text:
What is the relationship between Ha and Hb in the given compound, trans-1,3 dichlorocyclobutane. Are they enantiotopic or homotopic?
The question is to find the TOPICITY relationship between  Ha and Hb in the given compound, (trans 1,3 dichlorocyclobutane)

They seem to be in the same chemical environment, so I tried the substitution method, and replaced Ha and Hb with an X* atom in-turn (for convenience, I have used Br as the X* atom ). Then, I used CIP rules to assign priority, and got RR and SS configuration respectively. (See img below) I concluded them to be enantiotopic.
However, I am still confused because I cannot imagine any plane which would be used to show the mirror image relationship between the two.

Is my answer and reasoning correct?
If so, kindly demonstrate the mirror plane which makes the two compounds enantiomers.
If not, please explain.
Nb: I couldn't find an axis of symmetry that makes Ha and Hb equivalent. That led me to the conclusion that thay are not Homotopic.

I think you're overthinking the substitution method. Remove the a and b labels on the two hydrogen atoms. Then, one at a time, replace each previously labelled hydrogen with H*. How are the two molecules, which have no a/b labels and a single starred hydrogen each, related to each other? That will tell you the stereotopic relationship between the hydrogen atoms.

@Zhe I have labelled Ha and Hb just for convenience. I completely understand your point that they are the same. Still I couldn't figure out the answer, until I put forward this question to my professor. However, thanks for pointing out that info to the readers.

They are ENANTIOTOPIC.
According to IUPAC Gold Book,

Atoms or groups of a molecule which are related by an n-fold rotation axis (n = 2, 3, etc.) are called homotopic

In the given question, Ha and Hb are not related by an axis of rotation. However, they are related by a center of symmetry. Therefore, they are enentiotopic protons. The following chart is helpful.

Note: Enantiotopic protons are indistinguishable through NMR spectroscopy.

10 months later, we have a correct answer...

What is curious about the bromine-substituted molecule is that it does not matter whether the bromine "points up" or "points down", it is the same molecule:

For example, the top left and the top center molecules are identical, related by a two-fold rotation about the X-axis (which goes from left to right).

[OP] If so, kindly demonstrate the mirror plane which makes the two compounds enantiomers.

The two circled structures correspond to the molecules and orientations shown in the question. As Cyclopropanol's answer states, they are related by a center of inversion. A center of inversion can be realized by a two-fold rotation combined with a mirror plane. The structures on top and on the bottom are related by a mirror plane (in the plane of the paper). So you can take the top circled structure, rotate it 180 degrees about the Z-axis (perpendicular to paper) to arrive at the top right structure, and then apply the mirror plane to get the circled structure on the bottom. Or you can first apply the mirror operation and then the 180 degree rotation.
If you prefer a mirror plane swapping left and right, compare the top left structure with the bottom right structure.

[OP] Is my answer and reasoning correct?

Yes, because the two bromine-substituted structures are enantiomers, the hydrogen atoms are enantiotopic (indistinguishable in the NMR unless you add a chiral co-solvent). I was surprised that you did not label the stereochemistry of the bromine-substituted carbon, but now I understand why (not a chiral center).

Geometrically, the relationship between $\ce{H_a}$ and $\ce{H_b}$ is through a center of inversion. This is the same as a mirror reflection combined with a $180°$ rotation. Even without an explicit mirror plane, the implicit reflection component associated with a center of inversion is sufficient to distinguish enantiotopicity from homotopicity.
You may want to perform an excercise involving a simple chiral molecule, let us say 2-butanol. Initially place the substituents around the stereocentric carbon so that they have the R configuration. Thus when the molecule is viewed with the lowest-ranking hydrogen atom pointing away from you, the ranking of the other substituents ($\ce{OH > C2H5 > CH3}$) corresponds to a clockwise rotation. Now, instead of reflecting the molecule in a mirror, invert all the substituents through the stereocentric carbon. You must now rotate the molecule so as to get the lowest-ranking hydrogen back to the bottom, and this causes one of the other substituents to slip between the remaining two. The $\ce{OH > C2H5 > CH3}$ order has thereby become counterclockwise; the central inversion acts like a mirror reflection by inverting the chiral orientation.


text:
Variational (Rayleigh-Ritz) method and multielectron wavefunction
I am studying Rayleigh-Ritz method and I have some problem to understand what the final wavefunction looks like. Suppose that we have a system with $N$ electrons, atom or molecule. We are interested in solving the time indepedent Schrodinger equation:
$$H\Psi = E\Psi$$
We assume a trial wavefunction $$\Phi = \sum_{i=1}^k c_i\chi_i$$
and we are trying to find the parameters $c_i$ that correspond to critical points (I am omitting the mathematics and the general procedure).
After the procedure we obtain different sets of parameters, $\{c_i \}_0$, $\{c_i\}_1$, $\{c_i\}_2$, $\ldots$, $\{c_i\}_k$. When substituted to second equation these parameters give back $\Phi_0$, $\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_k$. I interpret these wavefunctions as the ground state, first excited state etc.
Is this interpretation correct? What troubles me is that if $\Phi$ is just a linear combination of single-electron orbitals then how can we use it to represent a multi-electron wavefunction? I mean we have $N$ electrons, so $\Phi$ must have the form $\Phi (\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)$ which I can't understand how it can be obtained by just taking a sum.

In general this interpretation is not correct. A Hamiltonian set up to give you the ground state will only ever give you the ground state. If all, these are excited configurations or expansions of the ground state; they don't have a physical meaning of interpretation.

One Electron
Starting with your trial wavefunction:
$$
\Phi = \sum^k_{i=1} c_i \chi_i,
$$
you seem to be thinking about one-electron orbitals $\chi_i(r)$ which will add up to a one-electron molecular orbital (MO) $\Phi$.
In that case, when you find $c_i$ by minimising the energy
$$
E = \frac{\langle\Phi \vert \hat{H} \vert \Phi\rangle}{\langle \Phi \vert \Phi \rangle}
$$
that vector of coefficients will give you the ground state one-electron wave function.
For one-electron wave functions, E can be written as a matrix equation with the one-particle operator $\hat{H}$, where the eigenvectors are the $c_i$ vectors.  In that case (exact factorisation) the eigenvectors corresponding to higher eigenvalues correspond to other eigenfunctions of the system which are excited one-electron states.
If your basis of input functions $\chi_i$ isn't complete for the relevant space, your final ground state wave function $\Phi$ will only be an approximation, but the "best" approximation possible (lowest energy) in a given basis.
For example, the Hartree-Fock (HF) approximation to the one-electron hydrogen atom is analogous to the Rayleigh-Ritz (RR) approximation to the ground state.  When using a standard gaussian basis, used in most quantum chemistry programs, the quality of that answer and wave function will improve as you increase the number on input basis functions $\chi_i$.
Many Electrons
The simplicity of the RR approximation/ansatz relies on the fact that the equation for $E$ is a matrix equation that can be solved for eigenvectors $c_i$ which describe $\Phi$ in terms of $\chi_i$.
Many electron wave functions are anti-symmetric with respect to exchange of electrons.  One method for creating an anti-symmetric ansatz from one particle wave functions is the Slater determinant $\Psi$.  Modern approximations of many electron wave functions rely on composing wave functions of Slater determinants, which are themselves built up of one electron wave functions.
If we want to recover the linear form of the original expression for $\Phi$ we would need to use a basis of Slater determinants for our trial basis $\chi_i$,
$$
\Phi\left(r_1,r_2,...,r_n\right) = \sum^k_{i=1} c_i \Psi_i\left(r_1,r_2,...,r_n\right)
$$
The question this begs, is how to choose a suitable many electron basis $\Psi_i$.
The Configuration Interaction method uses this approach - a Rayleigh-Ritz approximation using a basis of Slater determinants.  It selects the Slater determinants by first solving the system with the HF approximation.  This gives a set of one electron MOs, some of which are occupied and some unoccupied.  By building Slater determinants with the occupied MO, then all but one occupied MO and one unoccupied MO (single excited Slater determinants), then all but two (double excited Slater determinants) etc, you can build an arbitrarily large basis of Slater determinants.
For a system containing $N$ electrons, $O$ occupied MOs and $V$ unoccupied MOs and $M=O+V$ total MOs, the number of Slater determinants grows exponentially as higher excitations are added.

The ground state has 1 HF Slater determinant,
There are $VO \sim M^2$ single excited Slater determinants,
There are $\frac{V(V-1)O(O-1)}{4} \sim M^4$ double excited Slater determinants

and so on.
As the size of the basis rapidly explodes, often on the ground state and double excitations are included, giving the CID or configuration interaction doubles method, the simplest possible CI method.
Note: For symmetry reasons, the singly excited determinants aren't coupled to the HF ground state by the Hamiltonian, and would have no contribution ($c_i=0$) as the only excited determinant are included, the simplest CI method is CID, singly excited determinants can then be added to give CISD.  See Brillouin's theorem.
Excited states
If the whole CI matrix is diagonalised, rather than relying on a iterative method to find the lowest eigenvalue, you can get excited states represented as sum of determinants.  However by constraining them to be within the basis of determinants made with HF MO they are generally considered poor approximations.  The exponetially increasing size of the CI matrix typically makes calculation such eigenvectors prohibitive.
The Configuration Interaction Singles (CIS) method is essentially doing this with only the singly excited determinants to calculate approximate one-particle excitation energies.
Problems with CI
One notable problem with CI methods is size extensivity - the property that N non-interacting copies of the same systems, e.g. 10 water molecules 1 m away from each other, has the energy of $N$ times that of one copy of the system.  This property is true for HF for example.
Obviously this is physically true, but by limiting the total number of excitations in CI even as the system size increases, CI is not size extensive.  Methods such as Couple Cluster, which approximates $Phi$ as a product of Slater determinants rather than a sum, have since been developed to build on the deficencies of CI.
Further reading:
Szabo, Attila, and Neil S. Ostlund. Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory.

pg 60, 2.2.7: Form of the Exact Wave Function and Configuration
Interaction
pg 231, Chapter 4: Configuration Interaction

CIS

pg 297, 5.3: Many Electron Theories with Single Particle
Hamiltonians

esp, pg 305, 5.3.1.2

If I understand it correctly the procedure is still the same (when moving to many-electron problems) but we change the form of basis functions? Now the basis functions are not one electron wavefunctions but Slater-determinants, which means that the $$\sum_i c_i\chi_i $$ leads to a function of $n$ variables if the system is composed of $n$ electrons? Also the elements of the Slater matrix are what we call molecular orbitals?

Yes, if you do a one electron calculation with an basis of atomic orbitals, your solutions will be molecular orbitals - linear combinations of atomic orbitals.
If you take $n$ molecular orbitals you can make an $n$ electron Slater determinant.  You will take $m$ Slater determinants, where $m$ is typically much much larger than $n$, and solve for $m$ $c_i$ variables.

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How can I derive the ratio of concentrations?
I need to derive the equation
$$
  \frac{[\ce{In^-}]}{[\ce{HIn}]}
  = \frac{A_\mathrm{int} - A_\mathrm{lo}}{A_\mathrm{hi} - A_\mathrm{int}}
\tag1\label{eq1}
$$
from the material balance equation that expresses the conservation of matter principle:
$$c=[\ce{HIn}]+[\ce{In^-}].\tag2\label{eq2}$$
I need to use the above mentioned equation along with the equation
$$
  A^{\lambda}
  = a_\mathrm{HIn}^{\lambda} b[\ce{HIn}]
  + a_{\ce{In-}}^{\lambda} b[\ce{In^-}]
\tag3\label{eq3}
$$
where $A^{\lambda}$ is the absorbance of an indicator solution whose pH is within the effective pH range.
I will admit, I have no idea how to go about this. I've tried finding this derivation online but I don't even know what exactly to search. Can someone please help.
If I substitute \eqref{eq2} into
$$A_{\ce{HIn}} = a_{\ce{HIn}}^{\lambda} b c,$$
then I get
$$A_{\ce{HIn}} = a_{\ce{HIn}}^{\lambda} b([\ce{HIn}] + [\ce{In^-}]).$$
What do I do from here? Did I start this correctly?

This question needs more details and work you have done so faar to solve it. Otherwise, it is more like a homework question and will be close it sooner.

At least, what is absorbance denoted by $A_{int}, A_{lo},$ and $A_{hi}$? What are the conditions?

The signals of the two species add up (linear relationship). If species B absorbs more than species A, call the absorbance of B at a concentration $c_0$ $A_\mathrm{hi}$, the absorbance of A at a concentration $c_0$ $A_\mathrm{lo}$. A mixture of A and B, with a total concentration of $c_0$, will have an intermediate absorbance of $A_\mathrm{int}$. In the example above, 70% B and 30% A has an absorbance that is closer to the absorbance of pure B (70% of the total way from A to B).
$$\frac{A_\mathrm{int} - A_\mathrm{lo}}{A_\mathrm{hi} - A_\mathrm{lo}}$$ gives you the percentage of B.
$$\frac{A_\mathrm{hi} - A_\mathrm{int}}{A_\mathrm{hi} - A_\mathrm{lo}}$$ gives you the percentage of A.
Defining "hi" and "lo" is just to avoid negative numbers. You could use the absorbance of A and B straight up for the same type of formula.
To get the ratio, divide the two fractions and cancel the two denominators.
This gives the general idea. For the formal proof, you would calculate the absorbance of pure A, pure B and a mixture using the linear relationship, and then plug that into the expression for the ratio of A and B (rhs of first equation). If all goes well, the ratio of A to B will result after simplifying the longish expressions.

Please check your second formula ! Because, in the percentage of $\pu{A}$,  I read $\pu{A_{int} - A_{int}}$, but this difference is equal to zero. So the percentage is zero!! I would guess the first $\pu{A_{int}}$ should be $\pu{A_{hi}}$. Am I right ?

@Maurice Yes, oops, you are right.

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Work done by gases on each other in the process of mixing them
I was wondering if gases, which are in the process of mixing, do any work on each other. I haven’t found many resources on this.
We all know that gases, which have a higher pressure than their surroundings, expand. For example, suppose a (approx. massless) piston divides an ideal gas and the ambient $\pu{1 atm}$ pressure. If we heat the ideal gas, the pressure rises, and the piston gets displaced so that the ideal gas can re-equilibriate to $\pu{1 atm}$ pressure, so outside pressure and inside pressure are the same.
But what is up with the case when two gases at different pressure mix? For example, an ideal gas is in a closed container at $p_1 (\lt p_2), V_1 (\gt V_2),$ and $\pu{273 K}$. Now this container is connected to a smaller one, inside is a different ideal gas at $V_2$, $p_2$, and $\pu{273 K}$. Now, of course the gases will mix with each other, to maximize entropy by mixing.
But does the higher pressure gas perform work against the lower pressure gas while expanding into the other container, like the first example I gave? Or is it like a free expansion (a.k.a no work done), because ideal gases don't occupy volume by themselves? And how would we calculate the work?

"to maximize entropy by mixing" sounds like the gases have free-will, desires, and goals.

In ideal gas context, work is done/absorbed by expansion/compression of regions with different pressure. It is irrelevant if they are filled  by different gases or the same gas. Spontaneous mixing of 2 different gases of the same pressure does not do any work, it just increases entropy and decreases Gibbs energy.

Consider what happens to just the gas at higher pressure, if allowed to expand to a lower pressure -- what happens to its temperature? Then consider the gas at lower pressure, and the converse.
As an example, consider two containers at 300 K, one with helium at 10,000 kPa (~100 bar), another, of the same volume, with He at 100 kPa (~1 bar). When a valve opens between them, the equalized pressure would be ~5,000.5 kPa, with far more gas (by mass) expanding than being compressed. Would you expect the final temperature to be lower or higher?
Of course, this is a highly simplified example. One would need to take in to account the heat capacity of differing gases, the volumes of the containers, the mass of gas in each, initial temperature of each and, to be closer to a "real life" answer, departure from ideal gas laws, and thermal storage in the containers themselves. Oh... and if the gas expand through a venturi or through a turboexpander.

If I understood your example correctly, I would expect the final temperature to be lower. If we assume adiabatic conditions (dQ=0), only work is done, and as much more helium expands to the other side, it also needs to push against the 1 bar of helium on the other side, and needs to overcome more and more pressure as it builds up on the side that was originally 1 bar. How could we quantify this though?

@Mäßige, one gas does not compress another. If gas A in container A expands to fill both container A and B, its partial pressure is the same, regardless of whether B originally contained nothing, or He under 100,000 kPa. Each molecule of gas does its own "pushing", and all molecules can fit between each other. Just consider each separately -- easily calculated.

Basically, if the partial pressure is the same regardless, then no work should be done, because the ideal gas molecules fit between each other and have no interactions?

Is work done when a gas expands from one container to another? Could that move a piston, or turn a turboexpander, or change the temperature of the gas and its container?

No. As you put it: When a gas expands freely (that would mean against no outside pressure) to just fill another container, dW=0 and dQ=0 (adiabatic container), that means dU=0 and no change of temperature, just different pressure and volume. So no work done at all, if we consider each gas seperately while mixing, as per ideal gas law (molecules themselves occupy no volume). That means that generally no work is done when two ideal gases (with p₁, V₁, T₁; and p₂, V₂, T₂) are getting mixed. Correct? (May differ when dealing with real gases)


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Anodic stripping voltammetry linear sweep rate?
I am using an PSoCStat potentiostat to measure Pb, with a Pt counter electrode, a pencil lead working electrode and an Ag/AgCl reference electrode. This device doesn't not support Square Wave Voltammetry.  Adding Bi(III) and some Pb shows the peaks as expected.
The problem is that the PSoCStat paper demonstrates Pb detection using a linear sweep rate of 0.1V/S. This results is a very weak peak in my experimental setup. Increasing the sweep rate to 10 V/S results in a very large peak. This is 100X the sweep rate used by the PSoCStat example and other LSASV examples I see on the web. I read that the faster scan rates do indeed increase the current.
The question I have is why is the LSASV examples using such slow rates which appear to decrease sensitivity? It is because slower rates allow narrower peaks which allow simultaneous Cd and Pb detection?

I am not a voltammetry expert (aside of Czech chemists are supposed to be as followers of the founder of polarometry/voltammetry Jaroslav Heyrovský), so just pointing out general electrochemical principles:
If there are multiple redox systems in a sample, slower rate may improve selectivity. As at fast rates, redox systems are far from equilibrium state, with reaction kinetics and diffusion playing bigger role.
Hypothetically, one redox system can have its stripping preferred thermodynamically, while other one kinetically, so the latter can start forming its peak while the former is not finished yet (but would be already at slow rate).
As consequence, fast rates kinetically broaden peaks due kinetic overpotentials. Additionally, there can be more significant transient capacitance or double layer change effects.


text:
Dissolution Thermodynamics - What is the standard state in units of mole fraction?
I have recently been working on a project that involves the correlation of mole fraction solubility using the Apelblat equation (in simplified form: $\ln\chi = A +\frac{B}{T}$), which can allegedly be used to compute some of the standard state thermodynamic quantities associated with the dissolution process via the van't Hoff equation ($\Delta H_s^{\circ} = -R \frac{d \ln K_d}{dT^{-1}} = -R \frac{d \ln \chi}{dT^{-1}}$). This seems to be the common approach in the literature that I have seen so far, and doing a fit with some experimental data gives a fine fit and reasonable values for $\Delta H^{\circ}$ and $\Delta S^{\circ}$, but if we attempt to use the typical $\Delta G^{\circ} = -RT\ln K$ relation we run into some trouble due to the definition of the mole fraction unit. Essentially, since the mole fraction of solute in solution will always be less than unity, it appears that this "standard state" Gibb's free energy will always be positive. This appears to defy the obvious experimental observation that the solute dissolves at STP.
This causes questions to arise about what necessarily is meant for a standard thermodynamic change when working with mole fraction solubilities. Given how prolific this sort of analysis has been in the literature I have read, I am puzzled that this point seems to not have been explored very much. My initial instinct to resolve this issue to to convert the mole fraction to molarity, while multiplying the thermodynamic quantities found through the Apelblat coefficients  by the same factor to get something more reasonable, but I am not sure how much this throws the entire idea of Apelblat correlation  to the wind.
Anyways, my main question is what is meant by standard state when working with units of mole fraction in dissolution thermodynamics?


text:
Van der Waals Parameters (non-bonded parameter) in Amber?
I have a complex that contains transition metals. I am doing parametrization and building the force field.
I have one question about the frcmod file.
In the file,
There are non bond paramters that are written like that.
OH 1.7210 0.2104
CT 1.9080 0.1094

When I googled it, I found that the numbers are the radius and well depth parameter, respectively.
But I dont know, how can I measure these values, and I am not sure if I will need the van der waals equation or not ?
$$E_{VDW} =4 \sigma [1/4(r_0/r)^{12} - 1/2 (r_0/r)^{6}]$$

This is a duplicate - and yes, you'll need to get both from a potential energy curve.

@GeoffHutchison Thanks for your help.
I tried to look for all the answer related to this question, but I have some doubt, because they use water dimer or helium dimer.
1- In my case, I have a complex, and it contains transition metal in the center and this transition metal form 6 bonds with Nitrogen. Therefore, I don't know which distance should I use to do the scan and fit the equation ?
2- I saw the tutorial to use psi4numpy. But After I form the potential energy curve, I don't know what should I do next to get the values that it is written in frcmod (mentioned in the post)


text:
How can I get a breakdown of different salt contents in a fertilizer solution?
I am experimenting a lot with hydroponics i.e. growing plants in a liquid solution. From time to time, I would like to do a check on what is the current salt level for major salts (ions) in parts per million (ppm) of my fertilizer solution.
For instance, Calcium, Nitrate, Nitrite, Fe, etc. would be some nice measurements that I would like to tabulate. For instance, I use this device for measuring calcium level.
Is there a general method/ device that I could use to get a breakdown of commons ions in plant fertilizer (NH4, K, Ca, Mg, P, S, etc.). Even if I can just get N-P-K, that'd do too.


text:
2p wave function radial distribution
The following graph shows the values of the wave function $\psi_{210} $ (i.e. the $2\mathrm{p}_z$ orbital) versus the radius (divided by $a_{0}$).

I don't understand it, however, since the $2\mathrm{p}_z$ orbital is not spherically symmetric:

do these represent the values along the $z$-axis (perpendicular to the angular nodal plane), or

is it the average of the values all around the sphere (some areas are zero-valued, some are non-zero-valued...)?

If (1) is the case, then why isn't it specified by the graph? They could have said $\theta = 0, \phi = \pi/2$.

What you're seeing is not a wavefunction, thus, the title of this graph is quite misleading. As you correctly pointed out, a wavefunction has an angular dependency as well; in fact, wavefunctions can generally be factorised as the following:
$$\psi_{nlm}(r, \theta, \phi) = R_{nl}(r) \cdot Y_{lm}(\theta, \phi)$$
$R$ and $Y$ here are respectively called the radial and angular (components of the) wavefunction.
The graph that you've shown is actually only showing the $R$ component (technically speaking, it is showing $|R|^2 r^2$, which is what we call the radial distribution function).
Where does this expression $|R|^2r^2$ come from? Well, the electron density at any point is given by $|\psi|^2$ — we know this from quantum mechanics. The total electron density, let's call it $\rho_\mathrm{tot}$, is obtained by integrating $|\psi|^2$ over all space:
$$\begin{align}
\rho_\mathrm{tot} &= \int_0^{2\pi} \int_0^{\pi} \int_0^\infty |\psi|^2 r^2 \sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi \\
&= \int_0^{2\pi} \int_0^{\pi} \int_0^\infty |R(r)Y(\theta, \phi)|^2 r^2 \sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi  \\
&= 1.
\end{align}$$
[the $r^2 \sin\theta$ factor arises naturally when performing integration in spherical coordinates.] Of course, the total electron density in any single wavefunction is just 1, so that's not very interesting. What's more interesting is if we were to integrate out only the angular components, such that what's left only contains a radial dependence, and tells you how much electron density there is at a given $r$. Essentially, what happens is that all the angular bits above integrate out to 1 (by normalisation); and you're left with an integrand of $|R|^2 r^2$.
Because we are integrating over all angles, what you get at the end is pretty much an average (or more accurately a sum): so, to directly answer your question, it's more like the second option rather than the first.
For s-orbitals (i.e. $l = m = 0$), it turns out that the angular component is simply a constant: that is to say, there is no angular dependency, and hence we have $\psi \propto R$. In this case, it is OK to conflate the two. But for other orbitals this is not permissible.
I wrote more about this before, so for further information, please see: What is the exact definition of the radial distribution function?

I don't understand how we can have the choice to normalise the angular wave function? Is it a choice we have when solving the Shrödinger equation ?

Yes, it's conventional to separately normalise both the radial and the angular components (which then leads to the entire wavefunction being normalised).


text:
How to find the standard deviations of lattice parameters after Rietveld refinement with GSASII?
I am trying to learn how to use GSASII. It is different from GSAS/EXPGUI, so I feel hard to find some information from GSASII interface. I want to find what are the values for the standard deviations of lattice parameters after Rietveld refinement. How can I check it?

I assume you are looking for the values of "estimated standard deviation, e.s.d." (also called standard uncertainty, s.u.).
You may find the values by exporting the result into a cif file. Click Export > Entire project as > Full CIF


text:
How might one distinguish between two acyclic diastereomers other than by X-ray diffraction?
I have a mixture of two racemic diastereomers (R,R and R,S - and their enantiomers), and while I can fairly easily tell with 2D spectra which NMR signals correspond to one molecule and which signals correspond to the other, I can't find any way to tell which molecule is which (the system is acyclic and so rotation is unrestricted). The ratio is not quite 1:1 (I have several samples with slightly different ratios), so if another technique could determine that, say, the R,S compound is the major compound, that would allow me to assign the NMR spectrum.
I'm wrapping up my current project so it probably doesn't matter practically to me, but nevertheless I'm curious: is there any way to tell which is which, by NMR or another method? (Even if I could purify it fully, it's an oil, so XRD isn't on the cards).
I realise the answer is probably just "Obtain enantiopure starting materials and see what compound you get".

It is possible... to a certain extent. Broadly speaking, the strategy is as such:

calculate the chemical shifts of both diastereomers using DFT
match them against the experimentally determined chemical shifts for each peak
use some kind of statistical measure to determine how "confident" you are in the assignment

The calculated shifts of course don't match up 'perfectly' with the experimental shifts; but all things being equal, it is more likely that they match up than not, so the statistical measures essentially take into account the error (i.e. experimental shift minus calculated shift) in order to determine which NMR spectrum is likely to be which compound. This technique has been used quite successfully for structure (re-)assignment. In your case, where you have the spectra of both diastereomers and want to match them against only two possibilities, then the statistical measure to use would be 'CP3', see:

Smith, S. G.; Goodman, J. M. Assigning the Stereochemistry of Pairs of Diastereoisomers Using GIAO NMR Shift Calculation. J. Org. Chem. 2009, 74 (12), 4597–4607. DOI: 10.1021/jo900408d.

The alternative case would be where you have one spectrum only and you don't know which diastereomer it is, in which case you should use 'DP4' (or its derivatives):

Smith, S. G.; Goodman, J. M. Assigning Stereochemistry to Single Diastereoisomers by GIAO NMR Calculation: The DP4 Probability. J. Am. Chem. Soc. 2010, 132 (37), 12946–12959. DOI: 10.1021/ja105035r.

(there are many more followup papers on DP4 so please do a literature search if you're interested)

Now, the issue with this is that most of the compounds shown there (and virtually all of the compounds assigned with these methods) is that they are polycyclic or otherwise structurally rigid. The reason for this is fairly obvious: you don't need to worry about (too much) conformational flexibility & hence averaging. So, I'd be wary of naively applying CP3/DP4 to your molecule.

Generally speaking, I'd say that accurate prediction of NMR shifts for flexible molecules is still an open research question. Essentially, you need to generate an ensemble of conformers, rank them in terms of energy, and then calculate the NMR shifts as an average, weighted by the Boltzmann population $p \propto \exp(-E/k_\mathrm{B}T)$. I did some research on this in my PhD, and in my experience the main issue here is that getting sufficiently accurate energies is very difficult — if you have many conformers which are very close in energy, then you have to get the energies extremely accurate, otherwise the relative populations are way off, and you get inaccurate results.
However, there has been progress on this made in recent years. In particular, crest (along with xtb) is a really good programme for generating conformer ensembles. The energies should not be trusted too much as it uses a relatively low level of theory, but once you have the geometries you can recalculate the energies with something more expensive. https://github.com/grimme-lab/crest Here are some more references:

Wu, J.; Lorenzo, P.; Zhong, S.; Ali, M.; Butts, C. P.; Myers, E. L.; Aggarwal, V. K. Synergy of synthesis, computation and NMR reveals correct baulamycin structures. Nature 2017, 547 (7664), 436–440. DOI: 10.1038/nature23265.
Grimme, S.; Bannwarth, C.; Dohm, S.; Hansen, A.; Pisarek, J.; Pracht, P.; Seibert, J.; Neese, F. Fully Automated Quantum-Chemistry-Based Computation of Spin-Spin-Coupled Nuclear Magnetic Resonance Spectra. Angew. Chem., Int. Ed. 2017, 56 (46), 14763–14769. DOI: 10.1002/anie.201708266.
Grimme, S. Exploration of Chemical Compound, Conformer, and Reaction Space with Meta-Dynamics Simulations Based on Tight-Binding Quantum Chemical Calculations. J. Chem. Theory Comput. 2019, 15 (5), 2847–2862. DOI: 10.1021/acs.jctc.9b00143.
Bohle, F.; Grimme, S. Hydrocarbon Macrocycle Conformer Ensembles and
13C-NMR Spectra. Angew. Chem., Int. Ed. 2022, 61 (14) . DOI: 10.1002/anie.202113905.

I expect that some of these methods are already, or will soon be, implemented in the ORCA suite of programmes. (However, I'm not super up to date; it's been a couple of years since I really looked at computational NMR...)
So, the tools are all there, but I don't think it's a thoroughly proven / tried-and-tested method. If you actually manage to use these to assign your molecule (and get some hard proof using some other analytical method to supplement it), I expect it would be publishable!

Finally, if your compound has a particularly common structure (for example, if it's a polyketide), it may be possible to use empirical correlations between J-couplings and local stereochemistry to "map out" the stereochemistry along a chain. This is a slightly older, and IMO more crude, method — but it does exist. Some terms to look for are J-based configurational analysis, and Kishi's "universal NMR database".

If the computational approaches don't give a definite result, the historical approach is to react the mixture with a reagent that is selective for one structure. Without knowing the structures of your compounds, I cannot suggest anything specific, but there are many options. Chiral reagents (particularly enzymes) can even be used to distinguish between enantiomers. In your case, even altering one enantiomer of one of the diastereomers should be sufficient for you to tell which NMR peaks have been affected.
If you are willing to share the structures of the compounds of interest, you may be able to get specific recommendations of reagents to use.

Yeah, if it wasn't too precious to derivatise / react, I'd definitely go with old methods like Mosher ester instead of trying to use computation (which would be ambitious but no guaranteed result). (+1 obvs.)


text:
Example of reactions series A → B followed by A + B → 2 C
I'm currently writing about geometric numerical integration and I need an example of the following series of chemical reactions:
\begin{align}
\ce{A     &->[$k_1$] B} \tag{R1} \\
\ce{A + B &->[$k_2$] 2 C} \tag{R2}
\end{align}
The reason I want to use this form of the reaction is because
$$\frac{\mathrm d}{\mathrm dt}([\ce{A}] + [\ce{B}] + [\ce{C}]) = 0$$
and this would be an example of a linear invariant.

Not exactly the same, but what about O2 -> 2 O and  O2 + O -> O3 ? with d(2p(O2) + p(O) + 3p(O3))/dt=0.  Or the same with molar amounts or molar fractions. Formally with molar concentrations too, but that would be little unusual for gases. But possible.

Is that correct though? When I do my calculations I get $\frac{d[O_2]}{dt}=-k_1[O_2]-k_2[O_2][O]$, $\frac{d[O]}{dt}=2k_1[O_2]-k_2[O_2][O]$ and $\frac{d[O_3]}{dt}=k_2[O_2][O]$ and if you add these you get $\neq0$. Or am I doing something wrong?

I was not addressing the kinetics just the mass inventory in context of the invariant. As the total mass of O allotropes is constant the given derivative is zero.

Let's go back a bit further. Why do you think your invariant holds? The overall reaction is $\ce{2A -> 2C}$. Are you applying steady state?

I haven't worked with kinetics in quite some time so I might be wrong. But I guessed that our total molar concentration should be an invariant since $\frac{d[A]}{dt}=-k_1 [A] - k_2 [A][B]$, $\frac{d[B]}{dt}=  k_1 [A] - k_2 [A][B]$ and $\frac{d[C]}{dt}= 2k_2 [A][B]$. And then we get that $\frac{d}{dt}([A]+[B]+[C])=(-k_1 [A]- k_2 [A][B]) + (k_1 [A] - k_2 [A][B]) + 2k_2 [A][B]=0$ therefore we have a linear invariant.

@zhe The invariant of his original supposed schema holds because of trivial total amount inventory, which is constant.

Is there a reason you can’t have just $\ce{A -> B -> C}$? There are many examples of that scheme.

Yes that would satisfy it but since I'm writing a paper on mathematics primarily I want something more computationally interesting.

But it is not that important to find a real example. I just thought people with more chemical knowledge might know some thing spontaneously.


text:
Electrodeposition of aluminum
I wanted to electroplate aluminum onto a 3D printed part (treated with conductive paint). I found out that you can plate it with copper, zinc, gold, or palladium from an aqueous solution, but electroplating with aluminum seems to be difficult. I have two questions:

Why is that?
I found that you can use molten salt, benzene-tetrahydrofuran-AlCl3-LiAlH4 or  AlCl3 in acetamide to electroplate aluminum onto a surface. Would it work if I use AlCl3 in DMSO?

Aluminium has too negative electrochemical potential to be deposited from water solutions. Hydrogen obtained by reduction of H+(aq) or even H2O is preferably produced.

Exposing of 3D prints - or at least their conductive paint - to organic solvents may lead to damage or even destruction of the substrate aluminium is supposed to be deposited on. Check solvent - print-with-paint compatibility.

Aluminum coating are made with vapor deposition.

In my notes, I got a report from the year 1950 saying that the metal aluminum can be obtained by electrolysis in a solution of $\ce{AlCl3}$  in ether, of which a concentrated solution of 2-ethyl-hexylamine in ethanol has been added. In case of necessity, I could find this review. I only have the code of my university library.

BTW, plated aluminum, if pure, would be very soft, easily scratched by a fingernail. If it is possible to plate that part, and to do so wit a thick enough layer, it might require anodizing to withstand handling.

@Maurice I would be afraid of combination of electricity and diethylether. One small spark....

So, time went by and I got a little wiser. Answering my own question in the hopes that it may be useful to someone, somewhere, sometime.

The issue is that aluminum has an electrode potential way lower than hydrogen, so in an aqueous solution instead of Aluminum being reduced on the cathode, the production of hydrogen will be favored strongly.

Maybe. You can at least electroplate aluminium very well from DMSO2 (dimethylsulfone). The conditions reported were however 110-150 °C and the electrolyte has to be prepared in an argon-filled glove box (Source: https://doi.org/10.1016/j.surfcoat.2006.11.035), I suppose because AlCl3 is very hygroscopic.

Aluminum can be obtained by electrolysis of $\ce{AlCl3}$ in DMSO, as shown by you in your answer, iblue. But it should be stated that

the solution should be absolutely anhydrous

once deposited on any metallic support by electrolysis, aluminum must be kept and stored in an absolutely dry atmosphere. This allows the aluminum metal to be in contact with oxygen $\ce{O2}$ to produce a thin, waterproof, and continuous layer of aluminum oxide $\ce{Al2O3}$. This continuous layer prevents the aluminum metal to get in touch with water. All pieces of aluminum metal in the world are covered by this thin film.

Without this oxide layer, aluminum reacts quickly and violently with water, as can be demonstrated by dipping an aluminum piece in a solution of mercury chloride $\ce{HgCl2}$ : $\ce{HgCl2}$ is able to cross the aluminum oxide layer and react with the metal according to $$\ce{3 HgCl2 + 2 Al -> 3 Hg  +  2 AlCl3}$$ The deposited mercury produces an amalgam with aluminum $\ce{Al(Hg)}$. And then the continuous aluminum oxide layer is broken. As a consequence, the unprotected aluminum metal reacts violently with water according to $$\ce{2Al  +  6 H2O -> 2 Al(OH)3  +  3 H2}$$ This demonstration was popular in the middle of the previous century. Nowadays it is obviously not feasible anymore, because of the toxicity of mercury salts.


text:
What is the physical significance of the damping function in calculation of vdW interaction energy?
In order to take care for the singularity (also removing the double counting of exchange-correlation) obtained with this expression,
$$E_\mathrm{vdW} = \frac{C^6}{R^6}$$
many type of damping functions are utilized. Such as,
$$E_\mathrm{vdW} = \frac{C^6}{R^6} \cdot f_\mathrm{famp}(r)$$
One such damping function is, (Grimme type, Fermi function)
$$f_\mathrm{damp}(r) = \frac{s_6}{1+\exp\left[ -B\left(\frac{r^{AB}}{s_r R^{AB}_\mathrm{vdW}}\right)\right]}$$
when, $r$ = interatomic distance, $R^{AB}_\mathrm{vdW}$ or (R) is summation of vdW radius of two atoms $s_6$ and $s_r$ are two parameters.
$s_6$ depend upon the XC functional and $s_r$ is another scaling factor.
$B$ is the exponent.
Literature$^{[1]}$ suggests that value for $B$ is found by setting $f_\mathrm{damp} = 0.99$ when $r = 1.2 \cdot R$.
Also, the value of these parameters can also be found from here.
To my knowledge, these values are optimized for each functional.
Is there any physical meaning to these values of the parameters?
Actually, I am more interested in the $s_r$ value and if this has any physical significance...
[1] J Chem Phys, 116, 515(2002)


text:
Why does chickpea flour need to rest for several hours?
Many traditional Italian recipes make use of chickpea flour (e.g. farinata and cecina). Before cooking the recipes suggest to mix the flour with water and make it rest for several hours (in some recipes they suggest 6/7 hours).
What happens from a chemical point of view? Does such a long time really makes a difference?

Chickpeas are high in phytic acid.

This is good, because it's a nutritional anti-oxidant, helps prevent kidney stones, and has other health benefits.
But it can be bad because humans don't produce phytase, the enzyme that digests phytic acid, which will bind to other nutrients, such as iron, zinc, and calcium, making them nutritionally unavailable.

Long slow cooking, or long soaking will break down most of the phitic acid, leaving enough for the benefits while removing most of the detriments.

A closer look at phytic acid
Phytic acid, or phytates, is the stored form of phosphorus found in seeds, nuts, legumes, and unprocessed whole grains (concentrated in the outer bran layers).
The amount of phytic acid in these foods varies widely even within the same food, based on the seed type, environmental conditions, climate, and soil quality.
Phytic acid can have health benefits due to its antioxidant properties.
Laboratory and animal studies show that it can protect against DNA damage and cancer cell growth.
The chemical name for phytic acid, inositol hexakisphosphate or IP6, is very actively studied in cancer research and, though research is still early, IP6 has become a popular over-the-counter supplement.
Phytic acid can also prevent kidney stones from forming by inhibiting the buildup of calcium crystals, a component of kidney stones.
However, phytic acid is also labeled an antinutrient because humans lack the phytase enzyme needed to break it down.
As it passes through the gut, phytic acid binds to minerals like iron, zinc, and calcium so they are not well-absorbed in the intestine.
This will occur only when phytic acid is eaten with foods containing these minerals at the same meal.
The average Western diet usually has enough nutrients and variety to protect from a true deficiency, especially with the inclusion of some animal proteins (e.g., lean pork, poultry, shellfish) that are rich in zinc and heme-iron, a well-absorbed form of iron.
A nutrient deficiency more likely occurs in developing countries where a variety of food choices is limited, the risk of malnutrition is higher, and legumes or whole grains are staples that are eaten with every meal.
Also at risk are those who eat a vegan diet (which not only consists of plenty of seeds/nuts, legumes, and grains but these foods contain poorly absorbed non-heme iron), or those who already have an iron or zinc deficiency due to medical reasons.
How you prepare foods high in phytic acid can reduce the overall amount.
Cooking, soaking overnight in water, sprouting (germination), fermentation, and pickling can all break down phytic acid so that the phosphorus can be released and absorbed by the body.
Some natural bacteria in the colon contain the enzyme phytase and can also help to break it down.
— Are anti-nutrients harmful? | The Nutrition Source | Harvard T.H. Chan School of Public Health

I suppose such recipes are older than awareness of fytic acid metabolism, so they are probably based on long term empirical food experience.

About phytic acid = inositol hexakisphosphate,  according to its chemical structure, it does not seem to have antioxidant properties.

As chick pea is not a chemical, it is hard to say. 2 options come to my mind:

It is possible the hydration at natural pace at room temperature provides better results. Direct boiling and related protein denaturation may affect the hydration.
Legume frequently contains insoluble polysacharides like inulin(an analog of starch, based on fructose), which are not digestible by humans, but as consumed by bowel bacteria, causing flatulance. Long term soaking probably activates Inulinase, what is an enzyme hydrolyzing inulin to fructose.
About phytic acid (inositol hexakisphosphate), read the Rays answer. Note that according to its chemical structure, it does not seem to have antioxidant properties.


text:
Impact of temperature on an Energy profile diagram
What happens to the energy profile diagram if you increase the temperature of the system?
From my understanding, I think the whole diagram would shift upwards, as the reactants and products will both have more energy at this higher temperature. However I'm not too sure what would happen at the peak. Would that also increase? Or would it remain the same, if so doesn't this mean increasing temperature technically reduces the activation energy of the reaction?
ie in the following diagram, if black is the original energy profile diagram, at a higher temperature, will it look like red or blue?

A diagram seems to be missing. Energy can mean electronic energy, kinetic energy etc. Evidently it is a reaction schematic you are referring to. The activation energy depends on the energy of the intermediate wrt reactants.

Answer is depends. There is no "answer fits all" here. However, one universal principle is that higher T will result in population of more high energy states. The energy separation between activated states might be small compared to reactants, however.

Increasing temperature doesn't affect  the activation energy ,but it increase the number of molecules which having activation or more.

@AdnanAL-Amleh, this is incorrect.

@Martin-マーチンThanks .Referring to the diagram at the reference bellow :That activation energy can’t be altered by temperature.

https://www.chemguide.co.uk/physical/basicrates/temperature.html

@AdnanAL-Amleh That article is designed for beginning students, and thus paints a simplified picture in which the T-dependence of activation energy is ignored. This simplified picture makes sense for beginning students, particularly since the T-dependence of the activation energy is small. However, the picture the OP drew explicitly introduces a possible T-dependence in the energies of reactants, products, and activated complex. Hence this question, intentionally or not, goes beyond that simplified picture. E.g., suppose you arbitrarily set the energy of reactants as 10 and the activated ...

..complex as 20. Then the activation energy in the forward direction is 20 – 10 = 10. Now suppose we increase T, and suppose that increases the energy of the reactants by 5 (to 10 + 5 = 15) and of the activated complex by 4 (to 20 + 4 = 24). Then the activation energy in the forward direction is now 24 - 15 = 9. I.e., because the increase in T affects the energy of the reactants and the activated complex differently, the activation energy (which is the difference between them) changes as you change T. But, as I note in my answer, this is typically a small effect.

@theorist:Thank you for remediating my misconception, that's I taught my students all  the time that Activation energy independent of temperature but dependent on catalysis as written in the text .

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
Let's suppose that the reaction is taking place at constant pressure (for constant volume, see the last paragraph).  In that case, the most likely state functions to be represented on the vertical axis are H and G.  Then:
$$\pd{G}{T}{p} = -S$$
$$\pd{H}{T}{p} = C_p$$
From this we can see that the vertical coordinate can't be G. That's because your diagram has the dependent variable increase with temperature.  But, since $S>0$ (always), $\pd{G}{T}{p} <0$ (always).
Now, what about the sign of $C_p$?  Well, $C_p=\frac{\text{đ}q_p}{dT}$.  If you flow heat into a system ($\text{đ}q > 0$) without making any other changes, the temperature will always increase.  And vice versa if heat flows out of a system.  Thus the signs of $\text{đ}q_p$ and $dT$ will always be the same $ \implies C_p > 0 \implies \pd{H}{T}{p} >0$.
Thus, by elimination (and based on our starting assumption), we conclude the vertical coordinate is H.
In that case, we would expect the temperature-dependence of the enthalpies of the reactants, activation complex, and products to be relatively small—the main effect of increasing temperature would be to increase the rates of the forward and reverse reactions, by increasing the proportion of reactants and products able to pass over the activation barrier, as well as the collision frequency.
However, to the extent that you do increase the temperature enough to see a temperature-dependence in enthalpies, it would increase the enthalpies of everything—the reactants, products, and activation complex.  Thus, qualitatively, the blue curve would be correct.
I say "qualitatively" because the heat capacities of the reactants, products, and activation complex will all be different, and thus the extent to which their respective enthalpies are raised by an increase in temperature will likewise be different.
Note that the same argument would apply if we were at constant volume instead of constant pressure, in which case the most likely state functions to use for the vertical axis would be either U (the internal energy) or A (the Helmholtz free energy).  And here the only one of these two whose value could increase with temperature (at constant volume) would be U.
$$\pd{A}{T}{V} = -S < 0$$
$$\pd{U}{T}{V} = C_V > 0$$
$\endgroup$

@KarstenTheis  Thanks, I've added an explanation for why heat capacity must also be positive.  Regarding activation energy/enthalpy:  Yes, that restates what I said in my last paragraph. Consequently, the activation  enthalpy (at constant p) can either increase or decrease, since it's measured relative to the enthalpies of either the reactants or products (depending on which direction you're considering).  But, at constant p, the enthalpy of the activation complex (as opposed to the activation enthalpy) always  increases with temperature.

I really thought I have a quite solid grasp on this, but I don't get it. I do not understand any of your argument. I do not expect the OP to understand it either.

Why should it necessarily be enthalpy and not internal energy (not that it affects any conclusions)? [genuine question; not trying to be accusatory, because I'm sure I've seen these things plotted with just pure energies]

@orthocresol I added a paragraph at the end that should address your question.  Please let me know if it doesn't.

@Martin-マーチン I'm not sure what you're trying to say here.  Are you asking me to help you understand my answer, or just complaining that my answer is not understandable?  I don't wish to presume here, but I gather that Karsten and orthocresol understood it, and I would think with your background you'd have no problem understanding it as well.  Besides, it's not that complicated.  So I guess you could say that I am confused by your confusion :).

A reaction coordinate diagram has a horizontal axis that represents the progression of the reaction and a vertical axis for free energy. Temperature has little effect on the diagram.
Where temperature mostly comes into play is interpreting what a given difference in free energy means. In general, the relationship of interest is between free energy and either a ratio of concentrations or, indirectly, the rate of a reaction.
In such a relationship, we see a term that looks like $-\frac{\Delta G}{kT}$.
This basically means that for any of these relationships, at higher temperature, the effect of having different free energies is less. For example, rates are going to increase at higher temperatures (the height of the barrier is less important); and equilibria will approach equal preference for both sides (the intrinsic preference of the equilibrium matters less).

@Maurice :  Vertical axis represent free energy or enthalpy?

@Adnan AL-Amieh. Why do you ask me about the axis ? I am not the author of the question.

If the vertical axis is free energy, the difference between the product and reactant (ie the $\Delta G$ of the reaction) is temperature dependent, even assuming standard concentrations (as is normal in a reaction diagram). It is also common to use enthalpy for the vertical axis, which varies much less with temperature.

@Andrew That's a good point. In principle those values are not truly constant as functions of temperature...

@Andrew Interestingly, if the choices are blue or red, and assuming we're at constant p, the vertical axis cannot be G.

@theorist - good point. I was intending a more general statement that $\Delta G^\circ_T$ varies much more with T than $\Delta H^\circ_T$, not a reference to the specific red or blue options.


text:
Reverse Engineering of Cooling Wallet
I am looking to reverse engineer the Cooling wallet scenario. For e.g. FRIO has created technology where their inner pouch soaked in water and it cools down and keeps it there for 45h (as per their website). We can do this 1 time activity, but how come it becomes reusable. Does anyone know how create that chemical or what is that chemical they are using.
so basically i want to create a product like shown in this video:
https://www.youtube.com/watch?v=xjt19lvEwQY

Keep in mind CH SE readers are not big fans of watching youtube videos (related to the question :-) ).

Not speaking about what FRIO uses, but they are usually based on ordinary latent heat of melting of water, with melting temperature possibly shifted to freezing values by additives, like salts, ethanol, glycerol, ethylenglycol, diethylenglycol.
It is not a big deal to prepare a solution that freezes e.g. at -18 °C, so it keeps near that temperature until it is melted. (The temperature changes as the partial melting/freezing progress affects the liquid composition and the melting/freeing point.)
Similar packages (ice packs), stored in freezers, are used as cooling packs in insulated containers for transportation of temperature sensitive chemicals, samples or cargo.

Thanks @poutnik ! Do yo have any idea how to approach to solve this problem


text:
Is it possible to make a copolymer using Rice wax and Sebacic acid?
I would like to make a copolymer using Sebacic Acid, however I dont know which process of polymerization is required and which other monomer can be used, I would like to use Rice Wax or rice bran wax.
Is this possible to do?
[Edit]
It seems that rice bran wax is not suitable to produce a copolymer with sebacic acid - I am now in search of a naturally derived oil or wax that with polymerize with sebacic acid.

oils and waxes by definition are esters that would largely be unsuitable for polymerization with a diacid. There a a few relatively rare dihydroxylated fatty acids found in some oils and waxes, but you would have a hard time isolating enough to be useful. You would have to chemically convert a fatty acid into a diol, which may or may not fit your definition of "naturally derived"

Sebacic acid is a diacid, so the common compounds used to make a polymer with it are diamines (to make a polyamide) and diols (to make a polyester). Rice bran wax is composed of monoacids, so it is not going to make a polymer with sebacic acid.
The best known polymer using sebacic acid is made with 1,5-pentanediamine (aka cadaverine) and is known as nylon 5,10.

Thanks for the answer andrew, in this case is it possible to use another oil such as jojoba oil as the diol?

@RohamC no, jojoba oil is not composed of diols.

Thanks again for your insight Andrew. The reason I ask this is because I have come across a product that claims it is a copolymer used to seal the structure of broken hair. Now this is something im trying to achieve too and im looking for possible oils/waxes that would make a similar product. The one I came across was a copolymer made from hydrogenated castor oil and sebacic acid.... As Im reading more about these compounds I have even more questions so I do appreciate your knowledge... if you know of any naturally occuring oils that could be used to create copolymers then please let me know!


text:
Why does a carbon double bond "break readily" but this is not so for other elements (nitrogen double and triple bonds are stronger)?
A level chemistry student here, fairly basic question. We always learnt (from GCSE bond enthalpies) that "Bigger bonds are stronger", so to speak; my teacher often quoted $\ce{N#N}$ as very unreactive as the triple bond is very strong. However, now we are covering organic chemistry in a bit more depth at A-level, they are saying that the alkanes are generally unreactive as the $\sigma$-bond (which I read as: single bond) is fairly strong, but the $\pi$-bond (read: double bond) has more exposed electrons and so breaks more easily.
This is quite hard to accept, considering it contravenes everything I've ever learnt about bonds (more shared electrons will surely exert a stronger electrostatic attraction and have a higher bond enthalpy, no?) but then again we are not taught about bond energies in any great detail. I am hoping someone could explain or motivate this a little (I imagine a fully correct answer would be way over my head ;)): why are double/triple bonds sometimes weaker, but most of the time stronger, than single bonds? And what properties of the different elements (e.g. this holds for carbon but not for nitrogen) cause the discrepancy?
chem.libretexts.org Bond_Energies:

Bond
Bond energy [kJ/mol]
Incremental bond energy [kJ/mol]

$\ce{C-C}$
347
-

$\ce{C=C}$
614
264

$\ce{C#C}$
839
225

$\ce{N-N}$
160
-

$\ce{N=N}$
418
258

$\ce{N#N}$
941
523

Your teacher told you that nitrogen gas is a powerful explosive...?

Well, that the $\ce{N#N}$ bond is why nitrogen features in many explosives (TNT I think? Things like that) @orthocresol because when it is broken, it releases an unusually large amount of energy as its bond enthalpy is quite high (so we are told)

We are all doomed as atmosphere is full of $\ce{N#N}$ !!  :-) // No, TNT does not contain $\ce{N#N}$. //It is exactly the opposite, you need enermous amount of energy to break the $\ce{N#N}$ bond, that is why $\ce{N2}$ is so little reactive, not reacting directly with most of elements.

:) Oh ok in the converse then - I must have misremembered - $\ce{N#N}$ is very unreactive as the bond enthalpy is so high @Poutnik

But, the question stands (I'll edit out the explosive part)

https://chemistry.stackexchange.com/questions/77464/why-is-pi-bond-in-n-n-stronger-than-sigma-bond

https://chemistry.stackexchange.com/questions/100110/bond-dissociation-energy-order-for-c%e2%88%92c-n%e2%88%92n-o%e2%88%92o-and-f%e2%88%92f-single-bonds

Thanks for the links Mithoron, but I don't really understand most of them :) @Mithoron

The contexts are different here. Yes, nitrogen features in many explosive compounds because the explosion forms nitrogen gas. In this case, the N–N triple bond does not even exist in the starting material, so 'all three' of the bonds have to be formed at once, and that releases that huge amount of energy (941 kJ/mol according to your table).
Conversely, when one talks about alkenes or alkynes being relatively reactive, the issue is not about how strong the triple bond itself is. These reactions usually involve a double bond being converted to a single bond, or a triple being converted to a double. Thus, the energetic term under consideration is only the difference between the double and a single bond, which is much smaller.
So, it's not so much that nitrogen is special and carbon isn't. It's true that the nitrogen bonds are slightly stronger than the carbon bonds, but that is not really a factor here. The main issue is that you're comparing two different things (formation of a full triple bond vs the breaking of 'half' a double bond, or perhaps even less than half).

Right, and the difference between a $\pi$ and $\sigma$ bond's energy is less than the energy it takes to fully break a $\sigma$ bond, but if you were to attempt breaking both at the same time then the double bond is still stronger. Have I interpreted correctly?

Of course. Such reactions generally don't happen, and if they do, they happen stepwise in that the double bond is first converted to a single bond which is then broken.

@FShrike Yeah, seems like we all would like for your question to be better than that, but it seems to boil down to that a "breaking" of double bond only reduces it's order to one and new single bonds are introduced. That makes the title of question misleading and then post should just been closed as unclear.


text:
Why does low spin character predominate in hexafluoronickelate(IV)?
According to my book there are no unpaired electrons in $\ce{NiF6^2-}$. However, that should not be the case since fluoride is a weak field ligand. Why does hexafluoronickelate(IV) behave as a low spin complex?

Because the nickel is (IV), maybe? Metal centers with high oxidation states/charge tend to have high splitting energies.

d6 configuration with a high valent metal ion. Crystal field stabilisation energy is very much going to favour low spin. See https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy

Also see https://chemistry.stackexchange.com/questions/112675/is-conh%E2%82%83%E2%82%84cl%E2%82%82cl-paramagnetic-or-diamagnetic/115285 and then consider Nickel IV is even more likely to be low spin than Cobalt III

Does this answer your question? Why is hexafluoridonickelate(IV) diamagnetic?

This is almost a duplicate of this question. There it is shown that the Crystal Field Stabilization Energy (CFSE) is at a maximum for $\mathrm{d^6}$ ions, and low spin is strongly favoured unless there is a very small splitting energy, as is the case when you have

Low charge early first row transition metals - for instance $\ce{Fe^{2+}}$
Very weak field ligands

Without experimental evidence or high quality ab initio calculations you can't say a priori whether a given complex is high or low spin, but in practice for $\mathrm{d^6}$ species low spin massively dominates, the only exceptions being some $\ce{Fe^{2+}}$ species and a very small number of fluoro complexes of $\ce{Co^{3+}}$. Thus $\ce{NiF6^2-}$ is low spin for the same reasons that $\ce{Co^{3+}}$ is almost always low spin, namely that the large CFSE for $\mathrm{d^6}$ strongly favours low spin complexes, reinforced by the fact that $\ce{Ni^{4+}}$ is smaller and higher charged than $\ce{Co^{3+}}$, so resulting in a larger splitting energy.
[Yes, I know that $\ce{Ni^{4+}}$ and indeed $\ce{Co^{3+}}$ is unrealistic, but this is all argued within a crystal field framework, an ionic model]
Edit: I've just seen Why is hexafluoridonickelate(IV) diamagnetic? which is a duplicate of this question, and the answer there goes into the theory a bit more deeply than the very simple Crystal Field Theory I have used here


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why does sodium chloride in water dissolve/precipitate with little heat, or indeed, at all?
Sodium and chlorine presumably react together strongly, which is why the solid product is stable and takes a lot of energy to break the bonds formed, and to melt. According to the Wikipedia, melting point is about $\pu{800 ^\circ C}$.
So why is $\ce{NaCl}$ so very soluble in water, and when it does dissolve and precipitate, very little heat appears to be involved?
(please note if this seems a simple question, I'm a programmer not a chemist).

@Poutnik: re. your 1st coment, I'm not a chemist, I'm a programmer (https://stackoverflow.com/users/3779002/user3779002). I'm afraid my chem questions are going to be n00bish. re. the 2nd, I'd expect what you say would mean dissolving of NaCl would be possible but very slow.

I would also comment in saying that the problem of solubility is not yet completely understood. Nobody is able to calculate the solubility of a given substance. There are some rules working for a vast number of substances, but with always plenty of exceptions. You may use thermodynamics parameters, or electronegativities, or hydration degree, or whatever parameter you like, you will never be able to calculate a solubility. For example, why are $\ce{CaCl2}$ and $\ce{NaClO4}$ soluble in nearly their own mass of water, and why $\ce{CaF2}$ totally insoluble and $\ce{KClO4}$ nearly insoluble ? ?

As a generality, some chemicals evolve heat when dissolved, e.g., NaOH in water; others absorb heat, e.g., NH4NO3 in water. NaCl happens to be somewhere between those extremes.

@Poutnik: If you answer using your "There is balancing of energy.." or similar, I guess that's the closest we'll get to something I understand, so I'll accept that, thanks.

My relevant comments were converted with slight change into the answer and deleted.

During dissolution of ionic compounds, the thermal outcome is given by  balancing of the energy needed to break the ionic lattice and the energy released by the ion hydration, wrapping them by water molecule dipoles.
Nice example of this is exothermic (heat releaseing) dissolution of CaCl2 (where significant energy is released by $\ce{Ca^2+}$ hydration) versus endothermic (heat absorbing) dissolution of $\ce{CaCl2 . 6 H2O}$ where $\ce{Ca^2+}$ is  "pre-hydrated". This energy balance is about neutral for $\ce{NaCl}$.
See also Wikipedia: Gibbs free energy as thermodynamic parameter for evaluation of spontaneity of processes, involving both energy outcome and entropy change (in layman terms the level of disorder). Even energy consuming dissolution may happen, if the total entropy (disorder) change of the universe increases by that.
Note that thermodynamics ( energy and entropy changes ) and kinetics (rates of changes) are in large extent independent. Like 2 paths across mountains, one with bigger altitude change but through a lower pass and the other with smaller altitude change, but via a higher pass.


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Does the length of a salt bridge effect the voltage of a voltaic cell?
I am doing a Chemistry Lab in which it is completely student run. In this experiment, I basically set out the same exact voltaic cell, and used varying lengths of salt bridges as my independent variable. The lengths different by a couple cm (12, 9.5, 5, 2). And what I observed was that the smaller the salt bridge, than the higher the voltage. I am not 100% sure that what I observed was accurate, but it corresponds to the data I have. Is this right? and if it is then why? I know that increasing the total amount of salt bridges has an effect, but this is just one salt bridge cut into various lengths. My lab is due soon, and I'd love for some help, thank you.

For the same current, bridge elyte composition and cross section, there is a bigger potential drop along longer bridges, with is subtracted from the external cell voltage under the load.

I am still confused.

Intense searching is better cure for confusion than waiting for answers. Current is proportional to ion migration speed that is proportional to potential gradient, which mean value is equal to potential difference on the bridge divided by its length. Make your conclusion.

It's not like I haven't been searching, but I think the idea is a little too niche. I'm getting a lot of information on the amount of salt bridges on effect of voltage, and sure I can make an assumption, but I want to be 100% correct. I want to know why. It's interesting me to the point that its annoying that I don't understand, i've been searching for the past week, and this was kind of a last ditch effort kind of thing. I'm sorry that I have plagued this website with such a plebian question, like genuinely, if its common knowledge, then i will do more research. Thank you so much, sorry again

One learns more from being wrong (when noticed) than from being right. The direct analogy is you need higher water pressure to push water through longer pipes for the same flow, what causes lower terminal pressure at the same initial pressure.

You have your data, Presenting it and a more detailed description of your experimental technique would help.  All you can do is present the data and technique and possibly repeat the experiment to see if it is repeatable. you observe a voltage variance.  Is there a precise relationship with length?  Have you measured the current? For some reason you thought the length of the salt bridge should have an effect.  Why? You found that it  did.  Your job now is to verify the data, and define the conditions and possibly suggest experiments to explain the observations further.

If the cell is not delivering any current, the voltage does not depend on the length of the bridge. Let's call it $V_o$. It only depends on Nernst law. If the cell is working and delivers some current $I$, the measured voltage will decrease to $V$ < $V_o$, because of the internal resistance $R$ of the internal solution + bridge, with the following expression : ${V = V_o - RI}$. The internal resistance $R$ depends on a lot of parameters, like the nature and the concentration of the solutions, the distance between the electrodes, etc.

Hi, I am getting a little confused here. If I am using an analog voltmeter to measure the voltage being created, does that means the cell is delivering or not delivering any current? Thank you!

Using an analog voltmeter, the current delivered by the cell is so small, that it can be considered as zero. Its internal resistance $R$ is usually between 100 kilo ohms to 1 megohm.

Thank you for your answer. So if I am just making a Daniel cell and measuring the voltage created by the cell. It is considered not delivering some current and therefore the metal contact area will affect too much on the voltage we get, is it correct? Thank you!

I don't know exactly what you mean when you speak about the "metal contact area" . Anyway, this area does not affect the voltage measured by the voltmeter on a Daniel cell.


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Can two 1st order reactants in an overall 2nd order reaction have different rates?
I've recently done a saponification reaction of ethyl acetate with sodium hydroxide and used the initial rates method to find the order of reaction. (I change the concentration of one reactant and keep the concentration of the other one constant and find the initial rate for each concentration)
$$NaOH+CH_3COOC_2H_5 \to C_2H_5OH+CH_3COONa$$
My concentrations for each reactant is 0.5M-0.1M decreasing in 0.1M intervals and for the most part, the rates at different concentrations show that it is a second order reaction, and first order with respect to the reactants, which is correct. The only issue is, the initial rates that I got when I was changing the concentration of Ethyl acetate was much lower than the initial rates when I was changing concentration of NaOH even though both are first order. I've repeated this a few times and nothing changed.
Is there a reason for this happening or is this because of some methodological oversight?
Thanks in advance.

What's your solvent?

I am not too sure what you mean by this. If you mean what was used to dilute NaOH and ethyl acetate to 0.5M, then it is water. The reaction also starts when the ethyl acetate solution is mixed with the NaOH solution.


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SMILES vs. Graph for generating molecules
I have seen this question, but mine is slightly different, focusing on generating molecules.
There are many published methods for generating molecules; some of which generate SMILES and others directly generate graphs. Comparing these two approaches, I came to the following conclusions:

These two structures can be converted to each other. Obviously, we can obtaion SMILES from graphs by graph traversal. On the other hand, one can generate graph from SMILES using MolFromSmiles function in rdkit. One can say that the graph representation may contain more information represented in their nodes/edges feature vectors, but I assume that current generators just generate graph structures.
The main merit of directly generating graphs over SMILES is that we hope that the former generates better molecules. This is because, as mentioned in MolGAN paper, models that generate SMILES have to spend capacity on learning both the syntactic rules and the order ambiguity of the representation.

I appreciate if you can "validate" or "correct" the above conclusions.

Yes, I would prefer using a graph as it's easier to validate.
Also, if I were you I would take a look at SELFIES, which were made exactly to generate valid molecules.
https://arxiv.org/abs/1905.13741v1

I don't understand the machine learning aspect, but a graph (whether 3d or 2d) could theoretically encode (a lot) more information than smiles. Therefore, I'd conclude that you cannot convert smiles to graph without assumptions/prerequisites, which makes your point 1 wrong. For your second point, I'm missing what a better molecule would be. The inherent ambiguity of smiles is always going to be problematic. I guess in general, thus question would probably gain better suited responses on the more technical oriented site [mattermodeling.se].

@Martin-マーチン - SMILES encodes a molecular bond graph. So all SMILES can be encoded as that graph. While the syntax is ambiguous, there is the DeepSMILES variant (and SELFIES). Regardless, for both graphs and string representations, the order is ambiguous. That can be a good thing, since you can randomize atom order to augment your data set.

I'd agree with @Martin-マーチン that I don't know what "better" means. I can imagine invalid molecular graphs (e.g., 6 bonds to a carbon) as much as invalid SMILES. That's continued to be a problem with generative ML methods - much like generative image ML can produce really weird faces, animals, etc.


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A question regarding hydrogen emission and absorption spectrum
I have read in many articles as well as my textbook that when an electric discharge is passed through a sample of hydrogen gas it excites the electrons and glows emitting light.When this light is refracted through a prism we can see different wavelengths of visible light which is obtained from dexcitation of electrons from different energy levels like from n=4,n=5 etc.My question is that when an electric discharge is passed or the sample is heated shouldn't the electrons in all the hydrogen atoms excite to the same level and not to different levels and thus emit same wavelength of light?For example when a sample of hydrogen gas is put through an electric discharge since all atoms receive same energy shouldn't the electrons in all atoms excite to same energy level and dexcite from the same energy level like from n=2 to n=3 and back and not to different energy levels like one electron in one atom excites from n=2 to n=3 and the one electron in another atom excite from n=2 to n=4 shouldn't both excite to n=3?I hope you get an idea of what I am trying to say.

The atoms are excited by collisions with high energy electrons and do not all receive the same energy! Some get excited from n=1 to n=2, some from n=1 to n=4, etc. Then they de-excite by various pathways, i.e., in a single step or a cascade of steps. So you get the Balmer series, Lyman series, etc. See https://en.wikipedia.org/wiki/Hydrogen_spectral_series.

So you mean to say  based on the collision elastic or inelastic the electron in different atoms gets excited to different levels?

Ok so when I heat the sample of gas to high temperature it glows but in this case why do electrons in different atoms excite to different levels though they receive same energy?

Even when  the source of excitation is heating, the available energy for thermally induced excitation is spread over wide range. Remember Maxwell-Boltzmann statistical speed distribution for ideal gas molecules. At the given T, molecules have wide range of energies and therefore collision energies.

They do not receive the same energy! A hot gas or plasma has a distribution of energies of particles. Some are fast and highly energetic, others are slow and have low energies. Collisions are frequent and any given particle may be slow one instant and fast the next. And vice versa. Average energies are just that: averages. You need to drop the idea of “same energy”. Just saw what @Poutnik said. Same.

In an electric discharge, electrons may get any energy higher than $50$ electron-volt, depending on the type of device. With such energies, they hit hydrogen atoms. To send the 1s electron of hydrogen from the lowest level to the first excited level, $\pu{10.205 eV}$ is required. To send it to the second excited level, $\pu{12.084 eV}$ is needed. To extract the electron totally out of the atom, $\pu{13.606 eV}$ is required. All these transitions are possible in a collision electron-hydrogen, as the "missile" has enough energy to initiate all of them


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TLC monitoring of Buchwald-Hartwig coupling reaction using ninhydrin stain
I am conducting a synthesis involving Buchwald-Hartwig coupling reaction. The reagents are 2,3,5-tribromothiophene, 1-aza-18-crown-6, NaOtBu as base, $\ce{Pd_2(dba)_3}$ as palladium source, DavePhos as P ligand, and toluene as solvent. Because tribromothiophene is the limiting reagent in my case, so I monitored only the spot of it in TLC and concluded that the reaction went to completion. But some analysis didn't make sense, so I was advised to check the spot of aza crown also. Because aza crown doesn't absorb the 254 nm UV, I checked it through ninhydrin staining. Then I found that the stain of aza crown hardly fades as time goes by, even after 48 hrs. Initially I set the temperature to 100 ℃ and used 3 mol% of catalyst, and I have tried set the temperature to 80 ℃ and reduced the catalyst to 1 mol%, but it was not useful. Also, as far as I know, the ninhydrin stain only detects the amine compounds that have N-H bond. Because aza crown is secondary amine, any byproduct I expect cannot be detected in ninhydrin staining. The aza crown is stained with purple color with streak, but I found that some other spots are also stained in ninhydrin staining with no streak.

There are two possibilities:

Ninhydrin also stains other compounds other than amines.
There can be some byproducts that have N-H bond.

But I can't search any information about these two cases. Anyone who knows about this?


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Why does ethyl methanoate boil after methyl ethanoate?
Ethyl methanoate boils at 54.3C but methyl ethanoate boils at 57C.
These two organic compounds have the same molecular mass. In fact, they're just made by swapping who is the acid and who is the alcohol. Thus I would suspect them to have the exact same boiling point. Yet one is higher than the other by an appreciable margin.
I tried to reason it as that the ethanoic acid of methyl ethanoate having a second carbon might make it more polar, but I can't see the polarity difference.

Visual comparison of structures of $\ce{H-CO-O-CH2CH3}$ and $\ce{CH3-CO-O-CH3}$ suggests formyl differs to acetyl and the former molecule has overall more polar character.

It seems you mixed it up - boiling point of the acetate is 2 °C higher than of the formate... Not that it matters - bothering to rationalise such negligible difference is a waste of time.

@Poutnik Thanks, but can you elaborate why the former has a more polar character? I'm not seeing it

Polar groups at one end. But was rather speculation than statement. As the b.p. order is the opposite than originally stated, you may as well forget it, as overall effects go the opposite way and I was probably wrong. Similarly, acetone $\ce{CH3-CO-CH3}$ b.p. 56 °C is higher than for n-propanal $\ce{CH3-CH2-CO-H}$ b.p. 46-50 °C // BTW the same b.p of isomers, based solely on the same molar mass and composition, is unjustified. It happens only incidentally, as intermolecular van der Waals forces between different molecules are different.

A glance at the molecular formula is insufficient to explain something like a 2º difference in boiling points, or the corresponding (slight) difference in room temperature vapor pressures (173 mm vs 200 mm at 20 C), or the flash points (-10 C and -20 C), which are again related to the difference in boiling points or vapor pressure.
A difference in polarity, or polarizability, might be able to explain the difference, if we are able to assign polarities - but that needs some sort of quantification. One way would be to compare solubilities: ethyl formate is soluble in water (a polar solvent) to 9%, while methyl acetate is soluble to 25% (Wikipedia). This suggests that methyl acetate is more polar, and should therefore have a tighter bonding situation in the liquid. So polarity does seem to be a reasonable explanation.
Another way to examine the bonding of the molecules together in the liquid is to note that the two compounds have the exact same molecular weight, but significantly different (i.e., measurable) densities: EtOCHO has rho = 0.917, while CH3CO(OCH3) has rho = 0.932, a 1.6% difference. My conclusion is that when two oxygens, five hydrogens and three carbons are combined into methyl acetate, they pack tighter than if combined into ethyl formate. If they pack tighter, they would be more tightly bonded, and so would have less vapor pressure.
After going thru all this, I'm surprised there isn't a greater difference in the boiling points!


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Work done in thermodynamic process

Here the official answer key given by Jee main is 48 Joules which makes no sense I believe that they just blindly found the area under the curve . if I am correct the answer should be 48+2*10(the area of rectangle below the (2,2) point ) as work done area under the PV graph with volume axis
please correct me if I am wrong

If an authority said 3x4=15, would you call for checking your analysis that  3x4=12? I guess you would not as you would see they were wrong. This case is very similar. Why did you not see they were wrong here?

haha ik but even many teachers gave the wrong soln uk so i had to confirm and yeh jee is an all india level exam in which  usually they change the answer key if they commit any mistake

It is good if they at least acknowledge errors. But considering the particular values, the whole task is made up for the sake of the task, far from real problems.

Generally, you should not ask if your result is correct, but you should verify your calculation procedure and eventually ask about its details where you are unsure. You either know how to do it either you do not.

It looks like they have set the trap of shifted axis, but they have fallen for it themselves.
$$W=\int_{V_1}^{V_2}{p \cdot \mathrm{d}V},$$
so $$W=(\pu{8 m3}- \pu{2 m3}) \cdot \pu{8 Pa} + (\pu{12 m3} - \pu{8 m3}) \cdot \frac{\pu{8 Pa} + \pu{2 Pa}}{2}  = \pu{48 J} + \pu{20 J} = \pu{68 J}$$


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Can there be a molecule where every atom is covalently bonded to every other atom?
I was wondering if muliple atoms could bond in a ring, but share a pair of electrons with every other atom in the molecule so that every atom is connected to every other atom. Would it be possible for some amounts of atoms, but then become either geometrically impossible, or electron-wise impossible? e.g.
Also, can three atoms share three electrons with each-other, like this, or would the attraction be too weak for any atom?

Have a look at the structure of white phosphorus.

Each of 4 P atoms of $\ce{P4}$ in the corner of a tetraedr, bound to the other 3.

And than check out trihydrogen cation

https://en.wikipedia.org/wiki/Four-center_two-electron_bond

One thing to note here is: is valence body theory even "correct"? It's just a model. Are electrons really localized to a single bond between two atoms? What if you looked at the system via molecular orbitals instead?

As Nisarg Bhavsar stated in his answer, phosphorus can form tetrahedral tetramers.  Given this, and given that elements in the same group (same column of the periodic table) have similar properties, it's worth asking if arsenic, antimony, and bismuth might form tetrahedral tetramers as well.  And indeed they some of them do.  Specifically, tetrahedral tetramers have been found for both arsenic and antimony.  Whether they form with bismuth is still an open question.
arsenic ("yellow arsenic"):
https://pubs.acs.org/doi/10.1021/jacs.7b01890
https://pubs.acs.org/doi/10.1021/acs.chemrev.8b00713
antimony:
https://aip.scitation.org/doi/pdf/10.1063/1.463326
bismuth:
https://pubs.rsc.org/en/content/articlelanding/2018/cp/c8cp03529f

The proposition you propose is very much possible.
The most common example of such a phenomenon is the structure of white phosphorus. It contains four phosphorus atoms, each bound to the other three by single covalent bonds. It forms a regular tetrahedral structure with each $\ce{P-P}$ bond length being equal to $\pu{225 pm}$.
You can have a look at the structure below:

Image source:
(1) White phosphorus https://www.acs.org/content/acs/en/molecule-of-the-week/archive/w/white-phosphorus.html (accessed 2022 -04 -11).


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Effect of electron-donating and electron-withdrawing groups on the C=O bond
My notes states this:

An electronegative element tends to draw in the electrons between the carbon and oxygen atoms through its electron-withdrawing effect. This strengthens the C=O bond.
An electron-donating group, on the other hand, supplies electrons to the C=O bond, weakening the bond.

I just can't rationalise this. How does the electron-withdrawing group strengthen (and not weaken) the C=O bond and how does the electron-donating group weaken (but not strengthen) the C=O bond?

That depends if effect is mesomeric or inductive.

This is easiest to see when a mesomerically electron-donating group is attached to the carbonyl, as in the case of carboxylate:

The O atom attached to the carbonyl donates a lone pair of electrons via resonance. It does so by overlapping the p orbital containing the lone pair with the C=O $\pi^*$ LUMO. This donation of electrons into the $\pi^*$ LUMO weakens the C=O bond $\pi$ bond. This can be observed on the diagram above; in the lewis structure on the right the C=O $\pi$ bond is shown to be completely broken as a result of the conjugation of lone pair electrons. In reality the C=O $\pi$ bond is not completely broken but simply weakened, which is shown on the rightmost portion of the diagram.
This concept can be extended to inductively electron donating groups as well. For example, a CH3 group would donate electron density from its C-H $\sigma$ bonds via hyperconjugation into the C=O $\pi^*$ LUMO, weakening the $\pi$ bond as well.
For electron withdrawing groups, the the logic is the same just that the reasoning is opposite. If a halogen X was attached to the carbonyl then it would exert an electron withdrawing effect on the C=O C atom, giving it a delta positive charge. This makes breaking the C=O $\pi$ bond difficult as it would then intensify the positive charge on the C atom.
Side note
As a side note, perhaps you were confused because of the conflicting observation where the electron withdrawing groups that strengthen the C=O $\pi$ bond actually make the C=O more reactive, while electron donating groups that weaken it make it less reactive. This is because bond strength does not tell the whole picture of a molecule's reactivity.
The electron donating groups do weaken the C=O $\pi$ bond (as measured by FTIR) but at the same time reduce the kinetics of nucleophilic addition. The electron donating effect reduces the delta positive of the C=O C atom, decreasing the electrostatic attraction to electron rich nucleophiles. The overlap of electrons with the LUMO stabilises the lone pair of electrons but also increases the LUMO energy, increasing the HOMO-LUMO gap for any incoming nucleophiles. There are thermodynamic considerations as well: the electron donating and delocalisation of charge is generally stabilising. Overall this means that while the C=O bond is weakened, the reactivity of the molecule as a whole is also decreased.

Oh thanks, this is a comprehensive answer! Then if we link this back to IR spectroscopy, is it true that the absorption wavelength for, for instance, an ester tend to be higher than an aldehyde, due to mesomeric electron-donating effect?


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Why did my nickel + hydrogen peroxide turn red?
So I recently thought of doing some electroplating and bought some nickel strips from Amazon. As I was making the acetate solution I added some hydrogen peroxide (6%) because I wanted to see if nickel would form nickel acetate the same way copper acetate is made by reacting copper with acetic acid (I had disconnected power). To my absolute surprise the color of the solution changed from crystal clear to almost blood-like red.
Could someone tell me what products I may have made?
Could it be that my nickel was not pure and contained some impurities? The seller claims it's 96% pure and its the sort you'd use for spot welding but I'm not sure.

Nickel(II) acetate is green. // One of hypothesis is it could be a red-brown basic iron(III) acetate ( The formation of the red-brown complex was once used as a test for ferric ions. ). // Another question is, if $\ce{H2O2}$ forms Nickel(III,IV) or some peroxocomplex.

Nickel, especially in solution, is sort of unheathly, if you get my drift. I hope you wear gloves if there is any danger of getting it on your fingers, and have a environment-friendly plan for it's disposal.

Tell us exactly what you have done. All you say is that ... "as I was making the acetate solution". Well. When did the color change ? What else did you observe ? Any precipitate ?

Some nickel compounds are red, like nickel chlorate $\ce{Ni(ClO3)2·6H2O}$


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How can warm steel and wood merge together?
I need to describe the manufacturing of japanese knives. I struggle with the step when the steel blade is put together with the wooden handle.
Here are some pictures from the video https://www.youtube.com/watch?v=j2L_Ku47afo (at 6'42"):

After putting the hot blade inside the handle, you can see that the steel immediately cools down and after hitting the handle with a hammer, the two pieces are merged. I've been told that the welding could not function between two different materials, especially not wood. Could you please explain me how those two materials merge together in this situation please ?
Thanks in advance for your answers !
EDIT:

Sorry if this doesn't seem directly related to chemistry, I was more interested of what happened in the molecular scale than the physic reactions, if you think about a better forum to ask this do not hesitate to tell it to me;
For those who can't see the picture or the video, the blacksmith has made a very long, thin end to the steel blade which he has warmed up until it becomes orange and then he puts it inside the wooden handle (which has a drilled hole). After he did this you can see smoke going out the handle, and finally the blacksmith hits the handle with a hammer (in the direction of the length).

It looks to me that wood, heated up by steel, undergoes in various extend plastic mechanical deformation and thermal decomposition, releasing space for the steel object.

There is already a drilled hole in the wood when the steel is inserted in it, how do you think those reactions could prevent the blade from getting out this hole ? And why the hammer ?

Plasticity goes either way, together with possibility of making wood to stick to steel.

I think it is a "press fit" , mechanical. The pilot hole is smaller than the tang. The hot tang burns a hole that is just the right size.

As I mention in my comment to blacksmith37's answer, below, there is possbility that the heat creates some melting in the wood, and this causes a bonding to the metal.  We know that wood contains lignins and other components that, under the right conditions, can melt to form bonds to itself (see youtube.com/watch?v=X0k04hjdYuQ and link.springer.com/article/10.1007/s00226-006-0097-2). So I'm wondering if the wood might also be able to melt slightly to bond to the metal.

A common press fit , no chemistry involved. Most commonly used for files (in US). All my files (old) use this same press fit, I believe newer files have plastic handles. The use of heat to custom shape the hole for the tang is a good modification. A smaller pilot hole can be used because it will be enlarged to the right size by burning. This likely gives a tighter fit compared to a larger round hole

What about the possbility that there is welding involved? Wood contains lignins and other components that, under the right conditions, can melt to form bonds to itself (see youtube.com/watch?v=X0k04hjdYuQ and link.springer.com/article/10.1007/s00226-006-0097-2). So I'm wondering if the wood might also be able to melt slightly to bond to the metal.


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pKa of methylene protons in cycloheptatriene vs cyclopropene

The $\mathrm{p}K_\mathrm{a}$ values of methylene protons in cycloheptatriene vs cyclopropene are found to be 60 and 36, respectively.

From what I understand - both conjugate bases are non-aromatic (not anti-aromatic) as a result of some sort of Jahn-Teller distortion. I can't pin down the nature of the distortions but they appear to be a pyramidal geometry for the cyclopropenyl anion and either $\mathrm{sp^{2}}$ hybridisation or unequal bond lengths for the 'tropylium' anion.
In either case, if both species and both conjugate bases are non-aromatic what is the cause of the massive $\mathrm{p}K_\mathrm{a}$ difference? Is it something to do with the $\mathrm{sp^{2.68}}$ hybridisation in the three membered ring?

pKa = 60 looks like very rough estimate. Wouldn't be surprising if another source said 55 or even 50.

Would you not agree that in either case the difference seems strange?

Hmm, gonna check out the values, but it may be that tropyl is still stabilised by resonance without antiaromatic contribution and cyclopropenyl can't do the same. (Big ring could be only partially flat.)

I was thinking that too, but couldn't quite rationalise how it could be stabilised by resonance but not destabilised by the resulting antiaromaticity. Maybe unequal bond lengths remove the degeneracy (and thus the antiaromaticity) but it's still flat enough for p-orbital overlap and thus resonance?

I mean, tropyl is obviously not flat overall - not antiaromatic, but could be locally flat - just one allyl would be enough for such effect.

Plot thickens! While for methylene protons it might be about 60, cyclopropene's likely deprotonated on vinylic position and might be actually more acidic  https://chemistry.stackexchange.com/questions/105637/acidity-comparison-between-cyclopropene-and-cyclopropane

There might not be antiaromaticity. But very  harsh to remove allylic hydrogen. See here.

@mithoron I am thinking the same thing about "partial" conjugation. I suggest below a valence-bond contribution that could do that.

Not all "antiaromatic" rings are created equal.
Assuming the cyclopropenide anion is formed by deprotonating a methylene proton, forming a cyclopropen-3-ide ion (in reality, a vinylic hydrigen appears to be deprotonated instead, forming cyclopropen-1-ide ion; see Poutnik's comment to the question), it can remove the unfavorable antiaromatic coupling only by fully localizing the negative charge and pyramidalizing the bonding at the selected carbon, giving up any possibility of stabilizing the charge via delocalization.
Not so with the cycloheptatrien-7-ide ion. While delocalizing the charge over the full ring results in an antiaromatic coupling, a nonaromatic and resonance-stabilized alternative is available by forming an allyl-anion-butadiene contributing structure (see below). Thus with the seven-carbon anion, some stabilization is possible while still circumventing antiaromatic coupling. With multiple positions possible for the allyl-anion component, the $pK_a$ reported here for cycloheptatriene actually beats that for the simple allyl-anion former propene ($36$ versus $43$).

So -- what improvements are suggested? (I see your -1 vote).