text:
How do we check octet rule in resonance structures?
For example, this carbon has an incomplete octet:

Can we say this is because C has configuration $\mathrm{1s^2 2s^2 2p^2}$ and hence here its just forming 3 bonds whereas it should form 4 to make it $\mathrm{2p^6}$ and complete its octet? Or do we say that it's forming 3 bonds and hence has 6 electrons and therefore it's octet isn't complete?

Even though this structure "exists" on paper or computer screen, it is meaningless in the real world. In other words, this structure does not exist in the real world as drawn and has only come into existence as part of the human effort to describe the chemistry of this world using simple and straightforward symbolism. As you probably know, the true structure is a hybrid of all resonance structures, and the octet rule is satisfied for the central carbon. Therefore, I think this question has no actual answer.

Ok I agree with that. But my main concern is that how do we see the octet rule in atoms? Like I just want you explain this with respect to this particular structure. If you want I can present another example. I'm actually very confused how to check the octet rule in structures like these.

Like how can we say the octet isn't complete? By seeing the configuration and seeing the vacancy available or check how many bonds and lone pair an atom has? What're we supposed to do normally?

The octet rule is a measure that helps you to draw a correct and plausible Lewis structure. The rules for drawing correct Lewis structure are found in every basic chemistry book and should be followed in order. See section 5 in this link:  If the octets are incomplete, and more electrons remain to be shared, move one electron per bond per atom to make another bond... - if the octets are incomplete - if you follow the rules in sequence, you should not be left with an incomplete octet for carbon.

Also, traditionally, when you move (again, on paper...) the electrons from the double bond to the carbonyl oxygen, you should simultaneously move another pair of non-bonding electrons from the other oxygen in order to complete carbon's octet. Even on paper, you should never leave carbon with less than 4 bonds if they can be formed. This 3-bonded, positively-charged carbon is merely a methodological transition state for education purposes. Octet rule does not really apply in these circumstances.

Once you know the carbon involved, aren't the two assertions the same?

Then how do we say that an oxygen adjacent to carbon with a positive charge is highly stable if it's octet is complete and highly unstable if it's octet isn't while drawing the resonance structures?

I guess we could say that the farther you are from an octet, the more unstable you are. But again, all these drawing methods and resonance structures are for illustrative purposes. I don't think you could expect the octet rule (or any other rule, for that matter) to apply for a chemical structure that does not actually exist. The structure above does not float around in a stable manner. It does not exist independently...

Ok and just for educational purposes if we've an oxygen forming 3 bonds and having a lone pair with a positive charge and another again with a positive charge forming 1 bond and having 2 lone pairs which one will be considered closest to the octet rule and why? Can I say first because it will have 8 electrons and second will have 6? Just for the purpose of solving questions with ease.

Because if we go with the traditional way of 2p4 and it gaining just two electrons more then I think octet will be violated right? Or can I go with this way too and for the oxygen forming 3 bonds with a positive charge I can say that the valence configuration will be 2p3 and 3 bonds will form satisfying it's octet and the 2s2 electrons which don't participate will be shown as lone pair? Which approach is better according to you?

Octet rule means 8 electrons around an atom, doesn't matter if bonding or non-bonding. If an atom doesn't have 8 electrons around it, it violates the octet rule (but that does not always mean that you made a mistake; there are cases where the octet rule is not satisfied, for various reasons: expanded valence shells, hydrogen requiring only two electrons etc.)


text:
What is the fabric in heating mantles made of?
An ordinary lab heating mantle often has a white fabric coating for the flask compartment, which looks like an inorganic fiber:

Image source
I'm pretty sure that for some time in the past asbestos was used for that; these days I think it's some kind of woven fiberglass (assuming the heating element produces elevated temperatures up to about $\pu{400^\circ C}$), but I'm having hard times finding an exact chemical composition.
Also, what kind of chemicals can react with this coating? I guess something like molten $\ce{NaOH}$ and the melts of alkali metals can cause some damage, but I'm not sure.

Those older ones are often made of woven fibreglass, traditionally with aspestos behind, or nowadays usually some sort of ceramic.   More modern isomantles use an aluminium mesh instead of the fabric, again,with a ceramic insulator behind.

@NotEvans - Good answer - how about reposting your comment as the answer?

From one of the patents, "The textile-like materials referred to are generally fabrics made from quartz fibers or from glass fibers, or asbestos fibers, so to withstand the sometimes substantial temperatures developed by the heating blankets and heating mantles"

Early laboratory heating mantles use glass fiber yarn knitted material with a heat resistant naturally occurring fibrous mineral called asbestos.
Modern heating mantles have found their alternatives
among the metals like aluminium with ceramic or porcelain for insulation which helps in prevention of conduction of thermal shocks to glassware.

source: Stack Exchange

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text:
What explosive compound would be the most likely first explosive developed by a race?
Elaboration on title: What would most likely be the first explosive compound that some individual would create if they were just mixing together random things they had available? I'm not talking about our world history, where it was the discovery of black powder in China about 1200 years ago. I'm talking about some arbitrary race somewhere in the universe; what would most likely be the first explosive they develop, combining anything organic they could get? Would it be black powder? Nitroglycerin? Nitrocellulose?
(Keep in mind that this question is coming from someone who isn't really a chemistry nerd :p)

I don't think this is opinion-based, actually, so I decided against closing it unilaterally. The community might disagree. I think if you explain why you want this information, (is it for writing some book or something?) then it can probably be left open. You should use the tag chemistry-in-fiction instead of history-of-chemistry as well, since you're not asking about real history.

Are you serious, or is this info for a book (fiction/fantasy)? If it's the latter, then the choice of compound is really just up to you (the author). If it is the former, you'll have to specify what kind of environment this "race" (of what?) live in; throw in a little geochemistry and you may have your answer. Besides... how would YOU define an "explosive"?

@ortho Actually, I think it's better off on the WorldBuilding.SE...no... it is meant for the WorldBuilding.SE. So, migrate? O:)

@paracetamol It might very well do better on worldbuilding, but migration should only occur if it is definitely off-topic on the original site. Simply because it is better for site X does not necessarily make it off-topic for site Y; sometimes it does, but not always.

Voting to close as too broad. The first such compound discovered at some particular hypothetical planet will depend on the geology, geochemistry, ecology, plant biology, and other factors of that hypothetical planet. It could even be some compound not naturally occurring on Earth. There is no single answer here, so this is a poor question for Chem.SE; it is an ideal question for Worldbuilding.SE, though.

@hBy2Py I disagree. If we allow other geochemistry then you are right, but the question is OK if we interpret it as an alternative history of this planet. In fact it is interesting in that context and potentially offers useful chemical insight.

@matt_black From the post: "I'm talking about some arbitrary race somewhere in the universe". As written, off-topic.

I think the question is interesting, yet too broad and opinion based. For example, I could answer with "Definitely black powder," until or unless you remind me that on the planet in question there happens to be little to no sulfur and the atmosphere is comprised entirely of argon. I'd like to see it narrowed-down and will happily vote to reopen when the scope of the question is amended.

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text:
Hydrolysis of an ester

An organic compound X of formula $\ce{C5H10O2}$ forms two compounds when boiled in aqueous sodium hydroxide solution. One of these compounds forms a ketone when treated with acidified potassium dichromate.
Compound X could be
\begin{array}{ll}
\textbf{A} \quad \ce{CH3COOCH(CH3)2}. \qquad & \textbf{C} \quad \ce{CH3CH2COOCH2CH3}. \\
\textbf{B} \quad \ce{CH3CH2CH2CH2COOH}. \qquad & \textbf{D} \quad \ce{(CH3)2CH2COOCH3}.
\end{array}

Can someone confirm that the correct answer is option A instead of C?
Since one of the compounds (alcohol) forms a ketone, therefore it must be secondary alcohol. This means the ester must be branched, so it would be A?

Oxidation of an alcohol (with potassium dichromate) gives a ketone when the alcohol is secondary, like isopropanol.  So, which ester out of the given choices has a secondary alcohol component?
You see that A is correct as claimed.


text:
How do the electronic properties of a catalyst affect the rate of oxidative addition?
In my project I have a coupling that was planned to be done using Pd (Suzuki or Stille possibly).
Initially, I used an aryl chloride (the SM had already the chloride so I went with it), but the oxidative addition was slow. Changing to iodide helped but still slow (can't use triflate as would be destroyed in an earlier step if carry through).
In my lectures, changing the halide is always given as a way of gaining selectivity, this is also what is taught in Clayden if you look for the information on organometallics. But a postdoc suggested changing the catalyst, as apparently changing the electronics in palladium also helps. Is this the case? Or is OA just controlled by the strength of the C–X bond?

You should try small-scale reactions with half a dozen ligands as in my experience the choice of ligand is key to the success of a coupling reaction.

To quote Hartwig:

Oxidative addition to more electron-rich metal centres tends to be more favourable than oxidative addition at metal poor centres [1]

This is generally a thermodynamic issue, and can be rationalised quite easily if one considers that the change in electron density at the metal centre during the oxidative addition process (note, theres a difference between formal oxidation state and the 'actual' electronics about the metal centre).
As you point out, most catalysts systems used are either Pd0 to begin with, or generate Pd0 via a Pd2 pre catalyst (the palladium needs to be in its 0 oxidation state for oxidative addition to occur, usually, though 2 -> 4 is also possible in certain circumstances).
Your confusion seems to have came from the actual nature of the active catalyst, although you'll always end up with Pd0 , the metal centre has ligands attached, and these ligands are able to tune various properties at the metal centre.
The issue with trying to optimise a reaction ( as you seem to be doing) is that there is an almost infinite number of combinations of catalyst + ligand that you could consider, and indeed chemical companies such as Sigma-Aldrich even sell screening kits which provide many of the most common.
My honest advice in your situation would be to have a thorough search of the literature and try a few combinations of ligand + catalyst that you have available in your lab. Doing the optimisation rationally can be extremely interesting (doing it properly requires changing catalysts, ligands, solvents, temperatures, additives etc and is all time consuming, though often done in industry during optimisation before scale-up), but is of little use synthetically if you simply need to drag material through.

[1]: Hartwig, J. Organotransition Metal Chemistry- From Bonding to Catalysis; University Science Books:Herndon , 2010


text:
Technical name for the holes (missing electrons) in a p-type semi-conductor?
When a pure silicon crystal (giant lattice) is doped with boron atoms one of the four covalent bonds made between one boron atom and four surrounding silicon atoms is missing an electron as boron has three valence electrons. Is there a technical name for these holes (lack of electron in a covalent bond).
Is it acceptable to say that these holes are indirectly the charge carriers as they are allowing the movement of the electrons?
Furthermore, could you say that these holes are positively charged?(Hence the name p-type semi-conductor?).

You can call them holes and treat as positive charge carriers AFAIK.

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

Yes, the technical name is, in fact, an electron hole (typically referred to as simply a hole). It is considered suitable to describe these holes as electric charge carriers with a positive charge that is equal in magnitude to the negative charge of an electron. So, you have actually answered each one of your questions!

The holes are named electron holes or positive holes. Such holes are created when electrons go missing from a lattice site. In this case the ions or electrons from other lattice sites move into these holes one by one in direction of positive charge under the effect of electric field. Hence it can be said the positive holes move in the direction opposite to that of the electrons. Thus these holes increase conductivity of a solid. Since conductivity is due to "positively charged holes ", they are called "p type conductors ". A holes are created generally due to doping.


text:
calculate the number of shots of Captain Morgan’s rum
I am facing a challenge, and this is my first question in this stack exchange. I have worked out the following problem but the answer I got (number of shots) is ridiculous. Could any of the chemist here please advise me where I went wrong?
Question: Assume that you are on vacation in Florida and enjoying the company of friends at a bar. Using the information in Table 80-1 in your text, calculate the number of shots of Captain Morgan’s rum required to take your blood ethanol concentration above twice the legal limit to drive. [Note: The legal limit in Florida is 80 mg/dL]. Assume that the shots are consumed in rapid succession (the wisdom of which you were later unable to explain). Show your complete calculation, including all inputs and assumptions.
My calculation:
Table reference
Let us assume that my binge-drinking has led to a blood alcohol concentration (BAC) of 162 mg/dL.
The formula for BAC is as follows:
`BAC = A/(R x W) – (0.15 x H)` where

A = weight of pure ethanol consumed (g)
R = Widmark’s rho factor (0.68 and 0.55 L/kg for men and women, respectively)
W = body weight (kg)
0.15 = average ethanol elimination rate in humans
H = drinking period (hr)

Inserting the values of 162 mg/dL for BAC, R = 0.55 L/kg, W =  50 kg, and H = 1 hr, the equation becomes:
162 = A/(0.55 x 50) – (0.15 x 1)
162 = A/27.5 – 0.15

Therefore,
A = 27.5(162 + 0.15)
    = 4459.125 g

According to table 80-1 in the book, one shot is 30 mL and contains 40% ethanol.
So one shot contains 12 mL ethanol.
Density of ethanol = 0.789 g/cm3
The volume of pure ethanol which I consumed = 4459.125/0.789 = 5651.62 mL
So, no. of shots consumed = 5651.62/12 = 471, over the course of 1 hour
Conculsion: 471 shots of Captain Morgan Rum is absurd for any person. So I think I am missing something in my calculation. I'm not able to figure out what it is.

Note that some of the masses are in grams, others are in milligrams, and still some more are in kilograms. You need to perform the correct conversion. The main issue I see is that you didn't bother to write out units in your computation, so there's no way to tell that you are making this mistake as you are making it.

@Zhe I think I have mentioned all units and their conversions were already checked, yet I am pretty sure there some tiny mistake which I am not able to catch here. Please ask me which part of my question is not made clear, and I will correct it, as it will help the community as well.

OK. Average alcohol elimination. What are the units? Also 162 is in mg/dL. Yet somehow you magically multiplied some numbers that were either in unknown units or in kg and without any conversion.

@abhilash sukumari BAC is in mg/dl. You need to make each term on the right hand side match those units. Right away, the R factor being in kg/L will not lead to the right units. You also don't specify the units of 0.15 in the 2nd term, so its unclear if that term has the correct units.

As Zhe mentioned, you have inconsistent units in your calculation, so your numerical results are not physically meaningful.
Take for instance this expression:
162 = A/(0.55 x 50) – (0.15 x 1)
Here, you have different units for each term; from the values you supplied, the units are
162 mg/dL = A/(0.55 L/kg x 50 kg) - 0.15 g/(L•h) x 1 h
[Note that the units for the elimination rate weren't stated in the question, but a quick search of the literature gave studies measuring it at about 0.01-0.02 g/(dL•h), so I infer the 0.15 value refers to g/L]
If we use uniform units throughout, we get
1.62 g/L = A/27.5 L - 0.15 g/L
from where you get a more reasonable value for A:
A = (1.62 + 0.15) * 27.5 = 48.675 g
Each shot contains
V = 30 mL * 0.40 / 0.789 g/mL = 15.2 g
of ethanol, so you need to drink 48.675 / 15.2 = 3.2 shots of rhum over the course of an hour.
I'd suggest, however, that drinking shots in "rapid succession" implies a negligible drinking time (∆t ~ 0 h). In that case, A = 44.5 g and you need about 3 shots (2.93).


text:
Why is hydrogen evolved at the copper side of a lemon battery?
I have 2 quick questions about the principle of a lemon battery.
What I understand is that:

Zinc is oxidized, freeing up electrons that travel through an external wire to the copper electrode (to which they are pulled)
Once the electrons reach the copper electrode, they react with $\ce{H+}$ ions in the lemon, resulting in $\ce{H2}$ (gas).

Since the zinc electrode that generates electrons is already immersed in the lemon (containing $\ce{H+}$), why wouldn't the electrons just react at the zinc electrode with $\ce{H+}$ ions in the lemon? Why would they have to travel to the copper electrode before doing so?
Once the electrons reach the copper electrode, why wouldn't they react with the $\ce{Cu^2+}$ ions that have been eventually formed at the copper electrode (even though I understand there wouldn't be a lot of them since Copper is not very reactive)?

So you have a zinc electrode and a copper electrode and a lemon. The lemon is not a reagent grade chemical, so let's say it's the citric acid in the lemon that provides the corrosive action.
The copper essentially just sits there; as you indicated, the low reactivity of copper in acid means it doesn't dissolve readily. On the other hand, the zinc is much more reactive, and as you suggested, should just react all by itself. If it did, it would make a terrible battery; it would have a very short shelf life because the zinc would dissolve without providing an exterior current. But zinc has a high hydrogen overvoltage; that is, hydrogen does not readily evolve from zinc. The reasons are complex, but you may imagine that hydrogen ions or atoms are adhered to the zinc surface in such a way that they do not easily combine to form molecules which can form bubbles and go away. So the hydrogen evolution reaction is stifled.
However, the copper surface is different, and does permit hydrogen ions to react with an electron to form an atom which can then find another atom to form a molecule, and then more, to form bubbles. Alas! Copper has no extra electrons to spare; after you take off a few, the copper becomes positively charged and repels hydrogen atoms.
But the zinc electrode has gained a few electrons from the few zinc atoms that have dissolved in the citric acid. Let's connect the zinc electrode to the copper electrode: electrons flow from the zinc to the copper where they attach to hydrogen ions. The copper provides a surface for hydrogen ions to grab an electron and eventually become a bubble. The zinc provides electrons by dissolving as ions and leaving the electrons behind. And the battery could be useful for a long time if you disconnect the wire, because the zinc stops dissolving.
If this is still not clear, imagine a "battery" with a copper electrode and a sodium electrode. When you put the sodium into the lemon, it will react immediately without needing the copper to distribute its electrons. This "battery" might last about 2 seconds and would not be very useful.
And, as it turns out, impure zinc may have inclusions of different materials that provide a cathode surface right in the midst of the anode,  so the zinc could dissolve rather rapidly and make a poor anode. In the distant past, a tiny bit of mercury would be amalgamated onto the zinc, covering up the impurities. Mercury has an even higher hydrogen evolution overvoltage, so the zinc was protected. The electrochemical reaction then was the zinc dissolves in the mercury and travels to the mercury-solution interface before it ionizes and dissolves, but since the mercury layer was so thin, this was no problem. Eventually mercury was eliminated for potential human health reasons.

When the Zn-atom from the Zn-plate breaks down in the Zn++-ion and two electrons, this creates a equipotential, related to the electrochemical potential (ECP: which is typical for each type reaction).
At the other plate happens something alike: H+ steals an electron from the Cu-plate, but with another typical equipotential state. The difference between the two states of energy causes the flow of the electrons and therefore an electrical current.
The electrons coming from Zn don't actually travel through the wire to arrive at the Cu-plate so they can react there with H+. These chemical processes happen all at the same time.
So you can say that H+ needs special conditions to bond with an e- : the conditions at the Zn-plate don't suit these conditions, but these near the Cu-plate do. That why you can see the H2-bubbles appear near the Cu-plate.


text:
Energies of d orbitals relative to the barycentre

Rather than referring to the energy of an isolated metal atom, the weighted mean of these two sets of perturbed orbitals $(\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$ and $\mathrm d_{x^2 - y^2}$, $\mathrm d_{z^2} )$ is taken as zero. This sometimes is called the barycentre. The difference in energy between the two $\mathrm d$ levels is given either of symbols $\Delta_\mathrm O$ or $10~\mathrm {Dq}$. It follows that $\mathrm{e_g}$ orbitals have $+0.6 \Delta_\mathrm O$ above average level and $\mathrm{t_{2g}}$ orbitals are $-0.4\Delta_\mathrm O$ below the average.

It might be basic but why is orbitals split in $-0.4$ and $0.6$? If the barycentre is at the average then the orbitals should be split in $-0.5$ and $0.5$ below and above average, no?

No, there are 3 low and 2 high energy orbitals. You should put the barycenter at the average location.

Can you please add the source from which you quoted this passage.

@orthocresol Sure.

@PrittBalagopal  Do you mean something like $B = \dfrac{3L + 2H}{5}$ ?

@123 Precisely.

The passage you quoted is perfectly correct and contains one word which you may have overlooked:

the weighted mean of these two sets of perturbed orbitals is taken as zero [...]

In this case the weight corresponds to the degeneracy of the set. The lower-energy $\mathrm{t_{2g}}$ set contains three orbitals and the higher-energy $\mathrm{e_g}$ set contains two. If we place the $\mathrm{t_{2g}}$ set at $x \Delta_\mathrm{O}$ and the $\mathrm{e_g}$ set at $y\Delta_\mathrm{O}$, then
$$\begin{align}
y - x &= 1 & &\text{(difference must be equal to }\Delta_\mathrm{O}) \\
\frac{3x + 2y}{3+2} &= 0 & &(\textit{weighted}\text{ mean is equal to zero)}
\end{align}$$
and $(x,y) = (-0.4, 0.6)$.


text:
Calcium based Corrosion-Inhibitors used in Ethanol fuel, and their effect on the organic-solvent abilities of that alcohol?
Ethanol is a good organic solvent, often the preferred non-toxic choice for the processing of plant extracts. One of the cheapest sources of Ethanol is the kind used in fireplaces and racing fuel, but this has the corrosion inhibitor DCI-11 added. DCI-11 is added to the Ethanol at a rate of 86mg/litre and is made from Calcium Nitrate (max 1-10%) and Calcium Nitrite (max 25-50%). DCI-11 "reduces the corrosion and solvent effects of the alcohol, also buffering/stabilising the pH". My question is - to what extent if any, will this reduce its use as an organic solvent for the processing of plant extracts?

I expect the purchaser can accept or reject any addition to a commodity.


text:
Does selenium sulfate corrode or otherwise react with metals?
I was just watching a cartoon called Motorcity, and they featured a real compound called selenium sulfate.
While I'm sure that the compound blowing up a detached steel [car part] on contact is just cartoon wackiness, I'm curious whether it actually does anything bad to any metals, especially since a close compound, selenium sulfide, is commonly used in shampoos.

Yes, selenium sulfide is well known and it is used in anti-dandruff shampoo but other sulfur-containing selenium compounds are not really well known. Selenium sulfate, properly selenium(II) sulfate, $\ce{SeSO4}$ is unknown to the best of my knowledge. The analogous tellurium compound has been synthesized recently, but still it is not a proper sulfate compound, rather it is actually an oxide-sulfate mixture type of compound.
In case you are wondering whether selenium sulfate will be obtained by reacting selenium with sulfuric acid, then you are wrong. Selenium and sulfuric acid would react to form a mixture of acids (selenous and selenic acid) but they would all disproportionate to selenium dioxide, sulfur dioxide and water.

$$\ce{Se + 2H2SO4 ->[\Delta] SeO2 + 2SO2 + 2H2O}$$
Selenium sulfite or properly selenium sulfoxide, $\ce{SeSO3}$, is however known.
So, selenium sulfate is a fictitious compound and thus used in the cartoon.

Just as an additional detail: "selenium sulfide" is actually selenium disulfide, $\ce{S=Se=S}$, and not some salt of $\ce{Se^n+}$ and $\ce{S^2-}$. (Five years later edit: this is incorrect, please see below.)

@orthocresol Selenium (di)sulfide is not even that. Commercial material, as in shampoos, are actually a mixture of various molecules containing selenium and sulfur. See https://en.wikipedia.org/wiki/Selenium_disulfide.

@OscarLanzi Hmph, interesting. I don't even know what motivated me to post that comment, then.


text:
Making nylon experiment - Questions
I made nylon in a lab with these procedures -
1) Collect 30mL of sebacoyl Chloride dissolved in hexane
2) collect 30mL of 1,6-hexanediamine dissolved in water .
3) Transfer the beaker containing sebacoyl Chloride in hexane and carefully pour the solution down the side of the beaker containing 1,6-hexanediamine in water , such that the hexane layer floats on TOP of the water layer , without mixing the layers.,
I got these procedures from the Internet .
Now , I got a few doubts and questions ..

In the past , dichloromethane was used as the Solvent for the sebacoyl Chloride reagent. However dichloromethane is not environmentally friendly solvent and hexane was chosen to replace it in the experiment ...

a) how does switching it to hexane change the experimental procedures ?
b) What physical and chemical properties of the Solent must be considered when switching ?
c) why is hexane a good alternative in this experiment ?
Thanks in advance !

Nylon reaction:
$\ce{nH2N(CH2)6NH2 + nClOC(CH2)8COCl → H2N[(CH2)6NHCO(CH2)8]_{n}COCl + nHCl}$
Looking at the reagents, 1,6-hexanediamine is dissolved in water as it can form hydrogen bonds with the water molecules. Meanwhile the sebacoyl chloride is non-polar and thus dissolves best in non-polar solvents. The two solvents (polar and non-polar) cannot dissolve each other, just like oil and water.
The hexane is poured so it floats on top to minimize mixing -- the reaction only happens at the boundary between layers. Then we pull the thread causing more of the reagents to mix and react at the boundary. If we poured the hexane solution in quickly, we'd end up with a large lump of nylon instead of a thread.
To help with your other questions, some questions...

I've given some details about the solvents above. What kind of solvents would you choose to replace the hexane and water with?
Looking at the densities of hexane (0.659 g / mL) and water (1 g / mL), and looking at the procedure, why was the hexane poured? Why not pour the water onto the hexane?
Following on, do some research into the density of dichloromethane. That will give you your answer to (a)
You've mentioned "not enviromentally friendly" for dichloromethane. Are there any other issues? A good starting point are the CLEAPSS student safety sheets.


text:
Energy levels in non-hydrogen-like atoms
The energy $E_n$ for a hydrogen like atom is given as
$$E_n = -hcR_\ce{H}\frac{Z^2}{n^2}$$
However, aside from on wikipedia where there is
$$E_n = -hcR\frac{Z_\text{eff}^2}{n^2},$$
I can't find anywhere that relates to the energy of an atom that has more than one electron. Is there an equation that gives the specific (or approximate) energy or is there an 'easy' way to calcuate a value for $Z_\text{eff}$?

very approximate https://en.wikipedia.org/wiki/Slater%27s_rules

Thanks... seems to be along the lines of what I'm looking for.

Consider an atom to consist of an electron orbiting a positvely charged ion core (with one or more electrons associated to it). If you excite the electron to larger and larger values of the principal quantum number $n$ the classical electron orbit becomes larger and larger as well (in fact it scales as $n^2$). At sufficiently large distance, the postively charged ion core behaves as a point charge (like a single proton) and at long range you basically have the hydrogen problem with a different mass (assuming perfect shielding of the core electrons). In order to find the overall solution to this problem, one has to connect the long-range and short range behavior of the system. One way of doing this, is is by considering the system as a collision problem at negative energy (that is, a bound system). The effect of the nonhydrogenic ion core then results in a scattering phase shift between an incommming and outgoing electron. It can be shown that the solutions to this problem have a very similar appearance as the Bohr formula, namely:
$$
E_{n,\ell}=E_\text{IE}-\frac{hcR_A}{(n-\delta_\ell)^2}
$$
where $R_A=R_\infty(1-m_e/m_A)$ is the mass-corrected Rydberg constant, $E_\text{IE}$ is the ionization energy and $\delta_\ell$ is the so-called quantum defect for electrons with orbital angular momentum $\ell$. The quantum defect is related to the phase shift induced by the core and is different for $s, p, d, f$ etc. electrons. In principle, the quantum defect is a function of the binding energy of the electron to the ionic core, but in many cases this dependence may be neglected (in particular for large $n$).
What does this mean in practice? Let us for example look at the $n$s levels of the sodium atom. We find that $R_\text{Na}=109734.697205$ cm$^{-1}$ and $E_\text{IP}(\text{Na})=41449.451$ cm$^{-1}$. The second column in the table below gives the energy (in cm$^{-1}$ from the NIST website) and the third column gives the binding energy of the electron (also in cm$^{-1}$). The last column gives the so-called effective quantum number, that is $n^*=n-\delta_\ell$, with $\ell=0$ for $s$ orbitals
 3s         0.000   -41449.451  1.63
 4s     25739.999   -15709.452  2.64
 5s     33200.673    -8248.778  3.65
 6s     36372.618    -5076.833  4.65
 7s     38012.042    -3437.409  5.65
 8s     38968.510    -2480.941  6.65
 9s     39574.850    -1874.601  7.65
 10s    39983.270    -1466.181  8.65
 11s    40271.396    -1178.055  9.65

As you can see the quantum numbers take the form of the hydrogen solution if we start counting the principal quantum number form the 3$s$ levels and take $\delta_0\approx -0.65$.
To calculate the quantum defect is difficult as it requires ab initio calculations and then converting the energy levels to effective quantum numbers from which the defects can be extracted. However, experimentaly these numbers are very convenient, and one may use this quantum-defect theory for instance to determine the ionization energy of atoms and even simple molecules very accurately by extrapolating Rydberg series.


text:
What are the essential arrows for entering chemical equations in text?
I'm the author of a custom chemistry keyboard on iOS. I am looking at improving the selection of Unicode arrows I can provide. When the device is a phone in portrait mode, I don't have a lot of extra room on screen, so I give a flyout with:
→, ⇌, ↔, ⇢, ↛
When the phone is rotated or it's on a larger device, I have some extra room, so I was thinking about adding either more items in that flyout, or perhaps another button with an expanded set. Maybe something like:
⇄, ←, ↝, ↮⟿
I would like to make the most useful unicode arrows available, and not provide ones that nobody would use. I expect that this will vary from speciality to speciality, but I'd really appreciate some feedback.
I'm limited to what is in Unicode. (If you are on a Mac, you can bring up the Emoji and Symbols palette and look at the Arrows category)

I'd go with $\rightarrow,\leftarrow,\leftrightarrow,\text{ and }\rightleftarrows$. The rest are rarely needed anyway.

Still there are some usecases that cannot be covered with Unicode, e.g. I don't think there is an arrow for isolobality, which might come in handy sometimes, also how would one denote reaction conditions, which are typically written within the arrow? I would agree with @IvanNeretin that the basic set is good enough and for more advanced writings user will use other tools.

Yeah, conditions is a big thing. $\stackrel{\Delta}{\longrightarrow}$ is much more useful than $\rightsquigarrow$.

It seems like there is some distinction between the ⇄ and the ⇌ characters. Which is more useful?

I can see the value of this, but its highly subjective and dependant on the type/level of chemistry involved. It would perhaps be better off discussed on chat...

⇌ is for equilibriums, ⇄ is for forward/backward reactions.


text:
Selective sorption of toxins by polymethylsiloxane polyhydrate
Polymethylsiloxane polyhydrate (PMSPH) is used as enterosorbent for a couple of decades (nowadays under the trademark "Enterosgel"), intended for binding in the gastrointestinal tract and excretion of toxic substances of various nature, agents that can cause diseases and metabolites. Approximate matrix composition (2D) per micell looks like this:

Pharmacodynamics shows that PMSPH effectively bounds endogenous and exogenous toxic substances of various nature, including bacteria and bacterial toxins, antigens, food allergens, drugs and poisons, heavy metal salts, alcohol, metabolic products of the body (bilirubin, urea, cholesterol and lipid complexes).
What makes this gel so selective towards the adsorption of toxins only? Why nutrients (carbohydrates, fats, amino acids, vitamins) are mostly preserved and can be assimilated by the human body regardless?
I suspect it might have something to do with the pore size and geometry, involving a docking-alike behavior, but I don't have a solid enough background in biochemistry to prove it right or wrong. Manufacturer puts a warning "Protect from freezing!", so it might also have to do with the precisely controlled pore size (which is imminently changed after solidification of trapped water).


text:
Balmer and Paschen Series
If electron makes transition from, in $H$ atom, $6\rightarrow 2$ (to Balmer), then $\bar{\nu}_{{H}_{6\rightarrow 2}}=x$ (let).
Also if electron makes transition from, in $He^{+}$ ion, so that $\bar{\nu}_{{He^{+}}_{a\rightarrow b}}=x$, gives me $a=12$ to $b=4$, by numerical caulations.
Now, what is observed is if I multiply $Z_{He^{+}}=2$ to transition of $H$ i.e. by $2\times\big({6\rightarrow 2}\big)=12\rightarrow 4$, gives me same result without any nasty calculation, why this works I don't know can you help me with this.
Thanks.

Could you elaborate?

$$E_n = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
You can easily verify that if you replace $Z$ with $2Z$, then the energy of the corresponding transition will be four times the original energy. Now if you replace both $n_1$ and $n_2$ with $2n_1$ and $2n_2$, then you can take out a factor of $1/4$; so the transition $2n_1 \leftarrow 2n_2$ in an ion with nuclear charge $2Z$ occurs at the same energy/wavelength/wavenumber as the transition $n_1 \leftarrow n_2$ in an ion with nuclear charge $Z$.
Same applies for any other factor $k$: the $1 \leftarrow 2$ transition in hydrogen occurs at the same wavelength as the $4 \leftarrow 8$ transition in $\ce{Be^3+}$. Obviously this discussion ignores the variation in the Rydberg constants for different nuclei.


text:
Question about relation between two time constants
I am not a chemist, but a neuroscientist, so bear with me. I have struggled with this problem for over a week now and have realized it's a chemical question. I'll try to explain it so no neuroscience background is needed.
I am trying to fit a model to experimental data. The data describes how a certain channel opens and closes in response to a change in membrane potential of the cell the channel sits in. This is captured in two datasets:

The fraction of open channels at equilibrium for different values of the membrane potential (an example of a plot of this can be found here. This fraction I call G/G$_{max}$
The time constant describing the time it takes for the fraction to come to its equilibrium. This time constant I call $\tau_{G/G_{max}}$

It is thought that for each open channel, 4 gates have to be in the 'open' state. That is, it is thought that:
G/G$_{max}$ = n$_{\infty}^4$
The gate n, too, can be described by its fraction of open-state/total.
The rates of n moving between these two states are described by $\alpha_n$ and $\beta_n$. Once there are four gates in the open-state, the channel is instantly open.
 If I try to describe this in a chemical reaction, I would guess something like:
$$\ce{C_n <=>[\alpha_n][\beta_n] O_n}$$
where $C_n$ and $O_n$ depict 'closed' and 'open' n-gates respectively. And
$$\ce{4\cdot O_n <=>[instant][instant] O_{ch} }$$,
where $O_{ch}$ describes the whole channel being open.
Now is the data I've got for G/G$_{max}$(V) and $\tau_{G/G_{max}}$ (this is the only thing the experimentalists can measure). However, to model this behavior, I need $\alpha$(V) and $\beta$(V). I already know that
$n_{\infty} = (G/G_{max})^{1/4}$
but I cannot figure out how $\tau_n$ and $\tau_{G/G_{max}}$ are related to each other. Even though the reaction 4O --> open channel is instant, the shape of approaching the equilibrium differs and thus the point that they cross 1-1/e would as well. n will reach its equilibrium with a single exponential, while G/G$_{max}$ reaches it with an exponential to the fourth power.
I think the answer should be quite simple, and should only depend on the 4'th power.
I hope my question is clear, but if there are more details that will help solve this I am more than happy to clarify certain points.
Edit: Here is a figure that shows how the 'measured' time constant and the one for the differential variable n differ. Also, it shows that there is not a linear relationship.

Just to be clear about what I did:
The differential system describing this model looks like:
G/G$_{max}$ = n$^4$
with
$\frac{dn}{dt} = \alpha_n(V)(1 - n_{\infty}(V)) - \beta(V) n_{\infty}(V)$
As one of the answers pointed out (sort of), a solution to this system with initial condition V1 is:
$n(t) = n_{\infty}(V) \cdot e^{-(\alpha_n(V) + \beta_n(V))} + n_{\infty}(V) \cdot \Big(1 - e^{-(\alpha_n(V) + \beta_n(V))}\Big)$
and thus
$G/G_{max}(t) = n(t)^4$
To find the $\tau$ of either of these traces (the time constant being the time it takes to reach $1-1/e \cdot$(final value) (see link), I plotted:
$\frac{n - min(n)}{max(n) - min(n)}$
and
$\frac{n^4 - min(n^4)}{max(n^4) - min(n^4)}$

At the beginning of your post you stated 'I am trying to fit a model to experimental data', but then you wrote 'I cannot figure out how [...] are related to each other'. So it's the model itself that is missing, isn't it? If you have a complete set of equations relating all your variables and parameters, you can fit the data (i.e. find the parameters) even without an analytical solution, e.g. using R packages deSolve and FME. See this document and the specific package manuals for more info.

I think I understood your question now. I will reformulate it in more 'chemical' terms and give you my solution. Not nicely formatted for now, I will do that later if I find the time.
You have a cell with $G_{max}$ gates on it. They can open and close reversibly, depending on a voltage V. Closed gates open with rate constant $\alpha(V)$, and open gates close with rate constant $\beta(V)$.
This implies one differential equation. If $G_o$ is the number of open gates at time t, and $G_c$ is the number of closed gates at time t, then:
$\frac {dG_o}{dt} = \alpha G_c - \beta G_o$
And as the total number of gates is $G_{max} = G_o + G_c$:
$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o$
At steady-state, i.e. when $G_o$ stops changing, you have:
$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o = 0$
which, solved for $G_o/G_{max}$, yields:
$\frac {G_o}{G_{max}} = \frac {\alpha}{\alpha + \beta}$
The ratio $G_o/G_{max}$ is the probability that a single gate is open. As you say, a 'channel' is open when 4 of its gates are open. If we consider each gate being open or closed independently, the probability that 4 gates in a channel are open is indeed $(G_o/G_{max})^4$
What you measure is the time it takes, when the voltage is abruptly changed from an initial value to a final value, for the number of open channels to go from an initial value to 'equilibrium'.
This time is in theory infinite, but in practice it is probably the time it takes for the number of open gates to go from its steady-state value at voltage $V_1$, to a value sufficiently close to the new steady-state value at voltage $V_2$.
If so, it is sufficient to solve the above differential equation for $G_o$ with initial conditions $(t=0,G_o=G_{o,V_1})$. The solution is:
$\frac {G_o(t)}{G_{max}} = \frac {G_{o,V_1}}{G_{max}} \cdot e^{-{(\alpha+\beta)t}} + \frac {G_{o,V_2}}{G_{max}} \cdot [1-e^{-{(\alpha+\beta)t}}]$
where:
$\frac {G_{o,V_i}}{G_{max}} = \frac {\alpha(V_i)}{\alpha(V_i) + \beta(V_i)}$
The difficulty I see is the arbitrary nature of the 'time' you consider sufficient to reach steady-state. I would rather fit time points, i.e. number of open gates vs time. And my earlier suggestion to use numerical fitting may still be valid, actually.

First on the arbitrariyness of the time constant: I think people usually take the time where the variable crosses 1/e of the final (infinite) value

For time constant: https://en.wikipedia.org/wiki/Time_constant
About your solution: The thing is that this doesn't take into account the factor 4, does it? As in, G/Gmax will approach (1-1/e) exponentially, and the resulting shape is not correct. Instead the shape should be of exp^4 to achieve the shape depicted in this image: http://ecee.colorado.edu/~ecen4831/cnsweb/vcdata.gif (whereas without this power it would look like the blue line in: http://ecee.colorado.edu/~ecen4831/cnsweb/vclampact.gif).

The differential equation I'm solving is:
G = G/Gmax * n^4 together with
dn/dt = alpha_n * (1 - n) - beta_n*n

But the alpha and beta I am looking for are those that describe the equilibrium of n, and not that of G/Gmax (and same goes for the time constant). The problem is that I only have the data for G/Gmax.

Do you disagree with me?

I see that I actually used for gates the symbol you used for channels (G). Apart from that, I believe that the solution is there. Once you have an equation expressing the fraction of open gates at any time t as a function of the fraction of open gates at time 0 and the rate constants of gate opening and closing, you have all you need, haven't you? By your own equation you know that the fraction of open channels is a the 4th power of the fraction of open gates, so I'm not sure what else there is to know. The 4th power of the Go(t)/Gmax function I gave looks correct to me.

As for finding the $\alpha$ and $\beta$ constants from fraction of channels open vs time data, it it probably best approached as a non-linear fitting problem. R is again the tool I'd advise, for a nice statistical analysis. To start with however you could try Excel. If $f(V)$ is the fraction of open channels at steady-state at voltage V, then after an abrupt change of voltage from $V_1$ to $V_2$, $f = [f(V_1)^{(1/4)} \cdot e^{-{(\alpha(V_2)+\beta(V_2))t}} +(\frac {\alpha(V_2)}{\alpha(V_2)+\beta(V_2)})^{(1/4)} \cdot [1-e^{-{(\alpha(V_2)+\beta(V_2))t}}]]^4$.

You can plot your data (which I seem to understand are values of $f$ vs $t$) and the corresponding values of $f$ calculated with the above formula, using initial guesses of $\alpha(V_2)$ and $\beta(V_2)$, and then use Excel's solver to change these values until the sum of squared differences between each observed and calculated $f(t)$ is minimal.

The problem here is that I don't know what to fit that equation to, seeing that I don't have access directly to any information about the gate, but only to information about the channel. The conversion of the fraction information from channel to gate is easy: n = $G/G_{max}^4$. I don't know how to convert the measures time constants to the time constants that would describe the gates... (see figure I added). Thanks for your patience. As for the software, I use scipy's curve_fit

In the figure I added the orange trace would be the one I have access to (or actually only the value for the crossing point) and the blue one is the one I need to fit my model.

I don't understand your comments. If in your plots the x axis is time and the orange curve is the fraction of open channels, that's exactly what my last equation describes. $f$ in my last equation is the fraction of open channels. You don't need to fit the 'blue curve': you stated yourself that it is derived from the fraction of open channels vs time. So if you fit your original data to the orange curve, then you can calculate all the other parameters, in the same way as you did to go from the orange curve to the blue one. I can't see what you think is missing...


text:
Why does ozone disproportionate on reaction with potassium iodide?
In this redox reaction, iodide ion is oxidized to iodine and ozone reduces to hydroxide ion.
$$\ce{2KI + O3 + H2O -> I2 + O2 + 2KOH}$$
This seems like a redox reaction since iodide ion is getting oxidized and ozone is getting reduced. Then why is $O_{2}$ formed in this reaction?

To me, this hardly qualifies as disproportionation.

@Ivan Neretin yeah you are right. This isn't disproportionation since ozone is only getting reduced and not oxidized. But the question still remains. Why is oxygen being formed?

Well, $\ce{O2}$ is much less active than $\ce{O3}$, so the reduction just stops right there for kinetic reasons. If you apply a lot of $\ce{KI}$ and wait for some days, all oxygen will get reduced.

Do you mean that formation of oxygen is an intermediate step in the complete reduction of ozone? If that's the case, then why does some oxygen gets further reduced to hydroxide ion?

An intermediate step is still a step. If you reduce $\ce{O3}$, then one atom out of three gets reduced and the rest transform to the more stable $\ce{O2}$. If you don't reduce it, then there is no reaction and nothing to speak about.

If it's so easy to produce oxygen from ozone then why do we use electric discharge and maintain strict conditions of temperature and pressure to convert ozone to oxygen? All we can do is simply react it with a reducing agent and separate the so formed oxygen. Wouldn't that be more easy..

@Eloise I think you have that backward. We use electric discharge and strict conditions of temperature and pressure to convert oxygen to ozone. To reduce ozone to oxygen all you need is time; it self-disproportionates. If you don't have time, there are several catalysts that will make the disproportionation go faster.

The reaction equation does not really give information about the full mechanism, but one possibility is that the ozone is first reduced to ozonide ion, which is known in compounds with heavier alkali metals like potassium,; then the ozonide ion disprportionates to give diatomic oxygen:
$\ce{O3}+\ce{e-}\rightarrow \ce{O3-}$
$\ce{O3-}+\ce{e-}+\ce{H2O}\rightarrow 2 \ce{OH-}+\ce{O2}$
Note that the reducing agent is not specified as iodide ion.  Any mild reducing agent can act similarly.

As @Ivan said in the comment:

An intermediate step is still a step. If you reduce $\ce{O3}$, then
  one atom out of three gets reduced and the rest transform to the more
  stable $\ce{O2}$. If you don't reduce it, then there is no reaction
  and nothing to speak about.

This is the case. In an intermediate step, ozone gets reduced to diatomic oxygen and a monoatomic oxygen, also called nascent oxygen. Generally, this nascent oxygen is unstable so it reacts/gets used up to oxidize potassium iodide to iodine.

(Source)
Diatomic oxygen is stable and does not get used up in the reaction leaving it untouched during the reaction.


text:
Minimal basis CI of H2 dissociation
In Modern Quantum Chemistry by Szabo and Ostlund, they give an example of a minimal basis, full CI calculation of the dissociation of $\ce{H2}$ (on p.$241$ if you have the book). In explaining why this correctly describes the dissociation, they say:

To see this, recall that as $\mathrm{R}\rightarrow\infty$, ... that all molecular orbital two-electron integrals tend to $\frac{1}{2}(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$, where $\phi_{1}$ is a hydrogenic orbital.

I could not find where they justified this statement. The closest I could find was in a previous section describing the dissociation using RHF (p.$166$) they state:

[Besides $H^{\text{core}}_{11}$] All other integrals go to zero as $\mathrm{R}\rightarrow\infty$, except the one-center electron-electron repulsion integral $(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$.

What justifies this first statement? Is this unique to this case of a diatomic at infinite separation? These two comments seem to contradict rather than support each other.

Using Szabo and Ostlund's notation (page 85):
\begin{align}
J_{ij} &= (\psi_i\psi_i|\psi_j\psi_j)&
\text{and}&&
K_{ij} &= (\psi_i\psi_j|\psi_i\psi_j)\tag1\label{integrals}
\end{align}
For minimal basis set $\ce{H2}$ (page 162):
\begin{align}
\psi_1 &= [2(1+S_{12})]^{-1/2}(\phi_1+\phi_2)&
\text{and}&&
\psi_2 &= [2(1-S_{12})]^{-1/2}(\phi_1-\phi_2)\tag2\label{minimal}
\end{align}
Substitute equation $\eqref{minimal}$ into $\eqref{integrals}$, set $S_{12}$ and any integral involving both $\phi_1$ and $\phi_2$ to zero (large $R$) and you see that
$$J_{11} = K_{12} = \frac{1}{2}(\phi_1\phi_1|\phi_1\phi_1)$$
Remember that $(\phi_1\phi_1|\phi_1\phi_1)=(\phi_2\phi_2|\phi_2\phi_2)$ since you are dealing with a homonuclear diatomic.

$K_{ij}$ should be $(ij|ji)$, but I think it still works out the same. Thank you, this helped a lot! I guess part of my confusion was I was inadvertently equating R and r12.

if the orbitals are real then it's all the same.  Anyway, happy to hear the answer was of use to you

Here is my intuition. For two atomic orbitals $\phi_1$ and $\phi_2$ located on distinct atomic centers 1 and 2, respectively, there are two ways of justifying this.

The bra or the ket can be taken alone as an overlap integral over two Gaussians. At $r_{12} = \left|\mathbf{r}_1 - \mathbf{r}_2\right|$, $(\phi_1|\phi_1)$ and $(\phi_2|\phi_2)$ need not be zero because there is no "distance" to speak of; it's an integral over a squared Gaussian somewhere in space. However, because $\lim_{r_{12} \to \infty} r_{12} = \infty$, the overlap $(\phi_1|\phi_2)$ will approach (but not exactly be) zero. It is a matter of the overlap of two Gaussians at infinite separation being effectively zero.
Consider the full two-electron four-center integral, where the bra is on one center and the ket on the other: $$(\phi_{1}\phi_{1}|\phi_{2}\phi_{2}) = \int\!\int d\mathbf{r}_{1} d\mathbf{r}_{2} \, \phi_{1}^{*}(\mathbf{r}_{1})\phi_{1}(\mathbf{r}_{1})\frac{1}{r_{12}}\phi_{2}^{*}(\mathbf{r}_{2})\phi_{2}(\mathbf{r}_{2}).$$ Following the logic from above, the bra and ket taken separately as overlaps or charge distributions may be non-zero and large, however $\lim_{r_{12} \to \infty} \frac{1}{r_{12}} = 0$.

Now, if $\phi_1$ and $\phi_2$ are molecular orbitals that are delocalized, this will no longer necessarily hold; in that case, there is also the presence of the density $P_{\mu\nu}$. I am assuming that their hydrogenic orbitals are just like atomic orbital (Gaussian basis functions).

I might have been unclear with my question. The 2nd statement doesn't bother me, I think they explain it the same way in the book as you just did. My issue is with the first statement, which seems to contradict the 2nd.

I misread, sorry; I see why it's bothersome. Unless I'm missing something, it's worded poorly. I think it means that they all slowly disappear and you're only left with things that look like (11|11), so you "approach" only having (11|11). It isn't possible (by convention, at least) for the specific integral (11|22) to become (11|11). I don't think it's referring to the antibonding H2 molecular orbital, which would also go to 0 at infinite separation.

I would have thought so as well, but they use this statement to claim that the exchange integral K_12=1/2(11|11)

I'll have to find my copy of the book and remind myself of their operator notation then, because if J_12=(11|22), I would expect K_12=(12|12), which would go 0 under point 1.

K_12=(12|21), but I would think that should also go to zero simply due to the 1/r dependence.

I don't understand how $\lim_{r_{12} \to \infty} \frac{1}{r_{12}} = 0$  works. Isn't $\mathbf{r}_{1/2}$ the coordinates of the electron, not the "position" of the functions, therefore simply by increasing $R$, it does not say anything about the change for the electrons.

I was also inadvertently equating interatomic distance with interelectronic distance. This is true; all one can say is that the instantaneous interelectronic distance may be larger more often once $R_{AB}$ increases, but the $g_{12} = \frac{1}{r_{12}}$ operator is clearly not the reason for this going to zero. The wording for point 1 is correct but the mathematical reasoning is wrong there as well.


text:
What function is used to graph the potential curve of two hydrogens with parallel spins?
TL; DR
I am asking for

an analytic potential curve function for the triplet state of $\ce{H2}$ (preferred) with experimental parameters, or
experimental data (enough points to realise a smooth-looking curve).

Also see solid red curve pictured below.

To characterise the energy of a two-atom system, Morse potential is applied as a second approximation after the harmonic oscillator model. This is an analytic function of the form $$E=D_\mathrm{e}\left\{1-\exp[-a(r-r_\mathrm{c})]\right\}$$
where the parameters $D_e$, $a$ and $r_c$ depend on the atoms in play. Surely enough, better results are achievable. For examples on simple molecules, I found the paper by Hulburt and Hirschfelder, 'Potential Energy Functions for Diatomic Molecules', J. Phys. Chem., ($1940$) to be rather useful. Have a look at equations $(14)$ and ($15$) specifically (available more easily here). The one I used was

$$E_{\mathrm{P}}=\mathrm{D_e}[(1-\mathrm{e}^{-z})^2+\mathrm{c}x^3\mathrm{e}^{-2z}(1+\mathrm{b}z)],$$
where
$$z=\frac{\omega_\mathrm{e}}{{2}\sqrt{\mathrm{B_eD_e}}}\left(\frac{x-\mathrm{r_0}}{\mathrm{r_0}}\right)$$

This is for the case where spins are antiparallel.
If atoms of parallel spin approach one another, the qualitative picture is quite different. A smaller distance will equal higher energy regardless of separation, i.e., no stable equilibrium state exists. Unfortunately, finding a nice function for the antibonding situation proved less fruitful. An example of what I'm after is the solid red curve found in this Wolfram demonstration.

I did contact a representative of Wolfram Research Europe, but (for some reason) failed to receive an answer. Furthermore, I was initally reluctant to post this question due to the fact that it is essentially a data question... but here it is anyway. If it's off-topic, closing will not offend.

I believe that you wont be able to find a closed analytic expression for $\mathrm{H_2}$, since no closed analytic expresion is possible for a many-body problem (more than 2 bodies). That said you can calculate as many points on the triplet curve as you like by using a QM program able of doing unrestricted Hartree-Fock (UHF). Here you should be aware that UHF does not give very good potential curves, because the Hartree-Fock approximation lacks the electron correleation energy. $\mathrm{H_2}$ is a quite small system, so you will be able to make a potential curve using unrestricted MP2 or unrectriced CCSD. I do believe that PySCF is capable of doing such calculations. Regarding the experimental data, I am not aware of any databases for such kind of data. I would suggest finding a paper where a curve is presented and then use engauge digitizer to extract the points you need.

Thank you for the insight! Apologies for the delay, I was swamped. Didn't even think of the engauge digitiser approach. Will give it a go in the coming days, and depending on the outcome (probably) will accept the answer :).

Potentiology is a field that began in the early days of quantum mechanics:

1924: John Lennard-Jones publishes the Lennard-Jones pontial
1929: Philip M. Morse publishes the Morse potenital.

The most recent reference you gave was a potential from:

1940: by Hugh Hulburt and Joseph Hirschfelder

You seem to be hoping that in the last 80 years, someone might have come up with something better.
 The answer is the 2009 Morse/Long-range, which is the most flexible potential energy function known. It has the theoretically correct long-range behaviour built into it. You can easily do the anti-bonding potential, and also double-well potentials, potentials with a shelf, potentials with a shoulder, potentials with a barrier, etc.


text:
Why is gadolinium specifically used in MRI contrast agents?
Gadolinium(III) chelate complexes are routinely used as contrast agents in magnetic resonance imaging (MRI);1 the usual explanation is that paramagnetic species contain unpaired electrons, which cause relaxation of nearby $\ce{^1H}$ nuclei.
Why gadolinium, and not any of the other lanthanides?
In particular, $\ce{Tb^3+/Dy^3+/Ho^3+/Er^3+}$ have magnetic moments that are larger than that of $\ce{Gd^3+}$ (due to orbital angular momentum). At first glance, these should possess even higher relaxivity, since the rate of relaxation due to random fields is proportional to $\langle B_\mathrm{loc}^2 \rangle$.2
I do vaguely recall reading one sentence, about the fact that ground states that aren't S terms are not spherically symmetric, and there's some problem associated with this. But I can't find the reference anymore and I don't remember any other detail.

References

Werner, E. J.; Datta, A.; Jocher, C. J.; Raymond, K. N. High-Relaxivity MRI Contrast Agents: Where Coordination Chemistry Meets Medical Imaging. Angew. Chem. Int. Ed. 2008, 47 (45), 8568–8580. DOI: 10.1002/anie.200800212.
Hore, P. J. Nuclear Magnetic Resonance, 2nd ed.; Oxford University Press: Oxford, U.K., 2015; pp 65, 71.

Probably medical/economical reasons.

With very limited knowledge in this subject, I recommend reading a review 'Alternatives to Gadolinium-Based Metal Chelates for Magnetic Resonance Imaging' (Chem. Rev., $2010$, 110 (5), pp 2960–3018, DOI: 10.1021/cr900284a). It is also freely available as a manuscript here. Sections $3$ and $5$ should be especially pertinent. It specifically discusses $\ce{Gd^3+}$ vs other $\ce{Ln^3+}$ complexes.

You see a lot more variety in situations where people aren't bothered if the animal survives long term.. I think in human cases its simply that the Gd systems have been through trials and aren't horrendously toxic/can just about be gotten rid of by the body

I found this section on Wikipedia after a short browse: "The maximum number of unpaired electrons is 7, in $\ce{Gd^{3+}}$, with a magnetic moment of 7.94 B.M., but the largest magnetic moments, at 10.4–10.7 B.M., are exhibited by $\ce{Dy^{3+}}$ and $\ce{Ho^{3+}}$. However, in $\ce{Gd^{3+}}$ all the electrons have parallel spin and this property is important for the use of gadolinium complexes as contrast reagent in MRI scans."

The reasons may possibly be:

T1 shortening

Relatively less adverse reactions

Increased contrast enhancement

The most commonly used clinically approved contrast agents for MR
imaging are gadolinium-based compounds that produce T1 shortening.

Tissue relaxation results from interactions between the unpaired electron of gadolinium and tissue hydrogen protons, which significantly decrease the T1 of the blood relative to the surrounding tissues.
Additionally, adverse reactions to this agent are far less frequent than those seen with iodinated compounds, with common reactions including nausea, vomiting, headache, paresthesias, or dizziness.

T1weighted sequences are part of almost all MRI protocols and are
best thought of as the most 'anatomical' of images, resulting in
images that most closely approximate the appearances of tissues
macroscopically, although even this is a gross simplification.

Figure 1-10. MR enterography on a patient with Crohn disease. A. Coronal T2-weighted image shows wall thickening
and stenotic ileum (arrow). B. Coronal gadolinium-enhanced fat suppressed T1-weighted image shows increased
contrast enhancement of one thickened segment of ileum (arrow).
Credits Basic Radiology Chapter 1 page 11
Update
Most of the answer above was based on pharmacological/medical aspects but some chemistry aspects have been included below:
Also the article mentioned by@Linear Christmas contains valuable information on the subject
Contrast agents

Contrast agents can be divided into two groups depending on whether
they cause changes in either T1 (longitudinal relaxation – in simple
terms, the time taken for the protons to realign with the external
magnetic field) or T2 (transverse relaxation – in simple terms, the
time taken for the protons to exchange energy with other nuclei)
relaxation rates of the water protons, these being known as positive
or negative agents respectively.
The ability of an agent to affect T1 and T2is characterised by the
concentration-normalised relaxivities r1 and r2respectively. These
parameters refer to the amount of increase in 1/T1 and 1/T2
respectively, per millimole of agent, and are normally quoted as a
rate (mM21 s21).4 The values are used to determine the efficiency of a
contrast agent, and they consist of contributions from both inner
sphere and outer sphere relaxation mechanisms Negative contrast agents
influence the signal intensity by shortening transverse relaxation
(T2), thereby producing darker images as high T2 results in increased
brightness of the images.

Gadolinium (III) vs other lanthanide (III) ions

Gadolinium(III) reagents are commonly focused on, due to the coupling
of a large magnetic moment with a long electron spin relaxation time
of $\ce10^{-9}$s  at the magnetic field strengths used in MRI
techniques.

Gadolinium (III) ion is unique in a number of respects:
It has the ability to shorten both the longitudinal and transverse relaxation times of water protons approximately to the same extent by relaxing all nearby protons. Other Ln(III) ions with large magnetic moments are less efficient in shortening T1.

The gadolinium ion is also unique among other Ln(III) because its
symmetric seven electron ground state results in an electronic
relaxation rate that is six orders of magnitude slower than the
other Ln(III) ions. This in turn results in a strong magnetic
anisotropy and fast electronic relaxation states with very short T1,
on the order of $\ce10^{-13}$s.

Other Ln (III) ions tend to undergo Curie relaxation enhancement arising from the interaction of the nuclear spin with the thermal average of the electron spin. (The Curie-spin relaxation effect is significant at lower temperatures and higher magnetic fields and for ions with a large magnetic moment).
It has been observed that the Curie spin relaxation effect affects the transverse relaxation more than the longitudinal relaxation. Consequently paramagnetic Ln(III) other than Gd (III) are less efficient T1 relaxation agents.
Use of other Lanthanide (III) ions as MRI contrast agents
Dysprosium(III) is another lanthanide ion that has been used in
MRI, being classed as a negative contrast agent. In the use of high magnetic fields,

Gd(III) based contrast agents exhibit poor water relaxivity. As a
result, the interest in dysprosium-based complexes is increasing as
they display slow water exchange, due to the need to lengthen the
residence time in order to optimise the r2 relaxivity. Dy(III) has a
large magnetic susceptibility which induces local field gradients
resulting in a lowering of T2.

Generally, dysprosium complexes where the water molecules have a long residence time, possess potential application as negative contrast agents
at high magnetic fields due to their efficient transverse relaxivity. However it has been demonstrated that lengthening the residence time of water can actually be detrimental, because the transverse relaxivity
can then be limiting.
The overall conclusion from the use of dysprosium(III)
complexes as contrast agents is that due to the balancing of
factors required to optimise r2, design of a suitable molecular
structure is crucial. Fine tuning these residence times of the
water protons, and hence the relaxivity, may lead to promising
contrast agents for high field magnetic resonance imaging.

Europium(II) analogues have been proposed as alternatives to Gd(III)
because they are isoelectronic—each having seven unpaired electrons.

The only problem with this complex was that the stability of the Eu(II) chelate was 107 times higher than that of the Eu(III) chelate, which
meant that the Eu(III) complex could be susceptible to dissociation, possibly releasing toxic Eu(III) into the body.
Other strongly paramagnetic ions e.g Tb(III), Ho(III), Er(III) can have significant effect on line widths and such line broadening usually makes it difficult or even impossible to detect nuclei that are located within a certain distance of the lanthanide ion.
There is still promising use of other Lanthanide (III) compounds as contrast agents as alternatives to using T1 shortening contrast agents as contrast can originate from altering proton density or the total water signal that is detected (using another technique called Magnetisation Transfer (MT). In this case gadolinium(III) cannot be used in this technique because the T1 of water would be too short, but other lanthanides, such as Eu(III), can be used because they have smaller magnetic moments, relaxing bulk water much less efficiently.
Other potential candidates are still being tested under laboratory conditions, contrast agents used in the body clearly should be biocompatible, but there are also other requisites need to be addressed as well. These include the requirements of rapid renal excretion, water solubility, stability in aqueous conditions, and a low osmotic potential when in solution for clinical work.
References

Basic Radiology Michael Y. M. Chen, MD

'Alternatives to Gadolinium-Based Metal Chelates for Magnetic Resonance Imaging' (Chem. Rev., 20102010, 110 (5), pp 2960–3018,

Lanthanides in magnetic resonance imaging. Melanie Bottrill, Lilian Kwok and Nicholas J. Long
Chemistry Society Reviews, DOI: 10.1039/b516376p


text:
Nucleophilic Substitution with ethyl bromide and NaOH

Given is following nucleophilic reaction with product A. Draw the mechanism of the reaction and give the structure formula of A.

Here is my solution, I am sure it is not right. Could you help me, step by step to understand this reaction.

OH isn't a great nucleophile but is a good base. What is the most acidic proton in the starting material?

I think the H by OH,becauce it has a pKa from 10, and the N has a pKa bigger than 10. So the NaOH take the H and we have $O^-$

Correct! And phenoxide is a good nucleophile.

ok, so the phenoxide will replace the Br.

Exactly so. The reaction ethylates the phenol

With the help from Waylander (s.o.) we get the following mechanism.

Looks correct to me


text:
Understanding how to choose the right reaction for electrochemistry
I am quite new here and studying a bit of electrochemistry. In the course of my studies I encountered a problem I couldn't figure out:

A sample of iron ore weighing $\pu{0.2792 g}$ was dissolved in an excess of a dilute acid solution. All the iron was first converted to $\ce{Fe(II)}$ ions. The solution then required $\pu{23.30 mL}$ of $\pu{0.0194 M}$ $\ce{KMnO4}$ for oxidation to $\ce{Fe(III)}$ ions. Calculate the percent by mass of iron in the ore.

Having thought about this question somewhat, it is clear that I will have to use  the redox reactions for iron of
\begin{align}
  \ce{Fe^0(s) &-> Fe^2+ + 2e-} \\
  \ce{Fe^2+ &-> Fe^3+ + e-}
\end{align}
Furthermore, seeing as the reaction is occurring in an acidic environment, I thought to make the assumption that $\ce{KMnO4}$ will dissociate into its constituent ions, leaving me with picking between either the reaction of
$$\ce{MnO4^- + 2H2O + 3e- -> MnO2 + 4OH-}$$
or
$$\ce{MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$
When it comes to the technical working of this problem I think I can sort it out, but my issue resides in finding the correct mechanism. If anyone is willing to help me out with what I am sure is a simple problem for the experienced members of this community, I would be extremely grateful.
In addition to this I was wondering if anyone could recommend a resource to learn more holistically about these sorts of reactions, as this question in my textbook is part of the challenge portion, and hence there is not as much material regarding the exact mechanisms of these processes.

I wish there was a tag called classic (for titration and redox reactions)


text:
Potential energy surface when spin changes
I'm trying to compute potential energy surface of S4 molecule (in MOLPRO 2015) along S2-S2 bond (it has a trapezoidal shape) from equilibrium to the dissociation threshold, all the other coordinates are fixed. S4 is a singlet (0 unpaired electrons) in the ground state but a quintuplet once it's dissociated (2 independent S2 molecules in the ground state, each has 2 unpaired electrons). I tried to do CCSD(t) calculations with spin 0 and 2. With spin=0 I get the right energy at the equilibrium but too high energy at the dissociation limit and it's vice versa for spin=2. I need a single surface to solve the Schrodinger equation in so I was thinking about just taking minimum between the two surfaces but I'm not sure if it's a correct way to do it.
This problem doesn't seem to be that rare so I think there should be a well known way to tackle it. I think MRCI can use a mix of the two states, adjusting coefficients in the process to make the transition smooth, but I'm not sure how to properly do it.

So, as I realized later, two triplets can still give you a singlet in total when the spins have opposite directions, so my total spin remains 0 throughout the whole dissociation process. I didn't manage to fix spins orientation with CCSD(T) and I'm not sure if it's possible but MRCI worked just fine out of the box automatically orienting spins to make a singlet and it gave me good dissociation energies. But I had to use state average eventually (as was pointed out by @Feodoran) because points in the middle did not want to converge.

Dissoziation is a typical multi-reference problem, where single-reference methods like Coupled-Cluster usually fail. Therefore CASSCF+MRCI would be the better approach here anyway.
If you have a crossing Singlet and Quintuplet state, then you should calculate both. Different spin states do not mix, therefore there is no avoided crossing to be expected. The dissociation of the molecule will remain on its respective PES. To go from the at dissociation limit excited Singlet state to the Quintuplet you need some coupling. Something that switches the spin of the electrons. (I don't know what an actual process would be in this case, maybe some stimulated emission.)
So you should NOT just take the minimum of those two curves!
You should include both spin states (at least, maybe you need more) in the CASSCF state average, to get balanced orbitals. Make sure CASSCF results in smooth PESs for both states before attempting MRCI.
For MRCI you probably need to do independent calculations for both states, as Molpro does not allow different symmetries at the same time here. (But thats fine, as they should not mix anyway.)
Alternatively, you could try other multi-references approaches like MRCC or CASPT2, but I don't have any experience with those.

So if I do two independent MRCIs, I'll get two surfaces, is that right? How do I combine them into a single surface then?

Yes you get two. But you can NOT combine them! You can only consider transitions between them, which requires some kind of coupling, for example an electromagnetic field, like a laser. This depends on what you need the two PESs for.
Without a coupling, the molecule will remain in the state it started from!


text:
Free software to draw Newman projections
I need to draw Newman projection of a molecule. Particularly 1,2-dicloroethane rotamers, but have no idea if there is any program to do that kind of things.
Do you know about some program available for Linux?
I refer to this kind of sketch:

Do you use Latex? If so, the chemmacros package can do this.

Sure! You can make a standalone document, then convert the pdf to png using ImageMagick (convert file.pdf file.png). There is also a way to compile latex directly to png (run latex, then dvipng), but I have never used it.

My guess is, not enough people need this feature for the developers to implement it. It is however present in proprietary softwares like ChemDraw (which is, of course, not available on Linux).

Some versions of ChemDraw work on Linux via Wine.

@santimirandarp Wine is under GNU LGPL, ChemDraw is obviously proprietary.

@santimirandarp How do I put this - if you have ChemDraw already, you are good, Wine is free:)

You can use GChemPaint, which is available on the repositories of many popular distributions. Among its many components for drawing structures, it has a specific tool for Newman's projections.
You can add H, O and many other elements by simply right-clicking on the molecule, and it even allows you to search your molecule in databases like PubChem and ChemSpider.
A useful video tutorial can be found in spokentutorial.org

3.11 To Add a Newman Projection
To add a Newman projection, use the following tool:

This tool allows to add a Newman projection. The changes become permanent only when the mouse button is released.
To add a Newman projection, click on the background. Contrarily to other tools existing atoms are never used by this tool. It creates new atoms of the current element.
When the Newman projection tool is active, the toolbox looks like:

Options for the main bond

Bond length
The default length of the new projected bond. This has no influence ion the drawing, but might be important when exporting to a 3d model.

Bond order
The bond order.

Options for each atom

Bonds number
Number of other bonds for the atom.

First bond angle
The direction of the first bond. Angles are measured counter clockwise from horizontal right.

Angle between bonds
The angle between two consecutive bonds.

Warning
This version of GChemPaint does not correctly support Newman projections with eclipsed bonds.

Source: GChemPaint manual, pages 41-43.

You can visit ChemistrySteps.com and use the free drawing tool embedded in the text editor. It doesn't have the ring but you can try using the text and add "O" as the ring. It may not be the best drawing tool but it is free and you can definitely draw many different structures.


text:
Hybridisation and bond strength
I'm a high school student with minimal mathematical knowledge about the theory of chemical bond. (I just have a bit of theoretical background). I came across a statement which stated that "The more is the s character in a hybrid orbital, the stronger will be the bond. "
I also came across a statement the "The extent of overlap increases with increase in directionality of bond." (Example : p-p > s-p > s-s reflects the order of the extent of overlap of orbitals keeping the internuclear distance same). What I think is that the more will be the overlap , the stronger will be the bond formed. How can I account for these contradictory statements? I'm confused. Thanks in advance.

Hello and welcome to Chemistry.SE. Feel free to take the short [tour] or visit the [help] if you have any questions about how this site works. Best of luck with your question!

The statements are not contradictory when read both carefully. One is about overlapping and other is about strength. Also, you might be getting confused that more overlap leads to more strong bond but the actual statement is for a particular type orbital(s, p, d, etc.) greater extent of overlapping leads to more strong bond.

Thanks shubham for your consideration. So should I conclude that the second statement is only valid for pure atomic orbitals in reference to their bond strength?


text:
Unconventional hybridisation states
I just came across a table with interesting hybridisation schemes I have never seen before. The table is shown below. I was wondering if anyone could give me some examples of atoms in particular molecules which possess one of those weird hybridisation schemes (e.g. spd, p2d2 etc.).

 Image taken from: Shriver, D.; Weller, M.; Overton, T; Rourke, J.; Armstrong, F. Inorganic Chemistry; WH Freeman: New York, 2014

Long story very short: These examples do not make any sense, they are a post-rationalisation of geometries and will in most cases not be a good representation of the bonding in such molecules/complexes. Could you please add a citation for the book you are using.

https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes

Even shorter: find better book.


text:
Clarification of IUPAC name
Is the IUPAC name of $\ce{CH3CH=CH-C#CH}$ pent-3-en-1-yne or pent-2-en-4-yne?

Related: In nomenclature, does double or triple bond have higher priority?, Why is it -en-yne and not yn-ene? What is the explanation for this IUPAC specification?, Why doesn't the double bond get preference for suffix in IUPAC naming of an hydrocarbon when both double bond and triple bond are present together?, …

… Why do these two IUPAC naming examples not begin numbering the chain closest to the alkyne?, What is the order of nomenclature priority among an alkyl side group, a double bond, and a triple bond?

Subsection P-44.4.1.10.1 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads

For the endings 'ene' and 'yne' lower locants are assigned first to the endings as a set without regard to type and then to 'ene' endings.

Since the locant set "1,3" is lower than the locant set "2,4" (see here for a discussion of what that means), the appropriate choice of locants for the above compound leads to the name pent-3-en-1-yne.

As an aside, the configuration of the double bond, if unspecified, should be labelled with the symbol Ξ (P-91.2.1.2.1).

When the configuration is not known or must remain unspecified for lack of configurational homogeneity, the italicized symbol 'ξ' or 'Ξ' is used, preceded by the required locant. The symbol 'ξ' (small Greek letter 'xi') replaces noncapitalized CIP stereodescriptors such as 'r', 's', 'm', 'p'. The symbol 'Ξ' (capital Greek letter 'xi') replaces capitalized CIP stereodescriptors such as 'R', 'S', 'M', 'P', and nonCIP stereodescriptors 'E', 'Z', 'seqCis', and 'seqTrans'.

The preferred IUPAC name (PIN) for the compound with unspecified double bond geometry is therefore (3Ξ)-pent-3-en-1-yne.
Of course, if the configuration is specified, then the (E)/(Z) labels should be used instead.

So basically the "smallest sum of locants" again?!

@Martin-マーチン Please note that the widely-cited "sum of locants" rule is incorrect. See: IUPAC nomenclature: “Smallest sum of locants”? and the answers therein.

Alkenes are preferred for lower numbering over alkynes then there is sum locant rule.
So which rule is overtaken by the other?
Is it always applied when the same(grou ps) are present together in a compound?

I should not have tried to joke about this...

@AbdulKhan as the passage I quoted indicates: first you seek to minimise the locant set. Note that it is not the sum of locants that you minimise; see the link in my comment above. Only if the two possible locant sets are equivalent, do you assign a lower locant to the alkene.

behold the (minor) ERRATA (and feel free to delete me in the post-fix period)


text:
Why are fluoroalkyl chains hydrophobic/oleophobic?
I'm searching for an answer that explains the hydrophobicity / oleophobicity in terms of intermolecular forces, but can't really find one.
Below is an example fluoro-alkyl nano-particle, F-POSS. It has 8 ligands of a fluoroalkyl chain $\ce{(CH2)2(CF2)4CF3}$ with a silicon-oxygen cage in the center.

When grouped with other F-POSS molecules it forms a membrane with the fluoroalkyl chains:

Hydrogen Bonding - Wouldn't the hydrogen be attracted to the fluorine atoms? They're essentially electron sinks.
London Dispersal Forces - Why would other molecules not be attracted to these ligands? I was told that the ligands themselves aren't even attracted to each other.
Apologies for the lack of citation, but what's been given to me has mostly been word of mouth. Research articles I've read don't really go into the intermolecular forces.

Image source: http://pubs.acs.org/doi/abs/10.1021/la201545a?journalCode=langd5

C-F bonds tend not to form strong associations with anything else as they have very low polarisability. They certainly don't hydrogen bond.

You have identified a fairly common but counter-intuitive result. At least it's counter-intuitive based on the way the properties of fluorine are described in undergraduate chemistry classes. Specifically, we're told it's very electronegative and hence forms very strong hydrogen bonds, and that it is very reactive. This is, however, a poor description of what tends to dominate the behavior of fluorine when it is bonded to carbon.
Let's start with something simple but illustrative. If I told you that the boiling point of $\ce{CCl4}$ is $\pu{350 K}$ and the boiling point of $\ce{CBr4}$ is $\pu{463 K}$, what would be your guess for the boiling point of $\ce{CF4}$? Probably less based on the trend, but just how much less?

$\ce{CF4}$: b.p.=$\pu{145.5 K}$
$\ce{CH4}$: b.p.=$\pu{111.7 K}$

First time I've used one of these :)

Well hopefully you don't look this far, but you'll notice that the boiling point of $\ce{CF4}$ is much, much lower than that of $\ce{CCl4}$. Importantly, the difference between the fluorine and chlorine compounds is much larger than that of the chlorine and bromine compounds.
Additionally, the fact that $\ce{CF4}$ boils only slightly higher than $\ce{CH4}$ indicates that the molecule basically loses all interactions when going from the chlorine to the fluorine.
What does this actually illustrate? It doesn't tell us anything about hydrogen-bonding or dipole interactions (all have zero dipole), but it tells us about the dispersion forces present in the system. Namely, if this result generalizes, we have learned that a $\ce{C-F}$ bond is essentially non-polarizable. This is very important because polarization is most of what determines the strength of dispersion interactions. Okay, this basically answers the question, but let's be more systematic.
$\textbf{Dipole-Dipole Interactions:}$
The paper you linked is basically looking for a material which won't interact strongly with water and other polar substances. One important feature is to make sure the ligands on this material do not have an appreciable dipole then. It is true that the $\ce{C-F}$ is statically polarized towards the fluorine atom, so there is a dipole along the bond, but the molecular dipole is generally what we will care about. The ligands are basically cylindrical, so any dipole perpendicular to the cylinder ought to roughly cancel out. If anything there would be a dipole along the ligand backbone, but this is probably very small if it exists at all. So, dipole interactions don't particularly matter.
$\textbf{Hydrogen-Bonding:}$
Despite what you might think, it actually is quite possible for water (or other $\ce{O-H}$ containing solvents) to hydrogen-bond with the fluorine in a $\ce{C-F}$ bond. See [1] for some calculations demonstrating this.
So, if you're designing this material, you might have to worry about this a little bit, but leave it up to the experiment to tell if this matters.
$\textbf{Dispersion Forces:}$
Here is the key. In organic chemistry, you're usually sold a lie that the things determining the size of dispersion forces are the surface area and the molecular weight of a molecule. If this were strictly true, then we would expect the increase in boiling point we saw above going from $\ce{CH4}$ to $\ce{CF4}$ to be much more dramatic than it is based on the change going from $\ce{CCl4}$ to $\ce{CBr4}$. The reason this rule works is because generally organic compounds are just a bunch of $\ce{C-H}$ bonds, so most things are roughly as polarizable as the next thing.
Okay, but it's not like the $\ce{C-F}$ bonds are inherently unpolarizable. They are about as polarizable as a $\ce{C-H}$ bond, so for all consideration of physical interactions, we can basically imagine that these ligands are just lipids.
$\textbf{Why Do Lipids Form Membranes?}$
This is actually quite a tricky, but interesting question. Imagine a big fat lipid in a solvent of water. One thing we know is that water likes to hydrogen bond. This lipid disrupts the hydrogen-bonding network. Furthermore, because there has to be some water at the surface of the lipid, it is likely these waters will get stuck aligning their dipoles with whatever dipole is present in the bond it is near.
So, it seems like having the lipid there is both energetically unfavorable and entropically unfavorable. It is unclear without an experiment which of these things will dominate and if clumping all the lipids into a membrane is actually better because that's definitely a decrease in entropy for the lipid.
Well, [2] shows the experiment describing this same idea. Slightly different system but same idea. Basically the conclusion is that the energy plays almost no role ($\Delta H\approx 0$) but the entropy for forming a membrane dominates.
Shamelessly quoting from the abstract:

Possible mechanisms for the entropy increase include: (i) the attraction between discrete oppositely-charged areas, releasing counterions; (ii) the release of loosely-bound water molecules from the inter-membrane gap; (iii) the increased orientational freedom of previously-aligned water dipoles; and (iv) the lateral rearrangement of membrane components.

The lesson to learn here is that the entropy of the solvent can actually be very important when considering both the equilibrium arrangements of heterogeneous systems as well in chemical reactions.
$\textbf{Bringing it All Together:}$
So, I hope you see what I mean when I said that noticing $\ce{C-H}$ and $\ce{C-F}$ bonds have almost identical intermolecular forces behave basically answers the question.
When we consider explicitly the interactions of a substance with the ligands you have asked about (in comparison to a lipid), we see that dipole interactions might be a little stronger between the solvent and the ligand, but the ligand-ligand interactions should also be a bit stronger than in the lipid. So, again, $\Delta H\approx 0$. In contrast, the stronger dipole means that the solvent is more tied up, so the entropy change should increase (not accounting for the effect of the cube thing the ligands are mounted on).
Note that this argument for membrane formation based on entropy is not exclusive to strongly polar solvents such as water. That is, pretty much any solvent gains a lot of entropy when a membrane forms. So, as long as it does not interact strongly the ligand, the ligands should form a membrane.
Finally, based on what I've said, why not just use $\ce{C-H}$ bonds rather than have fluorine present? I only skimmed the paper, and they mention durability which makes sense because a $\ce{C-F}$ bond is much stronger than a $\ce{C-H}$ bond. Plus alkanes have the problem of being very combustible which is probably not good as these materials would likely be used at high temperatures.
Random semi-relevant note. Teflon is just an alkane but with all $\ce{C-F}$ bonds, and it too is quite slippery (it's the stuff on non-stick pans).

$\textbf{Edits:}$
I forgot to address the oleophobic nature of fluorocarbons, which is the most important part of the question, really.
$\textbf{Oleophobicity of Fluorocarbons}$
One interesting trend which I was not aware of, is that the boiling point of fluorocarbons increases more slowly than boiling point of corresponding hydrocarbons. For instance, the boiling point of perfluorooctane is $\pu{103 ^\circ C}$ while the boiling point of octane is $\pu{125 ^\circ C}$. So, this means that the dispersion forces in hydrocarbons grow more quickly than that in fluorocarbons. This makes some sense because the carbon in fluorocarbons will be more positively charged, and the fluorine atoms are comparable in polarizability to hydrogen, so the polarizability of the carbon atoms is diminished.
It seems, however, that the oleophobicity of this particular structure has a lot to do with the fact that the chains are not free to move about in solution. For instance in [3], the authors show that by attaching a fluorocarbon, which is ordinarily not oleophobic, to silica particles, the flurocarbon becomes oleophobic. This indicates to me (and this could be argued) that by restricting the translational motion of the fluorocarbons by mounting them on a large structure, as is done in the example you give, that the favorability of mixing is lost. This assumes that the energetics of having the fluorocarbons dispersed throughout the oil is roughly the same as having the fluorocarbons and oil separated into phases.
So, it seems like entropy is the dominant factor in both the hydrophobic and oleophobic characteristics of these fluorocarbons. The big difference is that basically any nonpolar organic molecule will form its own phase in the presence of a polar solvent like water. This is because the interactions between the polar substances are fairly strong, but there is also still a lot of free motion (hydrogen-bonds, etc. break and form again quite often). With the fluorocarbon, one has to manually restrict the entropy gain due to mixing by forcing the fluorocarbons onto a silica base. These things individually have the same effect of disrupting some of the degrees of freedom of the solvent, so the two separate to maximize entropy.
Obviously, some of the parts about oleophobicity are a little bit hand-wavy because really either effect could dominate, but the fact that putting it on the silica base makes the fluorocarbon oleophobic as ref. 3 points out convinces me this is due to entropy and not enthalpy.

References:
[1]: Andrew V. Razgulin, Sandro Mecozzi, "Binding properties of aromatic carbon-bound fluorine," Journal of medicinal chemistry 2006, 49(26), 7902-7906 (https://doi.org/10.1021/jm0600702).
[2]: Husen Jia, John R. Liggins, Wah Soon Chow, "Entropy and biological systems: Experimentally-investigated entropy-driven stacking of plant photosynthetic membranes," Scientific reports 2014, 4, Article number: 4142 (7 pages)(DOI: https://doi.org/10.1038/srep04142)(PDF).
[3]: H. F. Hoefnagels, D. Wu, G. de With, W. Ming, "Biomimetic superhydrophobic and highly oleophobic cotton textiles," Langmuir 2007, 23(26), 13158-13163 (https://doi.org/10.1021/la702174x).

This doesn't clearly explain why fluorocarbons are both oleophobic and hydrophobic.

Yes. I forgot about this. The answer is almost certainly going to be about entropy for the oleophobic part as well. I'll edit it tonight.

I did my best. Feel free to argue if you're not convinced :)

@Jiheindel this was an awesome answer! I think the part about entropy really starts to hit the nail on the head. I had a few more conversations with people, and surface energy of the system starts to come into play. I think I'll type up a complementary answer talking about that when I get the time.

@JohnSnow you should do that. I would be interested to hear about how surface energy is involved as I don't know very much about this. Looking at it on wikipedia I can imagine how this would be the case.

@jheindel just added it! Hope its accurate/an interesting read.

First it would help to define a term, surface energy: that is the energy that is required to create a surface per a given area. All surfaces have surface energy. This idea can be tied to Gibbs free energy of a system. Here's a quote from the wikipedia:

J.W. Gibbs developed the thermodynamic theory of capillarity based on the idea of surfaces of discontinuity. He introduced and studied thermodynamics of two-dimensional objects – surfaces. These surfaces have area, mass, entropy, energy and free energy. As stated above, the mechanical work needed to increase a surface area A is $\ce{dW = γ dA}$. Hence at constant temperature and pressure, surface tension equals Gibbs free energy per surface area:
$${\displaystyle \gamma =\left({\frac {\partial G}{\partial A}}\right)_{T,P,n}} \gamma = \left( \frac{\partial G}{\partial A} \right)_{T,P,n}$$
where G is Gibbs free energy and A is the area. γ is the surface energy
From this it is easy to understand why decreasing the surface area of a mass of liquid is always spontaneous (G < 0), provided it is not coupled to any other energy changes. It follows that in order to increase surface area, a certain amount of energy must be added.

So what they're getting at here is that it's always favorable for a liquid to take a spherical shape, unless if energy is added into a system. Now, surface energy is this added energy in the system.
When two different surfaces interact, they'll always reorder such that the total energy of the system is reduced. So If you have a high surface energy on a solid, such as glass, water which may have a lower surface energy, will wet and spread out covering the higher energy surfaces. Although the water spreading out created new water surface, which increased the waters surface energy, the total energy of the system was reduced.
Now with fluoroalkyl chains such as the material I gave an example of, they are an example of extremely low energy surfaces. So even if oils, which are low surface energy, try to wet these surfaces, it would raise the total surface energy of the system, which is energetically unfavorable.