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What is the physical significance of molecular partition function?
What is the physical significance of molecular partition function?
I reason, that the molecular partition function
\begin{align}
q &= \sum_i \exp(-\beta\varepsilon_i),&
\beta &= \frac{1}{k_\mathrm{B}T}
\end{align}
is found in the expression for the populations of each configuration, derived from the Boltzmann distribution:
$$p_i = \frac{\exp(-\beta\varepsilon_i)}{\sum_i \exp(-\beta\varepsilon_i)}$$

The partition function $q=\sum_i\exp(-E_i/k_BT)$ in your question can be regarded as the effective number of levels accessible to the molecule at a given temperature. It also means that in the equilibrium distribution the partition functions tells us how the systems are partitioned or divided up among the different energy levels.
In determining the form of the Boltzmann distribution the total number of 'systems' is N and these are divided up as $n_1$ systems have energy $E_1$, $n_2$ have energy $E_2$ etc. subject to conditions $\sum_i n_i=N$ and $\sum_in_iE_i=E$ where E is the total energy. It is found overwhelmingly that the chance that $n_i$ states have energy $E_i$ is given by $n_i = N\exp(-E_i/k_BT)/q$
In practice the partition function acts as a normalisation term in the calculation all sorts of thermodynamic properties, for example the average (internal) energy of N molecules with quantum energy levels $E_i$ is
$$ U=N\frac{\sum E_i e^{-E_i/k_BT}}{\sum e^{-E_i/k_BT} }$$
and the calculating the entropy $S=-Nk_BT\sum p_i \ln(p_i)$ where $p_i=n_i/N =\exp(-E_i/k_BT)/q$  is the probability that the ith state is occupied.
[ If the states are degenerate, with degeneracy $g_i$, then the partition function is $q=\sum_i g_i\exp(-E_i/k_BT)$.]

In the simplest terms and most convenient definitions, it represents the total amount of states that the energy can be in.
The probability expression you wrote emphasizes this. A probability is a part over a whole. So in the case of Boltzmann's distribution, it is one state over the sum of all the states.

I downvoted this answer because $Q$ doesn't represent the total number of possible states.

We first come across the canonical partition function $Q(\beta) := \sum_i e^{-\beta E_i}$ as a normalization constant for the probability that a given microstate is occupied, as given by the Boltzmann distribution. It is the Boltzmann distribution that allows for all the remarkable properties of the normalization constant $Q$.
We mention the two most important properties. For one, $Q$ is (proportional to) a moment-generating function of the distribution of microstate energies: $$Q(\beta) := \sum_i e^{-\beta E_i} = \sum_{E_i} \omega(E_i)\,e^{-\beta E_i} = \Omega\sum_{E_i}\frac{\omega(E_i)}{\Omega}\,e^{-\beta E_i} = \Omega\,\langle e^{-\beta E}\rangle, \quad \Omega := \sum_{E_i}\omega(E_i).$$ Recall that a moment-generating function is useful because it encodes all possible information about a given distribution; hence $Q$ contains all information about (the energy distribution of) our system. This extends to functions of $Q$, particularly the Helmholtz free energy $A = A(Q)$.
For another, differentiating $\ln Q$ with respect to $\beta$ will generate moments of the Boltzmann distribution. This is unique to the function whose derivative is itself, the exponential function. (We differentiate $\ln Q$ in place of $Q$ in order to get a factor of $1/Q$ to generate normalized probabilities.) This is a very useful and important technique, but I'm not sure that there's an associated physical significance.
Do note that I describe two different distributions in my answer: the distribution of microstate energies and the distribution of microstate occupancy. The canonical partition function is the moment-generating function of the former only. However, it is, in some sense, a generating function for the latter distribution, by the second property above.

It is a dimensionless quantity and provides the most convenient way for linking the microscopic property of individual molecules(such as their discrete energy level moment of inertia and dipole moment) with the macroscopic property (such as molar heat, entropy and polarisation)
From Maxwell Boltzmann distribution law, it can be written as
\begin{align*}
N_i/n &= \frac{g_i}{g_o} e^{-E_i/KT} \\
Q &=n/n_o
\end{align*}
The value of molecular partition function increases with increase in temperature.


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A glass of water with ice-cubes in it. Where's the water the coldest; at the top or bottom?
Suppose that I fill a glass with ice water.  As the ice melts, it cools the water around it.  Given that cold water is denser than hot water, I would presume that the cold water would sink to the bottom … but it would warm as it sinks, reducing the density.  Meanwhile, the ice is still melting and giving off its cold to the surrounding water.
So is drinking ice water with a straw going to get you cooler or warmer water than drinking from the lip of the glass?

https://en.wikipedia.org/wiki/Grashof_number

Hm, but the ice is near the top... shouldn't the water immediately surrounding it be colder than water farthest from it? Cold water isn't lead, it takes a while to sink...

No stirring with a significant amount of just water from melting ice and much ice floating on the surface. The temperature of the ice is 0 Celsius or lower. Since the ice is melting the surface T of the ice is 0 C and the water surrounding it is infinitesimally lower. Any water removed from the ice must again be slightly warmer the warmer water   reaching 4 C will be at the bottom. The water at the top in contact with the ice is the coldest water so sip from the top to get cooler water; a straw from the bottom to get 4 C water. [provided enough ice to maintain 4 C.] Stir well for cold thruout

The water surrounding the ice is slightly warmer not lower. It pays to proofread!

Interesting question!
A few things first:

As the ice melts, it cools the water around it.

Technically, the ice cube melts because the water cools down. This may sound ridiculous at first, but you must consider the fact that the ice melts because it has drawn "heat" (energy) from its surroundings. The "surroundings" being the air and water that surround it (but the water's more important since it's a better conductor of thermal energy).

Given that cold water is denser than hot water, I would presume that the cold water would sink to the bottom...but it would warm as it sinks, reducing the density.

You're right, cold water is denser than hot water. It is helpful to note that it shouldn't be too cold though. As the temperature of water drops to 4 °C, the density of water gradually increases. However, as the temperature drops below 4 °C the density of water actually begins to decrease and water in this range easily "floats" over water in the room temperature range.

Meanwhile, the ice is still melting and giving off its cold to the surrounding water.

Ice isn't giving off its "cold", rather, it takes in the water's "heat" (thermal energy).

Back to your question.
As Max mentions in his answer, you have done a particularly good job of indicating what physical parameters we're dealing with; the really important ones being the temperature of ice, temperature of water (at the time you put the ice in) and the quantity of ice used (at least with respect to the water).
But assuming you're drinking water (originally at room temperature) out of a 250 ml styrofoam or plastic cup, and you used two (normal-sized) ice cubes and that you began drinking the water a minute after you plonk in the ice cubes, the water should be colder at the top than at the bottom.
Consider minute, imaginary layers/regions/packets of water in the cup (thinking about this in terms of water "packets" rather than water molecules is easier to comprehend). Also, think of the cup as having three (crudely demarcated) regions: Top, middle, and bottom.
Packets of water immediately adjacent to the ice cubes are in thermal equilibrium with the outermost regions of the ice. However, these packets soon gain some thermal energy from other water packets that are adjacent to them. So as these packets slowly rise in temperature, from zero degrees to past 4 °C, they sink and new packets occupy locations adjacent to the ice. The cycle repeats for as long as the ice is there.
Now, as those packets of ice sink, they gain more thermal energy from the packets of water they come in contact with on their way down. This, coupled with the viscous effects of water results in the mild "warming up" of the sinking packets.
Now since they warm up a bit, they tend to rise back up. Back at the top, they're get cooled and sink again. This process repeats for as long as the ice remains in the water.
Take a step back, and you'll see that the middle of the cup ought to be cold, the bottom of the cup ought to be colder, and the top of the cup is the coldest.
So even if the ice cubes aren't actually touching your lips, you'll find sipping the water at the top to be colder than sucking out water from the bottom through a straw.

The convection to produce uniformity depends on a number of nebulous factors:

How much ice?
How tall is the glass?
Diameter of the glass?
Is the "glass" really a glass or paper cup, styrofoam cup, or perhaps a metal cup?
Initial temperature of the water.
Mass of water to mass of ice.

The gist is this. Lakes don't freeze solid in the winter. Without stirring, the lake water forms layers that mix very, very slowly. So the frozen ice floats on top of the lake.
So without stirring, the water at the lip is likely to be colder than the liquid at the bottom of the glass.

EDIT - This is an experiment that you can easily do at home. Fill a large clear glass (a real glass glass...)  about 2/3 full with tap water and add food coloring with stirring to make the liquid fairly dark. Then add enough ice to fill the glass without stirring. Let the glass sit on a solid counter until the ice melts. (Nice experiment to run overnight....) Because the water has layered, the layer of water on top will be noticeable lighter in color than the layer on the bottom.

Heat flows by radiation, conduction and convection. First, I have to quibble over the term "ice water". Ice water may or may not contain ice. What it clearly implies is that the water is at 0°C. Consider a system where there is minimal air flow (hence minimal air-water conduction). Consider a system where the water (here water = liquid water) is at 0°. Ice is typically created at between -15°F and +25°F, depending on the freezer. It is in all cases, below 0°C (obviously). Now what happens in an ideal case when I put a "tiny bit" of ice into water at 0°C? Yup, the water all freezes. Now of course because the real world is not continuous, this won't really happen and it is possible to construct a system in which water and ice are at equilibrium at 0°C - but it is definitely not "easy".
I mention the above to show you that you've failed to adequately specify your system. The masses and temperatures of the water and the ice matter. As does the temperature of the air, the temperature of the container, and the temperature of the bench/table. With a sufficiently insulating container, you still have to worry about the mass flow (currents) you've given to the water by pouring it. (These currents may take hours and possibly days to completely dissapate - even under near isothermal conditions!) And of course the shape of the pieces of ice matters. Consider a (perfectly insulating) capillary tube and a shallow tray with the same total volume capacity. The mass-flow and the thermal flow will be much different. Hydrodynamic calculations can be (and almost always are) enormously difficult. The water can't be any colder than 0° (ignoring supercooling). That clearly must occur in contact with the ice. But consider a cylindrical cup 10 cm in diameter (i.d.) Do you think that the answer would be different if I placed a 1 cm3 ice cube in 1 L of water in that cup OR if I placed a 1 cm thick disk with an diameter (o.d.) of 9.9cm into the cup? Yeah, under-specified system, big time. Another obvious missing piece is what you mean by temperature. Not in the sense that it's not a well defined (macroscopic!) property, but in the sense that you are (probably) thinking of the average temperature over some volume. As I said, the coldest water will be found "in contact" with the ice, but this doesn't mean the entire surface will be on average colder than the bottom 0.1 cm in the cup since it obviously depends on things you've not specified. Imagine as a final "monkey wrench" in the attempt to answer this, that the cup contains an insert and that metallic insert has been cooled to -200°C. You pour the 0°C water in, along with some unspecified amount of ice and you get some freezing of the water on the entire inside surface of the cup. Now even if we include this ice when we specify the volume and temperature of the ice, do you really think this ice would have the same effect as floating cubes? Or consider a ring of ice instead of a disk, and what about the surface roughness of the cups lining?

When the temperature goes down the kinetic energy decreases, the volume is decrease and then density goes up. Therefore solid is sink and liquid is on top.
It is related to the water but only up to the 4 °C. When there is water 10 °C, the temperature of the water goes down to 9 °C, the 9 °C water will go down because of the more density.
But when we take 4 °C water temperature goes down to 3 °C. It is different. You know that there is hydrogen bonds between each $\ce{H2O}$ molecule. The shape of $\ce{H2O}$ molecule is 'bent'. The two hydrogen atoms can make two hydrogen bonds with two oxygen atoms, and two lone pairs of electrons of the oxygen atom makes two hydrogen bonds with another two molecule's two hydrogens. Then it can have a tetrahedral shape. You know that solid has a regular structure. (Liquid) water has no regular shape, there are molecules and hydrogen bonds everywhere irregularly. When the temperature goes down from 4 °C as to the regular shape the tetrahedral $\ce{H2O}$ molecules tries to increase the distance and come to a regular shape. So, the volume goes increase and the density goes down. Then after 4 °C to down temperatures (3, 2, 1, 0, −1) the (water) floats on the top.

For the cold water to sink to the bottom of the glass, it has be colder than the water at the bottom that is displaced. So yes, the coldest water will be at the top.

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text:
Which organic compounds have the most strongly negative enthalpies of formation?
I've been using the NIST Webbook to list organic compounds and look at their heats of formation (kJ/g, not mol). If I limit myself to three carbons, I still get over 400 results.
I can use some heuristics to guess which ones have strong heats of formation. I'm interested in kJ/g rather than /mol so smaller molecules are more likely, and certain bonds like $\ce{C=O}$ help, but that's not good enough for my project because I need to be certain.
Oxamic acid, ammonium formate, and ammonium acetate are up there, but who knows how many I might be missing, limited as I am to looking up small molecules?
So I'm hoping someone with a lot of experience might tell me something off the top of their head.

Limiting yourself to three carbons seems like a way to bias yourself to things that contain large amounts of other atoms.  You may want to limit total atoms or total "heavy" (i.e. non-hydrogen) atoms instead of just C atoms.  The compounds you've found -- ammonium acetate, oxamic acid, etc. -- have more heavy atoms overall, and thus more bonds, and thus higher enthalpies of formation.

@CurtF. - Limiting the search at all is the very problem, because as I said, who knows how many I might be missing, limited as I am to looking up small molecules? (And by "small" I mean fewer than 12 or so atoms.)

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text:
Why does vibrational frequency influence quantum tunneling of a proton?

Furthermore, consideration of the tunneling in the framework of the Wigner theory assumes that the tunneling factor is small. However, the $\ce{O-H}$ frequency is very high ($\gg k_\mathrm{B}T/h$), and the H atom tunneling under a potential energy barrier when proceeding either from a ground or from other excited vibration levels may be significant.

Source: German, E. D.; Sheintuch, M. Kinetics of Catalytic OH Dissociation on Metal Surfaces. J. Phys. Chem. C 2012, 116 (9), 5700–5709. DOI:10.1021/jp2106499.
Why does a higher frequency influence the quantum tunneling effect?

It is not the frequency per se but the mass that is important,; the probability of tunnelling an energy E is given by
$$p(E)=\exp \left(-\frac{4\pi\sqrt{m}}{h^2}\int_{x_a}^{x_b}\sqrt{V(x)-E}\;dx \right)$$
where $x_a$ and $x_b$ are two points either side of the potential barrier through which tunnelling occurs. The integral is in effect a measure of the area of the potential $V$ above position $x_a$ to $x_b$ through which tunnelling can occur.
As m increases, for a given potential barrier and energy, the tunnelling probability falls.

But why does the author say it that way and compare it with kT/h?

@wangge Don't know other than $k_BT/h$ has units of $s^{-1}$ but in this case is not a frequency but units of a first order rate constant.

I believe the comparison to $k_BT/h$ is more commonly talked about as the dimensionless ratio $\hbar\omega/k_bT$. I have seen this comparison made quite a few times when discussing nuclear quantum effects, as it essentially measures whether or not we are in the classical regime where thermal effects will be most important. For something like $\ce{O-H}$, this ratio is quite large even at room temperature. Also note that this shows up in the tunneling rate given above by means of $E$ which will be quite close to the vibrational energy of the $\ce{O-H}$. Hence talk of the vibrational frequency.

@jheindel good point.


text:
Reactions of alcohol/ethers with hydrogen halides
General equation -
Alcohol - $\ce{ROH + HX -> RX + H2O} $
Ethers - $\ce{ROR + HX -> RX + ROH} $
I've been told that X = $\ce{Cl-}$  , $\ce{Br-}$ , $\ce{I-}$
Then I wonder ...

How Cl , Br and I nucleophile is going to Attack OH ?

All of the 3 are weaker nucleophile than OH as well as a better leaving group. So ...

How does it replace and substitute OH ?

Or

OR (with ethers)?

Is it $\ce{OH-}$ that leaves?

This reaction generally takes place in an environment where there are free $\ce{H3O+}$ ions. The oxygen atom in the alcohol or ether takes up $\ce{H+}$ from these ions. Now the leaving group is $\ce{H\bond{-}O+\bond{-}H}$ or $\ce{H\bond{-}O+\bond{-}R}$ which are much better leaving groups than $\ce{O\bond{-}H}$ or $\ce{O\bond{-}R}$. This makes the substitution of the halide ions easier.


text:
Why does mercury decolourise a gold ring?
I saw that a gold ring decolourised as it got in contact with mercury .
Why does this happen ? Is there any way to reverse this?

https://en.wikipedia.org/wiki/Amalgam_(chemistry) :3

If the ring was "decolourised" by mercury (amalgamating with the gold) then 1) The ring is gold plated, and 2) No, there's no simple way to "put back" the  gold on the ring. O:)

Can't I use nitric acid? Won't it atleast remove the grey colour ? The ring is actually gold and not gold plated.

@paracetamol I'm not sure why you think this is a sign the ring is gold-plated: mercury will damage any gold surface plated or not.

@matt That the amalgam would stick to the surface of the ring slipped my mind >_<. Hence I thought "discoloration of ring" = "Part of the gold surface is lost" :P

The grey colour is an amalgam of mercury and gold.
Mercury forms amalgams with many other metals. Some are used as chemical reagents in laboratory chemistry as they have different properties than the original metals involved. Gold amalgam is much greyer than gold. Silver amalgam has been used in dentistry.
Mercury has been used in the extraction of gold in mining as it can extract gold from low grade ore. But the recovery process involves distilling off the mercury which is dangerous and polluting.
Chemists used to work with mercury a lot and, when they did, they were recommended to remove their wedding rings or other gold jewellery as the mercury would damage the gold.
Once the amalgam is formed there is no easy way to reverse the process short of resmelting the ring. If the contact with mercury has been brief, you might be able to polish it to remove the amalgam layer, but this won't be easy.

Does immersing in acids or heating  upto 350 degrees in help anyway?like mercury alone vaporises and gol remains intact.

@Chit "Does immersing in acids or heating upto 350 degrees in help anyway?"  Uh...are you sure you want your ring back in one piece? ;)

As gold does not melt at that temperature, why can't this method be  used. Would it make the ring brittle?

@ChitraLekha you might drive off the mercury but you will still have a ring with a very damaged surface and you might poison yourself with the vapour. Try polishing first.

how do i polish it? And the damaged surface would be like  a eroded layer.Am i right?

If you're working with mercury, and your wedding ring is vulnerable, then you have a bigger problem than your wedding ring—you're working with mercury using your bare hands!

@ChitraLekha get a professional jeweller to do it if you are not already equipped to do so.

@CodyGray Working with mercury with your hands isn't that dangerous. In fact, you could probably drink it without much harm. The problem is mercury vapour which is absorbed into the body in ways liquid mercury is not. So the real risk comes from continuous exposure in a badly ventilated environment.

I'm no expert in metallic toxicology, but I'm pretty sure the claim that you could drink liquid mercury "without much harm" is completely bogus. It is true that elemental liquid mercury is less toxic than other forms, and that vapor represents the largest risk, but that hardly means it is "safe". (Plus, the liquid form does vaporize at room temp.) I don't know if I'm just unusual, but it is rare that my hands are completely free of cuts, so even if absorption through intact skin is very slow, I'd say the typical lab worker is still putting herself at undue risk for not taking precautions.

Once when I was doing an experiment I had the same experience; my ring was decolourised. I was afraid at first but we could reverse this process to achieve the gold colour once again.
There is no reaction between $\ce{Au}$ and $\ce{Hg}$; they form a mixture similar to dissolving $\ce{NaCl}$ in water. When the gold and mercury are mixed they make an amalgam (this process is called amalgamation). It is also used to allocate pure gold by colluvies.
Before I said that there is no reaction between gold and mercury, therefore there is an easy way to allocate the gold again. We know that the boiling point of gold is higher than that of mercury ($\ce{Hg}$: $\pu{356.7 ^\circ C}$ and $\ce{Au}$: $\pu{2700 ^\circ C}$). Therefore when you heat the amalgam the mercury will boil off and the gold surface will be restored. But don't do this in the house, give it to a professional goldsmith. (After my experience, I went to a gold shop, they heated my ring, and it became just like it was before.)

Actually mercury was used for volume measurement using displacement principle. If it were done in a muffle furnace would it work (I don't have the idea of trying it out , asked just out of curiosity)

did they just heat it to bring it back. After that was the layer smooth and still intact.

It is a little inaccurate to describe amalgamation as the same as salvation: it is a little more complex than that.

The boiling point of gold is irrelevant, here. If it //melts// your ring is lost already. Hopefully, 1064.18 °C for melting gold (wikipedia) is still far beyond the 356°C for boiling mercury. Not that it makes it a better idea to do at home.

Mercury is a stable element . It has the ability to amalgamate most metals. Whenever the grey alloy is heated, based on their boiling point with mercury having the smaller, it will boil first leaving the precious metal, gold behind.

The gold surface is simply "wet" by mercury and the gold is not decolorized. A tiny amount of mercury can quickly swarm across a gold surface making it appear that the color has changed almost instantly but the effect is superficial. When mercury was a common plaything, people would often coat silver coins with mercury. Mercury has the same effect on silver as it does on gold and the coins would become super shiny with a slick feel. After a few days, the mercury would either evaporate or oxidize leaving the coins badly tarnished. This is why it was common to find strangely dark looking silver coins in circulation. Mercury will leave the surface of a gold object in a few weeks, if not days, if left alone and the gold surface will return to normal. Heating the gold speeds the process but the gold is unharmed.

Nope, mercury doesn't just coat the surface, it forms an amalgam and that process isn't easy to reverse.

Thanks all for your active responses, actually the contact of the ring with mercury was very less. And it was just the top layer coated by amalgam .

Because the ring is starting to get back its original color after three full days. Actually no attempt was made to remove the layer like heating or dipping in acid. But when I saw the ring after 3 days(the ring is not mine) the layer was gone mostly.And I beleive if I leave it for some more days it will naturally get back its own color and lustre(doubtful about lustre)

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text:
Formation of zwitterion from amino acids
Why it is easier for the carbonyl group to lose a proton to become negatively charged and the amino group to accept a proton to become positively charged?
I know it has to do with the very polar bond due to the oxygen. And the $\ce{H+}$ ions attracted to the lone pair of electrons on $\ce{N}$ in the amino group.
But I don't understand why would an electronegative atom such as oxygen wants to lose a proton.
I was told that they will attract more electrons towards towards them. But that does not help in my understanding of why protons will leave.

It's not that the proton is "unhappy" on the oxygen, it's just happier somewhere else :)

The reason amino acids exist largely in their zwitterionic form at biological $\mathrm{pH}$ (usually around 7) is due to the $\mathrm{p}K_\mathrm{a}$ of the constituent groups. It's not that the oxygen 'wants' to lose a proton, but more that at that $\mathrm{pH}$ the equilibrium lies towards the deprotonated state (things are rarely 100% protonated/deprotonated).
In general, carboxylic acids easily lose their proton to form carboxylates (the anionic charge is delocalised over the three atoms making the proton loss 'favourable'), and amines act as bases, to pick up a proton.
At a $\mathrm{pH}$ of 7, the medium is basic enough to cause the carboxylic acid to be deprotonated, and the amine to be protonated. If you look at the diagram below, you can see that increasing or decreasing the pecan shift the equilibrium towards fully protonated (at low $\mathrm{pH}$) or fully deprotonated (at high $\mathrm{pH}$) amino acid.

Ionisation state of amino acids over 0-14 $\mathrm{pH}$ range, taken from Biochemistry, Berg (7 ed.)

You can see as, by the cleavage of proton the carboxylate ion becomes resonance stabilized which provides high stability and also due to the character of amino group as a lewis base it will be easier to attract the proton loosen by the carboxylic group.
So this type of reaction is proposed to explain the aspects such as high M.P.(Melting Point), B.P., etc. Because now it behaves as a salt( in the form of zwitter ion).

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text:
Chemistry of charging an insulator material
Earlier this year, I had learnt in Physics class that rubbing two different materials against each other can charge the materials (i.e. cause each material to either be positively-charged or negatively-charged). I had learnt that the charging process is an electron transfer phenomenon. The charging of conductors by rubbing (i.e. friction) is easily understood because conductors allow charge carriers to flow through them easily. Such as in the case of metals; adding more electrons to the sea of delocalised electrons is not difficult.
However, what happens during the charging of an insulator material, such as a polymer like poly(ethene), what happens to electrons which are transferred to these materials? Do they go into particular orbitals?
One thing I am sure of is that the electrons are surely localised to the region of the insulator they are transferred to. Thus, they are definitely not free moving.

This is a good question. However, I fear it may have to be closed as a duplicate of [https://chemistry.stackexchange.com/questions/76711/what-is-the-underlying-chemistry-behind-the-triboelectric-effect]. Your question is specifically about insulators, but the question I linked deals with triboelectricity in general (so your question could be a subset of that). If not this, then it is still a cross-site duplicate of this question [https://physics.stackexchange.com/questions/309653/charging-by-friction]. Like I said; good question...but not sure if it'll remain open as it stands. O:)

@Paracetamol I guess you are right that it is a subset of the question on triboelectricity. However, it is different from the one on Physics SE because mine discusses chemistry of the interaction, not of energy and forces.

You're getting sure too easily (last sentences). This isn't "normal" chemistry. Presence of charge means there are very reactive cation/anion radicals all the time.

@Mithoron Yes, I agree that there should be very reactive charged radicals and other reactive species in the insulator material. But we somehow cannot observe the reactions of these species. All we observe is that the negative charges present in the material repel negative charges in other materials. Also, regarding the part about localisation, I am indeed sure that the charges are localised. Because the teacher said that only the points of contact/rubbing were charged. He explained that the material lacked free charge carriers, since it is an insulator.


text:
Reaction mechanism for the synthesis of 9,10-phenanthrenequinone
In my lab course I had to synthesize 9,10-phenanthrenequinone (phenanthrene-9,10-dione) from phenanthrene and chromium(VI) oxide. The synthesis was made as explained in Organic Syntheses, Coll. Vol. 4, p.757 (1963)
The reactants are: phenanthrene (1.00 eq), chromium(VI) oxide (7.50 eq), sulfuric acid (15.0 eq), water (83.3 eq)
I now have to explain the reaction mechanism, but unfortunately I don't find it in any of my textbooks, neither on the internet.
Here is my assumption:

The chromium(VI) oxide reacts with water and forms chromic acid.
The chromic acid can oxidize an alcohol to a ketone (Jones-Oxidation)

So what misses is the alcohol. Somehow I have to get the 9,10-phenanthrenediol, which could react as alcohol in the Jones-Oxidation.
How could this possibly work. Remember, I don't have $\ce{OsO4}$, which could easily dihydroxilate an alkene.

EDIT 1: As an answer to the proposed mechanism by @Zhe:
Is this ok in your eyes?

EDIT 2: An other idea for the first step (dihydroxilation)
A person of the scientific staff of my university told me, that the $\ce{CrO3}$ should react in a very similar way as $\ce{OsO4}$ does. As I'm using 7.50 eq of chromium(VI) oxide, there is enough for the dihydroxilation and the Jones-Oxidation. He said that a radical mechanism is rather unlikely, because I'm working with water as solvent and water is an inhibitor for radical reactions. This is the mechanism I came up with, regarding these new information.

What do you think about this possibility?

This probably happens via a single electron oxidation  of that double bond to a radical cation as a first step. The cation can trap water and then we can oxidize the radical to a cation. Eventually, you get to a diol which you can oxidize to a diketone. I can suggest a more complete mechanism later if I time, but try that and see where you can take it.

@Zhe Thanks for your comment. Would you please explain this a bit more? I only know the single electron oxidation for carbonyl compounds and I don't find anything about alkenes. Thanks in advance!

That's not a bad idea. Pretty similar to how $\ce{OsO4}$ works. Again, without more mechanistic evidence, it's hard to know for sure, but I find this quite plausible.

Here's what I was thinking.

This probably happens via a single electron oxidation of that double bond to a radical cation as a first step. The cation can trap water and then we can oxidize the radical to a cation. Eventually, you get to a diol which you can oxidize to a diketone.

I ommited the steps for chromium oxidation from the alcohol to the ketone. Sounds like you are already familiar with those.

Thank you very much. How is the first step induced? How do you get the cation and the radical?

Chromium(VI) is a strong oxidant, so I assumed it can just pick up an electron. Granted, that is not how the oxidation of alcohols to ketones works, so I may want to come up with a mechanism that involves a $\pi$-complex with the chromium.

I edited my question. It would be great if you'd have a look at it.

Not sure that the SET would behave as drawn. There may be some significant changes regarding the oxo groups on chromium. Again, this is just a guess, and I think it's reasonable, but I wouldn't declare it correct since I don't have any mechanistic data. I'm also considering using the olefin a $\eta^{2}$ ligand, but I don't have a productive way of eliminating the chromium that I like. That's not to say that that mechanism isn't better. The best bet would be to look at the literature more thoroughly for mechanistic studies. Sadly I don't have journal access anymore to help there...

As an undergraduate student I don't have any access to those either :( And in my textbooks, there is nothing that seems to help with this.

I made another edit to my question. Your criticism is greatly appreciated.


text:
How many litres of liquid CCl4 (Density= 1.5g/cm^3 ) must be measured out to contain 1 * 10^25 Cl atoms?
Here's what I did-
No. of atoms of Cl - 10^25
No. of molecules of CCl₄ - 1/4 * 10^25 or 25 * 10^23
No. of moles = No . of particles/ Avogadro's No.
So (n) = 25*10^23/6.02(approx) *10^23
   (n) = 4.1
Also, No. of moles = Vol of gas at NTP/ 22.4 liters
so 4.1 = x/22.4 ======= x= 91.84 litres
But the answer in my textbook is .426 liters. What am I doing wrong?

Um, $\ce{CCl4}$ is a liquid. Why did you take volume of a gas? Besides, you're given the density of $\ce{CCl4}$. Find the mass first, then proceed.

I'm voting to close this question as off-topic because it's homework without sufficient input from OP

The question asks for how many litres of LIQUID $\ce{CCl4}$. Note what you have done is for GAS. So the steps would be:

Find the amount of substance of $\ce{CCl4}$ required.
Multiply it with its molecular weight (154 g/mol).
Use its density to find the volume.

Hope you understood your mistake.


text:
Can I obtain a different density of a recycled elemental material?
I am recycling Aluminum products/scrap. I melted it and then form cylindrical ingots. When I measure their density, I calculated $\pu{2.4 g/cm^3}$ but Al has density of $2.7$. I think all the materials are aluminum since they all melt in the same temperature using a home setup. Can I claim that the ingot is Al?

I think it is risky. The density observed lower than the expected density of pure Al is indicative for the presence of something of even lower density than Al itself.
Similar in appearance to aluminium (density $\pu{2.7 g/cm^3}$, melting point $\pu{660 ^\circ{}C}$, source) is magnesium (density $\pu{1.7 g/cm^3}$, melting point $\pu{650 ^\circ{}C}$, source).  Both are used, pure as well as in mixture of other metals (alloys), widely.  Even to the point that they are mixed to form alloys with each other, for example in the frames of laptops

(source)
or items that are used near the sea / in contact with sea water (source).


text:
Why photons with longer wave length demonstrate more active scatter and transmit that shorter length?
Why photons with longer wave length demonstrate more active scatter and transmit that shorter length?  This troubles me to understand why NIR and Infra-Red goes through diffused media when compared to other colors. My thought is lean to think that be the other way around. For example, X-ray are short waves but goes straight though a biological tissue.  Please, comment


text:
Do molecules with a hexagonal planar geometry exist?
I recently read an article on Wikipedia about pentagonal planar geometry.
There are only two molecules with this kind of geometry $\ce{XeF5-}$ and  $\ce{IF5^2-}$  and it has two electron pairs at the axial positions.
As soon as I looked at this, I thought that hexagonal molecular geometry can also exist much the same way i.e. 6 substituents bonded to the central atom and two lone pairs at the axial positions opposite to each other.
On the page, its given that arrangement of 6 atoms in a plane is found in higher coordination states but that does not help either as they have not given any examples of that.
Do hexagonal planar molecules exist, and if they do, what are they?

I'd suspect that sterically theres just no chance of getting groups around a central atom

You could say that in an fcc or hcp metal has atoms hexagonally bonded to any central atom, if you look in the proper plane.

I think it's nearly impossible to find or synthesize a "canonical" complex with hexagonal molecular geometry, but in the field of host-guest supramolecular chemistry there are numerous examples of "unusual" geometry. Probably the most well-established class of such compounds are torands ("hosts") incorporating alkali metal cations ("guests").
Check out, for example: (1) Bell, T. W.; Cragg, P. J.; Drew, M. G. B.; Firestone, A.; Kwok, D.-I. A. Angew. Chem. Int. Ed. Engl. 1992, 31 (3), 345–347, DOI 10.1002/anie.199203451.
Here is an example of the structure with potassium ion from (1), Tri-n-butyltorand-potassium picrate clathrate, which I quickly sketched in Olex2:
Top view:

Side view:

Unit cell and packing:

They're fascinating complexes, thinking about it more, I guess that this is somewhat an extension to 18-c-6 bonding with a cation, I suspect that adopts a similar geometr

Yep, they inherit a lot from crown ethers, though I picked potassium and this particular class of torands because their combination allows for a planar 6-coordinated geometry. Na would be too small to properly coordinate, Rb is sticking out from the plane quite a lot. Classic crowns are very flexible, and therefore achieving planar geometry (what OP required) using them would also be tricky.

In October 2019 a series of new transition metal complexes has been reported in an article published in Nature by Garçon et al. [1].
Among these, two palladium complexes have a near-perfect hexagonal arrangement of three hydride and three magnesium-diisopropylphenyl ligands according to single-crystal x-ray diffraction studies:

tris([2,6-di(propan-2-yl)phenyl][4-{[2,6-di(propan-2-yl)phenyl]imino}pent-2-en-2-yl]amido)-trihydrido-palladium-tri-magnesium n-hexane solvate (compound 1a, CCDC 1909687);
tris([(2,4,6-trimethylphenyl){4-[(2,4,6-trimethylphenyl)imino]pent-2-en-2-yl}amido)-trihydrido-tri-magnesium-palladium (compound 1b, CCDC 1909688).

Interatomic distance were also supported by DFT calculations, which also showed that $\ce{Pd–Mg}$ bond is primarily ionic in nature.

Fig. 1 […] Single-crystal X-ray derived model of 1a, annotated with selected experimental and calculated bond lengths.

Some news agencies such as C&EN called this structure "first hexagonal planar crystal structure", however the authors of the article make an important addition:

The hexagonal planar coordination environment is known, but it is restricted to condensed metallic phases, the hexagonal pores of coordination polymers, or clusters that contain more than one transition metal in close proximity. […]
Here we report the isolation and structural characterization of a simple coordination complex in which six ligands form bonds with a central transition metal in a hexagonal planar arrangement.

References

Garçon, M.; Bakewell, C.; Sackman, G. A.; White, A. J. P.; Cooper, R. I.; Edwards, A. J.; Crimmin, M. R. A Hexagonal Planar Transition-Metal Complex. Nature 2019, 574 (7778), 390–393. DOI: 10.1038/s41586-019-1616-2.

It looks as though hexagonal planar molecules are currently unknown, but you can find a hexagonal plane in higher coordination structures like a hexagonal bipyramid.
One example of a structure with hexagonal bipyramidal geometry is $\ce{UO_2(OAc)_3}$, which I found here on a site which gives similar info to your link, but includes example structures for known geometries.


text:
Rutherford's gold foil experiment
Rutherford projected $\alpha$-rays onto the gold foil. $\alpha$-rays are the same as a $\ce{He^2+}$ nucleus and they can ionize atoms. Therefore, why are gold atoms not ionized when the $\alpha$-rays collide with them?

Since electronic stopping (interaction of ions with electrons) is well known, the answer is the alphas did lose energy to electrons in the film. These low momentum interactions spread the beam. The spread of the transmitted beam was measured by Geiger and Marsden.

In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.
But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.

If you have a new question, please use the Ask Question button and link to this question if necessary.


text:
In brown ring test, how is NO^- formed?
$$\ce{NO3- + 3[Fe(H2O)6]^{2+} + 4H+  -> NO + 3[Fe(H2O)6]^{3+} + 2H2O}$$
$$\ce{NO + [Fe(H2O)6]^{2+} -> [Fe(NO)(H2O)5]^{2+} + H2O}$$
I read that in the brown ring complex $\ce{[Fe(NO)(H2O)5]^{2+}}$ , $\ce{NO^{-}}$ is the ligand.
What I don't understand is that :
How is $\ce{NO^{-}}$ formed? Nitrate ion got reduced to NO while $\ce{Fe^2+}$ was oxidized to $\ce{Fe^{3+}}$. The electrons that were released in the conversion of ferrous to ferric ion were used in converting nitrate ion to $\ce{NO}$. So where do these extra electrons come up that convert $\ce{NO}$ to $\ce{NO^{-}}$ ? I don't want to go into the mechanism; I just want to have an intuitive understanding of what's happening in the reaction.

Your second reaction has Fe(II) being oxidised to Fe(III), as well.

Related: https://chemistry.stackexchange.com/questions/10641/what-is-the-structure-of-feso₄-no

https://chemistry.stackexchange.com/questions/73971/what-compound-is-formed-in-this-case/73974#73974

https://chemistry.stackexchange.com/questions/374/why-is-iron-in-the-brown-ring-compound-in-a-1-oxidation-state


text:
Does the object blocking the cathode rays in Thomson's cathode Ray experiment need to be opaque?
Thomson kept an "opaque" object in the cathode ray tube and he saw the shadow of the object.then he decide that the cathode rays follow a rectilinear path. But Thomson also discovered that cathode rays actually consist of particles.
Suppose I keep a transparent object (such as a thick slab of glass) instead of the "opaque" object. If the cathode rays are particles, then we would still find a shadow formed behind the glass slab (the particles wouldn't be able to pass through the slab).
Numerous articles and books I've referred to insist that the experiment requires that an "opaque" object be placed inside the cathode ray tube.
However, in my example with the glass slab, "opacity" doesn't seem to be a pre-requisite for the successful implementation of the experiment. Am I right in thinking this?

Opaque to electrons, not to light. It is applied to both cases, although the second is the common usage since humans don't directly see electrons...


text:
How to calculate percent error for absolute zero?
In chemistry lab, we did an experiment with water of various temperatures to extrapolate a value for absolute zero based on experimental pressure values for water at different temperatures. Naturally, the data is a little off -- even though absolute zero is $0 K$, my extrapolated value is about $15 K$.
One of the questions in the post-lab asks us to calculate percent error, and I'm a little confused. I thought (and this post confirms) that percent error was calculated by dividing by the expected or "actual" value. In this case, however, the actual value is $0$, and you can't divide by zero.
How do you calculate percent error for absolute zero?
Using Celsius values doesn't really make much sense to me, because the Celsius scale is not absolute; $-273.15$ is essentially an arbitrary number.

It indeed doesn't seem to make sense, so I don't see much to add - it was a mistake in assignment.

@Mith Though I'm not in a position to answer this myself, I don't feel this was a mistake. What the OP did: 1) Conduct an experiment, 2) Expected (theoretical) value is x, 3) Experiment yields a value x', 4) Percent error must be {| x - x' |} 100/ x. Which is fundamentally no different from typical "percent-error" calculations... although the value of x (in Kelvin) is zero, which is tricky :P

@paracetamol And there's the rub. Is the percent error "undefined"?

What sort of data did you end up with? Perhaps you could take the percent error in your regression. For example if your data was a linear, take the error in slope and intercept of the line.

@iammax is correct perform a linear regression and use the errors produced to obtain your answer.

It sounds like this is one of those canned labs that ask you to do standard things, even when they do not apply.
As you noticed, the percent-error formula needs you to use a value in the denominator.  If you use 0K in the denominator, you get a divide by zero.  If you use your 15K extrapolation, the percent error will always be 100%, no matter what the error actually was!
In such cases, I would look at the percent error with respect to your data.  Why?  Well, because somebody said you need to calculate a percent error and you might as well use a real datapoint rather than divide by 0 or some other silly result.  If you choose your coldest datapoint, you can say something about the percent error of the extrapolation process at the least.  Thus, if your coldest temperature recorded was 2C, you could say "My extrapolation suggests that absolute zero is 260C below my coldest recorded value.  The actual absolute zero is 275C below my coldest recorded value.  This gives me a 5.7% error on my measure of absolute zero relative to my coldest recorded value."

But again, the Celsius scale is not absolute. Why is your proposal different than saying "absolute zero should be -273 C, and I got -258 C"? You can't do percent error with values that aren't absolute. (I may simply misunderstand your recommendation.)

In this case I'm not comparing absolute temperatures, I'm comparing relative temperatures.  I'm saying "My projected absolute zero is X degrees below my coldest point.  The known temperature of absolute zero is Y degrees below my coldest point.  Percent error is (Y-X)/Y"  By anchoring it to the coldest datapoint I have, I can use relative temperatures which give some meaningful percent error.  The truth is that it's kinda silly to ask people for percent errors regarding absolute zero, so all you can do is make it as minimally-silly as possible.

It's kinda like how the absolute Celsius scale is anchored to the freezing point of water.  As you pointed out, it's kinda arbitrary.  At least by using the coldest datapoint you pick an anchor that has some meaning.  This way it at least shows the error that crops up as you extrapolate (since you extrapolate from the last datapoint)

You should divide by the water's temperature in K. This answer is pretty close to Cort Ammon's (Cort Ammon seems to be dividing by how much you measured your water to be above absolute zero, rather than how much it actually is). My reasoning:
How can you calculate absolute zero to be 15K? 0K is defined as absolute zero. The only way your calculations makes sense is you're trying to calculate how far above absolute zero you are. For instance, suppose your water is at 2°C, and you're trying to use its properties to find AZ. You say that you found that AZ is at 15K. Well, first I'm going to reinterpret that as "I found that AZ is at -258°C", because, again, "I calculated AZ to be at 0K" is nonsensical. Next, I'll interpret "I found that AZ is at -258°C" as "I calculated AZ to be 260 degrees below 2°C". Note that although these last two statements are logically equivalent, the percent error calculations they suggest are quite different.  In comments to Cort Ammon, you ask why the temperature of your water isn't as arbitrary as the Celsuis scale. The answer is that this is how far you're trying to extrapolate out. Hopefully, if you had had water at -100°C (okay, that wouldn't be water anymore, but you get the idea), your error would be smaller. It is reasonable to compare how far off your extrapolation was to how far you extrapolated.  You were trying to extrapolate to something that was 275 degrees away, and you were 15/275 = 5.45% of the way off.
So in this example of water at 2°C, presumably at some point in your calculations, you found that you were 260 degrees above AZ, and then took the "known" number that your water is at 275K, and took the difference to find AZ = 15K. Or you did some equivalent calculation. You are looking at the final number of 15K and asking what percent off it is, but the more meaningful number is what percent off your intermediate calculation of 260 is.
Asking "How do I calculate percent error for absolute zero?" is a bit like asking "I tried to calculate the position of the center of the earth in a coordinate system in which the center of the earth is the origin. How do I calculate percent error?" If you calculated that the center of the earth is 6,356 km away, rather than 6,371 km, then your percent error is 15 km/6,371 km. Saying "I calculated that the center of the earth is at r = 15 km, but it's actually at r = 0 km. So the percent error is 15/0." is fallacious. Yes, your distance from the center of the earth is in some sense "arbitrary", but it's still the valid denominator, because that's what determining how far out you have to measure.
Any time you have a number that represents how far away you are from a point, whether it's Kelvin measuring distance from absolute zero, or r measuring distance from the center of the earth, measuring percent error with respect to that number will always get you infinity. So instead of measuring distance from that reference point, you should measure distance from the point you started from.


text:
Asking about definitions of word "sugar" in biochemistry
In a lecture about anabolic pathways of sugar, the lecturer was not clear when stating the name of a multiple sugar carrier & it sounded like "dolichol"so is it correct?

https://en.wikipedia.org/wiki/Dolichol

@xavier I'm not particularly looking forward to see this question tagged under [tag:biochemistry]. The tag wiki (excerpt) clearly states that the tag ought to be used if the post is about biochemical processes ;P

Apologies I was thinking of sugar transfer reactions in which dolichols take part in biochemistry, and I thought it was fitting

I'm no biochemist, but... aren't the title of the the post (definition of a "sugar"), and the post itself (is there a certain something called "dolichol"?), two different things? o_o

Maybe the title is slightly off, but I am sure OP is referring to anabolic pathways specifically sugar transfer reactions, in body text :)

Absolutely correct!

N-Linked oligosaccharides are initially synthesized as lipid-linked
  precursors. The lipid component in this process is dolichol, a
  long-chain polyisoprenol of 14 to 24 isoprene units (17–21 units in
  animals and 14–24 units in fungi and plants; isoprene units are C5
  units with the carbon skeleton of isoprene , which is linked to the
  oligosaccharide precursor via a pyrophosphate bridge.

Dolichols apparently anchor the growing oligosaccharide to the endoplasmic reticulum membrane. These compounds have strong hydrophobic interactions with membrane lipids, anchoring the attached sugars to the membrane, where they participate in sugar-transfer reactions.
The pathway of dolichol-PP-oligosaccharide synthesis involves stepwise addition of monosaccharide units to the growing glycolipid by specific glycosyltransferases to form a common “core” structure:

Dolichol pyrophosphate glycoside. The carbohydrate precursors of N-linked glycosides are synthesized as dolichol pyrophosphate glycosides. Dolichols are longchain polyisoprenols (n = 14–24) in which the α-isoprene unit is saturated.
References

Voet and Voet Biochemistry
Lehninger Principles of Biochemistry


text:
Do nitrogen and oxygen react to form nitric oxide or nitrous oxide?
Various sources claim different products for the reaction between nitrogen and oxygen:
$$
\begin{align}
\ce{N2 + O2 &-> 2 NO} \tag{R1} \\
\ce{2 N2 + O2 &-> 2 N2O} \tag{R2}
\end{align}
$$
Which nitrogen oxide is actually formed? Are nitric and nitrous oxides formed in variable amounts, but depending on certain reaction conditions one is produced predominantly?

Related: https://chemistry.stackexchange.com/questions/21684/why-does-n%E2%82%82-react-with-o%E2%82%82-to-form-no-at-high-temperatures?rq=1

and: https://chemistry.stackexchange.com/questions/49842/can-i-convert-elemental-nitrogen-to-nitrous-oxide?rq=1

Oxygen and nitrogen react to form nitric oxide. Following is from Wikipedia:

The uncatalyzed endothermic reaction of oxygen ($\ce{O2}$) and nitrogen ($\ce{N2}$), which is effected at high temperature ($\pu{>2000 °C}$) by lightning has not been developed into a practical commercial synthesis (see Birkeland–Eyde process):
$$\ce{N2 + O2 -> 2 NO}$$

Why nitrous oxide is not formed can be explained through thermodynamics. For any chemical reaction to take place spontaneously, we look at a factor known as free energy available for that reaction. Gibbs's Free Energy is given by the equation:
$$\mathrm{\Delta G = \Delta H − T\Delta S}$$
The quantity $\ce{\Delta G}$ must be negative for a reaction to take
place spontaneously. Now, most of the nitrogen oxides are unstable with respect to molecular nitrogen and molecular oxygen and have a positive value for the Gibbs free energy change of formation. For the reaction $\ce{2N2 + O2 −>2N2O}$, we put the values in the equation and calculate the Gibbs Free energy to confirm that this reaction can't take place.
Now coming to our actual reaction, the entropy change is small and positive.
$$
\begin{align}
\ce{N2 + O2 -> 2NO}   & \tag{ΔG_0 = 86.7 kJ/mol}\\
\end{align}
$$
The Gibbs free energy change can also relate to the equilibrium constant, K.
$$\mathrm{G = -RT ln K = -5.7 log K~~(in~kJ~at~298 K)}$$
At equilibrium, that is under thermodynamic control, the concentration of $\ce{NO}$ in the atmosphere should be $\pu{10^-15.5 atm}$ but actual concentrations of $\ce{NO}$ are significantly higher than this (approximately $\pu{10^-10 atm}$).
So, even though thermodynamics says that $\ce{NO}$ is unstable with respect to $\ce{N2}$ and $\ce{O2}$, it doesn't spontaneously break down to those elements. There is a high activation barrier that must be overcome between reactants and products. Nitric oxide is kinetically stable but thermodynamically unstable.
For more information, please read the article.
But the reaction doesn't stop here. It proceeds to form more more nitrogen oxides $\ce{NO_x}$. You can continue reading the article for more details.

Is the reaction $\ce{2N2 + O2 -> 2N2O}$
wrong ?

@Eloise well the reverse is true.

This is what I am confused out. I have seen the reaction in which $N_2O$ decomposes to $N_2$ and $O_2$ in many authentic sources but I have seen the reverse reaction on some online sources which may or may not be credulous. I need to be able to say with definiteness whether the reverse reaction is true or not.

@Eloise I edited the answer.

Throw in a little water vapor and you go all the way to nitric acid, which therefore exists in (unpoolluted) rainwater. Yuck, but also yum: the acid is both a source of fixed nitrogen and a reagent that releases other nutrients for plants by dissolving some minerals.


text:
Ring closing esterification
I've been taught that to make esters, the carboxylic acid has to be activated — for instance as the acid chloride or using DCC.
But, when I read papers on total synthesis, always they use more complex reactions such as a Yamaguchi in order to do a ring closing esterification.
Why are standard esterificaton conditions not suitable for complex molecules?   One though I had was that acid chlorides are so reactive that they're easily quenched by water, but this would just give back a carboxylic acid which could be reacted again (no damage done overall).

The reaction you're describing is known as a macrolactonisation, where a medium/large ring is closed by the formation of an lactone (a cyclic ester).
The real problem with macrolactonisation is that there is a somewhat significant entropic barrier to the reaction, you're taking something with a lot of flexibility and constraining it into a ring, reducing the conformational flexibility of the system.  As the ring size increases, entropy becomes significantly more important than enthalpy.
Because the reaction is so entropically unfavourable, the intermolecular reaction has a tendency of happening (rather than the desired intramolecular ring closure).
To avoid these issues, two things are commonly done:

macrolactonisations are ran with high dilution, high dilution favours the intramolecualar reaction by decreasing the probability of two molecules encountering one another
modern macrolactonisation procedures limit the amount of the activated carboxylic acid present (or indeed the activated alcohol) at any one time, again, preventing the likelihood of two molecules reacting together in an intermolecular sense.

The whole thing is maybe easiest explained with an example. The Yamaguchi macrolaconisation is one of the most common, but many others (i.e. the Shiina) follow the same principle.

In the first step, the seco acid (the molecule with the free OH and the free carboxylic acid) are reacted with the TCBC to form a mixed anhydride.
Although this mixed anhydride is 'activated' relative to the free carboxylic acid, it isn't usually reactive enough, meaning that the mixed anhydride can be isolated (no intra or intermolecular reaction takes place).
The isolated mixed anhydride is then added dropwise to a very dilute solution of DMAP (acting as a nucleophilic catalyst). The DMAP attacks the mixed anhydride, forming a highly reactive intermediate (much more reactive than the mixed anhydride), from which the alcohol may attack the carbonyl, displacing DMAP, and forming the desired macrolactone. By adding the mixed anhydride to the DMAP, it limits the amount of the molecule in the flask which is sufficiently activated to react at any given time, increasing the likelihood of intramolecular reaction further.


text:
Assigning absolute configuration when determining crystal structure isn't possible
When natural products are isolated, their structures are often determined by x-ray crystallography which gives a complete picture of the atom coordinates, the relative and the absolute stereochemical configurations. If however the isolated molecule won't crystallize, how can absolute stereochemistry be obtained?
I know that 2D NMR can tell about connectivity and also possibly relative stereochemistry ($J$-value analysis), but unless one stereocenter is known the others cannot be implied from it. Optical rotations also are useless unless the compound is known. How then can the absolute configuration be found without single-crystal x-ray diffraction?

A classic method is to carry out some reaction on the compound to transform it into a derivative with a known optical rotation.

@orthocresol.. doesn't this imply that its already a known compound in itself? A new... terpene... say, won't have a derivative with a know rotation.

For smaller molecules, you can potentially compute the absolute rotation. One of the computational chemist might be able to give you more details.

@NotEvans. I was thinking more of small molecules than huge natural products, but yeah, I suppose we'd run into problems with more novel stuff.

Theres always the good old days of 'heat it in acid' degradation studies.

@NotEvans. You might be able fragment the molecule and carry for a smaller piece to a compound with known rotation. Not clear how you would do this for a generic terpene though.

Possible duplicate of this question, though the answers don't really answer your question.

This question is possibly a little broad- there are entire books dedicated to methods of structure determination and assignment of stereochemistry.
The tl;dr answer is that there is no single method by which absolute stereochemisty can be assigned unambiguously, it depends on the nature of the natural product (terpene, polyketide, polypeptide, etc) and the exact structural features.
Some common methods which are often found in isolation papers:
1. Degradation
Historically, one of the most common ways of identifying new natural products was to attempt to degrade them to known compounds. If you imagine taking a new peptide natural product (this is actually a bad example as most peptides are made up of natural amino acids), you could easily hydrolyse it back to the constituent amino-acids (or smaller peptide chains) for which rotational data is available.
2. Derivatisation to make crystals
If your material doesn't crystallise, make small structural modifications to make it crystalline.
Being able to do this somewhat depends on knowing the 2D structure (by NMR). Once you identify the functional groups present, you can begin to modify the structure in order to obtain something crystalline (to do X-ray analysis on).
There are too many possibilities to go through in detail, but you could, for instance, take a carboxylic acid and form an ester, or some kind of salt.
This method is often carried out in the pharmaceutical industry, where they want their active pharmaceuticals to be in the solid state in a known form (different forms have different properties, such as solubility, so great lengths are taken to obtain uniform material).
3. Derivatisation for NMR studies
If the molecule of interest has an alcohol present, its possible to form esters with a special kind of carboxylic acid, this is the Mosher ester assignment.

The alcohol of interest is reaction with two enantiomeric acids (the Mosher acids), to form a pair of diasteromers. The Mosher esters shield the molecule with the phenyl group, causing shifts to change on the nearby protons around the chiral alcohol. By considering the relative changes, the assignment may be assigned.  This method is often found in structure elucidation papers, however there are some instances where it can give misleading results, so shouldn't really ever be used on its own without any other evidence.
It's worth noting that the Mosher esters themselves are often crystalline, allowing a crystal structure to be obtained in addition to the assignment by the NMR analysis discussed above.
4. Computational work
Computation is one of the newer methods of assigning absolute stereochemistry, so new in-fact that the FDA (the people who approve drugs) don't accept computational assignment on its own, unless accompanied by some other, more traditional method, of assigning configuration.
In theory, there are methods to predict optical rotation, or other properties such as circular dichroism, which can then be compared to the real experimental values. In practice, this rarely works well for complex molecules as there are too many variables involved which effect the computational result.


text:
Measure the complexity of a molecule
Back in April, there was an article in new scientist discussing the work of Lee Cronin, who postulated that molecules of a certain complexity must be biological.

Cronin has developed a way to measure the complexity of a molecule by
  counting the number of unique steps – adding chemical side groups or
  ring structures, for example – needed for its formation, without
  double-counting repeated steps. To draw an analogy, his metric would
  score the words “bana” and “banana” as equally complex, since once you
  can make one “na” it is trivial to add a second one.
Any structure requiring more than about 15 steps is so complex it must
  be biological in origin, he said this week at the Astrobiology Science
  Conference in Mesa, Arizona.

Does anyone know how to use Cronin's method to measure complexity?

Have you tried looking through the publications on his website?

There is a preprint of Cronin's article on Arxiv: https://arxiv.org/abs/1705.03460

Indeed, the paper includes some elements of graph theory linked with combinatorics, but in my opinion has nothing specific towards defining its title. With the same success one could utilize, lets say, big-O notation from the computer science field for describing complexity in chemistry. With all respect to Prof. Cronin he is famous for using buzzwords for popularizing his research. In late 2011 during his TED talk he basically announced (https://youtu.be/unNRCSj0igI?t=845) an advent of artificial (polyoxometalate-based) life in 2013.

@NotEvans I read through the paper which used English letters to illustrate complexity but was vague except to say that the number of unique elements seems to be the most important factor.  Meaning that in terms ofc omplexity CO2 is the same as CO.  However, surely some of the higher elements would indicate greater complexity?  For example does UO2 have the same complexity as CO2?

Pfft, remember that all organic compounds were once thought to origin only in life.

@Tyberius Thanks to the preprint you indicated I read the publication.  But I wonder about the references.  Maybe not the first (historically speaking) but at least among the first (as in first class) in categorizing organic reactions -- where is EJ Corey? With his formalizations like Logic of Synthesis, and software (e.g. LHASA) where reducing complexity became a criterion to select path B in preference over path A (https://www.scripps.edu/baran/images/grpmtgpdf/Maimone_Mar_06.pdf) which eventually lead him 1990 to Sweden?


text:
DFT: Can the calculated enthalpy of two systems that aren't isoelectronic be compared?
As a simple example, say I wish to calculate the product of a reaction between methyl iodide and triethylamine, then compare it with the product of benzyl iodide and triethylamine. My goal is to compare the products' relative stabilities (in a larger system this could be useful to compare reactivity options on a common substrate). I couldn't just calculate triethylmethylammonium's enthalpy and compare it with benzyltriethylammonium, right? I would expect the system with more electrons/bonds to always have a lower enthalpy value.
Alternatively, I could calculate all the reactants as well as products, including the iodide as a non-coordinating anion, then subtract the reactants from the products. This difference I could then use to compare the relative stability of the two products, am I correct in this? My worry with this approach is whether DFT can model the non-coordinating anion accurately, as well whether its result would mean anything without solvent effects. It feels like the rabbit hole can always go deeper, and I guess I am curious at what point I obtain qualitatively useful results.

Could you make the question more precise? The actual title question is unclear. Anything can be compared in principle. Perhaps you wanted to ask whether a difference of some energies has some particular physical significance? Then in the body you mix two other questions, the first one is not really DFT-specific, but general, and the other doesn't have much to do with the title question, but is about DFT accuracy.

Hi, sorry for the confusion. In DFT the energy of the system can be calculated, for example the enthalpy or free energy from a frequency calculation. I understand these to provide insight into the thermal stability of two compounds, for example in a complex to compare different coordination stereoisomers (eg. different ligands). My question is a) whether such values can be used if the molecules under comparison aren't isoelectronic and b) how I could otherwise go about comparing two molecules' relative stabilities.

+1. It's a great question. I would really love to invite you as a beta user for Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MWFmZGNmODU3OTY2ZDQ0MmVmZGE3NmM3ZWY2MThhMTIwNWMwMGE5NGViYmQyMDg4MzU2OTM5NzE2NDU0M2ZmMJRx4jTqf3qvU9JjTcsyQAIjWJ0uYYe7tfZf0kUs_8rG0. Can you please register there and click commit? There's numerous DFT questions in the example questions list, and I would be very delighted if you could help us out.


text:
How does the Mosher ester method allow determination of absolute configuration of secondary alcohols
One of the challenges in synthesis is the correct assignment of absolute stereochemistry. This has been asked previously on chem.SE (here and here), but this question specifically asks about the determination of hydroxyl stereocentres using the Mosher ester analysis.
Depending on the type of reaction carried out, this can often be done by inspection (for example, if a reaction forms a pair of diasteromers, they are often easily distinguished by NMR, assuming you know the stereochemistry of the initial stereocentre).
Where no chiral centres were previously present in the molecule (or where pre-existing chiral centres are too remote from the new centre to be relevant), assignment of absolute configuration can be challenging. Whilst x-ray crystallography can be used, many small molecules do not crystallise (or at least not without a significant amount of effort).
One method for assignment of absolute stereochemistry of newly formed hydroxyl centres is the Mosher ester analysis, which works by forming a pair of diastereomeric esters, the NMR shifts of which can be used to determine absolute stereochemistry.
How does this method work? There are many diasteromeric esters that could be formed, yet only a specific few are used for this purpose, and equally, how come there is a need to form both diasteromers of the Mosher esters, when the change in shifts could be observed by just forming a single diasteromer.

How are you going to know what the diastereomeric shifts are for? If you have both diastereomers, you can at least confirm which one is which.

Not strictly true, there are other esters (similar to the MPTA) where only one has to be formed (I just thought it was an interesting question that hadn't been asked here, and there haven't been a tonne of good questions recently)

Lots of literature out there about this. Try here and here for example. Do these not answer your question?

@long - I have read these. Apologies. I asked the question because I thought it was interesting, and given the complexity I thought it would be nice to get a variety of answers from the community here. I've answered, but given your clear experience with NMR, would appreciate anything you have to add. I find that the papers discussing this don't do a great job of explaining why (R) vs (S)

The NMR spectra of two enantiomeric molecules are necessarily identical (just as all properties of enantiomers are identical in the absence of an external chiral environment). In order to tell the enantiomers apart, some additional chirality must be introduced. Two methods exist for this in NMR spectroscopy:

Chiral solvating agents, such as lanthanide shift reagents
Chiral derivatising agents, such as Mosher esters

In modern organic synthesis, the use of chiral derivatising agents is far more common, and the process can be summarised as follows:

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)
a. The alcohol of unknown configuration is derivatised by forming both the (R) and (S) Mosher esters. This can either be done under Steiglic conditions (DCC,DMAP) or via initial formation of the acid chloride.
b. Proton NMR spectra are ran of the two diasteromeric esters. The analysis is often ran on the crude esters, therefore the acquisition of COSY/HSQC is also useful to aid in assignment.
c. The difference between the chemical shifts of the (R) and (S) esters are tabulated (by convention we consider R-S, it makes no fundamental difference to the outcome however).  Note that the method does not consider the shift of the underivatised alcohol at all, however the peaks for the (R) and (S) ester must be unambiguously assigned. Note also that we do not take into account the oxymethine proton (the shifts of this are erratic and don't contribute to the analysis).
d. The configuration is assigned by considering which side of the molecule is most effected by the presence of the Mosher ester.
Step d is often the hardest to grasp conceptually. When we make the Mosher esters, what we're counting on is that they adopt a specific conformation (which they do). In this conformation, the phenyl group of the Mosher ester will shield part of the molecule (which is different depending on which enantiomer of the Mosher ester we're looking at). Consider the following example, in which the side shielded by the phenyl ring has a lower delta overall:

Shielding by the Mosher ester. Taken from Harvard CHEM2016 lecture notes, Eugene Kwan
The following conformational diagram is also (possibly more) instructive:

Shielding by the Mosher ester. Source unknown

The only way to actually understand the Mosher ester method is to carry it out (ideally with something of unknown stereochemistry so you can't convince yourself of the answer simply because you know its the case.
The next best thing is a worked example:

Outline of the Mosher ester analysis. Taken from Organic Spectroscopy and Structure Elucidation lecture notes (not available online)
Looking at the two spectra, it becomes apparent that (R)-(S) is positive on the LHS methyl group, and negative on the protons on the RHS propyl group. Fitting this to the model, we can assign the sterocente of the secondary alcohol as (R).

The reason both Mosher esters should be formed is due to the fact that change in chemical shift between one Mosher ester and the free alcohol is often not huge. Other derivitising agents (such as AMA) can be carried out with formation of a single ester.

And a single Mosher ester can also tell you the e.e. of your product. This is often done with $\ce{^19F}$-NMR since there are usually no other fluorines around.

Yup. I was confining myself to determination of configuration. Though actually even for ee I find it convenient to make both just to make sure I know that the peaks I'm integrating are the right things!


text:
Is argon preferred over nitrogen for maintaining an inert atmosphere?
I recently started a project in a new lab-group over Summer that does medicinal chemistry. Unlike in the undergraduate labs, they use argon instead of nitrogen for their inert atmosphere work.  When I asked around, the lab members said that it was 'just because they'd always has argon', though a few people said that they thought argon was better, without explanation.
As far as I am aware, both gases are unreactive and can be dried, therefore it makes no sense to me why argon would be preferred given how cheap and abundant nitrogen is. There are some air sensitive steps in my route but generally nothing too unstable
Is there any real difference between the two that would make one better, or is it, just as my lab said, that it's just what they have and they've never bothered to change it?

Argon and nitrogen have different physical (i.e. viscosity) and chemical properties (i.e. nitrogen is generally more reactive than argon, depending on the situation). You can't really answer the question without knowing what you are using the gas for.

Nitrogen adsorbs to surfaces more readily than argon does, this may or may not be an issue - but nitrogen is usually cheaper

Nitrogen from cylinders can contain traces of oxygen and water vapour which may be important in your application

Speaking purely from an organic point of view, there isn't really a huge difference, labs tend to stick to one or the other, largely depending upon the availability (some universities use the boil off from the liquid nitrogen cylinders to furnish labs with nitrogen gas, whilst others have no such facilities meaning labs must buy cylinders of either nitrogen or argon).
For serious organic groups (organometallics, total synthesis) doing a lot of air-sensitive chemistry, there can be some advantage to argon due to its higher density [argon (1.714 g/L) nitrogen (1.251 g/L)], meaning it tends to blanket the reaction, and if you need to quickly open the flask to put a reagent in then the argon doesn't escape so quickly.
For inorganic chemists, there are situations where nitrogen will react with certain chemicals, and in these cases argon must be used.

Is argon more dry than nitrogen though ?

Both argon and nitrogen can be made as dry (or as wet) as you want them to be.

Nitrogen is cheaper. I have never worked in a lab that had argon and never come across a situation where I needed it.

The out-of-the-bottle purity isn't a big issues as labs can achieve higher purity easily with local processes. But nitrogen is more reactive than argon and for some reactions this matters. Also the density is higher so the blanking effect may be an added benefit for very air-sensitive reactions. I've had to switch to argon for some organometallic reactions for exactly these reasons.

I asked the same question while touring an organic lab, and they said the blanketing effect was the main reason they used argon instead of nitrogen.

My experience is largely colored by my days as a grad student in a organic chemistry lab that did some organometallics.
I believe for the applications we had (palladium / nickel cross coupling), argon and nitrogen were functionally equivalent.
For more air / moisture sensitive applications in inorganic chemistry, argon may be preferred because it is heavier than air and thus better forms a protective blanket around your sensitive samples.
However, there are two other downsides to argon that come to mind. The first is price. I was actually tasked by our PI to be our argon monitor to make sure people weren't wasting argon. It was significantly more expensive than nitrogen, and the PI didn't care if you were just blowing nitrogen around.
The second one which people some times forget is that argon has a higher boiling point than nitrogen. If you're using argon in your manifold and cooling your trap with liquid nitrogen, you may condense argon in your trap. If then you close off your manifold without an outlet and the trap loses cooling, the manifold will be under intense pressure from the argon boil off. This presents a serious explosion hazard if you're not paying attention.

I don't think either is generally preferred over another, its project and situation dependent.
For most organic reactions, it wont matter nitrogen is sufficiently non-reactive to be acceptable and the lower price makes it the go-to gas for most labs. However there is one consideration that makes argon attractive over nitrogen, and that is impurity levels. Although industrial grades for both gases are fairly pure, The specs for argon tend to have a lower impurity level of oxygen than for nitrogen, even though industrial grade nitrogen is typically purer than industrial grade argon.
For metallurgical work, argon is often preferred (actually vacuum is preferred) because nitrogen can react with metals at high temperature to form nitrides ($\ce{Ti3N4, Si3N4, AlN, BN,}\text{etc...}$) whereas argon will not react. Additionally nitrogen can more easily adsorb to surfaces which is the place where catalysts are active, thus it may be desired to reduce the amount of adsorbed material on a substrate (like Raney nickel) by using argon over nitrogen.

...unlike in the undergraduate labs, they use argon instead of nitrogen for their inert atmosphere work.

Given the Praxir argon has $\pu{1 ppm}$ less oxygen than the Praxair Nitrogen, it may be unconvincing that either gas matters. Consider though the cost of the reagents involved and the students time, it may not be that wasted of an investment to pay a little more for nitrogen to get more yield ever few reactions. Compare that to undergraduate labs, where there tends to be less serious research, the time scales of projects and are much shorter, the reactions are less sensitive and, are typically for pedagogical purposes, which brings me to my next point:

When I asked around, the lab members said that it was 'just because they'd always has argon'

Consider the implications of switching gases. Either 1) Everyone in the middle of research would be introducing a change in their experiment or 2) New Experiments would have to wait a year or two for current ones to complete or 3) a second gas system would have to be temporarily used to facilitate the long-term switch over. None of these options are tenable unless the lab is up for remodeling, moving or new ownership and projects, thus the gas first used is the gas they continue to use.

To be added: Nitrides form relatively easy with s-block metals.  Lithium, for example, already reacts with the nitrogen of air (and in contrast to other fires, these can't be extinguished with this gas).


text:
Pi molecular orbitals of polyenes
As an organic chemist, I'm comfortable deriving the pi molecular orbitals of linearly conjugate systems to give the following result:

In a qualitative sense, these molecular orbitals are easily arrived at. For a polyene of n-atoms, the lowest energy combination will have 0 nodes (all in-phase) and the highest energy combination will have n-1 nodes (in, out, in, out...).
Is there a way of arriving at the same result using a more rigorous, formal molecular orbital approach?

Is Hückel MO Theory what you are looking for?

Hückel MO theory  will also give you different absolute values for the linear combination coefficients, whereas your drawing suggests the same coefficients throughout. Note that all you need to perform HMO calculations are a way to set up the matrix (which is easily done by hand for these cases) and a diagonalizer, such as Wolfram Alpha. Interpretation is again easy by hand.

These pi-type MOs are most commonly derived using Hückel MO theory. I've never been a fan of Wikipedia's technical articles, so for further reading I'd suggest using the QMUL resources online, which are very comprehensive. $\require{begingroup}\begingroup \newcommand{ket}[1]{\left|#1\right>} \newcommand{bra}[1]{\left<#1\right|} \newcommand{braket}[1]{\left< #1 \right>}$

Theory
To begin with, MOs are obtained by solving the secular equations:
$$\mathbf{H}\mathbf{c} = E\mathbf{S}\mathbf{c}$$
The derivation of this will not be covered here (it can be found in e.g. Atkins' Molecular Quantum Mechanics), but it is worth mentioning what these things are. If we consider a polyene with $n$ atoms (e.g. for ethene, $n = 2$), then

$\mathbf{H}$ is the Hamiltonian matrix, an $n \times n$ matrix whose elements $H_{ij}$ refer to the the $i$-th and $j$-th p-orbitals: $H_{ij} = \braket{\mathrm{p}_i|H|\mathrm{p}_j}$.

$\mathbf{c}$ is a $1 \times n$ column vector of coefficients: the element $c_i$ corresponds to the coefficient of the $i$-th p-orbital. That is to say, the MO $|\psi\rangle$ corresponding to the vector $\mathbf{c}$ is given by
$$|\psi\rangle = c_1 |\mathrm{p}_1\rangle + c_2 |\mathrm{p}_1\rangle + \cdots + c_n |\mathrm{p}_n\rangle.$$
By solving the secular equations, we obtain a series of permitted vectors $\mathbf{c}$: each of these solutions correspond to one MO. In general, if there are $n$ p-orbitals used as inputs, then we will obtain $n$ permitted MOs.

Each solution $\mathbf{c}$ is associated with a particular energy $E$, which is a scalar.

$\mathbf{S}$ is the overlap matrix, whose elements are given by $S_{ij} = \braket{\mathrm{p}_i|\mathrm{p}_j}$.

Hereafter, we will simplify the notation and use $|i\rangle$ in place of $|\mathrm{p}_i\rangle$ to denote the p-orbitals.
Now, simple Hückel theory makes some key assumptions about the forms of the MOs as well as some of the integrals involved in a typical quantum chemical calculation:

π-Type MOs are linear combinations of p-orbitals, and no other orbitals contribute (so-called "sigma-pi separation")
The value of $\braket{a|b}$ (where $\ket{a},\ket{b}$ are p-type AOs on atoms $a$ and $b$) is $1$ if $a = b$ and $0$ otherwise
The value of $\braket{a|H|b}$ is

$$H_{ab} = \braket{a|H|b} = \begin{cases}\alpha \text{ if } a = b \\ \beta \text{ if atom }a\text{ is bonded to atom } b \\ 0 \text{ otherwise.}\end{cases}$$
Here, $\alpha$ and $\beta$ are just some constants. We don't have exact values for them (yet), but we can say that they are both negative. With these simplifications the secular equations can be readily solved. In fact, in simple Hückel theory the matrix $\mathbf{S}$ is simply equal to the identity matrix (see rule 2 above), so it can be completely ignored in the secular equations:
$$\mathbf{Hc} = E\mathbf{c}.$$
You may recognise this an eigenvalue equation for $\mathbf{H}$ (the original form, where $\mathbf{S}$ is not necessarily equal to the identity matrix, is called a generalised eigenvalue equation).
As mentioned previously, for $n$ atoms there will be $n$ vectors $\mathbf{c}$ that will satisfy the secular equations, and $n$ corresponding values of the energy $E$. Thus, for the allyl cation (for example), there will be three column vectors $\mathbf{c}^{(1)}, \mathbf{c}^{(2)}, \mathbf{c}^{(3)}$ and three associated energies $E^{(1)}, E^{(2)}, E^{(3)}$.
Note here that $\mathbf{c}^{(1)}$ refers to the first possible solution for $\mathbf{c}$, whereas $c_1$ refers to the first component of $\mathbf{c}$ (i.e. the coefficient of p-orbital number one).

The allyl cation

Using the atom numbering scheme as shown above, and the assumptions mentioned in the previous section, we find that the Hamiltonian matrix $\mathbf{H}$ is:
$$\mathbf{H} =
\begin{pmatrix}
\alpha & \beta & 0 \\
\beta & \alpha & \beta \\
0 & \beta & \alpha
\end{pmatrix}.$$
For example, the entry $\mathbf{H}_{12} = \beta$ because atoms 1 and 2 are bonded to each other, and the entry $\mathbf{H}_{13} = 0$ because atoms 1 and 3 are not bonded to each other.
The secular equations then take the following form:
$$\begin{pmatrix}
\alpha & \beta & 0 \\
\beta & \alpha & \beta \\
0 & \beta & \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix} = E\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix},$$
and the challenge is then to find the permissible vectors $\mathbf{c} = (c_1, c_2, c_3)$ which obey this equation, which are called eigenvectors. This process is outlined in many maths books, and is thus not covered here in full detail. Briefly, however, the idea is that we move the RHS over:
$$\begin{pmatrix}
\alpha - E & \beta & 0 \\
\beta & \alpha - E & \beta \\
0 & \beta & \alpha - E
\end{pmatrix}
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix} = \mathbf{0},
$$
and there can only be nontrivial solutions to this problem (i.e. solutions where $\mathbf{c} \neq \mathbf{0}$) if the secular determinant (i.e. the determinant of the matrix on the LHS) is zero:
$$\begin{vmatrix}
\alpha - E & \beta & 0 \\
\beta & \alpha - E & \beta \\
0 & \beta & \alpha - E
\end{vmatrix} = 0$$
Expanding this out allows you to obtain the eigenvalues. In increasing energy (recall that $\beta < 0$), they are
$$\begin{align}E^{(1)} &= \alpha + \sqrt{2}\beta & E^{(2)} &= \alpha & E^{(3)} &= \alpha - \sqrt{2}\beta \end{align},$$
and the corresponding eigenvectors are
$$\begin{align}
\mathbf{c}^{(1)} &= \begin{pmatrix} 1/2 \\ 1/\sqrt{2} \\ 1/2 \end{pmatrix}
&
\mathbf{c}^{(2)} &= \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}
&
\mathbf{c}^{(3)} &= \begin{pmatrix} 1/2 \\ -1/\sqrt{2} \\ 1/2 \end{pmatrix}.
\end{align}$$
In other words, the lowest-energy pi-type MO, $\mathbf{c}^{(1)}$, is expressed as
$$\ket{\psi^{(1)}} = \frac{1}{2}\ket{1} + \frac{1}{\sqrt{2}}\ket{2} + \frac{1}{2}\ket{3},$$
where $\ket{i}$ is the 2p atomic orbital on carbon $i$. Note that the coefficients all have the same sign, indicating that the p orbitals are all in phase with each other. However, the coefficient for carbon 2 (the middle carbon) is larger than the others. Hence, a sketch of this MO should technically show that the middle p-orbital makes a larger contribution than the outer two p-orbitals. This is what TAR86 meant by their comment on your question:

Hückel MO theory will also give you different absolute values for the linear combination coefficients, whereas your drawing suggests the same coefficients throughout.

Anyway, likewise for the LUMO we have
$$\ket{\psi^{(2)}} = \frac{1}{\sqrt{2}}\ket{1} - \frac{1}{\sqrt{2}}\ket{3}$$
Note that: (1) the p-orbitals have different phases (different sign of coefficient), and (2) the p-orbital on carbon 2 does not contribute to the MO. This leads to the node observed in the second MO. This is why the allyl cation is electrophilic on C-1 and C-3, but not on C-2: the LUMO is $\ket{\psi^{(2)}}$, which has zero coefficient on C-2.
All in all, the MO diagram for the allyl cation is as follows:

Results for other / larger polyenes may be obtained in a similar fashion, albeit perhaps with more complications in finding the eigenvalues and eigenvectors. It is recommended to use software which can perform symbolic maths: for example, if you type eigenvalues of [[a, b, 0], [b, a, b], [0, b, a]] into WolframAlpha, then it tells you the eigenvalues (i.e. the energies)

and the eigenvectors (i.e. the MOs), which is the same as what we found above, except that the solution above is normalised (i.e. every eigenvector is divided through by a constant such that $c_1^2 + c_2^2 + \cdots + c_n^2 = 1$):

I should like to extend @orthocresol's clear answer and so obtain results for any size of conjugated system. Starting with the determinant in $\alpha -E $ and $\beta$.
$$\begin{vmatrix}
\alpha - E & \beta & 0 \\
\beta & \alpha - E & \beta \\
0 & \beta & \alpha - E
\end{vmatrix} = 0$$
and substituting $(\alpha -E)/\beta = -x$ rewriting this in a more general form produces
$$D_n =\begin{vmatrix}
-x & 1 & 0 & 0 & 0&\cdots\\
1 & -x & 1 & 0 & 0&\cdots \\
0 & 1 & -x & 1 & 0 & \cdots\\
\vdots& & & \ddots& \\
 & & & & &-x
\end{vmatrix} = 0$$
Evaluating the determinant gives the results
$$ D_1 = -x , \quad D_2= x^2-1, \quad D_3= -x^3 +2x,\quad D_4=x^4-3x^2+1,\cdots \\ \cdots D_n=-xD_{n-1}-D_{n-2}$$
Now it is known that the solutions to these equations can be recast in a very convenient form because they are they are similar to the Chebyscheff polynomials.
The n'th term here is
$$T_n= \frac{\sin([n+1]\theta)}{\sin(\theta)}=2\cos(\theta)T_{n-1}-T_{n-2}$$
which has the same form as $D_n$ if $x=2\cos(\theta)$.
Thus
$$D_n= \frac{\sin([n+1]\theta)}{\sin(\theta)}$$
is satisfied if $\displaystyle (n+1)\theta = j\pi$ with $j=1,2,3,\cdots$.
Substituting  $(\alpha -E)/\beta = -x$  and $x=2\cos(\theta)$ produces for the energy
$$ E_j= \alpha + 2\beta\cos \left( \frac{j\pi}{n+1} \right)$$
where n is the number of carbon atoms and each j from 1 upwards indicates one of the solutions of the secular equation. Thus for butadiene, $n=4$ and $j=1,\,2,\,3,\,4$ only. It should also be remembered when ordering these energies that $\alpha$ and $\beta$ are negative numbers.
[In the case of circular polyenes the solutions are $\displaystyle E_j=\alpha + 2\beta \cos(2\pi j/n)$ ]


text:
Distillation of diethyl ether and tetrahydrofuran
In the lab, I'm working in over summer, there is a set of solvent stills.
Most of them use $\ce{CaH2}$ as a drying agent, however ether and THF use Na and K respectively in the presence of benzophenone.
What I don't understand is why the ether still uses Na, whilst the THF uses K. Equally, I'm not sure what the purpose of the benzophenone is — it makes the stills blue, but none of the other solvents (DCM, toluene, triethylamine) have this property.

Sodium and potassium are both drying agents. Being alkali metals they react (often vigorously) with water.
Sodium wire is often used in diethyl ether stills, whilst potassium is used in tetrahydrofuran stills. In general, potassium is more reactive, and hence a better drying agent.
The reason different desiccants are used is due to temperature. THF boils at a higher temperature, meaning a lump of potassium metal can be used in the still ( the lump of K melts once the THF reaches reflux ). Diethyl ether boils much lower, meaning potassium would not melt, for this reason sodium is often used in the form of a wire (giving a much larger surface area).
In respect to the use of benzophenone (or other indicators such), it generally indicates the presence (or absence) of oxygen in the solution. In the past, it was thought that the purple (blue is actually not great) colour was related to the moisture content, however we now know that the colour is related to the oxygen content. For this reason, using the indicator allows us to determine how oxygen-free the solvent is. A purple THF/ether still in which metal is still present indicates that the solvent is both dry and oxygen free,

But why are other solvents not distilled using sodium/potassium

Sodium/potassium are quite reactive, they cant be used with certain solvents (you mention Et3N for example) as they would simply react

And never, never use halogenated solvents with sodium/potassium. Best case scenario Wurtz, worst case total annihilation.

I would think that Na/K also serve to remove any etherperoxides already formed, though, ideally, you would not have any in the first place.

Toluene does not need a drying agent as it is dried by azeotropic distillation

These stills are also one of the major sourse of lab fires that I've heard of over the years. They are banned from our labs. We use the Sureseal bottled dry solvents and they work just as well.

I agree the stills can be dangerous, but for anything particularly sensitive I've never had much luck with the Aldrich bottles. We've done Karl fisher on them a few times and they're fairly wet and fairly inconsistent

To add to @NotEvans' answer, while I am not able to trace back the original "benzophenone study" he refers to, a (sort of) reprint may be found here, including a small range of molecules that may be generated by extended refluxing over benzophenone.
It may interesting for the OP to see that some of the more frequently used solvents may be dried by pressing across columns packed of $\ce{Al2O3}$, a concept popularised by Grubbs et al. (Organometallics, doi: 10.1021/om9503712).  Beside suggested modifications (like here), some groups have it installed instead of the distillations:

(source)