way to write that number.
I expect the text meant 0.400g, which is 10 times higher than the number actually printed, and this would explain why the book's answer is 10 times higher than the actual answer: they are working from the number they didn't print.


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Colloids dispersion classification
I've been teaching chemistry for 25 years, but now I'm confused.  Is a colloidal dispersion a heterogeneous or homogeneous mixture? I've seen both answers on the web.

+1 for the "teaching chemistry for 25 years and am still learning it" - one of the best things about chemistry

Words like "homogeneous" and "heterogeneous" do not have a well-defined meaning.  Different authors use them differently.  The fact is, all matter is made of atoms, and atoms are not homogeneous.  Thus, from a certain point of view, nothing is homogeneous.  Matter is lumpy!
But suppose we don't care about things on the scale of atoms, i.e. angstroms.  Then the question is "what scale do you care about?"  Nanometers?  Hundreds of nanometers?  Chemists are making materials that have lumpiness on more more less all of these scales.
Back before folks invented centrifuges, filtration, and microscopes, bacterial cultures, with cells as small as 1 μm in diameter, probably seemed homogeneous.  But van Leeuwenhoek peered through a microscope and dispelled that notion.

A colloid is a homogeneous substance, consisting of large molecules or ultramicroscopic particles of one substance dispersed through a second substance. Colloids are classified as homogenous mixtures because the particles (generally of sizes $1-1000\:\mathrm{nm}$) do not settle out and cannot be separated out by ordinary filtering or centrifuging like those in a suspension, though the particle sizes are still greater than those in a true solution.

"particles do not settle out and cannot be separated out by ordinary filtering or centrifuging" that is factually incorrect. Many unstable systems that considered textbook examples for colloids, yet can be separated.


text:
What does chemical bonding look like at the quantum level?
How does chemical bonding look like on the quantum scale per se since the electron is a wave, and a particle?

It's "per se", and you can't see it.

Electron is neither a wave nor a particle; so you are really mistaking by saying electron is a wave. Wave is probability distribution function which assigns probability to place for finding the electron.

I believe this question is way to broad. Theoretical and computational chemistry are the two major branches that look into these interactions. Chemical bonding is nothing you can cover in a few words, I am not even sure if it can be understood, I am almost certain there is no way to picture it or make an image that you can see (just as orthocresol already mentioned).

I believe that generally in chemistry the wave characteristics of the electron are far more prevalent/relevant that the particle characteristics. So it's usually easiest just to think of it as a wave (or I think "region of space" is easier).
A chemical bond looks like the electron wave or area spread out between two or more atoms. This shape of the wave has lower energy than any of the orbitals it could occupy on just one atom. Because it has lower energy, that means it needs energy from somewhere for the electrons & atoms to change back to a different configuration. So unless it gets some more energy it cannot change away from the bonded state. That explains why chemical bonds are stable and hold atoms together.


text:
Is there a general term for a vertex in a skeletal formula?
When a carbon atom is single bonded to a residue, it is referred to simply as "methyl." When it is single bonded to two attachments, it is methanediyl, or methylene bridge. When it is single bonded to one atom and double bonded to another, it is called methine. Is there a generic term for the class of single-carbon-atom groups of the form  $\ce{CH$_n$}$, or to an unmarked vertex in a skeletal formula with sufficiently many hydrogen atoms to satisfy valence?

The generic term is "carbon atom".

Even when the vertex also denotes accompanying single-bonded hydrogen atoms to the carbon atom? It is only referred to as a "carbon atom"? Couldn't that be potentially ambiguous (confused with a carbon atom without any such hydrogen atoms) and misleading in certain contexts (such as calculation of molecular weights, treatment of point groups, etc.)?

You might say methaneXyl, where for n>1: X is numerical prefix from n, for n=1: X is  "-1-" (to prevent "disharmony"); i.e. methane-1-yl, methanediyl, methanetriyl, but these certainly are not preferred, and probably not generic in a way you ask for. Do you have a real case context where you did want to use it?


text:
Excitation Source for visible emitting Quantum dots
I read that the best excitation source for visible emitting quantum dots is UV light of wavelength $\mathrm{360~nm}$ to $\mathrm{380~nm}$. But why should the excitation source have lower frequency than the desired emission maximum?
If the excitation source have lower frequency, the energy of photons in the source will be lower. If that is the case wouldn't it have insufficient energy to excite the electrons from valence to conduction band?

UV is higher energy photons than visible.

You're conflating frequency and wavelength. Shorter wavelengths have higher energy per photon.


text:
Boiling point and vapor pressure
Does a liquid having higher vapor pressure have a lower boiling point?
I think so because I read in wikipedia that, "A substance with a high vapor pressure at normal temperatures is often referred to as volatile." A volatile liquid has lower boiling point.

Do you know the definition of "boiling point"?

Temperature at which vapour pressure is equal to the external pressure.

So would a compound with a higher bp or lower bp reach this first?

consider two substance A and B.If A has more vapour pressure than B then you can think it as,A has more particles in vapour phase than that B has in the vapour phase.Which means A will attain a vapour pressure which is equal to external pressure more quickly than B.Since A and B are same temperature (say 356K) and since A attains a vapour pressure equal to external pressure more quickly than B =>(implies) A boils quicker than B which implies A has lower boiling point than B.

Not necessarily. The vapor pressure of a liquid (or solid) is the pressure of the vapor in equilibrium with its condensed phase in a closed system. The boiling point of a liquid at a given pressure is the temperature at which the vapor pressure is equal to the given pressure. This means the vapor pressure is a function of the temperature,
$ P^* = f(T) $, and the boiling point is a function of the pressure, $ T_{bp} = f(P) $.
So your question is a bit vague. In general, yes, liquids with higher vapor pressures will have lower boiling points. In any first year chemical engineering course, students are introduced to the Antoine Equation, an emperical relation between vapor pressure and temperature. Plotting the Antoine equation for a set of liquids lets one visualize the relationship between vapor pressure and temperature very quickly; these plots are called Cox charts. For example, examine the Cox chart below:

Now look at the relationship between ethanol and benzene. At 50 F, benzene has a higher vapor pressure than ethanol. At pressures below ~10 psi, the general trend holds and benzene has a lower boiling point. However, the lines for benzene and ethanol cross at ~10 psi, so above ~10 psi benzene actually has a higher boiling point than ethanol.
So in general, yes, liquids with higher vapor pressures will have lower boiling points. However, the question is vague and exceptions do exist.

Wait, so when I out toluene/water on high-vac, the toluene boils first?! Wow.

Yes, the higher the vapor pressure the more volatile it is and the lower the normal boiling point.


text:
Why can mercury(I) exist, but not zinc(I)?
Mercury shows variable valency while zinc does not.
Its electronic configuration is $\ce{[Xe]\:4f^14 5d^10 6s^2}$. So it can donate the $\ce{6s^2}$ electrons and should only be able to form $\ce{Hg^2+}$, right?

How can it show $+1$ oxidation state?
Why does not $\ce{Zn}$ show $+1$ oxidation state?

On the contrary, zinc(I) compounds do exist, though they are rare, and relatively unstable. Most zinc(I) compounds contain a $\ce{[Zn2]^{2+}}$ core, which is analogous to the $\ce{[Hg2]^{2+}}$ cation. The $\ce{[Zn2]^{2+}}$ ion does, however, rapidly disproportionate into zinc metal and zinc(II), and has only ever been obtained by cooling a solution of metallic zinc in molten $\ce{ZnCl2}$ and in a few compounds such as decamethyldizincocene.
This should show you that the chemistry of zinc and mercury really aren't all that different. The $\ce{[Hg2]^{2+}}$ ion's much greater stability relative to $\ce{[Zn2]^{2+}}$  is due to the atypically large ionization enthalpy of the $\ce{Hg}$ atom ($100.7\:\mathrm{kJ\:mol^{-1}}$ greater than $\ce{Zn}$), which in turn due to relativistic stabilization of the $\mathrm{6s}$ orbital. Relativistic effects account for a myriad of the properties that are characteristic of heavy metals, including why mercury is the only metal that is a liquid at room temperature, and even gold's color.

Why does $[Hg2]2+$ exist?

@ArchismanPanigrahi see this-http://chemistry.stackexchange.com/questions/37141/why-does-mercury-form-polycations/42131#42131


text:
Predicting the products of a redox reaction
How can you predict how a given reactant will be reduced/oxidized? How can you predict how a given reactant will be reduced/oxidized? I would like to be able to do this without being given an electrochemical series chart. How can I then complete the half reaction?

In effect, you are asking how to predict the products of an unknown reaction, given the reactants. This question is as wide as chemistry itself.

then can u suggest me a book or give me a reference link

https://www.khanacademy.org/science/chemistry/oxidation-reduction


text:
Is there a way to Determine if an Element can undergo Disproportionation?
Can we determine the lowest and highest possible oxidation numbers of an element?
Since if we know this, then it is easy to determine whether the element will undergo disproportionation.

Well, if you want to determine if an compound will undergo disproportionation, looking at a Latimer Diagram is a really quick way to find out. I don't think you can really look at a compound and say it will undergo disproportionation without any knowledge of its thermodynamic properties. This link provides a good introduction to Latimer Diagrams: http://chemwiki.ucdavis.edu/Textbook_Maps/Inorganic_Chemistry_Textbook_Maps/Map%3A_Inorganic_Chemistry_(Wikibook)/Chapter_04%3A_Redox_Stability_and_Redox_Reactions/4.3%3A_Latimer_and_Frost_diagrams

@Nanoputian You really should expand this into an answer.

Expanding on my original comment, I disagree with the OP's statement that knowing the possible oxidation states of an element will easily allow you determine if a disproportionate reaction will occur.
To predict such a reaction, thermodynamic data such as the $E_\mathrm{cell}^\circ$ of the two reactions must be known. Latimer diagrams are a really simple and compact way of showing the $E_\mathrm{cell}^\circ$ values for a variety of reduction and oxidation half-equations between different oxidation states of the same element. Below is an example of a Latimer diagram for manganese in acidic medium:

Just from this I can tell the $E_\mathrm{cell}^\circ$ potential for any oxidation or reduction reaction between any pair of manganese species (in fact there is extra information that isn't even required). For example by looking at this diagram I can tell what $E_\mathrm{cell}^\circ$ will be for the reduction of $\ce{Mn^7+}$ to $\ce{Mn^3+}$ (I won't go into how you would do that in this post, I suggest that if you want to know how to, have a look at the website that I have linked).
But wait, there is more! Now the neat thing about Latimer diagrams is that I can easily tell if a particular oxidation state of manganese is stable or not (in other words, will it undergo disproportionate reaction?). All you have to do is follow one very simple rule:
'if the number on the left is lower than the number in the right, the species is unstable to disproportionation'
For example, let's consider $\ce{Mn^3+}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO2}$ to $\ce{Mn^3+}$ is $0.95~\mathrm{V}$ (that is the number on the 'left') while the $E_\mathrm{cell}^\circ$ for the reduction of $\ce{Mn^3+}$ to $\ce{Mn^2+}$ is $1.51~\mathrm{V}$ (this is the number on the 'right'). Since the number on the left is smaller than the number on the right, $\ce{Mn^3+}$ will spontaneously disproportionate to form $\ce{MnO2}$ and $\ce{Mn^2+}$. The reaction is below:
$$\begin{align}
\ce{2Mn^3+ + 2H2O} &\ce{-> MnO2 + Mn^2+ + 4H+} & E_\mathrm{cell}^\circ = +0.56~\mathrm{V}
\end{align}$$
Note how for this reaction $E_\mathrm{cell}^\circ > 0$, hence it is spontaneous.
However these 3 species that are involved don't have to be right next to each other. For example, let's consider $\ce{MnO3-}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO4-}$ to $\ce{MnO3-}$ is $0.419~\mathrm{V}$ while the E cell for the reduction of $\ce{MnO3-}$ to $\ce{Mn^3+}$ is $2.61~\mathrm{V}$. Again, the number on the left is smaller than the number on the right, hence $\ce{MnO3-}$ will disproportionate spontaneously.
This may not be the answer that you were expecting originally, but hopefully this will help you with your original questions.

just a tip: mhchem on chem.SE is a little odd, so $\ce{Mn^{3+}}$ gives a weird space between the 3 and the +: $\ce{Mn^{3+}}$. If you omit the curly braces it looks better: $\ce{Mn^3+}$ renders as $\ce{Mn^3+}$.

@orthocresol thanks for the tip, I always thought you needed the curly brackets. It definitely looks much better now.

The primary and secondary question are in fact two separate issues since we can artificially bombard elements or ions with photons or electrons to try to get other oxidation states. So in principle I can force into existence Na2+ but whether or not it survives in nature is a separate issue. Hence the 'highest' and 'lowest' oxidation states are rather arbitrary, depending on how much energy you want to put into the endeavor.
However, we have experimentally determined the typical oxidation states of elements in the periodic table and Compound Interest has made a version of the Periodic Table which shows such common oxidation states available
Whether or not the said element undergoes disproportionation then becomes a question of thermodynamics and the specific reaction in question.

The concept may sound strange, but it does happen.  Alkali metals such as sodium can be combined with an appropriate chelating ligand which forms a stable complex with $\ce{Na+}$.   Then where is the negative charge? On sodium anions!


text:
How do the atomic orbitals look like in reality?
We all know about atomic orbitals and their 3D structure. But the structure in theoretical. What does these orbital really looks like

Well, what does a mathematical function look like in your idea?

I wanated to know how does the electrons spin in an atom. I mean what does it look like when the electrons spin... The theoretical models shows how they spin... But i wanna know how do they looks like when they spin

This theoretical model is the best we got. Sure, it would be nice to look at the real orbital with your own eyes, but you can't do that anyway.

The electrons don't spin around the nucleus. They just exist somewhere in the general vicinity of it and the probability of finding one at any particular point is quite well described by the quantum mechanical description.

You cannot "see" an electron to determine whether it is spinning or not. The "spinning" of the electron is not measurable; it makes no sense to speak of it in science. However, you can measure an electron's angular momentum; it makes sense to speak of angular momentum in science. Therefore, don't think of the electron as a "spinning" object (which we can never know or observe); think of it as simply having "intrinsic" angular momentum.

It's not just theoretical. That's like saying the shapes of protons and atoms is theoretical. We have seen them with scanning tunneling electron microscopy! Are you saying electrons are theoretical also? What these orbitals really look like is a hybrid of the basis orbitals (not a theory) that follow VSEPR guidelines.

Atomic orbitals are just the solution of Schrodinger's wave equation. They help us to understand better the behaviour of electrons. Electrons behave more like waves and atomic orbitals can be thought of as places where the probability of finding those electrons is maximum. In reality, there is no such place as an atomic orbital.

The question is not what equation is associated with the eigenvalue that describes orbitals. It does not ask about probability as you are assuming the question asked about exact shapes. And in reality there are such places as atomic orbitals, if not matter could collide and pass through itself.


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Using Dalton's law of partial pressure to find percentages of composition

A mixture of helium and neon gases is collected over water at $28.0 \rm~^\circ C$ and $745\rm~ mmHg$. If the partial pressure of helium is $368\rm~ mmHg$, what is the partial pressure of neon? (Vapor pressure of water at $28\rm~^\circ C =28.3~mmHg$). Calculate the weight percentage of He and Ne in the original mixture.

The question is based on Dalton's law of partial pressure. I got the first part right.In the second part, since we have to deal only with the original mixture, total pressure should be $$745-28.7=\pu{716.7 mmHg}$$ So we calculate the total moles in terms of "v" (total volume) and then moles of neon and helium in terms of "v" from which we calculate the weight of neon and helium and calculate the weight percentage. "v" will eventually cancel out and we get weight percentage of neon as $34.54\%$ and that of helium as $65.45\%$ but that is not the right answer.
In the solution to this question, they took the total pressure while dealing with the second part as $\pu{748 mmHg}$ only but why would we do that. We are only talking about the original mixture. Where am I wrong?
Answer in my textbook : Neon weight% = $82.58 \%$, helium weight% : $17.43\%$

Sorry, I wrote hydrogen in place of neon. The question does not mention any other detail besides these, and I don't think we need any more details. Rest of the variables can be assumed as they will eventually cancel out.

OK.  What do you get for the mole fractions of helium and neon in the original mixture (excluding the water)?

Weight % = 100*(weight of substance/ weight of SOLUTION),so whats your doubt?

@SujithSizon The question is not about simply applying the weight % formula. All I want to know is why should we take total pressure as 748 mm Hg when we are just concerned about the original mixture. The vapour pressure of water was just given to solve the first part, it should not have any relation with original mixture.

@ChesterMiller I already mentioned it; weight percentage of Neon=34.54% and Helium=65.45%

@bluebellae by 748 you mean 745 ?

@SujithSizon Oh sorry, that was a mistake. I mean 745.

@SujithSizon We will need the w/w relation but that is not the integral part of the question. The question, according to me, is not framed to test your knowledge about this formula.

@bluebellae the w/w formula is based on the "solution" of the mixture so shouldn't we take the total pressure of the solution (=745)? (Also while doing salt analysis you write O.S. instead of X solution right? )

@SujithSizon Even if I had not known this formula, to calculate the weight of a substance in a compound, I would have used : (weight of that substance /total weight)* 100 using some simple maths. Why you are so fascinated by that formula? Also, we will need the total pressure, but that will be the one in the original mixture because that is what is asked.

@bluebellae its a solution you have to include partial pressure of water

@SujithSizon Who said it is a solution? I said it is a mixture.

Let us continue this discussion in chat.

First find the mole fraction of each gas in the original mixture using the formula, P(gas) = X * P(tot). For He:
$$X = \frac{368}{716} = 0.514$$
For Ne:
$$X = 1 - X(\ce{He}) = 0.486$$
Say you have 1 mol of mixture then you have 0.514 mol He and 0.486 mol Ne.
Converting these to weights using periodic table:
$$\rm 0.514~mol~He = 2.06~g~He \\\
0.486~mol~Ne = 9.81~g~Ne$$
The total mass of your mixture is the sum of these two masses, 11.9 g. Divide the individual masses by the total mass to get mass percent.
$$\rm\% w. He = 2.06/11.9 = 17.3 \%$$
Subtracting this from one for Ne (or you could apply the formula again):
$$\rm\% w. Ne = 1 - 17.3\% = 82.7\%$$
Note: I didn't carefully carry sig figs but you should!


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Radicals' formation
If carbon atom is tetravalent while oxygen atom is divalent so when they combine together the result should be $\ce{C2O4}$ then how are there compounds of formulas with $\ce{CO2}$, $\ce{CO3}$  and others? How do they bond?

$\ce{CO2}$ is not a radical.

Yes sorry i got confused .. but i want to ask about their formation regarding valencies

Could you please tell me first what you would consider the definition of a radical?

Group of atoms that act like one atom during the chemical reaction .. ? I don't know much, i am asking so may someone correct my information about radicals

This is the lewis structure of $\ce{CO_2}$.Each bond symbolises of sharing of two electrons.One from oxygen and other from carbon.

You can imagine $\ce{CO_2}$ as such.
Your confusion

Are you thinking something like this should be happening?
Formation of $\ce{CO_2}$ can be calculted by transfer of valencies but cannot be explained.
And there are other compounds which cannot be calculated by valency transfer but can be explained  by Kossell-Lewis approach.

Transfer of valencies only calculate and predict the formation of a compound.It does not explain.
$\ce{CO_2}$ is not a radical

In chemistry, a radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electrons. Like this.