text:
Rearrangements of α-aminoketones in Clemmensen reduction

What is the possible mechanism for this reaction? Activated zinc in presence of concentrated hydrochloric acid is used for Clemmensen reduction, that is conversion of ketone group to methylene group. I'd expect the product to be 1-methyl-2-propylpyrrolidine:

However, ring expansion occurs leading to a piperidine derivative. Could anyone please explain or provide a mechanism for the conversion? Thanks!

Regarding this question, there are two things that should be stated beforehand.
Firstly, the mechanism of the Clemmensen reduction is not well-understood. There are two plausible pathways that have been proposed:1

so, the mechanism for this rearrangement is equally unclear.
Secondly, this compound does not actually undergo rearrangement in the Clemmensen reduction, according to studies by Leonard and Ruyle.2 Curiously, the authors could only isolate the alcohol product. No explanation was offered as to why the alkane was not obtained.

However, α-aminoketones are known to undergo rearrangement when the nitrogen and the carbonyl carbon are in the same ring. Two examples which Leonard and Ruyle studied are offered below:2,3

Their proposed mechanism occurs via migration of the C–N bond, following protonation of the carbonyl oxygen. The carbocation is captured by what appears to be hydride, although the existence of hydride is inconsistent with the proposed mechanisms we have seen above. It might just be their shorthand for sequential electron + proton transfer; I can't really tell.

The second half of the mechanism is not explained, but it is plausible to protonate the hydroxyl group and use the nitrogen lone pair to expel water.
The authors posit that the tendency to undergo rearrangement here is related to a "geometrical directing effect": in the case where the nitrogen and the carbonyl carbon are in the same ring, the nitrogen lone pair has a larger opportunity to approach the carbonyl carbon.2 In modern-day terms, this might be considered to be a stereoelectronic effect.
How can this be reconciled with the proposed mechanisms in Kürti-Czakó? Considering that the mechanism of the reduction itself is unclear, I do not want to make any assertions about what the exact mechanism of this rearrangement is. However, I think it would be plausible that C–N bond migration can occur at some stage in the above mechanism - especially in the intermediates where the carbon bears a positive charge.

References

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; p 92.
Leonard, N. J.; Ruyle, W. V. Rearrangement of α-Aminoketones During Clemmensen Reduction. II. Contraction of a Six-membered Ring in the Monocyclic Series. J. Am. Chem. Soc. 1949, 71 (9), 3094–3098. DOI: 10.1021/ja01177a043.
Leonard, N. J.; Wildman, W. C. Rearrangement of α-Aminoketones During Clemmensen Reduction. I. Bicyclic Compounds Containing a Bridge-head Nitrogen. J. Am. Chem. Soc. 1949, 71 (9), 3089–3093. DOI: 10.1021/ja01177a042.

Thanks! It's clear now. Just another thing, I also remember reading that 5,5-dimethylcyclohexane 1,3-dione on treatment with Zn-Hg in presence of HCl (Clemmensen reduction) gives 2,4,4-trimethylcyclopentanone. This is clearly an example of ring contraction. How does it occur? I'd like to know a mechanism for this conversion too. Also, as six membered rings are more stable than five membered rings, I'd predict that the reaction is not feasible, but it is. How?

@user28968, for your first question I'd suggest you ask a new question. Regarding the 6 vs 5 membered ring, you're making an implicit assumption that the reaction is thermodynamically controlled at some stage. I'm not sure if that is necessarily the case here. It is difficult to analyse it in any meaningful way, though, since the mechanism is not confirmed to begin with.

Orthocresol is correct. The mechanism of the Clemmensen Reduction is enigmatic. Leonard and co-authors provided a 1949-type mechanism where hydride or zinc-nascent hydrogens was the reducing agents of the carbocation while not addressing the issue that the amine is totally protonated in conc. HCl (pKa H3O+ ~-1.7; pKa "NH4+" 9.3]. The nitrogen electron pair is not available as described by Leonard. Orthocresol has provided both prevailing options for the Clemmensen Reduction. I'll focus on the carbenoid route. Structures 2 are three ways of thinking

about a carbene for mechanistic purposes: metal bound 2a, electron pair 2b and vacant and filled orbitals 2c. Bond migration in 2c affords 3a, which is equivalent to a nitrogen protonated enamine 3b. However, enamines1,2  such as 1-pyrrolidino cyclohexene undergo facile hydrolysis in aqueous acid. This means that C-protonation occurs to give an immonium salt and, if N-protonation occurs, it is reversible. Given an N-protonated enamine here, isomerization to the immonium salt 4 is expected followed by a 2-electron reduction to afford the tertiary amine 5.

In Memoriam: Professor Gilbert Stork (1921-2017)
Stork, G., et al., J. Am. Chem. Soc., 1963, 85, 207. http://pubs.acs.org/doi/abs/10.1021/ja00885a021


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Structural entity for the tattoo resembling spine surgery involving 6AL-4V ELI alloy
I have very little background in chemistry, so forgive me if I sound ridiculous and uneducated.
I want to have a tattoo designed in honor if my spine surgery that I had. I think that molecular structures (I think that's what its called -- the hexagonal structure stuff?) look pretty cool and I'd like to look into getting something like that.
Anyways, I dug some stuff up and came up with the alloy that orthopedic implants are usually made of, 6AL-4V ELI. Now, this is where I get stupid. Do alloys have molecular compositions at all? And if so, what would this one look like?
If I just sound dumb right now and there is none, is there an alternative structure that could be used?
Again, I'm sorry if this is a silly question and is just wasting time, but this surgery is something I'm really passionate about and I would really like to have something visual to represent it by.

No, alloys don't have molecular structure, only domains at best. People obsessed with tattoos usually prefer certain organic molecules instead. Also, congrats on a successful surgery!

a tattoo designed in honor if my spine surgery <--- That threw me off. I'm not saying it's a "stupid" question...but I'm not quite sure I understand it...

No prob; do you want something compact, or elongated, say, hexagon chain along the spine?

@andselisk  "People obsessed with tattoos usually prefer certain organic molecules instead." I wasn't aware that was a thing....besides, I'm marveling at how you managed to dig that up from Chem.SE :D

@paracetamol People want to imprint the stuff that makes them happy, or remind of something important. So molecules of drugs and hormones are the obvious choice. There was a related CSE question  on a tattoo resembling a steel alloy used for crossfit weights or something; a woman won a competition and decided to make a tattoo, so it's not all that uncommon:)

@andselisk Yup, I remember the steel-alloy-weight thingy. I also happen to remember one where the OP wanted to tattoo a molecule of LSD on her arm (think you might've seen that) ;)

@Julia 6AL-4V ELI alloy is mostly titanium, so maybe you can use it's electron shell configuration as a basis. Just an idea.

@andselisk Hang on...is the steel-alloy-weight thingy question still around? If so, I think that would make this post a duplicate...

@paracetamol I cannot find it anymore, probably deleted by OP. Still, this is a different alloy and case, so it would be related IMO.

@Julia In the meantime you can also find a decent tattoo master and ask this person for a concept art (inspiration) you explain your situation. If something will look off to you, you can always edit your question here and post a piece of graphics for the discussion.

The idea of Andselisk seems great!

The alloy is a mixture of hcp and bcc titanium grains, stabilized by the small amounts of Al and V. I'm sure a good artist could figure something out with a few unit cells, but I'm not an artist.

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In rovibrational spectra, when exactly is the Q branch forbidden?
I'm aware that the Q branch ($\Delta j = 0$) is usually not observed in rotational-vibrational spectra of diatomic molecules in the gas state, however, I've heard different things about when exactly this transition is forbidden.
I've sometimes been told that the transition is not allowed in general, for any diatomic molecule, but I've sometimes heard that this is only true if the electronic state has a $\Lambda \neq0$, so which is the correct selection rule here?
I'd love to have this clarified for me. If it is possible I would also like to know when is a Q branch observed in vibronic spectra of heteronuclear and homonuclear diatomics.

You are correct in that the electronic angular momentum $\Lambda \ne 0$ for a Q branch to be observed in a diatomic molecule, i.e. the molecule is not in a $\Sigma$ state. The rule is then $\Delta J =0; \Delta v = \pm1$. (Homo-nuclear diatomic molecules do not have an ir spectrum as $d\mu/dx = 0$, as the dipole is zero)

ahhh, I know homo nuclear diatomics don't have an IR spectrum, there was a typo in the question, thanks. Are you sure about the $ \Lambda$ selection rule? is a Q branch always permitted if $\Lambda \neq 0$? Is there some easy derivation of this or somewhere where I could find a derivation? I can't see where this comes from from a simple transition moment integral.

A photon carries one unit of angular momentum which is conserved.  When the photon is destroyed on absorption, this ang mom must be transferred to the molecule in one way or another, in your case $\Lambda_i +1=\Lambda_f $ where i and f are initial and final states. This is the basis of the transition moment integral and so selection rules.

I understand that argument, however, $L$ is not really a good quantum number in diatomic molecules and $\Lambda$ is not the angular momentum, but the projection of the angular momentum in the direction of the internuclear axis, so I don't think that the fact that the total anguar momentum is conserved helps a lot.

In Hydrogen, for instance, although $\Delta l$ is constrained to be $\pm 1$, $\Delta m_l $, the z projection, can be zero, and this doesn't break angular momentum conservation.

To determine the intensity of a rotation-vibration transition, you need to consider the transition momentum integral $\int \psi_f \hat{\mu} \psi_i d\tau$, where $\psi_f$ is the final state, $\psi_i$ is the initial state, $\hat{\mu}$ is the transition dipole moment operator, and $d\tau$ indicates integration over all space. Assuming separation of vibrational and rotational motion, this factors into the product of two terms: one that specifies the intensity of the vibrational transition and another that specifies the intensity of the rotational transition. The vibrational integral leads to a selection rule $\Delta v = \pm 1$ (in the harmonic approximation) and the rotational integral leads to a selection rule $\Delta J = \pm 1$. So no Q branch occurs. However, if there is some additional source of angular momentum (e.g., from vibrations or electronic motion) then it is possible to have a Q branch because now that extra motion can contribute to the total angular momentum in a way that allows the Q branch to occur.
So here is the rule that you ask for: a Q branch will not occur in transitions between $\Sigma$ electronic states of a diatomic molecule, but can occur in transitions between $\Sigma$ and $\Pi$ states. Since vibration-rotation transitions by definition occur within a single electronic state, Q branches do not occur in vibration-rotation spectra of diatomic molecules in a $\Sigma$ state. In a polyatomic molecule, however, vibrational modes can carry angular momentum so even within a single electronic state that carries no electronic angular momentum it may be possible to observe Q branches in a vibration-rotation spectrum so long as the vibration being excited is one which carries nuclear orbital angular momentum (e.g., from a vibrationless ground state to a vibrationally excited bending mode).

I think this is a solid answer. It might be nice if you can expand a bit on the vibrational and rotational integrals in the first paragraph. While I know Q branches can exist, I'd love to know if there are example molecules / spectra in which they have been observed.

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Formating chemical equations for proteins binding in multiple configurations
I am working on problems involving protein-protein binding, particularly ones in which two proteins may bind in two or more configurations, and where some of the resultant structures may also bind with other proteins subsequently, and I am wondering how best to format chemical equations.
I am used to writing chemical equations on a single line like:
$$\ce{[A] + [B] <=>[k_\mathrm{a}][k_\mathrm{d}] [AB]} \tag{1}$$
but formating a reaction where multiple dimers can arise from the same reaction on a single line looks confusing and ugly:
$$\ce{[AB]_1 <=>[k_\mathrm{d1}][k_\mathrm{a1}] [A] + [B] <=>[k_\mathrm{a2}][k_\mathrm{d2}] [AB]_2} \tag{2}$$
Worse than that, its not clear how one would write an equation where a third configuration could emerge from the reaction.
What is the best way to format equations like this?

FYI: I switched the constants' places in (2), assuming $a$ and $d$ stand for association and dissociation, respectively: $$\ce{[AB]1 <=>[\color{red}{k\mathrm{d1}}][\color{red}{k_\mathrm{a1}}] [A] + ...}$$ If this is wrong, feel free to adjust accordingly or roll back.

Use vertical space for 3. Fundamentally, there is a limitation in how many you can represent. Probably will need to do something more general:
$$\ce{[A] + [B] <=>[k_{\mathrm{a}i}][k_{\mathrm{d}i}] [AB]_{i}}$$ for $1 \leq i \leq n$

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Molar Conductivities of H+ and OH- V/S other Ionic Electrolytes
The following table gives the Ionic Molar Conductivities of a few electrolytes in water. What are the reasons for the unusually high values for both H$^+$ and OH$^-$ even though both undergo solvation in water? Why do they retain their mobility in water even after solvation?
\begin{array}{lr}
\text{Electrolyte} & \text{Ionic Molar Conductivity (mS m$^2$ mol$^{-1})$} \\
\hline
\text{H$^+$} & 34.96 \\
\text{Li$^+$} & 3.87\\
\text{Mg$^{+2}$} & 10.60 \\
\text{[N(CH$_3$)$_4$]$^+$} & 4.49 \\
\end{array}
\begin{array}{lr}
\text{Electrolyte}&\text{ Ionic Molar Conductivity (mS m$^2$ mol$^{-1})$} \\
\hline
\text{OH$^-$} & 19.91\\
\text{F$^-$} & 5.54\\
\text{CO$_3$$^{2-}$} & 13.86\\
\text{HCO$_2$$^-$} & 5.46\\
\end{array}

Your question looks better then old one, but there are answers already.


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Hoffman's Isocyanide test: How does addition of concentrated HCl degrade the isocyanide formed?
This question is regarding a laboratory procedure followed for Hoffman's Isocyanide test (also known as the "Carbylamine test")
I've been taught that the isocyanide formed in this reaction is highly toxic and must therefore be destroyed in situ.
I'm told that one method of carrying this out is to cool the test tube and carefully add an excess of concentrated $HCl$.
Though it wasn't mentioned, I'd like to know how exactly the addition of "excess concentrated $HCl$" results in the degradation of the isocyanide formed.
I've provided the reaction-mechanism for a primary amine subjected to the Hoffman Isocyanide test below;

What is a ‘carbyl amine’? What is its systematic name? I have never heard the term before.

@Jan [ https://en.wikipedia.org/wiki/Carbylamine_reaction ], as for "carbyl amine"...it needs editing  O:)

I wonder why that reaction has that weird name.

@Jan I was going to ask you that over at chat sometime...

A carbylamine is an isocyanide @Jan

I edited the question. Please look into it.

What exactly happens when we treat  iso cyanide with excess conc HCl ?

@Sriharsha Guess, you'll just have to wait for an organic chemist who finds the question interesting enough to answer. User Jan is one of the more esteemed organic chemists (a quick look at his profile will give you an idea about his credentials) on this site, apparently he didn't give this question much of a read (busy man, I guess?). Anyways, I'm editing your question (a bit) and I've up-voted...hopefully that'll draw some attention to it. Good luck!

Isocyanide is not as toxic as its isomer cyanide and can be tolerated. The purpose of its removal is due to its disagreeable odor which is not at all tolerable.

While some isocyanides (e.g., cyclohexyl isocyanide) are toxic, others
  "exhibit no appreciable toxicity for mammals". Referring to ethyl
  isocyanide, toxicological studies in the 1960s at Bayer showed that
  "oral and subcutaneous doses of 500-5000 mg/kg can be tolerated by
  mice".(Wikipedia)

Isocyanide is removed on the basis of its property of acidic hydrolysis.

Isocyanides are stable to strong base (they are often made under
  strongly basic conditions), but they are sensitive to acid. In the
  presence of aqueous acid, isocyanides hydrolyse to the corresponding
  formamides:
$$\ce{RNC + H2O → RN(H)C(O)H}$$
This reaction is used to destroy odorous isocyanide mixtures. Some
  isocyanides can polymerize in the presence of Lewis and Bronsted
  acids.

But the story does not here. The formamides are still pretty unstable and further degrades to form amines and formic acid in the presence of acidic medium.(Here)

Isocyanides are hydrolyzed by dilute mineral acids to give primary
  amine and formic acid (Isocyanides are not hydrolyzed by bases).
  Isocyanides are hydrolysed only by acids and not alkalies because the
  negative charge present on carbon atom in isocyanides initially
  attracts electrophiles ($\ce{H+}$) but repels nucleophiles ($\ce{OH–}$). When a
  proton gets attracted to negatively charged carbon atom, the tendency
  of this carbon atom to attract a nucleophile increases due to presence
  of positive charge on the N atom and hydrolysis is facilitated as
  shown below:
Mechanism:

Here is a research paper for a reference.


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Is it true that Enthalpy can not be defined?
My teacher told us that only the ""Change in Enthalpy "" is definable.
 Whereever I looked for it, I got only an "equivalent " type definition like:  It's a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume. But change in enthalpy also holds the same kind of definition.
Thanks

There is a reason why only change in enthalpy can be measured (experimentally).  See, you can conduct a chemical experiment and measure the heat produced / lost in the reaction. Then you do some algebraic manipulation and calculate the change of enthalpy. But see, for a constant change in enthalpy c, final state and initial state enthalpy can be anything as long as their difference is c. So, you can't really know absolute enthalpy but the chane of it.

Actually, enthalpy is definable. What your teacher really meant was that you can't know absolute enthalpy.

Can energy be defined? How about change in energy? ;)

@Paracetamol Energy is defined. Actually, every concept in science is defined whether you can measure it or not.

@Mockingbird That comment was supposed to be "food for thought" <---- Intended for OP to ponder over it :)

While it may be not immediately apparent, the problem boils down to this^


text:
Gaseous nitration of glycerin
Could nitric and nitrous oxides react with glycerin to produce nitroglycerin? What conditions would I need in order to make that work?


text:
A galvanic cell with a salt bridge for demonstration purposes (high school)
I want to design a meaningful galvanic cell experiment for my high school science class. I am tempted to do an experiment without a salt bridge. Why? Because the most common batteries (i.e. lead batteries) only use one electrolyte ($\ce{H2SO4}$) and no salt bridge.
I am not aware of technology or processes that would justify the salt bridge. Anyone who can enlighten me? If so -- very grateful :)


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Can water be a Lewis acid?
I asked this question to my teacher a few days ago. In a reaction such as
$$\ce{NH3 + H2O -> NH4+ + OH-},$$
$\ce{NH3}$ is a Lewis base, but is $\ce{H2O}$ a Lewis acid?
He said that it isn't. Now I fail to see the reason why. Can someone explain?

I would think that H2O can act as a Lewis acid. The bound hydrogens are electron-deficient and can act as lone pair acceptors.

Perhaps, your teacher thought that it could not be a Lewis acid because all the hydrogens had complete octets.

So the bound hydrogen ions that get separated then bond with the NH3 ? Does that make the hydrogen ion the lewis acid? Since it's the one accepting the electron pair

Sorry, I meant complete duets.

Not true... The hydrogen ion may not even form in the first place. A hydrogen bond can be formed between the ammonia and the water molecule. Then there is a transition state, with the hydrogen having partial covalent bonds with the nitrogen atom and the oxygen atom. Finally, the bond between N and H forms fully and the bond between O and H breaks fully.

I see. I remember him saying that it's because in an H2O molecule, all the atoms have full octets and duets, but wouldn't that same logic apply to say, an HCl molecule as well? He says those count as lewis acids

By agreeing that HCl is a Lewis acid and that H2O isn't one, your teacher is contradicting himself...

https://chemistry.stackexchange.com/q/141282/119245

If you agree that ammonia is a Lewis base (electron pair donor) and that this is an Lewis acid-base reaction, (donor-acceptor interaction) then water is a Lewis acid (electron pair acceptor), by definition and the fact that there is a reaction. Now, identifying the mechanism of acidity or the acidic site is more complicated, but you can't argue that it is acidic, by definition.

I asked my teacher again and he said that it was indeed acidic. Thanks!

Strictly speaking, a Lewis acid is something that forms a bond by accepting an electron pair form another molecule (Lewis base).  Water as such does not do that, rather it is a hydrogen ion from the water that does so.  When ammonia acts as a proton acceptor in water, the Lewis theory calls it a displacement reaction where the acid, $\ce{H+}$, is initially combined with one base ($\ce{OH-}$) and ends up combined with another base ($\ce{NH3}$).

Yes i thought so too, but the issue is that same logic can be applied to all the other acids like HCL, HF ect, and my teacher called those lewis acids. Though I did ask him the other day and he said that water was a lewis acid in that reaction all technical details aside ( or at least it'll be for our book)

Not even HCl.  No matter how strong, protic acids are sources of the actual acid, hydrogen ion, in the Lewis theory.

Yup he said that as well. It's just weird that he didnt say it from the start tbh

I see. An H+ ion gets dissociated from the H2O molecule, leaving two electrons behind, and gets attached to the NH3 molecule. The H2O molecule isn't accepting an electron pair, but rather it is the H+ ion from the H2O molecule that does so.

I have finally found a satisfactory answer to your question after much research. Jensen (1978) provides an excellent review of the Lewis acid-base theory and p. 4 of the article provides the answer that you seek$^1$. I shall quote from that page the relevant insights:

So rapid and striking were many of these neutralizations that Lewis went on to propose that that criterion 1 (i.e., rapid kinetics) was the salient feature of acid-base behavior, suggesting further that a fundamental subdivision of acids and bases be made on this basis...

Lewis classified those acids and bases which underwent acid-base reactions which showed "essentially zero activation energy" as primary, while those which had measurable activation energies were termed secondary. He further broke down this secondary class into two types (ref 1, p. 4):

The first of these involved species, such as $\ce {CO2}$, in which the slow kinetic behavior was apparently due to the necessity of the species undergoing some sort of internal activation before its primary acid or base properties became apparent.
The second class involved those species in which the finite activation energy was
due to the breaking of one or more auxiliary bonds upon neutralization, causing the initial $\ce {AB}$ complex to dissociate into several smaller fragments. Hence, Bronsted acids like $\ce {HCl}$ and $\ce {HNO3}$ were still acids, though now of the secondary variety, and their neutralizations could be thought of as initially resulting in an unstable hydrogen-bridged adduct which then underwent further decomposition.

To clarify, the Lewis acid-base reaction defined by Lewis is as such:
$\ce {A + :B -> AB}$
Back to your question... Essentially, $\ce {H2O}$ can be seen as the second class of secondary acids proposed by Lewis. During a reaction with Lewis base $\ce {:B}$, there is essentially some sort of complex formed, that looks like this $\ce {[B -- H -- OH]}$. The dotted lines indicate partial covalent bonds. This complex can be seen as sort of a "transition state". However, note that this was not in the original formulation of Lewis. The image below shows the reaction between pyridine and $\ce {HCl}$ viewed from Lewis' perspective. Lewis called this complex an "unstable adduct".

Consolidation
Based on the above, we can say that acids of the type $\ce {HA}$ (where $\ce {A}$ is an electronegative atom or group of atoms) are secondary Lewis acids, which participate in acid-base reactions with simultaneous bond breaking of auxiliary bonds. This is because the idea of complexation to form an adduct is still present.
Reference

Jensen, W. B. The Lewis acid-base definitions: a status report. Chem. Rev., 1978, 78(1), 1-22. doi:10.1021/cr60311a002

My friend, here is how i understand it:
H2O is a neutral medium, does not like to be acid or base, but if we mix acid with it, then the water will try to fight back by turning to Lewis base. Same thing goes with adding base to H2O, H2o will fight back by turning acid to naturalize the solution.
So yes, water can be both Lewis acid or base.
I hope that makes sense

It would be better if you used the definitions for Lewis acidity in your answer...

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text:
Method for measuring volume of gas
I would like to setup something up my office to accurately measure the amount of Carbon Dioxide (CO2) being released from beans.
I looked at the site http://examples.yourdictionary.com/examples-of-ways-to-measure-volume.html which suggests several methods, the first of which, fluid displacement, I was going to employ. I know that CO2 is slightly soluble in water so I was going to use a saline solution/brine to reduce solubility. Can anyone help me here in explaining additional things that I should be doing to reduce errors in measurement with this fluid displacement method, other than of course reducing the possibility of air leaks and CO2 solubility? For example I have been reading about the risk of water evaporation in the measuring cylinder so I wonder how this can be minimised? I've also heard about the vander waals methodology for accuracy in calculation.
What other methods should I be considering? I'm not asking for better methods as I realise that's subjective, but other methods that you believe I should be investigating as possible alternatives. I have looked at a gas syringe and a balloon as alternatives.

Hint: We closed your question as being too broad. It really doesn’t help to make it even broader in an edit …

@Jan I thought it sounded more refined, clearer, I still do, I'll put that down to inexperience :-(

Unless you have a way to separate CO2 from the rest of the gases, you won't be able to measure it properly. Which isn't easy. You could put a substance that absorbs carbon dioxide inside the container for the beans. However, the substance needs to be able to indicate that it has reached capacity. And also, the volume of CO2 that it can absorb must be known.
You could also set up a respirometer. You could make some assumptions about how the oxygen is consumed. Nevertheless, it should give you a good estimate. There are several videos and tutorials explaining them. And if you still have questions, you can ask Bio Stack Exchange.
The comment about the van-der-Waals equations: I think that you were probably thinking on using the ideal gas law to estimate the number of gas molecules in a container.
$$pV=nRT$$
According to the equation above, you can measure the number of particles if you know the pressure and the volume. This is true for ideal gases. For a gas molecule to behave like an ideal gas particle, it would have to have no volume and have no interactions with other gas molecules. Real gases, however, do not behave like an ideal gas.  Therefore, a better model to describe real gases is needed. The van-der-Waals equation is one of these attempts.

Thank you. I think I'll use total gas displacement as a proxy for CO2 because it's mainly CO2 that's produced. Thank you for the tip about the respirometer!


text:
Bridgehead carbocation: Violation of Bredt's Rule

This is the solution of a university's problem set (nucleophilic substitutions)
It says that the given compound undergoes SN1 type substitution.
What confuses me is that even though iodide is a good leaving group, bridgehead carbocations (sp² bridgeheads) are extremely unstable. According to me, this reaction shouldn't occur.
Could anyone explain what's going on here? If the explanation is just thermodynamic (hopefully not), could some data be provided? Thanks!

You can check a literature search, but, as far as I know, this does not happen.

The reaction in question is known and has been studied fairly extensively, both experimentally and computationally, and is accepted to go via the bridged carbocation that you propose.
To give some data, one experiment  measured the relative reaction rate of a series of such structures towards solvolysis in 80% aqueous ethanol (basically the reaction you describe above):

The intermediate carbocation is known as a non-classical carbocation- a carbocation that does not conform to the standard organic chemistry definition of a planar sp2 intermediate (as we often see with SN1 reactions).
In the case of the norbornyl cation, the carbocation can be observed spectroscopically by treating the parent halide with a superacid such as $\ce{SbF5.SO2ClF}$ at very low temperatures:

One issue with non-classical carbocations is their tendency to rearrange, such that you wouldn't get a single product as predicted, but instead end up with a mixture of products resulting from other classical carbocations formed through hydride shifts on the non-classical carbocation (the exact mix of products would depend upon the conditions of the reaction).
References
All figures taken from Carey, F. Advanced Organic Chemistry - Part A


text:
Which cup of tea cools down more quickly?
There are 2 cups (both cups are identical) of tea which are served at the same temperature. Assume that the walls of the cup to be non-heat conducting. Which cup of tea cools down faster, a full-filled glass or a half-filled glass?
According to what I think, The walls of the cup don't make a difference and neither do the top and bottom surfaces. The only determining factor would be the volume of the tea. So what I came up with is that, somehow the cooling process by convection will be affected by the volume of the tea in the cup but I'm not sure about my above assumption and even if it's true how will the volume of the tea affect the cooling process by convection within the volume of the liquid?
Taking the initial temperature higher than the room temperature.

Let the "same temperature" be 60 degree celsius and room temperature 25 degree celsius.

Useful read: [ https://en.wikipedia.org/wiki/Newton%27s_law_of_cooling ]

Anyway, what do you think determines the heat exchange rate in this case, and then what determines the resulting temperature change?

@Loong I think that there will be heat exchange between the open top surface and the hot tea due to which convection will occur within the liquid volume

What shape does the cup have?

@aventurin Cylindrical cup


text:
What is the physical state of a single atom?
Can a single atom on its own be either a solid, liquid, or a gas? Or is it none of them?

Solids and liquids are bulk systems of interacting atoms/molecules, so no, a single atom cannot be a solid or a liquid. The (largely hypothetical) situation of an isolated atom in a vacuum would probably be considered a gas.

@airhuff There isn’t much more to add to an answer, is there?

This answer has been written so it hopefully can be understood by people that do not have a degree. If something is not correct, let me know, but be aware that throwing around fancy words will not help anyone understanding this any better. Consider the other posts here concerning liquids and solids, I don't want to be redundant and repeat what has been said already.
The answer is a bit complicated. One would intuitively say no, but that doesn't tell the whole story.
Take the interstellar medium. We have about 1 atom per cm³ in there and it is called a "gas". Now that's what I would call an isolated atom in the gas phase.
An atom is an object that belongs into the realms of quantum mechanics. A solid, liquid or gas is something that belongs into our classical world. If you zoom in close enough, that means if we look at a problem microscopically, words like "solid" or "liquid" no longer make sense. If you now begin to zoom out, the quantum effects vanish. This is sometimes called quantum decoherence and we enter the world of classical physics where solids and and liquids exist again.
So I would argue it all depends on how closely you look at the problem. Isolated atoms in a large volume? A gas if you ask me. But if you have to zoom in to talk about an individual atom within a larger compound consisting of many more atoms, let's say one atom within a piece of iron, you no longer can say that this single atom is "solid".
Think of it like this:
Can a single person have a political system?
Yes, why not. I can proclaim my own nation and be the only person in it. But mostly, a nation is something that many people form together. If you look at one individual person, you are in the realm of psychology (or biology). If you look at an entire state, you are in the realm of (for example) politics. If you zoom in at one individual person, you can no longer say that person has a certain political system. But if you zoom out far enough, you suddenly get a political system. Zooming out of course is a bit more abstract here, but, as I said, this is the realm of psychology/politics ...

Both the liquid and solid phases are, on a molecular level, defined by some kind of regular arrangement. Solid phases typically have short-distance and long-distance regular arrangements (think the periodicity of a crystal), while liquid phases only have short-distance arrangements. This can be modelled in various computer simulations and I myself have done so in a physics practical course.
Gases are then defined as phases without any type of arrangement; neither short- nor long-distance.
A single atom of course has no arrangements of other atoms around it, so it cannot be considered a liquid or a solid. One would, most likely, consider it a gas — but remember that ‘gas’ is a macroscopic term and one may equally consider a single atom as ‘just a single atom’.

If the single atom isn't expect to undergo any gaseous interactions with other atoms or molecules, then I'd agree that it is just a single atom.

Do you have any references regarding the definition of solids vs. liquids vs. gases as the difference in the distance of regular arrangement? Certainly crystalline solids have long distance regular arrangements, but do amorphous solids (e.g. flash-frozen vitreous ice)? If they don't, does that mean they're really liquids?

@R.M. I don’t, just the verbal memory of being taught so in the physics lab course. In any case, all this phys chem stuff is not my strong point anyway ~

Solid, liquids, gases, plasmas, stars, neutron stars, black holes etc. are attempted descriptions of grouped or agglomerated matter. These are all descriptions of the interactions of matter particles not of the particles themselves. This means an individual matter particle is simply a relative energy minimum in a surrounding field be it subnuclear, nuclear, electric, gravitational, photon flux, or some unnamed field.
The phase of a particle depends on the field that it is immersed in. An isolated atom does not have a name for that field. Call it a Quantum Mechanical entity in a photon-gravitational flux.


text:
Is hydrazine a better nucleophile than ammonia?
Which is a better nucleophile, $\ce{NH3}$ or $\ce{NH2NH2}$? For the sake of limiting scope, I want an answer for behavior in a protic solvent though non-protic solvent answers are welcomed.
My logic is that since $\ce{NH2NH2}$  has 2 lone pairs it will be a better nucleophile. I would like to know if this conclusion and reasoning is correct, or if I'm correct for the wrong reason why, or if I'm wrong, why is ammonia a better nucleophile?

Yes, hydrazine is a better nucleophile than ammonia. This is the case in both water (a protic solvent) as well as acetonitrile (an aprotic solvent). A quantitative approach may be found in the Mayr nucleophilicity parameters (Ctrl-F "ammonia" and "hydrazine"):
$$\begin{array}{ccc} \hline
\text{Nucleophile} & N\text{ parameter in }\ce{H2O} & N\text{ parameter in }\ce{MeCN} \\ \hline
\ce{N2H4} & 13.46 & 16.45 \\
\ce{NH3}  & 9.48 & 11.39 \\ \hline
\end{array}$$
The larger, the more nucleophilic. It also makes sense that the protic solvent water reduces the nucleophilicity of both species.
For more details I suggest reading the articles cited in the database entries. The most relevant one is: J. Org. Chem. 2012, 77 (18), 8142–8155. Inside there is a good discussion of the methodology.
As for why hydrazine is a better nucleophile, the jury is still out. In general there is a phenomenon whereby nucleophiles with lone pairs on two adjacent atoms are more nucleophilic than expected, which is called the α-effect. [Note that I deliberately avoid saying that hydrazine is more nucleophilic "because of the α-effect", because that merely begs the question. The α-effect is not (yet) a theoretical explanation, but rather an empirical observation.]
I don't have the interest to summarise reviews on the α-effect (at this point in time, at least) so I will quote Mayr from the article above:

The α-effect, which has been investigated for various reactions of nucleophiles including acylations, Michael additions, SN2 reactions, and nucleophilic aromatic substitutions has been the topic of several reviews, but its origin and extent are still discussed controversially. [...] Several theories on the origin of the α-effect have emerged, which include the destabilization of the ground state by electron repulsions, the stabilization of the transition state, thermodynamic stabilization of the products, and solvent effects.

See also: Is there a general consensus on the causes of the alpha-effect? Unfortunately, that question does not have a proper answer, and I believe the direct answer to that question is "no". As Jan's comment indicates, too, the explanation must go beyond "hydrazine has two lone pairs but ammonia has one", because that alone doesn't actually explain anything.

Yes hydrazine is a better nucleophile than ammonia (ignoring solvent effects) and this is called the "alpha effect" which states that the nucleophilicity of an atom increases if it has an adjacent atom with a lone pair of electrons.


text:
Are there molecules that take the shape of every platonic solid?
Methane is the simplest molecule I know of that takes the shape of a platonic solid — in this case, the tetrahedron. Wikipedia has a page on platonic hydrocarbons that includes molecules with tetrahedron, cube, and dodecahedron shapes. There are lots of molecules that have octahedral geometry (e.g. sulfur hexafluoride).
Are there any molecules that form an icosahedron? The closest I've found is $\ce{C_60}$, buckminsterfullerene, which takes the form of an Archimedean solid, the truncated icosahedron.
To be explicit: I am interested in molecules with atoms positioned at the vertices of platonic solids; whether or not there are direct bonds between those atoms is not relevant.

Related: Which platonic solids have been built as carbon skeletons?

A former co-worker investigated interesting molecules with an icosahedral sub-structure: https://www.researchgate.net/publication/229985915_Twelve_One-Electron_Ligands_Coordinating_one_Metal_Center_Structure_and_Bonding_of_MoZnCH39ZnCp3

Well, non-carbon icosahedra are not too hard to find. $\ce{[B12H12]^2-}$ should do the job.

Also dodecaborate has an icosahedron as a sub-structure
https://en.m.wikipedia.org/wiki/Caesium_dodecaborate

Polyoxometalates also come in many forms and shapes; there is even a book titled "Polyoxometalates: From Platonic Solids to Anti-Retroviral Activity" by Pope P. T. and Müller A. (ISBN 978-94-011-0920-8)

I am a little bit confused by your question. Are you looking for molecules that have a central atom and ligands arranged around it in the manner of platonic solids (e.g. in methane or sulphur hecafluoride), meaning that you need to connect the outermost atoms to arrive at the platonic solid shape (there are no bonds there), or are you looking for compounds which have bonds in the shape of the platonic solids?

@Jan That distinction hadn't occurred to me. I'm looking for molecules with atoms at the vertices of a platonic solid, preferably the outer-most atoms.

Yes, for each of the 5 platonic solids, there is at least one molecule that takes its shape.
There exist tetrahedron-shaped molecules. Some instances are $\text{CH}_4$ or $\text{CCl}_4$.
There exist octahedron-shaped molecules. Some instances are $\text{SF}_6$ or $\text{Mo(CO)}_6$.
There exist hexahedron-shaped molecules. An instance would be $\text{C}_8\text{H}_8$ AKA "cubane". Perhaps simple cubic lattices (like NaCl) could be considered "hexahedral molecules" as well?
There exist icosahedron-shaped molecules. An instance is elemental boron, $\text{B}_{12}$.
There exist dodecahedron-shaped molecules. An instance is dodecahedrane ($\text{C}_{20}\text{H}_{20}$).

Wikipedia says that the isolated icosahedra in boron are not stable, they are only stable when part of a large covalent network so I'm not sure if this counts.

There is icosharderal $\ce{B_{12}H_{12}^{2-}}$ though.  Also octahedral $\ce{B_{6}H_{6}^{2-}}$.


text:
Can chloride attack the carbonyl bond in formaldehyde in the reaction with hydrochloric acid?
I already know that a cyclic trimer and polymer of formaldehyde is formed at lower temperatures, but would chloromethanol and dichloromethane be possible at higher temperatures?

I think the question is OK, even if what OP suggests probably doesn't work.

@Mithoron title suggests there is a specific reaction but body does not mention which.

@KennyLau Reaction leading to chloromethanol and dichloromethane...

@Mithoron Question ask for mechanism in the title but whether it is possible in the body.

We can both immediately agree on the first step of the reaction between formaldehyde and hydrochloric acid which is the protonation of the carbonyl oxygen. The mechanism of formaldehyde oligo- or polymerisation then suggests the attack of the oxygen atom of another formaldehyde moiety to form a hemiacetal. This would be a nucleophilic addition onto a π system.
One possible course this species can now take is the back reaction to reform formaldehyde. Another one is a subsequent attack of the newly-formed hydroxy moiety of the hemiacetal function onto another carbonyl group to form a full acetal. The latter reaction creates a product which is inert enough to not undergo back reactions.
You are indeed correct that chloride, which is also a nucleophile, can equally attack the π $\ce{C=O}$ bond of formaldehyde. And this attack would indeed result in chloromethanol. However, this reaction is also reversible as is the attack of oxygen. Furthermore, this molecule has no real way of stabilisation: chloride will always be a good leaving group while a formal alcoholate is not. Thus, while the attack of chloride may occur it is not productive and it is rapidly reversed.
Therefore, we do not see any chlorinated product — which would decompose on isolation at the latest anyway.


text:
Net ionic equation for a synthesis reaction?
In the following reaction
$$\ce{S (s) + O2 (g) -> SO2 (g)}$$
there are no spectator ions because both $\ce{S}$ and $\ce{O2}$ take part in the reaction. Does that mean the net ionic equation is the same as the molecular equation?
I think there should not be a net ionic equation at all, because there are no ions in the reactants.

You are right, you can't write ionic equation for that reaction. Quick search showed me that even the mechanism doesn't really involve any kind of ions, which, I'd say, is completely logical, considering reaction conditions, which favor homolytic, not heterolytic  bond cleavage.
You can only write ionic equations for reactions where ions are in charge, and they are preferably formed in polar medium.
P. S. The mechanism of S+O2 reaction isn't researched much, so here is the study of H2S+O2 reaction .


text:
Calclulation of MP2 first order wave function
My question is, how can I calculate the MP2 first order wavefunction, and thus construct the MP2 density matrix? Now here is what I understand so far:
The density matrix is given as:
$$P_{ij}=\sum_{a}C_{ia}(C_{ja})^*$$
Here $C$ is the wave function coefficient matrix. To get the density matrix for the MP2 first order wavefunction, their coefficients need to be found.
The first-order wave function is given as [1]:
$$\left|\left.\varPsi_{i}^{(1)}\right> \right.=\sum_{n}c_{n}^{(1)}\left|\left.\varPsi_{n}^{(0)}\right> \right.$$
Now this gives:
$$\left< \left.\varPsi_{n}^{(0)}\right|\varPsi_{i}^{(1)}\right> =c_{n}^{(1)}$$
Which by some more steps, that can be seen in Szabo and Ostlund leads to:
$$c_{n}^{(1)}=\left< \left.\varPsi_{n}^{(0)}\right|\varPsi_{i}^{(1)}\right> =\frac{\left< \varPsi_{n}^{(0)}\left|\mathscr{V}\right|\varPsi_{i}^{(0)}\right> }{E_{i}^{(0)}-E_{n}^{(0)}}$$
In Szabo and Ostlund the energy is derived to have the expression:
$$E_{i}^{(2)}=\sum_{n}'\frac{\left< \varPsi_{i}^{(0)}\left|\mathscr{V}\right|\varPsi_{n}^{(0)}\right> \left< \varPsi_{n}^{(0)}\left|\mathscr{V}\right|\varPsi_{i}^{(0)}\right> }{E_{i}^{(0)}-E_{n}^{(0)}}$$
Though it can also be written as:
$$E_{i}^{(2)}=\sum_{n}'c_{n}^{(1)}\left< \varPsi_{i}^{(0)}\left|\mathscr{V}\right|\varPsi_{n}^{(0)}\right>$$
Now forwarding to the energy expression for the MP2 energy:
$$E_{0}^{(2)}=\sum_{i<j,a<b}\frac{\left< \varPsi_{0}^{(0)}\left|\mathscr{V}\right|\varPsi_{ij}^{ab}\right>\left< \varPsi_{ij}^{ab}\left|\mathscr{V}\right|\varPsi_{0}^{0}\right>}{E_{0}^{(0)}-E_{ij}^{ab}}=\sum_{i<j,a<b}\frac{\left< ij\left|\right|ab\right>\left< ab\left|\right|ij\right>}{\varepsilon_{i}+\varepsilon_{j}-\varepsilon_{a}-\varepsilon_{b}}$$
Now I would like to think that I could identify the wave function coefficients by comparing the two above equations to get:
$$c_{ijab}^{(1)}=\frac{\left< ab\left|\right|ij\right>}{\varepsilon_{i}+\varepsilon_{j}-\varepsilon_{a}-\varepsilon_{b}}$$
I would have expected that I would have found a vector of $c$ such that I would be able to construct the coefficent matrix as:
$$C^{MP2}=c\cdot c^T \Rightarrow P^{MP2}_{ij}=\sum_{a}C^{MP2}_{ia}(C^{MP2}_{ja})$$
But instead I got a four index coefficent out. Where did my line of thinking go wrong? And how can I proceed to construct the first order wave function, and thus get the density matrix?
References to old papers where this have been done explicit is also appreciated, because I have not been able to find any thing but recent papers about MP2 and density matrix fitting, which is not what I am looking for.
[1] Szabo, A.; Ostlund, N. S. Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory; Dover: Mineola, NY, 1989.

You have two different kinds of coefficients here. The density matrix in your first equation is calculated from MO coefficients $C$. While the coefficients $c$ are from the development of the MP2 wave function in terms of eigenfunctions of the unpertubated Hamiltonian.
Thus, those are two different density matrices you are talking about.

I see I have made some bad notation. So my goal is to end up with the density matrix from the MP2 wave function. Now  I thought that all (not just MO) density matricies could be made by the first equation. Assuming that the equation to find the coefficient $c_{ijab}^{(1)}$ is correct, the thing I don't get it how I go from four indecies to a single index.

One might simply reshape it, but I don't know if this is helpful here. But why do you need a single index? Density matrices may have any number of indices (not sure about odd numbers, but at least even numbers).

I would have thought that the $c^{(1)}$ would be a vector, thus making $C^{MP2}$ a marix and thus making $P^{MP2}$ a matrix. Now I think I would like the denisty matrix to be of two dimension, because then it would fit directly into the equation to find the molecular dipole moment. And it could be diagonalized to find the Natural Orbital occupancy numbers.

Where did you get your Ansatz for $c^{(1)}$ from? I am not sure if you can do this ...

$P^{MP2}$ does not carry any information about the molecular orbitals. So you cannot calculate properties solely from this.

The main problem with the last statement,
$$
\begin{align}
C^{MP2} &= c\cdot c^T \\
\Rightarrow P^{MP2}_{ij} &= \sum_{a}C^{MP2}_{ia}(C^{MP2}_{ja}),
\end{align}
$$
is that one never has "MP2 MO coefficients" in this way; you would have MP2 natural orbital (NO) coefficients that come from diagonalizing this density, so it's the other way around:
$$
P^{\text{MP2}}C^{\text{MP2 NO}} = n^{\text{MP2 NO}}C^{\text{MP2 NO}},
$$
where $n^{\text{MP2 NO}}$ is the diagonal matrix of natural orbital occupation numbers. This still leaves the question of how to calculate $P^{\text{MP2}}$. Conceptually, the way I view it is that the set of $\{c_{ijab}^{(1)}\}$ are the amplitudes describing how to correct the uncorrelated ground-state wavefunction; they don't specify the wavefunction on their own, so $C^{\text{SCF}}$ will probably have to appear in the total density.
All of the following assumed a restricted wavefunction. $i,j,k,\dots$ are occupied MO indices, $a,b,c,\dots$ are virtual/unoccupied MO indices, $p,q,r,\dots$ are general (any) MO indices, and $\mu,\nu,\dots$ are general AO indices. The 2nd-order correction to the density, specifically the occupied-occupied and virtual-virtual blocks are given by Eqn. (11a) in Trucks [1]:
$$
\begin{align}
P_{ij}^{\text{MP2}} &= -\frac{1}{2} \sum_{kab} c_{jk}^{ab} c_{ik}^{ab} \\
P_{ab}^{\text{MP2}} &= +\frac{1}{2} \sum_{ijc} c_{ij}^{ac} c_{ij}^{bc},
\end{align}
$$
where I switched to using your notation. This is indeed similar to your idea from above; accumulate over both indices of the opposite space, then do a matrix multiplication.
The total density is then [Eqn. (7)]
$$
\mathscr{P} = P^{\text{SCF}} + P^{\text{MP2}},
$$
where $P_{pq}^{\text{SCF}} = \delta_{pq}$ because so far we're working in the MO basis. The back-transformation to the AO basis will be the same as for any other 2-index MO-basis quantity:
$$
\begin{align}
P_{\mu\nu} &= \sum_{pq}^{\text{all MOs}} C_{\mu p} P_{pq} (C_{\nu q})^{T} \\
\Rightarrow \mathbf{P}^{\text{AO}} &= \mathbf{C} \cdot \mathbf{P}^{\text{MO}} \cdot \mathbf{C}^{T}.
\end{align}
$$
Then, we can evaluate molecular properties that are represented as energy derivatives [Eqn. (6)]:
$$
\left.\frac{\partial \mathscr{E}}{\partial \lambda}\right|_{\lambda=0} = \sum_{\mu\nu} \mathscr{P}_{\mu\nu} Q_{\mu\nu}^{\lambda}
$$
where $\mathscr{E} = E_0^{(0)} + E_0^{(2)}$ and $Q_{\mu\nu}^{\lambda}$ is the matrix representation of the external perturbation $\lambda$ in the AO basis. This is analogous to the SCF result:
$$
\left.\frac{\partial E^{\text{SCF}}}{\partial \lambda}\right|_{\lambda=0} = \sum_{\mu\nu} P_{\mu\nu}^{\text{SCF}} Q_{\mu\nu}^{\lambda}
$$
Unrelaxed vs. relaxed density
The above definition of the density is incomplete, since it describes the unrelaxed correlated density. Our usual formulation of correlated methods is to describe correlation as some to-be-determined mixture of the virtual orbitals with the occupied orbitals. There is no such mixing in the above equations. In reality, $P_{ia}^{\text{MP2}} \overset{!}{\neq} 0$, so there is something that looks like a "transition density" with non-zero occupied-virtual blocks, in order to correct for the fact that our $C^{\text{SCF}}$ has no correlation. Allowing this mixing (which corresponds to orbital response) leads to the relaxed density. However, one has to solve the coupled-perturbed Hartree-Fock (CPHF or CPSCF) equations.
See between Eqns. (8) and (9) in [1] for the CPHF equations and how to solve them for the occupied-virtual mixing coefficients and finally the $P_{ia}^{\text{MP2}}$ block. Generally, note the similarities to the CIS density equations.
CPHF equations and first-order properties
As stated, when differentiating the energy with respect to some perturbation $\lambda$ to calculate first-order properties, one needs to contract the density with the operator property matrix. If you transform Eqn. (6) from the AO into the MO basis,
$$
\begin{align}
\left.\frac{\partial \mathscr{E}}{\partial \lambda}\right|_{\lambda=0} &= \sum_{pq} \mathscr{P}_{pq} Q_{pq}^{\lambda} \\
&= \sum_{pq} P_{pq}^{\text{HF}} Q_{pq}^{\lambda} + \sum_{pq} P_{pq}^{\text{MP2}} Q_{pq}^{\lambda},
\end{align}
$$
the SCF contribution simplifies due to the nature of the SCF density matrix in the MO basis,
$$
P_{pq}^{\text{HF}} = \left\{
\begin{array}{lr}
2 & \text{if }p = q\text{, p is an occupied MO} \\
0 & \text{otherwise}
\end{array}
\right.,
$$
leading to the expression
$$
\begin{align}
\left<\hat{\lambda}\right> &= \sum_{pq} P_{pq}^{\text{HF}} Q_{pq}^{\lambda} \\
&= 2 \sum_{i} Q_{ii}^{\lambda} \\
&= \mathrm{tr}\left(\mathbf{P}^{\text{AO}}\mathbf{Q}^{\text{AO}}\right) \\
&= \mathrm{tr}\left(\mathbf{P}^{\text{MO}}\mathbf{Q}^{\text{MO}}\right)
\end{align}
$$
for one-electron operators and wavefunctions that obey the Hellmann-Feynman theorem. I absorbed the 2 into $\mathbf{Q}^{\text{MO}}$, but this depends on the program implementations and their definition of the density. The point is that this contribution is relatively simple, and illustrates how for one-electron operators (dipole, quadrupole, angular momentum, ...), evaluating them for a converged Hellmann-Feynman wavefunction is very fast: one the density and property integrals are in the same basis, either do a matrix multiply and then trace, or do an element-wise (Schur) product and accumulate all the elements, so it scales as $O(N^3)$. No solution of CPHF equations is required. The same goes for the gradient with respect to nuclear coordinates $\frac{\partial E^{\text{SCF}}}{\partial x}$; it isn't a one-electron operator, and requires a mathematical trick to make $\frac{\partial P^{\text{SCF}}}{\partial x}$ disappear, but the SCF gradient doesn't require solving the CPHF equations either. This isn't true for $\frac{\partial E^{\text{MP2}}}{\partial x}$, which requires $\frac{\partial P^{\text{MP2}}}{\partial x}$, where you need to solve the CPHF equations to even form the full $P^{\text{MP2}}$.
Consider the remaining MP2 part of the property evaluation; the total density contains contributions from the occupied-occupied (oo) block, the virtual-virtual (vv) block, and the occupied-virtual (ov) block. I'm going to switch from $P^{\text{MP2}}$ and $c$ to $D$ and $t$ to make typing easier.
$$
\sum_{pq} D_{pq} Q_{pq}^{\lambda} = \sum_{ij} D_{ij} Q_{ij}^{\lambda} + \sum_{ab} D_{ab} Q_{ab}^{\lambda} + 2\sum_{ia} D_{ia} Q_{ia}^{\lambda}
$$
The first two terms can already be calculated from the unrelaxed density; $D_{ia}$ is the term that must come from solving the CPHF equations.
In general, the form of solving CPHF equations looks like $\mathbf{AU} = -\mathbf{V}$, where $\mathbf{A}$ is the orbital Hessian (2nd derivative of the energy with respect to orbital rotations), $\mathbf{V}$ is the gradient of some perturbation $\hat{V}$. In [1], it's $\lambda$, but Neese [3,4] and the Scandinavians usually use $V$ or $Q$ when the perturbation is associated with an observable. This perturbation gradient, termed the right-hand side (or RHS), describes how the perturbation might cause occupation of virtual orbitals, and the orbital Hessian describes how the response of one MO to some perturbation is coupled to the other MOs [5,6].
Start by rewriting $\sum_{ia} D_{ia} Q_{ia}^{\lambda}$ as the combination of a perturbation-dependent response $\mathbf{U}$ and the perturbation gradient $\mathbf{X}$:
$$
2\sum_{ia} D_{ia} Q_{ia}^{\lambda} = 2\sum_{ia} X_{ai} U_{ai}^{\lambda}
$$
$\mathbf{X}$ serves the same purpose as the $\mathbf{V}$ mentioned above, but rather than coming from an external perturbation, it is an intermediate quantity that describes the MP2 wavefunction.
First, the perturbation-dependent response. This part is not specific to MP2, and is also important for higher-order SCF-based molecular response [3].
$$
\begin{align}
U_{ai}^{\lambda} &= \sum_{bj} (\mathbf{A}^{-1})_{ai,bj} Q_{bj}^{\lambda} (\varepsilon_{j} - \varepsilon_{b})^{-1} \\
A_{ai,bj} &= 1 + \left( \left<ab||ij\right> - \left<aj||ib\right> \right) (\varepsilon_{a} - \varepsilon_{b})^{-1}
\end{align}
$$
The $\varepsilon$ are MO energies. If we don't assume the use of canonical MOs, this energy denominator is actually $F_{ab}S_{ij} - F_{ij}S_{ab}$.
Jumping ahead, the relationship between the ov density, the orbital Hessian, and the MP2 RHS is [Eqn. (9)]
$$
\sum_{bj} D_{bj} (\varepsilon_{j} - \varepsilon_{b}) A_{bj,ai} = X_{ai}
$$
See how this is structurally similar to solving for $U$, but the RHS is different. To show the equivalence between all quantities,
$$
\begin{align}
2\sum_{ai} X_{ai} U_{ai}^{\lambda} &= 2\sum_{ai} X_{ai} \sum_{bj} (\mathbf{A}^{-1})_{ai,bj} Q_{bj}^{\lambda} (\varepsilon_{j} - \varepsilon_{b})^{-1} \\
&= 2\sum_{bj} D_{bj} Q_{bj}^{\lambda}.
\end{align}
$$
The explicit equation for $X_{ai}$ is [Eqn. (11b)]
$$
X_{ai} = \sum_{jk} D_{kj} \left<ki||ja\right> + \sum_{bc} D_{bc} \left<bi||ca\right> - \frac{1}{2} \left[ t_{kj}^{ab} \left<ib||kj\right> + t_{ij}^{cb} \left<cb||aj\right> \right],
$$
which is in fact the MP2 Lagrangian, $L_{ai}$ [2, Eqn. (11)]. Again, notice the similarity to the CIS Lagrangian. The Lagrangian is required because the wavefunction doesn't need to be relaxed in order to calculate the correlation energy, so you pay the price here. These equations must be evaluated for MP2 geometry optimizations (during the gradient part) or for fully-relaxed MP2 molecular properties. To be clear, if one is only interested in the density and not molecular properties, only $\mathbf{DA} = \mathbf{X}$ needs to be solved for, and not the equations involving $\mathbf{U}$. In terms of scaling, SCF-based CPHF is about $O(N^4)$ (after back-transforming the $\mathbf{A}^{-1}\mathbf{Q}$ terms to be AO-direct) but might require fewer iterations than the SCF itself. According to [2], the MP2 gradient is something like $O(ON^4) + O(O^{2}V^{3})$, but they can do similar tricks during iterations to cut down on the number of fifth-order steps. In my experience, MP2 gradient calculations are about as expensive as an SCF polarizability calculation.

References

Trucks, G. W.; Salter, E. A.; Sosa, C.; Bartlett, R. J. Theory and implementation of the MBPT density matrix. An application to one-electron properties. Chem. Phys. Lett. 1988, 147 (4), 359-366. DOI: 10.1016/0009-2614(88)80249-5
Frisch, M. J.; Head-Gordon, M.; Pople, J. A. A Direct MP2 gradient method. Chem. Phys. Lett. 1990, 166 (3), 275-280. DOI: 10.1016/0009-2614(90)80029-D
Neese, F. Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling. Coordination Chemistry Reviews 2009, 253 (5-6), 526-563. DOI: 10.1016/j.ccr.2008.05.014
I cannot recommend the above paper enough; it contains excellent descriptions of how electric and magnetic response are calculated with explicit equations that are actually usable. There, you can see how CPHF is used for molecular response properties and the connection with excitation energies, found by solving $\mathbf{AU} = \mathbf{0}$ (diagonalization rather than solving a set of linear equations).
Toulouse, J. Introduction to the calculation of molecular properties by response theory.
Parrish, R. M. Psi4 Development: Theory Supplement, CPHF for Dipole Polarizabilities.

If you would like an elaboration on how the CPHF equations are solved and what each term means, let me know and that part can be expanded. Unlike correlated methods, I actually know how CPHF works.

An elaboration of how the CPHF equations are solved would be very appricaited!


text:
How can a potential energy function be used to sketch different levels of the wavefunction?
If the potential energy function is provided, say $V(x) = 5x^{2}$, how am I then able to use this function to visualize, graphically, the wavefunction where $V \neq \infty$?

Solve the Schrödinger equation and see. There is no shortcut.

Which one is supposed to be potential? $f$ or $V$?

@Feodoran, I clarified in the question; it should have been V(x) = 5x^2 when x > 0. I'm looking for more of a general process to go from having simply the function for potential to graphing the wavefunction.

I think this is a good question in the sense that if you haven't taken college-level physical chemistry, it isn't obvious that the eigenfunctions of $\hat{V}$ are the wavefunctions and the eigenvalues are the associated energies, and this is relatively complicated compared to just plotting the potential.

So, if I get this right, then the question is "How to solve the Schrödinger Equation?". This is a very broad question. There is a whole field trying to solve this problem in a general way.
For $V(x)=5x^2$ you have an harmonic oscillator, which can be solved analytically.

@Feodoran, I figured out what the question was asking and it simplifies somewhat. It was really asking, given a function for potential, graph a few wavefunctions at different energy levels, demonstrating that you understand where the particle will behave as it should and where it begins tunneling, and comprehend how the difference in energy levels is changing.

Is that a homework question?
I still don't get the full question. Is it about sketching eigenfunctions or some wavefunction in general (which may be any arbitrary superposition of eigenfunctions) ?
Also this $V\neq\infty$ is confusting, since we already have $V(x)=5x^2$.
About tunneling: That requires some barrier in the potential, which we don't have in the harmonic oscillator.


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Is it possible to calculate electronegativity of surface atoms?
As the title asks, is it possible to calculate the electronegativity (EN) of specific atoms on extended surfaces (slabs), more specifically using computational methods? There are a few issues here I am running into:
To find the Mulliken EN $\chi$, we can use the relation
$$\chi = \frac{E_\mathrm{i} + E_\mathrm{ea}}{2}$$
where $E_\mathrm{i} = E(\ce{M+}) - E(\ce{M})$ and $E_\mathrm{ea} = E(\ce{M})-E(\ce{M-})$. However, these energies are for the entire system, usually a molecule, and not the individual atoms. I hope I am not misunderstanding the concept here.
Alternatively certain definitions use $\chi = -\mu$, or the work function. Again, this is usually calculated on a surface basis, not per atom of site.
Is there no accurate determination method which is specific to the atom? Or is it conceptually impossible?

Have you looked at the Fermi surface? https://en.wikipedia.org/wiki/Fermi_surface


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Why is trimethylsilanol a stronger acid than tert-butanol?
The Wikipedia article about trimethylsilanol says the following:

Trimethylsilanol is a weak acid with a $\mathrm{p}K_\mathrm{a}$ value of 11. The acidity is comparable to that of orthosilicic acid, but much higher than the one of alcohols like tert-butanol ($\mathrm{p}K_\mathrm{a}$ 19).

Why is it so? A silicon atom is more electropositive than a carbon atom, therefore the conjugate base should be more destabilized by the +I effect of the $\ce{Si(CH3)3}$ group compared to the $\ce{t-Bu}$ group.

The involvement of d-orbitals in the stabilisation of silanol anions is minimal. The ions $\ce{H3CO-}$ and $\ce{H3SiO-}$ were investigated by Hopkinson and Lien,1 who concluded that:

Silicon is a more electropositive element than carbon and consequently the silyl group might be expected to be an electron donor. As such the silyl group should stabilise cations and destabilise anions. However, ab initio molecular orbital calculations show both the α-silyl and β-silyl to strongly stabilise anions and to weakly destabilise cations. Inclusion of d orbitals on  silicon does not significantly alter these conclusions. This contradictory behaviour can be largely understood in terms of n $\to$ π* delocalisation from the lone pair to π* of the silyl group.

These π* orbitals are antibonding orbitals derived from linear combination of Si 3p and H 1s. A partial MO diagram for the $\ce{H3Si}$ fragment is provided here (from the same paper):

Delocalisation of the oxygen lone pair into the $\pi_x^*$ and $\pi_y^*$ orbitals therefore leads to stabilisation. The authors write that the π* orbitals of the $\ce{H3Si}$ fragment are lower in energy than the corresponding π* orbitals of the $\ce{H3C}$ fragment, and hence the stabilisation is greater for silicon than it is for carbon.
A good test for this is to examine how the Si–H bond lengths change upon deprotonation. If the delocalisation is into Si–H antibonding orbitals, then the bond lengths should increase upon deprotonation. In contrast, if the delocalisation is into Si 3d orbitals, then there should be no change. Upon deprotonation, the Si–H bond length increases from 1.473 Å to 1.528 Å, which indicates it is the former.
However, there isn't a thorough explanation of why the energy levels are so. I would surmise that there are two factors: one is the fact that Si–H overlap is poor, compared to C–H overlap. Therefore, Si–H antibonding orbitals are lower in energy than C–H antibonding orbitals, and can act as better acceptor orbitals. Another factor (and one commonly taught in textbooks) is that since Si is more electropositive, the π* orbitals have a larger coefficient on Si; the π* orbitals of the $\ce{H3C}$ fragment have a smaller coefficient on C.
There is also an issue of why the stabilisation is larger in the anion than it is in the neutral molecule: if both the anion and the neutral molecule were equally stabilised, then there would not be any difference in the acidity, which is essentially a measure of the relative stability of the two species. It may be easily explained - at least partly - by the fact that the lone pairs in the anion are higher in energy than those in the neutral molecule, and consequently act as better donor orbitals.

Now, this explanation can be easily extended to the methylated compounds $\ce{Me3CO-}$ and $\ce{Me3SiO-}$. Due to the poorer acceptor capability of Si–C antibonding orbitals, we expect the factors above to play a smaller role in the preferential stabilisation of the silicon anion. As such, the difference in $\mathrm pK_\mathrm a$ between $\ce{X3COH}$ and $\ce{X3SiOH}$ should be smaller when $\ce{X = Me}$ compared to when $\ce{X = H}$.
Is there a way to test this? A crude analysis is as follows. I will take the value of $\pu{27.6 kcal/mol}$ given by Hopkinson and Lien1 for the reaction
$$\ce{H3SiO- + H3COH -> H3SiOH + H3CO-}$$
which corresponds to a $\mathrm pK_\mathrm a$ difference of 20 units when $\ce{X = H}$. This is obviously not rigorous, but serves as a good preliminary benchmark. Now, if Wikipedia's sources are to be believed, then when $\ce{X = Me}$, the $\mathrm pK_\mathrm a$ difference is decreased to a mere 8 units - which is in line with our prediction.

The topic of negative charge stabilisation by silicon has been studied much more thoroughly in the context of carbanions, because they are synthetically useful species (e.g. Peterson olefination). In this context, Schleyer et al.2 also conclude that d-orbital participation is negligible. The myth of d-orbital participation in main group bonding has been thoroughly debunked decades ago,3 so it is regrettable that it is still being taught nowadays.

References

Hopkinson, A. C.; Lien, M. H. A theoretical study of the effects of silyl and methyl groups on acid-base reactions at oxygen, nitrogen and carbon. J. Mol. Struct.: THEOCHEM 1983, 92 (1–2), 153–166. DOI: 10.1016/0166-1280(83)80065-7.
Schleyer, P. v. R.; Clark, T.; Kos, A. J.; Spitznagel, G. W.; Rohde, C.; Arad, D.; Houk, K. N.; Rondan, N. G. Structures and stabilities of α-hetero-substituted organolithium and organosodium compounds. Energetic unimportance of d-orbital effects. J. Am. Chem. Soc. 1984, 106 (22), 6467–6475. DOI: 10.1021/ja00334a001.
Reed, A. E.; Schleyer, P. v. R. Chemical bonding in hypervalent molecules. The dominance of ionic bonding and negative hyperconjugation over d-orbital participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. DOI: 10.1021/ja00160a022.

You are correct in saying that Carbon is more electronegative than Silicon, but due to presence of empty 3d-orbitals in silicon, it can engage in back bonding with the lone pair of oxygen.
(Following is a crude example of back bonding)
$\hskip2.5in$
This leads to delocalisation of the negative charge lending it a stability comparable to that of resonance, resulting in Trimethylsilanol being a better acid than tert-butyl alcohol.

This is incorrect, d-orbital participation is negligible. See for example JACS 1984, 106, 6467 (they talk about carbanions, a slightly different context, but equally applicable to the present one).


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Can you calculate the energy needed to ionize an electron in hydrogen?
If I have an electron in the 2s orbital,it would have different energy with the 2p orbital.Can we calculate the difference in the energy ?

Possible duplicate of Calculating the ionization energy for hydrogen

Because the single electron in the hydrogen atom feels a purely symmetric Coulomb potential, the states with different orbital angular momentum $\ell$ are degenerate in non-relativistic quantum mechanics and the binding energy of the electron is given by
$$
E_{n,\ell}=-\frac{\mathcal{R}_\text{H}}{n^2},
$$
where $\mathcal{R}_\text{H}$ is the mass-corrected Rydberg constant for the hydrogen atom and $n$ is the principal quantum number.
When Dirac introduced his relativistic theory for one-electron systems it was shown that the spin-orbit coupling in the hydrogen atom results in a splitting between the 2P$_{3/2}$ level on one hand and the 2S$_{1/2}$ and 2P$_{1/2}$ levels on the other hand (the latter two are degenerate). The subscript number indicates the total angular momentum $J=S+L$.
Later, Willis Lamb was able to measure a shift between the 2S$_{1/2}$ and 2P$_{1/2}$ levels that was not predicted by Dirac's theory and this splitting was explained by Hans Bethe (traveling back by train from the conference where he learned about Lamb's results) and marked the birth of quantum electrodynamics (QED).

Effect of Dirac theory and QED on the non-relativistic energy of 2S and 2P levels of atomic hydrogen (taken from H. C. Wolf H. Haken, The physics of atoms and quanta, 7 ed., Springer, 2005.).
To come back to your question, yes, it is possible to calculate the splitting between the electronic states of atomic hydrogen, however, it requires some advanced maths. If you are really interested in calculating it yourself, I recommend Quantum Mechanics of One- and Two-Electron Atoms by Bethe and Salperer.

Thanks for the explaantion sir ! It would be out of my e


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Why is the molecular dipole the main determinant of whether a substance dissolves in another substance?
Consider a covalent molecule such as carbon tetrafluoride. In $\ce{CF4}$, each $\ce{C-F}$ bond is polar covalent. However, this substance is considered overall non-polar because each bond dipole moment cancels out each other, resulting in a negligible molecular dipole moment. This explains why it is non-polar. But how can that be the case? Certainly, such a substance with such strong bond dipoles would have some sort of strong interaction with polar solvents? Why do we not take the individual bond dipole moments into consideration?

Related: Why is carbon dioxide nonpolar?; What are dipole moments in a molecule supposed to act upon?

@orthocresol Just to clarify, I completely understand why CF4 is nonpolar. What I don't understand is why the polar C-F bonds don't seem to affect it's interactions with polar solvents.

I'm voting to close this question as off-topic because it's based on false premise.

@Mithoron May I know what is that false premise?

@TanYongBoon Dipole moment is a vector, so it's all about the vector sum. Just as a satellite on a stationary orbit appears to be stationary for the observers on Earth, when centrifugal force and gravitational force equalize each other, same happens with four polar $\ce{C-F}$ bonds.

Basic idea of your question? Short answer to question in title is "It isn't".

I kinda get your question, and it's a good one. Basically, even though carbon dioxide is non-polar, it dissolves in water. And how is that? Basically the relatively positive carbon atom attracts the negative end of the water dipole, and the two negative oxygens attract the positive end of the dipole. But why isn't this the case for $\ce{CF4}$ is your question right?

@andselisk I already understand this concept. What I'm interested in is why inidividual bond dipoles do not contribute to the interactions of an overall nonpolar solute molecule with polar solvent molecules.

@PrittBalagopal You can actually extend my question to all nonpolar covalent molecules with polar individual bond dipole moments. It is not restricted to CF4 or CO2.

See discussion here: https://chemistry.stackexchange.com/questions/41952/is-the-triiodide-ion-polar

@Mithoron Thanks for sharing the insightful link. However, I don't think it answers my question

"inidividual bond dipoles do not contribute to the interactions of an overall nonpolar solute molecule with polar solvent molecules" They do!

@Mithoron Ok I get what u mean. An example would be the water-soluble carbohydrates with many polar hydroxyl groups. However, I am saying that the main determinant is still the molecular dipole. Not individual bond dipoles. For example, could u explain the interact of CF4 and CH4 in water. This is an example of my case. Because the C-F bond is hugely polar but the C-H bond isn't that polarised. How would this affect their solvations in water. Are you clearer now?

https://chemistry.stackexchange.com/questions/7147/why-distinguish-between-polarity-and-hydrophobicity https://chemistry.stackexchange.com/a/645/9961 and here is your duplicate: https://chemistry.stackexchange.com/questions/80695/why-are-fluoroalkyl-chains-hydrophobic-oleophobic


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Model CO2-C6F14 Vapor Liquid Equilibrium with an Equation of State
I'm trying to model the VLE for a binary mixture of CO2 and C6F14 using a cubic equation of state (e.g. Peng-Robinson) for both the liquid and the vapor phase (phi-phi approach).
From available data in literature I know that at 30°C and at a pressure ranging from 5 bar to 60 bar the system shows VLE.
In the first step I assume ideal mixture.
When I apply the equation of state to either C6F14 and CO2 to evaluate the fugacity coefficients for the liquid and for the gas phases I obtain only one root so that seems that there isn't any equilibrium.
Where am I wrong? Am I missing something from theory?
Any help is appreciated.


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Why is sodium acetate a good nucleophile?
As $\ce{CH3COONa}$ is bulky, I thought that $\ce{CH3COONa}$ is a weak nucleophile.
But I found out that it's a nucleophile good enough to perform $\mathrm{S_N2}$ reaction on a secondary carbon.
Why is it so?

The acetate anion is actually not particularly bulky. The negative charge is located on the two oxygens which are approximately at a $120^\circ$ angle away from the only residue that has the potential to be bulky. Whatever that residues does will be ‘behind’ the carbon and hardly noticeable. However, they are still rather bad nucleophiles since the negative charge is delocalised over two oxygen atoms in the π system.
I’m actually not sure which orbital attacks nucleophilically but it could equally be an oxygen lone pair rather than the π system. If that is the case, there is really very little steric hindrance that could inhibit the reaction from proceeding. Thus, acetate is able to attack nucleophilically, as it does for example in a Mitsunobu reaction.


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Predicting the generation of singlet oxygen
On Wikipedia, I saw the reaction $\ce{H2O2 + NaOCl -> O2 + NaCl + H2O}$, where the $\ce{O2}$ is in the $^1\Delta_g$ (singlet) state. Can anyone offer an explanation for why it isn't in the usual triplet state, maybe using the theory of molecular orbitals?

Molecular orbitals are actually, to the best of my knowledge, not really helpful in explaining this observation. Instead, a much more fundamental concept does a much better job: the conversion of (electronic) spin.
The reacion you mentioned can be analysed in terms of spin states. Nobody will disagree that hydrogen peroxide, sodium hypochlorite, sodium chloride and water are singlet compounds. Therefore, using generic Greek letters as placeholders for the different compound to illustrate the point that their true doesn’t matter much, we can write the equation as:
$$\ce{^1\unicode{x391}\ + ^1\unicode{x392}\ -> ^?\Gamma\ + ^1\Delta\ + ^1\unicode{x395}}\tag{1}$$
We see that we are reacting two singlet compounds two three compounds and two of the products are also singlets. This means, every spin on the left-hand side of the equation is paired and most of the spins on the right-hand side, too. To generate a triplet product, we would need to flip a few spins but that is a slow process on the timescale of a chemical reaction. (Note that the conversion from singlet oxygen back to triplet can be observed as a faint red glow demonstrating that it is not instant but delayed.) Thus, in the moment of the particle collision, generating all-singlet products from all-singlet reactants (if only a single product potentially can have an unpaired spin) is the only option. Thus:
$$\ce{^1\unicode{x391}\ + ^1\unicode{x392}\ -> ^1\Gamma\ + ^1\Delta\ + ^1\unicode{x395}}\tag{2}$$

A small point, but if there was enough energy then S+S$\rightarrow$ T + T is a possible spin allowed process. Presumably  to form water in a triplet excited state would need vast amounts more energy compared to what is available.

@porphyrin But it still requires two triplets to be formed, doesn’t it? So in an oxygen case, you would need a double collision at least, no?

not a double collision, just enough energy initially to form two triplets. Two triplets can be formed from two singlets in the collision just as can two singlets as you point out. The reverse process is more commonly observed; triplet-triplet annihilation.

@porphyrin No, what I meant was like — We would need to have two species that are triplets at the end of the equation, so if it’s the generation of oxygen as discussed here, we would need an additional species capable of reaching a triplet state to take part here? Most likely two sets of reactants to give two triplet oxygens?

yes you are correct :)


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Series of dilution without a constant dilution factor?
I have to make a series of dilutions though the dilution factor is not constant.
I'm trying to understand how to carry out this dilution in the most efficient way as not to waste much solution.
Series of dilution
Initial dilution: 1/250
1/250 -> 1/500 -> 1/1000 -> 1/2000 -> 1/3000 -> 1/4000 -> 1/5000 -> 1/6000
I need to end up with 25µl in each tube. I've attached my solution in an image below though i'm not sure i'm doing this correctly as i would have to discard so solution.
My solution as depicted in the image:

Well actually dilution or not, this is more related to mathematics than chemistry. Maybe you should ask this question in MSE :-)


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Are any natural waxes PH neutral?
I'm a collector of historical artifacts, and one of the most common methods to preserve them is waxing. And one of the best options out there is Renaissance Wax, a microcrystalline wax. The reason it is considered superior to natural options like beeswax is that it is PH neutral, meaning it will not corrode the artifact over time with its acid content.
My only hesitancy to continue using it would be that it is petroleum based. I'm curious if any natural waxes offer the PH neutrality that RW does, as well as being just as long lasting for protecting items from moisture.

Well, it's "pH" and more proper description would be doesn't contain any significantly acidic compounds, otherwise question seem kinda broad, but oh well...

Apologies, I'm no chemist.
So how do I check whether a specific ingredient is acidic? And what natural options would there be for waxes without acidity?

https://en.wikipedia.org/wiki/Beeswax says there's ca 12% of fatty acids in it, not so sure if it's really problem though.

Petroleum paraffin wax is a natural compound. It's been in the Earth a considerable time before mankind was thought of. Yes, it's purified, but so is beeswax (unless you want your beeswax mixed with larvae, honey and bee wastes).


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Can this happen to superheated water?
Superheated water refers to water that is heating well beyond its usual boiling point under extreme pressure. This cause it to display anomalous property.
I noticed that as temperature increases, so does the self-ionization of water, that is, the concentration of $\ce{H3O+}$ and $\ce{OH-}$.
It appears to me that the above phenomenon is caused by increased "freedom of movement" in hydrogen cation, or protons, allowing them to leap from one charge-neutral molecule to another, ionizing both their source and destination.
My question is, what would happen if the pressure and temperature of water is taken to the extreme? Would there be sufficient kinetic energy in the material to allow essentially free movement of $\ce{H+}$ among particles of $\ce{O^2-}$? Which makes it so that the water essentially behave like a plasma consisting of the two types of ions? Or would the energy required to reduce $\ce{OH-}$ be so enormous that some other phenomenon will take place before this is possible?

It all depends on what you mean by "taken to the extreme." If the temperature is hot enough and there is enough pressure (eg big bang...) even the nucleus comes apart.

Does superionic water count?  https://en.wikipedia.org/wiki/Superionic_water

https://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

https://chemistry.stackexchange.com/questions/20009/hypothetical-what-happens-to-water-as-pressure-increases-to-infinity

At some point before ionization your superheated water will become a supercritical fluid.


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Normal phase vs reverse phase HPLC
What is the difference between normal phase and reverse phase high-performance liquid chromatography?

Normal phase
Normal phase HPLC systems are similar to the flash-column chromatography that you might be familiar with.
A silica stationary phase is eluted with a non-polar solvent such as hexane, or a fairly non-polar solvent mixture such as 2-propanol in hexanes. In normal phase chromatography, only organic solvents are used.
In the normal phase, polar molecules elute slowly, and non-polar (greasy) molecules elute quickly.

Reverse phase
Reverse phase is essentially the opposite of normal-phase.
A non polar stationary phase (often silica in which the free hydroxyl groups are end-capped with something greasy, C18 chains are common but many many variants are possible) is eluted with a polar solvent such as acetonitrile/methanol, or a fairly polar solvent mixture (acetonitrile water mixtures are common, or methanol water mixtures).
In the reverse phase, polar molecules elute quickly, and non-polar (greasy) molecules elute slowly.

Normal phase HPLC means the stationary phase is polar and the mobile phase is non-polar; reversed phase means the stationary phase is non-polar and the mobile phase is polar.
See HPLC Separation Modes (from Waters, a manufacturer of chromatography equipment and supplies) for more insight.


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What's the actual difference between Formality and Molarity?
Today, I came across a question: "State the differences, in not less than 5 points, between Formality and Molarity".
Using my knowledge, I wrote down 2 points, and stretched it to 3 somehow, but that's all I could do... Also, in the process, I got confused between the two, because in pH calculations, even when Ionic solids and acids are dissolved in water, their concentration is expressed in terms of molarity, and not formality.
Please note that I am not asking this for the sake of my homework, but am asking for a better scientific knowledge of Formality, and how it differs from Molarity.
Here are the 3 points I wrote:

Molarity is the concentration of a molecular substance, expressed in terms of the ratio of moles of substance present to the volume of the solution it forms. Formality, on the other hand, is the concentration of an Ionic substance, that exists in the form of an Ionic Crystal Lattice, in terms of the ratio the gram formal masses present to the volume of the solution it forms.
Molarity is defined for molecular substances, whereas Formality is defined for Ionic substances.
The unit of molarity is $\pu{mol m-3}$, while that of formality is $\mathrm{form}~\pu{m-3}$ (where $\mathrm{form}$ is the number of gram formula mass units).

Can someone please explain the actual differences, and help with expanding the answer for more accuracy?

I have never come across formality before.

@Jan Really? I thought it was a standard form of concentration, used alongside Molarity, Molality, Normality, etc. No wonder there's no answers yet...

Molarity is what I always come across. I have come across normality in a few instances (the diluted acid stock solutions in Munich were all 2 N, meaning that sulphuric acid was half-concentrated with respect to hydrochloric and nitric acid). A chemistry high school course once briefly touched molality; I have since forgotten its definition. That’s how standard they are ;)


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Why is the melting point of BaCl2 > BeCl2 while CsCl<NaCl
I'd think that increasing ionic character by increasing electronegativity difference would result in strong bonds but in the case of NaCl and CsCl(and the other chloride of the group) the case is different.

Related: https://chemistry.stackexchange.com/questions/82246/which-has-the-highest-melting-point-naf-nacl-nabr-nai/82254#82254

Ionic compounds have no bonds. It's a question of efficient packing and electrostatics.

@Karl Ionic compounds have bonds, just constructed differently from those th things like water and methane.

@OscarLanzi A "bond" implies directional connection. No such thing in sodium chloride. The term is simply not helpful in explaining the properties of ionic compounds, as can be see by the above question, where it has mislead the OP.

@Karl A bond does not imply directionality. An ionic bond is a perfectly valid bond whether you are discussing the 'real' ionic bond (with non-zero covalent/metallic character), or the ideal 'pure' ionic bond. Since this is an issue of chemical definitions, I refer you to the IUPAC Gold Book.

also https://chemistry.stackexchange.com/questions/65914/why-magnesium-has-the-lowest-melting-and-boiling-point-in-its-group?rq=1

Sodium chloride and caesium chloride are both primarily ionic compounds. Within ionic compounds, a greater melting point can be caused by a higher ionic charge or a higher charge density/better interaction. Since both have the same charge while sodium has a smaller ionic radius, the sodium-based crystal is held together better due to more interaction than the caesium based one.
When comparing barium chloride and beryllium chloride, you are essentially comparing apples and oranges since beryllium chloride is a primarily molecular compound. In its gas phase, it forms $\ce{BeCl2}$ monomers and $\ce{Be2Cl4}$ dimers. The intramolecular attractions in molecular compounds are typically rather strong with the intermolecular ones are weaker. For ionic compounds such as barium chloride, we cannot distinguish between the two since we cannot define molecular entities.
When comparing ionic compounds to molecules, the ionic compounds typically have a much higher melting point (with exceptions; both low-melting ionic compounds and high-melting molecular compounds).

Arright, that explains it for BeCl2 but what about CaCl2 (I don't think this dimerises, does it?) ? It's smaller but has a lower melting point it looks like (does CaCl2 even exist in a crystal lattice?)

Well a quick google suggests they have different crystal structures, CaCl2 taking a rutile structure, and BaCl2 being calcite or PbCl2. It's not easy to compare different structures, especially when coordination numbers change as in this case, so periodic trends such as MP will be hard to spot. In fact the most obvious periodic trend in the alkaline earth chlorides is the change from covalent to ionic behaviour, from covalent BeCl2 through the layered MgCl2 to the typical ionic structure of BaCl2.


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How do you pronounce GAMESS
Do you pronounce the software's name, GAMESS, as ga-mess, gaim-z (as in plural of game) or something else.
I know it's sort of not the kind of question to ask here, but I couldn't think of anywhere else better.

"Hello, this is L̶i̶n̶u̶s̶ ̶T̶o̶r̶v̶a̶l̶d̶s̶ Frank Neese, and I pronounce L̶i̶n̶u̶x̶ Gamess as ORCA." To the topic, I more often hear people pronouncing GAMESS as [ɡeɪmz].

it's pronounced like the plural of game

I'm voting to close this question as off-topic because it's not about chemistry.

Well, since the official GAMESS documentation doesn't insist on any particular pronunciation, I'd argue you can call it anyway you like. :D As mentioned by Jan Jensen the absolute majority of computational chemists indeed call it [ɡeɪmz], but I habitually call it something like [gɑːmɛs] and colleagues always understand what I'm talking about. :)

And I agree with @Mithoron: such question are better asked, say, in our chat than on the site.


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Does the multiplication and division significant figures rule apply to unit conversions as well?
My chemistry teacher told our class that answers to calculations that involve multiplication and division should be given to same number of significant figures as the factor, dividend or divisor with the lowest no. of sig.figs.
For example, to calculate the no.of moles of calcium in a pure 40g sample of calcium, the dividend (the mass) has 1 sig.fig. and the divisor (the molar mass) has 3 sig.figs. Hence the no. of mole sis given to 1 sig.fig.
40/40.1 = 1 mol of Calcium
Does this same rule apply to unit conversions? For example if I wanted to convert 12.4 milligrams to grams, I would need to divide 12.4 milligrams by 1000. 12.4 has 3 sig.figs. whereas 1000 has only 1 sig.fig. Does that mean the figure in grams must be given to 1 sig.fig?

In regards to the calcium situation, it depends on whether you measured 40g of calcium (which would be 2 sig. fig.), or 40g was the stated value (1 sig. fig.). Yes, the number of sig figs in your answer should equal the number of sig figs in the least accurate piece of data.
When converting units, the 1000 is taken as the exact value, i.e. it has infinite accuracy. Here, it is not considered, so the answer should have 3 sig fig from 12.4 (which is also why scientific notation is used to decrease ambiguity).


text:
Why is phenolphthalein at pH 10 non-planar?
Phenolphthalein is colorless when pH = 0-8.2 and pink when pH = 8.2-12.
Why does the ring bearing the carboxylate (COO-) group not lying on the same plane as the other 2 rings?

Image source: https://www.learner.org/courses/chemistry/images/lrg_img/Pink.jpg

In fact, the other two rings aren't exactly in the same plane either.

I suggest drawing it out on paper with constant bond lengths and perfect $120^\circ$ angles. ;)

@Jan I suggest making this an answer.

@OscarLanzi Yes, but I didn’t have access to ChemDraw when I commented (and won’t until tomorrow morning). In the meanwhile, DSVA has provided a very nice answer that I could hardly add anything to …

Ok, let's look at this structure. I've drawn/calculated (B3LYP/6-31G) a 3D model of this structure and optimized the geometry.

So what do we see here? Actually no two rings are in one plane, but two are nearly planar (~40° of twist) while one is basically at 90° twist. Let's first look at the one with the carboxy group which is twisted a lot. The carboxy group is in ortho position. If we twist the ring to be in a plane with one of the other rings the oxygen would collide with a CH group. The carbons and oxygens are quite big, so is the repulsion between them and the angle has to be quite big for it to work out. Here's a picture of those groups colliding (I did put it at some degree angle because it's easier to see), keep in mind that this representation of atoms is not the truth. The electron "cloud" around them is actually much larger.

Now why do the other two rings have an angle of about 40° between them? If they are planar they can have conjugation which is favorable. Now What happens if we put them into one plane?

You can see that the two hydrogens in the top get very close. In fact, too close. So there are two ways to avoid this. Twist the two rings or make the angle around the central carbon atom between the rings bigger. In a sp2 carbon like we got here the optimal value would be 120°, we could make it bigger, let's say 145 degree:

Now the hydrogens would not clash any more, but it turns out this costs much more energy than we would lose if we twist the two rings to avoid the hydrogens clashing together. What happens in the end is that this angle gets a little bit bigger (~124° instead of 120) and we get that twist between the aromatic rings which is still low enough for good conjugation between them (see first structure).

You only had to explain why the last ring is on 90º and you didn't..there could be an arrange with another ring perpendicular, for example..anyway..the structures look good..

@HernanMiraola OP assumed that the other rings are in one plane. They most likely are not so it's almost necessary to explain it, otherwise there will be even more questions. And I don't get the second part. it is twisted almost 90° (perpendicular) to the other rings, I explained that this is because of steric reasons and there is no other way to do this.


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When do I use a Grignard reagent such as PhMgBr, instead of an organolithium such as PhLi to react with butanone
Essentially what I am asking is, how do I know whether I should use a Grignard in the place of an organolithium compound or vice versa when reacting it with aldehydes and ketones, or does it not matter as much for these carbonyl compounds?

I think it's down to what level of selectivity you need. If other groups are present on the ketone or aldehyde, which may be sensitive to an excessively reactive organometallic species, then you may want to tone it down. On the other hand, the more reactive one may be more appropriate in some cases. E.g. if you have an alpha-beta unsaturated ketone and you want the C=O carbon to be attacked rather than the C=C one, then you may be better off with PhLi. I seem to remember cuprates are appropriate for the C=C addition in this case. Then there is the problem of suppressing proton abstraction...

In case of 2-butanone it may be a question of what pre-prepared reagent you have available. For ease of preparation and scale-up, the Grignard is the way to go.