text:
About Fe(III) and Fe(II) coordination complexes
Why does Fe(III) form octahedral coordination complexes if it has 5 electrons in its d-orbitals? I understand that Fe(II) has 6 electrons in its d-orbitals and 6 lone pairs from 6 ligands as the 12 electrons fill up the 3d, 4s and 4p orbitals (18-electron rule, $6+12=18$). However, Fe(III) has 5 electrons and 6 lone pairs is not enough. I know basic crystal field theory and thought that some of the metal-ligand interactions were covalent bonds, but it doesn't add up for both low and high spin states.
Also, do the ligands need to line up with the orbitals form coordinate bonds? I know it sounds counterintuitive since they repel and that's the basis of the crystal field splitting, but how do coordinate bonds form in coordination complexes then?

Octahedral complexes are somewhat of a ‘standard’ for transition-metal complexes. Not only can they happen if there are too few electrons to satisfy the 18-electron rule (e.g. $\ce{[Ti(H2O)6]^3+}$, a $\mathrm d^1$ complex) but also if there are too many electrons to satisfy the 18-electron rule (Jahn-Teller distorted $\ce{[Cu(H2O)6]^2+}$, a $\mathrm d^9$ complex).
So how are they still stable? Well, coordination compounds quickly exceed the simplistic bonding pictures that students are taught at the beginning of their chemistry classes. And the ligand field (or crystal field) model—while it does a good job of explaining some observations at a low level—also doesn’t exactly capture what we consider the current knowledge of coordination compounds. A much better description of how they are bonded lies within molecular orbital theory and scheme 1 below.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
On the left of the scheme, the metal’s 3d, 4s and 4p orbitals are placed, while the right-hand half contains the ligand orbitals; simplified as one σ-type orbital per ligand in an octahedral complex. After taking a look at the scheme and finding it perplexing, we can attempt to find the stabilising and destabilising factors in forming a complex.
The most obvious stabilising effect is that all ligand orbitals form bonding interactions that lower their energy. For this, they make use of the typically unoccupied metal’s 4s and 4p orbitals alongside the often occupied 3d orbitals. As we are stabilising occupied orbitals at the expense of unoccupied orbitals, this already results in a net stabilisation.
The second bit of stabilisation results from the interaction of the ligands with parts of the 3d orbitals: precisely the $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals, often collectively termed $\mathrm{e_g^*}$ orbitals according to their symmetry group. These may be occupied, but these also have better overlap with the ligand orbitals so the bond formed is stronger than in the case of 4s and 4p.
Finally, the electrons in the other three metal d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$; collectively termed $\mathrm{t_{2g}}$ according to their symmetry group) are, in fact, non-bonding in this simplified picture as their symmetry does not match any ligand orbital. Thus, any electrons in here do not incur any stabilisation but also—arguably more importantly—are not destabilised.
The 18-electron rule derives from this simplified picture and assumes that all bonding and non-bonding orbitals are occupied (that means: all ligand orbitals and the $\mathrm{t_{2g}}$) while all antibonding orbitals (most notably: $\mathrm{e_g^*}$) remain unoccupied. If you count electrons and add up, this gives 18.
However, as I have established the $\mathrm{t_{2g}}$ electrons don’t actually destabilise the complex (under the simplified scheme). Thus, it is relatively irrelevant if there are 0, 2 or 4 of them; it doesn’t change the general scheme. Likewise, even though any $\mathrm{e_g^*}$ orbitals are antibonding, their overall effect is not extreme, so there can be electrons occupying them while still allowing for a stable complex.
The picture on the metal is further complexified (pun intended) by considering high spin versus low spin. In an uncomplexed metal ion, all five d orbitals are energetically equivalent, so they should be filled according to Hund’s rule. In a complex, you have two options:

Hund’s rule is observed across all five d orbitals regardless of the energy difference
the energy difference is large enough that spin pairing in the $\mathrm{t_{2g}}$ orbitals occurs before the $\mathrm{e_g^*}$ orbials are filled according to Hund’s rule.

The first case is termed high spin and should be treated as the standard case for 3d metals, while the second case is termed low spin and is the standard case in 4d and 5d metals. Note however, that clever choice of ligands may cause a 3d complex to adopt a low-spin configuration and some complexes are even capable of flipping between high and low spin.
While $\ce{Fe(H2O)6]^2+}$ may look like a prime example of the 18 electron rule, it is actually a high-spin complex so the outline I drew above is no longer valid. On the other hand, high-spin $\ce{[Fe(H2O)6]^3+}$ has 5 d-electrons which allows for every d orbital to be populated by exactly one electron—a ‘slightly more favourable’ state. Therefore, high-spin iron(II) complexes are easily oxidised in aquaeous solution but high-spin iron(III) complexes are not as easily reduced. (Obviously, the picture changes when you consider low-spin complexes such as $\ce{[Fe(CN)6]^3-}$, which is a good oxidising agent.)

tl;dr:

The 18-electron rule is more of a guideline than an actual rule; it states that more stable adducts can be formed, if, but that’s all;
All ligand–metal interactions can and should be considered covalent bonds;
Technically, the orbitals and the ligands do need to line up but because electrons are volatile we can assume they have lined up automatically;
Essentially, the crystal field model isn’t entirely wrong when it predicts the $\mathrm{e_g^*}$ orbitals to be raised to a higher energy level. It just provides a bad explanation. The orbitals end up being antibonding; that should equally explain.


text:
Right Raman Spectrum of Cyclohexane
In this site, https://www.princetoninstruments.com/userfiles/files/assetLibrary/Technical%20posters/Poster-Aberration-Free-Raman-Spectroscopy-with-the-IsoPlane.pdf the raman spectrum of cyclohexane shows higher peak at the right.

While in this site, http://group.nanovietech.com/products/advantage-raman.html, the raman is showing higher peak at the left:

Both are raman spectra of cyclohexane. My questions relate to the 2853 cm-1 peak and 802 cm-1 peak. Based on structure of cyclohexane. Which of them should be higher? Ignore the other peaks which I wasn't concerned.
I found this too:

Should the 802 cm-1 peak be higher than the 2853 cm-1 peak based on the molecular structure of cyclohexane?

Its not duplicate bec it doesnt show the exact spectra of cyclohexane. I was asking which of the above 2 is correct. Both are raman of cyclohexane.

I've already removed the comment. I have been too quick and reading left right I thought you mean the axis reversal, sorry. For your actual Q, this should be related to the isoplane spectrometer having different performance. As they say reduced aberration, we can't expect to be frequency independent. Look at their example on peak shape and resolution . Also the detector can have a different response. This is absolute photons counting, by the way.

For the most sophisticated raman spectrometer. Which does it show? It is the CC (at ring) that can quickly varies. Cyclohexane is the reference calibration for most raman spectrometer. So which one is the common spectrum? Have you seen one in your lab? Which is it? There must be a standard one in the library.

I an not aware of the calibration of shapes. You probably mean some internal calibration for quantitative purposes or otherwise Raman shift calibration. I don't even see the ground for a calibration in your actual sense. Can't judge FWHM from pictures, tough it seems that top spectrum has indeed high resolution.

When you go to the lab or university later. Please see what kind of spectrum the cyclohexine is showing in the most expensive spectrometer or available. I just want to have idea what an actual one would look like in other brands.

If you have to choose a spectrometer is another story. But to be clear, it is like looking at two photos. There is no the right, or at least it is not straightforward. There is a better photo with high resolution, clarity, etc. Choosing the camera would depend on what and when you shoot. There is no Raman here. Not of the kind you mean. Both spectra are ok. They might even have the same peaks intensity.

You still don't get my questions. I was not asking what particular counts would result, of course it depends on the settings. I was merely asking the particular peaks at 2853 cm-1 and 700 cm-1. Based on the chemical structure of cyclohexane, should 2853 cm-1 or 802 cm-1 be higher in peak with respect to each other. Ignore the other peaks which I was not questioning.

That is exactly the point.  Besides that I am not use to quantitative analysis by Raman, but it exists, it could be that the photons count is even the same internally. One company claims to get less aberration and high resolution with the isoplane line of spectrometer and apparently they are right. The intensity of a peak is area, and perhaps the ratio between the peaks in each spectrum is the same. You got focused on peak height rather but this could beneficial in terms of resolution, not itself. At least with no dramatic differences- that would be indeed point to something wrong.

I think I should change my question to "Should the 802 cm-1 peak be higher than the 2853 cm-1 peak based on the molecular structure of cyclohexane?" I added this in original post with added 3rd graph.  Hope someone else can answer it.

Last comment. Basically it is always the same..... https://www.laserfocusworld.com/test-measurement/spectroscopy/article/16556835/isoplane-and-vhgs-measure-lowfrequency-raman-spectra. Plus base line correction can be applied with different level of correctness. A background in Raman is not easily defined. ..

If someone has seen an actual raman spectrum of cyclohexane using any standard Czerny-Turner Raman spectrometers. Please share the spectrum. I just want to see how an actual one look like.

Possible duplicate of Wavenumber range reversed in IR spectroscopy

If you will read carefully. It's totally unrelated because nowhere was I talking about IR spectroscopy. I was all referring to Raman spectroscopy. So please delete the comment above and I'll delete this too.

First of all, you are extremely lucky in asking this question for cyclohexane: cyclohexane serves as reference material for checking your intensity calibration, so it is one of the few substances for which integrated band intensity ratios are available in the literature.
Integral intensity 2567 - 3068 cm$^{-1}$ is higher than integral intensity 700 - 900 (i.e. 801 cm$^{-1}$ band). At 532 nm excitation, the ratio is roughly 10.3 : 1.

You always have the fundamental $\nu^4$-dependence of Raman scattered intensity. If you look closely, that's $\nu_0 (\nu_0 - \nu_j)^3$ for the band with Raman shift $\nu_j$ for photon count and  $(\nu_0 - \nu_j)^4$ for power. Thus, the relative band area differs depending on excitation wavelength.
For dispersive Raman instruments, the sensitivity is highly wavelength-dependent and this varies from instrument to instrument and for instruments with moving grating also depending on the grating position.

Thus, for intensity comparisons across different instruments, you'll need to calibrate the intensity axis for the spectrometers (and possibly fold the higher-resolved spectrum with the difference in line shape if that isn't negligible due to integration of broad spectral ranges)
However, when training chemometric models (calibration, quantitation, classification, ...) for one instrument (or maybe very similar instruments: say, same manufacturer & model), as long as that instrument is stable intensity calibration is not necessary as it will automatically become part of the model.
Which is why many Raman spectra are shown as raw (uncalibrated) intensity.

This doesn't have much to do with "highly sophisticated" Raman spectrometer rather than doing sophisticated calibration or not. And among the highly sophisticated Raman spectrometers, you have tradeoffs that can make such a calibration more or less stable. In fact, different instruments are highly sophisticated in different aspects (e.g. spatial resolution, spectral resolution, signal stability/drift, versatility/possibility to adapt to different experimental setups, ...) some of these influence how easy intensity calibration is and/or how often it needs to be re-done.

Literature:

McCreery: Photometric Standards for Raman Spectroscopy, in: Handbook of Vibrational Spectroscopy, Wiley 2006
McCreery: Raman Spectroscopy for Chemical Analysis, Wiley 2000

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text:
How does the splitting in this molecule take place?

Consider the following molecule and data:

$$
\begin{align}
\mathrm{H_a} &= \pu{1.0 ppm} &\quad J_\mathrm{ab} &= \pu{5.0 Hz}\\
\mathrm{H_b} &= \pu{3.0 ppm} &\quad J_\mathrm{bc} &= \pu{8.0 Hz}\\
\mathrm{H_c} &= \pu{6.0 ppm} &\quad J_\mathrm{ac} &= \pu{1.0 Hz}\\
\end{align}
$$
Draw a splitting diagram for $\mathrm{H_b}.$ $(\pu{1 box} = \pu{1 Hz})$

Tried to draw the splitting diagram and ended up with this diagram:

I am also stuck as to how to draw the spectrum of the whole molecule as I feel $\mathrm{H_a}$ would be triplet of triplet. Am I on the right track and how to draw the sketch of the spectrum?

You missed a split. The central peak must also split on the last coupling.

According to given data, you may consider expected peak patterns are not complicated such as $\mathrm{ABX}$ or $\mathrm{A_2B}$  patterns (more reading look here. This is simply a  $\mathrm{A_2X}$  pattern ($\delta_X = \pu{6 ppm}$ and $\delta_A = \pu{1 ppm}$ without geminal coupling of two $\ce{H_b}$ protons), and hence, peak is a $dt$ for $\ce{H_b}$ as depicted below with relative heights of the peaks:

Similarly, $\ce{H_c}$ is also a $dt$ resonance with $J = 1 \text{ & } \pu{8 Hz}$ and $\ce{H_a}$ is a $tt$ with $J = 1 \text{ & } \pu{5 Hz}$. OP should able to figure out the intensities of relevant peaks.
A valuable reference:
Roy H. Bible, Jr., Ph.D., In Interpretation of NMR Spectra: An Empirical Approach; Springer Science: New York, NY, 1965 (https://doi.org/10.1007/978-1-4684-8288-1) (Note: Most relavent to this question is chapter 4 of the book titled "Higher-Order Spin Patterns and Multiple Resonance").

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text:
Can I use the quasi-equilibrium approximation when drug and ligand distribution volumes are different?
A drug $D$ binds a ligand $L$ to form a complex $C$ according to the law of mass action. Furthermore, mass balance should be achieved. Can I still assume quasi-equilibrium when the distribution volumes of the drug and the ligand are different?
From the quasi-equilibrium approximation it follows that
$$k_\mathrm{on} [D] [L]=k_\mathrm{off} [C],$$
from which the dissociation constant is
$$K_\mathrm{d} = \frac{[D][L]}{[C]} = \frac{\frac{D}{V_D}\frac{L}{V_L}}{\frac{C}{V_C}},$$
with $[x]$ describing the concentrations, $x$ the amounts, and $V_x$ the distribution volumes.
Mass balance now means that
\begin{align}
D_\mathrm{tot} &= D + C &
\text{and}&&
L_\mathrm{tot} &= L + C.
\end{align}
Can I now rewrite the equation for $K_\mathrm{d}$ as
$$K_\mathrm{d} = \frac{\frac{D_\mathrm{tot}-C}{V_D}\frac{L_\mathrm{tot}-C}{V_L}}{\frac{C}{V_C}} ?$$
Even though this might be very basic, I would really appreciate if someone could explain it in detail.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. (We prefer to not use MathJax in the title field, see here for details.)

Evidently you have a model that is a little more complicated than a homogeneous single phase. You may want to simply declare multiple phases with independent equilibria. Or do you have flow conditions to satisfy and that is why you invoke "quasi-equilibrium"?


text:
What gas would be suited best to form a density gradient in a long tube?
I want to set up an aero-optics experiment using a sealed ~5 m length of pipe with a gas inside. The gas must form a density gradient, and must be transparent enough to view a target at the other end of the pipe.
Barometers at the top and bottom of the pipe will measure density.
The purpose of the experiment is to replicate the density structure of earth's atmosphere to establish what affect density gradients alone have on light passing through.
Question: What gas would be best to use for this experiment?
Would a moderate amount of boiling water in the bottom of the pipe do the job? Are there better gasses that could be used? My thinking is once the gas is added and the pipe is shaken up to distribute the gas around, then as it settles back down it would form some kind of gradient for a short while.

Well, the best is air, and vertical dimension should be several km high. The most significant effect on density gradient has temperature gradient. The dry adiabatic gradient is -0.0098 K/m

....aside of pressure based density gradient ...

@Poutnik I agree this is futile on a length of 5m, but why would air be "best"?

To simulate air behaviour. But the true is, the size is critical. Density=pM/(RT), dp=-g.pM/(R(T0-a.h).dh. Mixing gases would always end in tiny density gradient ).

The only thing that comes to my mind is experimentimg with static  heavy vapour of a volatile organic solvent, but it would be much closer to light reflection conditions than to usual air gradients.


text:
Is enthalpy of combustion the 'heat released' or 'enthalpy change'?

In a constant volume calorimeter, $\pu{3.5 g}$ of a gas with molar weight $\pu{28 g}$ was burnt in excess $\ce{O2}$ at $\pu{298 K}.$ Temperature of calorimeter increased from $298$ to $\pu{298.45 K}$ due to combustion. If heat capacity of calorimeter is $\pu{2.5 J/K},$ then magnitude enthalpy of combustion of gas in $\pu{kJ/mol}$ is?

In this question, according to me $ΔU = 2.5 × 0.45 = 1.125,$ i.e. $1.125/(3.5/28) = \pu{9 kJ/mol}$, and $ΔH = ΔU + nRΔT = 1.59.$
But according to my teacher, we have to find the heat evolved (which in isobaric process equals ΔU). So answer is $9$.
This is a question from IIT-JEE so I think there is no error in question or answer.
Isn't enthalpy of combustion the change in enthalpy?

As a general note, never omit units (also during intermediate calculations); also, molar weight is measured in $\pu{g mol-1},$ not just $\pu{g}$.

The enthalpy of combustion (aka heat of combustion) is defined as the enthalpy change in going from pure reactants (each at 1 bar pressure) to pure products (again, each at 1 bar pressure) at 298 K.
In this problem, it is first of all necessary to assume ideal gas behavior to proceed.  For ideal gases, the change in enthalpy in going from pure reactants each at 1 bar pressure to the starting gas mixture in the calorimeter, there is no change in enthalpy at constant temperature.  Then, application of the first law of thermodynamics to the mixture before and after the reaction leads to the conclusion that $\Delta U=Q$, since the reaction volume is constant.  Since the temperature change between the beginning and end of the combustion in the calorimeter is essentially zero, this means the the calorimeter has measured the internal energy change of the reaction.  As Buck Thorn has pointed out, the change in enthalpy for the gases in the calorimeter is given for an ideal gas by:
$$\Delta H=\Delta U+\Delta (PV)=\Delta U+RT\Delta n$$where $\Delta n$ is the stoichiometric change in the number of moles of gas.  He also correctly pointed out that there is nothing said in the problem statement about the stoichometry of the reaction or the change in the number of moles of gas.  So, as he indicated, the problem statement is flawed.  However, if we arbitrarily assume that the change in the number of moles of gas is zero, we would have to conclude that the change in enthalpy for the process in the calorimeter is equal to the change in internal energy.  Then, since, again, the products form an ideal gas mixture, there is no change in enthalpy in going from the final gas mixture in the calorimeter to the pure products at 1 bar pressure.  So, based on these assumptions, the enthalpy of combustion is equal to the internal energy change in the calorimeter (also Q) per mole of species being burned.

For an isobaric process $$q_p=\Delta H$$ and, as you rightly write, for an isochoric process $$q_V=\Delta U$$ and the relation between $U$ and $H$ is $$\begin{align}\Delta H&=\Delta U + \Delta(PV)\\&=\Delta U+RT\Delta n\end{align}$$
The last equality holds only if T is constant, which it evidently wasn't during the experiment, but for our calculation of $\Delta H$ we are allowed to assume this since we are interested in the enthalpy change for the system (which comprises the reaction, not the calorimeter) at constant T, and we can safely assume that $\Delta U$ and $\Delta H$ of the system are approximately independent of temperature over the small range of T in the experiment, but possibly dependent on n. The same assumptions cannot be made for the calorimeter: its energy is assumed here to depend entirely on T.
The next step is figuring out $\Delta n$. The only way you can have $\Delta H=\Delta U$ for an isothermal ideal gas is if you end up with the same molar amount of gas you started with ($\Delta n=0$), which is necessary based on the stated solution but seems ungrounded. A hint might be the MW, which suggests the gas is perhaps $\ce{N2}$ or ethene, or even carbon monoxide, but none of these combust with the desired change in stoichiometry (well, maybe ethene, if you assume the water produced does not condense).  So I'd charge that the question is flawed.


text:
Why does the viscosity of honey decrease when mixed with other viscous fluids?
In order to make honey mustard, I mix honey and mustard.  But the resulting mixture is less viscous then both honey and mustard.  How does mixing two viscous substances give a less viscous one?

Would you say that the mix of mustard and honey is less viscous than the less viscous component (the mustard)? Also, what treatment do you use (do you whip the preparation)?

Mustard takes a long time to flow, but isn't dense or sticky like honey.  Honey mustard flows easier than both.  To mix them I whip them together with a fork.

Physical chemistry of soft matter and complex fluids are perfectly on topic for chem.SE.

Honey is a homogeneous substance, a solution of around 98% sugar and water.
Mustard is a heterogeneous mixture, containing about 1:1:1 oils, proteins and water, all in very small particles. The proteins act as a stabiliser to the small oil droplets. They cant slide past each other easily, and act more like a dense network of small solid particles. Like a heap of wet sand.
Now if the aqueous phase gets much larger due to the honey, the oil+protein droplets loose direct contact, and you no longer have the solid-like behaviour of wet sand.
Your aqueous honey-phase however now contains perhaps 25 or 30 percent of water instead of 15 or so in pure honey, and that lowers its viscosity a lot.
You have a major phase with reduced viscosity, and a disperse phase which is no longer continous (not "percolated"), and so the whole thing flows rather easily.
(Btw. the viscosity of mustard is not that high, however it usually shows a yield stress, that is it does not flow at all e.g. under its own weight. You open a mustard glass and the inside looks just the same as it did when you last closed it. Honey however is just a boring Newtonian fluid, no shear thinning, no thixotropy, nothing.)

Spot on, except I quibble about the composition of honey: (https://en.m.wikipedia.org/wiki/Honey): Fructose: 38.2%, Glucose: 31.3%, Maltose: 7.1%, Sucrose: 1.3%, Water: 17.2%, Higher sugars: 1.5%, Ash: 0.2%, Other/undetermined: 3.2%

@JamesGaidis OK, 97.5% ;)  But of course it is a natural product, and the numbers vary hugely. While we are on that wp artikel, this https://en.wikipedia.org/wiki/Honey#Rheology is of course highly relevant.


text:
Help understanding Electrochemistry diagram

I was reading some literature when i came across the following figure and was wondering if someone could help explain it to me. I see on the vertical axis it is potential vs Li/Li+ and see that Oxides are at ~1.2V since the potential of lithium ion batteries are ~4V and it is a transition metal oxide are they talking about other oxides? I am assuming by potential they mean Reduction potential? Graphite is at ~0.2V Silicon at~0.3V and TiO2~1.5V but they are all used as anodes in lithium ion batteries why are they not reduced as
$$
\Delta G = -zFE
 $$
and E is positive so dG is negative?

I understand that the cathode material is probably at an even higher potential and is thus preferentially reduced but is TiO2 also reduced?

I think the y- axis is oxidation potential rather than reduction potential, the way they are writing Li/Li+ suggests that. You have to check the context very well.

The standard reduction potential of Lithium is approx -3V and the standard reduction potential of TiO is approx -1.3V thats +1.7V like on the diagram https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

The symbolism of Li/Li+ as a half cell is wrong since electrochemical convention is to always show the half cell as reduction e.g., Li+/Li rather.

@M. Farooq It rather follows the usual electrode description schema like [/][/]. The measured electrode potential obviously does not depend on notation.

..as there is (electrode) potential(1) in electrostatic context and potential(2) in TD context, proportional to delta G. I suppose the diagram follows the potential(1), what is more common in applied chemistry.

@M.Farooq yeah "Vs Li/Li+" is a common way of writing it in Li-ion battery paper.


text:
Is neutralized and dried suphuric acid still a hazard? Is it still acidic to touch?
I mistakenly put a sulfuric acid drain cleaner in  my bathroom  sink. Some of it spilled into the bathroom in my basement. I cleaned it up using baking soda, but I'm concerned if I missed any, whether it will still be acidic after it dries. Does sulfuric acid lose its acidity after it dries?

Dried sulphuric acid is kind of nonsense. If you let a drop of the acid to be exposed to open air, it will never get dry and solid. It is still liquid, with acid concentration depending on air humidity, as it is hygroscopic. After acid neutralization ( i.e. not drying ) by solid baking soda, you can still neutralize the small residual amounts by washing contaminated surface by the baking soda solution.

It reacted with the baking soda and is no longer sulfuric acid, but much safer sodium sulfate. If you're unsure, add more baking soda until it no longer fizzes.
The remaining white powder is a mix of baking soda or $\ce{NaHCO3}$, and sodium sulfate, $\ce{Na2SO4}$. It safe to wipe up with a damp rag and fairly harmless to touch. Just avoid getting the powder in your eyes.

It appears that you've voted to close the question giving the reason "unclear what you're asking" and yet you've posted an answer, which implies that the question wasn't unclear to you after all. I understand it might have been an impulsive decision during reviewing, but please try to act more coherent in the future.

Actually, I believe the question does not necessarily belong in Chemistry StackEchange because of its quality, but gave an answer to assuage the fears of the inquirer.


text:
Reaction between Sodium Thiosulfate and Iron Chloride
I'm doing a paper on catalytic effectiveness in this reaction, and am trying to identify how the catalysts are speeding up the reaction. Based on what I can find,
$\ce{Fe}$ starts as $\ce{Fe^3+}$, and it will react to form a transitional violet compound, which I think is $\ce{Fe2(S2O3)2}$. This will then break down into $\ce{Fe^2+}$.
I am using various catalysts including iron sulphate, cobalt chloride, etc. However, as far as I can tell, the catalysts are not speeding up the first step ($\ce{Fe^3+}$ to violet compound), but the second (violet compound breaking down to $\ce{Fe^2+}$ and other ions). The catalysts I've used have varying levels of effectiveness. The main 3 factors that would change how catalysts could affect this, as far as I can tell, are electron transfer,complex shape, and ligands. I don't think they are acting to transfer electrons here, and I think all the catalysts would have the same complex shape (octahedral). Therefore, I want to know how the ligands formed from each catalyst (causing it to go from linear monodendate violet compound to octahedral ions) would affect this differently. The catalysts I'm using are copper sulphate, iron sulphate, cobalt chloride, and nickel sulphate. I can't find much information about what is happening in this reaction. If anyone knows what ligands are formed, and how the shape/electrons/ligands of the catalysts are affecting this reaction speed, I would really appreciate it.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you want to know more, please have a look here and here. As it stands your question is quite broad and might get closed. It would also be advisable to formulate a reaction mechanism (with equations) of the uncatalysed reaction as a starting point.

Possible source: https://edu.rsc.org/resources/catalysis-of-a-reaction-between-sodium-thiosulfate-and-ironiii-nitrate-solutions/442.article


text:
Sulfate and carbonate ions as ligands
The pentahydrate salt of copper (II) sulfate exists as a blue solid while the anhydrous and monohydrate salts appear as white solids. The structures of these salts are discussed here. The blue colour of the copper (II) sulfate pentahydrate salt arises due to a similar reason why $\ce {Cu^2+}$ in aqueous solution is blue in colour. Based on Crystal Field Theory, this is due to the phenomena of d orbital splitting in the octahedral complex. As discussed in the post linked, in the pentahydrate salt, the $\ce {Cu^2+}$ ion is octahedrally coordinated, with 4 $\ce {H2O}$ molecules forming dative covalent bonds with it while the other two sites are occupied by the oxygen atoms of the sulfate ion. This leads me to wonder if the sulfate ions contribute to any significant degree of splitting of the d orbitals of the $\ce {Cu^2+}$ ion. As the colour of the anhydrous $\ce {CuSO4}$ salt is white, we can infer that the sulfate ions do not contribute to the d orbital splitting. Consequently, we can infer that the sulfate ions do not really form dative covalent bonds with the $\ce {Cu^2+}$ ion.
But why would this be the case? In the sulfate ion, there is rather poor delocalisation of the lone pair due to long $\ce {S-O}$ bond lengths and thus, ineffective p orbital overlap between the p orbitals of the sulfur and oxygen atoms. Hence, the negative charges, as well as the lone pairs of electrons, are rather localised on the oxygen atoms. This analysis all points to the conclusion that the sulfate ion should be able to act as a ligand.
Similarly, the fact that anhydrous, neutral $\ce {CuCO3}$ exists as a gray solid, suggests that the $\ce {Cu^2+}$ ion does not have dative covalent bonds with the carbonate ion , i.e. $\ce {CO3^2-}$ does not act as a ligand. Although for the carbonate ion, due to the stronger delocalisation of electrons within the ion, it may be more convincing to say that it may not be able to act as a ligand effectively.

No, you can't infer that if anhydrous "salts" are colorless then there's no dative bonds, on contrary, every single transition metal compound has more or less covalent bonds, often more covalent then ionic.

First and hopefully obvious things first: sulphate ions do form (weak) coordinate bonds to metal centres and thus by definition cause some d orbital splitting. In a simplified crystal field model, you might imagine them as weaker negative charges but nonetheless the do interact or we would not see the structures we see.
Comparing the structures of anhydrous $\ce{CuSO4}$ and its pentahydrate is very much non-trivial. For starters, they don’t even share the same surrounding. While the copper in $\ce{CuSO4 . 5 H2O}$ is approximately in a $D_\mathrm{4h}$ environment with $\ce{Cu-OH2}$ bonds of $\pu{197pm}$ and approximately $\pu{195}$ for the two different copper atoms in the unit cell, the environment of copper in anhydrous sulphate is better described as $D_\mathrm{2h}$ because the $\ce{Cu-O}$ distances average out to $\pu{187pm}$, $\pu{215pm}$ and $\pu{236pm}$, respectively—around the same copper atom. Two symmetry-identical oxygens with a $\ce{O-Cu-O}$ angle of approximately $180^\circ$ will be equidistant, all others will not.
That already means the corresponding molecular orbital schemes will look very different, even if we treat the oxygen atoms identically. Group theory tables tell us, that while in the case of the pentahydrate at least some degenerate orbitals remain, all orbitals will be independent in the case of the anhydrous sulphate. This in turn means that we are dealing with a potentially much more complicated MO scheme and all excitation discussions will get progressively more complicated. I would not be willing to base an argument on low coordinative ability of sulphate only on the difference between these two crystal structures.
Finally note that the carbonate $\ce{CuCO3}$ itself opens another can of worms: as Seidel et al. show, the copper is in a pentacoordinated environment so we cannot even use our usual octahedral arguments there.

References:

Structure of $\ce{CuSO4 . 5 H2O}$: J. N. Varghese, E. N. Maslen, Acta Cryst. B 1985, 41, 184–190. DOI: 10.1107/S0108768185001914.
Structure of $\ce{CuSO4}$: P. A. Kokkoros, P. J. Rentzeperis, Acta Cryst. 1958, 11, 361–364. DOI: 10.1107/S0365110X58000955.
Structure of $\ce{CuCO3}$: H. Seidel, H. Ehrhardt, K. Viswanathan, W. Johannes, Z. Anorg. Allgem. Chem. 1974, 410, 138–148. DOI: 10.1002/zaac.19744100207.


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E2 towards secondary over tertiary and primary beta-carbons
I would like to convert this tertiary bromoalkane to this trisubstituted alkene via E2

Theoretically, using a strong base (NaOH) would result in the formation of the Zaistev product (tetrasubstituted) as major, which I do not want.

If instead I use a sterically hindered base such as t-BuOK I would get the Hofmann product (disubstituted), which I also do not want.

What would I need to use in order to obtain the 'intermediate' trisubstituted alkene? Would it work if I just used a base that's less sterically hindered than t-BuOK but still blocks formation of Zaistev? Like NaOEt for example?
Thank you

Your "Zaitsev product" cannot exist: https://en.wikipedia.org/wiki/Bredt%27s_rule

Also, you should remove the hydrogen when eliminating.

Your corrected Zaistev product (see orthocresol's and Jan's comments) might not form because it is less stable than expected product (thermodynamically). .


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Available datasets to study using Hückel theory
I'm trying to do some tests using graph Laplacians based on Hückel theory. This, however, is said to be only applicable for conjugated hydrocarbons. My questions are:

I have been working with QM9 dataset and I was wondering if QM9 passes the filter (being consist of conjugated hydrocarbons only). How about ZINC database?
Are there any datasets available that can be analyzed using Hückel theory (Preferably with their SMILES representations included, and some property value reported, for testing a prediction algorithm)?

No, neither of these sets will work with simple Hückel theory, since there are a large number of non-conjugated bonds.

Possible duplicate of Graph Laplacian based on Hückel theory

@GeoffHutchison Question edited.


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Electron Gain Enthalpy
When an electron ($e^-$) is added to $\ce{Cl}$ ,i.e.,
$$\ce{Cl + e^- -> Cl^- + ∆_{gh}H}$$
We can see that energy is released. My question is where is this energy coming from? What happens within the atom such that it loses energy and releases it to the surrounding?

Could you please explain why you downvoted it so that I can correct it?

I think that the energy is not released by atom. It is released by the system, consisting of atom, electron and the entity the electron came from, e.g. other atom.

https://chemistry.stackexchange.com/questions/78838/does-a-positive-electron-affinity-actually-correspond-to-the-release-of-energy https://chemistry.stackexchange.com/questions/98723/why-do-atoms-have-electron-affinity

I voted to close this question because of the many other duplicates, but I realized a little late that those might not actually address your question, which is why I added a brief answer. Your question seems to be twofold: "where does the emitted energy come from" and "how does the electron get rid of it".

The main question here is "why is it exothermic". The following diagram illustrates the conundrum:

This portrays a ball rolling over a track. Thanks to energy conservation, in frame 2 the ball has gained kinetic energy KE after descending into a well, KE diminishing as the ball reemerges again in frame 3. You can think of an electron as the ball and of the track as the potential generated by the atom. In this classical picture the electron will retain a fixed elliptical orbit or escape trajectory as it zooms past the nucleus.
If there was no way for an electron (or ion) to shed energy then there would be no bound state. One way for the electron to shed energy is by emitting a photon, which causes it to relax into a lower energy bound state (in the cartoon this would be a tighter oscillation within the well).
The cartoon above is admittedly a gross oversimplification, particularly inaccurate in suggesting that the system is classical and 1-dimensional. In reality binding of the electron to the atom with emission of a photon or some nonradiative process is best treated quantum-mechanically, but the essential principle of conservation of energy is retained.

Further reading
A deeper explanation of electron affinity involves some heavy-duty physics, but this presentation on the Breit-Wigner cross section might be useful. It explains:

Broad states which can be formed by collisions between the particles into which they decay are referred to as resonances. They are described by the Breit-Wigner formula.

Not the most recent reference, but the following article which mentions different ways of measuring and computing electron affinities may also be useful: Meunier et al, Molecular Simulation, 1999, Vol. 23. pp. 109-125

Is it the energy in the electron + nucleus system that is being shed when the outer electron enters the region where electrostatic force of nucleus is experienced? If yes, why does the interaction result is shedding of energy?

I cannot give you a detailed answer, an explanation involves pretty advanced quantum mechanics, but I added some refs to my answer.


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Vapor pressure vs. ambient pressure
I've managed to get myself thoroughly confused in thinking about the vapor pressure of a substance at a particular temperature $T$, and how it relates to the pressure of the substance if it is in a phase equilibrium at that same temperature $T$.
Consider two sealed, rigid containers.  All three containers are in thermal equilibrium with a large temperature reservoir at (say) 298 K.

We evacuate Container A completely, and place into it a sample of liquid water.  Some of the water boils away until the pressure of the water vapor attains the appropriate pressure on the phase boundary in the P-T diagram.  (We assume that the sample of water is appropriately sized so that some water is left in each phase once it reaches equilibrium.)  Once equilibrium is reached, the water vapor has reached a pressure $P_A$.
Into Container B (which is of the same size) we also place an indentical sample of water.  However, Container B also contains dry air at $P_\text{air} = 1$ atm.  The water evaporates, mixing with the air, until it reaches a particular vapor pressure.  (Again, assume that the sample of water is appropriately sized so that some water remains unevaporated in equilibrium.)  Once equilibrium is reached, the water vapor has partial pressure $P_B$.

Does $P_A = P_B$ exactly?  Or, to put it another way, does the value of $P_\text{air}$ affect the partial pressure of the water vapor in equilibrium?  If $P_A \neq P_B$ precisely, is there some limit or approximation under which $P_A = P_B$ exactly? If so, what is it?

I haven't been able to find a definitive answer in my searching online, and it seems plausible to me that the presence of an external source of pressure on the water (namely the air in container B) might affect the phase transition properties of the water.

The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$
The additional pressure compresses the liquid and thereby increases its chemical potential, resulting in an increased vapor pressure. However, since $V_{m,~l}$ is relatively small,  the effect on the vapor pressure is also small except at very high pressures. The effect of air at atmospheric pressure and $\pu{298 K}$ on the vapour pressure of water ($\pu{3.2 kPa}$ in the absence of inert gas) is negligible, an increase of $\pu{2.3 Pa}$.
So $P_A \neq P_B$, but they become nearly identical when $PV_{m,l}\ll RT$.


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Handling EPR spin-probes, specifically how to dissolve 5-DSA?
I want to do some EPR tests with 5-DSA (or related spin probe) dissolved into different crude oils.
It had not sunk into me that $100 of 5-DSA was a tiny amount of a very sticky solid. It's proving hard to deploy in each test. Better would be if I could dissolve my 5-DSA into, say, known volume of toluene and then pipette a few mL of that toluene mixture into the oil sample.
But is toluene the right answer? I don't want to destroy the free radical. How do other researchers handle their spin-probes?

Yeah, I guess you should make some solution of 5-doxylstearic acid , but toluene is somewhat prone to make benzyl radicals...

For the -DSA and -DSE spin-radicals, what worked really well was to dissolve in a bit of toluene and leave overnight. The left a very thin layer of the spin-probe in the test tube. I could then add water or whatever and the radicals would spin around desperately searching for something to latch onto and then I could do experiments (eg micelles in water). Putting them in toluene and adding the toluene mix was less interesting because they stayed in the toluene and didnt go away and latch onto anything else....


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Determining number of unique hydrogens in lactupicrinal 1H NMR
I'm having trouble understanding which hydrogens are magnetically distinct in the following molecule (lactupicrinal):

I've counted all the hydrogens (there are 20) and I know that hydrogens in a methyl group are all chemically the same (I think), but I don't know about the hydrogens in the ethyl group (double bonded on the lower right side). Am I right in that all other hydrogens outside of those groups are distinct, or are the mirrored hydrogens on the benzene ring also equivalent? Please help, I'm completely at a loss and I've been trying desperately to understand hydrogen equivalence for weeks.

Your question doesn't sound like you care about exact splitting, so I'm going to ignore magnetic equivalence/inequivalence. If you do care, you need to take that into account.
For the purposes of this discussion, equivalent means chemically equivalent.
You also need to keep in mind that NMR is achiral. In other words, opposite enantiomers will produce the same spectrum.
The standard procedure is to replace the atom of interest with something that is distinguishable from the regular atom. I usually replace an $\ce{H}$ atom with an $\ce{H}*$, but anything will do, provided it's still a hydrogen but different from the others.
With this replacement you've created a new molecule.
For two specific hydrogen atoms, you can perform the replacement operation to generate two molecules.
Now, compare these two molecules.

If they are different molecules (e.g., geometric isomers, (E)-/(Z)- isomers, cis-/trans- isomers), then the original hydrogen atoms are heterotopic and should have different chemical shifts. Their environments are different.
If the molecules are diastereomers, then the two hydrogens are diastereotopic. Since we can distinguish diastereomers by achiral methods, so can NMR. That means that the two original hydrogen atoms will have different shifts.
If the molecules are enantiomers, then the two hydrogens are enantiotopic. They are different, but because NMR can't distinguish between enantiomers, it can't tell that these two hydrogen atoms are different. They will have the same shift.
If the two molecules are identical, then the two hydrogens are homotopic. Nothing can distinguish them. They are chemically equivalent, and by NMR, they will have the same chemical shift.

Try applying this test to the hydrogen atoms on the ethyl group and the hydrogen atoms on the ring.

Thanks for you answer. I've used the H-replacement technique you're talking about in the past, but I can't always tell whether the molecules are equivalent; for example, if I replace each of the upper hydrogens on the benzene ring with H*, are those two new molecules equivalent because the single bond between the ring and the rest of the molecule rotates freely?

In the absence of a concrete example, your question would be difficult to answer because atropisomers (isomers due to restricted rotation) exist. However, free rotation seems very likely in this case, since the aryl ring is not very substituted (particularly in the ortho positions), it's not part of some larger ring, and the actual substituents are small.

@Zhe You didn't need to ping with @ - it was done automatically 'cause there was only 1 user to ping.

Your analysis is basically right. The three protons within a rapidly rotationaly averaged methyl group are magnetically and chemically equivalent, whereas the aromatic protons related by reflection symmetry are chemically but not magnetically equivalent. The ethylenic protons are not chemically equivalent and therefore not magnetically equivalent. The key to identifying magnetic equivalence is whether any proton outside of the group under consideration has identical couplings to each proton in the group. Magnetically inequivalent protons generate coupling patterns through J-couplings to each other, even though they might share a chemical shift due to chemical equivalence.


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Is there a connection between M1V1=M2V2 and Boyle’s Law?
Recently in Chemistry 101 we learned about $M_1V_1=M_2V_2$, where $M$ is the molarity of the two mixtures and $V$ is their respective volumes.
The first thing that popped into my mind when I saw this formula is, “Hey, that reminds me of Boyle’s Law from high school physics - $P_1V_1=P_2V_2$ - the relationship between the volumes of a gas at different pressures.”
Is it just a coincidence that these two formulae look the same, or is there a connection between the pressure variable in Boyle’s Law and the molarity variable in the first equation? Are there other formulae which are structured similarly which are connected?

Yes, they are related. The first comes directly from the conservation of number of moles of the solute in a dilution,
$$n_1 = n_2 $$
Since $n = MV$,
$$M_1 V_1 = M_2 V_2$$
The second is related to the conservation of the total number of moles in an isothermal compression or expansion,
$$n_1 = n_2$$
Using the ideal gas law $n = pV/RT$,
$$\frac{P_1 V_1}{R T} = \frac{P_2 V_2}{R T}$$
Or $P_1 V_1 = P_2 V_2$.
Again, both are related to the conservation of moles (which by the way, isn't an universal law, since it's broken in trivial situations like the event of a chemical reaction), although the first one refers to solute only and the second to total number of moles.

That’s a great way to look at it, thanks! Are there any other formulae that relate to conservation of moles in this fashion?


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diborane fragment orbitals energy of dihydrogen
I want to ask a question about the fragment orbitals of the $\ce{B2H6}$ system, specifically when I combine the $\ce{B2H4}$ with the $\ce{H2}$ fragments.
I am halfway through my inorganic chemistry molecular orbitals course and today we were discussing the application of MO theory in figuring out the structure of $\ce{B2H6}$.
So far, I have been able to combine the two $\ce{B2H4}$ FOs but I am struggling to justify the placement of the $\ce{H-H}$ FO to complete the task.
So far, I've been given this diagram.

And the professor said the following:

the $\ce{b2}$ orbitals are non-bonding FOs so it does has nothing to bond with in the $\ce{B2H6}$ molecule. We assume the H $\ce{1s}$ orbital is the reference and line up the hydrogen $\ce{H-H}$ FO with the $\ce{b2}$ FO, giving the symmetry groups $\ce{ag}$ and $\ce{b_{3u}}$ respectively.

I understand the use of Hydrogen as a reference and the assignment of point groups in this situation, but I can't seem to work out why Hydrogen lines up with the $\ce{b2}$ FOs and not, for instance, with the $\ce{a1}$ FOs.
I've searched across the internet for the past day and failed to find any conclusive answer.
Why is $\ce{H-H}$ given to have the same energy as the $\ce{b2}$ FOs of $\ce{B2H4}$?

I learned this as the precise ordering is somewhat arbitrary. You know the energy of the fragment is roughly near the energy of the fragment it's mixing with—otherwise they wouldn't mix to form a bond—and then place it within that range for whatever is convenient to make the drawing readable. These diagrams are qualitative, their purpose to serve as accessible, conceptual overviews, so the priority isn't precision. Only calculations can reveal more rigorously precise orderings.


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How to apply the parvo/amplo akamptisomeric stereodescriptors to this bicyclic system?
I'm looking for examples of akamptisomerism[1].
I'm try to find simple molecules able to (theoretically) show this kind of isomerism.
I thought about the isomers of (2R)-2-methyl-9-oxabicyclo[3.3.1]nonane.
Considering only the $\ce{(HC)O(CH)}$ group, I found the eight akamptisomers.

$\hspace{5em} \ce{(1$R$_p,2$S$,5$R$_a)} \hspace{7em} \ce{(1$R$_a,2$S$,5$R$_p)}$

$\hspace{5em} \ce{(1$R$_p,2$S$,5$S$_p)} \hspace{7em} \ce{(1$R$_a,2$S$,5$S$_a)}$

$\hspace{5em} \ce{(1$S$_p,2$S$,5$R$_p)} \hspace{7em} \ce{(1$S$_a,2$S$,5$R$_a)}$

$\hspace{5em} \ce{(1$S$_p,2$S$,5$S$_a)} \hspace{7em} \ce{(1$S$_a,2$S$,5$S$_p)}$
Could the (3S)-3-methyl-11,12-dioxabicyclo[4.4.2]dodecane show akamptisomerism too?
What descriptor to use? Is this also akamptisomerism or will be another (new?) kind of isomerism?

$\hspace{5em} \ce{(1$R$_?,3$S$,6$R$_?)} \hspace{7em} \ce{(1$R$_?,3$S$,6$R$_?)}$

$$\ce{(1$R$_?,3$S$,6$R$_?)}$$
Reference

Canfield, P. J.; Blake, I. M.; Cai, Z.-L.; Luck, I. J.; Krausz, E.; Kobayashi, R.; Reimers, J. R.; Crossley, M. J., A new fundamental type of conformational isomerism. Nature Chemistry 2018, 10, 615–624.


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Generating molecular graph Laplacian based on Hückel theory
While studying graphs and graph Laplacians from "Laplacian eigenvectors of graphs: Perron-Frobenius and Faber-Krahn type theorems.", I encountered a type of graph Laplacians that rise from quantum chemistry. This matrix is consists of off-diagonal elements called resonance integrals and diagonal terms called Coulomb integrals. This is a sparse matrix, meaning that the matrix elements corresponding to non-connected atoms are zero. It is further mentioned that

... the entries of this matrix, H, are tabulated for different atoms and
  bonds.

I was looking online to find some table that gives this information and write a package to compute this matrix given a SMILES entry. As an alternative, I also looked for an already existing package that provides this matrix. The only thing I found was the implementation of the extended Hückel method in RDKit, which requires molecule conformation as input (apparently, eHM needs atom coordinates).
I was wondering if anyone can point me to such a table or python package or let me know if I'm missing something.

You could always generate a conformer from the SMILES and use extended Hückel - my guess is that it's a lot more accurate and useful. (Simple Hückel is really only useful for $\pi$ conjugation)

See also: Comparing the Hückel and Extended Hückel methods

Not necessarily. Take 'CC12CCC1CC2' for instance, RDKit gives zero conformations. I tried using original QM9 database that you once referenced, but then for smiles like 'CC@@HC(=O)[O]' it rejected the entry.

I'm not quite sure why that particular SMILES wouldn't generate conformers from RDKit (sounds like a bug), but you can in Open Babel.

I'm not sure what you mean by 'rejected the entry' but if you have questions, feel free to contact me outside of Chem.SE. Suffice to say that neither of these compounds can be treated with regular Hückel.

You're describing simple Hückel theory. This is usually applied to $\pi$-conjugated systems to understand the stabilization in conjugated and aromatic molecules.
There are a variety of Python packages, for example:

Felix Plasser wrote one for his blog and the current location is at his homepage
Randle Taylor has a SHMO package on GitHub, including a GUI called Orbis

I believe the version from Plasser will read in files using Open Babel's python interface, so certainly you could handle SMILES.
People would perturb the $\alpha$ (atom site energies) and $\beta$ (bond interaction) parameters in various ways. It's hard to find good tabulations of these, but the best appears to be:
"A brief review and table of semiempirical parameters used in the Hueckel molecular orbital method" J. Chem. Eng. Data 1967 122 pp. 235-246
You mention the somewhat related Extended Hückel theory (EHT). This considers both $\sigma$ and $\pi$ bonds and is thus a lot more useful, although as you mention you need coordinates to calculate overlap values, etc.


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Stabilizing donor-acceptor interaction between orthogonal orbitals in the SN2 transition state?
For the $\mathrm{S_N2}$ reaction between chloroacetone and iodide, my professor has drawn a donor-acceptor interaction between the iodine lone pair attacking the Cl-substituted carbon and the π antibonding $\ce{C-O}$ orbital of the molecule.
If $\mathrm{S_N2}$ requires overlap between the σ* $\ce{C-Cl}$ orbital and the nucleophilic lone pair of I with σ symmetry, how can that lone pair participate in a donor-acceptor interaction with orbitals of π symmetry? I believed these orbitals did not interact as they are orthogonal to one another and thus experience no overlap.
Could it be that the lone pairs my professor is referring to are the other lone pairs on iodine in p orbitals that extend through the appropriate symmetry planes?


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Which of the following carbocations is more stable?

Which of the following carbocations is more stable?

I know two major factors in determining carbocation stability: resonance and the number of carbon atoms neighboring the positive carbon. Does the double bond in the  substituent of the first molecule participate in resonance?

The one with the aliphatic C=C bond may be more stable since there may neighbouring group participation, allowing dispersal of positive charge

@TanYongBoon There is an $\mathrm{sp^3}$ carbon in-between which breaks resonance so the shorter chain allows for more hyperconjugation.

@Jan I am not talking about resonance. When I mean "neighbouring group participation", I am referring to the $\pi$ electron density being donated directly to the positively charged carbon atom, forming a cyclopropyl or cyclobutyl structure

@TanYongBoon Oh, I think I see what you mean; analogous to a metal-ethene complex? If that’s the case, I’m not sure whether the single $\ce{CH2}$ group provides enough flexibility but I would have to build a model.

@TanYongBoon that mechanism would require a "six membered ring". In term or charge stabilisation/relative stability the smaller carbocation should be favoured.

@TanYongBoon actually now I feel stupid. The cation on the top is just the linear form of a cyclopropylmethyl cation … *head meets table*

@Jan Yes, that is one possible stable structure that can be derived from the "neighbouring group participation" I suggested

Are we talking about: What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

The second one will be more stable.
We usually first check for resonance as it is more effective.
But here it has to be explained on basis of hyper conjugation because the resonance due to the ring is same in both. But the second one will have more hyper conjugating structures as greater number of alpha hydrogens are present.
More about hyper conjugation concept:
https://en.wikipedia.org/wiki/Hyperconjugation
Edit :  No the double bond wont participate in resonance.

The Second One Will be More Stable as it has one of the resonating structure with octet configuration that is when the the double bond of the ring comes out and participates in the formation of double bond with the carbocation which is out of ring in no case the double bond from the first molecule will be involved in resonance ....


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Mathematical relation between bond angle and fraction of s or p character
In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 75, under the topic "Effect of Electronegativity - When the surrounding atom is same with different central atom having lone pair" it is given:

The effect of electronegativity as postulated in VSEPR theory explains the order of the angle for the above molecules (Hydrides of groups 15 and 16) but cannot rationalize very small angles (~$90^\circ$) in $\ce{PH3,AsH3,SbH3}$ and $\ce{H2S,H2Se,H2Te}$ with respect to $\ce{NH3}$ and $\ce{H2O}$ respectively.
To explain this, Drago suggested an empirical rule$^{[1]}$ which is compatible with the energitics of the hybridisation. It states that if the central atom is in the third row or below in the periodic table, the lone pair will occupy a stereochemically inactive s orbital$^{[2]}$, and the bonding will be through p orbitals, and bond angles will be nearly $90^\circ$ if the electronegativity of the surrounding atom is less than or equal to $2.5$.
The above rule is based upon the relation between hybridisation and bond angle for two or more equivalent s-p hybrid orbitals, where the fraction of s character (S) or fraction of p character (P) is given by the relationship:
$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$
for $\theta \in (90^\circ,180^\circ)$
The following are links to previous questions asked on Chemistry Stack Exchange regarding concepts in this quoted text.
[1] : Drago's Rule : What is Drago's rule? Does it really exist?
[2] : Stereochemically (in)active s orbital : What is a stereochemically active or inactive s orbital?

This formula,
$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$
relates the bond angle $\theta$ and the fraction of s/p characters. I tried to plug in some standard values.
For example, On substituting the expression with S=0.25 or P=0.75, I get $\frac{S}{S-1}=-0.33...$ and $\frac{P-1}{P}=-0.33...$. I know they both give the same as in sp$^n$ hybridisation where only s and p orbitals are involved, the sum of fractions of the s character and p character i.e., S+P=1, and I was able to interconvert the expressions $\frac{S}{S-1}$ and $\frac{P-1}{P}$ with simple substitution - S=1-P or P=1-S.
On finding the cosine inverse of -0.33...,
$\cos ^{-1} (-0.33...)=109.471^{\circ}$
which is exactly equal to the tetrahedral angle ($109.471^{\circ}$). From this, we can obtain the bond angle in sp3 hybridisation. And similarly, we can do it for sp, and sp2 hybridisation which I've skipped here.
It is also possible to the reverse with this formula, i.e., given the bond angle we can compute the fraction of s or p character. For example, for $\ce{AsH3}$, the H-As-H angle is $91.8 ^{\circ}$, and from the calculation, it can be shown that each As-H bond consists of almost 97% p character and 3% s character.
My doubt is how did they arrive at the given formula? Is there any derivation* for this or is that simply an experimental result? Will this work for all values of S and P (under the constraint S+P=1)? What is the logic behind this formula?
I am unable to find any useful information about this on the internet.

*Simple or easy to understand derivation which can be understood by a high school student.

https://chemistry.stackexchange.com/questions/33279/is-the-hybridization-only-related-to-atomic-orbitals-but-not-to-the-molecular-or https://chemistry.stackexchange.com/questions/26445/is-s-p-mixing-referring-to-hybridization-or-is-it-the-mixing-of-one-atoms-s-orbi

Possible duplicate of Sp5 hybridization in cyclopropane?

@Mithoron, The question you referred was useful but did not clarify my doubt completely. I am asking for the relation between the bond angles and the amount of s/p character, whereas that answer/question is about s/p character itself and doesn't derive the formula, just uses it.

@Intellex For future reference -- not limited to ChemSE -- quite often, it sufficies to state the tetrahedral angle rounded as equal to either 109.467 degrees, or rounded to equal to 109.5 degrees, especially if determined experimentally.  More often than not, to provide the inverse cosine with thirty decimals "just because" your computing engine is capable to do so does not provide additional significant information.

The full derivation should be available on Wikipedia: Bent's rule and is known as Coulson's theorem.

@Martin-マーチン, Thank you for specifying that theorem, which I haven't encountered until now. I read about Coulson's theorem from Wikipedia (which I never understood) and from this answer by ron. Even though it gives the same result, it is in a different mathematical form. Further ron has described it is only for carbon compounds, but the formula in my book is given for any compound. Is that really the theorem you mentioned? If yes could you please give a beginner level derivation (easy to understand)? Kindly reply when you find time.

I currently have little time, but I will try to get to it. I remember that ron's answers using the theorem always confused me, too. The theorem itself can be applied to all compounds, but it obviously only makes sense for those that can be described with spⁿ hybridisation. Also, since it is derived from Bent's rule, it is also an observation, and might lead to wrong conclusions if used outside its scope.

@Martin-マーチン, If possible, please answer before the bounty expires.

Does this answer your question? What is θ in Bent's rule equation (%s - character)?

Just like most other theories Bent's rule also has an formal theory which can be used to derive mathematical equations following from the rule.
Bents rule is based on quantum mechanics and thus derivation of its mathematical form inherently uses quantum mechanical nomenclature and thus might not fit your "understood by high school student" criteria.
But still I am referencing the derivation of the mathematical form of Bent's rule from Wikipedia[1] just in case you need it. Feel free to ignore the quoted text if it's too complex to comprehend.

To construct hybrid $s$ and $p$ orbitals, let the first hybrid orbital be given by $s + \sqrt{λ_i}p_i$, where $p_i$ is directed towards a bonding group and $λ_i$ determines the amount of p character this hybrid orbital has. This is a weighted sum of the wavefunctions. Now choose a second hybrid orbital $s + \sqrt{λ_j}p_j$, where $p_j$ is directed in some way and $λ_j$ is the amount of $p$ character in this second orbital. The value of $λ_j$ and direction of $p_j$ must be determined so that the resulting orbital can be normalized and so that it is orthogonal to the first hybrid orbital. The hybrid can certainly be normalized, as it is the sum of two normalized wavefunctions. Orthogonality must be established so that the two hybrid orbitals can be involved in separate covalent bonds. The inner product of orthogonal orbitals must be zero and computing the inner product of the constructed hybrids gives the following calculation.
$$
\begin{align}
\langle s + \sqrt{λ_i}p_i | s + \sqrt{λ_j}p_j \rangle &= \langle s |s \rangle + \sqrt{λ_i}\langle s | p_i\rangle + \sqrt{λ_j}\langle s | p_j\rangle + \sqrt{λ_iλ_j}\langle p_i | p_j\rangle\\
&= 1+0+0+\sqrt{λ_iλ_j}\cos{\omega_{ij}}=1+\sqrt{λ_iλ_j}\cos{\omega_{ij}}
\end{align}
$$
The s orbital is normalized and so the inner product $⟨ s | s ⟩ = 1$. Also, the $s$ orbital is orthogonal to the $p_i$ and $p_j$ orbitals, which leads to two terms in the above equaling zero. Finally, the last term is the inner product of two normalized functions that are at an angle of $ω_{ij}$ to each other, which gives $\cos{ω_{ij}}$ by definition. However, the orthogonality of bonding orbitals demands that $1 + \sqrt{λ_iλ_j}\cos{ω_{ij}} = 0$, so we get Coulson's theorem as a result:[15]
$$\cos{\omega_{ij}}=-\frac{1}{\sqrt{λ_iλ_j}}$$

Now at this point you need to know only one thing:
For any $\mathrm{sp^n}$ hybridized species with symmetrically hybridized orbitals, $n=\lambda_i=\lambda_j$.
Thus,
$$\cos\omega=-\frac{1}{n} \tag{2} \label{eqn:2}$$
Now as $n$ is the ratio of $p$ character to the $s$ character in the hybridized orbital.
$$n=\frac{S}{1-S}=\frac{1-P}{P}$$
Which when substituted in $(\ref{eqn:2})$ gives your desired results.

Reference:
(1)   Wikipedia contributors. Bent’s rule https://en.wikipedia.org/w/index.php?title=Bent%27s_rule&oldid=992423483 (accessed Jun 18, 2021).


text:
What is the effect of applied magnetic field on absorption spectrum?

Upon application of a weak magnetic field, a line in the microwave absorption spectrum of rigid rotor splits into $3$ lines. The quantum number $(J)$ of the rotational energy level from which the transition originates is
(a) $0$
(b) $1$
(c) $2$
(d) $3$

Can anyone explain the answer (a) for this?
For rotational levels, each level is $(2J+1)$ fold degenerate. According to that, the answer should be (b).

Transition occurs between levels

@Alchimista Do you mean that there is a transition from J=0 to J=1?

Yes else you would get more lines


text:
Cycloheptatrienyl carbanion aromaticity
Many sources including previous answers on Stack Exchange claim that it is antiaromatic.
But in a different yet similar question, I have also seen people saying that a monocyclic ring with 7 or more members can go out of plane to avoid antiaromaticity and become non-aromatic.
Shouldn't the same rule apply here as well?
Links mentioned:
https://chemistry.stackexchange.com/a/102606/74312

That is the general meaning of antiaromaticity. It is never enough stressed out that it is a hypothetical structure generally stabilised by the system arranging differently!

Food for thought and not a full answer: the big difference between oxepin and the cycloheptatrienyl anion is that the latter is charged so the molecule needs to strike a balance between structural changes to avoid conjugation and delocalisation of the negative charge.


text:
What's an appropriate ligand for Pd-Ag redox?
I want to reduce palladium with silver (in cellulose): $\ce{Pd^{2+} + 2Ag->2Ag+ +Pd}$. What is a good ligand and spectator ion for both metals? (The pH is not critical.) Ones I have considered:

$\ce{[Pd(H2O)4](NO3)2}$: Unstable; hydrolyzes to insoluble hydroxides over time, except in strong acid, where the nitrate will oxidize the silver.
$\ce{[PdCl4]^{2-}}$: Will precipitate $\ce{AgCl}$.
$\ce{[Pd(NH3)4]^{2+}}$: $\ce{[Ag(NH3)2]}$ can precipitate explosive $\ce{Ag3N}$.
$\ce{[Pd(en)2}]^{2+}$, other common organic ligands: These are all seem to be very poisonous.
Glycine-$\ce{(NO3)-}$: Maybe. I can't find any stability constant data.
$\ce{(NH4)5[Pd(S2O3)4]}$: Definitely stable with silver, although I can't find a stability constant for palladium. EDIT: "This complex is not thermodynamically stable, and S-bridged oligomers are ultimately formed" according to [1], although maybe it could be stabilized somehow.
$\ce{(NH4)+}$-$\ce{(SCN)-}$?

Are there any reasonable candidates that I haven't thought of? Any sources for stability constants, especially for thiosulfate?
[1] Thiosulfate Complexing of Platinum Group Elements.
Elizabeth Y. Anthony, Peter A. Williams. https://pubs.acs.org/doi/abs/10.1021/bk-1994-0550.ch033


text:
Negative Pressures in Van der Waals Equation of State
This is the plot of the Van der Waals equations:

As you can see, the Van der Waals equation predicts a "stretched liquid" area (MNS). Is this physically feasible and what exactly happens there? Are there any intuitive explanations for this?

There is no M-N-S, much like there is no F-G-H, but just the horizontal dotted line F-H.

Be aware the equation has limited applicability and is not applicable in the area below the triple point, when the formula extrapolation leads to not existing physical states. Sea also https://en.wikipedia.org/wiki/Real_gas#Models

Although you can reach the superheated liquids and supercooled vapours, these are not at equilibrium and tend not to last long.

I am aware that this is only a metastable state and that there is no M-N-S area in the chemical sense. What I'm inquiring about is the concept of negative pressure in a real gas context. There are papers out there that do say that there is negative pressure for liquids (http://ilm-perso.univ-lyon1.fr/~fcaupin/fichiersPDF/AdvChemPhys_Caupin-Stroock.pdf). I still do not understand what's happening with a gas/liquid under negative pressure conditions. It's not called "stretched liquid" for no reason in the graphic above...

Well, if you consider compressing a positive pressure, then stretching is negative, as simple as that. If you have viscous liquid then some measure of stretching is possible.

You can have a stretched liquid to some degree. The same way as pulling a solid body in all directions creates stresses that can have negative trace, you can stretch a liquid, which then holds together by cohesion forces.
The state is metastable and subject to cavitation, but not completely unstable. In thin capillary tubes (or rather very thin pores in some porous media) you can achieve negative pressures and hence be able to lift the water column above 10 m, as trees are able to do in their vascular tissue (Quenching Even the Tallest Thirst:
How trees suck up water from the ground and avoid embolisms) and as can be reproduced in experiments (The transpiration of water at negative pressures in a synthetic tree).
A lot more, with some thermodynamic diagrams: Negative pressure of stretched liquid water. Geochemistry of soil capillaries


text:
How can I make a reference humidity environment with NaOH pellets?
I have NaOH pellets at my disposal. Some of the pellets are degraded, but most are not, and I can provide a new batch if needed.
I would like to test several humidity sensors in an environment with a humidity value around 10%. NaOH should be able to reach about 7% humidity when pure.
How should I proceed? I have tupperware and a stone mortar and pestle, if needed.
Should I take some good pellets, crush them without making them too fine, and place the resulting powder in a container, which I will enclose in the tupperware together with the sensors? or should I place the pellets in water but leaving the solution oversaturated (undissolved pellets visible in the water)?
The pellets would probably have a much higher contact surface with air, but time is not an issue, I can leave the sensors in the container for a day, if needed. Accuracy of the result is more important.
I can tolerate 2-3% errors.

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

@Poutnik that question has the answer I am looking for: pellets in water should work perfectly.


text:
On mathematics behind molecular orbital energies
When explaining the mathematics behind calculating molecular orbital energy , namely HOMO and LUMO energies, using quantum chemistry, Chainer Chemistry documentation states that

From mathematical viewpoint it requires a solution of an internal
  eigenvalue problem for a Hamiltonian matrix.

I have never heard of internal eigenvalue problem in mathematics. Is this a term specific to quantum chemistry and what it means?
Is the Hamiltonian matrix same as Kohn-Sham Hamiltonian?

An internal or interior eigenvalue problem seeks a solution somewhere in the middle of the spectrum of eigenvalues as opposed to trying to find the smallest or largest eigenvalue. Its generally trickier to solve for a particular eigenvalue than it is to solve for the smallest/largest one. The HOMO and LUMO are necessarily somewhere in the middle of the eigenspectrum.
The Kohn-Sham Hamiltonian is the orbital-DFT approximation of the molecular Hamiltonian. The Hamiltonian describes the interactions within the molecule and allows one to solve for the molecular wavefunction through the Schrödinger equation. Most of electronic structure is focused on approximations of the Hamiltonian or the form of the wave-function that make solving this problem tractable. There is a lot more that can be said about Hamiltonians than would ever fit in an answer here, so you should try to find good textbooks and papers that discuss them.


text:
What is the Lewis dot structure for ozone?
Currently I studying about the Lewis dot structure for different compounds and my textbook says that the Lewis structure for ozone may be drawn as(as shown in the image):

And then it goes on to calculate the formal charges on different oxygen atoms in it(which are shown in figure with $+/-$ signs.) In justification of the structure it says:

Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms. The formal charge is a factor based on a pure covalent view of bonding in which electron pairs are shared equally by neighbouring atoms.

Now, my question is why can't the structure  be like this:

Clearly, if this isn't a possible structure then the flaw in my understanding arises because of not understanding the term 'smallest formal charge'(as I found out that the structure I gave has all oxygen atoms with zero formal charge )
Please suggest what does it mean by the smallest formal charge ? Also why energetically the structure isn't stable.
(For now please exclude any discussion about resonance hybrid of the ozone)

The reason the cyclic structure doesn't exist is due to the bond angles. The 60 degree bond angles puts the electrons in the covalent bonds too close to each other. The electron-electron repulsion makes the structure unstable.
Rarely will you find bond angles less than 90 degrees, making cyclic structures with less than 5 in a ring uncommon.

Why don't you post it as an answer.

I can't post as an answer, it's on hold.


text:
Do solids flow?
I am now confused whether solids flow or not. Some people say solidity is only a time scale notion. Some others say solids do not flow at all. While some authors say there is a threshold stress below which solids do not flow at all.
I came to know that glass do not flow which is against the common myth that old church window glasses are thicker at bottom because glass flows, which is actually due to the way it was made in medieval times.
Which is the reality? Do solids (example, metals) flow always very slowly? Is there something like a threshold stress for solids to flow? Do solids except some polymers have infinite viscosity?

Actually ... yes!  Even when stress is applied below the yield stress, over time it can cause solids to flow.  In industrial equipment, this process, called creep, significantly impacts the performance and lifetime of equipment.  The flow of glaciers under the stresses created by their own weight is also a form of creep.
During a majority of the lifetime of the stressed solid, creep is described by this equation taken from the referenced WP articke:
$\dfrac{d\epsilon}{dt}=\dfrac{C\sigma^m}{d^b}\exp(\dfrac{-Q}{kT})$
This says that creep rate is increased with increasing stress $\sigma$, with finer grain size $d$ (because more grain boundaries make it easier for atoms to slide by each other, even as fine grain size is often desirable for other properties), and especially with temperature as shown by the exponential term (analogous to the activation energy in chemical reaction rates, not surprising since what happens is atoms are breaking/making bonds and moving).  That temperature dependence means, in practice, creep will be extremely slow unless we are fairly close to the melting point of the solid, which in our ordinary experience isn't the case -- except, as alluded above, with high-temperature industrial equipment and with large, heavy ice formations.
Outside of such exceptions, we may have to wait longer than the current age of the Universe to see much of anything.  Yet, given appropriate time and conditions, solids flow.

Coarse grain is more resistant to flow in high temperature metals . Some high performance turbine blades are single crystals - no grain boundaries.

I think I said that fine grain structures creep more.  But I would have never guessed that about the turbines!


text:
Why is CO2 + Ca(OH)2 an acid-base reaction?

$\ce{CO_2 + Ca(OH)_2 -> CaCO_3 + H_2O}$

I was wondering why the reaction between carbon dioxide and calcium hydroxide is considered an acid-base reaction when there are no hydrogens which are donated or received (at least that's what I can see). It looks like a double displacement reaction to me.
I will be very thankful if anyone can clear up my confusion.

CO2 can act as Lewis acid... BTW are you talking about aqueous solution here?

Related to: https://chemistry.stackexchange.com/questions/121198/why-does-bubbling-carbon-dioxide-through-calcium-hydroxide-result-in-a-precipita and https://chemistry.stackexchange.com/questions/111118/reactions-between-caoh2-and-co2-and-sroh2-and-co2.

You could imagine this reaction taking place in the two steps $\ce{2OH^- -> H_2O + O^{2-}},\ce{O^{2-} + CO_2->CO_3^{2-}}$, then bring the calcium in to pair with the carbonate. That just isn't how it actually proceeds for the most part, which should make some sense since this reaction involves $\ce{OH^-}$ acting as an acid with another $\ce{OH^-}$, so it is analogous to the autodissociation of water. Instead, initially $\ce{OH^-}$ wants to act as a base (which is probably familiar) while the $\ce{CO_2}$ wants to act as a Lewis acid (which might not be familiar).

My guess is that your confusion stems from the fact that there are 3 main definitions of acid/base.
In introductory chemistry, we cover the concept of Arrhenius acid/base, where the acid is a proton donor and the base is a hydroxide donor.
$$\ce{HCl + NaOH -> H2O + NaCl}$$
As we learn more, we're introduced to the concept of Brønsted-Lowry acid/base, where the acid is a proton donor and the base is a proton acceptor.
$$\ce{HCl + NH3 -> NH4+ + Cl-}$$
Finally, we're introduced to the concept of Lewis acid/base, where the acid is an electron-pair acceptor and the base is an electron-pair donor.
Your example is confusing probably because it is one of these cases. Hydroxide has a lone pair (Lewis base). That lone pair interacts with an empty $\pi*$ orbital in carbon dioxide (a Lewis acid) to create a new bond (bicarbonate). The bicarbonate then reacts with water in a more familiar way, so technically, there are multiple acid/base interactions happening even in your simple example.

What about this ?
$$\ce{CO2(aq) + H2O <=> H2CO3}$$
$$\ce{H2CO3 <=> H+ + HCO3-}$$
$$\ce{HCO3- <=> H+ + CO3^2-}$$
$$\ce{Ca(OH)2 <=> Ca(OH)+ + OH-}$$
$$\ce{Ca(OH)+ <=> Ca^2+ + OH-}$$
$$\ce{H+ + OH- <=> H2O}$$
$$\ce{Ca^2+ + CO3^2- <=> CaCO3 v}$$

An acid-base reaction is a reaction where a base is destroyed by an acid.
Here the base OH- is present in the molecule Ca(OH)2. And it is destroyed by CO2, which in solution is acid as it produced the acid H2CO3 with water.
So the reaction CO2 + Ca(OH)2 is an acid-base reaction


text:
Is chirality a property of only molecules with exactly 1 chiral centre?
I have seen chirality being defined in a number of ways in terms of a molecule not being superimposable with its mirror image.
My syllabus has a statement: "know that optical isomerism is a result of of chirality in molecules with a single chiral centre". This seems odd because surely you can have a have optical isomerism between molecules with more than 1 chiral centre so why a single.
Overall, I am asking for either a confirmation that the statement is incorrect or an explanation of where my understanding is wrong.

No, it's property of virtually everything but simplest stuff.

Related: https://chemistry.stackexchange.com/a/122660/

I don't want to try to interpret the intent behind the wording of the phrase you quoted, so I'll just state the situation and you can read into it as you wish.
There is no truly general relationship between the number of chiral centres in a compound and whether it displays chirality overall.
However, if a compound has exactly one chiral centre, then it must be chiral. Compounds with more than one chiral centre may or may not be chiral, depending on what those chiral centres are and their relative stereochemical relationship. And as has already been rightly pointed out (4 seconds before me), compounds with zero chiral centres may still be chiral.

If by "chiral centre" you mean a carbon atom with 4 different substituents, then consider $\ce{C(H)(X)(G_{R})(G_{S})}$ (where $\ce{G_{R}}$ represents a chiral group with the "R" configuration) has 4 different substituents but is achiral (plane of symmetry).  Conversely, $\ce{C(H)(X)(G_{R})(G_{R})}$ with only 3 different substituents is chiral.

@ron I think those are technically called "pseudo-asymmetric centres"; it seems that IUPAC has defined a "chirality centre" in a way ("an atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image") that circumvents this case, which you are very right to bring up.

The statement is incorrect. You also don't even need a stereocenter (chiral center) to get chirality, as in the case of molecules with an axis of chirality.


text:
Intramolecular Reaction with carbocation rearrangement
I came across a question recently which is as follows,
I came up with a mechanism but I am stuck at a step as I am unable to reason out the driving force.
Step(3) has a hydride shift and its next step involves methyl shift. I want to know the driving force for step (3) or any other approach to this problem. Do let me know if I am going wrong with my approach.

General comment: don't use double-headed arrows like that for reactions. Those are intended for resonance structures. Alkyl- and hydride-shifts are not resonance.

In terms of driving force, ask yourself how much of what's written is reversible? Also look at the differences in energy between the reactant and product.

@Zhe Ok. I will see to that from next time

Your approach is feasible. Thermal cyclizations of appropriate dienes, enones, enynes, and other related unsaturated systems are known in literature. For example, there are extensive examples of intramolecular ene-reactions in Organic Synthesis and some of them are reviewed (Ref.1 and Ref.2). To justify your first step, I'd include the abstract of reference 1, which states that:

Thermal cyclizations of appropriate dienes, enynes and related unsaturated systems, some of them carried out on an industrial scale, demonstrate increasingly the preparative power of the intramolecular ene reaction. A variety of substituted, fused and bridged ring systems, including natural products, are thus easily accessible in a regio‐ and stereo‐selective manner. Numerous examples are discussed systematically illustrating the possibilities, limitations, and common features of this cyclization reaction and its reverse ring‐opening process.

Nonetheless, see following example given in reference 1 for thermal cyclization:

Although those are different than the problem in hand, the intramolecular cyclization has possible clear path. For example, various types of intramolecular cyclizations has been discussed in Ref.1. Now, I modified your mechanism using arrows to make it clear:

The electrophile here is the carbonyl carbon of 1, which would get more electron deficient by carbonyl oxygen’s complexation to a Lewis acid such as $\ce{AlCl3}$, or  $\ce{AlEtCl2}$, or even $\ce{AlEt2Cl}$ (Ref.2). This is evident in the intermediate I. The neucleophile, the alkene group of the molecule is in close proximity to make a bond with elecrophilic carbonyl carbon (intermediate II). The driving force here is making a less strained 5-membered ring and relatively stable $3^\circ$-cabocation. Keep in mind that all of these intermediates are reversible. The reversible hydride transfer is possible because energy different between resulting $3^\circ$-cabocations are minimal. Yet, this transfer facilitate pinacol-pinacolone type rearrangement of methide shift to give final stable product 2.
References:

Wolfgang Oppolzer, Victor Snieckus, “Intramolecular Ene Reactions in Organic Synthesis,” Angew. Chem., Int. Ed. Engl. 1978, 17(7), 476-486 (https://doi.org/10.1002/anie.197804761).
Barry B. Snider, “Lewis-acid catalyzed ene reactions,” Acc. Chem. Res. 1980, 13(11), 426-432 (https://doi.org/10.1021/ar50155a007).


text:
Is it safe to mix calcium carbonate and sodium chloride?
I know zero about chemistry, and Google has not offered up a plain answer, so I beeseech thee; Is it safe to mix calcium carbonate and sodium chloride to make a simple mouthwash?  Calcium carbonate for the acid neutralizing, and sodium chloride for disinfectant.

So, salt and chalk?

Ofc, without any issue. Nothing wud happen

I cannot even.$%$

@Jan but you have been beseeched, have you not?!

@UbaidHassan with a loud buzzing noise.

Safe in context of chemistry, physiology or long term health care prevention ?
With zero knowledge, it is safer to stay with commercial products, until you know, what you are doing.
For mouthwash, calcium carbonate is quite ineffective, being insoluble solid. Better would be solution of sodium bicarbonate ( baking soda ), present in some toothpastes as well.
Salt is not great desinfectant and must be used in really high concentration, that can be very disturbing if swallowed.


text:
Do magnetic fields generated from ion flow through a membrane channel have any physiological relevance?
Membrane pores and transporters see millions of ions flow through them per second. This creates a current and therefore a magnetic field. Do cells have any use for these fields (like maybe drawing charged receptors together?) or is it physiologically irrelevant?

Not that we know of.

You only have a current if the ions flow without an opposite flow of counterions.

@Karl I’m talking about local magnetic fields... like through a single uniporter not global B fields

For currents in biological systems the magnetic field is minuscule, and likely has no physiological effect.
That said, magnetoencephalography is based on measuring the tiny magnetic fields created when nerves fire. Detection requires an incredibly sensitive detector, a superconducting quantum interference device (SQUID), to detect the location of the currents. A SQUID can detect fields of just a few atoteslas (aT)!
Perhaps one could investigate how much ion flow in electric eel muscle is constrained by the induced magnetic field... It could lead to shocking revelations.

A better answer starts by being more precise in the type of magnetic field (static for example) and magnitude of exposure.
I have conducted one small crude experiment employing a 'Magnetizer', which is encased in plastic, which is reputedly claimed to produce an ordered collection of strong and weak fields to leave a residual magnet footprint when a metal object is inserted (or fluid passes). The Magnetizer was placed in an open vessel with H2O (tap water with oxygen), Fe as iron filings, some dissolved CO2 from seltzer water, and a bit of sea salt (see https://www.sciencemadness.org/whisper/viewthread.php?tid=77204#pid493712 ). There was also another identical vessel with the same reagents, but absent the Magnetizer, as a control. The results, following periodic pictures, seemingly indicated a jump start in the Magnetizer assisted iron oxidation experiment. In fact, I roughly suggested that the presence of the ordered magnetic field as supplied by the Magnetizer appeared to increase the reaction rate by 20%!
Reading comments in the alluded thread experiment does reveal some of the possible complexities in setting up a rigorous experiment. I do believe that an alleged criticism relating to clustering of Fe particles (from the magnetic field effect) is in error, as I would argue, it is not an advantage to bury Fe particles thereby reducing contact with surface water rich in needed oxygen.
Also, the literature further claims that radical based reactions are also accelerated. In fact, I cited a reference in that study relating to the use of a magnetic field to reduce the amount of free available chlorine (supplied as tablets) normally added to a swimming pool in experiment conduct in England. That study also claimed measurable positive benefits.
So, there are studies (which may be of suspect quality also) asserting the effects of a magnetic field to be apparently very real and certainly not marginal.
To answer you question: Do cells have any use for these fields (like maybe drawing charged receptors together? My answer is perhaps.

Sounds like pseudoscience. The magnet would align filings to have more surface area, a change in physical design from the control. Try it on one solid lump of iron, or on porous iron, and then report back.

Even if a magnet aligns filings to have more surface area, that is a reproducible attribute that should be recognized!

This is not a productive response... how is this relevant to the question


text:
Carbon Capture Absorption with Henry's Law

Assume the Henry’s coefficient for water at $\pu{20 °C}$ is $\pu{1.6E8 Pa}.$ What is the mole fraction of $\ce{CO2}$ that will be dissolved in water assuming that it is at equilibrium with a flue gas containing $0.11$ mole fraction $\ce{CO2}$ at atmospheric pressure $(\pu{10^5 Pa})?$

My understanding is that we can use Henry's law
$$y_i\cdot p = x_i\cdot H$$
where $H$ is the given Henry's coefficient. This yields a solution for $x(\ce{CO2}) = \pu{6.89E-5}.$

A coal plant emits $\pu{120 kg s-1}$ of $\ce{CO2}$ at a mole fraction of $0.11$ and atmospheric pressure $(\pu{10^5 Pa}).$ Assuming that the flue gas and water come fully into equilibrium during separation, what flow rate of water $[\pu{m3 s-1}]$ would be needed to capture $95\%$ of the $\ce{CO2}?$

For this part, I'm a bit confused as to where our $x(\ce{CO2})$ and the $95\%$ capture rate comes into the play. I assume after you calculate a new mole fraction, some dimensional analysis is involved to get from $\pu{kg s-1}$ to $\pu{m3 s-1}.$
Any insight to this would be appreciated.

I guess you must have heard about concentration kg/m3 and equation for ideal gas density=f(M,p,T)


text:
C=O bond length comparison among cyclohex‐2‐en‐1‐one and cyclohex‐3‐en‐1‐one

I can see that in the first case a resonance would occur, and in the second case, there would be no resonance effect. I have read the fact that resonance reduces bond length due to partial double bond character.
According to that, shouldn't the $\ce{C=O}$ bond length in 1 be lesser than that in 2? The answer in my book is that the $\ce{C=O}$ bond in 1 is longer (reason is simply given as: "Resonance"). Could someone help me understand what's going on here?

Think of it this way: resonance shortens some bonds and elongates others, all that due to partial double bond character.

@IvanNeretin How to know which are shortened and which become longer?

Why, that's simple. What is shorter, a single or a double bond?

@IvanNeretin Ahh i think i got it. In case 2, it is a double bond, no partial character, thus shorter and in case 1, it is a partial double bond character only, so it will be longer relative to the 2nd case. Is my reasoning accurate?

Yes.$\mathstrut$

The case 1 has partial double bond character (due to resonance) while case 2 has total double bond character of the C-O bond. So, relative to case 2, the C-O bond length in case 1 is longer.


text:
Separation of Noble Metals for scrap recovery
I am researching scrap metal recovery as a side job and am hitting a wall with the noble metals. Gold, Silver, Platinum, and Palladium seem like they're all fairly easy to recover. I just can't find a process to separate the rest (Re, Ru, Os, Ir, Rh), not even a description of the industrial processes.
I suppose what I'm looking for is a full reactivity series for these metals, and any lead on a separation process.
Eventually I'd like to expand this to all transition metals (Scandium is more valuable than Platinum from what I've read) and the Lanthanides. So any info down that rabbit hole would be appreciated too.
I appreciate the time of whoever answers here. I can get my business off the ground with the knowledge I already have, but adding that extra recovery of trace noble metals would add an element (pun intended) of fun as well. I'd hate to see all that dust go to waste.

Where would you get scrap that contains e.g. Os?

As it stands, the question looks quite vague and broad (this doesn't mean it's useless). Probably you could add more details as to what kind of scrap you are dealing with. Obsolete consumer electronics, foundry waste a fallen Soyuz rocket are quite different beasts.

Good point. Mainly jewelry, electronics parts, catalytic converters, & spark plugs.


text:
Basicity comparison between pyridine amines and other compounds

In cases 1, 2, 3, the compounds will resonate, thus giving them more stability.
Firstly, I want to know whether nitrogen has a +M effect? (due to it having a lone pair of electrons?)
In case 1, resonance occurs and the electron density above the $\ce{NH2}$ group is delocalised, even in the cases 2, and 3, this is true. But in 1 the $\ce{N}$ is at the para-position, while in 2 the $\ce{N}$ is at the meta-position. How would that affect the order?
It would help if someone could hint at the factors coming into play here.
Case 4 is non-resonating, so it would be the most basic, right? How do I arrange the rest in order?

The pyridine nitrogen does not involve in resonance but is boosted by para activation by the amine group. So, 1 is more basic than 2 since amine boosts ortho and para positions more than meta. We can put 3 to the end of the list. You are right about 4 being the most basic because there are just two conjugation structures which imply the lone pair is available more easily than 1 or 2 or 3. Hence I would go with 4>1>2>3.

@SurajS Could you please show the 2 conjugation structures of 4? I thought that there would be no resonance at all in 4.