text:
Does inductive effect operate through pi bonds?
I have been told that the inductive effect is only present in sigma bonds by my teacher, and the book I've referenced also says the same thing.
However, I can't think of why this is. Answers to this question seem to agree with me. But Wikipedia also says that inductive effect only operates through sigma bonds.
So which way is it and why?

(From Paula Bruice, page 47)

The presence of double bonds implies  1)a double faced electronegativity for the C atom 2) resonance (mesomeric effect) eventually at work. This might render I effect propagating through pi bonds not so important to be a subject in introductory chapters. Look at https://chemistry.stackexchange.com/questions/4904/why-inductive-effect-operates-only-through-single-bonds where at least one user think as you (and me).

Also, a couple of considerations I do: 1 statements you mentioned are about sigma and pi, not really about single & double. 2 a double bond is anyway a reservoir, it logically damps the I effect propagation through bonds. Like you need water or dump water in a tank placed between two pipe lines.

It's just terminology.  Electron donation or withdrawal through the sigma system is termed "inductive".  Electron donation or withdrawal through the p-orbitals of a pi system is termed "resonance".

@ron not really / not the point. In fact my damping example refer only to the inductive part,normally negligible. The answer is given by user Zhu in the thread I've linked just above.

Sorry, I see that you were aware of the other question. If it helps, I like the answer by Zhu.

@ Alchimista  I'm not sure I buy Zhu's answer.  Fluorobenzene undergoes EAS faster than chlorobenzene due to resonance effects. See this earlier answer.

@ron but inductive effect is about partial charges and formation of dipoles, while resonance is supposed to be a complete dislocation?

@Amadeus resonance can lead to partial charges on atoms.  Look at the allyl carbocation for example.

@ron oh yeah, good point. I'm most satisfied with your answer then; if you convert your comment into an answer I could accept it if you want

@ron. Resonance effect delocalise electrons, inductive ones "pin" them to an atom. A double bond includes a sigma one. This last line makes the question even superfluous for a yes or no answer. It remains to see how big the effect can be, if it is worth consideration, if something useful can be concluded. And the EAS discussion seems totally different to me. It is about the prevalent effect exerted by the halogen. I will read that later.

@Alchimista I don't think there is any reason to doubt that inductive effect would work through the sigma bond in a double bond as the physics involved is the same, so I've edited the question (and Zhu's answer also gives an example though ron has contested that, but I don't believe he disagrees either with the claim that inductive effect works through sigma bonds of double bond, just that the given example is wrong). Actually I had misinterpreted Zhu's answer earlier, now I understand that he was also only talking about inductive effect through the sigma bond in a double/multiple bond.

@Amadeus that is the point I am considering. I also think that, there is no physical reason for not having inductive effect through bonds. However, if you think carefully, the Double bond is effectively a "damper", especially because one side is already electron poor and one electron reach.

@Alchimista that's a very interesting point! I also thought some more about what Ron said and i no longer think it makes sense, resonance only acts in conjugated pi systems (or in presence of lone pairs or ± charges). What about induction in something like Cyclohexanone?

For me the final answer then would be that induction, which is just shifting electron densities due to electrostatic attraction, happens just fine for both sigma and double bonds since the physics involved is the same and there's nothing special about the pi bonds position, but as total electron shifting has to be pretty much the same the induction gets dampened, distributed between the two bonds. Thus in effect it doesn't matter and you can form a general rule (which is wrong but always gives the right result) that all induction only happens through sigma bonds

Creates more confusion than help but oh well

@Amadeus I think your second last comment although could be more elegantly formulated resume the situation. And now that you have changed double into pi in the title, even more. In a way one side of the C involved tends to "repel" electrons, while the other tend to "attract" them. This makes the unit involved a kind of stopper. But the physics is just buried, still at work.


text:
The relationship of the osmotic pressure of a polymeric solution and its the molecular mass

Why is the osmotic pressure of a polymeric solution decreases when the molecular mass is increasing?

I am honestly unsure if the above statement is true, at least it seems to me to be true. When I look at the equation of Osmometry it states that:
If ${\pi} = $ osmotic pressure; $R = $ ideal gas constant; $M = $ molecular mass; $ T = $ temperature; $A_2 = $ second viral coefficient;  $c = $ concentration;  $\chi = $ Flor-Huggins-Interaction parameter; and $V = $ volume, following equations are given:
$$\frac{\pi}{RTc} = \frac{1}{M}+ A_2 c \tag1$$
with
$$A_2 = \left(\frac {1}{2}-\chi\right) \left(\frac{V}{M} \right)^2 \tag2$$
2 in 1
$$
\frac{\pi}{RTc} = \frac{1}{M} + \left(\frac {1}{2}-\chi \right)\left(\frac{V}{M}\right)^2 c
$$
So if $M$ will now rise the osmotic pressure will decrease, this is what I understand from the equation. I also created some synthetic data, where I assumed the second viral coefficient to be 5, which seemed like a valid estimate based on a paper I was reading.
So far so good, but what I do not understand is when I now picture a big polymer chain that will occupy space in a solution and a small one, certainly the bigger chain has more chain solvent interactions. Also from my perspective when I think in occupied space as for concentration, and with higher concentration, I get a higher osmotic pressure, then certainly I thought with more occupied space due to higher molecular mass I should see the same effect.
Background - read-only if you are interested:
I observed that covalently bound polymers to a surface will detach more easily if they are bigger. Hence I thought this comes from the entropic pressure on the chain and I was hoping that osmotic pressure would back me up here.

The statement is true if the mass concentration (e.g., $\pu{g//L}$) of polymers in solution is constant, but false if the number concentration (e.g.,$\pu{mol//L}$) of polymers in solution is constant.
Here are some examples corresponding to the above.  Let's start with the second case:
Increase molar mass while keeping the number concentration of polymers constant.  An example of this would be going from $N$ polymers of length $n$ to $N$ polymers of length $2n$.  Longer polymers have more effect on osmotic pressure, because they provide more interactions with the solvent.  Thus increasing the molar mass at constant number concentration will increase the osmotic pressure.
And now the first case:  Increase molar mass while keeping the mass concentration of polymers constant.  An example of this would be to go from $2N$ polymers of length $n$ to $N$ polymers of length $2n$.  Note the mass would be the same in both cases.
The osmotic pressure will decrease because the former has a higher osmotic pressure than the latter.  That's because, while the number of possible exchange configurations with the water is higher for a longer polymer than a shorter polymer, the exchange configurations available to the monomers are constrained by the fact that they are connected to each other.  If you cleave the polymer into two pieces, you provide more degrees of freedom for the same number of monomers (i.e., for the same mass), and thus increase the osmotic pressure.
The limiting case would be if you divided the polymer into individual monomers.  That would give you the maximum possible osmotic pressure for a given mass of polymer (of course, it would no longer be a polymer at that point).
[Note: In my example I described what happens when going from long polymers to short polymers at constant mass concentration: The osmotic pressure increases. The question posits the reverse case (increasing the polymer length at constant mass), so there the osmotic pressure would decrease.]
Going back to the original statement, my guess is that the author was thinking of constant mass concentration, rather than constant number concentration.  Since mass concentration is often used for polymer solutions, that would not be surprising.

Okay, I think I understand it now. Thank you for your explanation!

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text:
BTX nitration work-up
I'm trying to understand washing step in a btx nitration process. In all these processes the work-up streams are very similar. The reaction mixture passes from the nitrator to a separator or centrifuge, and aqueous spent acid is fed to the recycle loop via a concentrator or strength adjustment stage as appropriate. Crude nitrobenzene, as
the top layer, is passed through a series of washer — separators.
In washer—separator 1, mineral acids and Nitrogen dioxide is removed with water.
In washer—separator 2, residual acid removed and weakly acidic byproducts like nitrophenols and nitrocresols with are removed with dilute alkali
In washer—separator 3, water is used for further extraction. And my question is why? nitrophenolate sodium, nitrocresolate sodium, Sodium sulfate and Sodium nitrate are all in the aqueous phase and should have been separated in the first two steps.

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text:
Error in calculating G-matrix
I have a question regarding the calculation of the $G$-matrix in the classic
"Molecular vibrations" textbook by Wilson (1955, McGraw Hill).
On page 63 it gives an example for the calculation of an off-diagonal term
coupling stretching with bending internal coordinates.
Unfortunately I get a slightly different result when calculating it by hand.
Assume a non-linear triatomic molecule.
Atoms 1 and 2 are on the outside, atom 3 is in the middle.
$s_1$ is the bond between Atoms 1 and 3,
$s_2$ is the bond between Atoms 2 and 3,
$s_3$ is the angle between atoms 1, 2, and 3.
The $s$-vectors are (page 63) given by:
Bond between Atom 1 and 3:
$$
  \vec{s}_{11} =  \vec{e}_{31}
$$
$$
  \vec{s}_{13} =  -\vec{e}_{31}
$$
$$
  \vec{s}_{12} = 0
$$
Bond between Atom 2 and 3:
$$
  \vec{s}_{21} =  0
$$
$$
  \vec{s}_{23} =  \vec{e}_{32}
$$
$$
  \vec{s}_{22} = -\vec{e}_{32}
$$
Angle over Atoms 1, 3, and 2:
$$
  \vec{s}_{31} = \frac{\cos(\phi) \vec{e_{31}} - \vec{e}_{32}}{r_{31} \sin(\phi)}
$$
$$
  \vec{s}_{32} = \frac{\cos(\phi) \vec{e_{32}} - \vec{e}_{31}}{r_{32} \sin(\phi)}
$$
$$
    \vec{s}_{33}
  =
    \frac{
      (r_{31} - r_{32} \cos(\phi))\vec{e}_{31}
      + (r_{32} - r_{31} \cos(\phi))\vec{e}_{32}
    }
      {r_{31} r_{32} \sin(\phi)}
$$
If I calculate $G_{13}$ I get:
$$
    G_{13}
  =
    \mu_1 \vec{s}_{1, 1} \cdot \vec{s}_{3, 1}
    + \mu_2 \vec{s}_{1, 2} \cdot \vec{s}_{3, 2}
    + \mu_3 \vec{s}_{1, 3} \cdot \vec{s}_{3, 3}
$$
$$
  =
    \mu_1 \vec{e}_{31}  \cdot \frac{\cos(\phi) \vec{e_{31}} - \vec{e}_{32}}{r_{31} \sin(\phi)}
    - \mu_3 \vec{e}_{31} \cdot \frac{
      (r_{31} - r_{32} \cos(\phi))\vec{e}_{31}
      + (r_{32} - r_{31} \cos(\phi))\vec{e}_{32}
    }
      {r_{31} r_{32} \sin(\phi)}
$$
(EDIT: in the following equality I wrongly assumed $\vec{e}_{31} \cdot \vec{e}_{32} = 0$)
$$
  =
    \mu_1 \frac{\cos(\phi)}{r_{31} \sin(\phi)}
    + \mu_3 \cdot \frac{
      (r_{32} \cos(\phi) - r_{31})
    }
      {r_{31} r_{32} \sin(\phi)}
$$
$$
  =
    \mu_1 \frac{\cos(\phi)}{r_{31} \sin(\phi)}
    + \frac{\mu_3 \cos(\phi)}{r_{31} \sin(\phi)}
    - \frac{\mu_3}{r_{32} \sin(\phi)}
$$
$$
  =
    (\mu_1 + \mu_3) \frac{\cos(\phi)}{r_{31} \sin(\phi)}
    -  \frac{ \mu_3 } {r_{32} \sin(\phi)}
$$
The result in the book is instead:
$$
G_{13} = -\frac{\mu_3 \sin(\phi) }{ r_{32} }
$$
What am I doing wrong?

My error was that I read the $$\vec{e}_{ij}$$ vectors as if they were canonical vectors.
Especially I implicitly assumed that
$$\vec{e}_{31} \cdot \vec{e}_{32} = 0$$
Which is of course wrong.
If one inserts
$$\vec{e}_{31} \cdot \vec{e}_{32} = \cos(\phi)$$
the correct expression is obtained.


text:
addition of HBr to a conjugated alkene

here we shift double bond and change the carbocation from 3 degree to 2 degree, by doing so we are saying that hindrance matters more than carbocation stability, is this always the case in such reactions ? the carbocation and the 3 degree carbon are both sp2, does hindrance play a major role in triangular planar structures?

Your thought is correct, but the application not quite. You are correct, the resonance structures vary from secondary and tertiary carbocation, because the proton will attack the double bond at the end of the molecule to form the more stable carbocation, but: The more substituted alkene will be considerably more stable. The product you have drawn is substituted three times, but the product you are thinking about is only substituted two times. (If the Bromide ion attacks the tertiary carbocation.).

You should not consider the two carbocations distinct. They are one and the same, the identical molecule. There will be a fractional positive charge observable on the tertiary carbon and a fractional positive charge observable on the secondary one. Looking at orbitals, there is just one LUMO (lowest unoccupied molecular orbital) which in first approximation has large lobes on the tertiary and secondary carbon and nowhere else – a simple allylic system.
There are three effects that influence where the bromine will attack. First, the trisubstituted double bond is more stable than a disubstituted one, meaning the product is thermodynamically favoured. Second, attack on the secondary carbon is less hindered sterically, meaning that the product is also kinetically favoured. And finally, bromide can re-dissociate to a certain extent regenerating the carbocation; it is more likely to do so as a tertiary bromide, again tipping the scales in the direction of product d.


text:
Heat capacity at constant volume and Gibbs free energy
I want to know if it is possible to derive heat capacities, in this case, in constant volume from another thermodynamic Potential which is not the  Helmholtz free energy $F$. I am aware of the following relationship between heat capacity at constant volume, entropy, and $F$:
$$
C_v
  = T\left(\frac{\partial S}{\partial T}\right)_{V,N}
  = -T \left(\frac{\partial^2 F}{\partial T^2}\right)_{V,N}.
$$
But I am trying to see if I can derive an expression for $C_v$ when I consider the Gibbs free energy (particle number constant $N$) and I want to express $C_v = C_v(T,P,n)$ where $n$ is taken as a constant and can be left out:
$$G = U - TS + PV = F + PV $$
The problem I have going forwards has to do with the variables. This is what I mean:
\begin{align}
  C_v &=
    -T\left(\frac{\partial S}{\partial T}\right)_{V,N} =
    -T\left(\frac{\partial^2 (G-PV)}{\partial T^2}\right)_{V,N}\\
  C_v &=
    -T\left[\left(\frac {\partial^2 G}{\partial T^2}\right)_{V,N}
    -\left(\frac{\partial^2 PV}{\partial T^2}\right)_{V,N}\right]\\
  C_v &=
    -T\left[\frac{\partial}{\partial T}
      \left(\frac{\partial G}{\partial T}\right)_{P,N}\right]_{V,N}
    -T\left[\frac{\partial}{\partial T}
      \left(\frac{\partial PV}{\partial T}\right)\right]_{V,N}
\end{align}
Then, I use the following relation:
$$
  \left(\frac{\partial P}{\partial T}\right)_{V,N} =
  -\left(\frac{\partial^2 G}{\partial P \partial T}\right)
   \left(\frac{\partial^2 G}{\partial p^2}\right)^{-1},
$$
which gives me
$$
  C_v =
  -T\left[\frac{\partial}{\partial T}
    \left(\frac{\partial G}{\partial T}\right)_{P,N}\right]_{V,N}
  -T\left[\frac{\partial}{\partial T}
    \left(-V
      \left(\frac{\partial^2 G}{\partial P \partial T}\right)
      \left(\frac{\partial^2 G}{\partial p^2}\right)^{-1}
    \right)\right]_{V,N}
$$
And this is as far as I can get. Here is the problem. If we focus on the first term and do the derivations, I get:
$$
-T\left(\frac{\partial(-S)}{\partial T}\right)_{V,N}
$$
Now I don't know what to do. The entropy was derived by focusing on the Gibbs equation, and now if I want to go further, I have to look at the entropy for the internal energy. But there is nothing I can do. All I can say is that the above term can be equal to:
$$
C_v = T\left(\frac{\partial(-S)}{\partial T}\right)_{V,N},
$$
where this $C_v$ has nothing to do with the one in the beginning. My problem is with the 2nd term. I don't know what to do there because $G$ is a function of $T$, $N$, $P$, and we also have $V$, $N$. If I were to write it explicitly, this is what I cannot solve:
$$\begin{multline}
  \left[
    \frac{\partial}{\partial T}\left(
      -V(\frac{\partial^2 G}{\partial P \partial T}
    \right)\left(
      \frac{\partial^2 G}{\partial p^2}
    \right)^{-1}
  \right]_{V,N}
  \\ = %split here
  \frac{\partial}{\partial T}\left(
    -V\left(
      \frac{\partial}{\partial P}\left(
        \left(
          \frac{\partial G}{\partial T}
        \right)_{P,N}
      \right)_{T,N}
    \right)_{V,N}\left(
      \frac{\partial^2 G}{\partial p^2}
    \right)^{-1}_{V,N}
  \right)
\end{multline}$$
This term:
$$
  \frac{\partial}{\partial T}\left(
    -V\left(
      \frac{\partial}{\partial P}\left(
        \left(
          \frac{\partial G}{\partial T}
        \right)_{P,N}
      \right)_{T,N}
    \right)_{V,N}
  \right)
$$
How do I try and solve this? The indexes confuse me. Can someone give me a hint?

Welcome to Chemistry! On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. There were a couple of parenthesis/brackets missing and I) hope I didn't put them in all the wrong places. Please check.

What you are looking for can be trivially obtained by referring to the well known general relation between constant-volume and constant-pressure heat capacity:
$$
C_v= C_p -\frac{\alpha^2TV}{\chi_T}.
$$
In this formula, present in almost all textbooks on thermodynamics and easily derivable by using Maxwell's relations and some partial derivative manipulation, all the quantities on the right-hand side of the equation can be represented as derivatives of the Gibbs free energy and the independent variable $T$:
$$
\begin{align}
V &= \left(\frac{\partial G}{\partial p}\right)_{T,n};\\
C_p &= T\left(\frac{\partial S}{\partial T}\right)_{p,n}=-T\left(\frac{\partial^2 G}{\partial T^2}\right)_{p,n};\\
\alpha &=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p,n}=
\frac{1}{V}\left(\frac{\partial^2 G}{\partial T\partial p}\right);\\
\chi_T&=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T,n}= -\left(\frac{\partial^2 G}{\partial p^2}\right)_{T,n}.
\end{align}
$$

@Poutnik Actually, in my case,  It was simply a typo. Thanks for pointing it out. I am going to correct it.


text:
What causes the amber/colourless/blue colours in the Briggs-Rauscher oscillating reaction?
I am using this commonly cited but simplified reaction mechanism of the Briggs-Rauscher reaction.

Reaction 1 & 2 occur simultaneously. For reaction 1, when the iodide concentration is low, the fast radical step occurs and when the iodide concentration is high, the slow non-radical step occurs.
The colourless -> amber -> dark blue oscillation of the BR reaction is due to the relative concentrations:

I2 = amber
I- = colourless
I2 + I- → I3 + starch indicator = dark blue

However, I do not understand in what situations the colours arise.

In the non-radical phase, [I-] is high but I am not sure what the
relative concentration of [I2] is. If it is high, I'd assume there
would be a high I3 concentration so the solution would be blue, and
if it is low, I'd assume the [I-] would dominate so it would be colourless.

In the radical-process, since [I-] is low, I assume the colour is amber, given by the high [I2] concentration? And because [I-] is low, there is little [I3] formed so the solution would not be blue.

During the non-radical phase, both [I-] and [I2] decreases (I think) so I'm guessing that is why the colourless stage visually appears to be much shorter than the blue or amber stage (in videos I've found online).
MIT video demonstration
Is this the right explanation for the colours?

Welcome to Chemistry! On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Your explanations are correct, except that the neutral species $\ce{I3}$ does not exist. Only the anion $\ce{I3^-}$ exist.

Also I am not sure the mechanism of the radical process is correct, because it starts from the iodous acid $\ce{HIO2}$ which is not present in the initial mixture

@Maurice that is true, I'm not sure where it came from. Does the reaction start with the radical process instead of the non-radical process? I'm guessing since [I-] starts from low concentration (?), the reaction starts with the radical process?

@SoySoy4444. Both processes start from absent compounds ! The radical process starts from $\ce{HIO2}$ and the non-radical process from $\ce{I-}$, and none of these ions or molecules are present in the initial mixture. So what ??? Is it to be understood that the initial products are not really pure, containing some reduced stuff ? Just enough to start the whole series of rections ? Who knows ?

@Maurice I was wondering about that too, I found it https://people.chem.ucsb.edu/feldwinn/darby/DemoLibrary/DemoPDFs/Demo005.pdf here, do you think you could take a look? I'm guessing it's just because the reaction mechanism is simplified?

@SoySoy4444. I have read your article. It contains exactly the same mechanism, and the same equations that you wrote in the very beginning of your post. So it is no use for explaining the origin of $\ce{HIO2}$ and of the ion $\ce{I-}$.


text:
Can ferrous ions exist in an acidic solution?
Can Fe2+ ions exist in an acidic solution or will they always pick up available protons and become Fe3+? My hunch is they can't exist but I'm not really sure.
If the latter, does that mean that dissolving iron in an acid will raise the pH since it ties up some of the H+? Or will the additional protons still be just as free to float around and interact with other stuff as they are before iron comes to the party?
If they can exist, however, is there some kind of ferric:ferrous equilibrium based on... something?
I ask because I have a solution of oxalic acid and water, with iron dissolved in it, and it is bright yellow-green (like a green or yellow highlighter), and the internet says Fe3+ ions are yellow-green. Also I'm pretty sure there can't be actual iron oxalate particles (which are also green) suspended in the solution (conclusion from this lesson). I'm trying to work out the actual contents of the solution on my own.
So I'm pretty sure I'm looking at ferrIC ions -- and it doesn't seem to make sense that ferrous ions could exist here -- but I just want a sanity check on that guess, since I'm still trying to wrap my head around this.

As an aside: Now after a little more reading, I'm wondering if maybe my oxalic acid solution contains ferrioxalate ions too (or instead), although that's specific to my oxalic acid experiments rather than acids in general. They're also bright green, it seems. Apparently those are light sensitive though, so maybe my UV laser will do something if the wavelength (405nm) is right and so I can test for them? That's an experiment for tomorrow.

Ferrous ions are easily oxidized by oxygen in the air: https://chemistry.stackexchange.com/q/31243/79678. Other questions and answers here may be helpful.

Thanks for digging that up! Sort of OT but does that mean iron ions will readily precipitate as iron oxides in an alkaline solution? Cause that is related to a question I was actually going to post later, to confirm my suspicions about why the acidic solution turns reddish brown as it presumably becomes saturated with iron (which is incidentally the real reason I asked about iron raising the pH in this question), and why adding more acid turns it bright green again. Also why it turns reddish brown when I mix it with sodium bicarbonate. (No spoilers please lol)

(I just needed to confirm my suspicion about the iron ions in acid first before posting the next question, to make sure that question even made sense.)

I think the answer to the first of your two questions is yes, but I am not an expert on iron chemistry and it is a bit complicated anyway. As for adding baking soda, someone else may have an answer or hunch.

Re: This question... now after a little more reading, I'm wondering if maybe my oxalic acid solution contains ferrioxalate ions too (or instead), although that's specific to my oxalic acid experiments rather than acids in general. But still...  Apparently those are light sensitive though, so maybe my UV laser will do something if the wavelength is right? That's an experiment for tomorrow.

@JasonC - Over several pH ranges you will get iron(III) precipitates. As a geochemist, I am most familiar with naturally-occurring goethite and limonite in areas I have studied. Have a look at the iron(III) oxide-hydroxides for more info, including precipitation of iron(III) oxyhydroxide from solution at pH ranging from 6.5 to 8 by addition of NaOH to a solution of ferric chloride/nitrate.

Consider bidental complexes of Fe(II/III) and oxalate have probably different colours than respective ion aquacomplexes.

Found a gold mine of info: Metal Recovery Using Oxalate Chemistry: A Technical Review. (Via https://www.semanticscholar.org/paper/Metal-Recovery-Using-Oxalate-Chemistry%3A-A-Technical-Verma-Kore/0f2c2ba513132d55427ea04dedc4684040df4af4)

Can Fe2+ ions exist in an acidic solution or will they
always pick up available protons and become Fe3+? My hunch
is they can't exist but I'm not really sure.

I was intrigued by your query because there is a hidden misconception. There are no available free floating protons in solution. If you want $\ce{Fe^2+}$ accept a proton, you are asking to cause a nuclear reaction. Of course this is far from reality. Notationally, even if you wanted $\ce{Fe^2+}$ to accept a proton (=from an acid), like the way organic chemists talk about it, it would be written as $\ce{[HFe]^3+}$. You don't want that either, because such a species does not exist.
In short, if you have an acidic solution, free from dissolved oxygen and kept in an inert atmosphere, $\ce{Fe^2+}$, will sit swim happily for a long time.
When we talk about the color of the transition metal ions, usually we are talking about the free ion surrounded by water, for example, $\ce{Fe^2+}$, which is a light green solution, chemists are then actually talking about $\ce{[Fe(H2O)_6]^2+}$.
Iron(II) oxalate, as you can see from Wikipedia is yellow... just like the dry yellow highlighter ink. When the metal ions are complexed (=surrounded) by something which can hold the metal ion like a crab's claw (chelate), you cannot predict the color beforehand.
Another example is that iron(III) thiocyanate is blood red. This has nothing to do with the yellow color of Fe(III).

Perhaps more importantly, even in an acidic solution with available protons (though they are not really free), $\ce{Fe^2+}$ is not going to become $\ce{Fe^3+}$, because, er, well, like you said, this is simply not a thing that can happen.

Now I have to do a lot more thinking to figure out the difference between the bright highlighter greens vs deeper yellow solutions I see. I ordered an electronic pH tester, hopefully that'll help me eliminate some guesses.

@JasonC, What are you trying to do with highlighter inks?

Nothing... it's just the words I used to describe the colors I've observed. :)

Check it: https://www.semanticscholar.org/paper/Metal-Recovery-Using-Oxalate-Chemistry%3A-A-Technical-Verma-Kore/0f2c2ba513132d55427ea04dedc4684040df4af4

Ferrous salts are stable in acids, there are two places in nuclear chemistry where acidic ferrous salts are commonly used.
The solvated protons in an aqueous acid solution are unable to act as oxidants to convert Fe(II) to Fe(III). The only time I would regard the solvated proton as an oxidant would be when considering the chemistry of the solvated electron.
There the reaction
e- + H3O+ --> H.  +  H2O is important.
This reaction generates hydrogen atoms from solvated electrons but it is important to understand that this is a very strong reducing agent being converted into a another strong reducing agent (but slightly less strong reducing agent).
One common high dose doseometer is the Freike which is ferrous sulfate in acid, even when the solution is saturated with oxygen in the absense of radiation the iron stays as ferrous (iron(II)).
The other place I am thinking of is in the classical PUREX process as used in B205, there ferrous sulfamate (Fe(SO3NH2)2) is used as a reducing agent to convert a solution of plutonium(IV) into plutonium(III). What is done is that the iron(II) solution is added to a mixtrure of uranium(VI) and plutonium(IV) in nitric acid. While the tributyl phosphate at 30 % (v/v) in kerosene will extract Pu(IV) and U(VI) from nitric acid it is unable to extract Pu(III) or other trivalent f block elements.


text:
The MOs computed with HF method by programs like Gaussian are symmetry adapted orbitals?
I know that some molecules have symmetry elements, and that we can associate operators to them. These operaors commute with the molecular electronic hamiltonian $H$, so the wavefunction can be chosen as a simultaneous eigenfunction of all these operators and $H$. When we use a basis set to expand the MOs, using, for exemple, gaussian-type functions, we can take linear combinations of these functions to construct symmetry adapted linear combinations (SALCs), that are eigenfunctions of the symmetry operators. My question is: when we do a HF calculation of the molecular orbitals of a molecule using programs like Gaussian, are the MOs SALCs MOs ( independent of the basis )? I am asking this because I was studyng the SFC proceddure using Szabo and I don't remember the book to comment about these questions about symmetry.

Gaussian provides an option to respect symmetry during a calculation. In this case Gaussian will use symmetry adapted orbitals based on the detected point group and will show you in the ouptput to which irreducible representation an orbital belongs.
But you don't have to use symmetry in your calculation and in most cases it is done without it simply due to the fact that small deviations in geometry "break" your symmetry. Especially during geometry optimization on larger molecules with flexible alkyl-residues. The Hartree-Fock method does not really depend on symmetry. Symmetry can speed up calculations but it is not a requirement to carry it out.


text:
Can the electronic configurations give any information about the actual structure of the atom?
The zeroth order wavefunction for an He atom in the ground state is $\psi^0_\text{He}=\psi_{1s}(1)\psi_{1s}(2)$ and hence the electronic configuration $1s^2$. But then we start making purterbations, we are no more talking about the non-interacting electrons as presented in the w.f. $\psi^0_\text{He}$. Hence, the electronic configuration doesn't really tell us about the actual atom, but gives a crude, starting-point-model of it. Same goes for all multi-electron atoms , so can we modify it to represent the actual wavefunctions (like the molecular MOs, they describe the actual molecular structure to a great extent)?

No electron configuration of atoms is just a book keeping. Nobody knows the micro structure of the atom. Orbitals are not observables.

The basis you used for constructing your ansatz wave function for a helium atom assigns the one-electron hydrogen wave functions to each of the two electrons of helium. By doing this, you neglect the interaction between the two electrons and using only your ansatz to describe your system gives poor quantitative results (some basic spectral patterns can be explained based on the electron configuration). One approach to take the electron correlation into account is to include a term in your trial wave function that depends on the coordinates of both electrons and minimize the energy with respect to one or several parameters in your wave function. This is basically the approach of Hylleraas and in more complicated form still the basis of state-of-the-art quantum mechanical calculations (search for example for correlated Gaussians).
Even when we do not take the electron correlation into account we can get more accurate results by using a large basis of products of hydrogen wave functions as you did. This basis corresponds to the ground state and (in principle) all possible excited states of the atom. In practice, you need to truncate your basis somewhere, but the more states you include, the more accurate your results in general will be. You can now evaluate the terms of your Hamiltonian for all cross terms of your basis functions. This results in a (Hermitian) symmetric matrix that can be diagonalized. The eigenvalues correspond to the energies of your system and the eigenvectors to the character of each of your trial wave functions in your state of interest. Mathematically, perturbing your system results in admixing other electronic states to your unperturbed state. What I just described corresponds to the Heisenberg representation of quantum mechanics and which is very convenient for doing numerical calculations.


text:
Isn't my book doing this problem about ionic product wrongly?
Problem:
After adding a $\pu{150 mL}$ $\pu{6.7\times10^{-5} M}$ $\ce{NaOH}$ solution to a $\pu{100 mL}$ $\pu{2.5\times10^{-5} M}$ $\ce{FeCl2}$ solution, green colored $\ce{Fe(OH)2}$ precipitate is found. What is the ionic product of $\ce{Fe(OH)2}$?
My book's solution:
The concentration of $\ce{Fe^2+}$ will be the same as the concentration of $\ce{FeCl2}$. So, using the formula for dilution we can get the concentration of $\ce{Fe^2+}$:
$$\begin{align}
    M_1V_1 &= M_2V_2 \\
    2.5\times10^5\times100 &= M_2\times250 \\
    M_2 &= \frac{100\times2.5\times10^{-5}}{250} \\
        &= \pu{1\times10^{-5} M}
\end{align}$$
In the same way, the concentration of $\ce{OH-}$ will be the same as the concentration of $\ce{NaOH}$. So, using the formula for dilution we can get the concentration of $\ce{OH-}$:
$$\begin{align}
    M_3V_3 &= M_4V_4 \\
    M_4    &= \frac{150 \times 6.7 \times 10^{-5}}{250} \\
           &= \pu{4.02\times10^-5 M}
\end{align}$$
Now, ionic product of $\ce{Fe(OH)2}$, $K_\mathrm{ip}$,
$$K_\mathrm{ip} = [\ce{Fe^2+}][\ce{OH-}]^2 =1.616\times10^{-14} \quad \text{[Ans.]}$$
My observations:
My book assumed that $\ce{Fe(OH)2}$ remains completely dissociated and thus the concentration of $\ce{Fe^2+}$ will be the same as the concentration of $\ce{FeCl2}$, and the concentration of $\ce{OH-}$ will be the same as the concentration of $\ce{NaOH}$. My book's answer is correct only at the very beginning of the process when any $\ce{Fe(OH)2}$ precipitate hasn't formed yet. However, after the precipitate has formed, isn't my book's answer wrong?

For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ).
See also upright vs italic

The textbook uses a simplified approach with a badly worded task. It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \le  \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive. The constant also does not count with ions trapped in precipitation.

@Poutnik Edited the question using \ce and \pu using the link you kindly provided. Moreover, if you post this comment as an answer, I'll accept it as my answer.

Poutnik is right. The problem should have been better written according to : When mixing small amounts of a $\ce{6.7·10^{-5} M}$ $\ce{NaOH}$ in $ 100$ mL $\ce{2.5·10^{-5}}$ $\ce{ FeCl2}$ solution, nothing happens. But when $150$ mL of this $\ce{NaOH}$ has been added, the first $\ce{Fe(OH)2}$ precipitate becomes visible. Calculate the ionic product of $\ce{Fe{OH}2}$ .

The textbook uses a simplified approach with a badly worded task.
It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \lt \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive.
The constant also does not count with ions trapped in precipitation.


text:
How to determine whether a solid has dissolved or no reaction occurred?
From this link: https://chemhaven.org/che111/lab/Lab_9_Unknown_Anion_F21.pdf

When dissolving a solid or precipitate,
only a portion is dissolved, thus leaving some behind, and fooling a student into incorrectly reporting that no reaction occurred,
or that a precipitate formed. This can be avoided by realizing that the original compounds are generally crystalline, and will
sit at the bottom of a test tube (or settle out quickly), while a precipitate will have a milky appearance, and tend to spread
evenly throughout the solution.

I am finding it difficult to understand what this quote means. Let's say I dissolve a solid in a reagent. How am I supposed to tell whether (a) a solid dissolved partially, (b) no reaction occurred, and (c) a precipitate has formed?

I would take it as a warning. Moreover a test involves tiny amount of solid. It basically means wait enough, shake the tube, look carefully. Of course there can be plenty of reactions happening with undetectable signs, in a test tube. It also depends on what liquid is in the tube and what one might expect. Ps the link isn't that useful, as such.

How am I supposed to tell whether (a) a solid dissolved partially, (b)
no reaction occurred, and (c) a precipitate has formed?

With your two eyes as the detector for (a) and (c). For (b) you use your intution/brain and previous chemistry knowledge.
The quote is basically advising that one should be very careful during scientific observations.*
(a) If a solid has dissolved partially, you will see a reduction in the amount of the solid. Practice this instruction by slowly stirring a tablespoon of sugar or salt in half glass of water and see how the amount of solid decreases slowly as you stir.
(b) Nobody can guess beforehand if it is a completely new and an unknown reaction. However, you know what you are adding beforehand and what is present in the test tube. So, this instruction asks you to brainstorm if there is no color change, ppt. formation etc and ask youself: Is a reaction expected in this scenario?
(c) Again use your eyes.

Several major discoveries have been made by good visual observations. The field of complexometric titration was invented by observing that when Eriochrome (dye) solution is discarded in the sink and tap water is opened to wash the dye, the dye changed its color to blue from wine red. What caused the color change? He thought. Calcium and magnesium ions water...this led Schwarzenbach to open a new area of research which is still useful today.


text:
Optimal lipophilic solvent for washing PEEK-lined reversed-phase HPLC columns
What would be the reasonable strategy for washing a reversed-phase (RP) octadecylsilyl (ODS) C18 HPLC columns from lipids and other non-polar residues? The columns are used for biochemical analysis of various matrices using acetonitrile, methanol and water (often with $~0.1\,\%$ formic acid), and it seems that lipids from cell membranes tend to cause clogging and increased pressure over time, despite the derivatized/extracted probes being filtered with 0.20 μm filters prior to injection. The process of washing with an excess (20× column volume) of 2-propanol recommended by manufacturer for removing lipid-soluble substances has only marginal effect, if any.
Usually, in this case it is recommended to try dichloromethane (DCM), dimethyl sulfoxide (DMSO) or tetrahydrofuran (THF) after replacing all plastic tubing and fittings with stainless steel ones. The problem is that the columns have polyether ether ketone (PEEK) lining that obviously cannot be switched, and all three solvents (DCM, DMSO, THF) are known for causing swelling in PEEK parts, possibly resulting in altered column geometry and traces of plasticizers.
Is there a reasonable solution for this issue involving another PEEK- and HPLC-friendly lipophilic solvent, or is there a totally different strategy for PEEK-lined columns?
Please note that I'm asking about the PEEK-compatible solvent in particular, not clogging issue in general:

The clogging occurs at the column, not anywhere else in the system. Replacing the problematic column with a brand new one or installing a resistance tube result in expected pressure values.

The retention times vary within acceptable limits. The columns are relatively new and were used for just several dozens of probes. There are, however, peaks of what appears to be post-eluting components despite a $\pu{40 min}$ washing cycle after each run.

Reversing flow direction and washing at low flow rates of about $\pu{0.1 ml min^-1}$ with either 2-propanol or acetonitrile didn't help at all. Although the pressure was a bit lower for the reversed flow (about $\pu{25 MPa}$ vs $\pu{30 MPa}$).

There is an dark side of HPLC with biological samples...with time, silica bonded column becomes a protein or a "bio-gunk" column, instead of a C18 column because of very strong adsorption of biomolecules on the surface. I gather your main problem is the pressure increase with these samples and the manufacturer's recommendation to wash with 2-propanol is not helping. You did not mention whether the retention time is also changing or not.
(0) The first test is to remove the column from the HPLC and connect the inlet and outlet tubings with a union. Start the flow and check the pressure of the system. If it is unusually high (>10 bar) at 1 mL/min chances are there is a cloggage in the system and the column is fine. Cloggage can occur in the injection system as well, just before the column.
(i) Run a QC test on this column with the manufacturer's suggested analytes and mobile phase. Are the retention times okay, within 0.2-0.4 min of the specified time? If not, this is indicating chemically changing silica surface.
(ii) Reverse the flow direction of the column. Connect the outlet side and make it the inlet. Do you see a significant change in the pressure? If yes, the inlet frit has some precipitated/strongly adsorbed stuff. Continue washing in this "opposite" direction for 3-4 hours and see if the pressure has dropped.
PEEK lined columns often have a PEEK or a titanium frit. With time, it happens that PEEK frit gets compressed due to pressure changes causing a permanent increase in back-pressure.
Edit: Try doing a step gradient of 95% H2O +5 % ACN to 95 % ACN + 5% H2O. Say 10 min of the first to suddenly switching the composition at a low flow rate say at 0.5 mL/min. Repeat this multiple times. ACN is pretty good for removing bio-stuff. Column beds hate viscosity shocks and they collapse fast. This could be tried at 0.4 mL/min overnight.

I edited the question addressing the points, of which we did pretty much everything already (except for QC — we need a new standard solution), that's why I'm asking specifically about the solvent, not the troubleshooting process in general. Thank you though for mentioning that there is a change of damaged plastic frit, that's something worth considering, although I don't know how to check its integrity.

You can also add another clarification: Is your HPLC is compatible with normal phase solvents or not. The pump seal and the injector rotor may die after being exposed to DCM and /or THF. I will be more worried about the pump compatiblity because the column is disposable.  Observing a change in the pressure by reversing the flow direction is indicating that the initial segment of the column is gooed including the frit.

Yes, there is no problem using chlorinated solvents or THF with the rest of the system, but we cannot change the column model because of SOP and method we have to follow. Again, the question is strictly about alternative PEEK-compatible lipophilic solvent or alternative washing procedure.

Okay, as a last resort: Try doing a step gradient of 95% H2O +5 % ACN to 95 % ACN - 5% H2O. Say 10 min of the first to suddenly switching to 95% at a low flow rate say 0.5 mL/min. Repeat this multiple times. ACN is pretty good for removing bio stuff. Column beds hate viscosity shocks and they collapse fast. This could be tried at 0.4 mL/min overnight.

Oh, I think I heard of this method (something along the lines of "shockwave therapy"), but haven't tried it yet, only gradual switching back-and-forth between ACN and aqueous phase. Will give it shot, feel free to add this one to your answer)

When this has happened to me, I usually give up and just buy new columns.  However, you could try LC-MS-grade pentane.  You'd probably want to wash the column in a "ladder" of to avoid miscibility issues.  Water -> ACN -> isopropanol -> acetone -> pentane and then back again might do the trick.  Supposedly, PEEK has a compatibility of "A - Excellenet" with all of these.

I've already done this with hexane in place of pentane — it helped to drop the pressure by a bit. What helped to restore the column's original performance in my case was washing with DMF, the solvent from the third region of solvent-selectivity triangle that hasn't been been covered by other solvents.


text:
Electronic configuration of excited states

Fill the following table:
$$
\begin{array}{lccr}
\hline
\text{Electronic configuration} & \text{Term} & J & E/\pu{cm^-1} \\
\hline
[\ce{Ne}]\mathrm{(3s)^2(3p)^5}   &     &     &     0 \\
                                 &     &     &   881 \\
\hline
[\ce{Ne}]\mathrm{(3s)^2(3p)^44s} &     &     & 71954 \\
                                 &     &     & 72484 \\
                                 &     &     & 72823 \\
                                 & ^2D & 5/2 & 74221 \\
                                 &     & 3/2 & 74861 \\
                                 & ^2P & 3/2 & 84116 \\
                                 &     & 1/2 & 84127 \\
                                 & ^2S & 1/2 & 88351 \\
\hline
\end{array}
$$

My questions are:

How do I compute the electronic configuration of the excited levels in general and in this case?
I know that when taking into account the spin-orbit interaction, the quantum number $l$ is no longer a good number, however we always use the notation with S, P, D, … orbitals, which corresponds to $l = 0, 1, 2, \ldots$ Does this mean that an electron can "move" between different values of $l$ although the occupied states with those values of $l$ remain the same?

"we always use the notation with S, P, D.. orbitals" That's not exactly right; the big letters S,P,D have not much to do with s, p, and d orbitals (small letters). Big letters denote the quantum number $L$ which reflects the total angular momentum of all the electrons, not the quantum number $l$ which reflects the angular momentum of one individual electron. Put another way: big letters describe the electronic state of the atom. Small letters describe the state of one individual electron.

Thank you very much! And is L a good number regardless of the spin-orbit interaction then?

Hmm, if your spin-orbit interaction is very strong (heavy atoms), then $L$ isn't good anymore. For small atoms it's good enough. The reason why term symbols use $L$ and not individual $l$ is not because of spin-orbit coupling (that couples orbital AM and spin AM). It's because of coupling between the orbital AM of different electrons. The steps are: (1) couple all the electron spins $s$ to form one total spin $S$; (2) couple all the electron orbital AM $l$ to form one total orbital AM $L$; (3) and then couple $S$ and $L$ to get the total AM $J$. The term symbol is then $^{2S+1}L_J$.

thank you so much, this clarifies a lot.


text:
Why do we need salt in a salt water battery?
I am having problems understanding why we do need salt added to the water to make this shown galvanic cell work.
I understand that Zinc wants to dissolve into the electrolyte, resulting in $\ce{Zn^2+}$. Since there is $\ce{Cl^-}$ present, it will bond with the Zinc.
At the cathode, the $\ce{H_2O}$ will be electrolysed, leaving $\ce{OH^-}$ behind. I assume the $\ce{OH^-}$ will bond with $\ce{Na^+}$.
I assume this is the reason, why we need the electrolyte: If there would be no electrolyte, $\ce{Zn^2+}$ would remain close to the anode, while $\ce{OH^-}$ would remain close to the cathode, resulting in an electric field that counteracts against the redox-reaction.
I think, adding NaCl results in negative Cl-ions moving towards the anode and positive Na-ions moving towards the cathode, resulting in a closed electric circuit, keeping the whole thing running. If we wouldn't add salt, the electric current should disappear very quickly.
Please let me know if my assumptions are correct. I wouldn't mind if you would explain the whole thing as easy and detailed as possible, since I have obviously not studied chemistry.

Little update: I have just built that galvanic cell and let it sit there for 15min. Voltage was around 0.7V. Current was in the mA-range, drastically increasing when adding table salt.
As you can see, the Zinc-plate got some whitish residue on it. I am unable to scratch it off. Any idea what that is? Could be areas where Zinc-ions left the metal?

The copper plate also shows some change below the waterline, but I have no possible explanation why it did change.

I was also able to see that the current through the wire was decreasing steadily. After adding salt and stirring the water, it showed ~30mA. 15min later it showed ~3mA. When I was stirring the water again, the current was increasing again to ~30mA and was decreasing again thereafter.
I assume after some time cations accumulate around the anode and anions accumulate around the cathode, inhibiting further flow of electric charge.

Any process  leading to cumulation of electric charge quickly forms strong coulombic forces stopping such a process.

The cations do not bond with the anions in solution and this is basically just a variant of the lemon ‘battery’. See here for another variant: https://chemistry.stackexchange.com/a/116181/79678.

Imagine you managed to take from such an electrolyteless cell the current 250 mA for 1 minute. It would mean there is the negative charge 15 C near the anode and the same but positive charge near the cathode. In air, it would be enough for 1 lightning. Something tells me the voltage of such a cell would drop faster than a falcon at attack.

Without the electrolyte, the cell would be a poor reference cell, capable of providing only tiny current and power. Even trying to measure the open circuit voltage, with an ordinary digital multimeter (DMM), would yield poor results because of the $\pu{10 M\Omega}$ input impedance of the DMM.

Okay, I think I now understand, why the salt is necessary. Without Na and Cl, the anode will very soon be surrounded by positive Zn-ions, while the cathode will be surrounded by negative OH-ions. This inhibits further current through our wire, because it will eventually get too hard for Zn-atoms to dissolve. And here Na and Cl come into play: Negative Cl will move to the anode, positive Na to the cathode. This balances out the accumulated electric charge on both sides. Correct?

I was able to notice the following: Once the current has decreased to a low value, carefully "stirring" the water very close to the Zinc-anode, nothing changed. The current was continuing to decrease. But if I did the same very close to the Copper-cathode, the current went up. I assume this means that in close proximity to the Zinc-anode, there is no net-charge present (probably  since the Zn is bonding with the Cl). But close to the Copper-cathode, there must be positive ions present. I assume Na-ions. Is that correct? It seems, by stirring the water, they get dragged away from the cathode.

I ran the experiment a few minutes ago. My photos and results are at this temporary link. So using distilled water gave poor results, as expected: declining voltage and 11 micro amperes of short circuit current. Using saltwater gave about 0.8 V, jumpy, and low mA short circuit current. Better, but still pathetic, as expected for such a simple cell with no salt bridge, etc.

Zinc ions are not bonding with the chloride ions: this is an aqueous solution!

If you recall, the earliest Voltaic cell also consisted of copper and zinc with a piece of paper soaked in salt. For a cell, you do need an electrolyte to maintain charge balance and reduce the resistance of the solution. Pure water has a resistivity of 18,000,000 $\Omega$.cm (~ million ohms!). This is too high and current will be wasted or may not be generated at all. Consider a similar case, remove the two electrodes from the solutions and hold them in air. Does your meter still register a current? Hopefully not, because air has very high resistance too.
Whenever direct current passes through a solution, there must be a chemical decomposition (by definition of an electrolytic cell). Consequently, in your salt water battery, in a closed circuit, zinc should lose electrons and become zinc ions. Zinc ion needs a counter balancing charge which is kindly provided by the salt.
Simultaneously, there is no free cation in water which can be reduced at the cathode, so water has to sacrifice itself to the electrons. As a result of water reduction hydrogen is produced, along with hydroxyl ions. Now hydroxyl ions cannot stay alone, you need a counter ion, $\ce{Na+}$, to form $\ce{NaOH}$.
You can use a pH indicator to see how pH is locally changing near the electrode. There are beautiful demos with a nail rusting in salt water. Prof. Ed did experiments to demonstrate this pH idea using Zn and Cu sitting in salt water. It is clear that Zn side is almost neutral and Cu side has become alkaline (Photo courtesy Ed V).

Updated showing the pH test strip on the copper cathode is deep blue: https://chemistry.meta.stackexchange.com/a/4757/79678. Took way under a minute to start showing blue. The photo was after about a minute, with short circuit condition. Maximum current was about 3 mA, but it dropped a great deal. Too bad this “salt water battery” gets its power, feeble as it is, from the zinc oxidation and not the sea water. And the deficiency of no salt bridge, so the salt water electrolyte has to be replaced continually.

Dear M. Farooq, thank you, this explanation helped me a lot. So Cl-ions balance the charge of the Zn-ions close to the anode (but do not bond to ZnCl_2). And Na-ions will bond with OH-ions close to the cathode. Makes sense. It also makes sense that the current increases when I stirr the water after some time, since new Cl-ions can get close to the anode and new Na-ions can get close to the cathode. Now I have just one last question: Why does stirring the water close to the zinc-anode increases the current temporarily, while stirring the water close to the copper-cathode does not increase it?

Sodium ions do not bond with hydroxide ions in aqueous solution! Please stop saying such things! All the ions are mobile in solution. Their local concentrations in this cell are not the same as their average concentrations. Stirring just averages the concentrations for a short time. You could always add a stirring mechanism.

@Rainer_Zoufal, Yes, remember these are ions, all you need is a charge balance. For each Zn ion, you need two chloride ions. I have not investigated the salt water battery formally but oxygen is known to cause a current enhancement, and so does nitrogen. While you are stirring you might be increasing the amount of dissolved air. Continue stirring, do you see a steady increase in current? I do not know the theoretical reasons.

@EdV, Seeing is believing. Beautiful demonstration of pH, proving the argument of water reduction.

Dear Ed V, dear Farooq, thanks for your help. I think I understand the basic principles now. Regarding the NaOH: I was citing Farooq. But I totally accept that they do not bond. Being spatially adjacent basically gives the same effect - a balanced electric charge.

@M. Farooq: I will do another test tomorrow. I'll let you know if continued stirring keeps the current up.

@Rainer_Zoufal Glad to be of a little help and it was fun to do the experiment! If either you or M. Farooq would have any use of the photos I posted, please download them: they are in a temporary location and I cannot leave them longer than a day or two. You can do what you want with them: I am not posting an answer and have already upvoted you and M. Farooq.

Pure Water, $\ce{H_2O}$, is a bad conductor of electricity. By adding $\ce{NaCl}$ to it, it becomes a good conductor of electricity and allows the electricity to pass through it. Water has only few ions in it that can move. $\ce{NaCl}$, when dissolved in water greatly increases the number of free-flowing ions in it, making salt water a good conductor of electricity.

Try to type $\ce{H2O}$ to get $\ce{H2O}$ // For better site experience, you can find useful these links: notation ,  expressions. ( Not to be applied to titles ) and
upright vs italic

Please note that grave accents in Markdown are reserved for source_code. Chemical name isn't capitalized unless the first word in a sentence. Instead of operating in terms of good-bad scenarios, just use numbers (i.e. cited physical quantities for conductivities). Finally, "NaCl makes water gain charge" doesn't seem to make much sense.

Sodium chloride does not make water gain charge: this is nonsense. Water is a poor electrolyte because it has few charge carriers, i.e., ions, that can move, i.e., constitute a current flow, in response to an applied electrical potential. Dissolving salt in the water greatly increases the concentration of mobile charge carriers, so the salt water is a good electrolyte.


text:
What are the wavefunctions for interacting electrons inside an atom?
The electron cloud for the single electron orbital $1s^1$ is radially symmetric in 3 dimensions.
This is perhaps because of the probability distribution function (p.d.f.) $|\psi|^2 = ce^{-\frac{2r}{a_0}}$ depending on a single variable $r$. On the other hand, the p.d.f. of He atom is the bivariate function depending on the positions of both electrons $|\psi_{\text{He}}|^2 = ce^{-\frac{r_1 + r_2}{a_0}}$. Moreover, both electrons have opposite spins. Will this modify the  distribution of dots in the electron cloud? If yes then how?
More exactly, what is $\psi^2 (x,y,z)$ for each electron in the actual He atom (not the unperturbed He atom in which both electrons don't interact)? Do the electrons lose their individuality so that it is impossible to give separate wavefunctions for each interacting electron?

That turns out to be quite a tricky problem to solve, but fortunately Bethe and Salpeter did in 65 years ago. See https://www.amazon.com/Quantum-Mechanics-One-Two-Electron-Atoms/dp/1614276226 (no endorsement of the shop, just a pointer to the book, likely in your nearby physics library).

I am surprised how difficult is finding data in the standard web to this very simple question. Some plots I saw suggest radial symmetry, however it is hard to know if e-e interaction were taken in account. You can also try Physics SE, eventually.

As you surmised, the two electrons do not have separate wave functions because of their interaction and interchangeability. The single wave function for helium which accounts for both electrons is unknown because we can't solve the Schrodinger equation for that case. See http://websites.umich.edu/~chem461/QMChap8.pdf for more detail.


text:
Help drying polyurethane sealent top coat
I've used a polyurethane water sealant coating for concrete that is called (vulkem 951 part A ) and it was supposed to mixed with part b the catalyst that starts the drying process but do to a miscommunication it was not ....the people from the company that makes it said it has to be removed as it won't ever dry ...we've removed as much as possible and also tried rolling the catalyst or (part b ) on top but it seems to have done little to nothing as the area is still tacky and wet and it's been over a week since it's been rolled out ...imagine a gooey grey paint spread out over a whole balcony that just won't dry and apparently will never dry ....that's my problem any help would be great ...I was thinking maybe ferric chloride powder  or concrete dust and roll it in to the paint substance ? not sure how safe it is for me to personal use ferric chloride ....please help
Best regards
A nervous maintenance man

Disclaimer: Not a chemist, and I'm not familiar with the specific polyurethane you are using. Although I've worked with urethane and epoxies a bit, you should probably get a second & third opinion.
A careful application of heat should help soften the material, hopefully enough to mix with the catalyst.  It might not cure properly but it could at least clear up the sticky situation without resorting to throwing on more chemicals. I've had luck with applying a hair dryer to epoxy that was insufficiently catalyzed. A heat gun might be too hot and even a hair dryer should be kept back at least a few inches from the surface. For my application, the heat helped the process along, however, you must be careful not to burn it and wear a respirator rated for organic solvents because  it will almost certainly out-gas some pretty toxic fumes when you heat it up.


text:
Will a liquid-liquid emulsion and the separated emulsion have same volume?
Let's assume that there is no dispersed gas (bubbles). Not sure about dissolved gas; maybe you can tell me if that matters. Ordinary liquids (aqueous solution and biological oil). Surfactants are present. Same uniform temperature for both emulsion and separated emulsion.
What I'm really asking about - homogenized milk. If it has no bubbles and you centrifuge it sufficiently to separate the phases, is there a volume change? What I need is a practical answer rather than an exactly correct answer - I would say a change <0.1% is negligible. I wouldn't mind also knowing the exactly correct answer also.

The change is negligible.

I think it might in principle depend on the size of the droplets (interphase surface sum) but always negligible.


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Hydrochloric acid from diaphragm electrolysis
I have a diaphragm electrolysis cell, which is made up of an outer container containing the cathode and a small terra cotta pot containing the anode. If sodium chloride solution is placed in the outer container, distilled water is placed in the pot, and current is run through the solution, will hydrochloric acid form at the anode or will hypochlorous acid form? I know that chlorine reacts with hydroxide ion, but would there be a significant amount of hydroxide at the anode if it is separated by a diaphragm?

What are your thoughts?

Putting pure water in the path of electric current is generally very bad idea.

I am only using 6v. Also, ions quickly travel across the diaphragm, so it doesn't stay pure water.


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Why does $\mathrm{NO_3}$ have charge $1-$?
I'm given the following problem:

Write the molecular equation for the reaction $\mathrm{Iron\;(III)\; Nitrate\; and\; Sodium \;Phosphate}$

I begin by attempting to determine the empirical formula of $\mathrm{Iron\;(III)\; Nitrate}$. In class, we were taught to memorize the following table:
Group     Charge
#1        +1
#2        +2
#3        +3
#5        -3
#6        -2
#7        -1

Additionally, we were asked to memorize the "formula" for Nitrate ($\mathrm{NO_3}$).
The $\mathrm{III}$ after Iron indicates that it has a $3+$ charge. From the table above, $\mathrm{N}$ has charge $3-$, and $\mathrm{O}$ has charge $2-$. There are $3$ $\mathrm{O's}$, giving a net charge of $-2 \cdot 3=-6$.
Unable to determine the empirical formula of $\mathrm{Iron\;(III)\; Nitrate}$, I looked it up to find $\mathrm{Fe(NO_3)_3}$.
If $\mathrm{NO_3}$ has charge $6-$, then $3$ $\mathrm{NO_3}$'s has charge $18-$. Meanwhile, $\mathrm{Fe}$ has only charge $3+$.
As it turns out, $NO_3$ actually has charge $1-$. How is this determined? I'm told its "just something that needs to be memorized" in order to progress in the course.

As it turns out, the simple "adding and subtracting" of charge values (from the table given in the question) we were taught doesn't always (ever?) work. Drawing the Lewis structure for $\mathrm{NO_3}$, I obtain:

...the charge is then clear.

The problem is the crooked table: it is not on the level! ;-) Best to toss it.


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Will concentration of sodium hydroxide affect the result of chicken feather hydrolyzate if i don't standardized it?
The title of the paper:
Isolation and characterization of keratin from chicken feathers https://doi.org/10.1063/5.0002792
Here is the method:
Keratin was isolated by dissolving 5g of chicken feather in 95 ml sodium hydroxide 1N solutions at 70°C for 3 hrs. Glacial acetic acid was added to neutralize the solution (pH 7) to obtain a solution of 5% w/v concentration. The chicken feather is then heated at 70°C in a water bath to form a gel. Then heated in an oven at 70°C until dry.
My overall  question whether the reaction between acetic acid and the chicken feather with 1N sodium hydroxide be affected if I didn't standardized the sodium hydroxide?

And the second one is that ..
how did the author ended up with 5% w/v concentration? Does this mean that the precipitated chicken hydrolyzate constitute only 5g of weight and 95 ml of acetic acid is added? I am confuse with how the author obtain a 5% w/v concentration.

The lye is the agent to push the hydrolysis of the peptide forward.  The initial concentration of the lye may affect the rate of this digestion (temperature, pressure, time of residence of the reagents in the reactor are equally factors to be considered), as well if the digestion is exhaustive.  As shown by the reaction scheme, the reaction yields water; water thus dilutes the lye and the concentration of lye decreases.
As stated in your quotation, glacial acetic acid is added after $\pu{3 h}$ of reaction.  This is to quench any excess of lye which did not react (yet). Using glacial acetic acid instead of just «acetic acid» as from the groceries has the advantage of its high concentration ($\pu{100\%}$ HOAc), thus the introduced change of volume caused by the addition of this reagent is much smaller.

Back on an envelope: $\pu{95E-3 L} \times \pu{1 mol/L} = \pu{95E-3 mol}$ of $\ce{NaOH}$ engaged. With data from Wikipedia's entry about acetic acid ($M = \pu{60 g/mol}$, and $d = \pu{1.049 g cm−3}$), one needs at maximum $\pu{5.7 g}$ (or $\pu{5.5 mL}$) of glacial acetic acid to neutralize it if there were no digestion.  (However, because of the digestion of the peptide consuming NaOH, the amount of HOAc needed for the neutralization likely is lower.)
Then, assuming the density of the the digest were $\pu{1 g cm-3}$, you add $\pu{95 mL}$ (from NaOH) and $\pu{5 mL}$ (from HOAc) to yield $\approx \pu{100 mL}$ (in total) containing $\pu{5 g}$ of digest.  Here you are, $\approx \pu{5\%wt}$ (m/v).


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Why does Phenylalanine to Tyrosine release H2O?
So I was reviewing Tyrosine and Phenylalanine metabolism (according to this video) and maybe I'm missing something but in the first step, 1:

Oxygen atom reacts with Phenylalanine to make up Tyrosine, but the remaining Oxygen atom reacts with what exactly to end up as H2O?

The remaining oxygen reacts with the cofactor (called tetrahydro biopterin) of the enzyme phenylalanine oxidase, which loses two hydrogens to become dihydro biopterin (the oxidized form of the co-factor):

As for why: this is a monooxygenase, and an alternative would be to have $\ce{H2O2}$ as a product, but that's a reactive oxygen species the cell is better off without.


text:
Chemical analysis of campholene nitrile
I have a problem. I am synthesizing an organic compound called campholene nitrile (2,2,3-trimethylcyclopent-3-en-1-ylacetonitrile).

I need to analyze it. But there's a problem. I do not work for a university or company, so I do not have access to many of the modern analytical methods. I do have access to:

Boiling point (it's a liquid)
Basic chemical tests

So you see, I don't have that much at my disposal. In case it helps, I'm synthesizing it by the isomerization of camphor oxime with 30% sulfuric acid.
Does anyone have any suggestions for how I could do this analysis using my resources and be reasonably sure of its identity?

It very much depends on what are your purity requisites. If you just need to know if the reaction went well, just follow the literature including the isolation step (likely distillation, if you have that setup). Explain what "I have to analyse it" means. If that is true, likely you have to contact an university service or a private lab.

Can you get TLC plates?


text:
What does “efficiency” refer to for a refrigerator?
A problem from my textbook mentions a refrigerator working with $35\,\%$ efficiency. I thought that in this case the coefficient of performance (COP) was used, specifically $Q_\mathrm{cold}/W,$ which is different from the COP of a heat pump.
Let's say it refers to $W/Q_\mathrm{hot} = 0.35,$ but the refrigerator works between two reservoirs: one at $\pu{0 °C}$ and the other at $\pu{20 °C}.$ If we use the temperature ratio for efficiency, it can't be above $35\,\%.$
So, what does $35\,\%$ refer to? Does it refer to $35\,\%$ of the maximum efficiency?

@ToddMinehardt Hmm, for thermal pumps like refrigerators, is not the thermal efficiency rather $\eta = \frac{Q_\mathrm{in}}{Q_\mathrm{out}}$ ? With $\lim_{T_1 \rightarrow T_2}{(Q_\mathrm{in})}=Q_\mathrm{out}$ for the ideal Carnot-like case? Your efficiency seems to me rather for heat engines which have rather complementary efficiency to the one of heat pumps.

P.S.: Like $\eta_\mathrm{pump} + \eta_\mathrm{engine} = 1$ for the same $T$ range and Carnot-like cycles.


text:
major product formed in wurtz reaction in the given molecule[ 1,3-dibromo-2,2-di(1-bromomethyl)propane ]

This is the reactant molecule which undergoes wurtz reaction. All possible products are-

What will be the most favorable in all of these? In my opinion that the last one should not be the one because it forms a three member ring which is quite unstable due to angle strain. Still it is the answer to the question.
why any other molecule is not the major product?

Reactions that can take place within a molecule generally occur faster than those between molecules

@Waylander And what about angle strain? To what extent we can ignore it's role in determining the product

According to https://doi.org/10.1021/jo01118a042, it is indeed the last one, spiropentane (approx. 50 % conversion). You can foster the search process knowing your precursor's more common name (pentaerythrityl tetrabromide) and CAS number (3229-00-3).

@ashank Yes, it is a strained species, but there are very many cyclopropanes known so clearly the strain is not a determining factor.

@Waylander if we had a molecule (apart from cyclic one) having exceptional stability, can we say that it will be formed more in comparison to the one formed by intramolecular reaction?

@andselisk That was too complicated  for me. I grasped a few things but most things were beyond my knowledge. btw thanks

@ashank I don't think you can make a general statement about this. It will be different for each situation.

@ashank No prob, this was just a published confirmation for the given answer. It doesn't seem authors provide an in-depth explanation as to why spiropentane is favored, so your question remains.

@Waylander I just want to convey that intramolecular reactions are fast, but they might lose in some cases, now that depends .  I believe that we can't really explain some things just by theory if things are happening so close in terms of stability. There are too many factors to look upon.

Spiropentane is preferred, because it does not contain any bromine atom. If enough sodium is available, all bromine atoms will be transformed into $\ce{NaBr}$, so that the final product must be a hydrocarbon with $5$ atoms, or a polymer.

@ashank  Exactly so. That is why chemistry remains an experimental science. You cannot predict the outcome of many reactions so you have the run the experiment


text:
Why SCF procedure in Hartree-Fock doesn't keep returning the same coefficients?
In the context of Hartree-Fock theory, using Roothaan formalism, we write:
FC=SCE
where F is the matrix of the Fock operator, C are the coefficients to be used in the construction of the wave function from the basis set, S is the overlap matrix and E are the energies.
The problem is that the coefficients in C are actually included in F, if we make it explicit. Using Roothaan approach, we just need to choose a basis set to be able to compute all the integrals in F and in S . . . but we need a "density matrix" (a part of F) that takes into account what molecular orbitals are occupied, in order to (partially) account for electron-electron Coulomb interaction.
So the approach is:

Choose basis set and molecular geometry
Compute all the integrals, as stated above
Guess initial density matrix
Solve the matrix equation to find the energies E

We are guided by the variational principle that ensures we cannot "overshoot" in the search for the minimum energy . . . the lower we find, the better (=closer to true ground state energy).
If up to this point i got it right, then my question is the following; after one cycle, as described above, i'd be tempted to say that the correct procedure is this:
I) Use some kind of minimization algorithm to make a new guess for the density matrix, which should be equivalent to saying 'make a new guess for the coefficients in C'
II) Repeat the cycle above
III) Evaluate if the new energy minimum is below the previous one (or check for another, more convenient, convergence criterion)
IV) Repeat until convergence
What instead seems to be the correct way is that, after one cycle, you use the energies to compute a new set of coefficients, which you then use to construct a new density matrix and so on . . . you are now in a cycle (SCF) and you can stop iterating once you achieve convergence according to the criteria you set.
What seems strange to me is that if you guess a density matrix, then solve the Roothaan matrix equation, i would expect the energies to be "consistent" with the guess we've made . . . meaning that we will NOT get new coefficients from the procedure. Could you give me an intuition why this is not the case? Is everything else i've stated above correct?
Thank you.

Can you explain why you "expect the energies to be "consistent" with the guess we've made" - I can't see any reason for this to be the case. Maths would be useful. Please edit the question to explain this.

BTW the minimisation procedure, or at least something conceptually related, is not uncommon in codes that employ a plane wave basis set.

@IanBush it might be that actually doing maths is required to understand what's going on, but i'm trying to understand the method from a bird's-eye view perspective. what i see is an equation: FC=SCe. If i guess C (and hence part of F) i can solve the equation and find 'e'. Then, how can 'e' correspond to a matrix C that's different from the one i used to find 'e'? It seems that this way the equation is no longer satisfied . . .

Can't see how you can do anything without maths. I really can't see why you think a matrix which is a complex non-linear mapping of a set of vectors Q will have eigenvectors C such that C=Q (or strictly C=QU, where U is a unitary matrix). This is effectively what you are asking.

@IanBush I'm not sure i follow you in your terminology. Do you mean that the fact I don't "get back" the coefficients I "put into" the matrix F, once i solve the eigenvalue equation, is due to the fact that those coefficients don't appear in F as a linear combination, but are multiplied together? btw I don't think you have to use mathematical notation, especially when you are trying to build an intuition for something, even though of course maths underlies the whole thing.

I think that the problem you have is that once that you introduce a guess density matrix (step 3) you go straight away to solve the Rothan-Hall equations (step 4) which will give you a new density matrix, a new set of molecular molecular orbitals and a new energy.
You could obtain the energy associated to a particular density matrix using the expression for the energy associated to that density matrix

where $P_{\mu\nu}$ represent the elements of the density matrix, $H^{core}_{\mu\nu}$ are the elements of the core hamiltonian matrixm and the intregals correspond to the value of the bielectronic integrals.

That's an interesting perspective, which puts emphasis on how the energy you get by guessing the density matrix is NOT the energy you get if you then solve the eigenvalue equation: FC=SCe. In fact, the eigenvalue equation comes out if we now put the derivative of the energy with respect to all molecular orbital coefficients to 0, and then try to solve the set of N equations that arise from this. So, yeah, I'd say from this angle it's clearer why the coefficients would change (are no longer equal to my first guess) once i solve the eigenvalue equation. I still need to think a bit about this.


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How are the energies of natural atomic orbitals calculated when they are, strictly speaking, not "meaningful" wavefunctions?
There are several programs that can calculate the "energies" of the natural atomic orbitals(NAO) that come from NBO analysis. However, I have several evidence that, unlike the energies of the "normal" MO's, the energies of NAO's are not trivially determined:

Practically, they're not always equal to the ground-state energies of the same orbital in the neutral atom, even when the oxidation states are the same; for example, the five 3d orbital energies are not the same in a lone palladium atom([Kr]3d¹⁰) and in tetrakis(phosphine)palladium(0).
Theoretically, NAO's are not true eigenfunctions of the relevant Hamiltonial operator, so their "energies", strictly speaking, should not be able to be well-defined at all.

My question now follows- exactly what methods are used to calculate the NAO "energies", assuming that a wavefunction (from HF, DFT, or any other method) is already given?

I would assume that the initial NBO paper discusses this. The newer versions of the program are probably a bit different, but the theory should remain largely the same. Have you read it?

Comparing  single energies of basis functions (or subsets) from different basis sets is not meaningful. There is no reason for the NAO energies of single orbitals to be identical with the energies of functions from a different basis set. The only energy that has to be identical is the energy expectation value the complete electronic wavefunction. This energy has to be the same no matter what basis you chose for the calculation.
Note that the NAO's are specific to the particular wavefunction of your system, since they are based on the concrete one-electron reduced density matrix of your system, which itself is based on the full multiparticle-wavefunction of the system.
This means that the NAO's change as soon as your electronic wavefunction changes. The same oxidation state does not imply the same electronic wavefunction, so there is no reason that the NAO's should be exactly the same in the palladium example of yours and neither will their energies be the same.


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Is the wavelength of light generated through a chemical reaction independent?
The chemical reaction in question is the reaction between Luminol and hydrogen peroxide. Is the wavelength of light generated through this reaction dependent on the concentration of either reactant?
I am inclined to believe that the answer is no but I do not know why. I could not find an experiment that tested this online either and I'm wondering if anyone else tried this. If I do turn out to be correct, what other factors affect the wavelength of the light emitted in this specific reaction (pressure, temperature, etc.)? or is it fixed?

Since the individual photons emitted are the result of electronic transactions in a specific molecule then only things that affect the structure of that molecule will affect the wavelength.

If the concentration of the reactants are changed, the light intensity will change, but not the wavelength of the emitted light.

Generally, absorption and emission spectrum of dissolved substance or intermediate product may change with concentration, directly, or indirectly via higher temperature because of reaction enthalpy. But it would at the best rather slight shift than significant change.

See the table here: https://www.researchgate.net/figure/Effect-of-solvent-on-the-maxi-mum-of-the-luminol-chemilumines_tbl3_230011888.

@EdV, I was exactly thinking the same thing- solvatochromism of luminol in the excited state will slightly change the emission maximum.

@M.Farooq Yeah, the freely downloadable paper at the link is a bit old, but has useful information. I probably should have given the full reference in my other comment. As for the shifts themselves, I still commit the sin of thinking primarily in terms of wavelengths rather than wavenumbers, but it is simple to translate them via an online converter! ;-)


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Filter out the SiO2 ultrafine particles from water solution to measure their total weight and size distribution?
The black soil sample mostly contains the organic matter, chalk and quartz.
It is necessary to measure the relative weight of SiO2 particles that are 2.5 μm or less.
To do that the organic matter and chalk were dissolved by additional chemicals and then the sample was filtered through the paper filter and dried.
Then the dried part of the sample was weighted and Mastersizer 3000E laser diffraction particle size analysis instrument was used to measure the sizes of particles in it.
The measured relative number of particles with size from 2.5 μm to 0.6 μm was 3.6%
But Mastersizer 3000E did not find any particles smaller then 0.6 μm perhaps because they went through the paper filter into the filtered out solution.
The question is how to catch that particles that are smaller then 0.6 μm in this case ?
Which filter or process can be used to collect and return them to that dried part of the sample ?

There are plently of commercial membrane filters which can readily filter out particles less than 0.6 $\mu$m. All you have to do is to buy the right one. Simplest and most common filters are HPLC filter for solvents. They can easily block > 0.2 $\mu$m particles. Search "Filter membrane, Nylon 47 mm, pore size 0.20 µm". They are not very expensive.
Note that you cannot quantitatively scrape out very small particles on nylon filters. Also pore size definition is "The validated retention/removal cut off for a membrane. Note that a 0.22 $\mu$m membrane does not have 0.22 $\mu$m “holes” but rather a pore structure validated to remove all particles or microbes above its stated pore size."

Most dangerous are so called PM0.1 that are smaller then 0.1 μm. But 0.20 µm HPLC can be helpful anyway because it should give more realistic number for PM2.5, thank you for suggestion.

There are 0.1 micron filters too. I am 100% sure there are standard protocols for all these things. Search the protocols to buy the right filters.

Could you recommend some protocol collections and keywords to search ?

Start searching ASTM https://www.astm.org/Standards/D5907.htm, ASTM D5907 - 18
Standard Test Methods for Filterable Matter (Total Dissolved Solids) and Nonfilterable Matter (Total Suspended Solids) in Water. Sadly not free. Also try EPA methods


text:
Racemization of hydrocarbon in presence of AlCl3
Our textbook mentioned that the compound in the diagram racemizes in presence of $\ce{AlCl3}$, but it didn't explain anything.
Any help regarding the mechanism of the reaction would be appreciated.

I think you'll get better results if you specifically ask the question you have, not just stating a situation you don't understand.

@StianYttervik I just want to know how the reaction proceeds. Was it that difficult to decipher from the post ?

@HelloWorld1729 Please cite your textbook, use proper compound name ((butan‐2‐yl)benzene), mention conditions and your thoughts on mechanism. Also, please refrain from using MathJax in titles and feel free  to check out this page, this page and this one on how to format your posts better with MathJax and Markdown.

A plausible reaction leading to your product is the following reaction:

i.e., a mixture of (R) and (S)-configurated product is formed.  In the presence of $\ce{AlCl3}$, the alkylhalogenide becomes something like a carba cation, where you potentially loose stereogenic information (-> racemic mixture). In the work-up, the addition of water destroys [quenches] $\ce{AlCl3}$, thus the forward and backward reaction no longer is catalyzed, and eventually, it is possible to separate the two isomers.
However, if you mix one of the two pure isomers with $\ce{AlCl3}$ (or an other Lewis acid, e.g., $\ce{FeCl3}$ and heat the mixture, you will again generate a mixture of alkylated benzene.  (I intentionally use mixture here; because the installation of an alkyl group increases benzene's character as a Lewis base [$\pi$ bonds!], eventually, you might end up with more than one alkyl group installed, too.)
The reversible character of the Friedel-Crafts alkylation may be used to your advantage, though to block e.g., the p-position relative to an already installed substitutent directing in (ortho) and (para)-position, especially with a tert-butyl group.


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What does Partial charge means in Dipole Dipole forces
Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of the dipoles possess “partial charges” and these charges are shown by Greek letter delta (δ). Partial charges are always less than the unit
electronic charge ($\pu{1.610e–19 C}$). The polar molecules interact with neighbouring molecules. The figure shows electron cloud distribution in the dipole of hydrogen chloride and thereby shows dipole-dipole interaction between two HCl molecules

I want to know how partial charges can exist as it would go against the quantization of charge?

It is simple: partial charge = measured dipole moment / measured distance of the dipole centers ( e.g. atom distance with a polar  bond ).

Partial charge is a formal, not quantized charge, that would cause for the given displacement the given value of dipole moment. // By other words, if there is probability 0.5 of occurance of charge 1 e, it is equivalent of probability 1.0  of presence of formal charge 0.5e.

The title of the question and the question within the body are different. It would be nice to clarify which one is to be answered.

Think about the probability of finding an electron is a specific region and it will be clearer
To understand where partial charges come from remember that electrons are not located in one specific place (Heisenberg!) but spread around in a cloud. In molecules, those clouds are called molecular orbitals which are more complex in shape and distribution than the atomic orbitals in isolated atoms.
In simple molecules with different atoms those orbitals are not, in general, symmetric around the two atoms. So in hydrogen chloride, for example, the electrons in the bonding orbital are more likely to be found near the chlorine. The bond still only contains 2 electrons each of charge -1. But the probability weighted distribution shows more negative charge around the chlorine.
So the partial charge is a function of the probability distribution of electron density which is dependent on the 3D shape of the bonding orbital. It isn't a violation of quantisation of charge, just a consequence of the quantum mechanics of bound electrons which can't be nailed down to being in a single specific place because of the uncertainty principle.
From the outside, this looks like a dipole with a partial charge and that simplification is good enough for most systems and avoids excessively complex quantum mechanical calculations that would add little to the picture.

Very good question but keep in mind that partial charge concept is a gross oversimplification. The very first sentence on Wikipedia on partial charge Partial charge is not fully correct either. Consider the partial charge as a book-keeping "number". Just like the oxidation state concept, a "partial" charge can be estimated for a diatomic molecule AB, on the basis of electronegativities
$\delta_A = G_A -L_A - B[\chi_A/(\chi_A+\chi_B)]$
where, $\delta_A$ is the partial charge, $G_A$ is the group number, $L_A$ is the unshared electron number on A, B is the number of bonding electrons, and $\chi$ is the electronegativity value proposed by Allen. Allen avoids the term "partial charge" and calls it this number "Lewis-Langmuir Atomic Charges." So don't treat this number as a fraction of an elementary charge.
The following references will help:
Ionization Energies, Electronegativity, Polar Bonds, and Partial Charges
Ronald J. Gillespie, James N. Spencer, and Richard S. Moog, J. Chem. Educ. 1996, 73, 7, 627.
"Lewis-Langmuir atomic charges." Allen, Leland C., Journal of the American Chemical Society 111.25 (1989): 9115-9116.
And finally if you are interested in who came up with this $\delta$ symbol see...
Jensen, W. B. (2009). The Origin of the" Delta" Symbol for Fractional Charges. Journal of Chemical Education, 86(5), 545.
An old 1959 article in the Journal of Chemical Education addresses this idea. See Models for demonstrating electronegativity and "partial charge" R. T. Sanderson, J. Chem. Educ. 1959, 36, 507.

Knowing molecules to be composed of two or more “point” positive
charges in relatively fixed positions, imbedded in a diffuse cloud of
electrons, one can recognize that the assignment of partial charges to
the individual atoms must at best be only a deliberate simplification.
It is indeed an attempt to describe an exceedingly complex system in
terms useful toward reaching a practical understanding of the
molecule.

Furthermore he writes,

All these physical data, when considered together with the properties
of compounds, strongly support the conclusion that the condition of a
combined atom does indeed depend on its environment, which to a large
and usually dominant extent results from electronegativities. One may
call this contribution “partial charge” or by any other name, as long
as its true significance is clearly understood: “Partial charge’ is a
quantitative estimate of the relative extent to which differences in
electronegativity of the atoms before compound formation have altered
the electronic field about the nucleus of an atom in a compound. In
the following discussion it will become apparent that as an index of
the condition of a combined atom, partial charge provides a uniquely
useful and satisfactorily reliable basis for explaining much of
chemistry


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Is there a molecule which has only lone pair electrons?
Lone pair electrons are "valence electrons that are not shared with another atom in a covalent bond" (Wikipedia).
Is there a molecule which has only lone pair electrons?

That wouldn't be a molecule. Lone atoms are fine with that, obviously, and so called van der Vaals molecules technically don't have bonding pairs.

Only if atoms can mutually bond and not to bond at the same time.

Thank you guys, I could not understand that part.

An ion pair, if the bonding can be described with good accuracy as fully ionic.

The answer is No ! If a molecule had only lone pairs, it cannot be a molecule. A molecule has bonds, and these bonds are made of paired electrons. Unpaired electrons cannot be used for bonding atoms. So a molecule must have more bonding electrons than non.bonding electrons.

Yeah, but these must be single-atom molecules. As soon as you have multiple atoms, you are going to have a bond, and the bonding electrons are inevitably no longer lone pair electrons.

When copy-pasting links from Wikipedia, please try to avoid using mobile links which go like en.m.wikipedia.org; instead change them to the usual desktop links en.wikipedia.org. If somebody is reading on mobile they will be directed to the correct version anyway. Also, it's better to quote whatever you need from the page, rather than just putting the link without explanation.

Aren't singlet carbenes not examples of this?

@ml in a singlet carbene, there is a lone pair on the carbenoid carbon but also two bonding pairs from that atom to the substituents.

What about nitrogen monoxide and dioxide?

Simple answer: Almost. It is nearly the case that all the valence electrons of the $\ce{CsF(g)}$ monomer can be considered as lone pairs.
At high temperatures you can have gas-phase salt monomers, i.e., single molecules of $\ce{MX(g)}$, where M is a metal ion and X is a non-metal ion (e.g., $\ce{NaCl(g)}$). [See: Gas-Phase Dimerization of Sodium Halides and Formation of Mixed Sodium/Phenyl Iodides] And these would meet IUPAC's definition of a molecule:

An electrically neutral entity consisting of more than one atom (n>1). Rigorously, a molecule, in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

https://goldbook.iupac.org/terms/view/M04002.
If the bond between the ions were purely ionic then, yes, you could have bonding in a molecule where there are only non-bonding electrons, i.e., only lone pairs.  However, even the most ionic bonds have some covalent character, which means there is some electron sharing.
There are many different ways to assess ionic character. For the purposes of answering this question, I'll (somewhat arbitrarily) use Pauling's formula applied to the difference in electronegativity, since it's simple and well-known.  However, note that there are other measures of ionic character besides electronegativity difference; and even when using the latter, there are many formulas besides Pauling's.
Among atoms whose electronegativities have been measured, the most ionic bond possible would be that between F and Cs, which have Pauling electronegativities of $3.98$ and $0.79$, respectively. [Source: Wikipedia: Electronegativities_of_the_elements].
Pauling proposed the following formula to estimate the amount of ionic character in a bond:
$$\text{Amount ionic character} = 1 - e^{-\frac{1}{4}(\chi_A-\chi_B)^2}$$
Plugging those electronegativities into the above formula, we find that the bond in $\ce{CsF(g)}$ has only a small amount (8%) of covalent character:
$$\text{amt. ionic character} = 1 - e^{-\frac{1}{4}(3.98-0.79)^2} = 0.92
\implies \text{amt. covalent character} = 0.08 $$
Thus we can conclude that, in the most ionic molecules, while there is some electron sharing, it is small.  Hence all the valence electrons can nearly be considered purely as lone pairs.
Note:  Using Pauling electronegativities and Pauling's formula gives a rough approximation of ionic character.  A more sophisticated analysis could yield an answer that, while qualitatively similar (it would still show very little covalent character in $\ce{CsF(g)}$), could differ quantitatively.
As an aside (and a caution):  The formula in the IUPAC Gold Book ( https://goldbook.iupac.org/terms/view/IT07058 ) is wildly incorrect, and has apparently been so since 2014.  It looks like some things got lost during the typesetting.  I emailed IUPAC to suggest a correction.

Addendum
It seems we shouldn't be too hard on IUPAC, since Pauling himself appears to have a typo in his own formula! Here's a screenshot from:  Pauling, Linus. The Nature of the Chemical Bond. Third Edition. Ithaca, NY: Cornell university press, 1960, p 98.  [Available from:  https://archive.org/details/natureofthechemicalbondpauling/page/n115/mode/2up ] You can see the electronegativity difference in eqn. 3-15 is properly exponentiated, but it also should be squared, and it's not:

To confirm that there's a typo in Pauling's book, take a look at the table at the bottom of the above page. If we don't square the electronegativity differences, we don't get the right values (calculations on left); if we do square them, we do (calculations on right) (calculations done in Mathematica; these are screenshots of the inputs and outputs):

"Fr should be slightly less electronegative than Cs, but its electronegativity hasn't been measured". Not everyone agrees. See the discussion under "Characteristics" here. Relativistic effects are entering in.

It seems that the original article (https://doi.org/10.1351/pac199971101919) from which the Gold Book entry was excerpted also got the formula wrong.

...Hmm, now I'm pretty intrigued; do you happen to know where Pauling originally proposed this formula? I've been taking a look at https://doi.org/10.1021/ja01348a011 and some other articles in the same series he published, but with no luck. Have to get back to real work now, but I might try looking again later.

With regard to reporting Gold Book errors, please see my comment here: https://chemistry.stackexchange.com/a/115884/79678. Stuart will dig into it!

@OscarLanzi Thanks, I accidentally reversed the statement about francium in the Wikipedia article I cited. It actually said "francium is expected and, to a small extent, observed to be more electronegative than caesium". I'll edit my post. BTW, do you know of more sophisticated assesments of the %ionic character of the Cs–F bond and, if so, what do they say? Also, for salts, do you know if the bond dissoc. energies on which the electronegnativity differences are based are for ion pairs or ions in crystals? I'm wondering about the diff. in %ionic character for ionic bonds in pairs vs. crystals.

@EdV Actually, Stuart was the one I emailed :)!   See also the addendum I added to the end of my post.