text:
The similarities and differences between electron transfer and charge transfer
Please explain the similarities and differences among electron transfer,electron transport, charge transfer, and charge transport from the molecular, atomic and electronic levels.
Is electron transfer different from electron transport?
Is charge transfer different from charge transport?

I'm not sure what you are asking but electron 'transfer' is different to 'transport' in the sense that transport ultimately involves many electron transfers events one after the other. Charge transport is the same as electron transport.
There is is difference between electron transfer and charge transfer. Charge transfer is usually associated with charge transfer complexes, such as TTF-TCNQ (tetrathiofulvalene-tetracyanoquinodimethane) but there are very many others. In these complexes the charge transfer from one species to the other occurs as the photon is absorbed so that the lowest excited state is $\ce{D + A + photon -> D+A-}$. This often means that these complexes are highly coloured.
In electron transfer no photon is needed and the process follows Marcus theory whereby a barrier is crossed between $\ce{D + A}$ and $\ce{D+ + A-}$, i.e. this is a thermal process with an activation barrier. The unusual feature of Marcus transfer is the that activation energy has a parabolic dependence on energy thus the rate constant can be slow in both endothermic and very exothermic conditions.


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Why is Lithium Fluoride much safer than Sodium Fluoride
I've looked at the MSDS of Lithium Fluoride and Sodium Fluoride. It really seems that Lithium Fluoride is much safer to work with. Why is that, if both are in the Alkali Group?

My gut feeling is that they must be in the same basket. If anything, LiF is less soluble; maybe that explains the difference.

Solid, or powder? Both LiF and NaF are used in specialty optics equipment as lenses and windows, with the only caveat being that coatings are desirable to prevent degradation from moisture. In this form (bulk pieces) neither is considered particularly dangerous.

@George Custer: Perhaps it's because Sodium and Potassium both are involved in neurotransmitting, and the Fluoride ion interferes with the Calcium ion which is needed in muscle contraction. Maybe together they're more lethal. While Lithium along with all the other heavier Alkali metals don't have significant biological roles, so they're not as lethal?? Also, Lithium is known to actually known to help with mania, perhaps by attenuating excessive  neurotransmitting. I dunno. The pathway for Lithium's action is still a mystery.

@Dehbop: the amount of anion is irrelevant in the concentration of fluoride ion that would be toxic. A Ivan Neretin states, LiF is far less soluble than NaF. http://pubs.acs.org/doi/abs/10.1021/ie50227a024?journalCode=iechad

It's possible that it's down to the solubility, but I think there's better explanations.

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Why does conductivity of electrolyte solutions peak at high concentrations?
I've recently done a lab about conductivity of brine over time (high concentration salt water). I passed current through salt water and measured the current over time.
The graphs I got showed that current through the solution rose to a peak before falling, and it looks almost like the positive x-value half of a quadratic function. I understand it falls because resistance increases with the progress of the redox reaction, but what causes the peak at the top?

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How many isotopologues of water are there?

If ordinary water molecules can contain any of the three hydrogen isotopes, $\ce{^1H}$, $\ce{^2H}$, $\ce{^3H}$, and any of the three oxygen isotopes, $\ce{^{16}O}$, $\ce{^{17}O}$, $\ce{^{18}O}$,  how many kinds of water molecules $\ce{H2O}$ are there?

I'm a little confused because since there are $\binom{3}{2}$ ways of choosing two hydrogen atoms and 3 ways of choosing oxygen atoms, shouldn't there be 9 different kinds of water molecule? Why would it be 18 (as the answer says)? In other words, why would the order of hydrogen atoms matter here?

3 choose 2 isn't correct here, as that neglects the possibility that you can have the same isotope of hydrogen twice.

In an example as small as this, you could hypothetically construct all the cases, or at least enough of them to see that $9$ is incorrect.
The only thing you need to change to get the right answer is instead of using $3\choose{2}$ which is a combination without replacement, you need to do combinations with replacement so ${2+3-1}\choose{2}$$=6$ ways to choose the hydrogens. We can very quickly verify this result by writing out the pairs of masses and seeing how many unique combinations we can make: 11 22 33 12 13 23. And then of course each of those combinations can have one of the three different types of oxygen, so 18 possibilities total. If you haven't seen combinations with replacement before, go here to learn more about them.

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text:
Can a solution of FeCl2 go bad over time?
My lab has been using solution of FeCl2 for an assay that was mixed about three years ago. Recently the assay has failed twice. I was wondering if it is possible that the iron ions in the solutions have changed oxidation over time?

Yes, that's quite probable.

@IvanNeretin any guesses on the self-life of 1 liter of 1M FeCl2?

Depends on how you store it, etc. but Fe(II) is oxidised fairly easily by oxygen.

I agree with Ortho but how you didn't notice color change?

@Mithoron so I'm just the lab comp-bio/bioinformatics guy I try to stay out of the wet-lab. This came up in discussion and I was wondering how an ionic solution would go bad. It's been a couple of years since I took general chemistry lol.

https://chemistry.stackexchange.com/questions/31243/what-is-formed-when-you-leave-ironii-sulfate-in-plain-air

Water evaporates. Water dissolves the fixed gases ($\ce{N2}$, $\ce{O2}$, $\ce{CO2}$). A well-sealed bottle of water will last for years - that's assuming that we understand that 'bottles' are made out of heavy glass*, and seals are permanent (in the sense that opening the bottle is irreversible). That's not a particularly useful way to store stuff you need access to. (In other words, most bottles allow only limited storage life.) $\ce{O2}$ can oxidize $\ce{Fe(II)}$. Unless working under (and stored under) inert atmospheres, the titer will inevitably change over time. Even when working in inert atmosphere, there will be evaporative losses. All that said and assuming that there wasn't any accident involving the reagents, nor changes to their storage conditions (temperature, location, etc.) then the decline in $\ce{Fe(II)}$ should be roughly proportional to the number of times the bottles are accessed. Assuming that that is regular, with no recent upsurge, then the change in titer should be gradual over time; no sudden change should have occurred.(*please don't tell me that they're storing reagents in plastic! yet still any changes should be gradual, not sudden.)

Generally, ferrous chloride gives ferric chloride and iron(III) oxide on heating in air (pubchem) but on long standing, ferrous chloride forms other Fe(III) species like iron hydroxide($\ce{Fe(OH)3}$) and iron oxide-hydroxide($\ce{FeOOH}$), iron oxychloride($\ce{FeOCl}$) etc. all present in the solution in equilibrium. The type of species present in the solution depends on the color of solution. Orange-red reveals ferric chloride, propably hydroxide and the oxide-hydroxide. More darker color like brownish reveals iron(III) oxide(magnetic?).
Giving a cursory search on Google, numerous references can be found which discuss on ferrous chloride oxidation. I am showing 2 of them:

Source 1 - propose formation of a binuclear complex having the composition $\ce{Cl2FeOOFeCl2}$
Source 2 - gives details explaination about the mechanism of this oxidation


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What's the difference between reactant, excess reactant and limited reactant in a balanced equation?
In some reactions, the reactant is mentioned as being "excess". For example, "phosphorus pentoxide forms in excess oxygen" or "pentahalides form with excess halogen".
But since the balanced equation shows exactly how many moles of a reactant are necessary, what is "excess" about the reactant? And how does it contrast with "limited reactant" and just "reactant"? (Ie: What's the difference in a balanced equation between oxygen, limited oxygen and excess oxygen?)
Does each balanced equation need to explicitly say somewhere that this is specifically an excess/limited reactant, or can this be inferred from the stoichiometric coefficients of the balanced equation? Do you need at least two equations side by side to see that one is excess and the other isn't?
Additionally, taking for example the following two reactions
$$\ce{P4 + 3O2 -> P4O6}\\[1em]
\ce{P4 + 5O2 -> P4O10}$$
it seems to me that mathematically on a mole basis, sure we can say that tetraphosphorus hexoxide will form with 3 moles of O2 available (per mole of P4), but tetraphosphorus decoxide needs 5 moles of O2, however, given the huge amount of actual molecules in the reaction, how do the molecules know when there's excess? I mean, why doesn't the phosphorus just form a lot of P4O10 and then just a few molecules of P4O6 with the few leftover O2 molecules, or would this be some kind of unstable situation and the phosphorus molecules are pretty considerate and don't start forming P4O10 until every P4 has at least 3 O2 attached to it?
To sum up, my questions are:

What's the difference between "reactant", "limited reactant" and "excess reactant" in a balanced equation?
Can the fact that the reactant is excess/limited be inferred from the balanced equation?
Why is it that the same reactant (such as P4) proceeds in two different ways depending on whether the second reactant is excess/limited?

Re. the two different equations: I'm not a chemist, but – in general –
which of multiple reactions takes place (assuming enough material for either) may depend on environmental conditions (temperature, pressure; possibly presence of catalysts) and the "ease" of the reaction taking place (not sure of the technical term). According to this page the first reaction happens first, using up all available phosphorous; if there's left-over oxygen [...]

... then the that is used to complete the process (P4O6 + 2O2 --> P4O10). (See also this page). Also, this page mentions the enthalpy involved; both are exothermic, but the first gives off less energy. Possibly this is why it happens in preference to the second (but I'm speculating).

Equation when balanced shows proportions absolutely not amounts.

Reagent is an overarching word and it covers everything that is added to a chemical reaction.[1] So all we need to define are limiting and excess reagents.
A limiting reagent is one that limits the maximum outcome of a reaction. Remember that a reaction needs all reagents to proceed from starting material to product, and that it will consume them in the ratios given in the balanced equation. On paper, we can write a fully balanced equation and know exactly how much we have. In the real world, we will not have added exactly the same amount of everything so some reagents will remain unreacting. Those reagents that can be used up completely — the scarcest ones — limit the maximum output, therefore they are limiting.
In a real world example, consider burning petrol (approximated as hexane, $\ce{C6H14}$). The full reaction equation will be:
$$\ce{2 C6H14 + 19 O2 -> 12 CO2 + 14 H2O}\tag{1}$$
Of course, the reaction will consume the hexane from the jar you have it in and oxygen from the surrounding air. After some time, the fire will go out. The petrol will have been completely consumed — but the air of course stil contains oxygen. Petrol was the limiting reagent.
If you do the same experiment but in a closed, transparent container, you will notice that the flame is extinguished before all the petrol was burnt. Here, there was only a limited supply of oxygen in the sealed container. After it was used up, the combustion cannot continue. Oxygen was the limiting reagent.
An excess reagent, on the other hand, will be present in excess, i.e. more than is needed to drive the reaction to completion. In the above examples, the reagent that was not limiting was automatically present in excess (oxygen in the first, petrol in the second example).
In examples actually performed in the chemistry lab, one defines one reagent as the ‘standard’ and its amount as ‘1 equivalent’ or $\pu{1eq}$. A limiting reagent would be one that is present in less than $\pu{1eq}$ (assuming equal stoichiometries; otherwise balanced accordingly)[2] while an excess reagent means that it is present in (significantly) more than $\pu{1eq}$ (same stoichiometry caveat).
In the examples I have shown you, both reagents can be limiting or excess, depending on the actual reaction conditions. Thus, you cannot infer which is which by looking at a balanced equation. If you read the experimental sections of journal articles, these will make it quite clear because rather than listing fully balanced equations they list methods and (in organic chemistry typically) always refer to equivalents used.
Now how do we explain different reaction products depending on how much in excess a certain reagent is? It is not always trivial but you can always consider it some sort of a stepwise process. In the phosphorus example you gave, you might consider a first reaction step (very rapid) that forms phosphorus trioxide from white phosphorus.
$$\ce{P4 + 3O2 -> P4O6}\tag{2}$$
After this reaction has run to completion, phosphorus turns out to be the limiting reagent so a second subsequent oxidation step is added to finally generate phosphorus pentoxide.
$$\ce{P4O6 + 2 O2 -> P4O10}\tag{3}$$
This may noy be a correct mechanism on microscopic scale but it does help explain how excess of one reagent can lead to different products.

Notes:
[1]: Unless you also count reactant as something different. I never understood the true difference between the two, as far as I can tell, they are synonyms.
[2]: Beneath a certain threshold, however, the term will be catalytic rather than limiting. The difference is that, in theory, one molecule of the catalytic reagent will be enough to forward the reaction while the limiting reagent will be used up and end the reaction.

reactant: "A substance that is consumed in the course of a chemical reaction. It is sometimes known, especially in the older literature, as a reagent, but this term is better used in a more specialized sense as a test substance that is added to a system in order to bring about a reaction or to see whether a reaction occurs (e.g. an analytical reagent)."

Suppose that you want to make a cake. You would first need to go your fridge to check whether you have the necessary ingredients to make one. So you go to your fridge. Luckily, you have all the ingredients: milk, eggs, flour, and frosting. So you decide to make one. You also decide to make the cake as big as possible to avoid the hassle of cleaning/preparing again. Now, what determines how big you can make the cake?
It is the quantity of ingredients of course. But what happens if you  have enough milk to make a 10 cm cake, enough eggs to make a 1000 cm cake, enough flour to make a 1000 cm cake, and enough frosting to make a 10000 cm cake? How big can you make the cake?
You would only be able to make a 10 cm cake. Because you would run out of milk. And since you no longer have milk, you cannot make anymore cake.
Lets go back to your question of excess reagents and limiting reagents. The limiting reagent is the milk in the example above. It is what determines how much of a product you will obtain. It is what limits the amount of product you get (hence the apt name limiting reagent). Everything else are excess reagents: the eggs, the flour, the frosting. They are leftovers; they are in excess.
Now question 2. You cannot tell which reactant will be in excess by just looking at the balanced  equation. Going back to the cake example, the balanced equation is sort of like a recipe for the cake. You cannot tell whether you will have eggs, flour, and frosting left over by just looking at the recipe. You also have to check your fridge. In other words, you have to know what you are starting with.
Now question 3. I think question 3 will be answered in more detail as you study more chemistry. A complete answer involves thermodynamics and statistical mechanics. Nevertheless, going back to the example of the cake: Whether or not you can make cake depends on whether you have all four ingredients. It also depends on how much of each ingredient you have. If you have no milk, you won't be able to make any cake.
But say that you want to make boiled eggs as well. You only need eggs to make boiled eggs so you would could turn all eggs leftover eggs from the cake into boiled eggs. The fact that you can make boiled eggs with the leftover eggs does not change the fact that you can make a 10 cm cake.
This is what the two equations are trying to get at. The major product, will be determined by how much of a certain reactant you have. It doesn't mean that the reaction is solely determined by what it is in excess or not.

"You would only be able to make a 10 cm cake. Because you would run out of eggs. And since you no longer have eggs, you cannot make anymore cake." Shouldn't that be milk twice?

Excellent  use of analogy. +1.


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What do the brackets mean in this inline structure representation?
Based on the way the molecular formula is written — $\ce{Br(CH2)4Br}$ — I'm assuming this means that there are $4\ \ce{-CH2}$ groups attached to $\ce{Br}$.  Am I correct? I've always thought that $\ce{Br}$ can't accept so many groups though.

Organic chemistry is starkly different from inorganic, even in the ways of writing the molecular formula. So in fact this is just Br-CH2-CH2-CH2-CH2-Br.

Really? I always thought adding the brackets signified another chain so as to differentiate between isomers.

Brackets may serve multiple purposes.

When you see a formula like the one in the question in organic chemistry, the person writing it down either already assumes a lot of chemical knowledge on the reader’s side or wants to teach them how to interpret this. So to a practising organic chemist, one inline formula may mean one thing while a similar one (with brackets in the same positions etc) may mean something completely different. Sometimes even, inline formulae are (and must be, due to their nature) potentially ambiguous. Therefore, organic chemistry relies a lot on actually drawing out the structures, whether it be each atom explicitly or in the sum formula formalism.
In the case of $\ce{Br(CH2)4Br}$, organic chemists will know that bromine will typically only form one single bond. Additionally, each $\ce{CH2}$ unit is known to be able to form two bonds — either two single bonds or one double bond. Therefore, an organic chemist will automatically expand the formula to $\ce{Br-CH2-CH2-CH2-CH2-Br}$ or 1,4-dibromobutane.
However, you might also sooner or later encounter inline formulae such as $\ce{BrC(CH3)2CH2Br}$ or $\ce{(CH3)3COH}$. Here, the brackets do actually mean repeated addition of said group to a single common atom. so the first of these two formulae describes 1,2-dibromo-2-methylpropane and could also be written $\ce{H3C-C(CH3)(Br)-CH2Br}$ while the second one represents tert-butanol or 2-methylpropan-2-ol.
Immediately realising which type of brackets is used requires practice and experience.

Although denoting linear repeating unit by parentheses (round brackets) in organic compound formulae is very very common, it's not the correct, or recommended way.
From Nomenclature of Organic Chemistry, IUPAC Recommendations and Preferred Names 2013, P-16.5 Name writing: Enclosing Marks:

P-16.5.1.11 Parentheses are used to enclose groups attached to a chain in linear formulas.
Examples:
$\ce{\overset{3}{C}H3-\overset{2}{C}H(SH)-\overset{1}{C}H3}$   propane-2-thiol (PIN)
$\ce{\overset{4}{C}H3-\overset{3}{C}H(OH)-\overset{2}{C}O-\overset{1}{C}H3}$   3-hydroxybutan-2-one (PIN)
$\ce{\overset{5}{C}H3-\overset{4-3}{[CH2]}_2-\overset{2}{C}(CH3)2-\overset{1}{C}H3}$   2,2-dimethylpentane (PIN)
…
P-16.5.2.6 Brackets are employed in formulae to indicate repetition of
  a group in a chain.
Example: $\ce{CH3-[CH2]68-CH3}$ heptacontane (PIN, P-21.2.1)

Or, Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), GR-2.3 Atom labels and other chemically significant text: Atom labels representing more than one non-hydrogen atom:

GR-2.3.4 Interpretation of contracted labels
…
Divalent structural fragments may be enclosed in brackets and followed by a repeat count to represent repeating fragments concisely.

So the correct recommended short representation of, obviously 1,4-dibromobutane, should be $\style{font-weight:bold}{\ce{Br[CH2]4Br}}$ (or $\ce{Br-[CH2]4-Br}$).


text:
How does the state of iodine affect its HOMO–LUMO gap?
My teacher told me that the HOMO–LUMO gap in solid iodine is smaller than that in gaseous iodine.
When I asked about the reason, he told me that it is due to higher interactions in solid iodine as compared to gaseous iodine, which I did not understand. How can the physical state of a compound affect the HOMO–LUMO gap, even though the chemical formula is the same?

Chemical formula of solid Na and gaseous Na is also the same: Na. But I suppose you aren't surprised that their HOMO/LUMO stuff is quite different.

I will broadly mention several mechanisms that may influence color of a solid vs. gas, not just the HOMO-LUMO transition.
Without going into the literature and/or performing (extensive) calculations, I cannot confirm or reject your teacher explanation.

In the solid state, you may find intermolecular MO interactions that will affect orbital energies. Depending on the strength, you may even have to abandon the concept of molecular orbitals in favor of crystal orbitals. Quantum dots and their dependence of color on size are somewhat of an example.
Molecules may alter their shape in solid state compared to gaseous state. This is less of an issue in $\ce{I2}$.
In the solid state, intermolecular excitations are more likely (i.e. going beyond HOMO-LUMO transitions). This is especially true if you do not have a pure substance, think doped solids.
Note that supermicroscopic spatial distribution of a substance will have influence on the color as well: finely distributed silver or mercury (as in calomel) may appear as black rather than shiny.

What are quantum dots ?


text:
Which has the highest melting point among NaF, NaCl, NaBr, NaI?
I've been trying to understand why $\ce{NaF}$ has the highest melting point among all of the given solids. I thought it would be $\ce{NaI}$ as it will have the biggest molar mass among them which would mean it has the strongest intermolecular forces. Is it electronegativity difference or molar mass that you look at when analysing intermolecular forces? Am I missing something or misunderstanding something? Thank you very much.

NaF has better packed crystal structure, which is harder to break.

Another point. There is no such thing as a NaCl "molecule" in solid NaCl. All the ionic charges interact. Lookup the "Madelung Constant."

A simple way of comparing the M.P.s of ionic solids is through calculating their lattice energies.

Evidently, NaF has the highest L.E. anong the four as fluoride has the smallest ionic radius among the four anions.

All those are ionic compounds, so it is just Coulomb forces that hold them together. Two charges are attracted with the same force regardless of their mass. The distance, however, does matter a great deal. And the distance depends on the ionic radii, and I'm sure you know how those change down the group.
Things could have turned out otherwise if the compounds had different crystal structures. Luckily for us, this is not the case.
So if you are confronted with an unknown compound, you first look at the electronegativity difference, and if it tells you the compound is ionic, you use the same line of thought as I just did. Otherwise it is probably a molecular compound held together with dispersion forces, and they tend to grow with molar mass. Oh, and there might be a covalent solid...
See, chemistry is complicated, and it is not always possible to get by with a few simple rules of thumb. But at least I think I've answered your original question in the narrow sense, so let's leave it at that.

As with boiling points, the melting point of a solid is dependent on the strength of intermolecular attractive forces. A strong attraction between molecules results in a higher melting point. In general, ionic compounds have higher melting points compared to covalent compounds, because the electrostatic forces connecting the ions (the ion-ion interaction) are stronger than molecular-molecular or polar interactions exist in covalent compound. Unlike covalent compounds, those interactions do not get stronger with increasing molecular weights. Actually, it is the contrary.
The high melting point of ionic compound reflects its high lattice energy. When ions are tightly packed together of an ionic substance, it has a higher melting point than another ionic substance with ions that do not pack well (recall that this packing depends on the ratio of ionic radii of positively and negatively charged ions, which changes the crystal structure).
Most alkali metal halides crystallize with the face-centered cubic lattices (FCC). In this structure both the metals and halides feature octahedral coordination geometry, in which each ion has a coordination number of six. The exception are cesium chloride, bromide, and iodide, which are crystallize in a body-centered cubic lattice (BCC) that accommodates coordination number of eight for the larger metal cation (and the anion as well).
Thus, the melting point of ionic solid is depends on lot of factors including ion packing. However, as Ivan Neretin's excellent answer pointed out, the major force is the Coulomb forces acting between oppositely charged ions $\left(F_{attract} = \dfrac{q_+q_-}{d^2}\right)$. Yet, the strength of these forces changes with other factors such as ionic character of the bond. This would be evident on following table:
$$
\begin{array}{c|ccc}
\text{Substance} & a \ (\pu{pm})^a & Z & r_+ + r_- \ (\pu{pm}) & V_m \ (\pu{cm3mol-1}) & d \ (\pu{g \! cm-3})^b & \text{m.p.} \ (\pu{^\circ C})^c \\
\hline
\ce{LiF} & 401.4 & 4 & 283.8 & 9.737 & 2.66 \ (2.64) & 845 \\
\ce{LiCl} & 513.2 & 4 & 362.9 & 20.349 & 2.08 \ (2.07) & 605 \\
\ce{LiBr} & 548.9 & 4 & 388.1 & 24.898 & 3.49 \ (3.46) & 550 \\
\ce{LiI} & 600.0 & 4 & 424.3 & 32.519 & 4.12 \ (4.08) & 449 \\
\ce{NaF} & 462.0 & 4 & 326.7 & 14.846 & 2.83 \ (2.56) & 993 \\
\ce{NaCl} & 564.8 & 4 & 399.4 & 27.125 & 2.15 \ (2.17) & 801 \\
\ce{NaBr} & 593.6 & 4 & 419.7 & 31.489 & 3.27 \ (3.21) & 747 \\
\ce{NaI} & 646.2 & 4 & 456.9 & 40.624 & 3.69 \ (3.67) & 661 \\
\ce{KF} & 532.8 & 4 & 376.7 & 22.771 & 2.55 \ (2.48) & 858 \\
\ce{KCl} & 627.6 & 4 & 443.8 & 37.216 & 2.00 \ (1.98) & 770 \\
\ce{KBr} & 657.0 & 4 & 464.6 & 42.695 & 2.79 \ (2.74) & 734 \\
\ce{KI} & 706.4 & 4 & 499.5 & 53.068 & 3.13 \ (3.12) & 681 \\
\ce{RbF} & - & 4 & - & - & - \ (3.56) & 795 \\
\ce{RbCl} & 653.4 & 4 & 462.0 & 41.997 & 2.88 \ (2.80) & 718 \\
\ce{RbBr} & 693.0 & 4 & 490.0 & 50.105 & 3.30 \ (3.35) & 693 \\
\ce{RbI} & 730.8 & 4 & 516.8 & 58.759 & 3.61 \ (3.55) & 647 \\
\ce{CsF} & 600.8 & 4 & 520.3 & 32.649 & 4.65 \ (4.64) & 682 \\
\ce{CsCl} & 411.8 & 1 & 356.6 & 42.053 & 4.00 \ (3.99) & 645 \\
\ce{CsBr} & 428.7 & 1 & 371.3 & 47.446 & 4.49 \ (4.43) & 636 \\
\ce{CsI} & 455.8 & 1 & 394.7 & 57.025 & 4.56 \ (4.51) & 626 \\
\hline
\end{array}
$$
$^a$: The lattice constant $a$ is the physical distance of the relevant unit cell. The values are from Ref.1; $^b$: The density $d$ is calculated from the equation $d = \text{Molar Mass}/V_m$. The values in the parenthesis are experimental values listed in various  sources; and $^c$: The melting points are from Ref.2.
As the table has shown, melting points of halide salts of each alkali metal decrease with the decreasing binding energies from fluorides to iodides. However, most striking feature I have noticed here is the each series of particular alkali metal halides (at least those of $\ce{Li}, \ \ce{Na}$, and $\ce{K}$) have linear relationship with their melting points and corresponding lattice constant $a$:

Unfortunately, $a$ value of $\ce{RbF}$ is absent from literature due to its highly hygroscopic nature (at least the time of Ref.1; nevertheless, I couldn't find that either). Therefore, whether this relationship exists with $\ce{RbX}$ is yet to be seen (the three remaining points are not good enough, even though it shows the linear relationship with $R^2 = 0.967$). All of those crystals have closely packed FCC geometry.
Strikingly different from this phenomena is melting points of the series of $\ce{CsX}$, which does not show the linear relationship. This behavior is expected since only $\ce{RbF}$ has FCC geometry while other three are Body Centered Cubic (BCC). Alike existing three $\ce{RbX}$ crystals, three $\ce{CsX}$ with BCC geometry also show the linear relationship, but three-point straight line is not so reliable.
However, although this is out of the question, I'd like to make a suggestion. Using the equation given in this three-point straight line, we can calculate approximate value of $a$ for $\ce{CsF}$, which is seldom found in literature. For the convenience, I plotted all four metal halide graphs ($\ce{LiX}, \ \ce{NaX}, \ \ce{KX}$, and $\ce{RbX}$) in one paper:

According to the equation: $y = -0.915 x + 1319.6$ for $\ce{RbX}$, when $y=\pu{795 ^\circ C}$ (m.p. of $\ce{RbF}$), $x=a=\pu{573 pm}$, which is a reliable number compared to the trend.

References:

Ralph W. G. Wyckoff, In The Structure of Crystals, Part 1; American Chemical Society Monograph Series; The Chemical Catalog Company, Inc.: New York, NY, 1924.
S. L. Baldochi, I. M. Ranieri, In Reference Module in Materials Science and Materials Engineering; Saleem Hashmi, Editor-in-Chief; Elsevier, Inc.: Dublin, Ireland, 2016 (The values are available online from https://www.sciencedirect.com/topics/chemical-engineering/alkali-halides).

The point that is crucial here is that ionic compounds are held together in a crystal lattice structure.

What you have misunderstood is that there are no intermolecular forces between these ions, as number one the forces are not between  molecules but ions, and number two if you break the bonds between a $\ce{Na+}$ and a $\ce{Cl-}$ ions, you have effectively broken an intramolecular bond resulting in a sodium ion and a chlorine ion.
Hence mass has no effect on the strength of the bond but factors affecting coulombs law will.
The reason why $\ce{NaF}$ would have the highest melting point is because, if you look at the electronegativity values fluorine has the highest hence it have the greatest pull or attraction towards the electrons in the ionic bond, resulting in a really strong sodium positive ion and a strong negative fluorine ion, this thereby results in a high melting point. You can also look up lattice enthalpy values to be sure.

I think it could be because the enthalpy of formation on NaF is the most negative and so NaF is the most stable compound among the alkali metal halides. Consequently the melting point would be the highest.
Also the melting and boiling point always follow the trend : fluoride>chloride>bromide>iodide.

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text:
What is the momentum of a particle described by the following equation?
$$\psi(x) = Ae^{ikx}  \ + \ Be^{-ikx}$$
The above is not an eigenfunction and to obtain its momentum my professor applied the momentum operator separately to the first and second parts. The final result was that he obtained two different values of momentum,
$$\hbar k \text{ and } -\hbar k\text{.}$$
My question: is this method correct? If yes, how to explain the separation of values?

What's to explain? Like you said, this is not an eigenfunction for momentum, hence it corresponds to multiple values of momentum. Likewise, it is not an eigenfunction for coordinate, hence it corresponds to multiple values of $x$.

But then it will be like applying momentum operator to  2 different wave functions.

Yes it will; so what? Any wave function which is not an eigenfunction can be thought of as a mixture of different eigenfunctions.

If you don't want to separate the two states, then what is the single momentum of this function? And if you can't answer this question, then I think it highlights the need to understand why and why the linear combination of eigenstates should be treated separately.

The idea of superposition of eigenstates is quantum 101, so you should probably go and read a proper book on QM. I personally recommend Griffiths.
In quantum mechanics, observables such as $p$ are represented by Hermitian operators such as $\hat{p} = -\mathrm{i}\hbar(\mathrm d/\mathrm dx)$.
The momentum of a particle is not well-defined unless it is in an eigenstate of the momentum operator $\hat{p}$. If it is, then a measurement of the momentum will always return the corresponding eigenvalue. For example, if the wavefunction of the particle is simply $\exp(ikx)$ , then a measurement of the momentum will always return the value $\hbar k$, and the particle is said to exist in a state of well-defined momentum.
Now, if the wavefunction of the particle is $A\exp(ikx) + B\exp(-ikx)$, then a measurement of the momentum will sometimes yield $\hbar k$ and sometimes yield $-\hbar k$. So, the actual momentum of the particle is not well-defined, as it could be either of the two values (before you observe it).
The expectation value of the momentum, though, is well-defined. It is simply
$$\langle p \rangle = P_+(\hbar k) + P_-(-\hbar k)$$
where $P_+$ and $P_-$ are the probabilities of measuring the momentum to be $\hbar k$ and $-\hbar k$ respectively. Assuming that your wavefunction is normalised, i.e. $|A|^2 + |B|^2 = 1$, basic quantum mechanics tells us
$$P_+ = |A|^2;\quad  P_- = |B|^2$$
and so the expectation value in this case is found to be
$$\langle p \rangle = \hbar k (|A|^2 - |B|^2)$$
However, that simply means that if you prepare many particles in the state above, measure their momenta, and average their momenta, the average will tend toward this value. Each individual measurement is guaranteed to give either the value $\hbar k$ or $-\hbar k$.
And yes, each individual particle does not have one single well-defined momentum. That's quantum mechanics for you.


text:
Why does acidity increase going from sulfide to sulfoxide to sulfone?
It is fairly well-known that the acidity of a proton alpha to a sulfide/sulfoxide/sulfone increases in that order. Some pKa values (in DMSO), courtesy of the Evans pKa table:

Traditionally this has been explained by resonance, or delocalisation of the carbon-based negative charge onto oxygen (left, below). But this necessarily invokes hypervalency and a S=O double bond. The major resonance form in a sulfoxide should be the charge-separated, octet-compliant form (right, below).

What is the proper explanation for the acidity trend above, then?

So if we were just considering inductive effects, how much would we expect the $\mathrm{p}K_{\mathrm{a}}$ value to drop for every extra oxygen in a $\beta$-position?

@Zhe I actually suspect it is perhaps more to do with the positive charge on sulfur, rather than the oxygen (don't think the oxygen really does much, cf. acidity of typical alkane vs typical ether) But I don't have a good source, it's just a hunch.

Isn't the positive charge on sulfur just caused by the electronegative oxygens?

@Zhe the positive formal charge, I mean.

The oxidation state of sulphur and its therefore increased inductive effect should be enough to explain this effect.
Besides the number of oxygens, the principle difference between the three structures is the oxidation state of sulphur that rises from $\mathrm{-II}$ to $\pm0$ to $\mathrm{+II}$. The higher an element’s oxidation state, the stronger it draws electrons towards itself since a higher oxidation state typically correlates with less electron density.
Unfortunately, we cannot draw good comparisons. Phosphorus would suggest itself, but neither the Evans table nor the Bordwell data have enough data points for scrutinisation. Furthermore, phosphorus is less electronegative than carbon, meaning the bonds would be polarised towards carbon.
Chlorine might be an option being on the other side of sulphur in the periodic table, but to the best of my knowledge no ‘chlorone’ or ‘chloroxide’ organic compounds are known let alone their acidities.
Nitrogen in its higher oxidation state (a nitro group) has a double bond which can resonate with the negative charge so it is not a good comparison. And the elements in the fourth period and below are again too electropositive in general. If anything, we may be able to draw conclusions based on the β-acidity of an ether, the α-acidity of an acetal/ketal or the α-acidity of an — if these are even stable enough to be deprotonated.

Well, there's for example Perchlorylbenzene, so some analogies might be drawn.

@Mithoron Didn’t know it existed!

Found this in J. Phys. Chem. A 2005, 109, 4966: "The acidity trend apparently stems from other factors [than resonance], such as the electrical stabilization of the anionic carbon center adjacent to polarized sulfenyl (RS), sulfinyl (RSO), and sulfonyl (RSO2) groups." I agree that sulfur electronegativity is probably the best explanation. For analogies, amine vs. N-oxide might work, but I wonder if the data exists.

@orthocresol Oooh, I wonder whether someone wants to add a literature-suppored answer to this Q.

As explained by Clayden (Reference 1), $\ce{ ^13 C}$ NMR can give insights into the acidity of the $\alpha$-carbon in sulfide/sulfoxide/sulfone compounds. As @orthocresol informs, the resonance is not essential. Then, as explained by @Jan, the inductive effects are important in the sequence n oxygen(s)  O (n=0, 1,2) to sulfur S to $\alpha$-carbon C and vice versa, considering mainly the electronegativity interaction of these three elements series (in der Aufeinanderfolge, in der Rehie). In this context, $\ce{ ^17 O}$ and $\ce{ ^33 S}$ NMR (Reference 2) would be useful to clarify these inductive effects. Also, the importance of the inductive effects of the resonance and inductive effects can be elucidated through the Hammet constants $\rho$, as shown in exercise 93 of Chapter 17 of Reference 3.
References.
Reference 1. J. Clayden, N. Greeves, S. Warren and P. Wothers. “Organic Chemistry”, Oxford University Press, Oxford, 2012, p.501 (figure of conjugated and not conjugated 13C NMR: 132, 143 ppm, respectively).
Reference 2. S. A. Evans Jr., Oxygen-17 and Sulfur-33 Nuclear Magnetic Resonance Spectroscopy of Organosulfur Compounds, NATO Series 124, Magnetic Resonance, 757-766,  1984.
Reference 3. P. Y. Bruice, “Organic Chemistry”, 6th Edition, Prentice Hall, p. 187, p. 770.


text:
Making elemental phosphorus from phosphoric acid
What is the most efficient way for making pure or semi-pure elemental phosphorus, by using phosphoric acid as an source of phosphorus?

You'd need to reduce it to elemental phosphorus. I'm afraid you didn't bother to check even Wikipedia. Also this is quite broad, it's not like there's only one method.

@Mithoron I searched for about 6 pages before wikipedia. Thanks.

Do you have a well-equipped chemical lab at your disposal, or is it more like a home/kitchen experiment? Or is it rather a theoretical question?

@andselisk home/kitchen experiment just for curiosity.

I wouldn't recommend doing this because this is not a safe and efficient method, but in theory one can mix sand and charcoal powder/grains, then soak the mixture with phosphoric acid in a can, and then heat it up in an oven. If  you are lucky, you can obtain white phosphorous powder on a bottom. Remember that white phosphorous and its vapors are highly toxic!

@andselisk Much appreciated, thanks, i'll do it outside and with a mask, it should have no problems using some metalurgy tools that i have here.

@andselisk I don't believe this can work at all. Reduction of phosphates with carbon requires heating well above 1000°C. Phosphoric acid will evaporate way before that.

http://www.sciencemadness.org/talk/viewthread.php?tid=13932 and https://www.google.com/patents/WO2010029570A1?cl=en

@IvanNeretin That's why it was a proposition, not an answer. Sure some H3PO4 will escape, but some will still react. Ideally you want to convert phosphoric acid to Ca3(PO4)2 by adding chalk/lime, but that will require even higher temperatures of approx. 1600°C to reduce phosphate with charcoal.


text:
What is the difference between "spontaneous" and "energetically favourable"?
I read somewhere that the difference between energetically favorable reactions and spontaneous reactions is that energetically favorable reactions are ones where energy is released, i.e., $\Delta H<0$, whereas spontaneous reactions are ones where the change in Gibbs free energy is negative, i.e., $\Delta G<0$. Thus it is possible for a reaction to be energetically favorable but not spontaneous if $\Delta H$ is negative, and yet $\Delta G$ is positive because $\Delta S$ is negative and the product of it and T is large enough to compensate for $\Delta H$ as per this equation: $\Delta G = \Delta H - T(\Delta S)$.
My questions are:
1) Is this true?
2) If so, what would be an example of a reaction that is energetically favorable but not spontaneous?

Here is a very simple example of an exothermic, nonspontaneous process at room temperature:
$$\ce{H2O_{(l)} -> H2O_{(s)}}$$
You need to extract heat from water to freeze it, but it is obvious that at $25\ ^{\circ}\mathrm{C}$, this does not happen spontaneously. Only if you decrease the temperature to decrease the contribution of the negative $\Delta S$ term will this reaction occur spontaneously.

An example of an exothermic non-spontaneous reaction would be a gas phase reaction such as, $\ce{ N2 + 3 H2 -> 2 NH3 }$, where $\Delta H$ is negative. As $\Delta S$ is also negative,  $-T\Delta S$ is positive and so there is a temperature at which this makes $\Delta G$ positive.
An endothermic spontaneous reaction example is $\ce{Ba(OH)_2.8(H2O)_{solid}  + 2NH4SCN_{solid}  ->  Ba(SCN)2 + 2NH3 + 10 H2O}$. The entropy change is large and positive so $-T\Delta S$ is large and negative. This reaction involves mixing two solids that react to form a solution cooling at the same time and so is an effective freezing mixture.

Thermodynamics
Generally speaking, the difference you have described is true. To be precise, it depends on the environmental parameters. What you said is true for constant pressure and constant temperature. For constant temperature and constant volume we would have $\Delta U$ instead of $\Delta H$, and Helmholtz free energy $\Delta A$ instead of $\Delta G$.
But the case you describe, the open beaker, is the standard case and your description is correct.
Kinetics
If a reaction is "spontaneous", it does not necessarily mean that the reaction actually runs at an observable speed, which is a totally different question. Spontaneity is only concerned with thermodynamics. Even a thermodynamically favorable reaction might run so slowly that we would say they don’t run at all. One example is the decomposition of diamond to graphite.
Example
2) That’s a little bit difficult since non spontaneous reactions do not run, we can’t experience them.
Some compounds, when dissolved in water, cause it to cool down (e.g. urea). These are examples of spontaneous but energetically unfavorable reactions.
Many biological reactions are energetically favorable but not spontaneous and need to be pushed or pulled by other reactions.
Generally energetically favorable equilibria ($\Delta H < 0$) will shift towards the product side if cooled. So the cold denaturation of proteins is one example. It’s not spontaneous at room temperature but is usually energetically favorable at room temperature.
Another example would be saturated solution of gypsum with a small amount of undissolved gypsum. No more gypsum is dissolved so the reaction is at the given conditions not spontaneous any more but if we would cool down the solution more gypsum would dissolve, indicating that this reaction is exothermal that is energetically favorable.
So the reaction of solvation in a stable gypsum solution with remaining solid gypsum is an example for a non spontaneous energetically favorable reaction.
Note that the last example is exactly the reverse of my first (spontaneous but energetically not favorable) example.
Of course the reverse reaction of any reaction that is energetically not favorable but spontaneous is an example of a reaction that is energetically favorable but not spontaneous. (see comment by Ivan Neretin below)

Once you have an example of reaction which is spontaneous but energetically not favorable, the reverse of it is inevitably non-spontaneous but energetically favorable.

Absolutely! Thank you for your comment, I added that to the answer.


text:
How to replace the 3 hydrogen atoms in phenylammomium with methyl groups

How would you replace those three hydrogens with methyl ($\ce{CH3}$) groups in its place to make trimethyl phenylammonim? It's been a real head stumper.

Adding excess $\ce{CH_3I}$ should work.

Adding excess $CH_3I$ to your compound above would result in the formation of a secondary amine. The electron-donating methyl group would increase electron density on the N atom and increase the availability of the lone pair for nucleophilic attack. Not only will the reaction will proceed all the way to form a quaternary ammonium salt, but it also will not be able to stop halfway at the secondary or tertiary amines. This is because with increasing levels of substitution, the lone pair on the N is more available for nucleophilic attack. I believe Master Organic Chemistry's website has some useful information on the mechanism:


text:
Calculating temperature change due to Joule-Thomson effect
I have a 10 L compressed argon cylinder with filling pressure of 10 MPa. According to Perry's Chemical Engineer's Handbook, argon's J-T coefficient (unit: K/MPa) is 3.7 at 0.1 MPa, 2.6 at 10 MPa. (Temperature is 300 K.)
When I open up the cylinder valve, 10 MPa argon in the cylinder will expand into 0.1 MPa atmosphere. So the $\Delta P$ is 9.9 MPa.

How much will the temperature change?
  The initial temperature is 300 K.

$9.9 \times 3.7 = 36.63$

(Which means the temperature of argon outside the cylinder will be 263.37 K)

$9.9 \times 2.6 = 25.74$
Something in between (like when the J-T coefficient in effect changes continuously as the gas expands).

It is going to be something in between.  The definition of the JT coefficient is $$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$If the temperature-dependence can be neglected, then this gives $$\Delta T=-\int_{0.1}^{10}{\mu(P)dP}$$

Unless I'm mistaken, won't they need to evaluate this numerically (thus presumably needing more points then they have) since there isn't a known functional form for the JT coefficient in terms of pressure for a real gas?

For a large change like this, it can be done using generalized corresponding states correlations for the 2nd virial coefficient to determine the residual enthalpy at the initial and final states.  See Smith and Van Ness, Chapter 7, Throttling Processes.

The initial state of the argon gas is completely defined by the given temperature and pressure:

Initial temperature $T_1=300\ \mathrm K$
Initial pressure $p_1=10\ \mathrm{MPa}$

The corresponding values of other properties of the gas can be looked up in so-called steam tables. For example, in REFPROP – NIST Standard Reference Database 23, Version 9.0 we find for the given initial temperature and pressure:

Initial density $\rho_1=167.60\ \mathrm{kg\ m^{-3}}$
Initial specific internal energy $u_1=78.313\ \mathrm{kJ\ kg^{-1}}$
Initial specific enthalpy $h_1=137.98\ \mathrm{kJ\ kg^{-1}}$
Initial specific entropy $s_1=2.8716\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$
Initial Joule–Thomson coefficient $\mu_{\mathrm{JT},1}=2.6066\ \mathrm{K\ MPa^{-1}}$

The final state of the gas is not completely defined by the given values since only the final pressure is given

Final pressure $p_2=0.1\ \mathrm{MPa}$

which is not enough to look up other values in steam tables.
However, we know that Joule–Thomson expansion is an isenthalpic process; i.e. the specific enthalpy $h$ remains constant. Thus

Final specific enthalpy $h_2=h_1=137.98\ \mathrm{kJ\ kg^{-1}}$

Now we have a second data point that can be used to look up other parameter values, for example

Final density $\rho_2=1.8105\ \mathrm{kg\ m^{-3}}$
Final specific internal energy $u_2=82.747\ \mathrm{kJ\ kg^{-1}}$
Final specific entropy $s_2=3.8155\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$
Final Joule–Thomson coefficient $\mu_{\mathrm{JT},2}=4.5226\ \mathrm{K\ MPa^{-1}}$

and also

Final temperature $T_2=265.65\ \mathrm K$

Since the pressure is not changing gradually, in the sense that it is either $p_1$ or $p_2$, not in between (it is not a reversible process), the temperature drop should be $\Delta T=\mu_\mathrm{JT}(p_2)\cdot p_2-\mu_\mathrm{JT}(p_1)\cdot p_1$.

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text:
Why does copper sulphate (CuSO4) in water increase surface tension?
So, recently for school, I did an experiment on surface-tension. We tested Water, Salt, Sugar, Dissolvable Panadol, Coffee Powder, Vinegar, Detergent, Flour, Methylated Spirits, Oil & Detergent, Copper Sulphate, Iron Sulphate, and
Ammonium Chloride.
Our theory at the time was:
Copper sulphate has a high surface tension when mixed with water because it has a lot of hydrogen bond acceptors that are not used. As more hydrogen bonds increase the surface tension, the increased amount of possible hydrogen bonds thus increasing the surface tension. What the copper sulphate is doing is being a whole bunch of extra links connecting the hydrogen atoms of water molecules together.
My Current Theory:
As Hydrogen likes Oxygen, the large amounts of Oxygen bonds in CuSO4 strongly attract to the H2O, which increases the pull force. Also, both H2O & CuSO4 are polar substances, helping them stick together.
I have no idea if any of this is true; our chemistry teacher had no idea either. Does anyone have a better theory/proof as to why this happens?
Note; We did not use capillary tubes as the order time would exceed the time our assignment due date. Also, our data might just all be noise too, as our method was not the most error-free design possible.
Our results are here.
(Relevant Results:)

                | Average | Test1 | Test2 | Test3 | Test4 | Test5 |
Control         |  3.21   |  3.25 |  3.4  |  3.1  |  3.1  |  3.2  |
Copper Sulphate |  3.94   |   3   |  4.7  |  3.5  |  4.5  |   4   |
Method:

Fill beaker with 40 mL of water (4th line)
Mix in approximately 3 mL amount of solute.
Using a pipette, drop water into a bottle cap until liquid is flat with the top.
Fill a pipette to 3 mL. Slowly drop in the water until it overflows, taking note of how much is used (i.e. emptied once & 2 mL used = 5 mL on top)
Record the results in a table. Repeat 4 more times.
Repeat the procedure with different solutes in different beakers.

It is good that you admit the possibility of your data being just noise. That being said, CuSO4 probably would indeed increase the surface tension, but not by much. Related: https://chemistry.stackexchange.com/questions/49570/does-nacl-reduce-the-surface-tension-of-water/49584#49584

Could you detail that as a question if possible? (Also, for some reason our noise was really big in the Copper Sulphate tests, not others.

From Sayok#6589 on Discord:

When dissolving CuSO4 in H2O, it does not remain as CuSO4. It's a soluble salt, so you make [Cu(H2O)6]2+ and [SO4]2-. The first coordination shell of either ions is filled with water, forming electrostatic/polar bonds between the water molecules and the ions.
Because the water molecule are forming bonds with charged ions, their polarity is slightly increased compared to a regular water molecule.
Surface tension of H2O is formed because of the polarity of the water molecules. They like each other and want to stick with each other. In general, the more polar a solution is, the more surface tension it can have.
Increasing the polarity of the water molecules because of the presence of the charged ions in solution thus results in a small increase in the surface tension.


text:
Problem in anode in experiment of electrolysis of brass
I am studying the electrolytic determination of copper content in brass (~58% copper, ~40% Zn and other trace elements) by using standard method BS EN 16117-1:2011.
Now the electrode used for the electrolysis are both platinum material electrode. Normally the copper ion in electrolyte moves to platinum cathode to form copper. After the electrolysis done, the electrolyte is colorless, the cathode is brown because of copper. Normally the platinum anode is unchange in its "silver" color however it is not in my experiment, the anode is brown too! I don't understand what happen in anode.
I try to put the "colorized" anode in a nitric acid solution (50%) to try to dissolve the "color" but it didn't work. Can anyone tell me why this happen and how to solve the problem in reserving the platinum anode? Please and thanks for reading my question.
(Some information in making electrolyte:
2.5g of brass was dissolved in (15mL 40g/L boric acid, 2mL HF, 30mL 50% nitric acid, neutralized with NH3 solution and reacidify with 50% nitric acid, make up to ~300mL with water and add one drop of HCl, as stated in the stardard method)
(Electrolysis environment: constant current, about 1-1.5A)


text:
Find atomic number given the group and period
I want to find the atomic number of an element in a short way, when I know the exact group and period of that element.
For instance; there is a question in our textbook to find the atomic number of the element which is placed in the 4th group and the 3rd period.

I think any formula for this is going to be more complicated then just checking the table.

This is the quickest way to do this.
Memorise the following table which contains the atomic number of all the noble gases. Noble gases are placed at the end of every period, so if you remember this table, then using a bit of common sense, we can easily locate the position of any element.

1st period :  helium - 2
2nd period :  neon-10
3rd period :  Argon- 18
4th period : Krypton -36
5th period : Xenon -54
6th period : Radon - 86

Now you are good to go.
Example 1) Say you want to find the atomic number of an element placed in group 4th and period 3.
Neon(10) is the last element of period 2. Come down to the next period ,i.e period 3. We need to reach four blocks to the right to get to group 4. Going each block to the right will increase the atomic number by 1. Therefore, going 4 blocks to the right will give us the atomic number = 10+4= 14.
Example 2) 4th period : 17th group
The last element of period 3 is Argon(18). To get to 4th period and 17th group, we need to come down to period 4 and go 17 blocks to the right. The required atomic number is 18+17=35.
Note : This also requires you to remember that there are 2 elements in the first period, 8 elements each in the second and third period, 18 elements each in third and fourth period.

Looking it up in the periodic table is the fastest way, because not all periods are equally long, and the groups are not always consecutive. The best I can give you is a piece-wise definition; given a period $p$ and a group $g$, the atomic number $n$ is given by:

$p=1$, $g=1$: $n=1$
$p=1$, $g=18$: $n=2$
$p=2$ or $p=3$, $g \leq 2$: $n = 8p + g - 14$
$p=2$ or $p=3$, $g \geq 13$: $n = 8p + g - 24$
$p=4$ or $p=5$, $n = 18p + g - 54$

$...$
Further down, you'll have to account for the rare earth elements which don't even have a group.


text:
How can you simulate the effect of a solvent on a polymer in an MD simulation?
I'm doing a molecular dynamics simulation of polymer melts and I want to include the effect of a good/bad solvent on the melt. How can this be done?
My simulation models polymers using a bead-spring model and the interactions forces are Lennard-Jones for pair interactions and FENE for bonding forces.
Edit:
FENE stands for Finite Extensible Nonlinear Elastic. It's a log based elastic force.
The potential function used, Lennard Jones + FENE is described here (http://lammps.sandia.gov/doc/bond_fene.html)

What's FENE? Can you provide a link or reference?

You have generally two choices a) implicit and b) explicit solvent. Presumably, you don't have good interaction parameters for explicit solvent, since you are likely doing a generic polymer melt.
There are a variety of solvation models; LAMMPS, for instance, implements a dielectric pair style for which you can attenuate the dielectric constant of your solvent (and a few other things). Other popular approximations are generalized Born models.
This paper uses GROMACS to simulate poly(styrene sulfonate) and poly(allylamine hydrochloride) in explicit and implicit solvent, as an example from the literature.


text:
What is a stereochemically active or inactive s orbital?
In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 73 under the section 3.6 (VSEPR Theory chapter Chemical Bonding):

...according to this theory, the position of the lone pair in $\ce{XeF6}$ molecule should be in axial position and the shape should be pentagonal pyramidal. But this is not observed and the actual shape of molecule is octahedral. This implies that the lone pair is placed at equatorial position, and this explains the structure of $\ce{XeF6}$ shown in figures 3.7 a,b,c.
According to Figure 3.7, $\mu_\text{expected}$ for $\ce{XeF6}$ is not equal to zero. But $\mu_\text{real}(\ce{XeF6})\sim0$ . To explain the above facts, it is considered that the actual molecule is in dynamic equilibrium of all the three structures shown in Figure 3.7. Here it is considered that the lone pair is present in the stereochemically active $s$ orbital and similar structure is observed as $\ce{IF6-}$. But anions like $[\ce{SbX6}]^{3-}$, $[\ce{TeX6}]^{2-}$ ($\ce{X}=\ce{Cl, Br, I}$), $[\ce{BrF6}]^-$,$[\ce{ICl6}]^-$ have been assigned perfectly octahedral structure on the basis of X-ray crystallography and the lone pair is present in stereochemically inactive $s$ orbital.

To explain the dipole moment, it is considered that the molecule is in dynamic equilibrium between three structures, where it is considered that the lone pair is present in the stereochemically active s orbital. However, I don't get what the author means by stereochemically active (or inactive) s orbital.

Generally, these terms are used to refer to lone pairs rather than s-orbitals.
In the context of VSEPR theory, lone pairs usually affect molecular geometry. For example $\ce{NH3}$ adopts a trigonal pyramidal structure, because of the presence of the lone pair on nitrogen; if there was no lone pair, then it would be trigonal planar, cf. $\ce{BF3}$. These are so-called stereochemically active lone pairs, since they affect the stereochemistry at nitrogen.
In a few rare cases lone pairs do not affect the molecular geometry, and these are termed stereochemically inactive. The only example I can think of, off the top of my head, is in $\ce{[XeF8]^2-}$, which should have 8 bond pairs and 1 lone pair around xenon. Despite the presence of the lone pair, the dianion adopts a square antiprismatic structure, without any obvious position in space for the lone pair to occupy. Looking in Shriver and Atkins' Inorganic Chemistry (6th ed.) there are a couple more example of $\ce{[SeF6]^2-}$ and $\ce{[TeCl6]^2-}$, which both have octahedral geometries despite having six bond pairs and one lone pair around the central atom.
In VSEPR theory, these stereochemically inactive lone pairs are rationalised by placing them in an s-orbital, which is non-directional. An electron pair in an s-orbital has a spherically symmetric electron density, and hence cannot influence the shape of a molecule. However, it is not exactly the orbital that is stereochemically (in)active, it is the electrons, and hence the more frequently seen term is "stereochemically (in)active lone pair".

There is one more context which should be mentioned, because it occurs frequently, and that is in tin(II) and lead(II) compounds. The electronic configurations of these two ions end with $n\mathrm{s^2}$, and this s-orbital lone pair tends to generate some distortions in stereochemistry.
A great example is that of the oxides. Solid $\ce{MgO}$ has a structure in which $\ce{Mg^2+}$ is octahedrally coordinated by six $\ce{O^2-}$ ions (rock salt/halite structure), but $\ce{SnO}$ and $\ce{PbO}$ have an unevenly distributed, square pyramidal coordination of $\ce{M^2+}$ by four $\ce{O^2-}$ ions. Traditionally this is attributed to the presence of a stereochemically active lone pair on Sn(II) or Pb(II).
This description cannot actually be complete, though, because if the geometry is solely determined by the lone pair on the metal, then one would expect PbS to have the same structure - and it doesn't. In fact, PbS has the same structure as MgO, with regular octahedral coordination, so clearly the anion has a role to play. A full explanation is beyond the scope of this answer, but one way of rationalising this is by invoking a second-order Jahn–Teller distortion. The interested reader may consult Chem. Soc. Rev. 2011, 40 (9), 4455–4463.
However, the "stereochemically active lone pair" is a recurrent theme in Sn(II) and Pb(II) structural chemistry (sometimes Tl(I) and Bi(III) also). For example, the metallocenes $\ce{SnCp2}$ and $\ce{PbCp2}$ adopt a bent structure, rather than the linear structure of $\ce{MgCp2}$ (and the more famous $\ce{FeCp2}$). Again, though, VSEPR theory is not the best way of explaining any of this; a proper explanation would involve MO theory.

In the second paragraph, it is given for XeF$_6$, "Here it is considered that the lone pair is present in the stereochemically active orbital". But there is a dynamic equilibrium between the structures and thus we can't ascertain whether the lone pair belongs to stereochemically active or inactive s orbital. Am I right? Kindly reply when you find time.

@Intellex Whether the molecular geometry is dynamic or not doesn't matter, I think; if you can detect some sort of asymmetry over a certain period of time, then it's enough to say that it's stereochemically active. However, note that (as I described above) it's usually the lone pair, not the orbital, that is described as being "stereochemically active".

@orthocresol the lone pair on $\ce{PH3}$ would perhaps be a more elementary example of a stereochemically inactive lone pair, isn't it?


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What causes steel rubbing on wood to leave grey or black marks?
In my own experience, running or rubbing a high-carbon steel on wood leaves a pencil-like trail. Is this simply carbon rubbing off, or is it a more complex reaction involving tannins in the wood, like another user suggested? EDIT (thanks Ivan Neretin): Or something else entirely, like iron (or its oxides) rubbing off? And to be more precise: is it lamellar hematite that rubs off and looks gray, or pure iron or a combination?

Carbon does not react (nor does it rub off). Iron does.

@IvanNeretin: well, that's the answer then!

The reaction with tannins requires moisture and some time.

Can you describe what kind of wood you are talking about ?  if it's a polished one or one with a brown colour coating it would be simply like rubbing of the color part by scratching

Not coated with anything, just regular softwood timber.


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Are "gas phase" calculations the same as "vacuum" calculations?
In both molecular mechanics/dynamics calculations and quantum mechanical calculations, the literature often refers to both gas-phase calculations and calculations done in vacuo. Is there actually a difference between these, or are they the same thing? If same, why would we call something a gas-phase calculation if there's no gas there?

Note that in-vacuo refers to an object surrounded by a vacuum, so it isn't that nothing is there, it's that nothing is there other than the object (i.e. gas molecule) of interest.

I don't know the etymology of the phrase, but yes, "gas phase" calculations are the same as "vacuum" calculations. There is no difference.
Think of it this way: the only option other than a "gas phase" calculation is one where solvent is included, either as explicit atoms or implicitly via a mathematical model or even a solid-state/bulk-model calculation. This indicates that the solute is in the gas phase, and hypothetically not interacting with the environment in any way.

What if someone wanted to model a solute in some environment with actual gas molecules? Such as to see the effects of a solute in N2? I'm not sure if that's an area of research, but am speaking hypothetically.

You would start off by building small clusters with explicit solvent: the solute surrounded by some number of $\ce{N2}$ molecules, represented via a force field or some ab initio-derived method. But, what constitutes "gas molecules"? Small clusters are weird models used to reduce computational cost. A true "gas-solute" simulation you're thinking of would be a bulk simulation with dozens or hundreds of solvent molecules that have the density and disordered structure of a gas.

That makes sense, and your description is what I would think of when someone says gas phase calculation (i.e., dozens or hundreds of solvent gas molecules). It doesn't make sense to me that calculations done in vacuum are also referred to as gas phase because there is nothing other than the solute.

@halcyon - you have to consider the density of the gas.. how far apart would the $\ce{N2}$ gas molecules be?

I didn't have anything in mind; this wasn't something I was planning on doing, but just musing on the terminology. For an ideal gas, we would have low pressure and high temperature, so in that case, there would be low density of gas, and perhaps the solute would act as though there are no gas molecules present (hence the interchange of terms)? What about for non-ideal gases?

"there would be low density of gas, and perhaps the solute would act as though there are no gas molecules present (hence the interchange of terms)?" - Yes, this is what I'm getting at; consider what the actual density of said gas is and the concentration of the solute, and the isolated molecule or small cluster model may be an acceptable approximation. One other thing to be aware of is that for non-dynamical simulations, such as geometry optimizations with QM methods or force fields, there is no concept of temperature, so we say that these calculations occur at 0 K.

If you actually wanted to study the behavior of real gases, I assume you'd need to perform non-zero temperature dynamical simulations with the appropriate solute and solvent concentrations, since the no-interaction approximation of ideal gases is no longer valid. Just to reiterate, keep in mind that this is not what is meant by "gas phase" calculation.


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What is the chemical name of the compound H2S4O?
I have been studying nomenclature and was asked to find the name of a strange compound: $\ce{H2S4O}$.
In my opinion, it should be dihydrogen-tetrasulfur oxide which seems very weird.
Can somebody advise me on how to name this compound (and whether this compound really exists?, I was able to come up with a structure similar to benzene...)

Could it by a type and it's H2SO4?

Yeah it could be but I dont think so because H2SO4 has appeared once and so had H2SO3 in the previous questions.

Your name is wrong, but there's too much isomers possible. Unless one of them is only relatively stable or otherwise obvious choice, this is too broad.

It looks like a typo to me, e.g. it's either a sulfuric acid $\ce{H2SO4}$, or one of the polysulfuric acids: $\ce{H2S4O6}$, $\ce{H2S4O13}$.

Section IR-5.4 of IUPAC’s Nomenclature of Inorganic Compounds (Red Book) of 2005 lists the rules that shall apply to naming inorganic compounds by a generalised stoichiometric name which does not carry any information about the compound’s structure. These rules dictate that:

The constituents of the compound to be named are divided into formally electropositive and formally electronegative constituents. There must be at least one electropositive and one electronegative constituent. Cations are electropositive and anions electronegative, by definition. Electropositive elements occur later in Table VI than electronegative elements by convention.
In principle, the division into electropositive and electronegative constituents is arbitrary if the compound contains more than two elements. In practice, however, there is often no problem in deciding where the division lies.
Section IR-5.4.1

In our case, oxygen must be considered electronegative and hydrogen must be considered electropositive since they have the highest and lowest electronegativity, respectively.
Within the classes of electronegative or electropositive substituents, ordering is strictly alphabetical. Electropositives come first, electronegatives last.
Section IR-5.4.2.1 allows the use of multiplicative prefixes as is the case with binary compounds. Further information such as charge, oxidation numbers or structure is not at our disposal.
Therefore, according to the principles laid out, we have two choices to name $\ce{H2S4O}$, depending on whether we wish to consider sulphur electropositive or electronegative:

Dihydrogen tetrasulphur oxide

Dihydrogen oxide tetrasulphide

To the best of my knowledge, this compound does not exist. But the principles of nomenclature are designed in a way to allow the naming of all compounds whether synthesised, hypothesised or fancily drawn on paper.

As noted in the comments, this is most likely a typo. In any event, if existent, this should probably regarded as an acid and thus could be called hydroxytetrasulfane acid, following its cousin dihydroxydisulfane.


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Keep two Mica Powders separate in oil and water mixture
First off, apologies if this isn't the appropriate place to ask this, let me know and I'll remove immediately.
I recently acquired some Mica Powder Pigments, and they are fun the play with.  I mixed some with mineral oil, and some with water.  On there own, that's neat.  However, if I mix the oil and water (with powder suspended), the pigments tend to mix in the oil, reducing the contrast between the oil and water.
Is there some process I can do that will make one of the pigments hydrophobic and the other oleophobic?
Preferably with "household" items, or easily attainable items, in a DIY/hobbyist environment.


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Chemical Thermodynamic Functions
I'm a high school student with basic knowledge about chemical thermodynamics. I have three questions regarding the thermodynamic state functions $H, S, G$. I have searched a lot in textbooks but they don't give the complete information as to whether the properties taken are that of the system or surroundings.
Q1):-Enthalpy is given by the formula
$H = U +PV$ .
 What I want to know is that whether $P$ mentioned above is the internal pressure of the system or the externally applied pressure?
Q2):- The change in entropy is given by the formula
$\Delta S= $$\int_1^2$${\frac{dq_{reversible}}{T}}$. Is the temperature mentioned of the system or the surroundings.
Q3)-On the same lines what temperature is taken to define the Gibbs free energy function:-$G=H-TS$.
Thanks in advance.

The three functions you mentioned are applicable only to thermodynamic equilibrium states of the system, for which the pressures and temperatures of the system and its surroundings match one another to within insignificantly small differences.  In the integral to get the entropy change (Q2), the same matching applies; that is, a reversible path consists of a continuous sequence of thermodynamic equilibrium states.
Note that, at the interface between the system and the surroundings, the temperature and pressure of the system always matches the temperature and pressure of the surroundings (irrespective of whether the system is at equilibrium).  It's just that, along an irreversible path of the system (for which the system passes through a sequence of non-equilibrium states), the pressure and temperature within the system vary with spatial position, and their average values (averaged over the volume of the system) do not match those of the surroundings at the interface.

Thanks for your enlightening answer.. Does it mean that I can apply these equations only during initial and final states for an irreversible process (when the system attains equilibrium after passing through a series of non euilibrium states )and not when the system is undergoing that process?

I also came across a statement that change in enthalpy when a gas undergoes free expansion process is equal to zero. How would you account for this statement sir? If I assume that the equation mentioned in q1 is applicable only in equilibrium states and the pressure is that of the gas it would give me a non zero change in enthalpy. Am I going wrong somewhere?

I forgot to mention that the aforementioned process is an adiabatic free expansion.

The answer to your first question is "yes" for the global system undergoing the irreversible change (because the pressure, temperature, and specific volume are varying with position within the system).  But, on a small scale, the answer depends on whether you are willing to approximate the thermodynamic functions locally (per unit mass) using the local conditions of temperature and pressure.  We engineers are, but some physicists, chemists, and other purists are not.

Regarding your questions about Joule Thompson and adiabatic free expansion, I feel it would be better to ask these questions in a new thread.


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Are there any sodium salts that are insoluble in water?
Are there any sodium salts (e.g. NaCl) that are insoluble in water? If so, what are their approximate $K_\mathrm{sp}$ values?

Quick Google search turns up a couple of examples: $\ce{NaBiO3}$ and $\ce{NaZn(UO2)3(CH3COO)9}$ (tangentially related: for potassium, $\ce{K(BPh4)}$)

Related (and interesting if you want to know more): https://chemistry.stackexchange.com/questions/175694/what-factors-cause-a-few-sodium-salts-to-be-water-insoluble

If your definition of a "sodium salt" is any solid compound that contains monovalent cationic sodium, then all you need to do is to look below your feet. The earth has abundant sodium bearing minerals (aka "salts").
Scroll down in The Mineralogy of Sodium to where it says Most widespread minerals containing Sodium. Out of all of these, only halite is soluble. The rest are mostly insoluble. I'm not sure their ksp values will mean anything here, because the dissolution of these minerals is up to kinetics and not thermodynamics at STP.
You can also have a look at Mineral Species containing Sodium (Na). Other than a few that are soluble (particularly some of the first on the list), the rest are not that soluble. Two reasons for that - they are silicates or phosphates which are in general insoluble, and they have other things in them.

Sodium tends to form water-soluble compounds such as halides, sulfate, nitrate, carboxylate and carbonates. There are only isolated examples of sodium compounds precipitating from water solution. However, nature provides examples of many insoluble sodium compounds such as the feldspars (aluminum silicates of sodium, potassium and calcium), sodium bismuthate ($\ce{NaBiO3}$), sodium octamolybdate ($\ce{Na2Mo8O25·4H2O}$), sodium thioplatinate ($\ce{Na4Pt3S6}$), sodium uranate ($\ce{Na2UO4}$). Both form of sodium antimonate (meta - $\ce{NaSbO3.7H2O}$, pyro - $\ce{Na2H2Sb2O7.H2O}$) has a solubility 0.7g/L. Sodium metaphosphate, $\ce{NaPO3}$ has both soluble and insoluble form.

Ref.: The Elements, PediaPress

Also, sodium fluorosilicate

So how strong the bonds are that water can't separate the ions!

Cryolite, $\ce{Na3AlF6}$, is one such compound.
 https://en.wikipedia.org/wiki/Cryolite

The salt $\ce{NaK2AlF6}$ ("elpasolite") is insoluble. Its exothermic precipitation is used as the basis for the thermometric titrimetric quantitative determination of sodium in a wide range of sample matrices. The determination is fast, accurate and highly precise.


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Comparing the mesomeric effects of amino and mercapto groups
I was quite surprised to find that $\ce{NH2}$ has much more $+M$ effect than $\ce{SH}$. But shouldn't we also consider the fact that $\ce{SH}$ has 2 lone pairs? Or is the greater electron density of nitrogen playing some role here?
How can I rationalise the stronger mesomeric effect of nitrogen?

For a mesomeric effect to occur, an orbital has to align with the π system it is supposed to resonate with. Since π systems have a defined $z$ axis — the orientation of the $\mathrm p_z$ orbitals that make up the π system — only orbitals that also feature such an alignment an interact. This limits any mesomeric interaction of any heteroatom with a π system to that heteroatom’s $\mathrm p_z$ orbital. Therefore, both nitrogen and sulphur can only have one orbital at their disposal for $+M$ effects.
The reason why the $+M$ effect of nitrogen is stronger despite its higher electronegativity is the period these elements are in. Nitrogen, being in the same period as carbon, has orbitals that fit together well size-wise with carbon orbitals. This is also reflected by the ease of forming $\ce{C=N}$ or $\ce{N=O}$ double bonds. Sulphur is one period down. Its orbitals are more diffuse than carbon’s or nitrogen’s; therefore, any π overlap is much weaker. This is also the reason why $\ce{S=O}$ double bonds are rare and $\ce{S=C}$ double bonds are typically more reactive than $\ce{O=C}$ double bonds.


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What is the most simple way to cut through a steel with very thin aluminum coating to obtain cross section without damaging the coating?
I needed to take cross section characterization of an aluminum coating. I tried to use a hand saw to cut through it, but it destroys the coating, and usually is separated with the substrate as I exerted force in using the tool. What can I do to get a cross section part without damaging the thin coating/film?

The positioning of your question at Chem.SE gives us a subtle hint that you might be hoping for a chemical solution. I don't think one exists. Aluminum is pretty active; once it starts reacting, it isn't going to stop, and the galvanic pair with steel only makes it worse. Then again, I might be missing something, so let's wait and see.

Use a good metalligraphy saw. Do a rough cut to get two pieces, glue them flush together with the coating on the inside, and do your clean cut through that. Or rough cut and polish back to an undamaged portion of the coating (doing this in a sandwich is good too). None of this is Chemistry per SE, but it wouldnt fit on Physics...

For metallography, cut with  a wet abrasive wheel ( as noted, you may need to place another surface against the aluminum for support)  ,. Then mount the sample in Bakelite or equal.  Then grind and polish in the traditional manor. It will require good technique not to round the aluminum edge, and excellent cleanliness to avoid scratching the aluminum.  This is not a "do it yourself" job.


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Rubbery surfaces degrading over time
I can think of 4 products I have with rubberized plastic surfaces on them that have degraded over time. (IKEA scissors and kitchen knife, Logitech wireless mouse, Rocketfish wireless keyboard.)
The rubberized surface has become gooey and sticky after years of (minimal) use, to the point where it leaves visible residue on my skin or I leave fingerprints in it.  The back of the keyboard became sticky like glue.
At first I thought it was just one product and the factory screwed up the plasticizer mix, etc.  But now that it's happened in multiple products from multiple companies, is there something in my environment that could be making this happen?  Humidity?  Sunlight?  (I don't think the mouse or keyboard were in the sun much.)  A chemical in the air?

I'm not sure if this is your exact issue, but here's a little information: https://www.polymersolutions.com/blog/stickiness-insulting-reality-rubber-reversion/ and also https://www.youtube.com/watch?v=R1Xmg5Gst0U

@Abhigyan Are these actually made out of rubber though?  I assumed it  was a special type of plastic.  (Or maybe those are the same thing.  I'm no chemist.)

Same doubt :) I hope a real chemist turns up soon... Meanwhile, you would do well to ask this question at Lifehacks.SE if you're looking for a solution to this menace...


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Cool air on sweat
When my face is full of sweat, I go near the A.C. and when the cool air touches my sweat it feels colder than dry parts? Why does the sweat feel colder than dry parts?

Because it evaporates and removes heat from your body.

Sweat is supposed to cool you down and has been evolutionarily ‘designed’ and kept that way.

The heat of evaporation of the water is $\pu{2260J/g}$. That means for every one gram of water, it takes $\pu{2260J}$ of energy to vaporize it. The water (evaporating) takes the needed heat energy from your $\pu{37 ^\circ C}$ body.
Source

And OP would feel the same effect if they used plain tap water instead of sweat.

@Jan Sorry but what does OP mean?

But then I should feel the same when near a fan, but A.C. seems to be much cooler. Why?

@RamKeswani OP is original poster; in this case the person who asked the question (you). And you do feel the same effect near a fan. The aircon is colder because it is blowing cool air at you rather than ambient air.


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Equations Used in Isothermal Titration Calorimetry
I have been attempting to understand and derive the formulas used for ITC which can be found here starting on page 310. I am particularly interested in the single set of identical sites. To first briefly summarize, my concern is their treatment for the stoichiometry. The equation they provide (not derived) for the equilibrium on page 312 is
$$K=\frac{\theta}{(1-\theta)[X]}\tag1$$
where $\theta$ is the fraction of sites occupied by the ligand $X$ and $[X]$ is the concentration of free unbound ligand at equilibrium.
However, when I attempt to derive the equation, I obtain something slightly different.
$$K=\frac{\theta}{(1-\theta)[X]^n}\tag2$$
where $n$ is the stoichiometry of the binding.
My Work
In the notation of the document, let $X$ be the ligand and $M$ be the macromolecule which $X$ binds to. Then the binding event can be written as the following chemical reaction.
$$\ce{nX + M <=> MX_n}\tag3$$
Let $X_t$ and $M_t$ be the initial concentrations of the ligand and macromolecule respectively. Define $x$ to be the amount of M that reacts with $X$. Upon reaching equilibrium, the concentrations will now be
$$\begin{align} [X]&=X_t-nx \\ [M]&=M_t-x\tag4 \\ [MX_n]&=x\end{align}$$
By definition of the equilibrium constant,
$$K=\frac{[MX_n]}{[M][X]^n}\tag5$$
Substituting in our equilibrium concentrations gives
$$K=\frac{x}{(M_t-x)[X]^n}\tag6$$
Now allow $x=\theta M_t$. This is just the definition of the fraction of occupied sites because the amount of complex made is equal to the fraction of sites times the concentration of the initial amount of macromolecule. So we thus now have
$$K=\frac{\theta M_t}{(M_t-\theta M_t)[X]^n}=\frac{\theta}{(1-\theta)[X]^n}\tag7$$
where $[X]=X_t-n\theta M_t$ as they have defined in the document. Equation $(7)$ is actually a rearrangement of the Hill Equation.
My General Question: It appears their treatment is exactly the same except they leave out the power of $n$ in the equilibrium constant expression without reason. What is the justification in ignoring the power in the equilibrium constant?
My analysis using the Origin-built ITC software tends to fail consistently when $n$ is not near unity. Could this be due to the gross oversimplification of the equilibrium? I can understand why they would leave out the $n$ mathematically because it is difficult to solve the equation. Though if the company knew how to solve equations numerically this would not prove to be as great as an issue.

It's not oversimplification so it's OK. Do you understand what it'd mean on molecular level if order was greater then 1?

@Mithoron It would mean that more than one binds simultaneously which is statistically not likely for an elementary reaction. I think I understand it now. Thanks.

Yeah, but you may have some point if reaction is more complicated then simple binding. Maybe binding of dimer... Usually such assumption should work, though.


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What is the molecular structure of xenon hexafluoride?
Xenon hexafluoride, $\ce{XeF6}$, can be obtained by reaction of xenon with excess fluorine gas.
What is the gas-phase structure of the molecule?
What experimental evidence is there to support it, and is there any theory which allows for an prediction of the structure?

[Note that solid $\ce{XeF6}$ has multiple polymeric forms, and is hence excluded from the scope of this question.]

TL;DR Xenon hexafluoride has a fluxional structure in the gas phase, with multiple rapidly interconverting conformers. The three most important conformers have $C_\mathrm{3v}$, $O_\mathrm{h}$, and $C_\mathrm{2v}$ symmetries. The minimum energy conformer is probably $C_\mathrm{3v}$.

Experimental evidence
The structure of xenon hexafluoride ($\ce{XeF6}$) has always been controversial; it is a famously strong oxidising agent, so experimental studies of it have been hampered by difficulties in the isolation and storage of pure samples.
Early studies indicated that the structure of $\ce{XeF6}$ showed deviations from octahedral symmetry. Gas-phase electron diffraction (dating from 1968) showed that the structure was approximately, but not exactly, octahedral: the experimental diffractogram differed from the theoretical diffractogram calculated for a hypothetical octahedral molecule.1 The authors considered a wide range of possible geometrical distortions, and found that $C_\mathrm{2v}$ or $C_\mathrm{3v}$ structures were relatively consistent with the spectrum obtained, with the latter being the most probable configuration.2 These could be interpreted in terms of a stereochemically active lone pair:

If we start with an undeformed octahedron we can characterize the deformation to the equilibrium structure in terms of the site of the repulsive lone pair [...] Symmetry considerations would seem to dictate that the lone pair be centered on the corner, face, or edge in the equilibrium structure. [...] If the lone pair protrudes through the center of a face, a $C_\mathrm{3v}$ structure results [...]  If the lone pair points toward an edge a $C_\mathrm{2v}$ configuration is induced.

This approach has stood the test of time, and is reproduced in multiple recent inorganic chemistry textbooks.3–5
At roughly the same time, experiments with molecular beams of $\ce{XeF6}$ were carried out by Falconer et al. These beams were not deflected in an inhomogeneous electric field, and the authors derived an upper bound of $\pu{0.03 D}$ for the electric dipole moment of a rigid structure.6
This was occasionally presented as evidence for an undistorted structure, but the calculations did not take into account the possibility of a highly fluxional process, and later Bernstein and Pitzer showed that depending on the barrier height, dipole moments up to $\pu{0.6 D}$ could be plausible.7 On top of that, the distorted conformers were not expected to have large dipole moments, as the dipole moment of the lone pair would oppose the net dipole moments of the $\ce{Xe-F}$ bonds.
Vibrational spectroscopy offers a unique possibility to eliminate, or at least reduce, the influence of dynamic processes, since the timescales involved are extremely short. In 1972, Claassen et al. carried out infrared (IR) and Raman analysis of $\ce{XeF6}$ at various temperatures.8 A group theoretical analysis leads to the prediction that an octahedral molecule should possess two IR-active bands, one $\mathrm{t_{1u}}$ stretch and one $\mathrm{t_{1u}}$ bend. However, the IR spectra at both room temperature and $\pu{4 K}$ (in an argon matrix) show at least three stretches alone. Similar inconsistencies with an octahedral structure were observed in the Raman spectra. However, the extremely broad peaks precluded a conclusive assignment to a specific distortion.
Based on all the experimental data, Pitzer and Bernstein concluded that $\ce{XeF6}$ could be described by a pseudorotational model with the potential minimum most likely being a $C_\mathrm{3v}$ structure.9 Following in the footsteps of Bartell and Gavin,2 they attributed this deformation to a second-order Jahn–Teller effect, which we will now turn to.

Theories of bonding
Clearly, traditional VSEPR theory cannot satisfactorily explain the structure of $\ce{XeF6}$. Often, the structure is described as having a stereochemically active but fluxional lone pair, which is generally accurate; however, VSEPR cannot be used to predict the region of space which the lone pair occupies.
We could reject the 2c-2e bonds utilised in VSEPR theory, and move instead to 3c-4e bonds, which works for the lower fluorides $\ce{XeF2}$ and $\ce{XeF4}$, and is briefly explained here. Essentially, each $\ce{F–Xe–F}$ interaction is modeled as a linear 3c-4e interaction. However, this model predicts a regular octahedral structure, which is again incorrect.
As far as simple bonding theories go, the best explanation is given by qualitative molecular orbital theory. A MO scheme for octahedral $\ce{XeF6}$ can be constructed from Xe 5s, Xe 5p, and F 2p orbitals (neglecting π-type interactions). It is similar to the one for $\ce{SF6}$ (see here), except that there are two more electrons:

This opens up the possibility of a second-order Jahn–Teller distortion, and I have described the mathematical basis of this elsewhere. By distortion to a lower symmetry, the HOMO and LUMO are allowed to mix, which leads to a stabilising interaction. The distortion should be mediated by a vibrational mode with symmetry $\mathrm{A_{1g} \otimes T_{1u} = T_{1u}}$, which does indeed lead to a $C_\mathrm{3v}$ structure. The Jahn–Teller effect, as well as the application to the case of $\ce{XeF6}$, is described in the textbook by Albright et al.10 I have reproduced one diagram from the book (p 364):

Equation 14.4 shows the HOMO-LUMO mixing upon lowering of symmetry; equation 14.5 depicts the particular linear combination of the traditional $2\mathrm{t_{1u}}$ orbitals that transforms as $\mathrm{a_1}$ in the lowered symmetry; and equation 14.6 shows the interaction of the $2\mathrm{a_{1g}}$ orbital with this linear combination.

Computational studies
Computational studies of $\ce{XeF6}$ have generally been plagued by a strong dependency on the level of theory and on the basis sets used, due to the very small energy differences between the relevant conformers. There is still a lack of agreement on the exact details of the potential energy surface.
Kutzelnigg et al. provided the first thorough in silico investigation of xenon hexafluoride, along with a number of valence-isoelectronic species.11 They concluded that the balance between an undistorted $O_\mathrm{h}$ structure and a distorted $C_\mathrm{3v}$ structure is governed by competing effects. Distortions are found to be present when nonrelativistic Hartree–Fock theory is used, and the addition of polarisation functions (e.g., Xe 4f) to the basis set favours the $C_\mathrm{3v}$ structure. However, relativistic effects as well as electron correlation both favour the $O_\mathrm{h}$ structure. In the case of $\ce{XeF6}$, these lead to very small differences between the two structures.12
Very sophisticated coupled-cluster calculations (CCSD(T)-F12b) were carried out by Peterson et al., and led to the conclusion that the $C_\mathrm{3v}$ structure was the ground state.13 The eight equivalent $C_\mathrm{3v}$ structures (an octahedron has eight faces) underwent interconversion via $C_\mathrm{2v}$ transition states $\pu{1.08 kcal/mol}$ higher in energy. The perfect $O_\mathrm{h}$ structure was found to be a saddle point $\pu{1.80 kcal/mol}$ above the ground state, with a threefold degenerate imaginary vibrational frequency.
However, the accuracy of these calculations was called into question by Cheng et al., who found that the $O_\mathrm{h}$ structure was in fact a local minimum (i.e. no imaginary frequencies).14 According to the authors, the imaginary frequency found by Peterson et al. arises due to basis set effects in the CCSD(T) triples correction.
All in all, computations broadly support the fact that $\ce{XeF6}$ is dynamically distorted from $O_\mathrm{h}$ symmetry. However, the exact details have yet to be ironed out, and given that the energies involved are so close to each other, "an unbalanced calculation may fortuitously give the right answer" (in Kutzelnigg's words11).

Notes and references
1. Gavin, R. M., Jr.; Bartell, L. S. Molecular Structure of XeF6. I. Analysis of Electron‐Diffraction Intensities. J. Chem. Phys. 1968, 48 (6), 2460–2465. DOI: 10.1063/1.1669470.
2. Bartell, L. S.; Gavin, R. M., Jr. Molecular Structure of XeF6. II. Internal Motion and Mean Geometry Deduced by Electron Diffraction. J. Chem. Phys. 1968, 48 (6), 2466–2483. DOI: 10.1063/1.1669471.
3. Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 4th ed.;  Pearson Education, Inc.: Upper Saddle River, NJ, 2012.
4. Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford University Press: Oxford, U.K., 2014.
5. Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th ed.;  Pearson Education, Inc.: Upper Saddle River, NJ, 2014.
6. Falconer, W. E.; Büchler, A.; Stauffer, J. L.; Klemperer, W. Molecular Structure of XeF6 and IF7. J. Chem. Phys. 1968, 48 (1), 312–318. DOI: 10.1063/1.1667921.
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From where did you get that MO diagram ? I want to derivation of it.

@A---B it is a standard treatment for a polyatomic molecule; any inorganic textbook (e.g. refs 3–5) will explain how to construct MO diagrams (find the symmetry labels of the atomic orbitals on the central atom; then construct group orbitals/so-called "symmetry-adapted linear combinations" of the terminal atomic orbitals; then overlap them to form MOs). There isn't a derivation of this MO diagram per se. I just took the general method and applied it to this molecule.

"electron diffraction (dating from 1968) showed that the structure was approximately, but not exactly, octahedral: the experimental spectrum differed" note that diffraction does not yield spectra, but rather "diffraction patterns" or "diffractograms". Otherwise very good overview about the state of the art!

@Rudi_Birnbaum Thank you for your comment and for the correction! I have changed it. Sorry, this must have slipped my mind when you posted it (I think I was working on my thesis).


text:
Allotropes of hydrogen
Is it possible to chemically bond 8 hydrogen atoms together to form an allotrope like $\ce{H8}$?

No, unless maybe in really extreme conditions. Why you ask?

I just thought of the idea, what would it create if its even possible to create...

OK, there was kinda similar thing with oxygen. It was found to exist in $\ce{O8}$ molecules under very high pressure,  so maybe hydrogen also does, but AFAIK such molecules were not detected.

Trihydrogen hydride would be the one I could think of

The only thing I found is Chemical Bonding of New Allotropes of Hydrogen and Its Implications to Fuel Science, A. C. Angus which screams opinion article but maybe someone has the time to look into it.

@Linear Christmas they really don't seen to back up their claims in any way. Also, even their suggestions seem to imply that these allotropes would only exist in the sun.

@Tyberius That what I was afraid of. The fact that they would exist soley in a sun is not a really a problem in itself because extreme conditions are required either way. But the credibility is disqualifying matter. Thanks for checking into it! Perhaps looking at some astrophysics / nuclear physics textbook is also warranted to be sure?