text:
Why are certain gases with higher molar masses less dense than the ones that have a lower one?
Neon, despite having a larger molar mass than nitrogen, oxygen and fluorine, is less dense than all of them. This also goes that fluorine (again) and chlorine are both denser than argon, even though argon has a higher molar mass than both of the halogens. The table below lists densities at standard pressure and temperature:
$$
\begin{array}{lrr}
\hline
\text{Gas} & M/\pu{g mol^-1} & \rho/\pu{g l^-1} \\
\hline
\text{Nitrogen} & 14.007 & 1.165 \\
\text{Oxygen} & 15.999 & 1.292 \\
\text{Fluorine} & 18.998 & 1.696 \\
\text{Neon} & 20.180 & \color{red}{0.900} \\
\text{Chlorine} & 35.453 & 3.214 \\
\text{Argon} & 39.498 & \color{red}{1.664} \\
\hline
\end{array}
$$
According to the ideal gas law, under equal pressure and temperature gases with higher molar mass should be denser:
$$\frac{M_1}{\rho_1}=\frac{M_2}{\rho_2}$$
Why isn't this the case for neon and argon?

Because a nitrogen molecule is N2, not N, and Ne is just plain Ne.

You list molar masses for elements, whereas nitrogen, oxygen, fluorine and chlorine are diatomic gases (noble gases are monoatomic).

In analogy, a molar mass is like a mass of a dozen ( $N_\mathrm{A}$ ) of the paper bags with the exactly same content, which may be 1 or more pieces of various fruits.

As the comments imply,the molar masses you quote are for atoms of each element, whereas the observed gas density is determined by the masses of molecules.
Most elemental gases have two atoms per molecule, so for example nitrogen is $\ce{N2}$ and thus, properly, the molar mass of the molecules is $2×14.007=28.014\text{ g/mol}$.
The noble gases helium, neon, argon, krypton , xenon and radon (we do not know the ambient state of oganesson, element 118) are exceptions: with their filled atomic valence shell these have only one atom per molecule in the gas phase, and so the molar mass of neon truly is $20.018\text{ g/mol}$.
Thus the molar mass of neon is less than that of nitrogen after all when you count molecules, so neon will come out less dense.
A more curious example is hydrogen fluoride, whose density at 25°C is $1.15\text{ g/L}$ despite apparently having a molar mass of only $20.006\text{ g/mol}$. Neon, as seen above, has nearly the same molar mass but its ambient density is $0.900\text{ g/L}$. This happens because some hydrogen fluoride molecules are joined together by hydrogen bonds, even in the low-pressure gas phase, to effectively make bigger molecules. It's a rare example of a gas showing significant non-ideal behavior at normal atmospheric pressure. But we may expect other gases with hydrogen bonding, such as steam, to act the same way at higher gas pressure (which means higher temperature at least for steam).

Ok,thank you. I never really thought of that. All the answers on google said " higher molar mass=higher density"

...these have only one atom per molecule in the gas phase...  By IUPAC Goldbook definition, molecules must consist of at least 2 atoms, so elementary noble gases do not consist of molecules. (I personally do not like it, thinking it could be formally like a molecule as the least amount of particular compound or allotrope. E.g. O3 as ozone molecule, O2 as molecule of "normal" oxygen and O as a molecule of atomary oxygen, being also an atom of element oxygen.)

@poutnik then please offer another way to describe the concept that fits with the definitions. I am apparently not as good at nomenclature as I am at physics

@OscarLanzi  Neither am I, nor it was my idea. I knew this because I was in past corrected myself. I have not received a positive answer to my question if there is any official common term for "an atom or molecule". Perhaps a particle, what can be an atom, or molecule, or a ion, or whatever.

The definition of molecule that is really appropriate is "The form of the compound or element that exists in the system under consideration" instead of making a Text book definition. It is what is used for n in PV= nRT or what should be used to calculate molality or mole fraction or partial pressures. A helium atom is a molecule until it is cold enough to become super fluid etc. Similarly, in space the hydrogen molecule is the H atom or possibly H+.


text:
What should be done to decrease the pressure of the HPLC column?
After using an HPLC (C18) column several times it doesn't do what it has to do. When we use it the HPLC instrument shows a very high pressure and the process stops. We have washed with methanol, water, and acetonitrile, but none of them have been effective. What are we overlooking? What should we do to decrease the pressure of the column?

Available pump heads Max. flow rate in ml/min is 10
Max back pressure in bar 400
Recommended particle size is 5 μm

What's the pressure, mode (isocratic/gradient), column geometry, analyte/matrix and mobile phase? Was there a downtime before you noticed an issue with pressure? What does the column's manual say regarding washing and storing conditions?

@andselisk with the flow of 1 the pressure reached to 380 bar, column C18 with nanometere column, the analyte is fructose, mobile phase has been water, we worked with that last week and it was good, this time it doesn't work.

You barely answered any of the questions. 1… what? ml/min, μl/s? What's the flow unit? What does "column C18 with nanometere column" mean? Are you doing nanoflow LC with a guard column, or what? What are the dimensions of the column (length, diameter) and particle size? Again, what mode you are working in? What are the specified maximum pressures for the column and for your LC system? What is the matrix and fructose concentration (could it crystallize)? What temperature is your column thermostat is set to? Feel free to [edit] these details into your question instead of posting comments.

@andselisk I'm really sorry if I can't answer your questions properly, I'm just a student and our professor suggested no option other than washing the column with solvent and it wasn't effective, We've been trying to determine the concentration of HMF made from fructose.

It's OK, the problem is that HPLC troubleshooting is a multivariant task and we need as much info as possible. If the washing procedure described in column's manual didn't help, or there is none, as the last resort you could try washing the column in reverse using low flowrate, say, 0.05 to 0.2 ml/min. Modern columns (which model is yours, BTW?) should handle this all right. I'd strongly recommend though first to talk with your supervisor/TA or anyone on campus who has experience with LC and could come by and have a look at your machine.

By the way, are you sure the problem lies with the column? Have you tried to run the solvent with the column disconnected? A bit silly suggestion I should've probably started with, but still…

@andselisk "washing the column in reverse using low flowrate, say, 0.05 to 0.2 ml/min" we have done, but the pressure is still high. Could the high pressure have any reason other than the column? " Have you tried to run the solvent with the column disconnected?" I'm not sure what should I do, but when we do such a thing the pressure is zero and we can't figure out what's wrong. When we use flowrate 0.5 the pressure is good but when the flowrate is 1 the machine stops the process. Our machine is AZURA Compact HPLC.

Well, if you disconnect the column and the pressure drops, then it is indeed the column giving you trouble. Alternatively, you could try to increase thermostat temperature to, say, 60 °C and wash with your MP again. Also, there are several regeneration guides for various columns such as one from Agilent (PDF).

@andselisk Thank you so much, I would be trying to do that again, and read your valuable link, hope it works, thanks again.

If you ran 100% water on a c18 column it possibly suffered phase collapse. Might or might not be recoverable

What your professor has to recognize is that an HPLC column is a consumable. It is not meant to last forever. The back-pressure shows that it has been mistreated in the past with careless handling. Also, the HPLC pump seal could be going bad and generating polymeric microparticles which are clogging the column. At both ends of a column you have very fine sieve called frits (micron sized, typically 2 micron pores).
If your solvent, sample had small particles they can easily clog the frits. Also, if someone were careless enough, a sample can precipitate at the head of the column due to solubility issues. It is always the inlet frit which causes problems.
Look at the schematic of a column:

Some standard ways to check the system:

Prime the pump, which just means that you ensure pump is filled with solvents and there are no air bubbles.

Remove the column and connect the injector to the detector with a long tubing. And start the flow, always start at low flow rates like 0.2 mL/min and then slowly change it to 1 mL/min. How much pressure is observed? There is negligible pressure like 5-6 bars, then the problem is in the column.

HPLC columns have an arrow to show the direction of flow. Intentionally use the opposite direction and start the flow, again slowly. Do not connect the column to the detector. Starting at 0.1 mL/min and then go up to 0.5 mL/min. If it still stays at the same high pressure even after an hour, it is time to either toss it in the garbage.

If your professor is more mechanically oriented, open the column (last resort). Sonicate the inlet frit in various solvents and fill in just bare silica if you accidently scraped a layer of stationary phase and put the cleaned frit back.

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text:
Why is the Einstein relation not valid for germanium?
In Elecronic Devices and Circuits, by J Millman and CC Halkias, it is written at one place:

.....coefficient of diffusion $D$ and mobility $\mu$ aren't independent. The relationship between them is given by the Einstein equation
$$
D_i=\mu_iV_T
$$
where $V_T=\bar kT/e = T/11600$ ..........

Whereas later on in the same text:

For germanium the diffusion constants $D_p$, and $D_n$ vary approximately inversely proportional to $T$.

Does this mean that the mobility varies as inverse squares of temperature:
$$\mu_i \propto T^{-2}$$
If yes then how, since lattice scattering mobility $\mu_L\propto T^{-3/2}$ whereas ionized impurity scattering at small $T$ values, $\mu_I\propto T^{-5/2}$ ( derived from equations given in The Dopant Density and Temperature Dependence of Electron Mobility and Resistivity in N-Type Silicon, by Sheng S. Li). If no, then how else? Does it have to do something with the low band gap of Ge? Please give a quantitative explanation.
Edit
I have come to know that an inverse square variation of mobility with temperature is resulted due to polar optical phonon scattering. However the exact derivation was missing.

Quantitative explanations seems hard to come by in textbooks.
In Andy Grove's (yes, the Intel guy) Physics and Technology of Semiconductor Devices (1967) one finds

Experimentally, mobilities have been found to follow a $T^{-2.5}$ dependence rather than the theoretically predicted $T^{-1.5}$ dependence in the lattice scattering range.

(Underneath that passage I wrote, many years ago, "Explanation?" - you are not alone!)
Moving to Sze's Physics of Semiconductor Devices (I've got the second edition, 1981), one finds on pp 28-29:

For lower impurity concentrations the mobility decreases with temperature as predicted... The measured slopes, however, are different from (-$3 \over 2$) because of other scattering mechanisms. For pure materials near room temperature the mobility varies as $T^{-1.66}$ and $T^{-2.33}$ for $n$- and $p$-type Ge, respectively; as $T^{-2.42}$ and $T^{-2.20}$ for $n$- and $p$-type Si, respectively; and as $T^{-1.0}$ and $T^{-2.1}$ for $n$- and $p$-type GaAs, respectively.

So, one is left with the rather vague reason of "other scattering mechanisms" which is not very helpful, to be sure. Without digging deeply into the literature, my feeling (even way back when) was that the combination of band structure (where carriers are and where they can scatter to, including things like intervalley scattering) combined with the phonon structure (phonon populations and $E$ vs $k$ available to scatter carriers) combine to make the actual scattering calculations really hard. In other words, the fairly glib and simple relationships for mobility that are presented just don't work very well.

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text:
Why doesn't XeF6 displace oxygen from water?
Consider the following reactions (unbalanced):
$$
\begin{align}
\ce{XeF2 + H2O &→ Xe + HF + O2} \tag{a} \\
\ce{XeF4 + H2O &→ XeO3 + Xe + O2 + HF} \tag{b} \\
\ce{XeF6 + H2O &→ XeO3 + HF} \tag{c}
\end{align}
$$
Considering that the oxidizing power order of the xenon fluorides is $\ce{XeF6} > \ce{XeF4} > \ce{XeF2},$ I don't understand why $\ce{XeF6}$ is unable to oxidize $\ce{O^2-}$ (in water) to $\ce{O2}.$ $\ce{XeF2}$ and $\ce{XeF4}$ are able to achieve this (rather difficult) oxidation, but $\ce{XeF6}$ is unable despite having greater oxidizing power.
Note: I have a primitive understanding on this topic and a deep and elaborate explanation will be appreciated. Also, an explanation of the reasoning behind each individual reaction ( for example, why only $\ce{XeF4}$ disproportionates?) will be even better.

XeO and XeO2 do not exist. Also you're oversimplifying what happens when XeF6 reacts with water - at least 3 equations would be needed.

XeO2 is known (https://pubs.acs.org/doi/10.1021/ja110618g) and its synthesis is via hydrolysis of XeF4

My guess: Xenon difluoride and xenon tetrafluoride produces xenon(II) oxide as intermediates which is unstable and disproportionates to xenon, xenon(VI) oxide and oxide and that's what you see in the products. Xenon hexafluoride also hydrolyses but as xenon oxide fluorides as intermediates which ultimately converts to xenon(VI) oxide. So, it is necessary for you to break down the reaction and balance them because the reaction are not that simple. See: https://pubs.acs.org/doi/pdf/10.1021/ja01065a009

Section 8.4 and 8.7 of Arrow Pushing in Inorganic Chemistry: A Logical Approach to the Chemistry of the Main‐Group Elements give mechanisms for these reactions. These are not mechanisms which have actually been proven to be true just like most of the book, but nevertheless they convincingly explain why we get the products that we do in these reactions.

@AshishAhuja Could you please convert the content of sections 8.4 and 8.7 into an answer as I can’t access the link you mentioned.

@ShashwatTomar Search for the book on zlib, I had downloaded a pdf from there. I don't think I have much to add to what is given in the book, so any answer I write will basically just be a blatant copy paste, which I think will fall under plagiarism.


text:
Chloro functional group is given number 4 in phenol but number 1 in nitrobenzene
In the IUPAC name 4-chlorophenol, we see that, the C atom where OH group is attached, is given number 1. But in the IUPAC name 1-chloro-4-nitrobenzene, we see that, the C atom where $NO_2$ group is attached, is given number 4. In the first case, phenol is the base. So C atom of OH group of the base is given number 1. Applying the same rule to the second case, C atom of the $NO_2$ group should be given number 1. Please help to clarify this point. Thanks.

"Applying the same rule to the second case" – which rule? Note that the parent structure of the second example is benzene and not nitrobenzene.

@Loong, in that case, the parent structure of the first example must also be benzene. We can not consider phenol as the parent structure. I must be missing some thing.

Why does it have to be benzene/phenol? In a way, the same 'rule' applies to 3-chloropropan-1-ol, for example. The alcohol gets the low locant.

@orthocresol Then NO2 must get the low locant in the second example in my question.

There is no rule that Cl must be the last.

Why should NO2 get the low locant? It's a plain old substituent prefix, just like chloro. If it was a bromo instead of a nitro, it would go first, because of alphabetical order. Alcohols, on the other hand, are generally suffixes, which tend to get priority over other prefixed substituents.


text:
Does hot potassium permanganate break benzene rings?
Benzene rings are broken during ozonolysis, and alkenes treated with hot $\ce{KMnO4}$ undergo oxidative ozonolysis.
So does that mean that the double bonds of benzene also are broken by hot $\ce{KMnO_4}$ ?
If this is the case then why doesn't benzene (pictured here) also break?

Can I make the conclusion that benzene rings are broken during ozonolysis but not by hot $\ce{KMnO_4}$ ?

Yes . Hot KMnO4 does not oxidise benzene ring but it can oxidise side chains attached to the ring.

Ozonolysis and oxidation by $\ce{KMnO4}$ cannot be compared.

See these questions: https://chemistry.stackexchange.com/questions/136156/why-do-alkylarenes-react-with-alkaline-kmno4-but-not-benzene ... https://chemistry.stackexchange.com/questions/162973/mechanism-of-alkyl-benzene-reaction-with-kmno4-koh-to-form-carboxylic-acids ... https://chemistry.stackexchange.com/questions/98966/action-of-hot-alkaline-kmno4-on-tert-butyl-benzene

As you are seeing in the pictured reaction, hot potassium permanganate does not break the benzene ring itself. It just reacts with the alkyl groups, oxidizing them to give the carboxylic acid functions shown.
Note carefully: an ordinary double bond can also be oxidized with potassium permanganate, but that does not work with benzene because you do not really have double bonds. Instead benzene rings have a special, delocalized bonding that confers extra stability (which shows up as resisting the potassium permanganate). You will learn more when you study aromaticity.


text:
Definition of amino acid side chain
Is the α-carbon considered a part of amino acid side chain?
Is the nitrogen atom on the backbone is considered a part of the side chain of the proline?

This depends on the context.
In a context where atoms are either classified as backbone or as sidechain, you would count the atoms C, O, N, CA (and connected hydrogens) as backbone, and all the other atoms as sidechain. This is how molecular viewers classify atoms.

(Jmol image of Pro22 and Glu23 of crambin, 1CRN, with sidechain atoms shown as spheres.)
According to IUPAC/IUB's JCBN,

2.2.5. Definition of Side Chain
When amino acids are combined in proteins and peptides, C-1 , C-2 and N-2 of each residue (the numbering being that of
aliphatic amino acids) form the repeating unit of the main chain ('backbone') and the remainder forms a 'side chain'. Hence the
words 'side chain' refer to C-3 and higher numbered carbon atoms_ and their substituents.

If you show the structure of a sidechain, it is highly likely you would include at least the alpha carbon for context, and in case of proline the backbone nitrogen as well to illustrate the ring structure.


text:
What is happening when I dissolve styrofoam in acetone? (And should this change temperature)
I’ve completed a science experiment on endothermic and exothermic reactions in which I measured the temperature of solvents before adding solutes and measuring again. I found that dissolving styrofoam peanuts in acetone lowers the temperature of the acetone, but my book says that should not happen. Can anyone tell me what is happening when acetone dissolves polystyrene? I thought the reaction would be endothermic because the acetone is breaking down bonds in the styrofoam and using energy to do so and my book did not even explain anything about the reaction.

When you measure the lowering temperature of the acetone, have you found that it lowers specifically from the addition of the styrofoam? Or is the temperature of the acetone lowering by itself, due to evaporation? Try a small experiment of introducing additional surface area (maybe just some wads of paper towels) and seeing if that has the same temperature effect as the styrofoam. If so, maybe the acetone is just evaporating away and cooling off?

based on this link, I can say that the reaction you're seeing is that the acetone is dissolving the carbon-hydrogen bonds within the styrofoam: https://chem.utah.edu/_resources/documents/pdf/undergrad_program/Disappearing%20styrofoam%20no%20pics.pdf

I tried introducing surface area using some starch based peanuts, but nothing happened. For clarity, this reaction should use outside energy to break/weaken the polystyrene bonds, right? I know that polar solvents need some outside energy to break polar/ionic bonds. Is this the same for non polar solvents? Also thanks for the link! Very helpful.

Theoretical background:https://en.wikipedia.org/wiki/Flory%E2%80%93Huggins_solution_theory

@TylerM : dissolving C-H bonds ?!? What utter nonsense.

The question can be reformulated "What happens when bulk polystyrene is being dissolved in acetone?". Perform 2 parallel experiments, 1 with piece of bulk polystyrene ( used for many various boxes or disposable vessels ) and the other with about the same piece of PE/PP. Compare the cooling. As acetone significantly cools itself by intense evaporation what I bet is what you observe (or at least interferes with real dissolution cooling ).

Mmmm… This seems like a good idea. I’ll do it once I get more acetone. Actually I’ll just reuse the acetone I previously used. Also why did you say that the resolving of C-H bonds was nonsense? That is what’s happening, is it not? Polystyrene is just C and H so what else could it be?

@Karl why is it nonsense? The paper I linked makes sense to me. To quote the paper I linked: "Remember that likes dissolve likes. Acetone is a relatively non-polar solvent (as compared to water, which is highly polar), and Styrofoam is made from polystyrene and foam. Due to their similar polarities, acetone can dissolve the carbon-hydrogen bonds of Styrofoam. In return, the starch based peanuts are soluble in water because of the polar hydroxyl groups. Water and starch share the same properties (they are both highly polar), while water and Styrofoam have opposing properties."

Dissolution rather just breaks relatively weak van der Waals intermolecular forces, which consists of various ways of interaction between charges, permanent dipoles and induced dipoles.

Styrofoam is made of polystyrene and foam is incorrect. Foam is formed by polystyrene as solid structure, filled by closed or open ( based on the product type ) gaseous phase created by evaporation of foaming agent.

Could the experiment be flawed, in that your acetone is evaporating and as a result taking in more energy to go into acetone vapor than the dissolving of the polystyrene gives? You might want to repeat the experiment under pressure and enclosed, resulting in much less evaporation.

@Poutnik or by adding gas into a liquid which creates foam directly. This foam might, or might not, cure into a solid, depending on the chemical in question.

@Trish Sure, these are technological technicalities that are not very relevant the question. The relevant is interaction polystyrene - acetone. The captured gas is a bystander.

@TylerM I checked that text, and the rest of it makes sense for a low level undergrad course (my profs wouldn't have dared insulting us with such triviality, but never mind), but that phrase about the C-H bonds is just ridiculous. Every highschool kid should know that covalent bonds are not broken by such dissolution process. It seems Utah University chemistry 101 is taught by their retired caretaker.

I re-did the experiment and the more styrofoam I added the cooler the acetone got. The polystyrene remains the same in composition so what exactly is happening if not a chemical reaction/C-H bond breaking? And why does it absorb energy?
P.s. This is for what I can only describe as a pre-introductory chemistry section of my physical science class.


text:
Reaction of phosphorus pentoxide with perchloric acid
I've come across the reaction of $\ce{P4O10}$ with $\ce{HClO4}$ (perchloric acid) which gives the products as metaphosphoric acid and $\ce{Cl2O7}$ but I'm unable to understand how does perchloric acid form and what causes the trimer of $\ce{HPO3}$ to be formed, also $\ce{P4O10}$ being an acidic oxide how does it react with a strong acid like perchloric acid?
$$\ce{12HClO4 + 3P4O10->6Cl2O7 + 4(HPO3)3}$$
Most importantly how exactly does this

Get converted to a structure like this

Is there some sort of mechanism behind this like we see in T3P coupling reaction (to be clear I'm not saying this reaction is exactly that, I'm just asking if this involves similar mechanism) where oxyphosphorous linkage follows an elimination? (Like the oxygen of $\ce{HClO4}$ form a linkage with phosphorus of $\ce{P4O10}$)
Even though phosphorus pentoxide acts as dehydrating agent how does HPO3 form as a product how does the cage structure there get broken down to $\ce{(HPO3)3}$?

I think it's double substitution by $\ce{ClO4-}$ - first at $\ce{P}$ then at $\ce{Cl}$, catalysed by the acid.

The result of the reaction can be any polymer of $\ce{HPO3}$, and not only $\ce{(HPO3)3}$.

@Maurice true but considering this case the product mentioned in the book J.D Lee , I want to know how that is formed?

Take any phosphorus atom. It is linked to three oxygen atoms $\ce{O}$ by three simple $\ce{P-O}$ bonds. These three bonds will be modified in a similar way, according to the following steps.
Step $1$. Approach one $\ce{HClO4}$ molecule. It provides one $\ce{H+}$ ion which will be attached to one of these bonding $\ce{O}$ atoms. The remaining $\ce{ClO4-}$ ion does not play a role presently.
Step $2$. After this attachment, the $\ce{P-O}$ bond coming from the same initial $\ce{P}$ atom is broken. The oxygen atom remains attached to the $\ce{H}$ atom and to the rest of the molecule. A positive charge appears on this initially chosen $\ce{P}$ atom.
Step $3$. Later on a water molecule will be attracted by this positive $\ce{P}$ atom, and will attach one $\ce{OH}$ group on this $\ce{P}$ atom, (plus release a free $\ce{H+}$ atom, that repeats the same game on the next bonding $\ce{O}$ atom). The whole is equivalent to providing one $ \ce{OH-}$ ion on the positive $\ce{P}$ atom.
Step $4$. Repeat the operations $1$ to $3$ on each $\ce{P-O}$ bonds leaving the initially chosen $\ce{P}$ atom. At the end, the $3$ $\ce{P-O}$ bonds (coming from the same $\ce{P}$ atom) are broken, and replaced by one $\ce{H}$ atom attached on each former bonding $\ce{O}$ atom, and $3$ $\ce{OH}$ bonds on the so modified $\ce{P}$ atom. This produces a lonely $\ce{H3PO4}$ molecule, plus the wanted cycle $\ce{(HPO3)3}$. The result is well described by $$\ce{P4O_{10} + 3 H2O -> H3PO4 + (HPO3)3}$$
Now the wanted substance $\ce{(HPO3)3}$ is obtained. It is easy to check that it has the same structure as drawn by Kafka yash in his original post.
Later on, both $\ce{H3PO4}$ and $\ce{HClO4}$ can be dehydrated to produce $\ce{HPO3}$ then $\ce{(HPO3)3}$ and $\ce{Cl2O7}$. The mechanism of these transformations may be studied by the same reasoning, but it is not required by the author.


text:
Best description of forces/ mechanisms that stabilize the interaction between a hydrophobic protein binding pocket and a hydrophobic ligand
This is a more theoretical or definition question that is related to the terms "hydrophobic effect", "hydrophilic interaction", and "van der Waals' forces" (and others I may have missed).
In case I would like to describe a hydrophobic ligand that is binding to a protein binding pocket that has also hydrophobic residues (such as Leucine, valine) that stabilize (or favor) the binding of the ligand. What would be the best description to use to describe the binding?
from this question and from this PNAS paper
Although it seems the definition for these terms is also confusing, generally these terms have the meaning of:

hydrophobic effect (the low solubility of hydrophobic solutes in
water)
hydrophobic interaction or force (the unusually strong attraction of
hydrophobic surfaces and groups in water, it is specifically
describing the repulsion of water)
van der Waal's force (attractions/repulsions between molecules or
atoms, other than attractions like ionic attractions, and covalent
attractions)

I also know that hydrophobic interaction is stronger than van der Waal's force (vdW). Given the "definition" of the hydrophobic interaction, I assume using this term to describe the interaction between the ligand and protein is not very appropriate because there is no necessary repulsion of the water molecules (that is usually used for forming the protein core). Please correct me if I am wrong.
Therefore vdW seems to be the answer. However, the problem is why the interaction between hydrophobic ligand and the hydrophobic protein surface is favorable? To be more specific, or in other words, whether there will be the same/ similar stability for a hydrophilic ligand to bind?
As also mentioned in this question, there is Deybe force or induced dipole if the ligand is polar or even charged. Should I assume the "stacking" of the London dispersal effect is much greater than Deybe force, although the London dispersal effect is weak if they are alone?

In discussing interactions in a biological context, often the most relevant measure of binding strength is the equilibrium binding constant (or more commonly dissociation constand $K_\mathrm{D}$), which is directly related to the Gibbs energy of binding at a given temperature.
On the other hand, when discussing intermolecular interactions, the focus often is on the enthalpy of binding, or the potential energy (or force) of the interaction at the molecular level.
The hydrophobic effect is described in the context of water as bulk solvent. If you split the Gibbs energy of binding of a hydrophobic ligand into enthalpy and entropy, the enthalpy contribution will be small (a hydrophobic ligand has slightly stronger interactions with water than with the hydrophobic binding pocket). The entropy contribution will drive the value of the binding constant. The hand-waiving explanation is that when a water molecule is at the interface between ligand and bulk water, it loses freedom of orientation because all hydrogen bonds are with the bulk water.

However, the problem is why the interaction between hydrophobic ligand and the hydrophobic protein surface is favorable? To be more specific, or in other words, whether there will be the same/ similar stability for a hydrophilic ligand to bind?

The hydrophobic ligand will have a lower dissociation constant ("tighter" binding) than the hydrophilic ligand. However, the enthalpy of binding will be larger (slightly) for the hydrophilic ligand if it fits snugly into the pocket. Trying to explain the binding constants (Gibbs energy) by just looking at the potential energy (enthalpy) will give an incorrect conclusion.
The distribution between entropic and enthalpic contributions might be different for different binding pockets. From the viewpoint of fitness, it does not matter whether binding is driven by one or the other, as long as the ligand binds. (Here is a paper that shows evidence for hydrophobic interactions driven by enthalpic contribution: https://pubs.acs.org/doi/pdf/10.1021/ct1003077).

Thanks, @karsten-theis for the detailed explanation. Could you please also confirm with me that we should use vdW force to describe the interaction rather than the hydrophobic effect?

@WilliamWong You would have to come up with a specific question you want to address. I don't know what is means to "use vdW force". Is this a conceptual use, or a quantitative use? Is this a question of measurement or of computational applications?

Indeed, I am going to talk to some biochemists. I just want to make sure my language is correct and precise. The thing I would like to describe is that ligand-binding (it is not entirely non-polar, but is trying to describe the part that has high hydrophobicity) is favorable because there is hydrophobic interaction. However, when I go over the "definition" of these terms as mentioned in my original question, I found that the term "hydrophobic interaction" might not be appropriate. Thus, I wonder if I should change it to vdW force instead.


text:
Resonance structure with allylic lone pair

Draw the appropriate resonance structure for 5-aminolevulinic acid:

As there is an allylic lone pair, my first arrow goes from the lone pair to form a π-bond, and the second goes from the π-bond to form a lone pair:

However, the solution selected a allylic lone pair in a different location in the molecule:

Is my answer also correct? If not, how would I choose the correct lone pair?

There's no "allylic lone pair" in there! It's a ketone not enol or enolate, or whatever you imagine.

Draw the hydrogen atoms and see what happens. If you add a proton (that carries the positive charge you have drawn), you would have a valid ionic resonance structure; one that's usually taken to explain hyperconjugation.


text:
Why, out of the main halogens, is bromine the most soluble in water?
Why, out of the main halogens, is bromine $(\ce{Br2})$ the most soluble in water?
Why is there such a drastic decrease for iodine $(\ce{I2})?$
Here is a table from my inorganic chemistry textbook with the numbers that prompted this question.
$$
\begin{array}{lr}
\hline
\text{Halogen} & s/\pu{g L^-1}~\ce{H2O} \\
\hline
\ce{F2} & \text{explos. reac.} \\
\ce{Cl2} & 6.5 \\
\ce{Br2} & 33.6 \\
\ce{I2} & 0.33 \\
\hline
\end{array}
$$

Bromine being a liquid helps things along, maybe?

And it might vary greatly with temperature.

Whatever the substance, the solubility is a problem that is not yet fully understood, quantitatively speaking. Nobody is able to calculate the solubility of a given substance. Maybe in some future ?

A citation for the textbook would be really nice.

In the case of iodine versus bromine, the sharp drop in solubility may be attributed in part to iodine showing incipient intermolecular bonding -- as a metal would have between its atoms. Such a shift towards metallic character would be expected to lead to reduced solubility in nonmetallic liquids such as water or even carbon tetrachloride (see figures quoted here).
When bromine is solidified, the bond length within a diatomic molecule is nearly the same as in the gas phase. Wikipedia, citing Reference [1], gives the solid phase bond length as 227 pm versus the gas phase bond length of 228 pm. With iodine, however, condensation leads to a lengthening of the intramolecular bond, indicating a delocalization of the bonding into intermolecular regions, From Wikipedia, also citing Reference [1]:

Gaseous iodine is composed of I2 molecules with an I–I bond length of 266.6 pm. ... It is even longer (271.5 pm) in solid orthorhombic crystalline iodine, which has the same crystal structure as chlorine and bromine. As such, within the iodine molecule, significant electronic interactions occur with the two next-nearest neighbours of each atom, and these interactions give rise, in bulk iodine, to a shiny appearance and semiconducting properties. Iodine is a two-dimensional semiconductor with a band gap of 1.3 eV (125 kJ/mol): it is a semiconductor in the plane of its crystalline layers and an insulator in the perpendicular direction.

The tendency of iodine towards adopting a metallic structure is also evident from studies of the transformation of the solid element under high pressure. Duan et al. [2] report that bromine is calculated to undergo an insulator-metal transition at about 55 GPa pressure. With iodine, Pasternak et al. [3] identify formation of zigzag chains and resultant netallization with 16 GPa pressure, less than one-third the calculated requirement for bromine metallization.
A final consideration: Given the role of incipient metallic character and intermolecular bond delocalization just discussed, the solubility of iodine in water might be improved by partially recovering these characteristics in the solute. This is where the well-known impact of an iodide salt comes in: iodine is much more soluble in water that contains iodide because the triiodide ion, $\ce{I3^-}$, with its delocalized three-center four-electron bond, is formed in the solution. Such a behavior is not significantly carried over to any lighter halogens.
References
1.
Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8. pp 800-804.
2.
Defang Duan, Yanhui Liu, Yanming Ma, Zhiming Liu, Tian Cui, Bingbing Liu, and Guangtian Zou (2002). "Ab initio studies of solid bromine under high pressure".
Phys. Rev. B 76, 104-113. https://doi.org/10.1103/PhysRevB.76.104113
3.
M. Pasternak, J. N. Farrell, and R. D. Taylor (1987). "Metallization and structural transformation of iodine under pressure: A microscopic view".
Phys. Rev. Lett. 58, 575. https://doi.org/10.1103/PhysRevLett.58.575

One thing I don’t see mentioned in your answer is the formation of hypo- acids and hypohalites (
H
X
O
and −
X
O
−
). Since these form when heavier halogens react with water, does their formation influence the overall solubility of the halogen?

Not much, really. The equilibrium constant for bromine hydrolysis is rather small and most bromine in aqueous solution remains unreached. See Beckwith et al., https://doi.org/10.1021/ic950909w. We would expect still less of this reactivity with iodine. I focus on heavier halogens specifically to minimize the role of the reaction you describe.

Solubility is the confluence of many factors and can be looked at through many chemical frameworks; it might first be a good idea to lay out our framework for understanding solubility before tackling the question:
One relatively simple-but-useful framework for solubility can be to see it as the thermodynamic stability of the substance when solvated compared to the thermodynamic stability of the substance when interacting only with itself. We'll start there.
Knowing that pressure and temperature affect the phases a substance or mixture—i.e. the way and degree to which it self-interacts—the conditions we are looking at are going to be an important consideration whenever we are asking a question like this. At any given pressure and temperature, if the substance energetically prefers self-intermolecular interactions (a.k.a. cohesion,) then it will not be favorable to solvate. If the molecule is energetically ambivalent to interactions with itself vs. interactions with the solvent, then this would be an example of miscibility. This ambivalence is what defines an "ideal solution" when generalizing the ideas of ideal gases to other phases. Following this trend, energetic favorability of a solvated state over a non-solvated state at a certain pressure and temperature will thus increase solubility.
So the question we are asking under this framework specifically is "Why, at standard temperature and pressure, are interactions between elemental bromine and water significantly more stabilizing than interactions between water and elemental chlorine or elemental iodine?"
This is where the molecular complexity begins to have a big impact on the answer: unlike more fundamental qualities like electronegativity or size, there is no singular trend across the halogens that can be used as a smoking gun for the solubility of their elemental forms. I'm not fully convinced that the phase of the elemental form aids in this solubility much ("bromine is a liquid, so it mixes easier" e.g.,) specifically because these figures define the concentration of a saturated solution and not the ease/rate of solvation which would be more of a kinetic effect than a thermodynamic effect.
Whenever I see an optimum in the middle of any trend (high solubility in the middle of the halogens, e.g.,) I usually look for two "forces" of similar magnitude that work in opposite directions to each other. The two in this question I'm most drawn to (there could be more, and probably is) are molecular size and the σ-hole size/halogen-bonding (XB) ability (we'll get to this part if you haven't heard of it, hang tight.)
Let's start with size. The larger the surface-area of a non-polar molecule is, the more susceptible it is to cohesion via van der Waals interactions. By this logic, chlorine should be the most soluble, followed by bromine, and then iodine. Because we see bromine breaking this trend, there must be a trend that works in the opposite direction.
A viable trend working in the opposite direction regards the (usually ignored) non-homogeneity/anisotropy of electron-density in singly bound halogens: it turns out that, instead of electrons smearing equally across the sphere of the halogen atom, there is a concentration of positive charge collinear with the axis of the σ-bond and the electron-density moves to form a belt orthogonal to the σ-bond.
This effect is more pronounced for larger halogens than smaller halogens (pictured below) for many reasons, but mostly it's a combination of: larger polarizability of the electron distribution; low s-orbital character in the σ-bond; and a lower stability of the σ-bond decreasing the energy of the σ* orbital, leading to some destructive interference in the molecular orbital. The effect this has on intermolecular interactions has led to the phenomenon being called "halogen-bonding (XB)", as to relate it to hydrogen-bonding.
Here are two examples of trifluoromethylated halogens that emphasize how the prominence in positive-charge (colored red) can differ between them, with the chlorine motif on the left and the bromine motif on the right:

The lone pair on a water molecule can interact with this area of positive charge (called a "σ-hole") similar to interacting with the partially positive charge of a hydrogen—hence the name halogen-bonding—and this can stabilize solvation. The belt of electron-density carries partial negative charge, and thus can also interact with the σ-hole; this results in water coordination to the σ-hole being in competition with the self-intermolecular interaction. The crystal structure of orthorhombic iodine suggests that this interaction is strong in the solid-phase, so it is likely at least partially responsibility for the lower solubility of iodine.
If you'd like to learn more about XB interactions, I'll leave the DOI for a quick 2007 overview I recommend here: 10.1007/s00894-006-0154-7. I pulled those images directly from it.
My current master's thesis work is on exploring XB interactions, so my response comes primed with the fact that I think about these interactions a lot. There are countless other factors that perturb solubility, which is why we compress it into one simple figure and measure it empirically—I do believe with my best intuition, however, that these are the factors of greatest influence.

Can you show the XB results for the diatomic halogen molecules, especially the difference between bromine and iodine? Would be more relevant to the question and avoid the confusing scenario of showing only three blobs in a five-atom molecule.

@OscarLanzi Let me try to find those or I'll simulate it myself—the diagram is mostly just demonstrating what a σ-hole is and how the prominence differs between them, but I agree the diatomics would be useful here. The σ-holes will not be as noticeable in the diatomics because the trifluoromethyl group increases the energy of the C—X σ-bond, but the trend is maintained.


text:
Are alkynes weakly polar?
I was reading about physical properties of alkynes in my Chemistry textbook.
One of the sentence states

Alkynes are weakly polar in nature.

I didn't understand how alkynes are weakly polar, so I tried to search about it on the internet but couldn't find any explanation.
Many sites (like this and this) rather state that alkynes are non-polar.
My question is: are alkynes weakly polar? If yes, why?

All chemical compounds are weakly polar anyway.

Your textbook should've defined such descriptive term. Or, better, avoid using nonquantitative terms for teaching crucial concepts [incorrectly].

@MissMulan Would an isolated homo-diatomic molecule like $\ce{H2}$ or $\ce{O2}$ also be weakly polar?

Yes due to London dispersion.The least polar molecule is F-F.

Are we talking about "polar" or "polarizable" here?

@IanBush even X-X bonds are polar because valence bond electrons always have the probability of being closer to one of the X.However  the higher the electronegativity of the X atom is , the less possible this is.This is why N2,02,CL2 can become liquids or even solids at very low temperatures.

That is not the usual definition of "polar" - https://en.wikipedia.org/wiki/Chemical_polarity https://www.britannica.com/science/polarity-chemistry https://biologydictionary.net/polar-molecule/

"Polar" means having electric dipole moment by any means possible.

Any molecule that doesn't have the right symmetry is at least slightly polar, even saturated hydrocarbons. acetylene isn't polar but only because it is highly symmetric. Propylene, though,  is (slightly) polar.

Nonpolar = not more polar than weakly. Polar = not less polar than very  :-)

We have talked about polarity too often on this site already. Usually without real conclusions. It's so fuzzy and ill-defined that not even the IUPAC Gold Book has a definition for it. Alas, for most intents and purposes, if a molecule has no permanent dipole, it's called nonpolar. The rest you can consider polar.

Yes, alkynes can display a (small) permanent electric dipole moment, and thus are polar molecules by the usual definition. For example see https://cccbdb.nist.gov/exp2x.asp?casno=74997&charge=0 where the dipole moment for propyne is given as 0.780 Debye. Given such a molecule is asymmetric is it shouldn't be a surprise that there is a dipole moment.


text:
How can the multiplicative number (n) of a chemical series like polyethylene glycol (PEG) be a non-integer?
https://en.m.wikipedia.org/wiki/PEG_400
Down at chemical or molecular formula, it says:
C_2n H_4n+2 O_n+1, like for all polyethylene glycols (PEG)s, but then says n=8.2 to 9.1....
Huh?

What surprises you, exactly? What do you think $n$ here is, and why should it necessarily be an integer number?

Please don't post links to the mobile versions of the website. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Is the 3D spacing arrangement of atoms in chemical bond always unique?
I would like to ask if the 3D spacing arrangement of atoms in chemical bond configurations are always unique, please?
For example, glycine consists of the set of atoms HHNCHHCOOH. Does there exist a different stable 3D arrangement of this collection of atoms, other than:

What causes this arrangement to be unique? If we throw in and react the ingredients in a different order, such as HHHHHNCCOO, will they all still end up as the same unique shape of regular glycine?
Thank you.

It is complicated. If we throw in the same atoms in different order, most likely we'll get a bunch of different molecules.

There are plenty of other possibilities, like $\ce{H2N-CHOH-CHO}$, or $\ce{HO-NH-CH2-CHO}$, or $\ce{H2N-CO-CH2OH}$, or $\ce{HO-NH-CH=CHOH}$ or $\ce{H2N-CO-CH3}$, etc.

The word you are looking for is isomerism: it is the phenomenon that given number of atoms can form different stable molecules. Depending on how they are different, there are different types of isomerism.

https://www.sigmaaldrich.com/GB/en/product/aldrich/480959

https://www.sigmaaldrich.com/GB/en/product/aldrich/159034

From a given number of atoms, you can usually construct multiple molecules (isomers), so we shall just concentrate on the case where the bond connectivity is the same.
The terms you are looking for are conformation and configuration. The latter is most usually related to double bonds and E/Z isomerism. If we take but-2-ene, for example, it can exist in 2 forms, E and Z. Both are stable and have the same atom connectivity. However, they cannot interconvert between each other - the activation energy is too high.
Two conformations of the same molecule, however, can interconvert between each other (at least at room temperature) and they are both stable (i.e., potential energy minima). A classic case of a molecule is butane, whose conformations can be seen in the picture below.

The anti- and gauche- conformations are both 'butane' and both are stable (even though the anti- is more stable), but have a different 3D arrangement of atoms.
As a sidenote, these can be distinguished at a low enough temperature, at which they cannot interconvert.
Another famous molecule with various conformations is cyclohaxane, with the two stable ones being the chair and the much less common twist-boat.

Please include a source and an image for the image. || Related: Conformations of butane

The arrangement doesn't have to be unique.
For instance, using your notation, propanol is HHHHHHHHCCCO, but it exists in two forms:
Propan-1-ol:

and Propan-2-ol:

The difference is where the OH is attached, to an end-C or the middle-C.


text:
Mechanism for the synthesis of benzimidazole
I have thought up a mechanism for the synthesis of benzimidazole from o-phenylenediamine and formic acid. I would like to know if it's correct. I have attached it below.
My rationale for the protonated intermediate is that methyl formate can run without an acidic catalyst (Lewis or Bronsted) and still provide fair yields. Formic acid must be protonating itself for this to work.

Such protonated carboxylic acid doesn't seem likely intermediate in reaction with amine. I'd rather write proton transfer after attack of amine on neutral acid.

Huh. I guess that makes sense, but then what kicks off water?

I don't see which part. So, would formic acid protonate the intermediate and then water get kicked off? Explain yourself a little more please.

Huh, I don't think I get your way of writing mech. Reaction is autocatalysed, you don't have some magical substrate that works out everything. There's surplus of the acid and every possible proton transfer happens all the time. So no, it wasn't OK :(

Okay. So no self-protonation is necessary?


text:
How to calculate vapor pressure of ethanol at given temperature?
The following is a part of the problem I am struggling with. I think what I need is a vapor pressure of ethanol at $\pu{25 °C},$ but I may be totally wrong.

A reactor is charged with $\pu{60 bar}$ of $\ce{C2H4(g)}$ and $\pu{40 bar}$ of $\ce{H2O(g)}$. The equilibrium mixture is cooled and the water and ethanol condense. The excess ethene is allowed to escape. What is the vapor pressure of the resulting liquid mixture at $\pu{25 °C}?$ What is the mole fraction of ethanol in the vapor in equilibrium with the condensed mixture at $\pu{25 °C}$?

First, I calculated the equilibrium constant $K$ at $T = \pu{298.15 K}$ and $\Delta_\mathrm{r}G = \pu{-8070 J mol^-1}$:
$$K= \exp\left(-\frac{\Delta_\mathrm{r}G}{RT}\right) \approx 25.93.\tag{1}$$
Using this $K$ I calculated the partial pressure of each gas assuming they are not condensed:
$$
K = \frac{p(\ce{C2H5OH})}{p(\ce{C2H4})\cdot p(\ce{H2O})}.\tag{2}
$$
$$
\begin{align}
p(\ce{H2O}) &= \pu{2.43 bar},\\
p(\ce{C2H5OH}) &= \pu{37.57 bar}.
\end{align}
$$
I know the water and the ethanol are actually condensed (liquid) at this temperature with the corresponding molar fractions
$$
\begin{align}
x(\ce{H2O}) &= \frac{\pu{2.43 bar}}{\pu{2.43 bar} + \pu{37.57 bar}} \approx 0.061,\tag{3}\\
x(\ce{C2H5OH}) &= \frac{\pu{37.57 bar}}{\pu{2.43 bar} + \pu{37.57 bar}} \approx 0.939\tag{4}.
\end{align}
$$
Now all I need is the vapor pressure of the water and the ethanol at $\pu{25 °C}.$
I was able to find the vapor pressure of $\ce{H2O}$ at $\pu{25 °C}$ is $\pu{23.76 Torr}.$ But I don't know how to calculate the vapor pressure of ethanol at $\pu{25 °C}.$
I found a formula to calculate the vapor pressure of ethanol on Wikipedia — Ethanol (data page):
$$p(\ce{C2H5OH}) = 10^{8.04494 - 1554.3/(222.65 + T)},\label{eqn:5}\tag{5}$$
where $T$ is in Celsius and $p$ is in torr. I plugged $T = 25$ to \eqref{eqn:5} and got $p = \pu{58.71 Torr}.$
$$
\begin{align}
p(\ce{mixture}) &= x(\ce{H2O}) \cdot p(\ce{H2O}) + x(\ce{C2H5OH}) \cdot p(\ce{C2H5OH}),\tag{6}\\
                &\approx 56.58 \pu{Torr}.
\end{align}
$$
But I don't think I am doing right. Is there a more appropriate way to find the vapor pressure of ethanol?

Why do you think ethanol vapour pressure is wrong and not other things?

The Handbook of Chemistry and Physics gives the following values for the vapor pressure of ethanol. $1$ Torr at -$31.3$°C ; $10$ Torr at - $2.3$ °C ; $40$ Torr at  $19$°C ; $100$ Torr at + $34.9$°C : $400$ Torr at$63.5$°C, and 760 Torr at $78.4$°C. So your calculation for $25$°C looks correct.

Please provide the rest of the problem statement.  The part you are looking at seems like a VLE calculation.

Thank you very much @poutnik for updating my question and pointing out my wrong calculation.

Thank you It's great to know the vapor pressure of ethanol is correct. I also added the rest of the question.


text:
Why don't universal indicator solutions react with themselves?
Universal indicator is a pH indicator made of a solution of

water, 1-propanol, phenolphthalein, sodium hydroxide, methyl red, bromothymol blue, sodium bisulfite, and thymol blue

that exhibits several smooth color changes over a wide range of pH values.
Won't the various components react with each other? Methyl red is a weak acid, and sodium hydroxide is a strong base, so won't the methyl red in the universal indicator react with the sodium hydroxide?
Why is it that the various components of the universal indicator are preserved in their neutral state?

Who says they do not react?

The methyl red does react with the sodium hydroxide, forming a salt in the indicator as made. If placed in an acidic solution, of course, the salt may reconvert to the acid.

In a solution containing multiple species that undergo acid/base chemistry, everything will react with everything (including water, which can act as acid or as base). Equilibrium is established quickly, at some pH. Ingredients with a $\mathrm{p}K_\mathrm{a}$ higher than the pH will be (mostly) protonated, while ingredients with a $\mathrm{p}K_\mathrm{a}$ lower than the pH will be (mostly) deprotonated.

I am guessing that the indicator stock solution is formulated to be neutral initially, judged from the yellowish greenish color shown and described on the internet.

Why is it that the various components of the universal indicator are preserved in their neutral state?

Indicators don't have a neutral state. They have a protonated and a deprotonated state, one of which predominates at neutral pH (unless the $\mathrm{p}K_\mathrm{a}$ is near 7, in which case both states will be present). To make the pH neutral after mixing all the indicators, you would titrate the solution until it is neutral. In this process, some indicators will get deprotonated (and change color).

Methyl red is a weak acid, and sodium hydroxide is a strong base, so won't the methyl red in the universal indicator react with the sodium hydroxide?

Methyl red is red in the protonated form and yellow in the deprotonated form.
The $\mathrm{p}K_\mathrm{a}$ is 5.1, so if you titrate a solution to neutral with NaOH, you will have the deprotonated yellow form. In fact, the indicator substances were chosen so that they are either yellow or colorless at neutral pH. As you change the pH to acidic or basic, you get the "darker" colors like blue, red or purple.
Here is an image showing the colors of the four indicators (bottom) along with the color of the universal indicator, which is a mixture of the four. If you pick a pH, you can rationalize the color of the mixture. For example, for neutral pH, you have one indicator in the colorless form, two in the yellow form and one in the green form; this mixes to a greenish yellow. To get blue at pH 10, the concentrations of indicators have to be chosen carefully such that the yellow from methyl red does not overpower the blue from thymol blue and bromothymol blue.

Source: Adapted from https://www.compoundchem.com/2014/04/04/the-colours-chemistry-of-ph-indicators/

Thanks for the constructive answer! I understand everything now, and I can't thank you enough


text:
Why isn't Beryllium a noble gas?
Somebody answered it before on physics StackExchange, but I couldn't follow the high-level parlance. All I could follow is that it is related to the p subshell. I cannot understand why elements form bonds when they have pairs of spins(If it matters: I may be wrong, but please apprise me about my misconceptions). An example would be Aluminium, though it's not entirely related. It losses all three of its electrons when it can conveniently lose one and have a pair of opposite spin numbers.
Please answer :-)

See the related possibly referred  Physics SE

Also: https://chemistry.stackexchange.com/questions/9636/do-lithium-and-beryllium-want-to-lose-electrons-to-be-like-helium/9643

@Poutnik I have mentioned that I went through the answer linked by you before! It doesn't make sense to me as I am a high school student.

@NilayGhosh I have gone through it, but it doesn't make sense. I don't need textbook answers.

I don't know why people downvote without reading the question. If it's idiotic or unclear, then tell me!

@user324713 "See the related possibly referred" means it is implied you are aware of it while readers may be not, so the exclamation mark is out of place. // Rudeness is not an attribute of a great scientist, unless you are the 2nd Fritz Zwicky.

If those answers doesn't make sense and you don't need textbook answers, then I am afraid there are no answers for you.

@Poutnik I assumed that you didn't read the question. I apologise-- I don't intend to be rude. I misinterpreted the answer. The thing is that somebody answered on Quora that it is about the spin numbers that electrons gain or lose electrons; when I ask about it on Physics/Chemistry StackExchange, many members comment that it isn't clear or there is no answer for it. It is a bit frustrating. Once again, sincere apologies.

@NilayGhosh No problem :). Thank you!

Pairing of electrons alone does not make atoms chemically inert.

Because beryllium is present in between the period table rather than at the extreme right of periodic table and it's outermost shell is not completely filled like inert gases

I interpret this as a more general question of why alkaline earth elements are electropositive metals and not noble gases.
The answer is basically they do not have enough electrons in their valence shells to be strongly stabilizedcas elements. All the electrons below the outer $s^2$ pair are in shells having fewer nodes in their quantum mechanical wavefunctions, and the presence of fewer nodes makes these inner-electron waves so compact they shield almost all of the nuclear charge from the outer pair. So the outer pair of electrons have low effective nuclear charge attracting them, which translates to their own wavefunctions being relatively diffuse and their ionization energies being relatively low.
Helium and, to a lesser extent, beryllium are exceptions because helium has no such inner-shell electrons and beryllium has only one pair, whereas magnesium and heavier $s^2$ elements have multiple pairs and more nodes in their outer-shell wavefunctions. Hence the failure of helium to readily form compounds and the lesser failure of beryllium to form predominantly ionic ones. By the time we get to magnesium we already have the shielded structure described above well-enough developed to enable the ready formation of predominantly ionic compounds, even with the relatively low-electronegativity elements hydrogen and boron.


text:
Why is the momentum of the wave function in a 1-D particle-in-a-box = 0?
Is this implying that a particle in the box for 1-dimension has no momentum? Or is the momentum average out to $0$ on throughout the region $0\leq x\leq a$? For reference, I was told the state was $n = 1$, and assuming the wave equation is therefore
$$\psi(x) = \sin\left(\frac{\pi}{a}x\right).$$
It just doesn't intuitively make sense to me. I'm thinking it has to be more in line with my latter thought that the momentum throughout the box would end up equaling $0$. Does that mean inside any box, whether it be 1, 2, or 3 dimensional, the momentum would always average out to $0$, since it is closed in a system?
I am calculating $\langle\hat{p}\rangle$.

The average momentum of this particular state is indeed 0. So it is for other states with $n=2,;3$, etc. The closedness of the system has nothing to do with this, nor is it a thing at all.

@IvanNeretin Ok thank you. So i'm just wandering, what does it mean when you have a momentum average of $0$ a system, like this one here? How would it be different if it had a non $0$ for $<\hat{p}>$?

You can have nonzero momentum if your particle is not in one of the stationary states. The states you listed are eigenstates of the Hamiltonian, i.e. stationary states, i.e. no momentum.

Also, it’s not a zero average momentum within a single system, but rather a zero expectation value for momentum. If you measure the momentum of any one system, it may well not be zero. However, if you prepare many, many different copies of the same state, then measure the momentum of every system and average it together, it will be zero. I suggest re-reading the fundamental interpretation of quantum mechanics (virtually every textbook will discuss this).

Why is the momentum of the wave function in a 1-D particle-in-a-box = 0?

There are a couple ways of looking at this. One is to churn through the maths with the integrals, and you find that the expectation value comes out to zero:
$$\int_0^a \psi^* \left(-\mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}x}\right)\psi \,\mathrm{d}x = 0$$
Another is to search for a more meaningful physical interpretation of this, which we will try to do so now. First of all, note that when you say the wavefunction, you are actually only referring to one, specific, wavefunction for the system.
In introductory QM courses, you are typically exposed only to stationary states, which are states that satisfy the time-independent Schrödinger equation (TISE):
$$\hat{H}\psi = E\psi$$
Only a certain set of states, let's denote them by $\{\psi_1, \psi_2, \ldots, \psi_n, \ldots \}$, obey this equation. You know already that these states are given by:
$$\psi_n = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$
However, these are not the only possible states of the system. These are actually the set of stationary states of the system: stationary in the sense that they do not change with time, or in other words, they are time-independent (hence TISE).
Formally speaking, the expectation values of all operators are time-independent. The proof for this is not too hard and can be found in standard QM textbooks such as Griffiths (see also the second box in https://en.wikipedia.org/wiki/Ehrenfest_theorem).
For all stationary states, it is thus impossible to have a nonzero momentum because that would imply that the particle is moving, i.e. the position of the particle is changing with time. This would directly contradict the fact that such states are time-independent.
Thus, it turns out that for all the states $\psi_n$ (regardless of what number $n$ is), the expectation value $\langle \hat{p}\rangle$ is always zero. This applies to any stationary state of any system, regardless of its dimension.

[How can a system have nonzero $\langle \hat{p}\rangle$, then?]

The system would need to be in a non-stationary state, i.e. a wavefunction that is not one of the $\psi_n$'s. In general, such a wavefunction can be constructed by linear combination of the $\psi_n$'s:
$$\psi = c_1 \psi_1 + c_2 \psi_2 + \cdots + c_n\psi_n + \cdots$$
Such a state no longer obeys the TISE, but that's not a problem, as we never set out to claim that it was a stationary state. These states will evolve in time: in particular, the coefficients $c_i$ are time-dependent, and therefore the expectation value of various operators can also vary with time.
The time evolution of these states is governed by the time-dependent Schrödinger equation, instead:†
$$\mathrm{i}\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi.$$
I wouldn't worry about this too much for now, but I just wanted to point out that the stationary states are special states which almost by definition have $\langle\hat{p}\rangle = 0$, and that it is totally possible to construct states which do not have this property.

Is this implying that a particle in the box for 1-dimension has no momentum?

Having a zero expectation value for momentum is not the same as having zero momentum. When you say zero momentum, what you are suggesting is that whenever the momentum of the particle is measured, it will always return the value zero.
However, that's not the case here, actually. The truth is that whenever you measure the momentum of the particle, it can return pretty much any value. However, if you prepare many, many copies of the same system and measure each of their momenta, the average of these will come out to zero.
One way of expressing this is using the standard deviation as explained in Hans Wurst's answer. A nonzero standard deviation (or "uncertainty", as it's often called in QM) means that you will in general get nonzero momentum measurements.
This probabilistic nature of quantum mechanics can be difficult to grasp, but I think the best solution is to (re)read a textbook: virtually all textbooks spend a good amount of time explaining this.
We could think of this somewhat classically (but don't try to stretch this analogy too much). If you roll a six-sided die, what is the expectation value? Well, you can roll a 1, 2, 3, 4, 5, or 6 with equal probability, so the expectation value is
$$E = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$$
However, that doesn't mean that every time you roll a die, you'll get 3.5 (that's obviously impossible). Likewise, with a particle in a box (in a stationary state $\psi_n$), each time you measure the momentum, it very well may not be zero; but if you do it lots of times, it will average to 0.

Footnotes
* Except for a global phase factor, $\exp(-\mathrm{i}E_nt/\hbar)$, which is meaningless when it comes to expectation values because it cancels out in the $\psi^*$ and $\psi$ terms whenever they are integrated.
† The time evolution of the stationary states is also governed by the TDSE, so it's not that it suddenly fails to apply. It's just that when you work it out, the stationary states have no time evolution.

You can calculate the variance of an observable to quantify the spread of measurements around the expectation value. Only if the variance is zero, every measurement will return the same value equal to the expectation value.
We can calculate the variance via
$$\text{Var}(\hat p)_{|\psi}= \langle \hat p^2\rangle _{|\psi} -\langle \hat p\rangle^2_{|\psi}$$
In your case we get for the properly normalized $\psi(x)=\sqrt{\frac{2}{a}}\sin(\frac{\pi}{a}x), 0\leq x \leq a$,
$$
\langle \hat p\rangle = \int ^a_0 \psi^*(x)\frac{\hbar }{i }\frac{d}{dx}\psi(x)dx=0\\
\langle \hat p^2\rangle = \int ^a_0 \psi^*(x)\left(-\hbar ^2\frac{d^2}{dx^2}\right)\psi(x)dx=\frac{2\pi^2\hbar^2}{a^2}
$$
The variance for this state is then
$$\begin{aligned}
\text{Var}(\hat p)_{|\psi} &= \langle \hat p^2\rangle _{|\psi} -\langle \hat p\rangle^2_{|\psi}\\
\text{Var}(\hat p)_{|\psi} &=\frac{2\pi^2\hbar^2}{a^2}-0 = \frac{2\pi^2\hbar^2}{a^2}\\
\sigma(\hat p)_{|\psi} &= \sqrt{\text{Var}} = \frac{\pi\hbar }{a}\sqrt 2
\end{aligned}$$
where I've also calcualted the standard deviation $\sigma$.
As you can see the variance isn't zero. This means that a single momentum measurement doesn't have to be zero, even if the expectation value is zero.


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Viscosity and repulsive forces
Suppose we let a ball move in a fluid with initial velocity $u_i$. Then as time goes on its velocity goes down and tends to $0$ after infinite time. I understand viscosity in the following way:
In order for the ball to move it must "create space", i.e. pull aside the molecules of the fluid. But to move them aside, it must break their intermolecular (attractive) interactions. The energy for breaking these interactions comes from the kinetic energy of the ball and is converted into potential energy that turns into heat.
This explanation though doesn't address what happens when the repulsive intermolecular forces are dominant. How does viscosity depend on these forces?

Note also that viscosity is not a single quantity -- it can vary with rate, as in thixotropic, such as mayonnaise and clay/water mixtures, or dilatant, such as oobleck and Slip Lock limited-slip lubricant. So rate should be a consideration in any theory.

In a liquid, the overall forces are attractive. If these forces diminish, or become repulsive, the liquid vaporizes and the viscosity drops (but gases have viscosity too!).

Intermolecular forces are attractive for 'big' distances between the molecules and repulsive for 'small' distances between the molecules.

In order for the ball to move it must "create space", i.e. pull aside the molecules of the fluid. But to move them aside, it must break their intermolecular (attractive) interactions.

The water molecules have a preferred distance between each other.At this distance the potential energy of the system is at its lowest.When a ball comes , the distance between 2 water molecules becomes shorter and the repulsive forces start to dominate.


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Permitted composition of lithium oxometallate in lithium-ion battery and its influence on ion transfer
The following scheme describes the overall reaction that takes place in a Li-ion battery:
$$
\begin{align}
\ce{Li_{$y$}C6 &-> $y$ Li+ + $y$ e- + C6} &\quad &\text{(anode)} \tag{R1}\\
\ce{Li_{$1-y$}MO2 + $y$ Li+ + $y$ e- &-> LiMO2} &\quad &\text{(cathode)} \tag{R2}\\
\hline
\ce{Li_{$y$}C6 + Li_{$1-y$}MO2 &-> LiMO2 + C6} \tag{R3}
\end{align}
$$
However, I'm confused by the $1-y$ subscript in the lithium oxometallate. While the formula makes sense for $y=1$ $(1 - 1 = 0),$ if $y=2,$ wouldn't the subscript for $\ce{Li}$ be negative?
What would this physically correspond to in terms of lithium ion transfer?

Note that "y" in $\ce{Li_{y}CoO2}$ must not reach anywhere close to 1 nor 0, as for 1 it tends to irreversibly separate to Li2O and Co2O3 oxides, for 0 CoO2 tends to release oxygen.

@Poutnik I was under the impression that the y referred to the number of Li+ ions that were transferred in the reaction. What does it actually represent?

y does not have fixed meaning. In my case it means non stoichiometric lithium coefficient in Li_yCoO2. I could have used x instead. I apologize for confusion.

Apparently the source you've borrowed the scheme from didn't mention $y\in[0;1]$ (well, they sorta did by using $1-y$ in place of plain $y$). The second question about the influence of $y$ on ion transfer doesn't appear that trivial though, +1 for that.

It is common that some mathematical expressions have physical sense just in some value intervals. The typical cases are solutions of a quadratic equation with one of the roots negative. In your case it is $0 \lt y \lt 1$, so does  $0 \lt x \lt 1$ for $\ce{Li_{x}MO2}$, as these materials within the cell operation context are kept avoiding stoichiometric composition. In context of this, your second question is unclear.


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Fluorophores with high quantum yield and low lifetime
Does anyone know of any fluorophore or family of fluorophores with a relatively high fluorescence quantum yield (larger than 0.1) and a short lifetime (below hundreds of picoseconds)?
I have been checking and rhodamines and coumarins, for example, have very large quantum yields, as Alexa Fluor dyes, but the lifetimes are of the order of the nanoseconds.

Good question. You could shorten the lifetime by adding quenchers, e.g., nitromethane, oxygen gas or various “heavy atoms”, but that reduces the quantum efficiencies as well. I hope you get a good answer!


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Finding Ka of an Acid from incomplete titration data
I'm studying for chem olympiad and have a question about a problem from a past year's test:
A pure sample of a monoprotic acid is dissolved in water.
The sample is titrated with sodium hydroxide solution.
At the point where $20.0$ mL of the NaOH solution has
been added, the pH is $4.15$. The phenolphthalein
endpoint of the titration is observed when $50.0$ mL of
NaOH have been added. What is the p$K_{\mathrm{a}}$ of the acid?
Here are my thoughts:
Since pH = $4.15$, pOH = $9.85$ $\implies$ $\pu{[OH^-] = 10^{-9.85}M}$, which I think must also be the molarity of NaOH (this might be wrong).
I don't understand otherwise how to solve this question. I have $K_a = \frac{ \mathrm{[H^+]^2}}{\mathrm{[HA]}}$ as concentration of protons equals concentration of conjugate base. I guess I have $\pu{[H^+] = 10^{-4.15}}$ from the pH but I'm not sure if that's right since we added $20$ mL of NaOH first, which would have changed the pH from what it originally was.
How would I solve the question?

What other formulas related to acids and bases do you know? Can you make a graph of a titration of a weak acid with a strong base when all concentrations and the pKa is known?

The answer is the same no matter what the concentration of the NaOH solution is. Does it help if I say the concentration is 0.1 mol/L? Once you have an answer, you can also try with 0.2 mol/L and check you get the same answer. Then, you can think about why the concentration of NaOH does not matter in terms of what the solution to the question is.

Did you discuss this with your faculty advisor or team coach?

The main formulas I know are that pH + pOH = 14, pKa + pKb = 14. I also understand how to calculate each of these quantities... At my school, there isn't really an advisor or a coach. They select the students w/ the top 5 grades and we are supposed to self study the material, so I don't have someone to go to. I'll try using 0.1 mol/L and will update if I make progress/get stuck.

@KarstenTheis   I have a question about titration: if I have HA and NaOH, does a neutralization occur? Or does the OH from the base become water and then react with the acid in acid-base reaction?

Is it a possibility that I haven't yet learned the correct material to solve this problem? I'm using Zumdahl textbook and just read through the part about pH and K_a and K_b (haven't yet gotten to the chapter on acid-base equilibria or buffering). I thought this problem would be doable as it seems to just be pH and K_a, but maybe it requires other concepts that I don't yet know??

pH=pKa + log([A-]/[HA]) // 4.15 = pKa + log (20/(50-20))

"haven't yet gotten to the chapter on acid-base equilibria or buffering". It would be hard, then, unless you derive the relationships yourself.

Paid answers are all over the internet.

Here is some context.

Thank you so much! I have a solution that I think is relatively similar to Poutnik's. I put it as an answer: how does it look?

@KarstenTheis As a quick aside, I have a question regarding my solution: would the method I used work for any point in the titration? Because I know how many mL it takes to complete the titration, I can figure out the concentration fraction at any point in the middle of the titration... Also, from this, couldn't I use the equation to figure out the pH at other times in the titration since I now know $K_a$?

The formula for $K_\mathrm{a}$ is : $$K_\mathrm{a} = [\ce{H+}] \frac{ [\ce{A^-}]}{[\ce{HA}]}.$$
We'll use the data from after 20mL of titration: the pH is 4.15, so $[\ce{H^+}]$ is $10^{-4.15}.$
Now, we need the fraction of concentrations. In the titration at 20mL, $\ce{NaOH}$ is clearly limiting, and at the titration at 50mL, the entire acid gets used up. This means that at 20mL, 2/5 of the acid is used (and thus, 3/5 is left).
the formula for titration is: $$\ce{HA + NaOH -> H_2O + Na^+ + A^-}$$ since $\ce{NaA}$ is always soluble.
At 20mL titration, the concentration of $\ce{A^-}$ is 2/5 the original concentration of acid and the concentration of acid is 3/5 the original. So, the fraction is 2/3, which means we get: $$K_\mathrm{a} = 10^{-4.15} \cdot \frac{2}{3}.$$
Thus, $$\mathrm{p}K_\mathrm{a} = -\log_{10} \left(10^{-4.15} \cdot \frac{2}{3} \right) = 4.33.$$

This method would work for all points in the titration where weak acid and conjugate base are major species, i.e. from 2 mL in to 48 mL or so.

You solved it without having to make an assumption about the concentration of NaOH. That shows a nice ability to reason in the abstract.

Great! Thank you so much for all of your help!


text:
Where in the apartment to store isopropyl alcohol
I want to have some 99% isopropyl alcohol for cleaning purposes and I'm wondering where to store it since in my apartment I don't have a storage room or much space.
I've read it needs to be air tight and away from light and heat.
Would it be ok to store it in the cupboard under the sink close to other regular cleaning products? There won't be much ventilation there.

You've probably bought that in a store. How do they keep it? I assume you're not storing hundreds of liters.

It's a small amout (< 1L), I just want to make sure it's safe. In the store they just have it on the shelves.

Unless you have a gallon or more, under the sink is good . I believe I have some under the kitchen sink and I know I have some under the bathroom sinks. My 16 oz isopropyl plastic bottles are 90 %.


text:
Rate equation for the decomposition of hydrogen peroxide with iron(III) chloride as catalyst
I am trying to determine the activation energy of the decomposition of hydrogen peroxide with iron(III) chloride added as a catalyst. To do that, I need to find the rate constant, but I am not sure what to do exactly.
What would the rate equation be in this case for me to be able to calculate this?

You should take steps to measure the volume ${V(t)}$ of the oxygen gas $\ce{O2}$ produced by the decomposition reaction vs. time. Do this up to the time when you consider that this volume will not increase any more. This final volume ${V_∞}$ is proportional to the initial $\ce{H2O2}$ concentration, called $c_o$. Now calculate all differences $\Delta V(t) = V_∞ - V(t)$ for all volume measurements. This $\Delta V(t)$ is proportional to the $\ce{H2O2}$ concentration at time $t$.
Then report the successive values of log$\Delta V(t)$ on the $Oy$ axis, and the time $t$ on $Ox$ axis. The points should be alined, and the slope gives you the first order rate constant $k_1$. Repeat the same measurements at different temperatures. If the points are not well alined, report the inverse $1/V(t)$ vs. the time $t$. If this time the points are alined, the reaction is second order with  a slope equal to the rate constant $k_2$
When these series of measurements are finished, report the log of the different $k_1$ or $k_2$ values vs. ${1/T}$, where $T$ is the absolute temperature in Kelvin. You should obtain a line with a slope equal to $E_a/R$, where $E_a$ is the activation energy of the reaction.


text:
Isopropyl Alcohol Resistant Paint
Which paint binders are most resilient to IPA?
Background:
I'm looking for an IPA-resistant spray paint to coat a 3D printed model composed of a high-performance stereolithography resin MSDS.
"Binder plays the role of an adhesive enabling the paint to stick to the surface and holds other components to itself. The binder dissolves in the solvents used and the thinners reduce the viscosity of the resultant mixture. Binders used in the paints are Alkyd Resins, Acrylic Resins, Poly Vinyl Acetate, Chloro Rubber, Epoxy Resins, Poly Urethane Resins, and so on." Quora
Paints I'm considering:

Modified Alkyd: Rust-Oleum High Performance Enamel Spray TDS
Modified Alkyd: Montana MTN94 TDS
Acrylic Modified Alkyd: Rust-Oleum Painter's Touch TDS
Epoxy Ester Modified Alkyd: Rust-Oleum Appliance Epoxy TDS
Polycarbonate: Tamiya Polycarbonate MSDS

Thank you so much for your time and help!

The Hansen Solubility Parameters is a useful tool in the context of solubility of paints, glues and polymers in general.
Basically every polymer is represented by a sphere in $\mathbb{R}^3$ where (most) solvents lie inside the sphere and (most) non-solvents lie outside.
The polymer you need should have its center far away from IPA coordinates, specifically farther than its radius.
The HSP parameters of 2-Propanol are $(15.8,6.1,16.4)$; to get an idea, PVC and PVAc (mentioned in the polycarbonate paint) spheres are respectively $(18,8.8,2.2,r=8)$ and $(17.6,2.2,4,r=8)$. IPA should be a non-solvent for both.
It's hard to say what would happen with proprietary blends, even if broadly categorized as (modified) alkyds: different alkyds can have wildly different spheres, the modification just adds even more complexity. That said, seeing Toluene and Acetone in the list of solvents for the same polymer makes me worry that its solubility sphere is big, and if the sphere is too big it will include IPA.


text:
Can organocuprates do SN2 reactions with tosylates?
In the presence of isolated ketones, can organocuprates add to secondary tosylates in an SN2 fashion without adding to the ketones?

Cuprates (also known as Gilman reagents) do not generally react with isolated ketones. If they did the widely used reaction of cuprates with alpha/beta unsaturated ketone would not work to give the substituted ketone product.
According to this review alkyl cuprates do react with secondary tosylates to give the expected SN2 product. I have not, however, been able to find an example of this being done in the presence of an isolated ketone


text:
How are the energies of, e.g. the 4p orbitals of the first transition series, determined?
Take a first row transition metal, say manganese. The 4s and 3d orbitals are occupied in Mn; each orbital's energy can can be definitely determined by ionising that orbital, obtaining the resulting Mn+ ion's energy, and comparing with the energy of the neutral atom. This can easily be done either empirically or in silico; there's no problem up to this point.
However, a problem occurs when one tries to obtain the energy of the 4p orbital. This orbital is not occupied in the ground state configuration of Mn, so ionisation is not possible here.
There seem to be two ways to solve this problem:

One by exciting, say, a 3d electron into a 4p shell and comparing the resulting (excited-state) Mn's energy with that of a 3d-ionised manganese cation(i.e. the "band-gap" method).
The other by pushing an electron into a 4p orbital of an Mn atom, creating an anion, and comparing its energy with that of a neutral Mn atom(i.e the "electron affinity" method.

But the two methods are not consistent with each other since the LUMO, and hence the band gap, of a chemical species is in general NOT correlated to its electron affinity.
There seem to be some data on the 4p orbital energies of first row transition metals, e.g. over here; which of the two (non-consistent) methods I mentioned was/is used in the literature to determine the 4p orbital energies of first row transition metals?
UPDATE: After further personal research on this issue, I found out on Wikipedia that the Kohn-Sham DFT theory exactly captures both the excitation energy("first way"), resp. the ionisation energy("second way"), by simply taking them to be the difference between the filled and unfilled orbitals' energy levels("first way"), resp. the filled orbital energy levels("second way").
So I'm going to re-ask it: are the "energy levels" of unoccupied valence orbitals of certain atoms, say transition metal's np orbitals, known in the literature derived from Kohn-Sham DFT theory, using the known (or calculatable) exact electron densities for these atoms?
UPDATE 2: The link basically says that, (even though the general functional is still unknown) the functional for a specific species can be derived exactly once the CI(configuration interaction) density for that species is known. And CI calculations are relatively straightforward for bare atoms.

Those 2 methods consistently provide different results,  as 4p energy of $\ce{Mn^{*}}$ is different than for $\ce{Mn-}$ , because of different screening effect and the Z_eff acting upon the 4p electron.

You may want to provide just the target link https://www.colby.edu/chemistry/PChem/notes/AOIE.pdf

Also note that in real atoms and molecules the ionization energy  is not equal to the orbital energy (whatever that may mean). Koopmans' theorem in such cases i only an approximation - https://en.wikipedia.org/wiki/Koopmans%27_theorem

It is exact for the DFT orbitals. As seen in your Wikipedia link.

But all implementations of DFT are approximate as we do not know the exact functional. Thus while the orbital energy will match the calculated ionization energy, it will not match the measured ionization energy, which is how I read your question. That is all.

Not merely "approximate" once one knows the CI electron density for that species(check link in post)- the calculation of which for bare atoms is relatively easy.

Kohn-Sham DFT cannot do anything exact. It uses orbitals for approximation, so it actually also doesn't even do real DFT. And by no means can you calculate the energy of an empty orbital.

(Assuming the functional is exact, vide supra) the energy of the LUMO can be indirectly calculated as the difference between (the energy of the HOMO) and (the band gap), both of which are exact for an exact functional


text:
Is the maximum work done by voltaic cell per mole of reaction?
From the equation $w_{elec,rev}=-nF\Delta E$, and n = the number of moles of electrons transferred in the overall redox reaction, does that make the work per mole of the redox reaction? I am confused because one mole of reaction does not necessarily happen, so how does w and ∆G adjust to account for fewer/more moles of electrons being transferred. Is this perhaps just a convention?

$W_\mathrm{m}=q_\mathrm{m} \Delta E=nF\Delta E$ and   $\Delta G_\mathrm{m} = -W_\mathrm{m} =-nF\Delta E$

Note that n is not the molar amount of electrons (We do not say number of kilograms either.), but literally the number of electrons exchanged with 1 particle (molecule/ion), so nF is the charge exchanged with 1 mol of electroactive particles. By dimension analysis, n is unitless and w has unit J/mol. // If we insist on n as molar amount of electrons than w has unit J, not J/mol.


text:
Delta U is a state function?
Can we say $\Delta U$ is a state function? According to me,
$\Delta U$ is not a state function.
$dU$ is an exact differential and $U$ is a state function.
Statement 1:
Basic meaning of state function (property) is something which is defined at state (equilibrium condition) like  $P, V , U, S$  i.e. if we change the state then the value of that property for the final state does not depend on how we reached that final state. i.e state function defined for the state
$\Delta U = Q +W$, its value is same between 2 states (initial and final) when we select any of the path joining the 2 states. The quantity $\Delta U$ is independent of path followed.
but it is not something that is associated with state. This is why $\Delta U$ is not a state function.
Statement:2
We are getting same value of $\Delta U$ from 2 different paths between same initial and final state. By saying this we can comment on $U$ that it is a state function.
I agree with this no issues.
Statement: 3
If we are still saying $\Delta U$ is a state function.
1. Then we should say this by saying $\Delta\Delta U$ is independent of path between 2 states. And $\Delta U$ must be defined for equilibrium state. Like $P$ is a state function. Then $\Delta P$ is independent of path between 2 states. And $P$ has its meaning for equilibrium state
But actually we don't say anything like this for $\Delta U$.
2. Does it mean $\Delta P$ , $\Delta V$ , $\Delta T$ all are state functions? Because we will get same value on every path between two states.
If yes then that bothers me that they do not have any meaning for a state (equilibrium condition).
If no, then what is different with $\Delta U$.

Is $\Delta U$ a function of the state and nothing but the state ?

$\Delta U$ depends on two states, not one.  So it is not a state function.

@ChetMiller It may be more grey zone that it seems. E.g. for a perfect gas we say, U = n.Cv.T, but in fact it is U(T)=U(T=0) + Delta U(T) = m(T=0)c^2 +  n.Cv.T

@Poutnik Quite - all energies are relative to an arbitrary zero. In your case the gas could also be in a box with translational energy. Thus one might claim that whenever we write U we really mean Delta U.

@IanBush That is what I wanted to say.  Delta U can be used to describe the state function in a relative way, but generally it is not a state function itself.

Definitionally, I think it's cleaner & more precise not to call $\Delta \pu{X}$ a state function (SF) since, as has already been mentioned, $\Delta \pu{X}$ is a function of the difference between states. Also: A comment on the converse of what you wrote in statement 3, #1: If X is a SF, $\Delta \Delta \pu{X}$ is independent of path. A familiar example would be the change in $\Delta \pu{H}^{\circ}$ for a chemical reaction when we change T at constant p. But that doesn't mean that $\Delta \pu{X}$ is itself a SF. Rather, $\Delta \Delta \pu{X}$ is path-independent simply because X is a SF.

Also, I wouldn't refer to p, V, and T as SF's.  That's typically reserved for thermodynamic potentials like U, G, A, H, S, and $\Omega$.  p, V, and T are, by contrast, considered state variables or state parameters.  I think to consider a state variable as a SF you would need to be treating it as a thermodynamic potential, i.e., you would need to be working in the "p representation", etc.  And a system couldn't be described using a p or T representation, because these are intensive, so you'd have a potential that is independent of system size.  Not sure if a "V" representation is possible.

As ChetMiller comments, $\Delta U$ is a function of two states, therefore not a state function.

$U$, like other state functions, is defined mathematically as a non-injective non-surjective function where the argument is the state of a system. In layman's terms non-injective non-surjective means that there can be many states that share a value of a thermodynamic property such as the energy, and values of the property that do not correspond to any known state, but each state has only one value of the property.
In classical thermodynamics a state function (such as $U$) is defined as a property of a static (or in practice quasi-static) system, not of a transition or difference between such systems (such as $\Delta U$). $\Delta U$ is also a non-injective non-surjective function, but it is a function of two states. Although you have found a feature that satisfies a mathematical property of a state function, the feature is not sufficient to render $\Delta U$ a state function. You might say the duck-type test includes that you limit yourself to features of a single system rather than of multiple systems.
Note that it is possible to define a reference state and the properties of all other states relative to the reference (as alluded to in the comments), but that does not change the basic argument.
Also, you can create systems in which a transition between two energy levels is an important property of the system (for instance in NMR or optical resonance). However those are time-dependent sytems. Classical thermodynamics focuses on static (time-independent) systems.

The reason why we use  ΔU instead of U is due to the fact we cannot measure the internal energy U of a gas but only the change of U.
For any process ΔU in a  closed system  ΔU can be calculated by the change of temperature of the gas(ΔT) thats why ΔU is a state function.

I'm reviewing this answer as "looks okay". That doesn't mean I think it is correct or good. In fact, I have chosen to down-vote it. But from a technical perspective, it is an attempt to answer the question.

When you say you can calculate $\Delta U$ for a change in T of a gas, what you mean is that you can calculate that for an ideal gas.  You can't typically calculate $\Delta U$ for real gases, since you would need an equation of state.  That's only available for some gases, and they're always, to some extent, an approximation.


text:
Crystal Field Splitting Energy - Tetrachlorocuprate and Tetrachlorocobaltate
I'm a high schooler from India and it's my first time on this stack exchange. I was wondering why the crystal field splitting energy of tetrachlorocuprate differs so much from tetrachlorocobaltate.
Peaks at  ~410 nm
253.457 kJ/mol for the copper(II) complex

Peaks at ~690nm at 175.5 kJ/mol for the cobalt (II) complex.
The general trend amongst octahedral cobalt and copper complexes is that cobalt experiences a greater crystal field splitting energy. But when it comes to tetrahedrals, how do I explain this? I have considered the Jahn-Teller distortion but I suppose there must be more to it. Any help would be appreciated!

I think it can be attributed to the fact that copper (II) forms a square planar geometry usually since it would be d8.

What am I seeing here? What are these images? How do you come from a wavelength to a splitting energy?

E = hv --> E = hc/λ where λ is the wavelength of maximum absorbance. The images I've attached are the absorption spectra of the relevant complexes. The wavelength at which the absorption is the maximum is λmax.


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Hydrogen Dispersion in room
I need to measure the concentrarion of pure hydrogen in a room caused by leakage or exhaust gases to evaluate the risk of ignition. Maybe there is some model available for indoor locations, similar to the Gaussian plume model for gas dispersion.
The density of hydrogen is very low compared to air, buoyancy is high and diffusion is high. So the diffusion formula for other gases may not work for hydrogen.
Does somebody know a formula, method or reference to calculate the concentration of hydrogen in a room?

One or both of Fick's laws should do the trick.

Diffusion is simple. Otherwise you won't get away without full 3D flow modelling of your room.

Diffusion is too simple for this case?
I'm not familiar with a full 3D flow model. Is it correct that that model should include the shape of the ceiling, the HVAC pipes at the ceiling, the air flows, open windows and objects around the room?

What such model is, is impossible. No way to tell what happens if someone leaves a door open in the end of corridor, or day is very windy.

First you should decide if buoyancy is relevant for your problem. Then you install a ventilation system, and close the case.

Hydrogen rises to the high points; not "pure", mixed with air so there is a high risk of ignition. So, those who work with hydrogen risk plan for it, they do not calculate possible concentrations. Explosion proof lights and other electrical devices are used. Also exhaust ventilation. Some organization like Underwriters Labs will have recommendations. All I know is when I designed a lab to work with hydrogen , the company safety department increased the estimated costs with various precautions.

Quite interesting that you designed a lab to work with hydrogen.
With explosion proof lights do you mean hydrogen concentration meters around the lab?
Was the exhaust ventilation local over the hydrogen devices and/or global on the room?

The testing was for high temperature hydrogen attack of steel. There was both hoods over test equipment and room ventilation. The safety group  did all the design so I do not know exactly what they did. I do not recall any hydrogen test meters in the room.

Look up Class 1 Div 1 equipment. Specifically for potentially explosive atmosphere.


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Feynman books analogous in chemistry
Is there a analogous to feynman lectures in physics but for chemistry? Like a book ir series of books that covers all the basics?

I like Linus Pauling's General Chemistry. It is lucidly written. Unfortunately no Feynman equivalenr! Chemistry and chemical reactions do readily succumb to relatively simple mathematical models which are used in undergraduate physics. And most of the general chemistry taught these days is half early 20th century physics (atomic structure sections, gas laws, etc.).

@M.Farooq Actually everything you learn in undergrad chemistry is from the first half of the 20th century. :D

@Karl, The same goes for undergraduate physics and mathematics (BC to early 20 th century)

@M.Farooq In biology, it's a bit the other way round. Before Watson, Crick, and the electron microscope, they were stabbing in the dark, or satisfied with collecting stamps. ;) Figuratively speaking.

All the math I know is probably at least 150 years old.

Cannot agree more! Math is the oldest.


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Using NaCl to change the relative humidity in the humidity chamber
The saturated NaCl will give around 75% RH in various temperature. Is there a way to use different concentration of NaCl in the water to generate different values of RH, i.e., 10% to 90% in the humidity chamber?

You can't go lower than 75%, but everything between 75% and 100% (pure water) is possible.

You need to use other salts. You can search for table of constant humidities of saturated solutions on internet and I remember I have posted some here in past too. // You can theoretically use NaCl for whole 75-100% range, but it would not be buffered nor the values are easily available.(can be non linear).

See https://chemistry.stackexchange.com/a/112915/35806


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Ambiguity on determining a relativistic Dirac orbital's "shape" when it has two angular components
Wikipedia says that a relativistic Dirac atomic orbital is a 4-component vector, with two angular parts and two spin parts. It also says, indirectly, that 'the top two components of a 2p(1/2) orbital are not spherical, while the bottom two components are', and vice versa for the 2s(1/2) one. So the 2p(1/2) orbital is both spherical and non-spherical at the same time.
However, it is known that a p(1/2) orbital is completely spherical, a contradiction. This can only happen if only one of the components of the 4-component vector is taken to represent its "shape"; that component is the top two in s(1/2) and the bottom two in p(1/2), and is hence arbitrary.
My question now follows- is there a definite algorithm to "pull out" two of the four components of a Dirac relativistic orbital, to determine the orbital's "shape"? It all seems arbitrary to me...
EDIT: Forgot that I was myself a mathematics undergrad student... maybe it's the dot product of that 4-component vector with itself that's spherical. Is it the case?