text:
Formic Acid vs Formica - Which came first?

Source: Google(Oxford)

So, we know that 'Formic Acid' was derived in formic acid in the late 18th century from the Latin word 'formica'. From Wikipedia:

Formica is a genus of ants of the family Formicidae, commonly known as wood ants, mound ants, thatching ants, and field ants. Formica is the type genus of the Formicidae, and of the subfamily Formicinae.

I think I've read in some junior classes that the biological name 'Formica' was given to ant because of the secretion of 'Formic Acid' by it which is the cause of the irritation at the stung area.
So, I want to know - Is formic acid derived from 'Formica' (ants) or are ants' (biological name) derived from the presence of 'formic acid' in them?

The acid was named after the ants it was isolated from. The Romans did not understand chemistry in the way we do.

The name 'formic acid' has been derived from the name 'formica' as it is derived literally from ants. Wikipedia clearly states:

The word "formic" comes from the Latin word for ant, formica,
  referring to its early isolation by the distillation of ant bodies.

If you type 'formic acid etymology' in Google, it will give two links that proves the above statement:

1791 (in formic acid), literally "from ants," coined from Latin
  formica "ant". The acid first was obtained in a fairly pure form in 1749 by German chemist Andreas Sigismund Marggraf (1709-1782), who
  prepared it by distilling red ants.(etymoline)

As early as the fifteenth century, some alchemists and naturalists
  were aware that ant hills gave off an acidic vapor. In 1671, English
  naturalist John Ray became the first person to describe the isolation
  of formic acid by the distillation of large numbers of ants. These
  insects secrete the substance for attack and defense purposes. Thus
  the name "formic acid" was coined from the Latin word for ant,
  formica.(newworldencyclopedia)

I'm now envisioning a new foodie trend: distilled red ant liquor...

Formic acid is the main culprit behind causing blindness from methanol exposure. Nature is not that forgiving.

And how did the name 'formica' for ant came by ?

@Sristy. Not clear. Some relate to a sanskrit word, meaning to give away (already knew they release a substance). Others to for (carrier) and mica (little speck).

@Sristy Formica is the Latin name for ant since the chemical was produced by distilling large number of ants, it was named formic acid. It is not the other way round.


text:
Unexpected result (sodium hypochlorite & steel wool)
So I got myself a large Erlenmeyer flask, some steel wool and the bleach from the bathroom, mixing them under safe conditions in small quantities to see the sparky reaction I'd heard about. Trouble is, I didn't know the bleach we had was non-chlorine!
EDIT: My silly mind assumed that no reaction meant I had non-chlorine bleach, but staring me in the face is hypoCHLORITE. My apologies, I'd not meant to mislead. The brand was Shaw's at the time, or "Signature Selections" which is their store brand. I got no sparks and no visible fumes nor immediate discoloration.
Disappointed, I left the mixture alone and uhm.. forgot about it.
Going back to check on it, I found the steel wool dissolved into a powder at the bottom, and the rest of the fluid was purple.
Curious, I used a giant coffee filter and strained the rusty cake-mix residue out of the rest of it, which was now a lighter purple.
I didn't know what I'd made and thought it best to simply leave it be for a while again. I kept it in the basement, visible when the door was open and kept an eye on it. Slowly the purple went away and the fluid decreased leaving a white powder and bits of clear-quartz like appearing crystals. After several months the crystals were larger, and the white powder has climbed the entire container and spread down the sides. It's now been about 18 months since the initial experiment, and the white powder (which is similar to efflorescence in appearance) has climbed about halfway down the outside of the container, and the inside crystals have grown slightly.
What in the world have I made?

Well, purple was most probably ferrate, but you got colorless stuff later? That's weird...

Interesting observation and reminds me of my own childhood experiments. I think that steel wool has a small amount of other transition metals which are causing this purplish color. One could have formed ferrate or even permanganate. This can happen with chlorine bleaches only. On the other hand, ferrates cannot form with oxygen bleaches. Ferrates can form when you react iron salts with hypochlorites (while gently heating). I rather suspect chromium alum type compounds (purple) or perhaps even a trace of vanadium. It is certainly not iron. What is the brand of your oxygen bleach?

[It changed my number for some reason when I created the account/this post, but I am the OP) Shaw's store brand (Signature Selection) Also I should clarify I never got sparks. I checked the container today, the crystals are still there and the white powder is still up the side, though it has peeled away a bit (about a half inch) and is cracked down to the bottom. There is still clear fluid under the crystal at the bottom, and the fluid comes through when tilted, so it is not solid. The crystal looks somewhat like clear ice, with bits of snow on it.

I sent a request to merge your accounts. In the meantime, this looks to me more like an edit/update to the question, not an answer.

It might help to tell us what the bleach actually was (read the ingredients on the label).

The title mentioned "hypochlorite" and the text said that there is no chlorine in the bleach. This is contradictory. Chlorine or no Chlorine ?

@OscarLanzi, I feel the oxygen bleaches are too mild as an oxidizing agent and as far as I remember ferrate and peroxide cannot co-exist just like permanganate ion.

I just confirmed from a very long review on ferrates from 1897 by Moser (Journal für Praktische Chemie): Die freiwillige Zersetzung der Kaliumferratlosung wird ausser durch Kalium- und Natriumhydroxyd auch durch die Gegenwart anderer Kalium- und Natriumverbindungen , wie Kaliumchlorid, - bromid, -nitrat, -carbonat, Natriumcarbonat, Kalium- und Natriumhypochlorit verzogert.

Dagegen beschleunigen Calcium-, Strontium- und Magnesiumsalze die Zersetzung, ebenso Metalle, Metalloxyde, Superoxyde und organische Stoffe. Eine sofortige Zersetzung des Ferrats wird bewirkt bei Zusatz von Sauren, auch Kohlensaure und sauer reagirenden Salzen, ferner durch Ammoniak, Ammoniumsalze und Wasserstoffsuperoxyd.

[Translation of the last para DeepL] "In contrast, calcium, strontium and magnesium salts accelerate decomposition, as do metals, metal oxides, super oxides and organic substances. An immediate decomposition of ferrate is caused by the addition of acids, including carbonic acid and acidic salts, as well as ammonia, ammonium salts and hydrogen peroxide." So it is very clear that oxygen bleaches will never form a ferrate as I was suspecting. The original poster did not bother to clarify the name of the oxygen bleach. The question title and the text are self-contradictory.

We can use proxy compounds in the solid state, as here.

Right, even that 1897 paper mentions heating iron with potassium superoxide to obtain ferrate in the solid state. The key point is that $aqueous$ ferrate and hydrogen peroxide are not compatible. I do vaguely remember that - I must have done this experiment two decades ago. That is why I feel OP's description is not correct. You can see that review mentions immediate decomposition of ferrate with H2O2.

After the edits by the original post, the OP clarified that they used a hypochlorite (not an oxygen bleach as stated earlier) and left a piece of steel wool in it. Due to corrosive nature of chloride/hypochlorite, the steel wool crumbled to iron (III) oxides. When you gently heat rust in highly alkaline medium in the presence of hypochlorite ion, it forms a purplish sodium ferrate. This has iron in the (+6) oxidation state.  Over time the purple color disappeared because ferrate slowly decomposes back to lower iron oxidation states in water solution, releasing its excess oxygen.  Ferrates are more stable in alkaline solution than in neutral or acidic ones, so absorption of carbon dioxide over time, making the solution less alkaline, could have promoted this decomposition.
This decomposition of ferrates by acids, carbonic acids, and acidic salts was known as early as 1890s as noted in a review by Moser in Journal für Praktische Chemie 1897, 56, 425 (see relevant comments). Ferrates were known in 1700s.
Also note that hydrogen peroxide and ferrate decompose each other immediately, just like permanganate ion in aqueous solution.
The quartz like crystals are nothing but sodium chloride with perhaps some sodium carbonate.
Image from: https://www.youtube.com/watch?v=5u8LOfTrO6w

Added a brief discussion of tge ferrate disappearing.  Please attribute the photo if not your own.

My mistake, I added the link in image description which was not visible. The Youtube video is in German & English from where the image was taken.

What is it about the alkaline conditions that stabilises ferrate?

I think there is no simple answer as to why alkalis stabilize it. The half life of ferrate is less than hour in neutral pH. It can oxidize water. The why part of science is often the most difficult part of a phenomenon.

@M.Farooq that’s interesting. I’m at a high pressure and temperature lab, and we noticed that alkalis and primarily sodium tend to stabilise higher oxidation states of transition metals in silicate melts and glasses. We have some explanation based on the structure of the silicate framework, and it’s really interesting that these things also happen at STP in aqueous chemistry.

@Gimelist, Interesting, but this this stabilization is not a general phenomenon. Think of KMnO4, the moment you add excess base, oxidation state of Mn goes down from +7 to +6.

source: Stack Exchange

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

sort: votes


text:
Order of coordinates in a wavefunction - why do they matter?
For the past few weeks, I have been studying Quantum Chemistry and lately in these lectures something has bugged me:
If I do a sum:

$1 \times 2 \times 3 \times {...} $

This is the same as simply writing the order in reverse:

$ {...} \times 2 \times 1 $

a bit like $ab = ba$
and for wave functions, I've seen that the order of coordinates inside a wave functions seem to imply that no two wave functions are the same:
$$\psi({r_1},{r_2}) \neq \psi({r_2},{r_1})$$
and from there on there is a discussion that there must be a change in sign and so on and so forth.
But I'm struggling to work out why if I change the order of my variables, that this makes any effect.
I'm just confused by why the order is such a fuss, if that makes sense!
Why are the two wave functions different by the order of $r_1$ and $r_2$?

You mean if you do the product. The order matters because maybe $\psi(r_1,r_2) = \phi_a(r_1) \times \phi_b(r_2)-\phi_a(r_2) \times \phi_b(r_1)$ (for instance) in which case swapping labels will give you a different function.

So are you saying that in reference to your equation, $\psi\left(r_{2}, r_{1}\right)=\phi_{a}\left(r_{2}\right) \times \phi_{b}\left(r_{1}\right)-\phi_{a}\left(r_{1}\right) \times \phi_{b}\left(r_{2}\right)$ as the order that I represent the coordinates in  $\psi\left(r_{2}, r_{1}\right)$ is important here?

Hopefully the example I presented (for which $\psi(r_1,r_2)=-\psi(r_2,r_1)$ shows you why the order of labels matters. For instance, electrons have this property, swapping the electrons inverts the sign of the wavefunction.

I removed my comment about commutativity since your focus is on wavefunctions, not operators, by the way.

@BuckThorn thanks! Is there a mathematical term that describes this property that I can look up on the internet too for reference?

search for "antisymmetric wavefunction" or even "fermion" (note: removed link to libretext - after reading that text noticed some glaring errors in it that could confuse a beginner)

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


text:
Computing force fields using Open Babel
I am trying to use Open Babel (OB) to compute universal force field (UFF) given an xyz file. I read each xyz file and turn it into a pybel molecule. Here is the code I have so far
def xyz_to_smiles(fname: str) -> str:
    mol = next(pybel.readfile("xyz", fname))
    smi = mol.write(format="smi")

    return smi.split()[0].strip(), mol

def get_qm9_OB(target_index=np.arange(0, 133884)):

    smiles = []
    molecule = []
    for i in target_index:
        mol_idx = f"{i+1:06d}"

        tmp_smiles, tmp_molecule = xyz_to_smiles('../dsgdb9nsd.xyz/dsgdb9nsd_' + mol_idx + '.xyz')

        smiles.append(tmp_smiles)
        molecule.append(tmp_molecule)

    return molecule, smiles

molecules, smiles = get_qm9_OB(target_index=np.arange(100,200))

for i in range(len(smiles)):
    mol = molecules[i].OBMol

    ff = ob.OBForceField.FindForceField('UFF')
    ff.Setup(mol)

    ff.GetAtomTypes(mol)
    ff.GetCoordinates(mol)
    for at in ob.OBMolAtomIter(mol):
        type = at.GetData("FFAtomType") # an OBPairData object
        print(type.GetValue())

Now, I am trying to compute bond stretching interactions. My questions are:

How can I get the UFF constants for bond stretching? (IOW, is there an OB equivalent for RDKit function GetUFFBondStretchParams?)
Is there a function to calculate bond stretching interactions for each bond? (IOW, is there an OB implementation of either harmonic oscillator or the Morse function?)
How can I get coordinates from OBAtom object? I found GetCoordinate() but the output is
<Swig Object of type 'double *' at 0x1a1eb4a9f0>

Alternatively, GetVector() outputs
<openbabel.vector3; proxy of <Swig Object of type 'OpenBabel::vector3 *' at 0x1a1fbddbd0> >

How can I get the actual values?

P.S. I posted the problem on OB mailing list, with no answers.

I honestly don't understand your question, and your code seems incomplete; I don't see how you are actually reading in an xyz file. I'm not sure how well this question is actually suited here, as it seems to be rather a programming issue. Maybe it is more on topic at another site, e.g. scientific computing it even stack overflow.

@Martin-マーチン Oh, I forgot to include xyz_to_smiles. Now you can see how it reads the file. Let me know what part you don't understand so that I can clarify.

While people have asked, at the moment, there's no API call to get parameters for various force fields. It's possible to hack the C++ code, of course. (It hasn't been clear to me what such an API call should look like - feel free to suggest something.)
There isn't an implementation because a harmonic oscillator and Morse potential are fairly easy to implement yourself. (Incidentally, despite mention of Morse in the UFF paper, the parameterization involves a harmonic oscillator, not a Morse potential for bonds.)
IIRC there's a SWIG convenience function to cast from vector3 to Python vectors, but you see that there are multiple ways to do this.

Coordinates can be accessed using
[OBAtom.GetX(),OBAtom.GetY(),OBAtom.GetZ()]

look at this Github discussion.

If you have a better answer please don't hesitate to answer the question.


text:
Theoretically predict the density of water
Is it possible to predict the density of water having just its molecular structure (basically obtain 1 g/cm3 at 3.96 °C)?
Are you aware of research/textbook that dive into this topic?
Is it able to predict also the variation of the density with the temperature?
An equivalent question will be: how is it possible to estimate the average distance between two different molecules?

In short: yes and no, due to quantum chemistry.

https://chemistry.stackexchange.com/questions/81756/why-is-density-of-water-maximum-at-4-degree-celsius/105390#105390 may help.

https://chemistry.stackexchange.com/questions/22024/density-of-water-and-temperature

It's hard to put this into an answer: one can create or choose a suitable method (such as a force field) and run molecular dynamic simulations of fluids at a given, finite temperature. (All aspects of that are hard, btw.) You can look into radial distribution functions for the average distance between particles. It can be determined experimentally and is used to validate computational results.


text:
Measurement of the lattice energy
The lattice energy of a solid ionic compound is the energy released when one mole of the solid compound is formed from its constituent gaseous ions at $\ce {298 K}$ and $\ce {1 bar}$. However, gaseous ions (e.g. $\ce {K^+ (g), Cl^- (g)}$) do not exist at such temperatures. Thus, how is it possible that the lattice energy of an ionic compound can be measured? A more reasonable definition would be to define the state of the compound formed as the standard state, rather than defining the state of the ions as the standard state. Wouldn't that make much more sense?

No, that wouldn't be as useful. Going from the gaseous ions to the solid is one step in the Born–Haber cycle.

@MaxW Yes I understand that it is precisely the fact that we can't measure the lattice energy that we invented the Born-Haber cycle. However, I am just puzzled at the condition for the standard state in the definition of the lattice energy.

What do you understand as standard state? What standard state are you referring to?

@BuckThorn 298 K, partial pressure of all gaseous species at 1 bar, concentration of all aqueous species at 1 M

I'd just say that the "standard lattice energy" refers to both reagents and products being in their standard states. I may misunderstand your question, but I'm guessing you are wondering about a reference state for computation of the chemical potential?


text:
Order of features in chemical descriptors
ECFP are small-molecule descriptors, which are usually represented as a bit vector. I wonder if order of elements of this vector matters, in other words, if this vector encodes any positional information. I couldn't find this information in the papers.
Thank you

This article suggests they are arranged in a particular way to prevent order dependence: https://pubs.acs.org/doi/10.1021/ci100050t "...(the neighbors
are ordered using their identifiers, and the order of the
attaching bonds, to avoid order-dependence)"

thanks, but I am still confused, is there positional information or not? will I get another compound if I randomly shuffle the bits in the vector?


text:
Can buta-1,3-diyne show conjugation in both the unhybridised p orbitals?
As 2nd carbon and 3rd carbon of butadiyne are sp hybridized so there are two unhybridised p orbital mutually perpendicular to each other in each of carbon atom  also these two orbitals are separated by a sigma bond.
Is There a possibility that it could show conjugation from the both the unhybridised p orbitals as opposed just conjunction takes place in either  one plane  ?
I'm not able to confirm this hypothesis as I don't have reliable resources.

You know for sure that butadiene has some conjugation. How is butadiyne different, really?

Its not mentioned under conditions for conjugation to occur in many references

What conditions? You have two $\pi$ bonds, one $\sigma$ bond apart, much like in butadiene.

Read this.

It is asking me to purchase the article

@Chemist What are the site policies surrounding paywall circumvention?

"For butadiene...the formal (single) C−C bond length is 1.454 Å.... It is shown for the first time that π-electron delocalization has the structural consequences of increasing the length of the formal double bond by 0.007 Å and decreasing the length of the formal single bond by 0.016 Å." https://pubs.acs.org/doi/pdf/10.1021/jp060695b
"An empirical equilibrium structure based on experimental rotational constants for 13 isotopic species of diacetylene and computed zero-point vibrational corrections is determined: ... r CC (triple bond) = 1.2085 Å, r C-C = 1.3727 Å and in good agreement with the best theoretical structure." https://www.sciencedirect.com/science/article/abs/pii/S0022285208000969
The first reference notes that conjugation in butadiene shortens the middle C-C single bond by 0.016 Å; the second reference notes the middle single C-C bond in diacetylene (butadiyne) is shortened to 1.3727 Å from 1.454 Å in butadiene, a reduction of 0.0813 Å, about five times as much as occurs when going from no conjugation on either side of the single bond to double bonds on both sides.
This bond shortening is a strong case for extra conjugation in diacetylene (butadiyne). The absolute linearity of diacetylene is a big help; in butadiene, the rotation around the C-C single bond would reduce conjugation.
Look at a flat ring; maybe there is more conjugation in cyclopentadiene: "The microwave spectra of the three monosubstituted 13C species of cyclopentadiene have been investigated. The effective rotational constants of these species together with those of the main isotopic species give the structural parameters for the ring. Labeling the methylene carbon as C1 and numbering the other carbons sequentially we obtain: r(C1C2)=1.509 Å, r(C2C3)=1.342 Å, r(C3C4)=1.469 Å, ∠C1C2C3=109.3°, ∠C2C3C4=109.4°, ∠C2C1C5=102.8°." https://aip.scitation.org/doi/10.1063/1.1697207 The C-C single bond between the two double bonds is r(C3C4)=1.469 Å, actually a little longer than in butadiene (1.454 Å)! Steric  restrictions in the ring may account for the difference, but it's small, so may be insignificant.


text:
Ionization isomerism in coordination compounds
Ionization isomerism is defined as complexes which have same empirical formula but ionize to different ions in solution.
Consider the complex $\ce{[Pt(NH3)4Cl2]I2}$. The other possible ionization isomers are:
$\ce{[Pt(NH3)4I2]Cl2}$ and $\ce{[Pt(NH3)4ICl]ICl}$.
But I have not come across the complexes similar to the last one with two different anions outside the coordination sphere.
So, do these sort of complexes exist?
Moreover, what can be the IUPAC name for this compound?
tetraamminechloridoiodoplatinum(IV) chlorineiodine seems wrong.

-iodido-, -chloride iodide.

I think it should be -
tetraamminechloridoiodoplatinum(IV) chloride iodide.
The rules of the naming of counter ions are pretty much the same as in general Inorganic Chemistry. I do know of a compound AgClI - named as Silver chloride iodide. So, I think the counter ions here - chloride and iodide should be named as by the nomenclature rules of anions i.e with an -ide suffix and alphabetically if there is the presence of more than one anion
Proof of Reference - Silver chloride iodide (NCBI PubChem)
EDIT: My teacher did give an example of it being a valid complex. I think it is not generally stated in the preliminary books of this topic to avoid confusion among students.I imagine it would be present in some more in-depth books of this topic.

OK... but my original query was... Is this complex possible? Seems affirmative to me. So why there are no such examples in the books?

@Apurvium I think its a matter of the level of the books. The book I mentioned was of STD 12 level-which is the first time Coordination compound is taught.I think such examples would be found in some higher level books.


text:
How is a phase equilibrium defined for a one-component system?
A question on this site asked whether a one-component system is at equilibrium when melting or boiling, and the disparate answers were somewhat dependent on the definition of phase equilibrium. Another question asks whether two phases can ever exist in equilibrium at the boiling point. The term "phase equilibrium" seems to imply that under certain conditions, a one-component system with two phases is indeed at equilibrium.
Searching for the definition of phase equilibrium, I found these two:
Definition 1

Phase equilibrium is the study of the equilibrium which exists between or within different states of matter namely solid, liquid and gas. Equilibrium is defined as a stage when chemical potential of any component present in the system stays steady with time.

Source: https://www.wattco.com/casestudy/phase-equilibrium
Definition 2

Phase equilibrium is the state of thermodynamic system, in which the different phases of the substance having common boundary surfaces do not vary quantitatively.

Source: https://www.elsevier.com/books/computational-materials-science/ovrutsky/978-0-12-420143-9
Implications of the distinct definitions
According to definition 1, water present as two phases at the melting point would be at equilibrium even if slowly melting isothermally. According to definition 2, the same situation would not be called equilibrium because the amount of ice is decreasing over time (and thermodynamic parameters such as entropy and inner energy are changing over time).
Phase rule
For the mentioned system, the phase rule specifies one independent degree of freedom (you can change one intensive parameter, say temperature, and all other intensive parameters, in this case just pressure, will be determined, given that the system is supposed to be at equilibrium). Because it is a one-component system, the concentrations (or activities) are constant and are not a degree of freedom. If you are used applying the equilibrium concept to solution chemistry, this is somewhat unusual.
Question: What is the official definition of phase equilibrium, and how does it apply to a one-component system where the chemical potentials are independent of the mole ratios?

Melting is a process. A process is not a state. Since only states can be "at equilibrium", it doesn't make sense to speak of a real melting substance as being in an equilibrium state. However, melting is deemed reversible if performed infinitely slowly, and during such a transition the system passes through a series of equilibrium (or "quasi-equilibrium") states. But such a process is an idealization.  In reality, you can't "melt ice isothermally". Real transitions are therefore irreversible because system and surroundings are not actually in thermal equilibrium during heat transfer$^\dagger$.
Note that if you place in contact two of the intermediate states sampled during the transition neither will change, that is, each remains in its original equilibrium state.
The second definition encompasses the first one (equality of chemical potential across phases). The second definition therefore better captures what is meant by phase equilibrium. The conditions for phase equilibrium are those used to derive the phase rule, namely mechanical, chemical and thermal balance between phases. A phase is a region of uniform composition, temperature and pressure.

$^\dagger$ We say that it is possible to melt a pure substance using an ideal reversible process that leads the system infinitely slowly through a series of equilibrium states (each intermediate stage in that reversible process is an equilibrium state).  In order to move between intermediate states, an infinitesimally small amount of heat has to be transferred from the infinitesimally warmer surroundings (exploiting, say, something like a Maxwell demon). In what is a logical sleight of hand, we say that the temperature difference is so small however, that the system and surroundings are effectively isothermal. This does not reflect what happens in a real process.


text:
Why does the emission wavelength of sodium appear to be an exception despite the trend for ionisation enthalpies among alkali metals?
The wavelengths for emissions in the visible region for Group 1 elements is as follows:
$$
\begin{array}{lc}
\hline
\text{Element} & λ/\pu{nm} \\
\hline
\ce{Li} & 670.8 \\
\ce{Na} & 589.2 \\
\ce{K} & 766.5 \\
\ce{Rb} & 780.0 \\
\ce{Cs} & 455.5 \\
\hline
\end{array}
$$
Now, considering that the ionisation enthalpies (I.E.s) decrease going down the group, it should be relatively easier to remove an electron, which should reflect in the energy band gaps and ultimately in the wavelengths. Thus, there should be increase in this trend. However, $\ce{Na}$ is an exception. Why so?
I think the $\mathrm{d}$-electrons play a role in the increased band gap for $\ce{Cs}$, which shows in the trend. But I am missing something for $\ce{Na}$. Is its I.E. decreases because size outweighs nuclear charge along with fair amount of screening? But $\ce{Na}$ has a unique balance wherein size and screening are not enough to tip over.

It hardly has to do with ionization energy. It's a matter or specific transitions, or more precisely spectral lines associated with them. Find which lines are these^ and then you can get into comparing.

On the NIST website you can look up the transitions corresponding to the wavelengths you list:
$$
\begin{array}{l c c c c}
\hline
\text{Element} & λ/\pu{nm}& |i\rangle & |f\rangle & \text{I.E./cm$^{-1}$}\\
\hline
\ce{Li} & 670.8 & 1s^22s & 1s^22p & 43\,487\\
\ce{Na} & 589.2 & 2p^63s & 2p^63p & 41\,449\\
\ce{K} & 766.5 & 3p^64s & 3p^64p & 35\,010\\
\ce{Rb} & 780.0 & 4p^65s & 4p^65p & 33\,691 \\
\ce{Cs} & 455.5 & 5p^66s & 5p^67p & 31\,406\\
\hline
\end{array}
$$
You see that the transition in Cs that you give in the table corresponds to a $7p\leftarrow 6s$ transition and that for consistency you should compare it with a $6p\leftarrow 6s$ transition which has a wavelength around 860 nm (no $d$ electrons involved).
The term values of the electronic states of multi-electron atoms are relatively well described by Rydberg's formula
$$
\tilde{\nu}_{n,\ell}=\text{I.E.} - \frac{\mathcal{R}_M}{(n-\delta_\ell)^2},
$$
with $\mathcal{R}_M$ the mass-corrected Rydberg constant and where the quantum defect $\delta_\ell$ depends on the orbital angular momentum of the electron $\ell$ and on the properties of the ion core. The transition $n'p\leftarrow n''s$ can therefore be approximated by
$$
\tilde{\nu}_{n',\ell=1}-\tilde{\nu}_{n'',\ell=0}=\frac{\mathcal{R}_M}{(n''-\delta_{\ell=0})}-\frac{\mathcal{R}_M}{(n'-\delta_{\ell=1})},
$$
and thus depend on the quantum defects for the $s$ and $p$ states. The quantum defects depend on the scattering properties of the Rydberg electron with the ion core and it is difficult to predict their values for different elements. It is therefore difficult to make quantitative predictions. Also for the heavier elements (higher nuclear charge $Z$), relativistic effects become more important, which influences the electronic structure.
Note that these lowest states are not really Rydberg states and the Rydberg formula is a bad approximation. For high values of $n$ the Rydberg formula gets much more accurate. In fact, it allows for a very accurate determination of the ionization energies of these elements.


text:
Osazone crystal
How does maltose form its osazone crystal? If osazone is to be formed, then there must be aldehyde form for phenylhydrazine to react, but maltose is formed after hemiacetal formation. How do we address this?

What happens if you tautomerize the hydroxyl proton on the far right to the nearby ring oxygen?

Then you have to break the ring.

And you then get what?  Sugars easily do this tautomerization.

The hemiacetal is in equilibrium with the aldehyde unless it reacts further to form a glycosidic bond (like the glucose unit on the left). When analyzing carbohydrates, you can distinguish between reducing sugars (aldehyde group is available, will react with Benedict's reagent like glucose unit on the right) and non-reducting sugars (aldehyde group is not available, like glucose unit on the left).
"Reducing" sugars will react to form osazone derivatives while "non-reducing" will not.


text:
Which would have a more intense transition: a low spin d6 complex or a high spin d5 complex?
I have an assignment question for second year inorganic which asks to rank the intensity of d-d transitions for a number of complexes. Two are $\ce{[Fe(OH2)6]^3+}$ and $\ce{[Fe(CN)6]^4-}.$ Both will have Laporte-forbidden transitions, but which will be 'more forbidden', so to speak?

You see that both cases are Laporte-forbidden, now how about spin-forbidden?
Assuming the complex vibrates so it's no longer exactly octahedral, can you make a spin-allowed transition for $\ce{[Fe(CN)_6]^{4-}}$?  How about $\ce{[Fe(H_2O)_6]^{3+}}$?  When you answer this you will answer your question.

 You should find that $\ce{[Fe(H_2O)_6]^{3+}}$, with all orbitals occupied by electrons having the same spin, has only spin-forbidden d-d transitions.  Therefore it will undergo weaker transitions than $\ce{[Fe(CN)_6]^{4-}}$ which has some spin-allowed transitions.

The chance of a transition being allowed or 'forbidden', and hence its intensity,  depends on both the symmetry of the states involved and the spin change that occurs, if any.
The first step, however, is to find out how  the orbitals $e_g,\; t_{2g}$ are occupied with the electrons.  The Fe(II) is $d^6$ and has a 'strong' ligand in CN$^-$ compared to H$_2$O (see spectrochemical series) which means that the CN splits the orbital energies by more than the spin pairing energy. This makes the Fe(II)(CN)$_6^{4-}$ low spin so that the $t_{2g}$ are filled and the $e_g$ empty. Thus a transition on spin terms is allowed, i.e. promoting one electron into the $e_g$ orbital can be done without an overall spin change. The ground state has no unpaired electrons so the total spin $S=0$ so the multiplicity is $2S+1=1$, a singlet. The excited state has 5 electrons in $t_{2g}$ and 1 in $e_g$ this can also be spin zero. If there were a spin change to $S= 1$, thus making a triplet, then this would be spin forbidden.
In the Fe(III) complex there are 5 electrons and a 'weak' ligand so the $t_{2g}$ to $e_g$ energy gap is smaller than the spin pairing energy and this complex is high spin $S=5/2$ with multiplicity $2S+1=6$, i.e. each orbital has a single electron with the same spin.  Exciting an electron can only mean that spin pairing must occur and this will change the overall spin multiplicity to $S=3/2, \;2S+1=4$ which corresponds to a change in (spin) angular momentum, and as as this must be conserved the transition is 'forbidden'.
Because the photon does not interact with the electron spin any spin changing transitions are very weak indeed. They do become partly allowed by spin-orbit coupling which is the interaction of the electron spin with the angular momentum of the electron in its orbital.
The second part of the allowed/forbidden transition involves the symmetry of the states. This occurs because the chance of a transition depends on the transition moment integral. This has a spatial and spin part, we never need to work out its value but use symmetry and spin arguments, as above, to determine which, is any, part is exactly zero, i.e. not small but exactly zero. The integral is $\int \psi_i \mu \psi_g d\tau \int \alpha_i\alpha_f ds$. The terms are $i,\;f$ for initial and final states, $\mu$ is the transition dipole operator (see later) and  $\psi$ are the wavefunctions for the two states involved and $\alpha$ the electron spin wavefunction. As these latter are orthogonal if they are the same $\alpha_i=\alpha_f$ then the integral is finite and if different it is zero. This is what is explained in words above in the spin case.
The transition dipole operator is a measure of the interaction of the photon with the molecule and it is an 'odd' function. Suppose $f(x)$ is the function, if it is odd, such as $f(x)=x$, then  $f(-x)=-f(x) $ and if even such as $f(x)=x^2$ then $f(-x)=f(x) $. To work out the $\int_{-\infty}^\infty \psi_i \mu \psi_g d\tau$ integral we can use the odd-even property of the wavefunctions. This is usually done by reference to their symmetry species as defined in a point group table. However, from the properties of odd even functions we can determine a lot. An odd function integrated from $-\infty \to \infty$ is always zero. The negative part always exactly cancels the positive one. If the two wavefunctions $\psi_i,\;\psi_f$ are both even, i.e. have even symmetry species, and as $\mu$ is odd then the product $\psi_i\mu\psi_f$ is odd and the integral is exactly zero and a transition is forbidden, and its intensity zero.
However, because a molecule can vibrate (or be distorted by Jahn-Teller interactions) it is quite easy to change the molecules symmetry and so make the transition a little 'allowed'. The strength of this symmetry 'forbidden' transition (The Laporte Rule) is about 100 to 1000 times stronger than any spin-forbidden transition.

(I wish I could find pictures of the UV/Vis spectra to support this answer but alas …)
In the very narrow context of your question that only considers $\mathrm{d{\rightarrow}d}$ transitions, I would like to refer entirely to Oscar Lanzi’s answer. However, there is an important aspect to the colours of both ions that $\mathrm{d{\rightarrow}d}$ transitions cannot explain.
As you correctly note, both complexes should have Laporte-forbidden transitions. This, and the expected answer Oscar Lanzi gives, can be demonstrated most beautifully by adding fluoride to a solution of $\ce{[Fe(H2O)6]^3+}$ to give the $\ce{[FeF6]^3-}$ complex—the solution will become entirely colourless as is expected based on $\mathrm{d{\rightarrow}d}$ transitions.
However, iron(III) solutions tend to be distinctly yellow as is $\ce{[Fe(CN)6]^4-}$; the latter even carries the German name gelbes Blutlaugensalz (rough translation: yellow alkaline blood salt) highlighting its yellow colour. Neither can be explained by $\mathrm{d{\rightarrow}d}$ transitions which, as they are Laporte-forbidden, should result in weak colours at best. Instead, the colours we observe originate from metal-to-ligand charge transfer transitions. For reference, please see figure 1 below which contains the full molecular orbital scheme of an octahedral $\ce{[ML6]}$ complex.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
The transitions which result in the strong yellow colour of both complexes originate not from transitions between the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ complexes (the transition labelled $\pu{10Dq}$ in figure 1) but from those labelled $\mathrm{t_{1u}}$. These correspond to orbitals mostly centred on the ligands, roughly describing a subset of how electron density is donated to the metal in dative bonds, and thus the transition could be described as oxidising the ligand and reducing the metal briefly; compare equation $(1)$.
$$\ce{[\overset{+III}{Fe}(H2\overset{-II}{O})6]^3+ -> [\overset{+II}{Fe}(H2\overset{-II}{O})5(H2\overset{-I}{O})1]^3+}\tag{1}$$
It is important to note that this should be considered a mere formality or thought experiment; it is an intramolecular transition and the species overall does not change. But this depiction may be helpful to imagine what is going on.
In any case, these transitions are from an ungerade orbital to a gerade orbital—the original $\mathrm{t_{1u}}$ orbital is antisymmetric with respect to inversion while the target $\mathrm{t_{2g}}$ or $\mathrm{e_g^*}$ orbitals are both symmetric with respect to inversion. Thus, this transition is not covered by Laporte’s rule (which only applies to gerade to gerade or ungerade to ungerade); the resulting absorption is strong.
These charge transfer transitions are orders of magnitude more intense than any Laporte-forbidden $\mathrm{d{\rightarrow}d}$ transitions and thus they dominate both the UV/Vis spectra and the resulting colour we perceive.


text:
Catalyst that combines oxygen and hydrogen
Are there any known catalysts that can combine hydrogen and oxygen into water?
Does this type of reaction have a name? Trying to search for information, it seems to be all about splitting hydrogen and oxygen, is there any work on combining these and some catalysts that could be used?

This looks like catalytic hydrogen combustion (CHC). Rarely needed, unless one needs to burn hydrogen in a more controlled manner without producing flames or explosion. I suggest searching Google Scholar with this keyword to get an idea what transition metals and their oxides are used in which cases.

@andselisk "Rarely needed" unless you want to generate clean electricity. This is what fuel cells do. Using exactly this reaction.

I have removed some of the comments, please discuss in [chat]. @matt_black

Yes
Not only is this possible it can be done while extracting electricity from the reaction. This is the key reaction in a hydrogen/oxygen fuel cell.
Fuel cells have been known for well over a century and Hydrogen oxygen fuel cells have been used commercially since the 1930s. Platinum is often used as a catalyst for the key reaction in the cells.

Platinum is a good catalyst for this reaction. This technique is well known. It has been developed in the 19th century by Döbereiner.


text:
Can oxygen gas in air react with Mn3O4 to form MnO2?
I think $\ce{MnO2}$ at high temperature can be used as a catalyst for $\ce{SO2 + O2 -> SO3}$:
$$
\begin{align}
\ce{MnO2 + SO2 &-> MnSO4}\tag{1}\\
\ce{3 MnSO4 &-> Mn3O4 + SO2 + 2 SO3}\tag{2}\\
\ce{Mn3O4 + O2 &-> 3 MnO2}\tag{3}
\end{align}
$$
I think the first two reactions are possible, but I don't know about the third one.

Intuitively, I suspect rather $\ce{Mn2O3},$ not $\ce{MnO2}.$ And yes, two first reactions look OK to me except you probably should've mentioned the temperatures.

$\ce{Mn2O3 -> Mn3O4}$ reaction has been discussed here: https://chemistry.stackexchange.com/questions/121339/how-to-convert-mn2o3-to-mn3o4

Manganese(IV) oxide decomposes around 500 °C (German Wikipedia lists $\pu{450 °C},$ [1, p. 396] lists $\pu{530–585 °C}):$
$$\ce{4 MnO2 ->[\sim\pu{500 °C}] 2 Mn2O3 + O2}\label{rxn:1}\tag{R1}$$
On the other hand, according to [2, p. 12], $\ce{Mn3O4}$ reacts with oxygen only above $\pu{500 °C}$ and forms manganese(III) oxide:
$$\ce{4 Mn3O4 + O2 ->[>\pu{500 °C}] 6 Mn2O3}\label{rxn:2}\tag{R2}$$
Upon further heating, the reaction in the opposite direction takes place [1, p. 396]:
$$\ce{6 Mn2O3 ->[>\pu{940–1090 °C}] 4 Mn3O4 + O2}\label{rxn:3}\tag{R3}$$
All in all, the third reaction you've proposed $(\ce{Mn3O4 + O2 -> 3 MnO2})$ is unlikely to occur and the product will be $\ce{Mn2O3},$ which unfortunately would break the catalytic cycle.
As for the first two reactions, they are described in the literature all right [1, pp. 397, 399]:
$$
\begin{align}
\ce{MnO2 + SO2 &->[\pu{450 °C}] MnSO4}\tag{1}\\
\ce{3 MnSO4 &->[\pu{850–1155 °C}] Mn3O4 + SO2 + 2 SO3}\tag{2}\\
\end{align}
$$
Reactions \eqref{rxn:1}–\eqref{rxn:3} are also described in Greenwood's Chemistry of the elements: [3, pp. 1048–1049]:

It is $\ce{MnO2},$ however, which is by far the most important oxide in this group, though it is not the most stable oxide of manganese, decomposing to $\ce{Mn2O3}$ above about $\pu{530 °C}$ and being a useful oxidizing agent. […]
$\ce{Mn3O4}$ is formed when any oxide of manganese is heated to about $\pu{1000 °C}$ in air […].
If $\ce{MnO2}$ is heated less strongly (say $,\pu{800 °C})$ than is required to produce $\ce{Mn3O4},$ then the more stable α-form of $\ce{Mn2O3}$ results […].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)
Turova, N. Inorganic Chemistry in Tables; Springer: Heidelberg ; New York, 2011. ISBN 978-3-642-20486-9
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.


text:
How are σ+ values in Hammett Plots calculated?
First of all, I know a similar question has already been asked & answered about σ values in general, however, I am not 100 % sure this answer is correct for the following situation.
So for the σ values, we know we have the following formula:
$$\log \left(\frac{K_x}{K_h} \right)=\sigma_x$$
with the equilibrium values being determined for the deprotonation of benzoic acid for different substituents. So, most logically we would find electron-withdrawing groups (EWG) having positive $\sigma_x$ values and electron-donating groups (EDG) having negative $\sigma_x$ values.
However, for the Hammett plot corrected for the stabilization of positive charge through resonance ($\sigma^+$) with the reference reaction being another reaction (see figure below), we still see the same trend for $\sigma_x$ values whereas we would expect the exact opposite.
$S_N1$ heterolysis reaction of para-substituted phenyl dimethyl chloromethane">
According to the literature, this inverse trend for EWG and EDG is being done purposely by changing the formula but I cannot find the formula. Does anyone know how this is done? Thank you in advance!


text:
Optical activity of cis coordination compounds
I came across a statement saying,

"Consider a Octahedral coordination compound of type [MA2X2]. Then,
  if the compound is in a cis form, it shows optical activity as there
  is no plane of symmetry and hence is non-superimposable on its mirror
  image. On the other hand, a trans form is symmetric and superimposable
  on its mirror image and so is optically inactive."

I do not see why the cis form is not superimposable on its mirror image. If all the ligands lie on the same plane, then isn't there a line of symmetry passing diagonally? Does that not make it superimposable on its mirror image?

$\ce{[CoCl2(en)2]+}$ isn't square planar. Please add a reference for the quoted part.

Point 1. The compound is not planar(it is octahedral). Point 2. Line/Axis of symmetry doesn't affect chirality. Only POS,COS,alternating Axis of symmetry does. But for any planar molecule, molecular plane is itself a plane of symmetry so planar compounds are always achiral

ah duly noted. Turns out I got the structures messed up. I will edit my question for future viewers.

It is not always as easy to see from a standard octahedral depiction. Optical activity in octahedral complexes can be understood better, if you draw the octahedron as looking onto a triangular face rather than along an edge; this results in a hexagonal arrangement of ligands around the central atom with ligands alternately pointing into and out of the paper plane.
If we do that with $\ce{[Co(en)2Cl2]}$, we get the following two enantiomers which you cannot convert into each other:

The complex is chiral because the $\ce{en}$ ligands necessarily cross the paper plane and can do so either in a right-turn screw motion or in a left-turn screw motion.


text:
Type of reaction between (+)-bis salt of (+)-phenylsuccinic acid and HCl
Im confused about what type of reaction occurs between the (+)-bis salt of (+)-phenylsuccinic acid and HCl. Would this be considered an acid base reaction or a neautralization reaction?
MY THOUGHTS: I know that an acid-base reaction occurs between (+)-phenysuccinic acid and (-)-proline to make the (bis)+ salt, im confused at to whether such salt is an acid or a base, or neither... If its an acid and an acid, it would be a neutralization reaction mechanism. If it were a base, it would follow an acid-base reaction mechanism. Any thoughts on if either of these make sense based on such info?

The reaction between the bis salt of proline and phenylsuccinic acid will give free phenyl succinic acid and proline hydrochloride as HCl is a much stronger acid than phenylsuccinic acid. All you are really doing is exchanging one acid for another so it is hardly an acid/base reaction.


text:
Second Ionisation Energy Confusion
It is intuitive that the second ionisation will require more energy than the first. However, I'm having trouble explaining exactly why. I came across this other question on a similar topic: Why second ionisation energy greater than the first?
The answer made sense until the answer said:

Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus.

Considering, say, chlorine, when a valence electron is removed, it makes sense that the electrostatic repulsion between valence electrons would decrease. However, by removing 1 electron, why does this increase the attractive force? I don't see how would increase the coulombic force between the nucleus and a valence electron, since the magnitude of both charges are constant.
If it was the radius changing, why should the electrons be closer to the nucleus just because 1 electron is removed?

Just to get a sense: Think in term of net charge. There is no charge for the first ionisation. After your test charge, ie the second electron to be removed, sense a net one

@Alchimista but why does having a net charge matter? Why would that affect the energy to remove 1 electron? The electron still has the same force of attraction to the nucleus essentially right?

Beside that quantum mechanics applies, just think of a distribution of n positive and n negative charges. Upon removal of the first one you get a net charge. Obviously the field is now different and the force experienced by each charge is now greater. Don't look at protons and the electron to be removed only, look at the entire distribution. It can be an exercise in the electrostatic chapter.

@Alchimista No I don't see why the force experienced by each charge would be greater. Say we have 10 protons and 10 electrons in your arrangement. Each electron will experience 10 times the coloumbic force of 1 proton attracting it and 9 times the coloumbic force of 1 electron repelling it. If we remove 9 electrons, such that there's only one, then the attractive force felt must be the same. All that has decreased is the repulsion force. Is this what you mean by an "increase in force felt"?

Well yes. Force is the resulting vector. That is why I invite you to look at a bunch of charges. Two plus and two minus suffice, they don't have to be specifically arranged. Just do it.

Beside that the proportional relationship proposed could be discussed further, as r rearrange, all this statement are related to the sums of forces. They stated that there would be a greater attractive force because they discuss ionization and this is the case. But the attractive force is the result of all attractions and repulsions. Any distribution of electric charges would lead to this situation and is generally an exercise treated after Coulomb law. Any reader can do that providing the dynamics is not involved.

In classical models of the atom, the radius of an electron subshell changes with the number of electrons in it (and with nuclear charge). This is essentially because the radius at which a set of electrons "settle" is a result of a balance involving the electrical forces, the magnetic forces and centrifugal forces.
For example, the radius of $\ce{He}$ is 0.567 (in Bohr units). When it ionizes to $\ce{He+}$, one electron moves to infinity and the other settles at a radius of 0.500 because the force balance has changed with the departure of the first electron.
Having said that, for the larger elements the main driver for increasing ionization energy (within a given subshell) is the increase of effective nuclear charge as you peel away electrons. In the case of Chlorine, the first electron to ionize effectively feels a +1 nuclear charge as it speeds away from its orbit to infinity (17 protons screened by the remaining 16 electrons). If the second electron departs, it will feel a +2 nuclear charge since the nucleus is now only screened by the remaining 15 electrons, leading to an ionization energy that is roughly twice as high, and so on.
The 3p subshell radius only changes by a few percent as a result of these successive departures, so the contribution of the change in radius to the ionization energy is small. It's only when you start peeling away the 7th electron (from the 2p shell) that you see a significant increase in ionization energies due to the much smaller radius, followed by another step-change when you reach the 1s electrons.


text:
Internal energy change for reactions at constant volume vs constant pressure

The reaction of cyanamide, $\ce{NH2CN(s)},$ with dioxygen was carried
  out in a bomb calorimeter, and $∆U$ was found to be $\pu{–742.7 kJ mol-1}$
  at $\pu{298 K}.$ Calculate enthalpy change for the reaction at $\pu{298 K}.$

$$\ce{NH2CN(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)}$$
$ΔH$ could be found from
$$ΔH = ΔU + ΔnRT$$
I want to know whether $ΔU$ found for constant volume conditions (bomb calorimeter) can be used in constant pressure conditions (while calculating $ΔH$ pressure is constant). In other words, is $ΔU$ same for both constant pressure and constant volume conditions, and why?
I searched the entire net and several standard physical chemistry texts but couldn't find any explanation.

The key point to keep in mind in this type of problem is that the energy of an ideal gas is only a function of temperature, not of volume or pressure$^\ast$. We also assume that the dependence of the energy of the condensed phases on p and V is negligible. Therefore the final pressure or volume is not going to affect $\Delta U$ for the reaction, provided n and T are constant. Since here the reaction refers to a conversion of a stoichiometric amount of reactants into products at the specified T, we therefore do not expect much of a change in $\Delta U$ with change in p or V.
$^\ast$Note that the total differential for U can be written as
$$\begin{align} dU &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV + C_VdT \end{align}$$ Therefore at constant T,
$$\begin{align} dU = &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV  \end{align}$$
The term in parentheses is zero for an ideal gas.


text:
How to recast Henderson–Hasselbalch when fitting data on a chemical shift vs. pH graph for NMR
I've just been getting into organic and been reading about H-H related to $\mathrm{NMR}$. I've been seeing articles state how recasting of H-H is necessary and want to know how equations like below are derived.

$$σ_\mathrm{peak} = σ_\ce{HA} + \frac{Δσ}{1 + 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}\tag{1}$$
$$y = y_\mathrm{min} + \frac{y_\mathrm{max} - y_\mathrm{min}}{1 + 10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}}\tag{2}$$

Eq. 1: https://www.nature.com/articles/srep43748
"In the equation, $\sigma_\mathrm{peak}$ is the measured $\mathrm{NMR}$ chemical shift of the peak of interest, $\sigma_\ce{HA}$ is the chemical shift of the protonated form and $\Delta\sigma$ is the difference between $\sigma_\ce{HA}$ and the chemical shift of the deprotonated form of $\ce{HA}$ ($\sigma_\ce{A}$)."
Eq. 2: https://pubs.acs.org/doi/full/10.1021/ed3000028?mobileUi=0
"$y$ is the experimentally measured parameter (chemical shift, in this case), and $(y_{max} - y)$ and $(y - y_{min})$ are proportional to the concentrations of the deprotonated and protonated forms of the acid, respectively."

Am I right in assuming that organic stands for organic chemistry and not organic food or agriculture; can I further assume that H-H stands for Henderson-Hasselbalch equation. Don't the articles you link to explain (or cite) how to derive these equations?

The derivation of these expressions begins from the premise that the observed value of the NMR parameter (chemical shift) is an average over values for two states weighted by their mole fractions: $$\sigma = \sigma_1 \chi_1 + \sigma_2 \chi_2= \sigma_1 \chi_1 + \sigma_2 (1-\chi_1)$$ This is a reasonable assumption in NMR experiments if there are two states (protonated and deprotonated) with distinct shifts and exchange between the two states is fast compared to the chemical shift difference such that the observable signal is a single time average. This allows us to write that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma_2)} = \chi_1 $$ Similarly we can write that $$\frac{(\sigma-\sigma_1)}{(\sigma_2-\sigma_1)} = \chi_2 $$ such that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{\chi_1}{\chi_2} $$ or, in terms of concentrations, that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{c_1}{c_2} $$ Introducing this last expression into the Henderson-Hasselbalch equation $$pK_a=pH+\log_{10}\left(\frac{c_{HA}}{c_{A}}\right)$$ gives upon rearrangement
$$10^{(pK_a-pH)}=\left(\frac{c_{HA}}{c_A}\right)=\frac{(\sigma-\sigma_A)}{(\sigma_{HA}-\sigma)}$$ which can be further rearranged into $$1+10^{(pK_a-pH)}=\frac{(\sigma_{HA}-\sigma_A)}{(\sigma_{HA}-\sigma)}=\frac{\Delta \sigma}{(\sigma-\sigma_{HA})}$$ and then further into the expressions in the articles (note the sign convention for $\Delta \sigma$).

Could you further elaborate on the last equation, specifically how it was rearranged to get 1? Is it just a matter of ratios?

@cam I just added 1 to both sides (farmost left and farmost right) of the preceding equation and used the identities $1=(\sigma_{HA}-\sigma)/(\sigma_{HA}-\sigma)$ and $\Delta \sigma = (\sigma_{A}-\sigma_{HA})$

Fraction vs Ratio
If you have two species with the concentrations $c_1$ and $c_2$ and you want to know the fraction of species 2, $\chi_2$, you can to the following:
$$c_\mathrm{total} =  c_1 + c_2$$
$$\chi_2 = c_2 / c_\mathrm{total} = \frac{c_2}{c_1 + c_2} = \frac{1}{1 + \frac{c_1}{c_2}} = \frac{1}{1 + \mathrm{ratio}} $$
In the last part, we divided both numerator and denominator by $c_2$ so it doesn't appear twice. This is how the mole fraction (or the percentage of one species, where percentages of species add up to 100%) is related to the ratio of the two species. This expression is commonly found in many contexts. (Try plugging in ratios such as 0:1 = 0, 1:1 = 1, 3:1 = 3 and 1:0 = infinity to verify it results in 100%, 50%, 25% and 0% species 2.)
NMR shift vs fraction of species 2
The NMR chemical shift (under conditions of fast exchange) will vary linearly between the shift of pure species 1 and that of pure species 2. So the lower the mole fraction of species 2, the closer the signal to that of pure species 1, and the higher the mole fraction of species 2, the closer the signal to that of pure species 2.
Putting it all together
We can chunk the equation given by the OP into the following:

The observed shift is equal to the shift of species 1 plus the difference in shift between the species times the mole fraction of species 2. (Try some examples like 0%, 50% and 100% species 2 to see whether that makes sense to you.)
$$σ_\mathrm{peak} = σ_\ce{HA} + Δσ \cdot \chi_2$$
The mole fraction of species 2 is given by adding 1 to the ratio and taking the reciprocal, as shown above.
$$\chi_2 = \frac{1}{1 + \mathrm{ratio}} $$
The ratio of species is given by the Henderson-Hasselbalch expression, leading to the power of 10 expression found in the denominator (try pH = pKa or much more acidic or basic than the pKa to see how this expression evaluates in the range from zero to infinity).
$$ \text{ratio} = 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}$$

Now look at the expression again and see if you can identify those three chunks.

$$σ_\mathrm{peak} = σ_\ce{HA} + Δσ \frac{1}{1 + 10^{(\mathrm{p}K_\mathrm{a} - \mathrm{pH})}}$$


text:
Does H2SO or H2CO have a higher dipole moment?
Given these two molecules, I realize that the molecules are pretty identical in terms of the individual atoms' electronegativities (with Sulfur only being a small bit higher than Carbon). So then, the only factors that I can think of contributing to differences in dipole moment would be shape (tetrahedral vs trigonal planar), the lone pair, and the double bond. However, I don't know how to look at these three factors collectively to determine the net dipole moment.
How would I go about finding which one has a higher dipole moment?

I suspect H2SO is supposed to be hydrogen thioperoxide, which molecule isn't tetrahedral and closely resembles the structure of H2O2. Maybe it would be helpful to directly put this structure together with the one for formaldehyde and draw a dipole vector for each.

Even though H2SO is not really a common chemical usually found in the shelves of labs, the molecule definitely can exist as a stable species as there is nothing wrong with it at all. In fact, just typing "H2SO" into Google instantly gives scientific papers related to it that are published even in high-quality journals. And even if the molecule were not to exist, the question still would make sense as the user is interested in understanding basic concepts, and for this, this way of thinking in chemical analogues is perfect. So I see no reason for cherry-picking here, this is a nice question.

@andselisk Or hydrogen sulfoxide?


text:
Energy required for obtaining required mass of aluminium from alumina via electrolysis
How do I know how much energy I need to obtain 20 g alumunium from alumina with 5 V?
I calculated the quantity of electricity is 2.22 F.
Any help appreciated!

Required energy $E$ can be found as a product between voltage $U,$ current $I$ and time $t$:
$$E = UIt\label{eqn:1}\tag{1}$$
Unknown multiplier $It$ can be found using united Faraday law:
$$m = \frac{MIt}{zF}\quad\implies\quad It = \frac{mzF}{M}\label{eqn:2}\tag{2}$$
where $m$ is the mass of the substance liberated at the electrode; $M$ is the molar mass; $z$ is the number of electrons transferred per ion; $F$ is Faraday constant.
Considering aluminium is produced via Hall–Héroult process or similar, at the cathode the following reaction takes place:
$$\ce{Al^3+ + 3 e− → Al}$$
and $z = 3.$
Plugging \eqref{eqn:2} into \eqref{eqn:1} allows to finally find the energy:
$$
\begin{align}
E &= \frac{mzUF}{M}\\
  &= \frac{\pu{20 g} × 3 × \pu{5 V} × \pu{96485.33 C mol-1}}{\pu{26.98 g mol-1}}\\
  &= \pu{1.07 MJ}\tag{3}
\end{align}
$$

With much respect, can you assist me to understand chemistry? Can i have your email or whatever platform, so i can ask simple chemistry question to you? Basically, i dont have privilege to ask many questions in this forum. @andselisk

@Lifeforbetter I prefer not to share my personal information and I don't do private teaching. If you have a question, the better way would be to ask community and post a new question. Also, please note this is not really a forum, but a Q&A site with somewhat restricted back-and-forth communication capabilities. In the meantime, I'd suggest to look at Resources for learning Chemistry, or enroll in free chemistry courses online.


text:
Role of tautomerism and its difference from hyperconjugation
I would like to know the role of tautomerism, if any, in increasing stability of an organic compound. Moreover, how is it different from that of hyperconjugation?
I know that hyperconjugation involves shifting of protons between adjacent
atoms in different hypothetical structures to form a practical hybrid structure.
I am not clear with tautomerism, so kindly deliver a precise explanation.

Tautomerism basicially means that you have different isomers that are able to convert into each-other. This is not really similar to hyperconjugation which is an electronic effect in a given molecular geometry. The main difference between them is that hyperconjugation is thought to be a property of a given molecular geometry while tautomerization is an equlibrium between separate, distinct, stable configurations.
There are many types of tautomerizations, as explained for example on the Wikipedia. Probably the most well-known type of tautomerisation is the keto-enol tautomerisation, which is an equilibrium between a ketone molecule and an enol (an alcohol connected to an $sp^2$ carbon). The equilibrium between two isomers is the explanation why enols are only stable under special scenarios: they tautomerise into ketons! Another important example for tautomerism is the equlibrium between the chain and the ring isomers for sugars.
So again: tautomerism is an equilibrium between different (real, existing) isomers as opposed to hyperconjugation which is just defined within a single structure.

Relevant definitions in the Gold Book: tautomerism, hyperconjugation

It would probably be good to mention that tautomerism cannot enhance stability of a given molecular structure as it is dependent on its relative instability towards another isomer.

@Martin-マーチン I was thinking about mentioning that, but then I realized that it probably affects the stability of the given tautomer by some second-order topological resonance effects. So I believe that pent-1-ene-1-ol is more stable than pent-2-ene-1-ol due to the possibility of the former one to tautomerise. Since this is just my gut feeling and would be hard to rationalize (at least in a StackOverflow post), I thought I would just not mention it.

Stability is (and cannot be) an absolute concept, you have to know the common state you are referring it to. That is why questions like tho one posed are generally hard to answer completely. However, we know in case of hyperconjugation, which is a special form of resonance, that a description with this effect is energetically lower. We cannot say that for tautomerism. Like you said this is an equilibrium and we have to asses stability for each structure itself. It may happen that other isomers are even lower (or higher) in energy, but that has nothing to do with tautomerism.

I pray to disagree: even for a simple symmetric two-state quantum system one can easily see that there is a coupling between the two separate energy levels, and the Rabi oscillation between them causes the system to be dynamically more robust. If we are not talking about a cryogenically stabilized single molecular entity (and even there we got tunnelig splitting!) then there is always an interaction between the two regions of phase space. And this results in a phase space volume that is bigger for a given energy, which, in my understanding, is a pretty absolute characterization of stability.

Let's agree to disagree then. I think this discussion is not helpful to the OP; if you really want to continue it, you can find me in [chat].

Just for OP: as in the comments by Martin. Retain that tautomers ARE compounds while mesomers ARE NOT. Teachers like to stress that but if their pupils have a clear idea of resonance (whatever the theory used) there shouldn't really be confusion. If the simbol used among mesomers would be ¤ instead of a double arrow ←→ one can't really mix up the two things.


text:
How is the distance between m/z axis values in mass spectrums called?
The output data of mass spectrometers is relative intensity (y axis) over m/z (x axis). How is the distance between m/z values on the x-axis called (in case of centroided data and profile/raw data respectively)?

I think using "distance between $m/z$" or "distance between peaks" is already fine. Alternatively, maybe use $Δ(m/z)$.

Probably the question was not clear. I mean the distance between m/z values on the x-axis, not the distance between m/z values of actual samples.

I still stand by my answer. The term 'neutral loss' is only applicable inside a single fragmentation series. When you have a larger molecule, you are ought to have parallel, independent fragmentation series, and he distance between two random peaks is then generally not due to a loss of a neutral species. The resolution, also mentioned above, is only defined between two peaks that are just slightly separated by your MS instrument. In general, the distance between two arbitrary peaks in a complicated spectrum has no name.

@Ezza After discussion with a collegue of mine the answer is still not completely clear. Pls let me some more time for more internal discussions.

I wanted to back up my comment and make sure it's justified, so I decided to flip through Gross'Mass spectrometry: a textbook [1] and it looks like there are indeed basically two possible variations, including distance between $m/z$ values, as you suggested:

{Distance | difference | mass difference} between {peaks | signals | peak tips}
{Difference in $m/z$ | $m/z$ difference | $Δ(m/z)$}

Both refer to describing isotopic distribution as well as to reading and interpreting the spectra.
Selected parts touching standardization (emphasis mine):
From [1, p. 11]:

The distance between peaks on that axis has the meaning of a neutral loss from
  the ion at higher $m/z$ to produce the fragment ion at lower $m/z.$ Therefore, the
  amount of this neutral loss is given as “$x~\pu{u}$”, where the symbol $\pu{u}$ stands for unified atomic mass. It is important to notice that the mass of the neutral is only reflected by the difference between the corresponding $m/z$ values, i.e., $Δ(m/z).$ This is because the mass spectrometer detects only charged species, i.e., the charge-retaining group of a fragmenting ion.

Further summary from [1, p. 20]:

1.7.1 Basic Terminology in Describing Mass Spectra
[…]

Ranges in spectra or ranges set in operating a mass spectrometer are to be
  referred to in the form of “$m/z~10–100$” or “$m/z~10$ to $m/z~100$”.

[…]

Neutral losses (rarely called dark matter […]) are exclusively recognized from
  the distance between peaks expressed in terms of the difference $Δ(m/z).$ The
  mass of the corresponding neutral is then given in units of $\mathrm{u}.$

References

Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.

Right, but this neutral loss is only meaningful if the smaller peak is produced from the larger peak, which is not true for all peaks in a spectrum. But this actually might be the definition that the poster is looking for.

@Ezze I suspect this is just an example where using atomic mass units makes sense, but I guess the terminology can be used beyond this case.

Hmm, I haven't heard about such a term. After a quick search in the IUPAC suggested terminilogy I haven't found anything either. Perhaps there simply is no name for this quantity. I guess you can simply call it separation of the peaks.

I've looked into IUPAC as well without success. I don't think "separation of peaks" is suitable cause not every intensity-m/z sample needs to be a peak. The distance depends on the mass spectrometer specs (mass resolution, mass accuracy, etc.), means it's performance for sure. However I don't know to which spec it relates and how.

@thinwybk I do not believe that the separation actually depends on the instrument as if one ion has an m/z value of 23 and a different one has an m/z value of 90 then their separation is 67 m/z on all instruments, right? Unless you are considering peaks that are really close to each other, in which case, the resolution comes into play - but this does have an accurate IUPAC definition.

The question was not clearly stated enough. I'm interested in the difference of the m/z values on the x-axis, not the distance between samples. I'm also considering high resolution data. As far as I know the resolution does influence the minimal distance between distinguishable peaks, not the "granularity" of the m/z axis values. Which IUPAC definition do you mean?

But the distance on the x axis really depends on the size of the paper you print it on, doesn't it? And as for the definition of resolution: http://goldbook.iupac.org/terms/view/R05318

The distance between two peaks is characterized not by $\Delta m$ but rather by mass resolution. The distance between the peaks is often referred to as $\Delta m$ and assumes that the $z = 1$, but this is not always the case.
IUPAC addresses this in detail: Resolution, DOI: 10.1351/goldbook.R05318

(10 per cent valley definition): Let two peaks of equal height in a
  mass spectrum at masses m and m−Δm be separated by a valley which at
  its lowest point is just 10 per cent of the height of either peak. For
  similar peaks at a mass exceeding m, let the height of the valley at
  its lowest point be more (by any amount) than ten per cent of either
  peak height. Then the resolution (10 per cent valley definition) is
  m/Δm. It is usually a function of m. The ratio m/Δm should be given
  for a number of values of m.

Since it can be expected that the UI of the Gold Book may change (like it did last year, which broke quite a few links), I've switched to the issued DOI and encourage you to do the same. Please also note, that I have wrapped MathJax around the whole expression, as this ensures the post doesn't break on other devices than your own.


text:
What is the difference between ideal stages and transfer units in a packed bed gas absorber?
I am having a hard time grasping what these two concepts are supposed to mean and how they differ. What I know so far is:
In a counter-current gas absorber, if we set the inlet and outlet concentrations of solute in gas and liquid, and also have control on the ratio of gas to liquid flow rates, we can find the outlet concentrations by making a "step" like a staircase (with the number of steps equaling the number of stages) from the operating line to the equilibrium curve.

I understand where this comes from. When we draw a "step", we are equating the two outlet concentrations to be in equilibrium.
Now, for the number of transfer units, we usually find this out by using the integral formula:
$$\text{NTU} = \int_{y_1}^{y_2}{\frac{dy}{y-y*}}$$
where $y$ is the operating line solute concentration in gas phase for a given $x$, i.e. solute concentration in liquid phase, and $y*$ is the equilibrium curve solute concentration for the same $x$.
What I have noticed is that often, the NTU is not an integer, it can also be a fraction. And this fraction, multiplied by the height of transfer unit (HTU), gives the total height of the bed needed for such a transition.
I am confused. I dont understand what NTU is exactly, or better, what a transfer unit is. I would greatly appreciate some guidance.

A transfer unit gives the change of composition of one of the phases equal to the average driving force producing the change
Source: http://www.separationprocesses.com/Absorption/GA_Chp04c.htm

So to phrase it differently, for some ideal zone in the absorber that has a particular value to the average mass transfer driving force, there will be a fixed change in the phases present at that ideal zone.
NTU is somewhat analogous to the total number of theoretical stages if the operating and equilibrium line are straight and parallel.
There is no direct physical interpretation of what a transfer unit. If you design any appreciable size packed bed tower, you will find that the vendors of the packing will use a lot of empirical correlations for their own products to determine separation efficiencies and other related parameters.


text:
Energy of mercury(II) oxide decomposition
Study check 9.11 from [1, p. 298]:

Mercury(II) oxide decomposes to mercury and oxygen.
$$\ce{2 HgO -> 2Hg + O2}\quad ΔH = \pu{+182 kJ}$$
[…]
b. How many kilojoules are needed when $\pu{25.0 g}$ of mercury(II) oxide reacts?

Correct answer is $\pu{20.5 kJ}.$ I get $\pu{10.5 kJ}.$ Here's how I did it:

Grams to moles: $200.59 + 16 = \pu{216.59 g}$ (using molar masses)
$$\pu{25 g} × \frac{\pu{1 mol}}{\pu{216.59 g}} = \pu{0.12 mol}$$
Moles to kJ:
$$\pu{0.12 mol} × \frac{\pu{182 kJ}}{\pu{2 mol}} = \pu{10.5 kJ}$$

What did I do wrong?
References

Timberlake, K. C. Basic Chemistry, 5th ed.; Pearson: Boston, 2016. ISBN 978-0-13-413804-6.

Your calculation is correct, although you forgot to use the correct units. Delta H given in the tables is always in kJ/mol, and not in kJ. I don't know why the correct answer should be 20.5 kJ. Is it not a typing error ? 2 instead of 1 ?

This question is taken from a chemistry book (by Timberlake) and they've written the answer is 20.5 kJ

I agree with @Maurice that it's likely a typo and should be 10.5 kJ; you followed the procedure described on a previous page pretty closely. I flipped through the rest and the way textbook handles units, presents calculations and its overall structure left a sour taste in my mouth.

Alright thank you guys, i'd consider this question solved then


text:
What is the type of orbital depicted in this diagram?
The following image is from this answer for the question Why is one lobe of an sp3 hybridized orbital smaller than its other half?:

I understood everything from that answer, but couldn't identify what is the type of orbital marked by the red circle. It could not be a $d_{z^2}$ orbital with $x$ axis as it's $z$ axis, as it has a big spherical node and not a circular ring? I haven't encountered such an orbital till now.

That's probably a very poorly drawn cartoon of a $d_{x^2-y^2}$ see eg https://en.wikipedia.org/wiki/Atomic_orbital . The "empty" lobes are along y and cancel the density in the annulus on the xy plane of the $d_{z^2}$ orbital at left.

@BuckThorn, Thank you. Could you please explain your second sentence - How can two lobes having the same sign of wave function cancel each other destructively to give $d_{z^2-x^2}$ orbital?

There is a negative sign on front of the orbital, if you look closely, which inverts the sign (I guess that means phase, actually) of the wavefunction.


text:
Does sulfoxide H₂SO exist?
The question Does H2SO or H2CO have a higher dipole moment? left me wondering whether $\ce{H2SO}$ has already been synthesized.
There are experimental evidences that both hydrogen thioperoxide $\ce{HSOH}$ (1) and  thiooxonium ylide $\ce{H2OS}$ (3) exist, whereas sulfoxide $\ce{H2SO}$, despite possessing global energy level lower than (3), is not [1]:

The global minimum corresponds to hydrogen thioperoxide $\ce{HSOH}$ (1). The sulfoxide $\ce{H2SO}$ (2) is predicted to be $> \pu{25 kcal/mol}$ higher in energy (relative to 1) and the thiooxonium ylide 3 is the least stable
  isomer $(> \pu{34 kcal/mol}$ relative to 1).

There were numerous theoretical investigations (ab initio) suggesting that (2) is possible, but one of the most recent ones I found [2] still claims there is no experimental evidences to support its existence.
One of the problems with its structure the paper [2] refers to is that predicted $\ce{S-O}$ bond is abnormally short (1.4749 Å) and the valence state of sulfur is too high.
It would be nice if someone could point me towards a reputable source proofing/refuting existence of (2).
Additionally, what would be the complete preferred name of $\ce{H2SO}$?
MarvinSketch generates quite an amusing one: oxidanethione.
References

Iraqi, M.; Schwarz, H. Experimental Evidence for the Gas Phase Existence of $\ce{HSOH}$ (Hydrogen Thioperoxide) and $\ce{SOH2}$ (Thiooxonium Ylide). Chemical Physics Letters 1994, 221 (5), 359–362. https://doi.org/10/ct4m2m.
Denis, P. Theoretical Characterization of the $\ce{HSOH},$ $\ce{H2SO}$ and $\ce{H2OS}$ Isomers. Mol. Phys. 2008, 106 (21), 2557–2567. https://doi.org/10/dnrk2v.

Here's another paper that computationally characterizes R2O=S: J. Am. Chem. Soc. 1998, 120, 33, 8461-8471 (https://doi.org/10.1021/ja980141p).

This is not an answer to the question "can the sulfoxide be synthesized in significant yield" (and under which conditions). Instead, this is a "naive" prediction based on the data you provide and two assumptions:

The predicted thermodynamic values in the articles you cite are accurate at $\pu{298 K}.$
The heat and entropy of formation of (2) and (3) from (1) are constant over $T.$

From the references we can build a data table of energies and entropies relative to "the ground state" $\ce{HSOH}$ assuming the $\Delta G^\circ$ values correspond to $T = \pu{298 K}:$
$$\begin{array}{lccc}
\hline
\text{Compound} & \Delta G/\pu{kcal mol-1} & \Delta H/\pu{kcal mol-1}^* & \Delta S/\pu{cal mol-1 K-1} & K_\mathrm{eq}\\
\hline
\ce{H2SO} & 25 & 15.8 & -31 & 10^{-19} \\
\ce{H2OS} & 34 & 38.2 & 14 & 10^{-25} \\
\hline
\end{array}$$
I include the order of magnitude of the equilibrium constants $K_\mathrm{eq}$ only to illustrate their ludicrously small values (and these are upper estimates!).
Could you find a temperature that would make $\Delta G^\circ<0$ for the reaction to $\ce{H2SO}$? The answer is no.  To see why we turn to the enthalpies. Since the reaction is endothermic you might hope that an increase in $T$ would help, but the accompanying entropy change is negative, so there is no way you can sufficiently increase the temperature to drive the reaction forward. The highest equilibrium constant you could hope for is $K_\mathrm{eq} = \pu{1.7E-7}$.
Things are not so bleak for $\ce{H2OS}$: the reaction is also endothermic but the entropy change is positive. At about $\pu{2800 K}$ $K_\mathrm{eq} = 1,$ and if we increase $T$ sufficiently, we can crank up $K_\mathrm{eq}$ to any arbitrarily large value.
This is of course not remotely exhaustive (therefore "naive") but hopefully a start to answering your question.

* Quoting your second reference:

At the CCSD(T)/CBS limit and including corrections for scalar relativistic, spin orbit and core-valence correlation effects, the estimated enthalpies of formation are $\pu{−28.1 ± 1},$ $\pu{−12.3 ± 1},$ and $\pu{10.1 ± 1 kcal/mol}$ for $\ce{HSOH},$ $\ce{H2SO}$ and $\ce{H2OS},$ respectively.


text:
What is the justification for using the Marcus rate equation to model charge transport in organic semiconductors?
I am trying to understand the history of how Marcus theory moved from describing electron exchange reactions in aqueous ionic solutions to also being applied to describe transport in organic semiconductors. As in who and when was the connection made and justified?

I believe the connections between Marcus Theory and electron transfer in organic semi conductors was made in the 80's in the context of understanding electron transfer in the photosynthetic reaction centers. The only reference I can think of off the top of my head right now is:
https://doi.org/10.1021/ja00002a076
If you can, I'd suggest mining through the references in there.

No sure about the history but basically the Marcus theory can be derived from the spin-boson model see 'Chemical Dynamics in Condensed Phases, Nitzan'. The spin-boson model itself is very closely related to another model called the Holstein model the only difference is the summation of the vibrational states which is in wavevector space for the Holstein model while in the spin-boson model its in real space.
The Holstein model is well known in solid states physics and I used model polaron transport.


text:
Salt analysis of Ag salts
So I was doing analysis of silver salts and I found $\ce{AgBr}$ is curry white partially soluble in ammonia, $\ce{AgI}$ is yellowish and insoluble but silver phosphate is also yellow. So can anyone tell whether or is soluble in ammonia or not?

Silver phosphate is soluble in aqueous ammonia. AgBr is only slightly soluble in ammonia. AgI is not soluble in ammonia.

According to Firsching [1],

silver phosphate is indeed soluble in ammoniacal solution due to formation of silver-ammonia complex.

The following is the solubility data of $\ce{Ag3PO4}$ in $\ce{NH4OH}$ at $\pu{35.5 °C}$ presented by Saraswat [2]:
\begin{array}{rr}
\hline
c(\ce{NH4OH})/ & c(\ce{Ag3PO4})/ \\
\pu{mmol l^-1} & \pu{mmol l^-1} \\
\hline
0 & 0 \\
4.05 & 0.133 \\
8.09 & 0.265 \\
16.18 & 0.531 \\
32.36 & 1.061 \\
64.72 & 2.057 \\
129.45 & 4.113 \\
258.90 & 8.292 \\
517.80 & 26.478 \\
\hline
\end{array}
References

Firsching, F. H. Precipitation of Silver Phosphate from Homogenous Solution. Anal. Chem. 1961, 33 (7), 873–874. DOI: 10.1021/ac60175a018.
Saraswat, H. C. Formula of Complex Compounds from Solubility Data. Proc. Indian Acad. Sci. (Math. Sci.) 1949, 30 (6), 329–332. DOI: 10.1007/BF03048752.


text:
Why n-alkanes can be found in organic solvent extract of rubber?
Sorry for the confusing title, I didn't know how to describe this better. So, if we douse rubber, say, a bicycle tire or a certain type of rubber glove, with organic solvent (n-heptane), we can find n-alkanes being dissolved in the solvent. Why is this? I'm talking especially about gas chromatography to analyse the paraffinic hydrocarbons here.
I guess this is quite a trivial question but I can't figure it out. As far as I know, natural rubber consists mainly of isoprene polymer chains. But isoprene has a double bond and in GC chromatogram we can see n-alkanes being the largest peaks. And not only that, we can also find straight chain hydrocarbons in synthetic rubber (like NBR or SBR) solvent extracts. SBR has a benzene ring in its structure but still n-alkanes are the dominant finding in its solvent extract chromatogram.
This is a major subject in my bachelor's thesis but I'm an analytical chemistry major and struggling a bit with materials science in this depth. Any advice and ideas would be greatly appreciated.
To make my point clearer I present these two chromatograms:

(GC-FID run of pure n-heptane)

(GC-FID run of n-heptane extract of bicycle tire)
This question still haunts me. I found this. Could that be a partial explanation for the aliphatic hydrocarbon finding?

In case you're still interested in this - I'm curious what part of the tire you did the extraction of? Tires are made of multiple layers that are adhered together, and the alkanes might be from the adhesion areas.

It was a sector of the whole tire (about 10 cm piece cut off of the length of the tire). As I have gained more experience in the field of chemistry I now strongly believe the alkanes are from paraffin wax.

Although natural rubber and synthetic rubber display a poor chemical resistance to alkanes among others, they do not  dissolve in alkanes. Your GC/MS traces (I believe that so because of alkane/alkene identification) have confirmed that since they have shown n-heptane as the main peak (I believe everything else are trace amounts exhibiting below your set-up threshold).
There is a reason for your finding of straight chain hydrocarbons in synthetic rubber such as SBR in your solvent extracts. For example, SBR is the shorten form for synthetic material, styrene butadiene rubber, which is made by anionic copolymerization of styrene & butadiene. For this anionic copolymerization, frequently used initiator is a solution of lithium alkyl compound in small paraffins, for instance, n-butyllithium in hexanes. The rubber may still have few molecules of these small paraffins, which may diffuse into the n-alkane you use to extract, and hence, would show up in your GC/MS chromatogram.

Your link almost certainly explains the alkanes. It says

commercial paraffins are mostly in the 22 to 38 range

That's exactly the range we are seeing in the chromatogram.
I'd consider this the answer.


text:
Ground state term symbol, why is $L = |M_L|$, if L must be a maximum?
I'm trying to understand how to predict ground state term symbol of atoms.
After finding the biggest S, why the biggest L will be $L = |M_L|$, where $M_L = \sum m_l$?
I know this rule works to predict the total L, but it doesn't make sense to me that it will give the biggest L, to me it seems like this is the minimum L.
For example, the nitrogen atom has three electrons in 2p.
Following the first rule, the spin $S$ of the ground state will be the maximum possible, which is S = 3/2.
The second rule, according to some books is to maximize $M_L$, in the case of Nitrogen, according to Pauli principle, since all the spins are up, we can only get $M_L = 0$.
Now, all material I read says that L of the ground state will be the maximum one, so L = $|M_L|$, but if we expand the Clebsch-Gordan coefficients of the three states:
$|1 \,\,\,\, 0> |1 -1> |1 \,\,\,\, 1>$ we get the possibility of L = 3,2,1 and 0, so why is L = 0 the maximum possible value?
There is an amazing answer to this here, but he doesn't explain why he cuts out the the terms with bigger L.

Formatting tip: use \langle and \rangle in math mode for bra-ket notation and avoid spamming \,. If you really need space that big, try single \quad or \qquad.

I am not sure I understand your question. For the ground state of nitrogen, you are right, you got three electrons on the p orbitals. Then, we maximize then spin, so we get that each orbital is half-filled. The orbitals have magnetic quantum number l=-1,0 and 1 in order. Each of them has one electron, so L=(-1)+0+1=0 and there is no other possibility here.

P means l = 1 right? So three eletrons with l = 1, means we can get a total momentum of 3, 2, 1, 0. My question is why does it have to be 0.


text:
Using AutoDock Vina to dock multiple ligands to multiple receptors?
I am trying do that what the title depicts, but I keep running into errors creating the executable script. Is it even possible to automate the docking of multiple ligands to multiple receptors?
Below is the script that I have. Right now I tried to get it to read from a configuration file, because the software could not read all of the arguments when they were in the executable. I am using
#! /bin/bash -i
for f in *lig.pdbqt ; do
  for n in nolig* ; do
    b=`basename $f .pdbqt`
    echo "Processing ligand $b"
    #(commented since I already have a directory for it)
    # mkdir -p $b
    vina --ligand $f --config avppar_conf.txt --out outputs/${b}_out.pdbqt --log outputs/${b}_log.txt
 done
done

This is the configuration file as it currently is:
--receptor $n (I thought this would not work, but I had to try anyway).

--center_x = -36.456

--center_y = 51.1

--center_z = 47.157

--size_x = 70

--size_y = 70

--size_z = 70

--num_modes = 9

--exhaustiveness = 8

I understand this looks messy; I would even appreciate it if I was directed to a more appropriate forum if needed.

I'm voting to close this question as off-topic because it's about how to use software.

Please read the manual, there is an example of the configuration. Your bash script also needs some love; it contains some deprecated code and mistakes. See shellcheck (https://www.shellcheck.net/#) for testing it.

I have tried to fix your script although I'm pretty sure it still won't work. You are not using $n. You should also try to use more descriptive variable names. this was very hard to decipher, and the done were missing completely.

I actually found the answer to my question. In order to link configuration and script files the way I wanted, I needed the 'seb' command.

I'm afraid this is not an answer.


text:
Make Na or K using carbonate + charcoal?
In the 19th century, Na metal was isolated by heating up Na2CO3 + carbon to 1200 C and distilling and capturing the Na vapor under oil.
I tried it a few years ago with a small steel tube welded at the end as a  retort and indeed vapors appeared burning with bright yellow (or purple in the case of K2CO3) flame, but could not capture it.
Did anyone isolate these alkali metals with this method ?

Danger, Mister Robinson! Your statement of burning vapours of sodium indicates that the distillation was not air tight.  While much more sodium is obtained via an electrolysis of a melt, you may distill it (like magnesium) at high temperatures and high vacuum.  The two-page preview (https://link.springer.com/article/10.1023/A:1012738010844) stops where the details start, but «forevacuum and oildiffusion pumps» (p. 556 right bottom) already tells it is considerably less than 0.1 mbar (to start an oildiffdusion pump) to somewhere down to 10E-8 mbar, not accessed earlier than in 1915 (Gaede).


text:
Is a weak deactivator considered a stronger activator than a strong deactivator for the purposes of directing effects
A nitro group is a strong deactivator whereas a halogen is a weak deactivator. The directing effects of these groups oppose each other: Nitro is meta directing whereas halogens are ortho/para directing. It is said that the strongest activator determines the directing effects in a disubstituted benzene ring. In a ring with no activators, such as one containing a halogen and a nitro group, is the weakest deactivator considered the "strongest activator", so that in the example I have just given the halogen would control the directing effects and said ring would be ortho/para directed?

How about an example of a "nitro halo" compound?

@user55119 Could one not nitrate phenyl chloride?

o, m, or p-chloronitrobenzene?

@user55119 This is not a homework question. It is frustrating that you would make that assumption. I was giving a concrete example to indicate an instance in which my question would have relevance. I was far more interested in understanding the rule at issue than the answer to the specific question. All persons who voted to put this question on hold have committed a cardinal sin of science: Never make unwarranted assumptions.

You may want to look at this: https://chemistry.stackexchange.com/questions/91636/why-is-the-effect-of-ring-activators-more-dominant-than-ring-deactivators?rq=1

I think you are reaching the limitations of directing group rules. These rules are not perfect and can often break.
In this example, chloride and nitro meta to each other, the group should add meta to the nitro group because the nitro group has a much more significant effect on the electronics of the molecule.

This was my initial impression but the book has several examples in which a weak activator trumps a nitro group (i.e. the ring will be ortho/para directed)

@David Reed: Why shouldn't a weak activator take precedent over a nitro group? Still looking for a concrete example from your book.

@user55119 A concrete example from book: Regarding direction of o-isopropylnitrobenzene: "In this case, a weak activator is competing with a strong deactivator for directing effects. Recall the more powerful activator controls the directing effects, so the isopropyl group determines the outcome in this case. The isopropyl group is ortho/para directing, so we must consider the positions that are ortho and para to the isopropyl group". My question is if a weak activator controls directing effects against a strong deactivator (because the strongest activating group controls), will a weak...

@user55119 ..deactivator similarly control directing effects against a strong deactivator, as one could reason that a weak deactivator is "more activating" than a strong deactivator, and hence would be the "strongest activator"

@David Reed: I can't speak to your logic but CH3O in anisole is a stronger o,p-activator than Cl in chlorobenzene because a positive charge is more readily sustained on oxygen than on chlorine. A Friedel-Crafts (FC) acylation of a mixture of the 2 aromatics with an acyl halide as the limiting reagent will give more of the anisole-derived product. The FC reaction of either anisole or chlorobenzene can be conducted in nitrobenzene as the solvent!!

@user55119 Here is my fundamental confusion: I don't know how to predict the directing effects in disubstituted benzene rings in which both substituents are deactivators. My book tells me to find the "strongest activator" to determine what the directing effects will be. It doesn't speak to when the ring has no activators at all. I'm hesitant to accept Curtis' answer here because it asserts the nitro group has a stronger electronic effect as being the reason. However, in o-isopropylnitrobenzene it also has a stronger electronic effect than isopropyl but isopropyl still determines direction.

@user55119 This is frustrating as according to my book the answer to #2 here (#2 from the 2nd question set) should be o/p to the methyl group instead of o/p to chlorine:
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Benzene/Electrophilic_Substitution_of_Disubstituted_Benzene_Rings
I may buy another text...that is what I generally do when I find one book is weak in explaining a concept. Thanks for your help, I appreciate how my question may be confusing.

@David Reed: Yes, example 2 in section 2 (o-chlorotoluene) is interesting in that this link https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm claims that the toluene is more reactive than chlorobenzene, which it is. Perhaps some effect having chlorine and methyl adjacent to each other. Perhaps it is a typo.

@David Reed: I just did a quick chem. Abst. search for the nitration of o-chlorotoluene. All examples give nitration para to the methyl group, which makes perfect sense. The example 2 discussed above is an error. Hope that restores your faith!