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Why iron reacts with steam to give Fe3O4 not Fe2O3?
Why $\ce{Fe}$ with $\ce{H2O}$ gives $\ce{Fe3O4}$ not $\ce{Fe2O3}$?
Like, $\ce{Zn + H2O \rightarrow ZnO + H2}$, why $\ce{Fe}$ doesn't simply gives $\ce{Fe2O3}$ when it reacts with steam?

Another similar question is : why does a strange compound like $\ce{Fe3O4}$ exist ? Why is Iron (plus one or two other metals) able to make an oxide containing the same element at two different oxidation states ? Why is it not the case for other metals like copper for example ?

An idea : did you compute the Gibbs enthalpy for either one of the two reactions?

Five words: Steam is not oxidizing enough.

And https://chemistry.stackexchange.com/questions/73275/why-is-the-product-of-the-reaction-between-iron-and-steam-is-iron-ii-iii-oxid

@NilayGhosh It seems to me neither of linked answers addresses the  reason why Fe2O3 is not formed.

Quoted picture from Researchgate.net for Ellingham diagram for iron oxides and hydrogen:

On the diagram above, you can see equilibrium partial pressures (expressed as oxygen chemical potential} of oxygen for particular redox systems, depending on temperature.
Each system has a line $\mu(\ce{O2})=f(T)$. A system  is able to oxidize systems with their respective lines below and vice versa.
We can conclude these consequences:
This reaction is thermodynamically favorable at all temperatures for the given range:
$$\ce{3 Fe2O3 + H2 -> 2 Fe3O4 + H2O}$$
This reaction is thermodynamically favorable below $\pu{900 K}$.
$$\ce{3 FeO + H2O-> Fe3O4 + H2}$$
The reaction  is thermodynamically favorable below roughly $\pu{1700 K}$.
$$\ce{Fe + H2O -> FeO + H2}$$
It should be obvious that

The reduction of water by iron stops at $\ce{Fe3O4}$.
Above $\pu{900 K}$, it stops even at $\ce{FeO}$.
Above $\pu{1700 K}$, hydrogen reduces $\ce{FeO}$ to iron.

Note that the respective temperatures are related to equilibrium states. The open system with the flow of steam reactant and removal of hydrogen somewhat shifts the reactions toward hydrogen as the product, so the effective decisive temperatures would be higher.
OTOH, for reduction of oxides by hydrogen, it is the opposite case and needed temperature would be lower.
I assume for $\ce{Fe + H2O <=> FeO + H2}$, for temperature near $\ce{1200 K}$ and above, the reaction  would go in large extend forward if hot water steam goes over hot iron, or backward, if hydrogen goes over iron oxides.

Consider the Schikorr reaction, proceeding in two stages:  "[A]naerobic oxidation of two Fe(II) into Fe(III) by the protons of water... [then] loss of two water molecules from the iron(II) and iron(III) hydroxides giving rise to... the formation of a thermodynamically more stable phase iron(II,III) oxide. Ma et al state,  that crystalline $\ce{Fe3O4}$  (magnetite) is thermodynamically more stable than amorphous $\ce{Fe(OH)2}$.
BTW, the stability of magnetite, and its adherence to iron, is used to protect steel from rusting. "Blueing... creates a layer of magnetite over the metal to prevent rust."
See also this question.

If protons of water were able to oxidize Fe(II) into Fe(III), then all of it would end up in Fe(III), either as Fe(OH)3 or as Fe2O3.

I  think the OP has in mind rather high T chemistry without hydrated forms of iron oxides,


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How to determine the value of the NOE from the NOESY spectrum?
I have 1D NOESY spectra from which I need to find out the value of the NOE (%).  Is it enough to simply measure the intensity of the resulting peaks or is it necessary to subtract it from the intensity of the standard spectrum?

What does your 1D NOESY look like? Is it one big peak (the proton that was irradiated) plus several small peaks (which often have a different sign compared to the original peak)?

Percentage NOE enhancements are very rarely used nowadays -- this refers to an older version of the 1D NOE where the irradiated proton is saturated and then a 1D spectrum recorded. You have to compare this against a normal 1D where the proton isn't saturated -- this is where the term 'enhancement' comes from. But modern 1D NOE experiments are transient NOE experiments which selectively excite the target proton, and any other peaks that appear arise solely from the NOE. There is no 'enhancement' to be measured here, only the presence/absence of a peak.

(Just for completeness' sake: of course, you can get quantitative information from transient NOE experiments, in particular, by repeating the experiment with multiple values of d8 — but I assume that is not what OP is looking for.)

From documentation (https://chemnmrlab.uchicago.edu/2021/03/10/1d-noe_1/):

Magnetization transfer by NOE between spins can be quantified. NOESY with D8 of just 10 msec (very short) yields almost no magnetization transfer, so the integral of the excited peak in this 1D spectrum can be used for quantitative reference when compared to peaks in spectra acquired with the same parameters (but longer D8 values).

So the denominator is the signal strength of the proton where magnetization originates (the peak you have to pick before running the experiment). You divide, not subtract.

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Which hydrogens are the most acidic in a THF molecule?
Which hydrogens are the most acidic in a THF molecule? I understand the main factors that influence acidity, but I can't seem to figure this one out and I have not gotten any answers on the internet.
My friend and I were joking around about chemistry and something about deprotonating an ether came up. That question really caught my attention. I was split between two answers for deprotonating THF:

One of the carbons bonded to the oxygen atom would be deprotonated due to a stronger inductive effect by the oxygen atom.

One of the carbons further away from the oxygen atom because the oxygen has a partial negative charge and some lone pairs, so deprotonating a carbon bonded to the oxygen atom would cause some electrostatic repulsion and lone-pair repulsion.

Is any of this correct or is there another explanation?

This is both pretty obvious and not really important, soo not a surprise you wouldn't find a ready answer.

According to this paper

The principal pathway for decomposition of THF by alkyllithiums is deprotonation at C-2, α to oxygen, to give 2, followed by a reverse [3 ю 2] cycloaddition of the resulting anion to yield one molecule of ethylene plus one molecule of the lithium enolate of acetaldehyde

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Does electroplating with copper takes less time than electroplating with zinc?
is electroplating with copper quicker than electroplating with zinc -- assuming that in both scenerios the cathode is steel, the surface area, the current, the thickness are the same, and the electrolyte is either zinc sulfate or copper sulfate? Thank you in advance.

There is an old saying "It can be quick, good and cheap. But not all at the same time."  Seriously, it depends much more on used particular technology than on used metal. There are dozens or hundreds of possible surface preparations, electrolyte compositions, geometry of the cell, used current densities..... Some are faster or better than others. You would have to compare particular techniques. But if everything is the same, plating by Zn is slightly faster (Zn molar mass, density)// More seriously, do not use StackExchange as an alternative to search engines. Do your search before asking.


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Can I make orthophosphoric acid from sodium dihydrogen phosphate and an acid which I can use to make ethyl iodide?
For my project I needed to di-substitute benzamide with ethyl group at N position to create DEB as a mosquito repellent. The first part of creating benzamide I was able to do, but the second part of substituting is a bit hard for me. I wanted to substitute using ethyl iodide as it was not a gas while the bromide and chloride was a gas.
Our lab apparently did not have the orthophosphoric acid to create hydroiodic acid in situ, so I resorted to using chloroethane, but I wasn't able to do it due to complications in my bubbling setup. Then I thought I could prepare orthophosphoric acid from sodium dihydrogen phosphate as it is available in our lab. Can someone help me on doing it,
or is there any other method to do it without the above methods?
The available materials are ethanol, $\ce{KI}$, $\ce{NaH2PO4}$, $\ce{HCl}$, $\ce{HNO3}$, $\ce{H2SO4}$, and there is no hydroiodic acid.

Isn't it easier to react benzoyl chloride with diethylamine?

I wanted to do it but I wasn't sure how to exactly di-ethylate ammonia and went ahead with making the amide.

Diethylamine is widely commercially available

Ethyl bromide is liquid, unless you need high temperature. Even cold chloride is.

Thanks @Mithoron but I was able to do it by the method given by Maurice. i didn't know it was EtBr was a liquid. but i wasn't able to EtCl as the benzamide wouldn't dissolve so i had to heat it and due to this i wan't able to use it in cold condition.

@Waylander it wasn't available in my lab as it was a pretty basic one.

Phosphoric acid can be produced from sulfuric acid $\ce{H2SO4}$, as it reacts with sodium dihydrogen phosphate $\ce{NaH2PO4}$ to produce phosphoric acid $\ce{H3PO4}$ according to :  $$\ce{NaH2PO4 + H2SO4 -> NaHSO4 + H3PO4}$$ Phosphoric acid can be used to produce $\ce{HI}$ even if it is mixed with $\ce{NaHSO4}$.

Thank you @Maurice. I was able to do it using the above method. The output came out with good yield and it even resulted in good looking crystals of DEB.


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Selective formation of enolates
I saw the following reaction on Modern methods of Organic Synthesis by Carruthers. The reaction is about the selective alkylation of trimethylsilyl enolates, and is shown below.

However, I have some doubts regarding the formation of the enolate in the first step.

Why does deprotonation occur from the alpha carbon on the left, and not from the gamma carbon? Is it due to steric repulsion between lone pair of oxygen and negative charge on Nitrogen? But there two gamma carbons, and only one would be hindered?

The condition shown is the kinetic condition (LDA,THF, -78 °C). So the given enolate is the kinetic enolate. Is it classified as a kinetic enolate due to the lesser number of resonance structures (less stable alkene)?
(Normally, if the enolate is less substituted, it is termed as kinetic, and more substituted one is the thermodynamic enolate)

Should the carbanion stability of each resonance hybrid be accounted to determine the type of enolate?

Kindly provide answers to all questions precisely. Otherwise, please do not attempt. I really need clear answers. Thanks in advance.

A few comments are in order as to the way in which the enolate formation is conducted. Lithium diisopropylamide (LDA) is formed in situ from the free amine and n-BuLi at -78oC at which temperature the ketone, in this case mesityl oxide, is added slowly as the minimal reactant (excess LDA). Subsequent addition of trimethylsilyl chloride produces the enol silyl ether. This mode of addition minimizes enolate exchange (α $\rightarrow$ γ). Excess ketone runs the risk of proton exchange with formation of  the thermodynamic enolate.
Question 1:

Why does deprotonation occur from the alpha carbon on the left, and not from the gamma carbon? Is it due to steric repulsion between lone pair of oxygen and negative charge on Nitrogen? But there two gamma carbons, and only one would be hindered?

While the electron pairs on oxygen and LDA are bases, the lithium cation can coordinate with the oxygen just as it does with the solvent tetrahydrofuran (THF). Incidently, the rate of the deprotonation of 2,2,6-trimethylcyclohexanone has been shown1 to be first order in ketone and one-half order in LDA.
Question 2:

The condition shown is the kinetic condition (LDA,THF, -78 °C). So the given enolate is the kinetic enolate. Is it classified as a kinetic enolate due to the lesser number of resonance structures (less stable alkene)?
(Normally, if the enolate is less substituted, it is termed as kinetic, and more substituted one is the thermodynamic enolate).

The less substituted enolate is the kinetic one but not because the $\ce{C1}$ methyl is less substituted. Kinetic has to with the rate of formation. Both the kinetic and thermodynamic enolates have single resonance structures but two canonical ones in the former case and three in the latter.
Question 3:

Should the carbanion stability of each resonance hybrid be accounted to determine the type of enolate?

In a sense, yes. The γ-enolate (thermodynamic) has greater delocalization of the negative charge than does the α-enolate (kinetic).

Why is the methyl group deprotonated kinetically? Comparison of the liquid film infrared carbonyl of the α,β-unsaturated ketones 1-3 provides a clue.3 The smaller the carbonyl stretching frequency the more single bond character in the carbonyl group. The enone moiety in cyclohexenone 3 has strong conjugation between the carbonyl and the double bond owing to restricted rotation about the $\ce{C1-C2}$ bond. The methylene group is deprotonated kinetically.2 Enones 1 and 2 have a degree of free rotation about their respective $\ce{C2-C3}$ bonds. Steric interactions between $\ce{C6}$ and the carbonyl oxygen or $\ce{C1}$ causes mesityl oxide 1 to have a higher frequency than pentenone 2, which is missing the $\ce{C6}$ methyl. Thus, in 1 and 2, the γ C-H bonds are out of conjugation to a greater extent than the methyl α C-H bonds, which are always available for overlap with the carbonyl group.

M. Majewski and P. Nowak, Tetrahedron Letters, 1998, 39, 1661.

W. D. Shipe and E. J. Sorensen, Organic Letters, 2002, 4, 2063.

SDBS DataBase of Organic Compounds

Please check the corrections "Enones 1 and 2 have a degree of free rotation about their respective C1-C2 bonds". Also, "Steric interactions between C6 and carbonyl oxygen on C1 causes mesityl oxide 1 to have a higher frequency than pentenone 2"


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Why can s and p orbitals of one atom form hybrid orbitals but the overlaping of s & p orbitals of two different atoms don't form a molecular orbital?
So my question is in the title : Why can s and p orbitals of one atom form hybrid orbitals but the overlapping of one s orbital and one p orbital (perpendicular to the bond axis) of two different atoms don't form a molecular orbital (MO)?
I'm confused because I see both resulting orbitals (hybrid and non-existent MO) as a linear
combination of atomic orbitals (LCAO) and that in the case of Molecular orbital theory (MOT), I learned that atomic orbitals of different atoms can interact to form molecular orbital only if they have the same symmetry (and are close in energy) so that their overall overlapping is positive or negative (=non zero).
If someone could explain where I'm wrong or what I see incorrectly, it would be very helpful. Thanks

This actually isn't a trivial question. Hybridisation is just mathematical tool and shouldn't be used within the MOT context. MO are by design delocalised. With this in mind, of course s and "perpendicular" p orbitals will overlap and my form a MO. This, however, might not be the most stable state, or might not lead to a lower lying state. The reason for this is basically the cancellation of two effects.

$s$ and $p$ atomic orbitals can make hybrids if and only if their energies are identical or nearly identical. It can happen in a given atom. It cannot happen between different atoms.

Thanks but do you have an explanation of why the symmetry of the atomic orbitals don't matter to fhe hybridization but do for bonding ?

The symmetry has nothing to do in this problem. It is only a question of energy. The energy of the two orbitals to be hybridized must be equal, or nearly equal, as they are both solutions of the Schrödinger equation.

This is really not a good answer, if any at all, for this question. First and foremost there is the issue, that hybridisation is a mathematical concept that is completely independent of the energy. Then that there are no hybrid orbitals in MOT. It does not even consider the part about the the overlap of orbitals between atoms, which I think is key.

@Martin Yes, Hybridization is a mathematical concept, that is independent of the energy. But hybridization can only occur in combining wave function which has the same energy as solution of the Schrödinger equation. Orbitals 2s and 2p can hybridize, because they produce the same or nearly the same value of energy in their solutions of Schrödinger equation. Orbitals 2s and 3p can never make hybrids because their energies are rather different

The way you describe this is very fuzzy. Hybridisation is a concept from Valence Bond Theory, so using this in the same context of MO Theory is an approximation at best. If we stay in the framework of MOT, then Molecular Orbitals will/ay form. You can later use localisation models to approximate contributions and infer hybridisation from this. But that is not the same as starting with it from scratch like in VBT. And it is certainly not the reason for anything. When we stay within VBT then yes, we could only describe meaningfully sp^x hybrids from A= similar in energy.

@Martin. Sorry I have never spoken of MO theory. Hybridization happens in an atom, not in a molecule. It does help explaining the shape of the molecule. But it is an atomic property.

No. That is completely wrong. Hybridisation does not happen. It is a only a mathematical description. And it does not make sense to use it at all if there is no external field. And you do not have an external field, if you're not viewing the atom in the molecular context.

Of course hybridization is a mathematical description that does not happen really. The geometry of the molecule can be expressed without hybridization, with just s and p orbitals. I remember having heard the famous E. B. Wilson speaking about this, a long time ago. But he also showed that the geometry is much simpler to explain with hybridization. What do you mean with external field ?


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How does the concept of the electron shell (also called energy level) and the orbital coexist?
Running into a bit of confusion trying to understand the atomic structure and the root cause is mostly related to orbitals and energy levels.
At a basic level, I was taught that electrons orbit the nucleus in fixed, circular orbits called “electron shells.” New information that I’ve learned states that there is some uncertainty in the positions of electrons at a given time. For example, although the one electron in a Hydrogen atom orbits the nucleus at the first energy level, its position is uncertain, that is, it can be further away from the nucleus or much closer.
To show where electrons are most likely to be found, orbitals are used which depict a region in which an electron will be 90% of the time.
The problem in my understanding lies with energy levels. The first energy level has an orbital called 1s and has a spherical shape with the nucleus at its center. How could an electron be classified as being in the first energy level when it’s possible that it can be found right near the nucleus? Similarly, the second energy level contains a spherical orbital as well called 2s. How could we say an electron is at the second energy level when it can very well be found much lower than that energy level according to the 2s orbital? It’s hard to see how orbitals and energy levels exist at once: one says an electron can be found near the nucleus whilst the other says it must be orbiting in the second energy level.
I’d greatly appreciate it if the explanation is sufficient enough for an As level student.

There are no circular orbit for electrons. Electrons have no orbit, no movement, no speed. If they had a movement, they would progressively loose their energy by radiation, and finally fall on the nucleus. Apparently you have never heard about Schrödinger equation, eigenfunctions and eigenvalues. This is necessary to understand what is an electron, an orbital, and the energy associated to this orbital. I could of course explain it to you. But this would take time. And it is quite a whole chemistry course that would be required to develop all this theory. So I stop here. Unfortunately.

Welcome to the strange world of quantum mechanics! Although not zero, the probability of the electron striking the nucleus is very small, in part because the nucleus is very small. But it is not zero and when it happens it can trigger a transformation of the nucleus: https://en.wikipedia.org/wiki/Electron_capture This was studied by Luis Alvarez, won a Nobel prize for this work. His son happened to lay down the foundation for the currently accepted theory explaining the extinction of the dinosaurs.

Mathematically, the probability of finding the electron at one point is vanishingly small, due to the delocalized ("uncertain distribution") nature of the wavefunction (a name for the function that describes an orbital). Uncertainty in quantum mechanics is a strange thing, by the way, not quite the same as we are usually familiar with.

@Maurice If electrons did not have speed then there would not be needed relativistic corrections in relativistic quantum chemistry. Nonrelativistic prediction of Hg melting point is +80 deg C. Relativistic one is about -38/-39 deg C. Aside of that Schr. wave eq. model of H atom starts with the classical model of energy of charged particle (1/2mv^2 - e^2/(4.pi.eps0.r) near a central quadratic force.

@Poutnik. When I say that the electron has no speed, it does not mean that its speed is zero. It means that the electron has no speed at all, a little like the fact that an electron has no religion. If the electron has no speed, it still has a kinetic energy, and this kinetic energy is sensitive to relativistic effects. This kinetic energy is not (1/2)mv^2. No, after Schrödinger, the kinetic energy is the second derivate of the wave function multiplied by constants.

@Maurice But that is rather the consequence of using Hamiltonian mechanics instead of the Newton one. In the latter, E_k=0.5mv^2. In the former, v=sqrt(2E_k/m)  (all nonrelativistically, in Sch.Eq. context)

@Poutnik. Your definition of E and v are similar. But in fact, the electron has no speed. It is not sqrt(2E_k/m) as written in your comment. It is not any other formula. Electron has no speed at all. So it has no use to calculate something that does not exist.

Funny thing is Dirac admitted relativistic corrections for electrons in light atoms are not needed because their speed is about 0.01c.

I agree, Dirac and also Pekka Pyykkö were able to state that the speed of the electron was much smaller than $c$. But this speed is not related to a point (x,y,z) like in Newtonian mechanics.

There always coexist several models for historical and educational reasons. Especially  the latter is based on several levels of trade off between accuracy and simplicity.
If we take mechanical work:

I was taught in elementary school physics lessons $W=F \cdot L$.
I was taught in high school physics lessons $W=F \cdot L \cdot  \cos{\text{(alpha)}}$.
I was taught in university physics lessons $W=\int_{L}{\vec{F}\cdot \text{d}\vec{L}}$.

The electron shell education model is based on obsolete Rutherford and Bohr atomic models, with simple energy levels indexed by natural numbers. The notion of orbits and shells is easily understandable by young students.
For a hydrogen atom, energy levels for the same main quantum number $n$ ($\text{1s}$),($\text{2s 2p}$), ($\text{3d 3p 3d}$)... are almost exactly the same (Originally thought the same but there are slight deviations explained later).
For atoms with more than one electrons, energies of "sub shells" $\text{s, p, d, f}$ differ and the concept of energy shells $n=1,2,3,4,..$ breaks down. Energy levels further differ even withing sub shells in presence of external electromagnetic fields.
The orbital model is based on more advanced quantum mechanics, using solving of Schroedinger wave equations of electrons. Orbitals can be understood in three different but related meanings as:

orbital(1) as a particular solution of the above equation, introducing particular integer parameters $n$,$l$,$m_\text{l}$ as "quantum numbers".
orbital(2) as a particular quantum state of an electron, described by orbital(1).
orbital(3) as 3D region with probable ooccurance of electrons, described by orbital(1) and orbital(2).

There is one important fact. Orbitals(3) as 3D regions in large extent overlap. Purely mathematically, considering probability density converging to zero in infinity, they fully overlap.
It means, no point around a nucleus belongs to a particular orbit nor a particular energy level nor a particular orbital. Electrons from any orbital can occur at any point, but with different probabilities (with exception of 3D node surfaces where their probability of occurrence is zero).


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Sign of ∆S in conversion of diamond to graphite
It is said that ∆S is positive as randomness is increasing from diamond to graphite as in diamond molecules are tightly packed together but we know that ∆S=∆H/T for phase transition and that conversion of diamond to graphite is exothermic which indicates ∆H as negative so shouldn't ∆S be negative?
(Correct me if I am wrong.)

The standard expressions for the Gibbs free energy of carbon are found in Per Gustafson, Carbon 24(2) 169-176 (1986), and collected (with many other elements) in AT Dinsdale, Calphad 15(4) 317-425 (1991). In general they are in the form of a power series in temperature $T$,
$G = a + bT + cT\ln(T) + \sum dT^n$
Then, using
$S = -b -c -c\ln(T)-\sum ndT^{n-1}$
one can evaluate $S$ from the expression for $G$.
In Dinsdale the Gibbs free energy for diamond relative to graphite is
$1009 + 4.88T -0.01 T \ln(T) +135400T^{-1} +33.0\times 10^5 T^{-2} -9\times 10^8 T^{-3}$
Then the difference in entropy $S$ is given by
$-4.88 + 0.01 + 0.01\ln(T) +135400T^{-2} +66.0\times 10^5 T^{-3} -27\times 10^{8} T^{-4}$
Plotting this in your favorite plotting program will show that the entropy of diamond is less than that of graphite at all reasonable temperatures at standard pressure.


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When does "always add acid" not apply?
I've been taught a LONG time ago to AAA "Always Add Acid". I've recently come across what feels like an exception to that rule when making piranha solution (you add the H2O2 to the acid).  I'm not a chemist so this surprised me. I'm wondering, in general, when should I not add acid to solution and instead add solution to acid?
Of course the safest thing is to always read SOPs when doing something I've never done, but I'd like to understand the exceptions so I can sanity check things too.

This rule of thumb is usually irrelevant, unless you're diluting highly concentrated acids. It shouldn't be repeated in such vague form, but I guess it's catchy.

https://chemistry.stackexchange.com/questions/155710/why-should-hydrogen-peroxide-be-added-to-sulfuric-acid-and-not-vice-versa https://chemistry.stackexchange.com/questions/893/why-add-water-first-then-acid

AAA alludes to concentrated (98%) sulfuric acid. Other mineral acids already contain significant amounts of water. You would also normally add them to water, but with H2SO4, it's not just dangerous, but guaranteed to go wrong, badly, every time. Other mineral acids, it's not even really dangerous.

The said rule does not apply when the reasons behind it fail to apply, and understanding those reasons is a mandatory prerequisite to actually doing such things in practice.

Most mineral acid solutions are denser than water. Adding tge acid -> it tends to sink through, which improves mixing.

The true duplicate is rather one of the Mithoron links
why-should-hydrogen-peroxide-be-added-to-sulfuric-acid-and-not-vice-versa

For the major case for application of the "Always Add Acid" rule, the reasons are obvious. Adding water to concentrated sulfuric acid usually leads to local overheating of water. That causes local violent boiling and splashes of concentrated acid. That is avoided by adding the acid to water as there is a lot of heat capacity to avoid boiling.
From personal experience, it is good to do it in two steps, if applicable, as it is less challenging, considering cooling and manipulation with hot acid solutions.

Pouring the rest of the older diluted acid to water and cooling it down.
Pouring the concentrated acid to this new mixture.

When concentrated sulfuric acid and $\text{30+ %}$ hydrogen peroxide are to be mixed, it is choosing the smaller evil of pouring the hydrogen peroxide to the acid. This still can cause eventual smaller splashes of the acid or surface hydrogen peroxide decomposition. But it avoids local overheating and dehydrating of hydrogen peroxide solution. That could destabilize it and making it violently decompose in the whole volume, splashing most of the liquid.
Note that sulfuric acid has kinetically stabilization effect due acidic environment and at the same time destabilization effect by heating up and bonding with water. That latter increases the thermodynamic activity of the peroxide.
Preparation and manipulation of this piranha solution is always dangerous and should be reserved to experienced chemists, considering all needed precautions. It is not called so without a reason.


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How to find the formula of a binary compound if we know the mass of solid product and the volume of escaped gas?

During the decomposition of a binary compound two simple substances
were formed: 44.8 grams of metal and 47.04 liters of a non-flammable
gas (conditions are normal). Find chemical formula of the reagent.

Dear community, of course I tried to solve it myself, so I think I did put some effort. Firstly, I calculated the quantity of substance of the escaped gas:
ν(gas) = V(gas) : Vm(gases) = 47.04 l : 22.4 l/mol = 2.1 mol.
I have a strong hypothesis that the escaped gas must have diatomic molecules. We know noble gases have monoatomic molecules, but they can hardly be products of any reaction.
So I tried writing my equation as: kMₐXb -----> akM + (bk/2)X₂↑
(Sorry, I have no ideas about how to write a subscript letter b.)
After this step, it became more like a game of guessing. I tried to divide the number 44.8 by different quotients of 2.1 over multiple natural numbers:
44.8 : 2.1 = 21.333 — apparently no such element in sign;
44.8 : (2.1 : 2) = 42.666;
44.8 : (2.1 : 3) = 64 — here is at least something good, 64 is copper. However, copper cannot have a valence of six.
44.8 : (2.1 : 4) = 85.333 — rubidium?
I am not sure at all if I must continue. May rubidium be the correct M? No clear ideas how to find  X though.
What am I missing and how to solve my problem? Help really needed.


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Difference between acid-catalysed iodination of acetone and the iodoform reaction
I am curious about what makes the iodoform reaction (in alkaline conditions) substitute 3 times whereas the iodination of propanone (acidic conditions) only substitutes once. Or am I just completely misinformed regarding this and the acid catalysed reaction can actually substitute multiple times? Thanks a lot from a student who is just very curious about why in this reactions the acidic or alkaline conditions can change the reaction so much.


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How to figure out concentration of ingredients in solution?
I have a solution that contains 3 inactive ingredients: ethanol, propylene glycol, purified water and 5% of an active ingredient.
What would be the smartest way to figure out the concentrations of the 3 inactive ingredients? Ideally something I can do at home without spending too much money.

Smartest way would be to ask. Getting  the composition of 4-component mixture at home is not probable.
But some approximate estimations can be done if some basic measurement techniques are available.
If the active ingredient is not volatile, careful evaporation may give by the mass decrease the mass sum (alcohol + water) and (propylene glycol + the active one).
If the latter is (assumed) small enough, you can estimate alcohol content from the density as the ration of the mixture mass and volume, with some density correction for the nonvolatile rest(the best if its volume can be measured too).
Illustrative example: Let assume you were able to measure the required parameters:

Let assign symbols P=product, W=water, A=alcohol, PG=propylene glycol, I = active ingredience.

Let assign symbols $m_\text{P}$, $V_\text{P}$, $\rho_\text{P}$ to mass, volume and density of the product.

Let assign symbols $m_\text{W}$, $V_\text{W}$, $\rho_\text{W}$ to mass, volume and density of the W+A product part.

Let assign symbols $m_\text{I}$, $V_\text{I}$, $\rho_\text{I}$ to mass, volume and density of the PG+I product part.

The volume $V_\text{P}=\pu{100 mL}$

The mass $m_\text{P}=\pu{95 g}$

The density $\rho_\text{P} =\frac{\pu{95 g}}{\pu{100 mL}} = \pu{0.95 g mL-1}$

You have let it evaporate, freely or in water bath until it get about constant mass $m_\text{I}=\pu{10 g}$.

That means, assuming the active ingredient is not volatile enough, the mass $m_\text{W}=m_\text{P} - m_\text{I}=\pu{95 g} - \pu{10 g}=\pu{85 g}$.

Propylene glycol has density $\pu{1.036 g mL-1}$. For simplicity, let assume the same for the active ingredient, so $\rho_\text{I}=\pu{1.036 g mL-1}$

Let assume approximation of volume additivity so if the product is mixed, then $V_\text{P} = V_\text{W} + V_\text{I}$.

$V_\text{W} = V_\text{P} - V_\text{I} = V_\text{P} - m_\text{I}/\rho_\text{I}=\pu{100 mL} - \pu{10 g}/\pu{1.036 g mL-1} \approx \pu{90.3 mL}$

Density of the W+A portion is then $\rho_\text{W}=m_\mathrm{W}/V_\mathrm{W}=\frac{\pu{85 g}}{\pu{90.3 ml}} \approx \pu{0.94 g mL-1}$

From the Ethanol-water density table, it has weight percentage $w \approx 40\ \%$.

The contexct of the ingredients is then:

Name
Mass

Water
$m_W \cdot 60 \%/100=\pu{51 g}$

Alcohol
$m_W \cdot 40 \%/100=\pu{34 g}$

Propylene glycol
$m_\text{I} - \pu{4.75 g} = \pu{5.25 g}$

Active ingredient
$m_\text{P} \cdot 5 \%/100=\pu{4.75 g}$

Thanks for the elaborate response! Just to recap: The idea is basically that PG has a lower evaporation rate than alcohol+water, so we can reduce the solution to PG + I only by evaporating all the A+W, correct? If so, how will I know when to stop the evaporation, i.e. when will I know I have reached the point where all A+W is gone? Btw, the active ingredient is Minoxidil, so 3g in a 60ml solution.

Recap: correct. It is expected to weight it periodically, when evaporation slows down, until it is like stops.

As it looks this is about medical application I advice against experimentation. I reserve myself right to refuse responsibility to any misuse and undesired effects, as I have warned you not to do so. Personal medical advices are explicit off topic on this site for these reasons.


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Is ΔH = ΔU + ΔPV only valid for gases?

In this question, formula ΔH = ΔU + ΔPV is used. In the solution, ΔPV is replaced for RTΔng. As there are solid products also in this equation, don't we consider ΔPV for them. Also if there were liquid products will we consider it's ΔPV?

We usually consider the volume of pure solids and liquids to be constant while solving such problems.

We often, even usually, consider the volume constant. But not always - e.g. for some phase changes in solids the volume change can be quite significant.

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$\Delta H = \Delta U + \Delta (pV)$ is always valid, as it is the alternative way of the definition $H = U + pV$.
But $\Delta (pV) = RT\Delta n_g$ is at constant T exactly valid only for reactions of ideal gases.
For real gases, regardless  of if all or just some components are gaseous, it is valid approximately. Contribution of condensed phases is then neglected.
If a reaction has no gaseous component, or if there is no change in total amount of gas, the approximation $\Delta (pV) \approx RT\Delta n_g$ is not applicable. OTOH, the term $\Delta (pV)$ is often negligible. But at specific cases like involving thermal dilation and compressibilty of condensed phases at constant volume, the $\Delta (pV)$ term can be very significant.


text:
Semiconductors and their electronic bands
While studying colors of sulfides and searching for the reason why they are black I came across this question:
https://chemistry.stackexchange.com/questions/138309/the-color-of-the-most-sulfides-of-p-and-d-block-elements#:~:text=The%20colours%20tend%20to%20be,of%20the%20transitions%20is%20small.
In it, the top commenter explained the fact that those sulfides are black with the valence band model, because their band gap is so small, they can absorb every visible wavelength of light.
I thought the valence band model only applied to metals, whose atomic orbitals mix with every other atomic orbital in the solid and therefore have an infinite amount of close lying orbitals, thus forming a band. How are ionic solids then able to do the same, when they are composed of different ions? Or does the theory work for all kinds of condensed matter.

Energy bands in solids apply to all crystals.

Even to organic frameworks for example??

Yes, but their conduction band is empty and sits pretty high above.

Broadly speaking, any solid bonded together has overlapping electron wavefunctions with often small energy differences between them. For silicon, the electrons involved with bonding most certainly are delocalized, as are the free carriers (electrons and holes) - the same solid state physics explains insulators, semiconductors, and metals.

Oh okay that makes sense. But in organic molecules, there are different kind of orbitals; for example consider solid ethylene, so the sigma bonds to the hydrogens form a band and the pi bonds form another or do the sigma and pi bonds form the same band together?

As indicated in some comments, all solids, and for that matter all condensed matter, has a band structure. If atoms are packed densely enough, the orbital overlaps that produce "molecular" orbitals become interconnected throughout the volume of the material, and you have bands of energy levels that are filled with closely spaced orbitals. It is true that having discrete molecules, or an ionic structure where the anions tend to hold all the valence electrons, can limit the delocalization of electrons and thus produce narrow bands with wide gaps between them. But the band structure is still there. In the case of water, understanding the band structure, even of a molecular material in the liquid state, is essential to understanding electrochemical reactions commonly carried out in that medium (seehere).
The impact of band structure on color is illustrated in black and white, literally, when we compare magnesium silicide ($\ce{Mg2Si}$) with sodium sulfide ($\ce{Na2S}$). Both have eight valence electrons per formula unit and an antifluorite structure, yet their colors are as shown below. Can you tell which is which (without clicking the attributions) in the pictures below?

Compound A

Compound B
Sodium sulfide, with a large difference in energy between the atomic valence orbitals of the two elements, has its valence electrons largely localized onto the more electronegative sulfur atoms, as in a classically ionic compound; with little delocalization the bands are narrow and the gaps so wide that sodium sulfide can barely absorb visible light. In fact the yellow color seen here actually comes from polysulfides, whose sulfur-sulfur bonding broadens the sulfur-based band and narrows the bandgap somewhat. Pure sodium (mono)sulfide is white.
Not so with magnesium silicide, whose much smaller difference in energy levels between magnesium and silicon atoms facilitates a great amount of electron delocalization and a narrow bandgap. The bandgap of magnesium silicide, like other black semiconductors, is so narrow that it actually falls into the IR range; such materials,including magnesium silicide itself, are candidates for thermoelectric applications.

Nice answer and I didn’t even know that band theory also applied to liquids! In university courses, for example ochem, one only looks at MO‘s of single molecules, thats why I was confused. One question though, how can one determine which orbitals are in the bands exactly. For example, there are many different bonds in hydrocarbons (sigma-H bonds, heteroatom bonds, pi bonds), or do those bonds independently form their own bands?

@Mäßige No, their molecular orbitals.

Okay then, their molecular orbitals then. Do those form the bands independently

Metals have a "sea" of electrons at the lowest level, i.e., even at room temperature or close to absolute zero. A potential of even 1 µV will cause a flow of charge in Cu. Metallic conductivity decreases with increasing temperature due to vibration disturbing the regularity of the structure. At operating temperature, a tungsten filament might have ten times its cold resistance.
Semiconductors' electrons need a "boost" to move up and out. In Si diodes, for example, at room temperature, it might take 0.7 V to overcome the barrier between p- and n-type (doped) Si. However, conductivity of semiconductors increases with increasing temperature, as electrons are knocked loose.
It also depends on allotrope, too. For example, in a diamond lattice, undoped C is a nonconductor at room temperature (though doped, it is a semiconductor), but in planar graphite, it's a conductor.
BTW, there are many colored sulfides (and oxides). Cinnabar ($\ce{HgS}$) was used to make vermilion, a bright orange-red. Covellite ($\ce{CuS}$) is blue. Orpiment ($\ce{As2S3}$) is yellow.

And this band structure applies to all crystals, be it organic compounds, silicium, carbon etc.?

Graphite is just a conductor, no semi. Also doped diamond can be semiconducting.

@Mithoron, the answer has been edited. The band-gap for diamond is ~5.5 eV, so it has high potential (P.I.) for high temp devices. Interesting to think of oil-well loggers using diamond chips -- rather carbon intensive.


text:
Analyzing Effusion Rates in Gas Mixtures: A Comparative Approach to He and CH4 and problems
Let's start it with a question :

Suppose there is a mixture of $\ce{He}$ and $\ce{CH4}$ Gas in a container of volume $V$  and a certain temperature with a small hole at a wall of the container, the ratio of moles of $\ce{He}$ to $\ce{CH4}$ is $3:1$. What would be the ratio of rate of effusion of helium( $R_He $) to the rate of effusion of the whole
gaseous mixture ($R_mi_x)$) ? ( assuming ideal gases and Molar Mass of the gaseous mixture (avg) is 7 g/mol)
Although it can be resolved by applying the rules on He gas and the gaseous mixture directly, I choose to solve it using a different method.
So, We know that
$(R_A/R_B) = (n_a /n_b)$ $\sqrt{M_B/M_A} $
where molar masses $(M_A and M_B)$ and the number of moles $(n_A and n_B)$ of each gas .
So
$R_\ce{He}/ R_\ce{CH4} = 6/1 $
Now, Let at the initial moment,
$R_\ce{He}$ =  $6x$ moles/sec and
$R_\ce{CH4}$ = $x$ moles/sec ( where x is any +ve number)
Now, in a very small time interval "$dt$",
Mass effused by $\ce{He}$ = $4 (6x)(dt)$
Mass effused by $\ce{CH4}$ = $16(x)(dt)$
$\ce{Total mass effused = 4(6x)(dt) + 16(x)(dt) = 40x (dt) gm}$
and Molar Mass of the gaseous mixture(avg) (let's call it Gas B)  is 7 g/mol
so The moles of the Gas B (gaseous mixture) effused in dt are simply = $40xdt/7$
Therefore, Rate of Effusion of the gaseous mixture must be $40x/7$ moles/sec
and Hence,
$R_\ce{He}/R_\ce{mix} = 6x/(40x/7) = 42/40$
right ?
but this answer doesn't matches with the given answer key :
$R_\ce{He}/R_\ce{mix} = \sqrt{7}$ * $(3/8)$ and which is the correct answer
I get what they did, which makes sense (they directly applied the ratio law to $\ce{He}$ and the mixture)
but My doubt is  what's wrong with solving with above approach ? Am I missing something ?
I'd value any response that attempts to increase my understanding in this.

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@Poutnik I've made edits accordingly.

You can also use $\frac{R_\text{He}}{R_\text{mix}}$ for $\frac{R_\text{He}}{R_\text{mix}}$ or $\dfrac{R_\text{He}}{R_\text{mix}}$ for $\dfrac{R_\text{He}}{R_\text{mix}}$

The supposedly "correct" answer is actually wrong. The answer key claims that the helium effuses at $3\sqrt{7}/8 = 0.99$ times the rate of the overall mixture. Applying the same logic to methane, we would calculate that methane effuses at a rate of $\sqrt{7}/16 = 0.17$ times the rate of the overall mixture. This is nonsense. These are the only two gases in the mixture, so the rates of effusion should sum to $1.0$ times the rate of the mixture. But $0.99 + 0.17 > 1.0$, so this method can't be valid.
When there are nonlinear effects (for example, in diffusion or effusion), don't expect a mixture to behave like a simple weighted average of its components.
Your proposed answer is also wrong, though much of your reasoning is correct. You made an error with the units. You calculated the helium effuses at a rate of $6x\ \mathrm{mol/s}$. The overall mixture effused at a rate of $(40/7)x\ \mathrm{g/s}$. You then divided the first number by the second, giving an answer in mol/g, which is not the pure molar ratio you're looking for.
The problem is actually quite a bit simpler than that. You identified that in this mixture, the molar rate of effusion of helium is 6 times that of methane. In other words, for every 7 molecules effusing, 6 are helium and 1 is methane. Therefore, the molar rate of effusion of helium is 6/7 that of the overall mixture.


text:
What's this reaction converting 9,10-dihydroanthracene into anthracene
I have no idea what this reaction converting 9,10-dihydroanthracene into anthracene is. Could someone help me with the name of the reaction/ or explain the mechanism of the formation of the product? 9,10-Dihydroanthracene has been heated in the presence of sulphur (S) to give anthracene. How?

It's likely a radical mechanism.

The rxn. is a2-electron oxidation of the substrate. What happens to sulfur?

https://chemistry.stackexchange.com/questions/77342/synthesis-of-anthracene-mechanism

https://chemistry.stackexchange.com/questions/61545/birch-reduction-of-anthracene https://chemistry.stackexchange.com/questions/173843/oxidation-mechanism-of-a-peroxide-reacting-with-a-sulfide-to-form-anthracene


text:
A reduction of an ester with DIBAL-H to alcohol
I will do a reduction with DIBAL-H and after I need to do the workup of this reaction. I have to quench the reaction with MeOH at 0°C.And I will let the reaction return to rt. As shown below :

I'll then add satured NaCl and Et2O and I'll then filter that in celite.
The thing is that I don't understand why I should filter in celite and what is the goal of that. Is it to remove subproducts ?
Here is the link to the work that I am trying to reproduce : Gropp, C.; Fischer, S.; Husch, T.; Trapp, N.; Carreira, E. M.; Diederich, F. Molecular Recognition and Cocrystallization of Methylated and Halogenated Fragments of Danicalipin A by Enantiopure Alleno-Acetylenic Cage Receptors. J. Am. Chem. Soc. 2020, 142 (10), 4749–4755. https://doi.org/10.1021/jacs.9b13217.
Thank you in advance for your indication

Aqueous work up of aluminium-based reducing agents results in the formation of aluminium hydroxide residues which form fine suspensions/gels. These do not separate easily from your organic phase unless you take measures to remove them. Celite filtration should give you a clear two phase system that you can separate.


text:
Can pure hydronium exist?
Can pure hydronium exist? If not, why not? It seems to me (and I am no chemistry expert) like pure hydronium should have the theoretical maximum acidity or minimum pH that a substance can reach.
Can't, for example, we obtain pure hydronium if we could somehow produce a container with a certain amount of pure ionized hydrogen (H⁺), and mixed it with an equal number of moles of water (H₂O) to produce pure hydronium (H₃O⁺)? Or if that example is too theoretical, couldn't we also simply mix an equal number of moles of hydrochloric acid (HCl) and water (H₂O) to produce something close enough, hydronium (H₃O⁺) with chloride ions (Cl⁻)?

Sure, good intuition! read about it here:  https://en.wikipedia.org/wiki/Hydronium_perchlorate An analogous compound might be ammonium chloride. This can be made as a cloud mixing HCl and NH3 vapors. I am not certain that water vapor isn't a catalyst for the reaction because the vapors are usually generated from concentrated hydrochloric acid and concentrated ammonia aqueous solutions.

What do you mean by "pure"? Electrostatic repulsion might be a problem.

See: https://en.wikipedia.org/wiki/Hydronium#Solvation

Does this answer your question? Solid salts of hydronium ion

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

$\ce{H3O+}$ is a ion – a charged particle.
As an isolated ion, "naked" in vacuum or gas, or hydrated in water, it is a common thing. But with macroscopic quantities, it is more complicated.
If we hypothetically collected about $\ce{20 g}$ of hydronium, it would contain electrostatic charge of about 1000 CG+ lightnings 10 km long. It would cause a gigantic Coulombic explosion.
OTOH, hydronium balanced with counter anion of a strong acid, dissolved in water, is a common molecular entity chemists works with. There can be easily $\pu{200 g/L}$ of hydronium ions.
Note that crystalline monohydrate of some very strong acids is de facto  a hydronium salt $\ce{[H3O+][A-]}$. It happens with some exceptions for acidity constants $\mathrm{p}K_\mathrm{a} \le –9$.

It's theoretically possible to create a diffuse plasma with H3O+ and e- being the dominating particles. Not sure if there is a range of conditions where it would be thermodynamics stable.

It is possible. But I bet OP is interested in more standard conditions.

Absolutely, and it's not even difficult. Sure, in the dilute gas phase naked $\ce{H3O+}$ is commonly found simply due to the relative ease of formation and abundance of $\ce{H+}$ and $\ce{H2O}$, in space and in your friendly neighbourhood mass spectrometry facility. But you don't need anything nearly as exotic.
As long as you have a sufficiently strong acid, its ionization in the presence of water is so favourable that in the right conditions it will force just about every single water molecule it touches to accept a proton and become a hydronium ion. Clear evidence of this is found in solid hydronium salts, where you can even directly determine the presence of hydronium cations through x-ray crystallography. These salts will often form with an exact 1:1 stoichiometric mixture of water and acid molecules.
Now, you might think it takes some crazily strong acid to do it. However, a nice example is para-toluenesulfonic acid ($\ce{TsOH}$), one of the most common organic acids in a chemistry laboratory, which is almost always encountered as the monohydrate, $\ce{TsOH.H2O}$. What not many people seem aware of is that "p-toluenesulfonic acid monohydrate" is perhaps more precisely called hydronium p-toluenesulfonate, $\ce{H3O+TsO^-}$. It's an easy-to-handle, very well-behaved solid with a stable composition in ambient conditions. It is indeed about as strong of an acid as you can get (in water), though this is not a particularly great achievement, as solvent levelling means a considerable number of acids, even those not containing hydronium ions, will form solutions of almost equal acidity when mixed in the same molar amount.
Lastly, as it turns out, hydrogen chloride ($\ce{HCl}$) is a fairly strong acid, but it cannot form hydronium chloride $\ce{H3O+Cl^-}$ in ambient conditions. This ultimately comes down to the fact that $\ce{HCl}$ is actually a gas in ambient conditions, with a very low boiling point of -85 °C, so before you can even get a 1:1 molar mixture of $\ce{H2O}$ and $\ce{HCl}$ the acid becomes too volatile and just boils away. However, if you lower the temperature, you can generate hydronium chloride $\ce{H3O+Cl^-}$ - it's a solid which happens to decompose at temperatures above -15°C, as shown by the rightmost point in this graph:

Source: Wikipedia

Acids reportedly need to have pKa <=-9 to be able to form solid hydronium salts ( CH SE answer )

@Poutnik That's a rule of thumb which applies in ambient conditions, but details such as the energetics of specific hydronium crystal lattices play a role to soften that boundary.

Good to know. Interesting fact.

As a curiosity, p-toluenesulfonic acid monohydrate melts at 105-107 °C (whereupon it presumably loses its stoichiometric hydronium nature and becomes an equilibrium including some free water molecules), while sulfuric acid forms a stable monohydrate  (i.e. hydronium bisulfate, or amusingly hydronium hydrogen sulfate) only below around 10 °C, even though it is a moderately stronger acid in principle.

@NicolauSakerNeto hydronium bisulfate melts under ambient conditions but retains its ionic structure, which leads to a maximal viscosity and minimal ionic conductivity of sulfuric acid solutions at this composition. See here.

Neutral hydronium may also exist. Huang et al. [1] calculate that $\ce{H3O}$ is stable as a liquid metal phase under high pressure and temperature conditions existing in the interiors of Uranus and Neptune. They hypothesize that this metallic fluid is a source of the peculiar magnetic fields seen on these planets.
Reference
Peihao Huang, Hanyu Liu, Jian Lv, Quan Li, Chunhong Long, Yanchao Wang, Changfeng Chen, Russell J. Hemley  and Yanming Ma (March 3, 2020). "Stability of H3O at extreme conditions and implications for the magnetic fields of Uranus and Neptune". Proc. Nat. Acad. Sci. 117 (11), 5638-5643.
https://doi.org/10.1073/pnas.1921811117.

Although the recent papers on Ice-XIX may also impact Uranus and Neptune in a similar fashion.

True, hence I (and the authors) say a "hypothesis". We would expect Ice XIX, with its mobile protons, to contain some mixture of species having the form $\ce{H_nO^{n-2}}$ with $n=1,2,3$ possibly also $n=0,4$.

Not that it was not obvious before, but water really is the weirdest stuff in the universe.

Other weird stuff: helium (needs pressure to form a solid, two liquud phases), silicon (beats water in percent contraction upon melting), many humans.

Plutonium's 6 allotropes (3 unique to Pu) at 1 atmosphere (and an additional one at moderate pressures) is cool, but 19 solid ice phases across the P-T diagram is hard to beat, and their diversity is broad as well.


text:
Can fats be composed of fatty acid esters other than triglycerides?
Fats are mainly referred to triglycerides. In triglycerides, a glycerol molecule form three ester linkages with fatty acids.
I think it is also possible for a butane-1,2,3,4-tetraol molecule to form four ester linkages with fatty acids, or for a ethane-1,2-diol to form two ester linkages with fatty acids. But why they not considered a form of fats, or to make my question clearer, why should fat be triglycerides?

There is nothing impossible in your hypothetical compounds. It is just that natural fats are mostly triglycerides.

Indeed there's nothing stopping such molecules from existing. In fact it is possible to make dendrimeric polyol polyesters such as Olestra, where sucrose (a disaccharide containing eight free hydroxyl groups) has 6-8 hydroxyls esterified with long-chain fatty acids. Olestra (and presumably a number of other similar polyol polyesters) behaves much like regular triglyceride cooking oils, and it was used in the recent past in commercial products for human consumption.

However, these do not arise from natural biochemistry, and Olestra at least does not appear to be significantly metabolized by humans. As such, considering them to be fats may be technically correct by slightly extrapolating their chemistry and material properties, but biochemically speaking is probably a discouraged classification.
Ben Krasnow of Applied Science has a nice video with the story, properties and synthesis of Olestra, and he even fries some chips in it!


text:
Band structure of solids in general
As my last question (Semiconductors and their electronic bands) was badly structured, I decided to elaborate my questions a bit.
As I now know, every solid/liquid forms a band structure, so all condensed matter. For simple elements this is easy enough to illustrate.
But what happens when there are complicated molecules like organic molecules with many different bonds? Are those bands vastly different than when those molecules are in solution?
For example, azo dyes come to mind whose color is the same in the solid state as in solution; also what about complexes for example Cu(II)-hexahydrate, whose color we usually interpret with CFT. Is there even crystal field splitting in a solid or does is form crystal field ‚bands‘? Somehow I am at a loss connecting the band concept to this.

This question isn't the greatest, too. In general, you always have some anti-bonding, or higher atomic orbitals, available for electron transiting between molecules or atoms. In context of band structure, they all are treated as bands.

I've found somewhat generally that trying to interpret the solid state physics of crystal energy bands with chemistry bonding principles can rapidly lead to incomplete or false analogies. Even for supposedly "simple" elements.

So the two theories can’t be combined or?

Actually I disagree to an extent with Jon Custer. You can give a perfectly adequate description of the bands in  materials using LCAO and symmetry adaption of the basis (aka k point sampling in the solid state world) - all bonding is ultimately the same. But you do have to be careful, a simple A bonds to B picture is often incomplete; delocalisation is rife, and a simple bond/anitbonding pair is not always a complete picture.

Basic DFT in general is ill-suited for calculation of one-electron properties and excited states, it requires patches. Plane-wave expansion does not account for localized states and defects. The common approach to model them is supercells. However, even this approach fails to model the so-called topological insulators, which requires further patching such as DFT+U method. All this is to be expected for formalisms derived from description of isotropic electron gas.

@permeakra Basis sets and the effective one-electron Hamiltonian are different things. You can perfectly well model localized states with plane-waves,  you might just need a lot of them. The problem is that a LDA Hamiltonian has a spurious self-interaction from the lack of exact exchange (or similar) to cancel the coulomb term; the system tends to minimize the cost of this by spreading out the electron - hence the problem with localization. But this is not directly a problem of a plane-wave basis set. DFT+U is a way to try and introduce the required exchange terms.

Secondly plane-waves can perfectly well do defects. You need supercells because otherwise the defect concentration would be unphysically high, not because you are using plane-waves. Consider an F centre in an NaCl cell. If you just used one cell the system would consist of the Na+ centre and an F centre - no Cl- at all. Hence the need for a supercell to bring  the concentration of defects down towards something sensible - remember the maximum doping of Si used is around 0.01%

@IanBush >You can perfectly well model localized states with plane-waves, you might just need a lot of them. || I would say you need infinitely many of them. Like continuum many. Plane waves are not square-integrable. Hence, you cannot use them to expand a square-integrable function in a countable series. At best, you can use them to expand a periodic function and hope it is a good enough model for your aperiodic system. This is a fundamental flaw in a plane-wave approximation. It's math, arguing with it is pointless.

@IanBush Also, if you point at 'gaussian basis set' in programs like Elk code, they still use plane waves implicitly.

I am an author of CRYSTAL, a Gaussian basis set based program for periodic systems. Please point me to the plane-waves in CRYSTAL.

@IanBush "The Bloch function" . It is by definition a periodic plane wave.

OK, last thing I will say, The Bloch function is purely a reflection of the translational symmetry of the crytsalline material being modelled. The eigenstates of the Hamiltonian of a periodic system are necessarily Bloch States. They are not being used by choice, they are a fundamental result of the mathematics underlying the model of the system being employed. F. Bloch, Zeitschrift für Physik 52, 555 (1928)

I don’t know about those softwares.

Not my field but given the lack of translational coherence, I would answer "Are those bands vastly different than when those molecules are in solution" with a resounding "of course", and would argue that being in solution (highly mobile) is not conducive to band structures.

I personally think and I could be wrong, that this question is a mixture of very inhomogeneous concepts. There is indeed a correlation between quantum chemistry and solid state physics. Quantum chemistry in the simple formulation evoked in the comments deals mainly with localised states that are not very too far from atomic orbitals, the expansion of the unknown WF is a set of localised orbitals (LCAO). Solid-state physics is concerned with delocalized states. Their difference is a matter if size and the symmetry of the system under study.
In solid-state physics, the most crucial symmetry is translation, and the single-particle translational operator $\hat{T}$ linked to the linear moment operator $\hat{p}=i\hbar \nabla \hat{T}$. The eigenfunctions here are often plane-waves (PW) or a combinaison of PW. $\hat{T}$ produces the band structure and Bloch states by assuming that it commutes with the Hamiltonian. Everything that affects translational symmetry leads directly to chemical localized states : pairwise $U$ Coulomb interactions, defects ...
In the chemical limit, the operator position $\hat{x}$ is important and the symmetries such as rotations or mirrors that commute with the Hamiltonian.
I don't know where this argument about band structure in liquids where there is no translational symmetry comes from.
CFT in solids like transition metal oxides (cupper oxide) is in my opinion more of a chemistry problem, I could go into mathematical details using the hubbard model to show that in this configuration the metal cation has a quasi-atomic structure for the d-orbitals simply because the overlap of the d-orbitals is too small due to impurities compared to the
coulombic interaction $U$.
The number of bonds is not the generator of a band structure but the symmetry of these bonds. Large molecules such as fullerenes $C_{200}$ do not have a band structure but energy levels because there is not enough translational symmetry. The band gap is just due to how this translational symmetry is affected by the potential, but this gap may differ from the splitting responsible for the colors of these oxides due to CFT.
Even if it's not the direct answer to the question I hope it helps


text:
How to choose an appropriate electrolyte
I am looking to electroplate using a zinc-nickel coating using an ionic liquid as the electrolyte. How should I go about selecting the proper ionic liquid? I choose the metal coating due to it increasing the resistance to corrosion of the substrate being plated. Is it best to look at the properties of the possible electrolyte itself or how it reacts with the metal alloy?

Why would it react with the alloy?

Why a ionic liquid? It may cause significant restrictions.

The properties of an ionic liquid look like a potential alternative substitution for the electrolyte in electroplating. From what I've read this is due to it's properties such as having a low melting point, high conductivity, thermal stability, low vapor range, possibly lower toxicity if compared to other electrolyte elements, as well as a couple of other characteristics.


text:
Hybridization and lone pairs
I understand how hybridization works but have trouble understanding it conceptually. I understand that for bonds to form, the electrons must be in orbitals with equivalent energy, but why does this also apply to lone pairs? In NH3, for example, there are three N-H bonds and one lone pair. Hence, according to the rules, we need to mix four orbitals and get an sp3 hybridization, but why could we not form the three hydrogen bonds with the three half-filled p orbitals of equivalent energy that nitrogen has already, and then leave the fully filled s orbital as the lone pair? I think it might have to do with the geometry somehow, but I'd like to clarify this.

Well, it happens like you say with PH3. Hybridization is a mathematical operation to describe how molecules are, not some rule.

https://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h


text:
Is -NR3+ really meta directing?
For electrophilic aromatic substitution, it is said that -NR3+ is meta directing, but since it only shows electron withdrawing inductive effect and inductive effect decreases with distance so it must para directing since at para position inductive effect will be minimum.So is -NR3+ meta or para directing?


text:
Why doesn't breaking hydrogen bridges in water increase the temperature?
Why, when hydrogen bridges are broken in water as during a phase transition, the input energy does not increase the temperature? Is it because some part of the energy is converted into a rotational and vibrational movement? Or because part of the energy is transferred to the electron cloud and re-emitted? I thought that to break a chemical bonds it's necessary to increase the kinetic energy of molecules in order for them to stretch and not be attracted any more.

It costs energy to break a hydrogen bond. Melting and boiling are endothermic processes.

Why doesn't breaking hydrogen bridges in water increase the temperature?

It does, or can, just not during the phase transitions we are familiar with (melting and boiling, which are first-order phase transitions). Such transitions occur at constant temperature until the starting phase has been completely converted into the other, with the two phases coexisting in the interim. Rather than T changing it is the internal energy that changes as exchanged heat is stored (or released). That internal energy is shared among all the possible motional degrees of freedom of the particles in the phase according to the equipartition principle.
To break bonds in general requires an input of energy (added to the substance in the form of heat if no work is done on the substance). When a phase in which particles locked into a constraining arrangement due to bonds absorbs heat, the additional energy allows the particles to overcome attractive interactions and move more freely relative to each other. Thus for instance ice becomes water as the regular lattice due to strong intermolecular hydrogen bonds breaks down.

Breaking a chemical bond requires always some energy (heat or work). ALWAYS.
Making a chemical bond produces some energy (heat or work). ALWAYS.


text:
Help with Calcium Alginate Precipitation
I am a high school student working on a research project that involves producing alginate gel directly from wet seaweed. I am starting the first phase next week, which is using a small batch to practice the full procedure. The guide I am using is somewhat vague, and it mentions having to add calcium chloride to a sodium alginate solution and then mixing "carefully" to ensure a solid precipitate. The exact quote from the paper is "If the calcium solution and filtered extract are mixed carefully, the calcium alginate can be formed as fibres - bad mixing gives a gelatinous solid". (After that, I am supposed to collect the fibers and add acid and then sodium carbonate to get solid sodium alginate.) Any tips on how exactly to go about the mixing?
Here is the link to the full paper if clarification is needed: https://www.fao.org/3/y4765e/y4765e08.htm#bm08.1

Though I have not tried this myself, it might mean that decanting one solution gently on top of the more dense one (likely the $\ce{CaCl2}$) would form two distinct layers, with the precipitate forming as a sheet where the layers adjoin.
Gently pushing the sheet down at one spot, or lifting the sheet at an edge could extrude the film, creating a long strip, and subsequent stirring would break the sheet into strips or fibers (or fibres, if done across the ocean).
This SWAG is based on the nylon synthesis "rope trick". In that video, note use of a glass stirring-rod to reduce mixing of the second solution as it is poured.

Note the description of the procedure by the OP: "mixing "carefully" to ensure a solid precipitate". That seems inconsistent with gentle layering of one solution atop the other.


text:
Moments of inertia of benzene and ammonia as symmetrial tops
Hello I was reading about symmetrical tops and how their properties relate to their rotational spectra. According to the 8th edition of Atkins, those molecules have one unique principal axis which includes a moment of inertia (denoted $I_{∥}$). They also have two orthogonal moments of inertia, which are equal in magnitude (denoted  $I_{⊥}$).
Unfortunately, the book doesn't give any pictural representation, so my question remains: Do all moments of inertia need to be orthogonal to one another or just to the principal axis? On the internet I find conflicting information for the moments of inertia for benzene, as seen here.

Here, you can see that the two $I_{⊥}$ (a,b) are not orthogonal to each other, but in this illustration they are (A, B):

So, which is it? And would this representation of ammonia be correct for the three moments of inertia?

The three I axes are always orthogonal and along x, y, z, but the I values along these directions can differ depending on geometry. You can choose any set of xyz orientation wrt to the molecule but its easier to use the molecules symmetry axes if possible. The molecule rotates about the total angular momentum which is the vector sum the 3 components. See chapter 7 of https://applying-maths-book.com/ for a discussion and examples of the 3x3 matrix you may need to diagonalise in the general case if you choose arbitrary xyz. Diagonalising rotates the axes to be along molecules own MOI direction.

There is non fundamental need for the axes to be orthogonal, merely non-coplanar is enough for a complete description of the object. This is so the unit vectors in the directions of the axes form a complete basis for the three dimensional space. But certainly if you can choose your axes an orthogonal choice will almost always make your life easier.

So @IanBush both options I mentioned for the two inertia axis (a,b) for benzene that I mentioned can be considered correct?

All the above are valid choices. They just may not have the same values. See @porphyrin comment on diagonalisation about relating them.

@IanBush But we can only measure two rotational constants for the molecule can’t we?  So there should be only two or am I wrong? And about porphyrins comment, I am not that deep into the topic as my expertise is in bioorganic chemistry so its a bit difficult to understand

We can measure one transition energy when exciting an electron from the 1s state in Hydrogen to the set of 2p orbitals. So are you claiming there is only a single 2p state, and not 3, namely 2px, 2py and 2pz? Just because we measure 1 number doesn't mean it corresponds to just 1 thing. This is what degeneracy is. At least in quantum land.

Yeah that is true, I see what you mean. The thing I was not sure about, is that the moments of inertia would correspond to the same energies, for example in the benzene example, it was not obvious to me that I(a) in the first example would have the same energy as I(b) in the second..


text:
Why are different Aluminium valencies observed from NMR and what logic can be used to normalise molar ratios of different valency atoms?
Question 1:
I've observed different chemical shifts in 27Al Nuclear Magnetic Resonance (NMR) spectra of commercial aqueous sodium aluminate, indicating the presence of both trivalent [Al(III)] and tetravalent [Al(IV)] aluminium bonds with different numbers of other atoms which is consistent with 27Al NMR seen in literature (Source1, Source2). However, considering aluminium's position in the periodic table with the $Al^{3+}$ charge (which is not 4+), why isn't Al(III) the only existing valence state for aluminium? What factors lead to the coexistence of both trivalent and tetravalent forms of Al?
Question 2:
An experimentally inferred (and generally accepted within literature) geopolymer structure can be observed below from Barbosa et al where an ideal geopolymer has $Al^{3+}$ in four-fold coordination but that the observed Al coordination can change depending on how the "activator solution" (essentially a chemical catalyst of sorts) is synthesised. Their proposed geopolymer structure in Figure 8 (below) had both $SiQ^4(3Al)$ and $SiQ^4(2Al)$.

It is conventional in geopolymer systems to specify a system molar Si/Al ratio when combining feedstocks and assumes that the covalent bonds formed is as dictated by an element's valence electron number and charge as follows:
$$\\\left(\frac{n_{Si}}{n_{Al}}\right)_{Conventional} = \frac{n_{Si^{4+}}}{n_{Al^{3+}}} \approx \frac{n_{Si(IV)}}{n_{Al(III)}}\\\\$$
Usually, $Na^+$ is added to the system via a $\frac{Na}{Al}$ ratio to provide charge balance with the $Si^{4-}$ and $Al^{3+}$ for the system to theoretically neutralise. However, the information from Q1 above shows that some of the $Al$ can have different valencies and thus form covalent bonds differently in the system. Assuming the molar trivalent and tetravalent Al can be quantified via quantitative NMR, does it make sense for the $\frac{Si}{Al}$ ratio instead be normalised as follows:
$$\\\left(\frac{n_{Si}}{n_{Al}}\right)_{Normalised} \approx \frac
{n_{Si(IV)}}
{(n_{Al(III)} \times \frac{3 bonds Al}{4 bonds Si}) + (n_{Al(IV)} \times \frac{4 bonds Al}{4 bonds Si})}
=
\frac{n_{Si(IV)}}
{(n_{Al(III)} \times \frac{3}{4}) + n_{Al(IV)}}\\\\$$
So, with the information above, is it reasonable to presume that the molar ratios should be normalised to account for different Al valencies which lead to different coordinations in the final geopolymer structure?
If the above is true then the other molar ratios, such as the $\frac{Na}{Al}$ would have to also be normalised.

'Tetravalent' doesn't mean $\ce{Al^{4+}}$, it just means aluminium bonded to four things. $\ce{[Al(OH)4]^-}$, for example, is tetravalent $\ce{Al^3+}$. Quoting from your Source 2: "The oxygen tetrahedra holding $\ce{Al^3+}$ ions hold a negative charge that is neutralized by an alkali metal cation."

Thank you for the correction, what is the term I'm looking for then? Or should I just be referencing the numerical charge?

Well, what are you trying to describe? If you're talking about aluminium bound to four things, then tetravalent is just fine. If you're talking about $\ce{Al^4+}$ then yes, use the charge. But the latter doesn't exist, so maybe it's worth reworking the question a bit — in particular, your first question — because there's no contradiction between Al being in Group 13 and Al being tetravalent.

I have edited the question with the Al(III) and Al(IV) (etc.) terminology as appropriate for valency, thanks for letting me know. I was getting mixed up with charge. :)

I'm getting requests to close the thread but no feedback onto why specifically it should be closed. What can I be more specific on?

Are you sure that the different 27Al resonances you observed are due to varying oxidation numbers? While I don't have a complete understanding of the materials you are using, I do have experience with zeolites and other Al oxides, and in my experience, distinct regions of the spectra usually suggest different coordination of Al. For instance, 4-coordinated Al is typically present in the range of 70-50 ppm, 5-coordinated Al is found in the range of 40-20 ppm, and 6-coordinated Al is usually in the range of 10 to -10 ppm. Here is a nice review on 27Al NMR of materials, in Figure 2 they discuss the chemical shift range of different Al species.
Bear in mind that 27Al is a quadrupolar nucleus, which means that you may have a quadrupolar broadening of your resonances. Additionally, what may appear to be a broad resonance can often be the result of a distribution of chemical environments. Czjzek's distribution is commonly used to deconvolute 27Al spectra of glasses and other Al oxides, separating the contributions from local coordination environments. You can apply this to your spectra using freely available software such as DMFit and ssNake.

I just made an edit to my question for clarity. The feedstock I'm using is commercially available aqueous sodium aluminate (as the primary source of soluble Al). Yes I think your answer is on the right track and answers the first part of my question.

I would not expect to have Al$^{3+}$ on the geopolymer. I would assume that you have a chemical analysis of the polymer so the Si/Al ratio should be just the molar amount of each element, independently of their coordination/valence. Then if you can make some quantification for $^{27}$Al NMR you can just decompose the total amount of Al, however, that should not change the total Si/Al ratio.

A good description of the geopolymer species is given by Barbosa et al (https://www.sciencedirect.com/science/article/pii/S1466604900000416) where an ideal geopolymer has $Al^{3+}$ in four-fold coordination but that it can change depending on how the "activator solution" (essentially a chemical catalyst of sorts) is synthesised. Their proposed geopolymer structure in Figure 8 had both $SiQ^4(3Al)$ and $SiQ^4(2Al)$. So, with this information, shouldn't the Si/Al be normalised?

Sorry I don't think I understand your question. Those SiQ4 species you indicate are just Si atoms surrounded by either 2 or 3 Al, nothing related to the oxidation state of Al. If that is the case maybe you can use the equation used in molecular sieves to determine Si/Al ratio from $^{29}$Si NMR
$$Si/Al = \frac{I_{Q4(0Al)} + I_{Q4(1Al)} + I_{Q4(2Al)} + I_{Q4(3Al)} + I_{Q4(4Al)}}{0.25I_{Q4(1Al)} + 0.5I_{Q4(2Al)} + 0.75*I_{Q4(3Al)} + I_{Q4(4Al)}}$$

Interesting comment! This might be what I'm looking for. I'm essentially just trying to more accurately define the Si and Al (and ideally also OH but this is out of scope for this question) that is available for reaction so that I can more accurately define my system; bigger $Q^n$ species are harder to breakdown and thus not available for reaction (sources). Could you please provide more information (either via these comments or an answer to the thread) and a reference of some sort so I can follow up and research this for better understanding before I come back with another comment?

You can find it for example in this paper


text:
Silica gel reheating (orange)
I am a non-chemist looking for some advice! I am an archaeologist and am currently using moisture-indicating silica gel (orange) to dry out some metal finds from our excavations. I have reheated silica gel before but I am looking for a refresher as it is many years I've done it!
I intend to heat a small amount at a high temperature (can anyone advise what temperature?) in an oven that is used solely for this purpose. How long do I need to heat it for? If I heat it for longer, is it likely to retain its colour for longer?
Additionally, how well ventilated does the space I am in need to be? I intend to wear gloves and a light-weight mask.

Consult Preparing Silica Gel for Contained Storage of Metal Objects – Canadian Conservation Institute (CCI) Notes 9/14

@Poutnik, perfect answer!


text:
How to remove the baseline in the Raman spectrum without falsifying the spectral data. Measurement of anisotropy using SERS
My goal is to study the anisotropy of the sample based on changes in the Raman spectrum as a function of angle. The sample was rotated around the Z-axis and changes in the spectra are visible in the raw data.
The sample was rotated every 10 degrees resulting in a total of 18 spectra.
The important parameters for me are the intensity/width/position of each peak for each angle. The baseline subtraction process has a significant effect on the intensity values and a moderate effect on the peak width.
It is obvious that when subtracting the baseline the absolute values of intensity change, but when studying anisotropy the relative changes between measurements for successive angles are important.
For each spectrum, the subtracted baseline will vary, and I am concerned that (especially for peaks close together) the subtracted baseline may disturb the relative differences between intensities in measurements for successive angles.
I process my data in Wire 3.3 software provided by Renishaw and subtract baselines using cubic spline interpolation.
I would appreciate any advice on how to prepare the data for curve fitting without falsifying the resulting measurement data.

Can you at least show how the baseline or the spectrum looks like before one can suggest you anything? Also, what is the main concern behind using cubic splines?


text:
Find the missing chemical structures in the reaction sequence
I came across the following question, where a set of reagents are used, starting from a carbohydrate derivative.

I understand that this is some total synthesis reaction, but I cannot find which one.

Please explain the structures of A, B, C, D, and E. A clear mechanism would be greatly appreciated. (If you get the original research paper, it would be very helpful)

From the total synthesis of reserpine and morphine that I have studied, Acetic anhydride is used for esterification. But, since there are three -OH groups here, will there be any selectivity in the esterification?

Ref: Reserpine total synthesis - /https://renaud.dcb.unibe.ch/journal-club/journal-club-2015/p2015_23-reserpine---woodwa.pdf (page-10)

Don't worry about selctive acetylation, you have excess acetic anhydride in the first reaction so that is designed to acetylate all three alcohols.

This may help you with the second reaction  https://en.wikipedia.org/wiki/Azide-alkyne_Huisgen_cycloaddition

You shouldn't search for some random synthesis, but specific reactions.


text:
Why don't Zn, Cd, Mn and Hg have more than one metallic structure at normal temperature, while other d-block elements do?
It's been given in my book, that "With the exceptions of Zn,
Cd, Hg and Mn, they have one or more typical metallic structures at
normal temperatures."
But, why this exception?

Exceptions are the rule in chemistry. There must be reason not to have them. Except for Mn, all 3 have fully filled d orbitals, like no other d block metal. Mn has it at least fully filled. Hg is liquid, so obviously just 1 form. BTW, the textbook statement has more possible interpretations.

See last paragraph of this answer: https://chemistry.stackexchange.com/a/91488/17368

@Poutnik Well said. Even Mn is prone to exception: https://chemistry.stackexchange.com/questions/39786/what-is-the-lattice-structure-of-manganese?rq=1

Manganese has 4 allotropes between room temperature and the melting point. And it isn’t like any of the others have phase transitions particularly near room temperature.

@Poutnik "Obviously just 1 form"? Liquids can show phases e.g. Wikipedia article on Polyamorphism , A recentish article in J. Chem Phys. Liquid–liquid transition and polyamorphism, and probably the best known Liquid Helium. Admittedly I can find nothing about Hg, but I don't think it obvious that it doesn't have more than one phase.

@IanBush What have I said about exceptions? :-) Helium is very special exception. Also, normal T was mentioned and helium is not a metal.

Well obviously there are exceptions :) Therefore it's not obvious. I'm confused.

@IanBush Well, obviously there are rules and obviously there are exceptions of the rules. Therefore, liquids have obviously 1 phase, except obvious  cases they have more  :-P

I challenge the basis for the textbook's statement.
Below I give a table of of the 31 elements in the middle sections of a standard periodic table, with information pulled from A.T. Dinsdale's SGTE Data for Pure Elements that is commonly used for thermodynamic modeling. One can quibble about Lanthanum vs Cerium so I included both. The table lists the element, the stable crystal structures at standard pressure, and the transition temperature from one crystal phase to the next, or to the liquid.
There are very few transitions between crystal structures at "normal" temperatures. Lanthanum goes from DHCP to FCC at 550K as the lowest temperature crystal-crystal phase transition. Cobalt going from HCP to FCC at 694K is next. Then Manganese going from CBCC to CUB at 980K.
Nineteen of the 31 elements listed have only one stable crystal phase at standard pressure. That is, the exhibit only a phase change to the liquid (i.e. melting).
I don't know about you, but none of those transitions is at a "normally" human experienced temperature.
Element Crystal   Transition
        Structure Temperature(K)
Sc        HCP     1608
          BCC     1814
Ti        HCP     1155
          BCC     1941
V         BCC     2183
Cr        BCC     2180
Mn        CBCC     980
          CUB     1360
          FCC     1411
          BCC     1519
Fe        BCC     1184
          FCC     1667
          BCC     1810
Co        HCP      694
          FCC     1768
Ni        FCC     1728
Cu        FCC     1357
Zn        HCP      692
Y         HCP     1751
          BCC     1795
Zr        HCP     1138
          BCC     2127
Nb        BCC     2750
Mo        BCC     2896
Tc       (HCP)    -
Ru        HCP     2607
Rh        FCC     2237
Pd        FCC     1828
Ag        FCC     1234
Cd        HCP      594
Ce        FCC     1000
          BCC     1072
La        DHCP     550
          FCC     1134
          BCC     1193
Hf        HCP     2016
          BCC     2506
Ta        BCC     3290
W         BCC     3694
Re        HCP     3458
Os        HCP     3306
Ir        FCC     2719
Pt        FCC     2041
Au        FCC     1337
Hg        Liquid  -

and regarding Technetium: The crystal structure of the bulk pure metal is hexagonal close-packed, and crystal structures of the nanodisperse pure metal are cubic.

@NilayGhosh - done. Technetium is not a common binary alloy component, so while there is some thermodynamic data available it was not included in the compilation above. Apparently the high pressure phase is tetragonal.

Tin transitions from the ductile white phase to brittle gray tin on cooling to about 13C.

@OscarLanzi - that is true, but it is below 25C that is considered the normal reference state. The thermodynamic data for the beta phase of tin is included in that compilation.


text:
What determines the location of nucleophilic attack on a conjugated ketone or aldehyde?
In a conjugated ketone or aldehyde, nucleophiles could attack the double bond or the carbonyl carbon. I have two examples that differ in the location of initial attack, and I am wondering what determines the outcome.
I came across this question:

The answer is given as 2nd options. I don't know why it is the case so I dug deeper and came across this post and inferred that nucleophiles prefer to attack the conjugated double bonds and also the carbonyl position.
But however this question:

I thought the hydride ion will attack the conjugated double bond, but looks like it does not affect the double bond instead attacks the carbonyl position to get the product in 1st option. So why it is the case? Is it based on the fact that carbocation formed in the benzylic position is more stable than the carbocation in the carbonyl position in the first question, which leads to the attack of nucleophile towards the benzylic carbocation ?

Well, "normal" Grignard would also only affect carbonyl. It's the first reaction that I'd wonder about, not the second.

@Mithoron Hi !. I thought it is based on the fact that benzylic carbocation is more stable than the one that would form at the carbonyl position. But in the links I have posted, the Lithium Aluminium Hydride attacks the conjugated double bond, in cinnamaldehyde. So it is an exception case only for cinnamaldehyde ?

@Mithoron https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not?noredirect=1&lq=1 This one

Well, that certainly looks unusual.

@Mithoron So it is based on the fact that benzylic carbocation is more stable than the one that would form at the carbonyl position ?

No, it's not. More like carbanion, if anything.

[Mithoron in the comments] Well, "normal" Grignard would also only affect carbonyl. It's the first reaction that I'd wonder about, not the second.

Without the copper catalyst, the Grignard would attack the carbonyl. With it, it attacks the conjugated carbon-carbon double bond instead.
See Wikipedia article, which gives a 1941 reference (Kharash) referring to the copper-catalyzed version as a 1,4 addition.
A modern review article shows the current applications, and explains that you first obtain an 1,4 adduct (with the metal bound to oxygen) which then is quenched to get the aldehyde back.

This is an example of the competition between conjugate (Michael) and direct addition to a conjugated carbonyl, which is itself an example of thermodynamic vs kinetic control.
You can read more about this in Clayden chapter 22, but in short,
The conjugate addition is the thermodynamic product and its formation is irreversible.
The direct addition is the kinetic product and its formation is reversible.
The conjugate addition can be favoured by:

Heating-as it is kinetically disfavoured, heating reduces the kinetic barrier to conjugate addition.

Using a soft nucelophile favours the conjugate addition, a harder nucleophile favours the direct addition.

Soft nucleophiles are ones good at attacking sp3 carbons, but not always sp2 carbons, even though in this case they would attack the sp2 alkene. They involve being polarisable, large, having a delocalised charge and having a high energy HOMO (not all of these characteristics are needed but they all help make a nucleiphile soft). Examples of soft nucleophiles include Thiolates(-SH), organnocopper reagents and deprotonated aromatic heterocycles.
Hard nucleophiles are small and densely charged, and examples include hydrides and gringards or organolithium derivatives.
There is no better example than the ones in your textbook. In the first example the copper is provided to transmetallate the gringard into an organocopper to favour the conjugate addition. Without the copper, a grigard would do a direct addition.
In the second example, the hydride is a hard nucleophile and so favours the direct addition.
Note that some nucleophiles are in between hard and soft and so if the stoichiometry allows, they will do both. A nice example is a soft hydride, like sodium borohydride.
There are chemoselective ways to reduce either the ketone or the conjugate double bond without reducing the other. Luche he reduction with NaBH4 and a catalytic Titanium (III) chloride would be used to reduce only the carbonyl. K-selectride would reduce only the double bond.

Thank you for the Nice answer. ! :-)


text:
Does it make any sense to graph 1/T versus ln(Kc)?
I calculated $K_c$ experimentally for the reaction $$\ce{CO2 + H2O -> H2CO3}$$ at different temperatures. Because $\Delta G =-RT\ln(K)$, I thought that graphing 1/T against $\ln(K_c)$ would give me a gradient $\Delta G/-R$, that I could calculate googling the Gibbs free energy of formation of the compounds of my reaction to compare to the gradient I get to check accuracy. Does this make any sense? (because the values I calculated were wrong)

Note this equation Delta G =-RTln(K) is wrong. It should be $\Delta G^\circ =-RT\ln(K)$. The circle (plimsoll) symbol is important, it means "in the standard state".

How do you measure the concentration of $\ce{H2CO3}$ ?

@Maurice by titrating with NaOH

As
\begin{align}
Δ_\text{r}G°&=-RT\ln{(K)}\\
Δ_\text{r}G°&= Δ_\text{r}H° - TΔ_\text{r}S°
\end{align}
then
\begin{align}
\ln{(K)} = -\frac{Δ_\text{r}G°}{R}\cdot \frac{1}{T}  &= - \frac{Δ_\text{r}H°}{R} \cdot \frac{1}{T} + \frac{Δ_\text{r}S°}{R}\\
\ln {\frac{K_2}{K_1}}&=   \frac{Δ_\text{r}H°}{R}\left(\frac{1}{T1}-\frac{1}{T2} \right)
\end{align}
with the $Δ_\text{r}H°$ and $Δ_\text{r}S°$ are assumed approximately constant in the range $T_1–T_2$ and $K_i$ is the equilibrium constant for the respective temperature $T_i$.
The slope value $\dfrac{\text{d}\ln{K}}{\text{d}\left(\frac{1}{T}\right)}$  could be used directly to determine  $Δ_\text{r}G°$ only if $|Δ_\mathrm{r}S°| \approx 0$ and therefore the regression line passes the origin, what is generally not the case. But it can still be determined from the values of both the slope and the intercept, as below:
Therefore:
\begin{align}
Δ_\text{r}H° &= −R \cdot   \text{slope}\\
Δ_\text{r}S° &= R \cdot   \text{intercept}\\
Δ_\text{r}G° &= −R \cdot   \left(\text{slope} + T \cdot \text{intercept} \right)
\end{align}
For more, see the Wikipedia page about the van 't Hoff equation.

Thank you for the information! Just a small correction to the last equation: it should be slope + T*intercept. I think.

Yeah, you are right. Fixed


text:
Why is the statement regarding the strength of ionic bonds reasonable?
In largely ionic compounds (e.g. NaBr and NaI), it seems to be generally true that, the greater the differences of electronegativity between the forming substances (i.e. The compound has more "ionic character" or has lower "covalent character"), the stronger its bond would be. (e.g. melting points: NaF>NaCl>NaBr>NaI with all of them having the same crystal structure. Pradyot, Patnaik (2003). Handbook of Inorganic Chemicals. The McGraw-Hill Companies, Inc. ISBN 978-0-07-049439-8.)
What I do not understand is that, as covalent bonds are not necessarily weaker than ionic bonds (For example, diamond and silicon both has very strong covalent bonds), why is it proper to claim that "more ionic bonds have greater strength" for such salts?

"It is often said that": Please edit, adding one or more sources for this claim.

When you say 'stronger than' you should qualify it by comparing with other molecules, it may not always be true. You can check by finding lists of bond formation/dissociation energy.

If a bond is ionic, the strength depends on the charge of the bonding partners and the distance of the ions. Strength here refers to the energy required to increase the distance between the two ions to infinity (in a vacuum).
For a covalent bond, the term "strength" often refers to the (homolytic) bond dissociation energy. In trying to rationalize bond dissociation energies, it does make sense to consider differences in electronegativity as one aspect among many others.

I think it is useful to distinguish between bonding in an ion pair, and bonding in a macroscopic lump of an ionic material. In the former there is a clear pairwise bond. In the latter each ion sits in the potential due to the whole of the rest of the lattice, and to my mind there is no simple pair bond.

@IanBush Yes, and you need the Madelung constant to figure out the lattice energy, not an easy calculation.


text:
Cyclovoltamogram of silver in solutions of its insoluble salts
Hello I was dealing with the following question in my electrochemistry course:

Sketch cyclic voltammograms for the silver oxidation and reduction in different
electrolytes. Assume electrochemical reversibility and a concentration of
50 mM of either one of the respective salt in the solutions. The solubility
products of the different silver salts in are: $K_{sp}(AgBr)=5.4\cdot 10^{-13}$, $K_{sp}(AgCl)=1.6\cdot 10^{-10}$, $K_{sp}(AgI)=8.5\cdot 10^{-17}$, $K_{sp}(AgSCN)=1.0\cdot 10^{-12}$. Sketch for (a) bromide solution, (b) chloride solution, (c) iodide solution, (d) thiocyanate solution. Hint: The concentration of silver ions in the Nernst equation can be substituted
with the rearranged equation of the solubility product.

So, to tackle this problem, I thought about two things. First, how to derive the expression for the Nernst equation in this situation, and I assume it is largely how the Ag/AgCl-formal potential is derived:
$E(Ag/Ag^{+})=E^{0}(Ag)+\frac{RT}{F} \cdot ln(a(Ag^{+}))$ (1)
But as the silver cation potential is controlled by the $AgX$ salt solubility product, $K_{sp}$, we get:
$E(Ag/AgX)=E^{0}(Ag)+\frac{RT}{F} \cdot ln \left (\frac{K_{sp}}{a(X^{-})}  \right )$ (2)
Then we seperate the logarithmic terms:
$E(Ag/AgX)=E^{0}(Ag)+\frac{RT}{F} \cdot ln (K_{sp}) - \frac{RT}{F} \cdot ln (a(X^{-}))$ (3)
As the solubility product and $E^{0}(Ag)$ are constant, we define those two as $E^{0}(Ag/AgX)$ and get:
$E(Ag/AgX)=E^{0}(Ag/AgX) - \frac{RT}{F} \cdot ln (a(X^{-}))$ (4)
So, as we have an expression for the $E^{0}(Ag/AgX)$, and the reaction is reversible, we know that the peak potentials of the cyclic voltammograms need to be at this value, which can be calculated from the solubility products using equation (3) (the peaks will be seperated by 57 mV from this value, $\frac{E_{pc}+E_{pa}}{2}$).
However, I am unsure how to draw the cyclic voltammogram, for you see, the potential actually increases when $Ag^{+}$ is formed due to oxidation, because the $X^{-}$ crashes out of solution because of the solubility product, in contrast to e.g the $Fc/Fc^{+}$ couple. It just increases and never reaches $E^{0}(Ag/AgX)$ for the peak in its CV diagram (eq. 4)
In the $Fc/Fc^{+}$ redox couple, as $Fc$ is oxidized  to $Fc^{+}$, its potential increases until the ratio (Q) of $Fc/Fc^{+}$ is equal to one (take a look at eq. 5) and $ E^{0} (Fc/Fc^{+})$ is reached for the peak of the CV-curve.
$E(Fc/Fc^{+})=E^{0}(Fc/Fc^{+}) - \frac{RT}{F} \cdot ln (\frac{a(Fc)}{a(Fc^{+})})$

So how can I draw the CV here?