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Ground state term symbol for Cerium (Ce)
I need to calculate the term symbol of the ground state configuration of Cerium.
The electron configuration is: $[Xe] 4f^1 5d^1 6s^2$
Since the d and f electrons are in different subshelles, they are non-equivalent electrons and Pauli exclusion principle doesn't apply, so:
$l_1=2$;$l_2=3$
using addition of angular momentum:
$L=5,4,3,2,1$ and $S=1,0$
Using Hund's rules, the ground state term symbol should be the one with $S=1$ and $L=5$, this is $^3H$.
However, NIST lists $^1G$ as the ground state.
Any help would be really apreciated.

Iván, you are right on your determination using Hund's rules. However, for heavy elements, with strong spin-orbit coupling, the LS coupling assumed by Hund's rules may not apply. In this case, the j-j coupling may give better agreement with experiment.
There are extensions of Hund's rules for j-j coupling [1].
If you check NIST Atomic Spectra Database Levels Data, you will see that the wavefunction for the ground state of the Ce atom is 55% $^1\text{G}^\circ$ and 29% $^3\text{H}^\circ$. This mixture between singlet and triplet is due to the spin-orbit coupling.
[1] T. E. H. Walker, J. T. Waber, Modified Hund’s Rule for jj Coupling. Physical Review A. 7, 1218–1224 (1973). https://dx.doi.org/10.1103/PhysRevA.7.1218


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Do all metal salts in aqueous solution contain metal aquo complexes?
I know that nickel nitrate in aqueous solution contains the metal aquo complex ion $\ce{[Ni(H2O)6]^2+}$. But do all metal salts in aqueous solution contain such complex ions? If not, on what factors does the formation of metal aquo complex ions depend? Is the metal required to be a transition metal?

Hint: If you dissolve an ionic substance in water, the total solution enthalpy has several components. One of them is the ... ?

@Karl sorry , I'm not a chemist at all. I'm asking because I read about it and am interested in it, and trying to understand if complexes always appear or if it's a special thing

Your question is awfully chemically specific, for a non-chemist. ;-) What I meant is the hydration enthalpy of each of the ions involved.

Umm, you're underestimating things there. Even in nickel nitrate there's also gonna be some coordination by nitrate, and with other "salts" it's usually more pronounced. In general there's no such thing as bare monoatomic cation  (or even polyatomic, to some extent) outside of stuff like plasma.

@Karl I'm an electrical engineer who had a bit of chemistry in university and watches chemistry videos. I watched a video on transition metal complex chemistry, and the person used nickel nitrate to make nickel carbonato complexes using ammonia starting from the nickel aquo complex. This made me wonder if all metal salts contain these complex ions. I don't know what else to tell you, I just want satisfy my curiousity, and this is diffcult to google for. I don't even know what hydration enthalpy is, I am sorry my knowledge is super spotty and not academic in nature.

OK, this is really hard to explain in a few sentences. Every ion in water is surrounded by shells of water molecules. Those that are closest are bound more strongly,  those in a "second shell" less so, etc, The further out, the more they are like pure water, and exchange freely. Where to draw the line is somewhat arbitrary. Typically some transition metals can include electron pairs from e.g. water into their own electronic structure, these classic "complex bonds" are usually stronger than what an alkali metal has in its hydrate shell. Of course your Ni complex ion has another hydrate shell.

[OP] But do all metal salts in aqueous solution contain such complex ions?

This is a good question and in many cases, the exact structure and dynamics of aqueous metal ions is still being studied. In the simplest terms, if something is in aqueous solution, it will be surrounded by water. To start getting some more details, take a look at the Wikipedia article on metal aquo complexes. It says that metals surrounded by a first shell of six waters (with an octahedral geometry) is very common. They also point out that some of these complexes are long-lived while others are not:

Rates vary over many orders of magnitude. The main factor affecting rates is charge: highly charged metal aquo cations exchange their water more slowly than singly charged species. Thus, the exchange rates for $\ce{[Na(H2O)6]+ and [Al(H2O)6]^3+}$ differ by a factor of $\pu{e9}$. Electron configuration is also a major factor, illustrated by the fact that the rates of water exchange for $\ce{[Al(H2O)6]3+ and [Ir(H2O)6]^3+}$ differ by a factor of $\pu{e9}$ also.[3] Water exchange usually follows a dissociative substitution pathway, so the rate constants indicate first order reactions.

Type of metal

[OP] Is the metal required to be a transition metal?

For sodium ions in aqueous solutions, just calling them $\ce{Na+(aq)}$ is probably sufficient because the waters change places so rapidly. For a nickel complex with waters in the first shell not exchanging that quickly, it makes more sense to talk of a aquo complex and to study its geometry.
Other coordination numbers and geometries
If you dig deeper, you will encounter other coordination numbers and geometries. For example, according to this report, copper(II) likes to surround itself with five ligands in a square pyramidal fashion. Here is an excerpt of the abstract:

Cu K-edge extended X-ray absorption fine structure (EXAFS) and Minuit X-ray absorption near-edge structure (MXAN) analyses were combined to evaluate the structure of the copper(II) imidazole complex ion in liquid aqueous solution. [...] This core square-pyramidal motif has persisted through [Cu(H2O)5]2+, [Cu(NH3)4(NH3,H2O)]2+,(1, 2) and now [Cu(Im)4Lax)]2+ and appears to be the geometry preferred by unconstrained aqueous-phase copper(II) complex ions.

I think for geometries that are not the optimal space-filling ones, you need transition metals, but I have not done a careful survey of the available literature.

I would note that in some compounds dissolved in aqueous solution, such as $\ce{K3[Fe(CN)6]}$, some metal ions will stay coordinated by ligands other than water. Of course, these are probably not the salts the question was talking about. However, we should not let the notion arise that aqueous solution equals aqua ligands.

In certain cases the metal cation is complexed (at least predominantly) with a soft-base solute that is also in solution.
As a graduate student, I had some copper (II) salts and thiourea in the lab, and I mixed them together in water solution to see how thiourea might chemically alter a copper electroplating bath (where the thiourea is used an an additive).
Thiourea + copper(II) sulfate --> solution turned colorless, copper (II) was reduced to copper(I)
Thiourea + copper(II) chloride -> white precipitate, copper(I) chloride sparingly soluble when only an aquo complex could be formed; but more thiourea gave a clear solution indicating copper(I) redissolved as a thiourea complex.

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Wave function of a harmonic oscillator
In a question it is given that consider $H-X$ chemical bond and find $H-X$ bond distance for which there is zero probability density of finding the proton. It is easy to solve we have just to calculate the position of nodes of wave function of corresponding harmonic oscillator.
But my question is does the wave function of harmonic oscillator gives the probability density of both electrons and protons of bonded atoms which behaves as harmonic oscillator? As while I was reading the chapter, initially I thought that $\psi$ only gives the probability amplitude of electrons if we extend harmonic oscillator concept to bonded atoms.

Hint: The vibration describes the motion of H and X. We haven't mentioned electrons at all. So what does that tell you about what the harmonic oscillator wavefunction is telling you in this case?

The wave function of two particle oscillating from their equilibrium position(having minimum PE) tells us about the probability amplitude of finding the oscillating particles. But the question which I describe firstly, ask specifically about proton, if it ask for H them it make sense. As two bonded atoms also make a harmonic oscillator system as the get stretched and compressed about their mean position. So does in this case if we use same $\psi$ as that of Harmonic oscillator then it tells specifically about the oscillating atoms or their electrons and protons?

Complicated problem ultimately - look up Born-Oppenheimer approximation and see if that helps you, strictly there is no such thing as a vibrational wavefunction for a molecule, it is all coupled up together, but it is normally a good approximation to facto the wavefunction into terms that a human can understand, and within  that approximation it only describes the motion of the nuclei. Further for H-X for heavy X almost all the motion is the H, so we can further approximate and say X is static, and not the wavefunction is (assuming springs connect the atoms) for SHM of the H atom

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Benzylation and acetylation of amines and alcohols
Question

Attempt
Since the nitrogen site is more nucleophilic it should react first. But still Oxygen can react and hence I believe A should be right. But the book says C. Where did I go wrong? Please help.

Well, this 100 deg. is to suggest the reaction is done is rather selective way.

But does temperature affect the reactivity of OH?

Phenol is a pretty good leaving group so phenyl acetate is an acetylating agent and fairly labile in aqueous base. The reaction conditions are not clear enough IMO to say whether you would get a or c, but with a basic work-up the OAc can hydrolyse back to phenol. If the amount of Ac2O is limited then any PhOAc formed will acetylate the aniline giving back PhOH.

Why would the OAc hydrolyse? Is it that weak? Also can PhOAc undergo intramolecular acetylation of Nh2? I have never read anything like that?

Phenol is a fairly strong acid hence its anion is stable which makes it a good leaving group. I think the transacylation is more likely intermolecular. You are on this site to learn new stuff aren't you?

@Waylander Sir, why carbonyl group is not producing schiff bases by nucleophilic addition as it happens with aldehyde and ketones?

Is it because medium is basic?

Because it is not a ketone or aldehyde, it is at a different oxidation level. If you want to form an amidine you need different chemistry.


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Adiabatic isobaric irreversible process
I cannot understand that

in an irreversible adiabatic isobaric process the enthalpy change is $ \Delta H <0$.

Could anyone please help me how to prove it.
Any help is appreciated.

For an adiabatic isobaric process the work done by the system during a spontaneous expansion is $w= -p_{\text{ext}} \Delta V $.  Since the heat $q=0$ the work is also equal to the change in internal energy, ie $w= \Delta U$. The associated enthalpy change is then
$$\begin{align} \Delta H &= \Delta U + \Delta (pV) \\ &= w + p_fV_f-p_iV_i \\ &=  -p_{\text{ext}}\Delta V+ p_{\text{ext}} V_f-p_iV_i \\ &=  p_{\text{ext}}V_i-p_iV_i\\&=(p_{\text{ext}}-p_i)V_i \end{align}$$
In order for the expansion to proceed spontaneously the initial pressure of the system $p_i$ must exceed that of the surroundings $p_{\text{ext}}$, which means $\Delta H<0$.
Since you are computing $H$, a property of the system, when computing the change in $H=U+pV$ you should use the pressure of the system, not of the surroundings. Before the expansion the pressure of the system (and surroundings) is $p_i$, while at the end of the expansion it is $p_f=p_\text{ext}$, since the system is then again in mechanical equilibrium with the surroundings. Note that previous to the expansion the system and surroundings are also in mechanical equilibrium. The pressure of the surroundings is then suddenly dropped to a lower constant value. The following figure illustrates the process:

Thanks but I've few doubts. If the process is isobaric then Pf and Pi are equal also you have written v that delta(pv)= PfVf-PiVi however i think delta(pv)= p del (v) -v del(p),since its isobaric so del(p)=0,which reduces delta(pv) to p dv. Where am i going wrong.

@YasirSadiq Please see my commentary at the end of the answer.

Thank you again, but since it's stated that the process is isobaric how come the pressure changes from Pi to Pext?

@YasirSadiq I edited my answer again, hopefully it is clear now.


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How value of energy of elecron becomes more negative and radius of orbit decreases with increase in the value of atomic number?
I know how energy of elecron and radius of orbit depends upon principal quantum number......But how it depends upon atomic number?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumblestone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Atomic number is proportional to the atomic kernel charge. From the classical electrostatics we know that potential around a charge $Q$ is:
$$U=-\frac{Q}{4\pi \epsilon_0 r}$$
So it's obvious it the electron energy must depend on this charge.
From Schroedinger wave equation we can see the charge is involved:

$$ \psi_{n\ell m}(r,\theta,\varphi) = \sqrt {\left (  \frac{2}{n a_0} \right )^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^\ell L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^m(\theta, \varphi ) $$
where:

$ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_q q^2}$ is the Bohr radius,

$ L_{n-\ell-1}^{2\ell+1}(\cdots) $ are the generalized Laguerre polynomials of degree $ n - \ell -1 $.

$ n, \ell, m $ are the principal, azimuthal quantum and magnetic quantum numbers.

The Rydberg formula for hydrogen

....can be extended for use with any Hydrogen-like (1 electron only) chemical element with
$$\frac{1}{\lambda_{\mathrm{vac}}} = RZ^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
where $R$ is the Rydberg constant and $Z$ is the atomic number.

For multielectron atoms, things get complicated by shielding of kernels by inner electrons and by mutual electron repulsion.

Can u pls answer in a simple statement form....

I would turn the table: Can you ask by not a simple sentence question ? Do your figural homework first.


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How to calculate the variation of enthalpy in a reaction that does not occur in standard conditions?
Standard enthalpy of formation is a value that represents the enthalpy variation in the formation of compounds with the purpose of using this values to know the enthalpy of a reaction where these compounds participate.
Those values are referred to the formation of said compounts at 273 K and 1 atm (known as standard conditions).
My question is:
Does the value change, for distinct pressures or temperatures? If yes, how?
Why am I asking this?
Let's suppose I have a reaction that goes like this:
$A + B \rightarrow C$
At 1 atm and 273 K (standard conditions).
I can say that in this reaction $\Delta H = Hf^0(C) - Hf^0(A) - Hf^0(B)$
Where $Hf^0$ means standard enthalpy of formation.
Because the reaction happens at 1 atm and 273 K.
But, what if the reaction happens at 3 atm and 350 K?
I asked my University teacher and he said: "You don't need that because we will always be working at 1 atm and 273 K".
What tool do I have to know the variation of enthalpy in this case?
A small extra confusion is the fact that in some examples I see standard conditions being 1 atm and 273 K, and some other times it's at 298 K.

Not my downvote, but generally on the site you are expected to show what you have tried to solve the problem. This not only avoids answers repeating information that is readily available elsewhere (or is already here on the site), but it also helps us better understand what exactly you want clarified. In regards to your question, this previous answer may get you started on the relationship of enthalpy and pressure https://chemistry.stackexchange.com/a/73226/41556

@Tyberius Thanks, I can understand the point. Maybe I did not make the question clear enough, but I am surely not the kind of person who just asks without researching first. Before I asked, I really did my research, but have not been able to find an answer. I updated the question's title and content to try to make it more accurate. Your linked answer is good information and I thank you for it, but I still cannot undertsand how I would face the problem I mention in the question. Always appreciate the effort people do on StackExchange, probably the most useful websites in the net.

What do you mean by "tool"?

@BuckThorn Thanks for your attention. By "tool" I mean a set of true affirmations accompained by some math, to be able to get the result I am looking for. I updated my question to make clearer what it is that I want to calculate. (I saw your edit to your answer, I am trying to dive into it and learn from it).

Yes, standard states can vary (see e.g. these wikipedia pages. The important thing is to know what standard state is associated with the data which sometimes requires a little care.

For a pure substance you can write generally that the enthalpy depends on pressure as follows:
$$\left(\frac{\partial H}{\partial p}\right)_T = V - T \left(\frac{\partial V}{\partial T}\right)_p \tag{1}$$
If you evaluate this for an ideal gas you find that  $\left(\frac{\partial H}{\partial p}\right)_T = 0$. If the substance is present in a condensed phase with a density (or molar volume) that is weakly dependent on the temperature, then roughly
$$\left(\frac{\partial H}{\partial p}\right)_T = V$$
If you speak of standard enthalpies then by definition there is no pressure dependence, as they are defined at a particular pressure.
If your reaction involves gases which can be assumed ideal, then you can use the standard enthalpy for all reagents and products, since according to my answer the enthalpy of an ideal gas is independent of pressure. On the other hand if it involves condensed phases you need to evaluate the molar volume and perhaps also the thermal expansion coefficient $\alpha$, since Eq. (1) can be rewritten as
$$\left(\frac{\partial H}{\partial p}\right)_T = V (1- T\alpha)$$
For the enthalpy of reaction you might write
$$\left(\frac{\partial \Delta_r H}{\partial p}\right)_T = \sum_i \nu_i V_{m,i} (1- T\alpha_i)$$
where $\nu_i$ are stoichiometric coefficients. In the summation in the rhs account should be taken for sign changes depending on whether species are reactants or products.
If you want to evaluate the enthalpy of reaction at one pressure given the value is known at another pressure, and the molar volumes are known, you can integrate the previous equation, giving (at constant T)
$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + \int_{p^\circ}^{p} \sum \nu_i V_{m ,i}(1- T\alpha_i) dp$$
where $p^\circ$ is the reference pressure at which the enthalpy is already known. This last expression may be simplified in various ways depending on the state of the species and the pressure integration range. If the molar volumes and expansion coefficients are approximately constant in the pressure range then
$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ) \sum \nu_i V_{m,i} (1- T\alpha_i)$$
If the species have negligible thermal expansion coefficients then
$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ)\sum \nu_i V_{m,i}$$
and so on. For instance, for a reaction converting a solid between two forms $\text{react}$ and $\text{prod}$ having constant molar volume and negligible thermal expansion coefficients in the pressure range $p^\circ$ to $p$,
$$\ce{\text{react}<=>\text{prod}}$$
$$\Delta_r H(p)  =  \Delta_r H(p^\circ) + (p -p^\circ)(V_m(\text{prod})-V_m(\text{react}))$$


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The effects of acetic acid on porous concrete brick and mortar joints
I mistakenly applied a patio cleaner containing acetic acid to my porous concrete facing brick thinking it was biocide. The concentrate solution has a pH of 3 and I diluted it to one part concentrate and three parts water.
About three hours later, having realised my mistake, I flushed it with water. I repeated this process the following morning and also on the morning after that. The bricks now appear to be dry. I have been told that the pore structure of the type of bricks I have is vital in protecting them from frost attack.
I am very concerned that the acid has damaged the brick and mortar joints. I am wondering if the acid has now completely evaporated — one day after the last rinse. Is there anything more I can do to neutralise any residual acid? I would be very grateful for any information for someone who has little understanding of chemistry.

Usually bricks are made of impure aluminum silicate. This substance is not attacked by acids, even if the acid is stronger than the acetic acid. The only condition for them to be attacked by an acid is the calcium carbonate (chalk) that they may contain as an impurity. But you will soon know that it is the case, if you observe the surface of your brick after having applied the acid. If there is some foam or froth on the brick, it means that there is some chalk in the brick. But don't worry. The effect will soon disappear if you solution had a pH 3, or even more. The result is negligible.

I've heard that vinegar is a mold killer, so maybe you did your bricks a good thing. If they were susceptible to acetic acid, I suppose acid rain would be terrible!

Maurice, is calcium carbonate not one of the ingredients of Portland cement, which is used in making concrete bricks and the mortar in which they are bedded?

@Brendan Little. Calcium carbonate is used to produce Portland cement. But when heated with clay, it loses its Carbon atoms as $\ce{CO2}$ and is transformed into a calcium  aluminosilicate. So there is no calcium carbonate any more in the cement

The quantity of acetic acid will dissolve very little brick before it is consumed.  An application or two will dissolve very little brick.

Per Wikipedia on Fired Bricks, to quote their composition;

Normally, bricks contain the following ingredients:[21]
Silica (sand) – 50% to 60% by weight
Alumina (clay) – 20% to 30% by weight
Lime – 2 to 5% by weight
Iron oxide – ≤ 7% by weight
Magnesia – less than 1% by weight

I would expect slowly, with time, that the acetic acid could attack all of the above with the exception of Silica.
Flushing with water would reduce the magnitude of the attack, but I would not categorically claim that the brick structure is completely sound.

Is acetic acid not totally miscible with water and, therefore, would evaporate at roughly the same rate? Even if the acid evaporate a a slightly slower rate, I understand that this type of acid evaporates completely leaving no residue or crystals. Surly copious dilution with water would cause its corrosive effects to be negligible during the evaporation precess?

"Contain" is a wrong word to use. These things could have been taken initially, but when heated together, they form something different.

Upvote for your last sentence.  Muriatic acid is used to clean bricks, but acid readily attacks mortar.  Try adding baking soda to the water flush, and rinse until no "fizzing" is observed.


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Desilylation with potassium fluoride and methanol
Recently, I have done following reaction with KF adn methanol:

I don't understand why the sp-carbon silicon bond breaks and why no other sp3 carbon silicon bond breaks.
Thanks in advance.

Which is more stable: $\ce{RC#C^-}$ or $\ce{H3C^-}$?

Does this answer your question? Desilylation mechanism with fluoride pay particualr attention to the last paragraph

So the sp carbon-silicon bond breaks because it forms a more stable anion than the other sp3 carbons?

That is exactly it.


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Reactivity of alkyl halides
I am unable to understand how to decide reactivity of alkyl halides based on inductive effect in a nucleophilic substitution reaction. I have 2 explanation for deciding. Consider 2 alkyl halides:

(A) 2-chloro-2-methylbutane
(B) 2-chloro-3-methylbutane

First explanation:
In (A) we have a $3^\circ$ carbon so if we break the $\ce{C--Cl}$ bond a $3^\circ$ carbocation will be formed, which is stabilized by +Inductive effect of surrounding  groups. Therefore we can say that as  breaking of $\ce{C--X}$ bond in (A) leads to the formation of a more stable carbocation than that formed in (B){$2^\circ$ carbocation}. And due to this the reactivity will be more of (A).
Second explanation:
As in (A) $3^\circ$ carbocation is formed, it will get more stabilized than $2^\circ$ carbocation formed in (B). So this means that $2^\circ$ carbocation in (B) becomes more positive than $3^\circ$ carbocation in (A), making it easier for a nucleophile to attack on (B)'s carbocation. This would make (B) more reactive.
Which one of them is correct?

Assuming your concern is on $\mathrm{S_N1}$ type nucleophilic substitution reactions, you are in the correct path with your first explanation. However, everything has gone south from that point on. As you found out, the stability of carbocations in $\mathrm{S_N1}$ reactions are: $3^\circ \gt 2^\circ \gt 1^\circ$. However, reaction rates of $\mathrm{S_N1}$ reactions do not depend on electron density (or lack of it) on carbocation, because rate determine step is rate of carbocation formation step:

As depicted in the diagram, $3^\circ$-carbocation from 2-chloro-2-methylbutane forms faster than $2^\circ$-carbocation from 2-chloro-3-methylbutane because the stability of $3^\circ$-carbocation is much larger than that of $2^\circ$-carbocation as you predicted in your first explanation. Also, you need to understand that in these reactions, 1,2-hydride (or methide) transfer to give more stable $3^\circ$-carbocation from $2^\circ$-carbocation is possible if there are available $\alpha$-hydrogen or $\alpha$-methyl group in the sought $2^\circ$-carbocation, alike in the case of 2-chloro-3-methylbutane here. Regardless of this rearrangement, the rate of carbocation formation of 2-chloro-2-methylbutane is still faster than that of 2-chloro-3-methylbutane, because it is strictly the stability of $3^\circ$- versus $2^\circ$-carbocations.
The second reaction (attack of the nucleophile to available carbocation) is very fast compared to chloride dissociation, thus it is not affecting the overall rate of the reaction. Thus, the product formation from 2-chloro-2-methylbutane is faster than the product formation from 2-chloro-3-methylbutane, and hence, 2-chloro-2-methylbutane is more reactive than 2-chloro-3-methylbutane.

@Ameya M: Good catch.

In tertiary alkyl halides,
carbocation formation occurs and reaction proceeds via SN1 Mechanism. Formation of carbocation is a slow step. It requires energy,(endothermic).
Carbocation as well as the leaving group (halide ion) is stabilised by solvation by a polar protic solvent
 Carbocation formation is ofcourse the Rate Determining step which means that stability of carbocation determines the rate at which reaction will proceed and not better positive centre for nucleophilic attack. Nucleophilic attack will be very fast and won't affect the rate of reaction

And in the said secondary alkyl halide it depends on nucleophile whether it will proceed via SN1 or SN2 mechanism as secondary alkyl halides are less hindered and use of a strong nucleophile will favour SN2 mechanism.


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Indirectly Measuring Diffusion of NO2 into Water
I am working on a data science project with the central hypothesis is that air-born pollutants can become water-born through diffusion. The example I am using as a proof of concept is $\ce{NO2}$. The problem I am having is that I do not understand the chemistry of how it interacts with $\ce{H2O}$ outside a laboratory setting that well.
$\ce{NO2}$ reacts with $\ce{H2O}$ to form $\ce{HNO3}$, $\ce{HNO2}$, $\ce{H+}$, and $\ce{NO3-}$. In addition, $\ce{HNO2}$ also reacts with water to form $\ce{NO}$, $\ce{H+}$, and $\ce{NO3-}$.
Do these reaction happen so fast that measuring dissolved $\ce{NO2}$ or $\ce{HNO2}$ in a sample an unreliable measurement?
If so, what is the best measurement I can use? $\ce{H+}$ and $\ce{NO3-}$ are both part of a lot of other very complicated biological pathways and have even more possible sources, so there is no way they can be relied upon. That leaves $\ce{HNO3}$, $\ce{HNO2}$, or $\ce{NO}$ as possibilities. Which of those would be the best to use as an indirect measurement?
Thanks for any help you can offer in advance. I am decent enough at chemistry in a lab, but application to real world systems is way beyond my abilities.

https://pubs.acs.org/doi/pdf/10.1021/ie50552a048

NO2 is a free stable radical, better express as $\ce{.NO2}$, and being a radical, does not directly react with H2O, or H+ or OH-.
The reaction mechanism likely proceed via a self-reaction creating nitrogen tetroxide:
$\ce{.NO2 + .NO2 <=> N2O4 }$
In a diluting medium of air, depending on the concentration of nitrogen dioxide, this reaction is kinetically expected to be slow.
When it does occur, it can be followed by:
$\ce{N2O4 +  H2O -> HNO2 + HNO3 }$
Source: See Reactions (9) and (10) on Page 28 of this paper.
Now, one can ignore the intermediate reaction mechanics, for simplicity, however, this does present a problem for some, as yourselves, as clearly the rate determining step is not even cited.
Also, if you review the paper, there are other reaction equilibriums that you may have to take into consideration. As a sidebar, what the paper does not mention is that aqueous nitrite is a photo-sensitive producing products, so you may wish to control the amount of lab light exposure as well.


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Synthetic route to a spirocyclic amide
I have a task to come up with a synthesis of the following molecule (2-azaspiro[5.5]undecan-3-one), but no starting materials have been explicitly specified:

This is my idea. I think that all the steps shown are quite reasonable.Do you have any critics/suggestion?

Look up quaternary carbon (https://en.wikipedia.org/wiki/Quaternary_carbon#Synthesis) or Spiro Compound (https://en.wikipedia.org/wiki/Spiro_compound#Preparation)

The aminomethylene grouping is a major signal for the retro-analysis. that just screams "I started life as a nitrile".

@Waylander: Indeed! Can hear it on this side of  the pond.

A quick note - although it wasn't explicitly stated, here on SE we generally expect people to show their thoughts on the question at hand, or at least engage in some way. The downvotes + question closure were almost certainly because that wasn't evident at the beginning. Your edit is just what was needed, so please keep that up with questions in future (I have reopened this one and tidied it up a little bit). Btw, please don't repost questions; just edit them and they will pop back up to the top of the front page.

I think your reaction of the hydroxy acid with HCN/KCN more likely to give you the lactone

I think this can be done in 2 steps though I can't find a literature example.

Alkylate the anion of the commercially available cyclohexane carbonitrile (cyclohexyl cyanide) with ethyl acrylate.
Hydrogenate the resulting cyano-ester over Pt to reduce the -CN to -CH2NH2 which with spontaneously cyclise to give the required product.


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What is the substance turning salty water green when electrolysed?
I am a student and I take O.L. Chemistry, in my knowledge I know if I electrolysed concentrated aqueous NaCl solution I should get Hydrogen evolved at the cathode and chlorine at the anode.
I did an experiment, I add a lot of table slat to tap water to get the electrolyte ready (aq. NaCl) then I used graphite electrodes (just got it from a mechanical pencil) and passed electricity through the salty water. I used of course direct current using a 6 Volts battery I got.
Well my suggestions were half correct, Hydrogen really evolved and bubbled violently at the cathode, but at anode I didn't see any yellowish green gas indicating presence of chlorine, yet only colorless bubbles continue to evolve from the anode but was less violently than the cathode. I thought this is Oxygen evolving so I increased the concentration of solution by adding table salt but also only colorless bubbles evolves at the anode. Well I didn't really calculated how much precisely I should add salt but I added insanely a lot of table salt.
Only to discover after around 5 minutes of electrolyzing the salty water, it became darkish green in color, I didn't really knew what is this and this is my question, I used graphite electrodes so those doesn't react like if I used iron, the solution was just nice tap water and not really bad or muddy. My only suggestion is that this is actually chlorine dissolved in water ... So is this correct ?? I knew that chlorine is soulable in water but I didn't thought it will turn the whole water greenish in color....
Can anyone help me in that ? And please if you're going to answer me, answer me as if I am somebody who knows nothing in chimestry .....
Thanks in advance.

@Mithoron  unfortunately no, it doesn't and I already saw this answer, I am using graphite (carbon) electrodes there is no other ions except Na ions and Cl ions and H and OH there is no copper or any other metal.

Your "graphite" can be leaking who knows what, then. Maybe even copper ;)

https://chemistry.stackexchange.com/questions/90529/what-is-the-greenish-substance-formed-after-the-electrolysis-of-two-graphite-ele

And your cables are still copper, aren't they?

@Mithoron well yes they are copper cables connecting the graphite electrodes to the battery ....

I think this is some metallic impurity coming from the copper cables or metallic cables. Show a picture of the set-up. Do you see corrosion on the copper cable? You can also try a different brand of pencil lead and see what happens.

@M.Farooq I thank you very much for the advice .. I will try another pencil lead and see . And will try to attach image of my setup If I can.

Also. it could be chlorine dissolved in $\ce{NaOH}$ solution. That means the solution could be bleach. See whether it bleached out color papers.

A very similar question with a suggested answer (paralleling mine) involving a ferrous salt formation was asked here https://chemistry.stackexchange.com/questions/35381/green-color-substance-forming-during-electrolysis .

My speculation is based on the words specifying the use of 'tap water' during the electrolysis of aqueous sodium chloride.
In particular, the interaction of formed chlorine and existing ferrous bicarbonate (found in tap water) could form small amounts of aqueous FeCl2. [EDIT TO QUOTE SOURCE]

Bacterial Iron in Well Water – Iron bacteria is usually identified by slime in places such as toilet reservoirs or by the presence of a slimy mass fouling softeners or filters.
Ferric Iron in Well Water – Also known as red water iron, ferric iron in well water is essentially clear water iron that has been exposed to oxygen – usually from the air, thereby oxidizing.  Carbon dioxide leaves the water and the oxygen combines with the iron to form ferric ions (Fe+++).  This gives the water a red rust coloring.
Ferrous Iron in Well Water – Often called clear-water iron because it is clear when poured, this substance is found in water that contains no oxygen. Typically, it comes from deeper wells and groundwater sources. Carbon dioxide acts on iron in the ground to form soluble ferrous bicarbonate. In water this forms ferrous ions (Fe++)

Per Wikipedia on Ferrous chloride, to quote:

The compound is highly soluble in water, giving pale green solutions.

Repeat the electrolysis experiment with distilled water and see if you get the same coloration issue.
[EDIT][EDIT] Interestingly, low solubility Fe(OH)2 is clear and only acquires a greenish tint in the presence of oxygen (per Wikipedia, to quote: "Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge.). Further, well-water is associated with high levels of bicarbonate and low O2 levels (see this significant work). To claim that Fe(HCO3)2 is at all green is without basis, as I also could not find any source anywhere asserting a coloration, other than that it is, in effect, 'clear' as was claimed above per the comment: 'Often called clear-water iron because it is clear when poured'. Theoretical formation of the usual hydration sphere for Fe(OH)2 interacting with bicarbonate in low solubility conditions with high HCO3-, low O2, significant CO2 presence (and its associated partial pressure) suggest reasons why it may not be a valid argument. If I am wrong, please provide a single source citing color, else it should be likely assumed that my supposition is correct and Fe(HCO3)2 is clear.  By the way, Atomistry on Ferrous bicarbonate, to quote:

Ferrous hydrogen carbonate or ferrous bicarbonate, FeH2(CO3)2, exists in solutions obtained by dissolving ferrous oxide or carbonate in water charged with excess of carbon dioxide. On exposure to air carbon dioxide is evolved, and a precipitate is obtained consisting essentially of ferric hydroxide.

which one could interpret as that clear-water iron, upon exiting the pipe, liberates CO2 and forms Fe(OH)2, which, with further oxygen exposure, may display a green tint (or, per the introduction of electrolysis formed chlorine).

Cl2 + H2O = HCl + HOCl.  Fe(HCO3)2+ 2HCl = FeCl2 + 2 H2O + 2 CO2. Further, HOCl will react in a fenton-type reaction with Fe(2+). Namely: Fe(2+) + HOCl = Fe(3+) + .OH + Cl-

I have EDITED my thread to make very explicit the sources of iron common to well-water. However, city water, with iron pipes, is also an apparent source of Fe ions even if the water is free iron-free sources. The decomposition of iron pipes is likely sourced via galvanic based corrosion .

I do thanks those who have doubts on the content of their water supply as I find the chemistry interesting and rich. I was personally shocked to find my water was sourced, per reading of my water bill, from 3 sources, reservoir, rivers, and well-water. The latter, I view, as more of a disclosure to explain the range of possible contaminants likely to be found in my tap water.

Note, my source reference to 'clear-water iron'  explicitly stated to be ferrous bicarbonate.

Rewriting my comment to be more clear and concise, since it was apparently not understood: If FeCl2 is the source of the green color, the color should be apparent BEFORE the electrolysis has begun, since the concentration of neither ion should increase during electrolysis. If anything, they will decrease because Fe(2+) may be oxidized to Fe(3+) and Cl- to Cl2.

Point 2 was that the ferrous ion itself provides the green color as an aquo complex even in the absence of chloride. The reason that ferrous bicarbonate is described as clear water is because the concentration is much to low to see the impact of the color, so the presence of iron is only apparent when oxidization to Fe(III) creates precipitates which cloud the water.

Sorry, don't believe it,  not regular conditions here, well-water exiting a pipe. See my cited sources and results of my research.


text:
Constructing a partition function for a gas-solid equilibrium in canonical ensemeble
For a closed system at fixed N,V,T there is a solid in equilibrium with its vapor. I'm told that I can minimize the helmholtz free energy to find that the equilibrium condition is given by
$N_g = \frac{q_g}{q_s} $ Where $N_g$ is the Number of particles in the gas phase, and $q_s$ and $q_g$ are the partition functions for the solid and gas phase respectively.
I'm also given that $Q_s = q_s^{N_s}$.
I've attempted to solve $ \frac{\partial A}{\partial N_g} = 0$ to show this.
Since N,V,T are fixed I know it's a canonical ensemble and that $A = -kTln(Q)$.
What I'm unsure on is how to find the partition function Q.
In the Atkins and DePaula textbook there's a brief example of factorizing the molecular partition function to isolate the different contributions from vibration, rotation, etc. I'm unsure if this applies here. The textbook starts from finding the total energy of the system from a sum of its parts. I don't intuit that a solid-gas equilibrium should follow the same logic but nevertheless I tried continuing assuming this would work so then
$Q = Q_sQ_g$
I first tried with the gas particles distinguishable (so I wouldn't have to deal with a factorial in the derivative) and found
$Q_g = q_g^{N_g},  Q_s = q_s^{N-N_g}, Q = q_g^{N_g}q_s^{N-N_g}$ (since the total number of particles N is constant, $N = N_g + N_s$)
which gives
$\frac{\partial A}{\partial N_g} = 0 = -kT(ln(\frac{q_g}{q_s})) => \frac{q_g}{q_s} = 1 $
I then tried with an indistinguishable partition function for the gas phase
$Q_g = \frac{q_g^{N_g}}{N_g!}$
I'm less familiar with this math so I relied on wolfram alpha for the derivative which gives
$\frac{\partial A}{\partial N_g} = 0 = \frac{\Gamma(N_g+1)\psi^{(0)}(N_g+1)}{N_g!} - ln(\frac{q_g}{q_s})$
Before I go any further, is this how I should construct the partition function? Is there a text online that someone could point me to?

If I wanted to study this I'd look at the solid-state hydrogen storage of metal hydrides. I know metal hydrides have been extensively studied for making hydrogen powered vehicles. One nice feature is that if the pressure tank is ruptured all of the hydrogen isn't released and once to create a massive BOOM! However the hydrides do release the hydrogen fast enough to provide sufficient fuel to the vehicle.


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If every radioisotope atom in your body decayed at once, would you explode? What are the odds of it happening spontaneously?
The title pretty much sums it all.

An atomic bomb explodes in about 1 microsecond which is $3.2\times 10^{-14}$ of a year.

Question titles should be like book titles. They distinguish the book from other books, give a hint about content, but do not tell the full story of the book. The content then should elaborate the topic to full depth the author is able to do.

Let's look at just the carbon-14 atoms.
Assune 18.5% of your body mass is carbon and you weigh 80 kg.  One carbon atom in a trillion is carbon-14.  Working out the resulting mass of carbon-14 atoms in grams, dividing by 12.01 g/mol and multiplying by Avogadro's Number leads to roughly $7.4×10^{14}$ atoms.
Carbon-14 has a half-life of about 5730 years.  Dividing $\ln 2$ by that figure you find that the probability of a specific atom of carbon-14 decaying in a year is 0.000176.  For all the carbon-14 atoms to do it within one year you're looking at roughly:
$0.000176^{7.4×10^{14}} \approx 10^{-2.78×10^{15}}$
And that's just the carbon-14.
Did I mention that only one carbon atom in a trillion has this potency?  Assuming the above scenario comes to pass with that proportion of atoms over the course of a year is going to make for a pretty wimpy explosion.

Oscar, I'm sorry for the delay in marking this answer as accepted, I completely forgot.


text:
Iron Out and CLR mixed - toxic gas?
Someone I know mixed the liquid forms of Iron Out and CLR (Calcium, Lime, & Rust Remover) together to clean out a plastic bathtub.
I have the SDS sheets for each:
https://images-na.ssl-images-amazon.com/images/I/61Dkhh0JXRL.pdf - CLR
https://summitbrands.com/wp-content/uploads/2018/03/Iron-OUT-liquid-English-GHS-Rev-01-29-18.pdf - Iron Out
The bathroom smells terrible, has produced a cough in the person using it, and now my eye burns a little from very brief contact. I was wondering what kind of gas (if any) was created when these two chemical cleaners were mixed?

Lactic Acid, Gluconic Acid, Lauramine Oxide for the CLR. Sodium hydrosulfite, Sodium carbonate, Sodium metabisulfite, Citric Acid, Sodium sulfite for the Iron Out which was actually the powder version rather than liquid version.

Most likely, sulfur dioxide. $\ce{SO2}$. was produced. The ingredients sodium hydrosulfite, $\ce{Na2S2O4}$, and sodium metabisulfite, $\ce{Na2S2O5}$, when mixed with acid, release $\ce{SO2}$, which oxidizes to $\ce{SO3}$ eventually.
$\ce{SO2}$ is quite irritating, but not immediately damaging in low doses, <100 ppm, though it can certainly induce a life-threatening asthma attack. "Sulfur dioxide is mildly toxic and can be hazardous in high concentrations. Long-term exposure to low concentrations is also problematic." $\ce{SO3}$ is even more dangerous, since it forms sulfuric acid when it meets water, e.g. in tears in the eye.
Air the room *thoroughly. There is not much else to do: theoretically, ammonia could be used to form a salt, removing the gas from air, but there is no way you could balance the reaction, leaving you with an excess of one irritant or another.

I will try my best to air it out, the room is closed off and only has a bathroom air exhaust fan, which is mostly just for removing moisture. Given that the Iron Out and CLR is approximately 15 years or older, I am hoping that they lost potency. In addition, not a lot of it was used, though enough to notice. I will try running a fan in the bathroom and a second fan in the hallway to the living room with open windows. Thank you for the help, hopefully this works.

At least SO2 gives ample warning -- it's not subtle. Once the annoying odor is gone, the room should be safe for long-term residency. My guess is that the exhaust fan should be effective in less than a day.


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Did I accidentally create phosgene gas
I accidentally mixed bleach and acetone trying to remove hair dye stains from my shower wall. The mixture turned orange when the bleach and acetone touched. Could I have created phosgene gas unintentionally? I’m really concerned for my families safety. I opened the window in the bathroom and closed the door. I poured water on the surfaces where the bleach and acetone mixture was a few times and all around the tub to make it sure there was none left. Should I be concerned or no?
I see the post was closed. I appreciate the feedback I received. Unfortunately poison control did absolutely nothing for me so that’s why I had to resort to online help. Thanks.

https://www.chemicalforums.com/index.php?topic=26330.0 this link might be of interest to you

Sodium hypochlorite (the active ingredient in bleach) and acetone react to produce chloroform.
Chloroform can slowly decompose into phosgene in air with exposure to oxygen
$$\ce{2CHCl3 + O2 ->[h\nu] 2COCl2 + 2HCl}$$
but not in significant quantities under the conditions you describe.
You did the right thing by opening the window for ventilation and isolating the fumes to the bathroom and rinsing with copious water.
It is best not to mix bleach with anything except water and especially not household cleaning products.

When someone says it SLOWLY turns into phosgene gas how slow are we talking? Could it be days, weeks or even hours? I’m moving from this house anyways so I really just want to make sure we are safe for the next 4 days till we move.

The quantity is negligible, especially with your thorough rinsing with water.

If the smell is gone, you should be fine. Be sure to ventilate well, as it seems you have done.

@Williams fyi the odor would be that of "musty" or cut hay (no i did not randomly go around sniffing it, that's just what it said here -- https://www.cdc.gov/niosh/npg/npgd0504.html )

Unfortunately my husband just weed whacked all around the property to get it ready for our move out so it smelled like that all inside. But it does not smell like it anymore since he weed whacked earlier so I have a feeling I didn’t accidentally create the gas. I’m still going to be on the lookout for symptoms of poisoning from the gas though.

So you’re telling me it’s possible to make a chemical weapon by mixing bleach and nail polish remover?!


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The equation representing the dissolution of cycasin
Is the following equation correct?
$$\ce{{C_8H_{16}N_2O_7}_{(s)} + H2O_{(l)} <=>  {C_8H_{16}N_2O_7}_{(aq)}}$$

Usually solubility is not represented by chemical equations. But most chemists will understand it and perhaps be okay with it.

Balanced equation: $$\ce{C8H16N2O7(s) <=>  C8H16N2O7(aq)}$$

Water Solubility of cycasin is $\pu{56.6 g/L}$ (it is a glucoside). Thus, it is fair to use a forward arrow instead of equlibrium sign.

Yes but I want to say it is an application of solubitity equlibria, that when the cycasin solution is washed away, the equilibrium is shifted to the right, producing more cycasin solution.

@M.Farooq Solubility was taught partly in terms of chemical equations to me in highschool, college, and university. That doesn't give or take merit to the practice, but problem or not, it has been systemic in my experience.


text:
Which one will undergo sn2 most easily?
[!well i thought that sterically least hindered compound will undergo sn2 most easily but in this case it the compound having benzene ring is undergoing sn2 more easily although it will repel the attacking nucleophile due to its electron dense nature.why isnt that happening here??][1]][1]
[1]: https://i.stack.imgur.com/28e4y.jpg

@Anket Aulakh Check this link: https://chemistry.stackexchange.com/questions/48872/how-are-sn2-transition-states-stabilised-by-adjacent-double-bonds-and-carbonyl-g

https://chemistry.stackexchange.com/questions/87827/why-are-alpha-carbonyl-halides-most-reactive-towards-sn2-reactions


text:
Ionic compound formation
Why do oxygen atoms differ in number in the compund Aluminate and Arsenate even they have the same suffixes ... Where Aluminate has 3 oxygen and Arsenate has 4 oxygen atoms.

The suffix -ate depends on the relative number of Oxygen. It does not depend on the absolute number of O atoms. This was decided a lot time ago,  in the beginning of the 19th century :
An acid must contain at least one H atom, and its symbol $\ce{H}$ is written first. It reacts with any hydroxide giving a salt and water.

If the acid contains only one other element X, its name would be hydro-X-ic acid, and its salts are named X-ide. Example :
$\ce{HCl}$ is hydrochloric acid, and its sodium salt $\ce{NaCl}$ is named sodium chloride.
Acids containing $\ce{H}$, $\ce{1 X}$ and $\ce{O}$ atom have a name depending on the number of different acids known containing these elements.

2a. If only one acid formula is known with $\ce{H}$, $1$ $\ce{X}$ and $\ce{O}$, its name is X-ic acid, and its salts (or ions) are  X-ate. Today, the criteria of the "oxidation number" of $\ce{X}$ has replaced the "number of Oxygen atoms included in the formula".
Ex.: Only one acid formula is known containing $1$ $\ce{C}$ atom : $\ce{H_2CO_3}$. This is the carbonic acid, and its sodium salt $\ce{Na_2CO_3}$ is sodium carbonate : its name does not depend on the number of $\ce{O}$ atoms in the formula of the substance.
This also the case of "your" aluminate. As Aluminum has only one oxidation number, namely +3,  the ion $\ce{AlO_3^{3-}$ is called aluminate
2b. If two acids are known with H, $1$ $\ce{X}$ and $\ce{O}$, the acid with the highest number of $\ce{O}$ atoms (today : the highest oxidation number of X) is named after the rule $2$a. The acid with the least number of $\ce{O}$ atoms is named by changing -ic to -ous, and its salts by changing -ate by -ite. Today, the concept of "oxidation number" has replaced the number of Oxygen atoms.
Example. In the beginning of the 19th century, only two acids were known with Nitrogen or Phosphorus, namely $\ce{HNO_3}$ and $\ce{HNO_2}$, or $\ce{H_3PO_4}$ and $\ce{H_3PO_3}$. Well ! $\ce{HNO_3}$ was called nitric acid, and $\ce{HNO_2}$ was called nitrous acid. And of course $\ce{H_3PO_4}$ was called phosphoric acid and $\ce{H_3PO_3}$ was called phosphorous acid. Their salts are called nitrate, nitrite, phosphate and phosphite.
This is the case for "your" Arsenic, which may have oxidation number +3 or +5. With +5, it produces the ion arsenate $\ce{AsO_4^{3-}}$, where the number of Oxygen atoms is not directly related to its name.
2c. If there are more than $2$ different acids with $\ce{H}$, $1$ $\ce{X}$ and some $\ce{O}$ atoms, prefixes are used for the extremes. The acid with the biggest number of $\ce{O}$ gets prefix hypo- for the smallest and per- for the highest.
Example. Four acids are known if $\ce{X}$ is Chlorine : $\ce{HClO, HClO_2, HClO_3, HClO_4}$. Their names are respectively : hypochlorous, chlorous, chloric and perchloric acid Their sodium salts are called hypochlorite, chlorite, chlorate and perchlorate.
There exists other prefixes for the cases where the acid contains more than 1 $\ce{X}$ atom, and for the case where $\ce{2}$ acids have the same $\ce{X}$ atom with the same oxidation number, but a different number of $\ce{H}$ and $\ce{O}$ atoms.


text:
How can I model cellular membranes in terms of topology?
I'm reading from Principles of Physical Biochemistry by Kensal E van Holde (1998-01-06) in order to review some biochemistry and work my way toward a more mathematical approach to understanding chemistry.
On page 18, chapter 1 (Biological Macromolecules), I read

"The structure of an integral membrane protein can be thought of as being inverted relative to the structure of a water soluble-protein, with the hydrophobic groups now exposed to the solvent, while the hydrophilic atoms form the core. An example of this inverted topology is an ionic channel (Figure 1.12). The polar groups that line the internal surface of the channel mimic the polar water solvent, thus alloweding charged ions to pass readily through an otherwise impenetrable bilayer."

I don't have a picture handy to show you Figure 1.12, but essentially it is a channel protein.
What's interesting to me about this passage is their use of the word topology. It reminded me of algrabraic topology's notion of a topology on a set, but I can't seem to state this membrane chemistry in terms of the mathematical formalism.
In algebraic topology, a topology $\tau$ on a set $X$ is a subset of the powerset $\mathcal{P}(X)$ where the elements of $\tau$ are open, meaning they satisfy

Containing the original set and the empty set:$$X \in \tau, \emptyset \in \tau$$
Unions of sets in the topology are closed. $$\{ U_i \}_{i \in I} \subseteq \tau \implies \bigcup_{i \in I} U_i \in \tau$$
Closure under finite union. $$\{ U_i \}_{i=1}^n \subseteq \tau \implies \bigcap_{i=1}^n U_i \in \tau$$

My initial thinking on this is that the membrane could be one set, the extracellular matrix a second set, and the intracellular matrix as a third set. Each of these sets might be thought of as their own topologies, and that their mutual intersections should be empty so long as you cleanly partition the spaces. But then again, membranes have a lot of intermolecular spaces within them; seems structurally complicated.
While knowingly extrapolating beyond the original intended meaning of the authors, how can I model  cellular membranes in terms of topology?

While an analysis in terms of formal topology is certainly possible, the word is here used more to mean "shape" and specifically where hydrophilic and hydrophobic residues are located relative to the surrounding medium. You have in essence two sets of residues (hydrophilic and hydrophobic) and one topological property (inside or outside). You might be overthinking this, depending on your objective.

I would agree that you're over thinking this. As Buck indicated the only topological property is inside or outside, as for a circle.

As Buck Thorn and MaxW already said in the comments, there is less to it than the textbook excerpt might suggest. The use of the term topology in this context is non-mathematical, similar to when someone says there has been a quantum leap in product development.
The actual topology of most proteins is the same as that of a sphere, and the topology of the typical cartoon channel is that of a donut. Having hydrophilic residues in the channel or outside of the channel is not an inversion because those parts of the surface are topologically equivalent.

[OP] I can't seem to state this membrane chemistry in terms of the mathematical formalism.

That's because it is non-mathematical. There is some real topology associated with membranes, and that is the concept of inside and outside. Membrane proteins insert in one direction, and they don't "flip" directions over time, even though they are perfectly mobile to move around in the membrane (or have lipids move around them) while pointing in one direction or the other one. During endocytosis and exocytosis, the relative orientation of membrane proteins is conserved but inverted, so that parts that are inside organelles (such as glycosylation) end up on the outside of the plasma membrane.

"actual topology of most proteins is the same as that of a sphere" - you are right of course, but sometimes "topology" is used instead of "fold" for proteins, again a little confusingly. In fact I've also seen the connectivity of small molecules called the "topology" for certain fields (like force field calculation)

Using topology for connectivity is great. Using it for the protein fold is not a rigorous use (unless you take the hydrogen bonds between strands in a beta sheet as fixed, or start discussing disulfide cross-links). However, the term has been widely used outside its mathematical definition, as you stated. A prominent example is the TOPS server: "The topology of a protein structure is a highly simplified description of its fold including only the sequence of secondary structure elements, and their relative spatial positions and approximate orientations." is far away from the original definition

I agree about TOPS  :) (I used to work on that project) I suspect that was originally driven by the 'T' in CATH : https://www.cathdb.info/wiki/doku/?id=glossary:topology

The original context of the textbook is almost surely being used in a non-mathematical sense. But since the question is looking for some mathematical way of looking at it, I will suggestion one.
Given a point cloud of the locations of each membrane's atoms, or alternatively of meshes approximately something like a space-filling model, you could induce a topological space using a metric. Considering the unions of $\epsilon$-close atoms will only be as interpretable as your choice of $\epsilon$ is meaningful.
Just considering such unions will inform you more about the induced connected components at some distance more than anything else. If you are interested identifying holes (like the hierarchy of compartments found in cells), you will need to go further. Given a similar input as above, you can compute can compute a sequence of simplicial complexes of a chosen type (e.g. Čech or Vietoris-Rips) induced by an interval of $\epsilon \in [a,b]$. From these simplicial complexes the persistent Betti numbers can be estimated. For more information see An Introduction to Topological Data Analysis: Fundamental and Practical Aspects for Data Scientists.
A potential application can be found in (rather computationally expensive) simulations. If you simulate the positions of atoms in a membrane being disrupted by a force, you could track their persistent homology. This approach could also be applied to simulations of protein folding to track whether transient pockets form. Sufficiently small systems have to be chosen in order for such calculations to be feasible.


text:
Problem detection in reaction procedure
We all know that oxidation of secondary alcohols leads to ketones. So, with that logic, in order to form acetone $CH_3-CO-CH_3$ through oxidation, we need a secondary alcohol like isopropyl alcohol. Now there is a problem I am trying to figure out, where I need to identify each reactants + reactions involved.
It is given that acetone is the end product , now with that information I need to traverse backwards/in backward direction and determine A, B and C. First let me mention the reaction procedure:
$A+HI\rightarrow B+KOH\rightarrow C+[O]\rightarrow Acetone$
Now, it is quite clear that C represents isopropyl alcohol (oxidation of secondary alcohols). Now, isopropyl alcohol has OH group in the second carbon atom, and that must have come from KOH. Also, since C (as a product) has 3 carbon atoms, hence B must also have 3 carbon atoms. Now, what is confusing me is this part:
$A+HI\rightarrow B+KOH$
Since, in the reactant side there is I (iodine) group, hence, B (as a product) must have iodine in it. But considering C, there is no iodine (I) group in it, so does it mean that I group was substituted by OH group? Because, normally, i have seen $OH$ group being substituted by I group. Also, I don't think that A will be alcohol (because C already is an alcohol), moreover it is quite apparent that even A must have 3 carbon atoms...or does it not? (I am not sure). Also, considering B, it is sure to have I group on second carbon atom. Is my approach correct? (Please correct me if I am wrong). What according to you should be A and B?

Your thoughts on the identity of B and C are correct. How would concentrated HI react with an alkene?

Oh yeah, it works, Markonikoff's rule into play!


text:
Why is distilled water such a poor conductor of electricity?
Water is composed of hydronium ions and hydroxide ions. Since ions do conduct electricity, why is distilled water such a poor conductor of electricity?

You are mistaken.
Water is not composed of these ions. There is ongoing autodissociation equilibrium reaction$$\ce{2 H2O <<=> H3O+ + OH- }$$ shifted strongly toward left, with concentration of ions  $\pu{1e-7 mol/L}$ at $\pu{25 ^{\circ}C}$ .
Therefore conductivity of water itself is very low.


text:
Reaction of ketones with cyclic diols
Question

In this reaction I am able to figure all reaction except first. What confuses me is the fact than in solution, the cyclic diol reacts only with the aldehydic group. But It can also react with the ketone right?

This is another poorly worded question in that the conditions for A to B require more detail to be certain of the identity of B. From the rest of the question I surmise that B is 2-(1,3-Dioxolan-2-yl)cyclohexanone i.e. the keto-aldehyde with dioxane protection on the aldehyde. Incidently I can find no reference to this compound by googling its name. Can someone with database access cast any further light on this?

@Waylander: For the ketoaldehyde (hydroxymethylene cyclohexanone), the dioxolane does not appear to have been formed in the usual way. I didn't look at all 214 pages in Chem Abst. But it has been made from the TMS enol ether of cyclohexanone, 2-methoxy-1,3-dioxolane and Lewis acid. The traditional ketalization with, H+, ethylene glycol and the hydroxymethylene (CHO) derivative of alpha-tetralone appears to be doable.

Interesting! Thank you for that. The enol mix of the alpha tetralone CHO analogue may be different to cyclohexyl equivalent which may influence the viability of this reaction.


text:
Complex Esterification reaction?
Question

Attempt
On seeing the question it seems like simple esterification reaction and the answer seems A. But the answer in book says C. How is it so
I also tried to develop a mechanism as follows

Driven by the formation of the highly stable triphenylmethyl (trityl) cation. If you want the methyl ester add the cH2SO4 of a MeOH soln of the acid.

I am sorry but i didnt understand the last part. How do I get easter for above reaction?

To get the ester dissolve the acid in MeOH and add a catalytic amount of cH2SO4.


text:
Under what conditions can epoxide undergo oxidation by HIO4
Question

Doubt
Can epoxide react with HIO4. I have  seen that in same cases they do. But in this given question they don't.  If they do react , then how  and what will be the products?Please help

You might be mistaken. The sugar you have depicted has no epoxide, it's an acetal.

This may assist you http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/ox_spal_diol.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/entfernen_wasserstoff/ox_spaltung_12_diole/anwendung.vscml.html


text:
How to fix 'error: hangup' in Gaussian 16 modredundant scan calcuation?
I have got a problem during my calculations with Gaussian 16. Everytime I start a job to scan the energy about 2 dihydrals I run into the following Error: hangup.
I am trying to calculate a 3D energy hyper-surface.
 Cycle   1  Pass 1  IDiag  1:
 FoFJK:  IHMeth= 1 ICntrl=       0 DoSepK=F KAlg= 0 I1Cent=           0 FoldK=F
 IRaf= 840000000 NMat=       1 IRICut=       1 DoRegI=T DoRafI=F ISym2E= 0 IDoP0=0 IntGTp=1.
 FoFCou: FMM=F IPFlag=           0 FMFlag=      100000 FMFlg1=           0
         NFxFlg=           0 DoJE=F BraDBF=F KetDBF=F FulRan=T
         wScrn=  0.000000 ICntrl=         0 IOpCl=  0 I1Cent=           0 NGrid=           0
         NMat0=    1 NMatS0=      1 NMatT0=    0 NMatD0=    1 NMtDS0=    0 NMtDT0=    0
 Symmetry not used in FoFCou.
 Inv3:  Mode=1 IEnd=    15309243.
 Iteration    1 A*A^-1 deviation from unit magnitude is 2.78D-15 for   1609.
 Iteration    1 A*A^-1 deviation from orthogonality  is 1.95D-15 for    692    229.
 Iteration    1 A^-1*A deviation from unit magnitude is 3.11D-15 for   2166.
 Iteration    1 A^-1*A deviation from orthogonality  is 2.64D-15 for   1559    734.

Error: hangup
Error: hangup
rax 00000000000001a8, rbx 00000000000000a5, rcx 00002b6798f00f70
rdx 00002b6798ea69d0, rsp 00007ffc4e691c10, rbp 00007ffc4e691c10
rsi 00002b6798ea1980, rdi 00007ffc4e691cf0, r8  00000000000001a8
r9  0000000000000035, r10 0000000000005050, r11 4000000000000000
rax fffffffffffffffc, rbx 00007ffe638fe4a0, rcx 00002b8d41cdfc6a
r12 0000000000000455, r13 00002b6798f00f70, r14 00002b6798ea69d0
rdx 0000000000000000, rsp 00007ffe638fe468, rbp 0000000000000001
r15 00002b6798ea1980
rsi 00007ffe638fe4a0, rdi 000000000000fbdc, r8  00002b8d41fc54a0
r9  00007ffe638fe3c0, r10 0000000000000000, r11 0000000000000246
r12 00007ffe638fe4c0, r13 00007ffe63903e90, r14 0000000000000000
r15 00007ffe63904e90

 /lib64/libpthread.so.0(+0x10b10) [0x2b67907f1b10]
   /opt/g16/l502.exe() [0xf49b9f]
   /opt/g16/l502.exe() [0x10b382f]
   /opt/g16/l502.exe() [0xe275b9]
   /opt/g16/l502.exe() [0x92dcbd]
   /opt/g16/l502.exe() [0x6db95b]
   /opt/g16/l502.exe() [0x4ced25]
  /opt/g16/l502.exe() [0x78d47a]
  /opt/g16/l502.exe() [0x8ed63b]
  /opt/g16/l502.exe() [0xe1a4d2]
  /opt/g16/l502.exe() [0xdf53c8]
  /opt/g16/l502.exe() [0xddee8d]
  /opt/g16/l502.exe() [0x8ee6db]
  /opt/g16/l502.exe() [0x574625]
  /opt/g16/l502.exe() [0x56e0c1]
  /opt/g16/l502.exe() [0x525430]
  /opt/g16/l502.exe() [0x429326]
  /opt/g16/l502.exe() [0x41b189]
  /opt/g16/l502.exe() [0x412b1b]
  /opt/g16/l502.exe() [0x412a24]
  /lib64/libc.so.6(__libc_start_main+0xf5) [0x2b6790d1b6e5]
  /opt/g16/l502.exe() [0x40d169]

I am quite new to Gaussian, but I don't see any error message I can work with. Is the error caused by overloading the server I run my calculation on?
I have started the calculation several times. It always ends up with this error message, just at another point.
May someone of you got an idea what my mistake is?
This is my input file:
%cpu=0-15
%chk=imida_endo0_scan_water.chk
%mem=50GB
#p opt=modredundant b3lyp/6-31g(d) scrf=(smd,solvent=water) nosymm
geom=connectivity

imida_endo0_scan_water

0 1
 C                  0.96689500   -0.23888100   -0.04910500
 C                 -0.27289900    0.15800900    0.46267600
 C                 -0.75272200    1.41154300    0.07611100
 N                 -0.09386700    2.24670700   -0.74436400
 C                  1.07709000    1.83312000   -1.19945400
 C                  1.66999300    0.60940200   -0.90121000
 Cl                 1.93440800    2.94665700   -2.27959000
 C                 -1.07173200   -0.72702800    1.39450000
 N                 -0.89853800   -0.34820200    2.80265100
 C                  0.40075500   -0.65358100    3.44244400
 C                  0.01604200   -0.70482400    4.93399800
 N                 -1.45280300   -0.87816600    4.96871800
 C                 -1.83690300   -0.67012200    3.76282500
 N                 -3.20632800   -0.83379300    3.41712100
 N                 -3.96819700    0.25054800    3.02514400
 O                 -5.18680300    0.11392700    3.12641500
 O                 -3.39241300    1.23438300    2.56573100
 H                 -3.74439800   -1.40887700    4.06216800
 H                  0.26577400    0.23186900    5.44495100
 H                  0.51524300   -1.51478500    5.47028100
 H                  1.14935100    0.10719800    3.21207400
 H                  0.76438500   -1.62496900    3.07813000
 H                 -2.13169900   -0.65335000    1.14736800
 H                 -0.76634600   -1.77367700    1.25314400
 H                 -1.71595300    1.76379600    0.43634900
 H                  1.38158100   -1.20867900    0.21188600
 H                  2.63192200    0.33177400   -1.31570100

1 2 1.5 6 1.5 26 1.0
 2 3 1.5 8 1.0
 3 4 1.5 25 1.0
 4 5 2.0
 5 6 1.5 7 1.0
 6 27 1.0
 7
 8 9 1.0 23 1.0 24 1.0
 9 10 1.0 13 1.5
 10 11 1.0 21 1.0 22 1.0
 11 12 1.0 19 1.0 20 1.0
 12 13 2.0
 13 14 1.0
 14 15 1.0 18 1.0
 15 16 1.5 17 1.5
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27

D 1 2 8 9 S 11 30.0
D 2 8 9 10 S 11 30.0

It's too hard to determine with out knowing more about the calculation. There can be several issues, From the dihedral scan, to scf convergence.

I bet this is a server error of some kind, or you may be asking for too many resources you don't actually have. Have a look at the files the queuing system may write, i.e. error messages or the output stream, or if you run locally, have a look at the messages printed to the terminal, you might find information there. Further analyse what's happening, make some tests. Try a pm6 scan to see whether the same error appears, if the calculation runs through. This needs small resources, so you can test in a few hours. [cont.]

You are doing a 2-dimensional scan of $(1 + 11)\times(1 + 11) = 144$ single point calculations, with a quite extensive method, i.e. hybrid dft. Are you running out of the time limit? Is b3lyp really necessary, or could you do with something a little less demanding? This is only an organic molecule, so you should be fine with pure dft for structure calculations, ie. #P BP86/def2SVP/W06   EmpiricalDispersion=GD3BJ DenFit. I'd also advise against Pople Basis Sets - they don't scale up well and don't perform good either.

The scan runs fine with pm6 on minimal settings, this must be an error related to you system settings and size.

@Martin-マーチン I tried with your advise with the PM6. Have you an idea where I can check if I am running out of the time limit?

This should be in the log files of your queueing system, some also send an e-mail with the job statistics, in some cases you can retrieve them with a command from a database. You may want to ask you administrator about these problems.


text:
Darzens reaction on allylic position
Question

Attempt
I thought the products would be from simple darzens reaction. So I replaced the $\ce{OH}$ by $\ce{Cl}$. But the answer of mine seems way wrong. Do give some tips on what might be happening in the reacton.

Your idea of a Darzens reaction seems to be rather different from the usual idea of one: https://en.wikipedia.org/wiki/Darzens_reaction I've never heard SOCl2 chlorination being referred to as a named reaction.

@orthocresol I think he is talking  about Darzen's halogenation

Yes the halogenation one which proceeds through SN2 ( or SNi)

This reaction is a special case of Darzen's halogenation. I think you are aware of mechanism of Darzen's reaction or else you can check it here. It occurs through a 6 membererd cyclic intermediate. I'll show you the first reaction you would be able to answer second one in succession.

The mechanism involving the chlorination of $2^\circ$-alcohols has been studied tremendously. Based on these studies, it was revealed that if nucleophilic solvents such as 1,4-dioxane or ether has been used in the reaction the stereochemistry of the original alcohol would be retained in the resultant chloroalkane product (retention). However, if a base such as pyridine or trialkylamine is used as a catalyst, the stereochemistry is reversed to give $\mathrm{S_N2}$ product (inversion):

The mechanism of former type is called $\mathrm{S_Ni}$ type mechanism (nucleophilic substitution with internal return). This mechanism believed to be progress through cyclic intermediate (Ref.1). The authors have suggested that, if the decomposition of the chlorosulfite intermediate does occur by the concerted mechanism, then there should be several ion pair combinations contributing to the transition state. This was supported by later research by Cram (Ref.2). Neighboring group participation (Ref.2 & 3) and 1,2-hydride transfer (Ref.4) are also evident, implying carbocation formation during the reaction. All of this suggest that the possibility of rearrangement of original structure (through the dissociation of alkyl chlorosulfite intermediate to form an ion pair).
The rearrangement of allyl alcohols upon chlorination by $\ce{SOCl2}$ was first discovered in 1960 (Ref.5). Accordingly, in dilute ether solutions, almost complete allylic rearrangement occurs such that $\alpha$-methyl allyl alcohol ($\ce{CH3CH=CHCH2OH}$) gives $\gamma$-methyl allyl chloride $(\ce{CH2=CHCH(Cl)CH3}; \ P)$, while $\gamma$-methyl allyl alcohol ($\ce{CH2=CHCH(OH)CH3}$) gives $\alpha$-methyl allyl chloride $(\ce{CH3CH=CHCH2Cl}; \ Q)$. The mechanism of the rearrangement was explained by the formation of cyclic (six-membered) intermediate through $\mathrm{S_Ni}$ process or through rigidly-oriented undissociated ion-pairs (the ion-pair mechanism is favored). This 100% rearrangement is happen only in dilute solutions, while concentrated solutions have given mixtures of products.
Assuming the reaction is in dilute ether solution, the best answer for the question is (c).
Note: Historically the use of $\ce{SOCl2}$ in combination with a tertiary amine such as pyridine or triethylamine was called the Darzens halogenation. Therefore, using that name for given reaction is completely wrong because it did not include a base.
References:

Edward S. Lewis, Charles E. Boozer, “The Kinetics and Stereochemistry of the Decomposition of Secondary Alkyl Chlorosulfites,”  J. Am. Chem. Soc. 1952, 74(2), 308-311 (https://doi.org/10.1021/ja01122a005).
Donald J. Cram, “Studies in Stereochemistry. XVI. Ionic Intermediates in the Decomposition of Certain Alkyl Chlorosulfites,”  J. Am. Chem. Soc. 1953, 75(2), 332-338 (https://doi.org/10.1021/ja01098a024).
C. C. Lee, Donna Newman,  D. P. Thornhill, “Rearrangement Studies with $\ce{C^{14}}$: XIV. The Reaction of 2-p-Anisylethanol-1-$\ce{C^{14}}$ with Thionyl Chloride,”  Canadian Journal of Chemistry 1963, 41(3), 620-624 (https://doi.org/10.1139/v63-087).
C. C. Lee, A. J. Finlayson, “Rearrangement in the reaction between thionyl chloride and 3-methyl-2-Butanol,” Canadian Journal of Chemistry 1961, 39(1), 260−261 (https://doi.org/10.1139/v61-030).
William G. Young, Frederick F. Caserio Jr., David D. Brandon Jr., “Allylic Rearrangements. XLIX. The Controlled Conversion of α- and γ-Methylallyl Alcohols to Chlorides with Thionyl Chloride,”  J. Am. Chem. Soc. 1960, 82(23), 6163-6168 (https://doi.org/10.1021/ja01508a047).

Well i dont know but my school books have often referred them as darzens reaction that is why I used them

Nowadays, we don't use that name anyway. ;-)


text:
How does the mass composition affect the empirical formula?
I have to do an assignment of empirical formula, and how different percentage mass compositions affect the formula obtained, from experimental data.
We did three trials and 2 of which of the same compounds (copper chloride hydrate), has different mass compositions.
I can't find any information about how mass composition affect the empirical formula, although I do understand i need extra proof to explain all the little detail behind it all.
I would really appreciate it if someone could help me thank you so much !!!

From Googling "mass composition to empirical formula": https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/empirical-molecular-formula/v/formula-mass-composition  There are a decent number of other examples if this isn't clear, but Khan academy generally does a good job.


text:
Bonds in R-H3N+ group

Source: Wikipedia
I am confused about the ammonium group in this molecule.
I can see that we have a nitrogen atom and 3 hydrogens. The group is attached to a carbon.
Nitrogen has atomic number 7 so my understanding is it can form up to 3 covalent bonds, and therefore cannot have covalent bonds with the carbon and all 3 hydrogens.
I can see a + on the nitrogen, suggesting it is an ion, and maybe some ionic bond?

Nitrogen along with it's 3 bonds has a lone pair. That's what comes into play. The H+ ion has no electrons of it's own and therefore when it forms bond with N, nitrogen coordinates it's lone pair to H+ completing it's duplet and at the same time it's octet and forms a coordinate bond also known as dative-bond.
You can find more about coordinate bond here

What do you mean with 'pure orbital'?

Pure orbital means that the pone pair doesn't resides in any of the sp3 hybridized orbitals instead the lone pair is in the pure p-orbital.

If you mean that you should probably also write pure p orbital. In this case, however, it would not be correct. An amino group has a pyramidal structure, so the resulting lone pair orbital is not pure p; this would require a planar coordination around nitrogen.

Does that means that lone pair goes in an hybridized orbital ?

Yes, in the hybridisation description it does.


text:
Can iodine replace other halogen as well?
While reading about halogen family I encountered the reaction:
$$\ce{NaClO3 + I2 -> NaIO3 + Cl2}$$
I know that that the oxidizing tendency of halogens decreases down the group, so I think that the reverse reaction should have taken place.
So why is this reaction taking place?
Is the reason for the above reaction taking place, the higher tendency of iodine to bond with oxygen?

True, iodine is a much weaker oxidizer than chlorine. Now see, iodine is not the oxidizer at all in this reaction. Quite the opposite, in fact.

So , Can I generalize that stability of higher oxidation states of halogen increases down their group...( Also is it because of their decreasing ionisation enthalpy )

Yes, you can say so.

But it was mentioned in my book that the stability of oxides of halogens is I > Cl > Br...Can you please comment on this as well...

I don't see any contradiction. True, the stability of compounds with halogens in positive oxidation states (in particular, oxides) increases down the group.

The order of Br and Cl is opposite of what it should be on the  basis of ionisation enthalpy......

The action of iodine is part of a long reaction chain with chlorate, so not a direct action as implied.
I start with iodine and water equilibrium:
$\ce{I2 + H2O <=> HI + HOI }$
Next more complex part of the reaction sequence is discussed here, to quote:

Chlorate ion oxidizes iodide ion to hypoiodous acid and chlorous acid in the slow and rate-determining step:
$\ce{ClO3− + I− + 2 H+ → HOI + HClO2}$
Chlorate consumption is accelerated by reaction of hypoiodous acid to iodous acid and more chlorous acid:
$\ce{ClO3− + HOI + H+ → HIO2 + HClO2}$
More autocatalysis when newly generated iodous acid also converts chlorate in the fastest reaction step:
$\ce{ClO3− + HIO2 → IO3− + HClO2}$

Chlorous acid is unstable, disproportionating to hypochlorous acid and chloric acid:
$\ce{2 HClO2 → HOCl + H+ + ClO3- }$
Also, the known action of hypochlorous acid (even H2O2) on HI liberating iodine:
$\ce{2 HOCl + 2 HI -> 2 HCl + I2}$
Finally, some chlorine evolution:
$\ce{HCl + HOCl <=> Cl2 + H2O }$
Further reference on the complexity of the reaction system, see Page 94, Table 5.2 in this ebook.


text:
Benzene as a pi-acceptor?
In coordination chemistry, we classify ligands as sigma donors, pi acceptors and donors. Benzene is a pi donor, as is the cyclopentadienyl anion, but do these two possess any pi acceptor character? (Can a ligand be both pi acceptor and pi donor?)

Of course they do.

Mithoron, won't the pi-acceptance hinder the aromatic stabilisation? The LUMO is antibonding, right?

It's antibonding like, what, 90 % of all LUMOs ? Still, they're used for backbonding all the time.

Aromatic rings that complex "face on" to a transition element are acting as pi acceptors all the time, including cyclopentadienide complexes.  And they act simultaneously as pi donors too.
Let's look at the familiar cyclopentadiende pi orbitals from a different perspective.  We know that there are three bonding orbitals that should be occupied and two antibonding orbitals that shouldn't be -- until we introduce the d orbitals of a transition element.  As illustrated by https://www.chemtube3d.com/orbitalsm-cyclopentadienyl/, now overlap occurs with the possibility of all the ring orbitals bonding to the metal:

So, in a complex such as ferrocene, the filled cyclopentadienyl pi orbitals share their electrons with three of the transition element d orbitals and the latter element, in turn, donates electrons back to the two unoccupied pi orbitals of the ring.
Of course these interactions sacrifice aromaticity in the cyclopentadienyl rings, but it's a bargain with the bonds forming between carbon and the transition element.  Aromaticity is not so much broken as it is superseded by a further bond deolcalization.  In the especially well-known case of ferrocene, the energy bargain is especially good because when all is said and done with the iron/cyclopentadienyl electron exchange, you've exactly filled the bonding orbitals and left the antibonding orbitals empty.  The diagram from this source shows the molecular orbital structure in the case of a $3d$ metal.  For ferrocene where the iron(II) center provides six additional electron, they fill the orbitals in the box right up to the metal $3d$ orbital level (the HOMO is the $a_{1g}$ orbital inside the box) but not beyond.  Note that the filled $e_{2g}$ orbital has a component from the antibonding orbitals of the cyclopentadienide, showing the pi acceptor action of the latter.

Now compare this outcome with magnesocene.  Magnesium has no valence d orbitals or d electrons.  So the pi electron exchange between pi and d orbitals is not happening.  In this case the cyclopentadienyl ion can still act as a donor to the metal s abd p orbitals, but this does not make for as good bonding because (1) the overlap is not as good and (2) magnesium, like most Group 1 and 2 metals, has relatively diffuse valence orbitals that make for relatively poor overlap anyway.  Unable to develop the full covalent bonding network seen in transition element cyclopentadienide complexes such as ferrocene, magnesocene has greater ionic character -- so much as to form ions in some solutions -- but much less stability.

So if there is both pi-acceptor and pi-donor capacity, what happens to the gap between $\ce{t_2_g}$ and $\ce{e_g^*}$ levels. Is it increased or decreased?

Not octahedral, therefore we don't really have $t_{2g}$or $e_g$. See here for a proper description of the molecular orbitals of metallocenes.


text:
What happens when a piece of copper is placed in 1M HCl?
I know that the answer is that nothing happens, but the explanation is confusing me in some places. Firstly, we are given that:
\begin{align}
\ce{Cl2 + 2e- &-> 2Cl-} & (E &= \pu{1.36 V})\\
\ce{Cu^{2+} + 2e- &-> Cu(s)} & (E &= \pu{0.34 V})\\
\ce{2H+ + 2e- &-> H2} & (E &= \pu{0 V})
\end{align}
What's confusing me is that in the solution, we'll have $\ce{H+}$, $\ce{Cl-}$ and $\ce{Cu(s)}$, which are all products in the equations (except for $\ce{H+}$). I get that for the reverse equation (oxidation), the oxidation potential is given by the negative of the reduction potential. What I'm wondering is, if we look at copper and chlorine first, is the only option for the $\ce{Cl-}$ in solution to be oxidised (with $E = \pu{-1.36 V}$)? Also, can the copper also only be oxidised (with $E = \pu{-0.34 V}$)? And if both of these are true, then does that mean that these two can't react at all, not because of the total $E$ being negative, but because they can both only be oxidised?

Copper metal cannot oxidize chloride ion to chlorine gas, so forget the chloride ions. As the reduction potentials show, copper ions are spontaneously reduced to copper metal, relative to hydrogen ions being reduced to hydrogen gas. So the spontaneous reaction would be copper (II) ion plus hydrogen gas => copper metal plus hydrogen ions. But you have neither reactant: you only have the products. So the net result is nothing happens.

One of your previous questions had a good answer by @titaniummorro, so why not do exactly what they did on this question? Simply compare standard cell potentials for $\ce{Cu (s) + 2H^+ (aq) => Cu^2+ (aq) + H_2 (g)}$ and the reverse reaction.

Whatever the concentration of HCl, Copper does not react with HCl.

@EdV thanks very much for the explanation. Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?

Correct: copper metal cannot be reduced, i.e., gain electrons. It is already reduced, as it were. So it can serve as a reducing agent for something else, like silver ion in aqueous solution. That is energetically favorable and spontaneous, e.g., the famous “silver tree” demo. But reducing hydrogen ions in aqueous solution is not energetically favorable, as shown by the reduction potentials. There is nothing in 1 M HCl that can oxidize copper metal and that acid is not an oxidizing acid. In contrast, 1 M nitric acid would oxidize copper metal.

This answer actually does the relevant calculations: https://chemistry.stackexchange.com/a/95375/79678.

@Ed V Actually copper actually reacts with conc.HCl and liberates hydrogen gas due to the formation of tetrachloridocuprate(ll) anion

@Adithya, Copper does not react with HCl in the absence of oxygen nor it will liberate hydrogen even if it is concentrated. Do you know a reference which says so?

@Adithya The link does not support what you say: it deals with copper complexes and copper compounds, not copper metal in hydrochloric acid that does not have an extrinsic oxidizing agent, e.g., dissolved oxygen from the air.

Iam sure that the statement was taken from a trusted source by my lecturer. The formation of the complex makes the equilibrium to move in forward direction. But failed to find an online source. I just gave that link to support the stability of the complex. I'll let you know if I find out any online source.

@Adithya The reference you provided is interesting, but the question at hand is solely about clean copper metal (no oxides, patina, etc.) and 1 M HCl with no dissolved oxygen from the air or any other additional oxidizing agent, suspected or not. This is turning into the "X-Y" problem.

@Adithya, Even if your lecturer told you, he himself might have a misconception. I have done these experiments plenty of times. Copper does not dissolve in HCl in the absence of an oxidizing agent. It has nothing to do with an equilibrium. The tetrachloro complex is only formed $when$ an oxidizing agent is present in HCl.

Related: Why doesn't copper react with hydrochloric acid while the other metals do?

Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?

Let us take a longer approach to address your query. For electrochemical problems like these, first you have to make a list of starting materials and possible products. Do a thought experiment even before even looking up the tables. It is long way but it will help you in doing future problems.
Your question is: What happens when a piece of copper is placed in 1M HCl?
Do a thought experiment +  use some basic intuitive chemistry.
Making a list of starting materials, a) Cu metal, b) H$^+$, c) Cl$^-$.
Now there are only two possibilities for each substance.
a) Copper Cu can be oxidized to Cu$^+$ or Cu$^{2+}$ or reduced to Cu$^-$. Basic chemistry tells us that metals like to form cations so Cu$^-$ can be eliminated. Similarly, there is possibility for Cu$^+$ or Cu$^{2+}$. However, say, you are just interested in Cu$^{2+}$.
Now you have the problem well defined: Will Cu -> Cu$^{2+}$?
b) Follow the same reasoning: H$^+$, it can be oxidized to  H$^{2+}$ or reduced to H$_2$. Basic chemistry will tell you that H$^{2+}$ is not possible. Now your question is more well defined: Will H$^+$ -> H$_2$
c) In the same way, ask the same question for chloride ion. It can be oxidized to Cl$_2$ or reduced to  Cl$^{2-}$. Basic chemistry would tell you that Cl$^{2-}$ is not feasible. So your only concern is: Will Cl$^-$ -> Cl$_2$
Now you can ask only two questions:
Can copper metal reduce H$^+$ to H$_2$?
or
Can copper metal oxidize Cl$^-$ to Cl$_2$? This can be easily eliminated because in order for copper to oxidize it must reduce itself further, which is not possible.
At this stage you can utilize the electrode potentials. Recall cathode refers to reduction and anode refers to oxidation
Ecell= E$_{cathode}$-E$_{anode}$
Ecell= E(for hydrogen half cell bc it is being reduced) - E (copper half cell)
Ecell= 0.00- (+0.34) = - 0.34 V
The negative sign shows this not possible under these conditions.
Do the same for chloride ion.
Ecell= E(Reduction of Cu to Cu$^-$) - E (chlorine half cell)
Ecell= Undefined- (+1.36) = Undefined because Cu$^-$ does not exist in solution.
In short nothing will happen to copper in HCl.
P.S. Practically, HCl slowly dissolves copper in the presence of oxygen. This not relevant here.

The expansive thought-provoking  answer still does not address what actually occurs in the conditions presented. The copper metal is 'cleaned' of its oxide coating (natural formed), and a little known cuprous chloro complex is created, as I noted in my answer.

Pardon my criticism. Note, if there is a chemist who has performed a reaction, and I get some of the details wrong, please speak up, as I, and perhaps others, should welcome the constructive advancement of the science.

Your point is a classical X-Y problem. You are addressing a problem which was not asked and it does not address the main problem, what happens to pure copper metal when it is dipped in 1 M HCl. You are right, a small amount of copper oxide dissolves, but the chloro complex is only formed in concentrated acid not in dilute 1 M HCl.

Without oxygen or oxidating agent, nothing happen to copper in hydrochloric acid, but dissolution of surface oxides. The only oxidant there is $\ce{H+}$, which is with $E^{\circ}=\pu{0.00 V}$ too weak oxidant to oxidize $\ce{Cu}$ with $E^{\circ}=\pu{+0.34 V}$.
If oxygen is present, then it gets slowly dissolved to tetrachlorocuprate ( similarly as to copper acetate in vinegar ):
$$\ce{2 Cu + O2 + 4 H+ + 8 Cl- -> 2 CuCl4^2- + 2 H2O}$$
The dissolution is very quick, if oxidant as hydrogen peroxide is present( it dissolves gold as well )
$$\ce{ Cu + H2O2 + 2 H+ + 4 Cl- ->  CuCl4^2- +  H2O}$$

Excuse me, but 2/3 of your 'answer'  addresses possible reaction with O2 or H2O2 creating a cupric complex that will not be actually created as per the experiment described. The actual, little known, cuprous complex created is detailed in my response. Your not knowing this is understandable, but should not be upvoted, in my opinion.

@AJKOER Well, the OP has been interested in Cu itself.

OK, if you takes a piece of copper metal (I employ plumbing copper, high purity) in concentrated HCl the copper, in time, will be cleaned of its Cu2O coating. [EDIT] To quote a source:

Copper has excellent atmospheric corrosion resistance. It becomes naturally covered with an oxide film, changing its color to dark brown/black in normal atmospheric conditions. Later, a green patina forms on copper outdoors with a varying intensity depending on where and how the surface is exposed.

So, perform an experiment on copper metal taking before and after pictures, and see if it is noticeably different.
Underlying chemistry, per Wikipedia, to quote:

Copper(I) chloride...is a white solid sparingly soluble in water, but very soluble in concentrated hydrochloric acid. Impure samples appear green due to the presence of copper(II) chloride (CuCl2).

Also, forms a chloro-complex with concentrated HCl:

It forms complexes with halide ions, for example forming H3O+ CuCl2− in concentrated hydrochloric acid.

So, expected reactions in concentrated HCl, a cleaning off of any Cu2O coating:
$\ce{HCl + H2O <=> H3O+ + Cl-}$
$\ce{Cu2O + 2 HCl -> 2 CuCl (s) + H2O}$
$\ce{CuCl (s) + H3O+ + Cl- <=> (H3O)CuCl2 (aq)}$
I have performed experiments where one adds much NaCl creating a soluble cuprous presence (in the form of NaCuCl2). Wikipedia also cites a preparation to Cupric oxychloride based on this solubility conversion of cuprous per Eq(6) and Eq(7) here, where:
$\ce{ Cu (s) + CuCl2 (aq) -> 2 CuCl (s)}$
$\ce{ 2 CuCl (s) + 2 NaCl (aq) -> 2 NaCuCl2 (aq)}$
There is no other apparent reaction occurring with the freshly cleaned Copper.

The three equations you list are not enough to answer the question. The electrochemical potentials are for standard conditions, unless otherwise specified. Now when you put a piece of copper into 1 M HCl, the [H+] is 1 M (standard), the [Cl-] is 1 M (standard) and the [Cu++] is zero (nonstandard). The equation to consider is
2 H+ + Cu --> Cu++ + H2
Since the copper concentration initially is exactly zero, a few copper atoms will dissolve; the driving force can be calculated from the equation:
E = 0.34V - (0.0591V)/2 log(1.0 M/x M) = 0.34V - 0.02955V log(1/x)
0.34/0.02955 = log(1/x) = 11.5
x = 10^-11.5 = 3 x 10^-12 M
The final concentration of copper from that reaction is very small (3 trillionths of a mole per liter, 190 ppt by weight), but not zero. And I haven't taken into account that the H2 is not at standard pressure (1 atm), which will drive the voltage up a bit more, and the eventual [Cu++] also. So copper is not totally inert in 1 M HCl. And that's assuming you are talking about an inert atmosphere, like 100% N2.
In real life, this would be done in air (O2 = 0.2 atm). And oxygen will pick up electrons and give hydroxyl:
O2 + 2H2O + 4e-  --> 4 OH-        E = 0.40 V
Remember Cu++ + 2 e-  --> Cu      E = 0.34 V
Now, in 1M HCl, the [OH-] is 10^-14, so the equation for the oxygen cell voltage becomes:
E = 0.40 -0.0591V/4 log((10^-14)^4))/0.2) = 0.40 - 0.014775 - 55.3 = 1.217
Wow! All of a sudden, the cell voltage for oxygen reduction is driven high because of the acid (i.e., [OH-] = 10^-14, not 1 M). So now, copper can be oxidized and will dissolve.
You don't believe me? Drop some HCl (1 M or higher) onto a penny and watch it corrode - and when it breaks thru into the zinc interior, bubbles ahoy!

Why is this downvoted? Formatting aside, this is the correct answer. A tiny amount of the copper will dissolve until the system reaches equilibrium. There is no such thing as an equilibrium constant of 0 or infinity.


text:
Can two metals combine to form a compound?
what I mean is can a metal react with another metal to form a compound? non-metals react with each other and form compounds with covalent bonds can the same happen with metals.

https://www.quora.com/Do-metals-react-to-other-metals

There are all kinds of binary, ternary, and higher order inter metallic compounds.

E.G. BaNa2, https://materialsproject.org/materials/mp-11820/ ... which is not banana.

https://chemistry.stackexchange.com/a/26548/9961 and its discussion in comments.

In short, yes they can. What exactly forms however is quite complicated.
Certain rules have been created to estimate structures but in general it's not as easiy as predicting simple inorganic compounds.
It's part of a topic that we call intermetallic compounds. Sometimes also called alloys, although this term should not be used. We destinguish four different type of compounds that might arise from two metals.

Intermetallic Compound: This has a clear stoichiometry (Sodium thallide)
Intermetallic Phase: There is a certain range
Solid Solution: Two metals can be mixed in all ratios
Alloys: A compound that has more than one of these phases present

Why these compounds form however is often attributed to certain abilities like radii, sometimes electronegativity and something we call valence electron concentration.
In general we seperate the periodic table into four groups:

Group I + II
Group III - XI
Group XII - XIV (parts)
The rest (if metallic)

Now we can consider each combination of two, either from the same group or from different groups. For example group I + group I will often yield simply a solid solution are both compounds simply mix. Group I + say group XIII however have quite some difference in electronegativity. Here we get stronly polar or perhaps even more like ionic bonds, as compounds like NaTl form.
In between stuff like Laves-phases can form, compounds that mostly form deltahedra only, or Hume-Rothery-Phases like brass. At some point you get a lot of possible phases and some means to explain why they form and not something else depending on size, electronegativity, valence electron concentration, etc.
I had a few lectures on these topics and every new lecture would basically introduce a new set of compounds with new rules. You could relate them back to this simple scheme I gave you above but it would always have it's own limitations and set of rules.

You can add Ionic Salts to your list - see https://en.wikipedia.org/wiki/Caesium_auride


text:
Can you give me a mechanism for the following reaction?
Can you give me a mechanism for the following reaction? Here methyl is not shifting. Other ester group is shifting.

What do think? Does the methyl jump --- or not? I believe this has been answered. Edit your query with some ideas. Then I'll look for it.

Ethoxide is a pretty strong base, where do you think the most acidic proton is?

This question answers your question https://chemistry.stackexchange.com/questions/89790/regioselectivity-in-claisen-condensation-and-aldol-reaction

It is claissen condensation happening in equillibrium

Thank you Waylander, I think H next to carbonyl is more acidic and the first ethoxide grab acidic H.But I have doubt, is there any possibility attack ethoxide to the carbonyl carbon of ring? if it is,next step should go through the ring-opening to get product. I am confused about these two ways.