text:
R-S configuration or Cahn-Ingold-Prelog configuration
How to assign R and S configurations in alkenes and more complicated structures ?
I know that the groups attached to the sp3 hybridized carbon is first prioritized by atomic number then we see the next set of atoms in the unassigned groups.
But it is said that groups containing double or triple bonds are assigned priorities as if they are duplicated or triplicated. What does this means ?

My suggestion - Google for "cip rules double and triple bonds" - you will find more examples than anybody here can list.

Duplicated means that if there is double bond between say C and O then consider that two O are  linked to C through single bond and with that O two C are linked to it each.
The following figure will help more.


text:
How to determine the iron content of a screw?
In Analytical Chemistry, I was asked to determine the iron content of a screw.
I thought it could be solved in the following way:

Treat the screw with a solution of sulfuric acid or hydrochloric acid.
As the iron in solution can be found in the form of $\ce{Fe^{II}}$ or $\ce{Fe^{III}}$, I would reduce the possible $\ce{Fe^{III}}$ ions with $\ce{SnCl2}$ or Walden's reducer.
Titrate the solution with potassium permanganate.
From the volume spent, the amount of iron present could be determined.

I don't know if it will be OK, it is the first time I have faced a problem of this type.

Weigh the screw.

Yes, you're right on that! @imalipusram

First, use sulfuric acid, to avoid Cl- ions. Second, If you titrate the solution within one hour after the end of the acidic dissolution, the amount of ferric ions formed will be negligible. And you can immediately titrate with $\ce{KMnO_4}$ without using Walden's reducer. I have done it in my classes for years without trouble about ferric oxidation.

Weigh the screw and deduct 0.7 % for Mn, Si, C, and few tramp elements.

There is another problem with this.
Potassium permanganate is a very strong oxidizing agent, which is capable of oxidizing Chloride ions to Chlorine. This will result in additional consumption of $\ce{KMnO4}$ solution, which makes you think there is more iron in your screw than there actually is.
To solve this, you should look up the titration of Iron using the Reinhardt-Zimmermann solution. It contains sulfuric acid, Manganese (II) sulfate and phosphoric acid.
The $\ce{MnSO4}$ changes the redox potential so the acidic $\ce{KMnO4}$ solution can no longer oxidize Chloride ions. Because Iron(III) gives an intense yellow solution, the addition of Phosphoric acid is necessary since it will react with Iron(III) to colorless Iron phosphates. That makes it easier to detect the end point.
So the way I did this back in my first semester was to make sure we have all Iron as Fe(II) by adding an excess of $\ce{SnCl2}$. The excessive $\ce{SnCl2}$ was quenched using $\ce{HgCl2}$. Then the Reinhard-Zimmermann solution was added and titration with $\ce{KMnO4}$ could start.

Thank you very much. I really like analytical chemistry. I am surprised by the amount of knowledge you have, would you recommend me any book to deepen? Because, for example, in the same problem I am also asked to explain how to determine the lactose content of a commercial milk using titrations or gravimetries.

Nice method, which textbook did you use? Today, it will generate a lot of fuss if someone teaches to use mercury salts in titrations.

You almost have the chemistry right, you miss the part about radical formations. For those interested in the likely actual reaction mechanics and not century-old chemistry, see my now edited answer on the suggested underlying chemistry below.

A simple, safe and effective electrochemical based path (absence strong acids which, acting by themselves, are generally inert on stainless steel at intermediate temperatures) to gather surface metal salts for testing is, per a source, to quote:

The iodine uptake, and corrosion of the stainless steel surfaces was extensive. The observed time-dependent behaviour suggests that the iodine adsorption on the stainless steel surface occurred in two stages, a slow induction period followed by a fast linear adsorption rate region. Iodine deposited on the stainless steel was desorbed during purging with air, but not with N2. These observations could be explained by the formation of metal-oxy-iodide scale, in which iodine is initially incorporated into but subsequently replaced by oxygen. This paper presents the preliminary results of the study, and their potential implication for "calibration" of iodine transmission through stainless steel sampling lines.

Some hot I2 vapor based corrosion on a stainless steel screw in an atmosphere of oxygen is a path to gathering surface formed oxy-iodides. The latter can be used to assess metal content of the alloy which is mainly Fe, but significant alloyed elements of Ni, Cr, Cu,..., also possible.
Also, in time, even very dilute iodized water is problematic on stainless steel, to quote NASA:

The effects of stainless steel exposure to iodinated water is a concern in developing the Integrated Water System (IWS) for Space Station Freedom.

Interestingly, the rate of corrosion may be reduced (or, even possibly accelerated) in the presence of chromium and nickel, to quote a source:

Iodine is capable of reacting with iron in the steel, as well as with nickel and chromium. This serves to complicate the situation. As well, there may exist synergistic relationships between these metals. This means that the presence of one metal may inhibit or enhance the reactions of iodine with one of the other component metals. Hellmann and Funke [19] performed a literature review of relevant reactions on this topic.

[EDIT] I found a good albeit dated (like 1913) reference of on the Zimmermann-Reinhardt (ZM) method here. I must correct, however, my misperception above that, somehow, a combination of H2SO4 and 85% H3PO4 in the presence of ferrous or stannous constitutes a 'strong' acid mix that attacks stainless steel. The less appealing truth, having an advanced knowledge of the reaction mechanics, KMnO4 (a substitute for H2O2), fosters powerful acidic radical formations. This is the actual basis of ZM method (which some may cite as historical, and others, as inefficient, reagent exhausting, and possibly dangerous route to the sulfate radical anion ($\ce{.SO4-}$) and likely the phosphate radical anion as well).
Now while, I have proclaimed elsewhere that the nitrate radical anion (see comments and source here), for example, exceeds the nitrating ability of concentrated Nitric acid, this was not well received. Here, the sulfate radical anion apparently assists in the attack of stainless steel and its creation, for the record, does not require strong H2SO4 or HgCl2 or KMnO4, just a source of hydroxyl radicals via a Fenton/Fenton-like reaction with transition metals in their lower valence states at an appropriate pH, or a photocatalyst creating an electron-hole in the presence of water, introducing the powerful hydroxyl radical ($\ce{.OH}$), which furthers reacts with the nitrate or sulfate or chloride ion as follows:
$\ce{ .OH + SO4(2-) -> OH- + .SO4- }$
$\ce{ .OH + NO3- -> OH- + .NO3 }$
$\ce{ .OH + Cl- -> OH- + .Cl }$
$\ce{ .Cl + .Cl -> Cl2 (g) }$
However, apparently couched in the veil of nostalgia ZM is applauded, congratulations.
For those actually interested in modern chemistry, I quote some revealing passages from the cited ZM source:

If dilute permanganate is allowed to run into a cold dilute solution of ferrous chloride containing hydrochloric acid,  the permanganate is decolorized and the iron is oxidized, but there is a noticeable evolution of chlorine...If, however, permanganate is run into cold dilute hydrochloric acid in the absence of ferrous chloride, there is no evolution of chlorine.

So, an acidic promoted Fenton/Fenton-like reaction with ferrous (or stannous, also mentioned in this source work):
$\ce{ Fe(2+) + H2O2 (or KMnO4) + H+ -> Fe(3+) + .OH + H2O }$
and the creation of chlorine as detailed above. The sulfate radical itself is likely a stabilized transport for the powerful, but transient, hydroxyl radical, to quote source:

It has also been suggested that the sulfate radical anion may be converted to the hydroxyl radical in aqueous solution.

So, I speculate that the attack of steel may be due to an aqueous sulfate radical assisted release of the hydroxyl radical on the iron surface:
$\ce{Fe + 2 .OH -> Fe(OH)2}$

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text:
Constraint optimization of the linear H4 and H6 chain with gaussian
I want to optimize the linear hydrogen chain ($d_{\infty h}$, equal bond lengths) with gaussian 16.
Using z-matrix
# opt=z-matrix hf/sto-3g

H6 chain

0 1
 H
 H  1  r1
 H  2  r1  1  a1
 H  3  r1  2  a1  1  d1
 H  4  r1  3  a1  2  d1
 H  5  r1  4  a1  3  d1
Variables:
r1=0.9
Constants:
a1=180.0
d1=0.0

is not possible, since angles of 0° and 180° are not allowed.
 Error on Z-matrix line number      3:  angle Alpha is outside the valid range of 0 to 180.
 Error on Z-matrix line number      4:  angle Alpha is outside the valid range of 0 to 180.
 Error on Z-matrix line number      5:  angle Alpha is outside the valid range of 0 to 180.
 Error on Z-matrix line number      6:  angle Alpha is outside the valid range of 0 to 180.
 Conversion from Z-matrix to cartesian coordinates failed:

Using generalized internal coordinate optimization (GIC) module
#p hf/sto-3g
opt geom(AddGIC)

H4 chain

0 1
 H
 H  1  r1
 H  2  r1  1  a1
 H  3  r1  2  a1  1  d1
Variables:
r1=0.9
Constants:
a1=180.0
d1=0.0

R12(active)=R(1,2)
R23(active)=R(2,3)
R34(active)=R(3,4)
R23(modify)=R12
R34(modify)=R12

LinBend1H1H2H3(freeze)=L(1,2,3,0,-1)
LinBend2H1H2H3(freeze)=L(1,2,3,0,-2)
LinBend1H2H2H4(freeze)=L(2,3,4,0,-1)
LinBend2H2H3H4(freeze)=L(2,3,4,0,-2)

does not work either.
 ***** Axes restored to original set *****
 -------------------------------------------------------------------
 Center     Atomic                   Forces (Hartrees/Bohr)
 Number     Number              X              Y              Z
 -------------------------------------------------------------------
      1        1          -0.000000000    0.000000000    0.093021603
      2        1           0.000000000   -0.000000000   -0.205327093
      3        1           0.000000000    0.000000000    0.205327093
      4        1          -0.000000000   -0.000000000   -0.093021603
 -------------------------------------------------------------------
 Cartesian Forces:  Max     0.205327093 RMS     0.092025570
 Leave Link  716 at Fri Apr 24 15:09:36 2020, MaxMem=  1342177280 cpu:               0.4 elap:               0.1
 (Enter /usr/local_rwth/sw/gaussian/LINUX_emt64/g16.c01_bin_avx2/l103.exe)

 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
 Berny optimization.
 NTrRot=    -1 NTRed=     6 NAtoms=     4 NSkip=    -1 IsLin=T
 Problem with coordinate system.

How can this be done?

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text:
Internal energy of a gas
My professor says that the internal energy of a gas is the sum of its kinetic energy and potential energy: $U = E_k + E_p$.
It also says that the root-mean-square speed of gas molecules is $v_{rms}=\sqrt{\frac{3RT}{M}}$. Rearranging this formula gives: $E_k = 1/2*M*v_{rms}^2=3/2*R*T$.
Further, my textbook also says that the potential energy of an ideal gas is 0, because there are no attractive or repulsive forces in the gas.
Then, it seems as though the first equation becomes: $U = 3/2*R*T + 0 = 3/2*R*T$, but we know that the internal energy of an ideal gas is not always $U = 3/2*R*T$; it changes based on the state (constant temperature or constant pressure) and the geometry of the gas molecule (monatomic, diatomic, or nonlinear).
What am I missing?

Yeah, internal energy of ideal gas changes with geometry of gas molecule, ig, it's $f/2 , ,  nRT$ where $f$ is degree of freedom. For ideal gas internal energy is function of temperature alone, i.e. $f(T)$. Maxwell-Boltzman distribution

@Zenix I understand that the internal energy changes with geometry. However, what is wrong the "derivation" that U always = 3/2RT in my post?

U=3/2RT IS valid only for ideal monoatomic gas. For other ideal gasses, rotational and vibrational modes have to be included. For real gases, U=f(T) is not linear, mainly due vibration quantization.

You should distinguish between translational and vibrational KE.

@BuckThorn can't we ignore vibrational energy at low temperatures?

The factor $\ce{3/2}$ that you use in the expression of the internal energy, is only valid for rare gases. For diatomic molecules like $\ce{O_2}$ or $\ce{N_2}$, this factor is $\ce{5/2}$

The only intensive variable that affects the internal energy of an ideal gas is the temperature.  The pressure has no effect.   Thus, for one mole of monatomic ideal gas, your equation (U = 3/2 RT) is correct.  And, for the same reason, your statement that the internal energy "changes based on the state (constant temperature or constant pressure)" is not correct. You might be confusing internal energy with the heat capacity, which is dependent on the constraints (constant p vs constant V).

@theorist what is the internal energy of a diatomic ideal gas then? is it 5/2RT?

Yes.   A monatomic ideal gas has 3 translational degrees of freedom (x, y, and z), and thus has $U = 3/2 N k T = 3/2 nRT$ (where $N$ is the no. of particles, and $n$ is the no. of moles).   A diatomic ideal gas adds two rotational degrees of freedom (no chemical bonds, so no vibrational degrees of freedom), giving $U=5/2NkT=5/2nRT$.

@Maurice  The 5/2 factor only applies to real gases like $\ce{O_2}$ or $\ce{N_2}$ at low temperatures.   At higher temperatures, since they are real gases, you need to add the vibrational contribution.


text:
Particle in a box with different barriers between two molecules
I'm trying to understand different ways to solve a PIB model.
I'm given a molecule with a tetrahedral geometry and there are two related ground states - electron pair pointing up or down. I'm being asked to draw the n=1 and n=2 states in three separate cases of a finite potential barrier in the middle of the box -> VBarrier = 0, VBarrier = V0, and VBarrier = ∞. So that left of the barrier are the wavefunctions of the up orientation and right is the down orientation.
I know that for the case of VBarrier = ∞, it'll just be the same sin(n pi x/L) wavefunctions on both sides of the barrier. I don't understand how they would change for V=0 and V=V0, and why these barriers would have different heights? I'd be grateful for any help.

Some things seem very strange about your question, PIB questions never deal with any kind of tetrahedral molecule or spin of any kind. What you are dealing with is a 1 dimensional problem, or in other words a line which doesn't correspond to any molecule. In any case, for the no barrier you have a box of L, rather than 1/2L. for the barrier in the middle you will have tunneling of the electron from one box to the next

Sorry, I may have made the question too confusing by paraphrasing. The actual question deals with ammonia's tetrahedral geometry, which inverts at a certain frequency. So (I think) I'm being asked to model a PIB by drawing and finding wavefunctions for each geometry, which will lead to the wavefunction for the composite system

for the infinite barrier I'd put an electron on just one side of the box with length L/2, meaning that electron is trapped on one side/face of the ammonia.
For the no barrier part treat the electron as a box with length L meaning it can easily go from one side to the other without barrier (for example maybe the ammonia is at the planar transition state).
For the finite barrier, I'd show tunneling from one side to the second, however, normally for PIB the tunneling region has finite length.
Assuming you have a tetrahedral geometry, according to the Born-Oppenheimer separation of electron and nuclear motion the finite barrier would extend from L/2 to L (or equivalently from 0 to L/2). Therefore the tunneling would be from the middle to the edge.
Image source

In the case of ammonia the N atom inversion can be considered as a 1D problem as it moves from one side of the H atoms to the other. There is a barrier of about $2000$ cm$^{-1}$ to cross. The figure shows a calculation which approximates that for ammonia.

You can see that the energy levels are doubled below the potential and this is because of tunnelling from one side to the other, i.e. one side feels the potential from the other and this splits the levels. (The lowest two levels are so close they appear as one in the figure.) The wavefunctions take on 'odd' end 'even' character, i..e are inverted (odd) or reflected about the middle of the potential.  Even above the barrier the wavefunction is distorted compared to that at far higher energy where it approximates to a sine wave as seen in a normal particle in a box.
Also notice that as the potential is not infinitely high the wavefunction tunnels into the potential either side. When tunnelling the wavefunction decays exponentially.
(A hartree is $\approx 27.211$ eV, which is $\approx 2.195\cdot 10^5$ cm$^{-1}$ and $a_0$ is the Bohr radius $0.529\cdot 10^{-10}$ m.)


text:
Do SALC-AOs really belong to their symmetry species?
I'm working through a molecular symmetry textbook and something keeps nagging at me. If I derive the SALC-AOs for NH3 (using the projection operator method), I'll get
A1:  $  \frac{1}{\sqrt{3}}(\phi_1+\phi_2+\phi_3)$
Doubly-degenerate E:
$  \frac{1}{\sqrt{6}}(2\phi_1-\phi_2-\phi_3)$
$  \frac{1}{\sqrt{2}}(\phi_2-\phi_3)$
where the $\phi$s are the H 1s orbitals, ie.:

(This was the best image I could find, but in the second E orbital the black s-orbital should be enlarged)
I'm having trouble seeing how these SALC-AOs would be a member of their symmetry species. The A1 SALC-AO is identical under each operation, so it makes sense that it would belong to A1. But if I want to confirm that the two E SALC-AOs belong to E, how would I do that?
My intuition is that I should be able to apply each C3v operation to the SALC-AO and get back the E row of the character table (Below). But if you apply a C3 rotation to the e2 orbital, you get the black orbital taking the place of the white orbital, the white taking the place of the node, and the node taking the place of the black orbital. This doesn't seem like it could be expressed as a number in a character table, since the SALC-AO isn't being taken to itself or -itself.
Am I thinking about this all wrong? How should I understand the SALC-AO as a whole? Any guidance is appreciated, the textbook seems evasive on this.

To show that your two functions do represent the $\mathrm{E}$ irreducible representation, we can approach the problem algebraically.
We start by representing the operators of the $C_{3v}$ point group in the basis of the AOs. The identity element is easy in any basis:
$$E=\pmatrix{1 &0 &0\\0 &1 &0\\0 &0 &1\\}$$
We can see that applying this operation to $e_1$ and $e_2$ will just return these functions, so $\chi(E)=2$. We can also represent this operation in the basis of $a_1$, $e_1$, and $e_2$, which will be important for explaining the remaining operations.
For the $C_3$ rotation, we can notice that this just has the effect of moving each AO to the next site (i.e. $\phi_1\to\phi_2$, $\phi_2\to\phi_3$, and $\phi_3\to\phi_1$), so we can write it in matrix form as
$$C_3=\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}$$
Lets see what happens when we apply this to $e_1$:
$$C_3e_1=\frac{1}{\sqrt{6}}\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}\pmatrix{2\\-1\\-1}=\frac{1}{\sqrt{6}}\pmatrix{-1\\-1\\2}=e_1'$$
Writing $e_1'$ in the basis of the original functions, we obtain:
$$e_1'=0a_1+\frac{-1}{2}e_1+\frac{-\sqrt{3}}{4}e_2$$
Doing the same with $e_2$:
$$e_2'=0a_1+\frac{-\sqrt{3}}{4}e_1+\frac{-1}{2}e_2$$
We can then take the trace of this matrix (or, to simplify, the 2x2 submatrix spanned by $e_1$ and $e_2$) to get the character of $C_3$:
$$\chi(C_3)=\mathrm{Tr}\pmatrix{\frac{-1}{2} & \frac{-\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4} & \frac{-1}{2}}=-1$$
You can go through the same sort of exercise with $\sigma_v$ (determine form in AO basis, apply to each function, write resulting functions in original function basis, take the trace to get the character).

Thanks for the answer! I'm not sure I understand (mathematically or conceptually) what you did when you wrote e1'  "in the basis" of the other SALC-AOs. Could you explain how you got there? Also, because you take the Trace of a matrix representing transformations of both e1 and e2, it seems like you're saying that e1 and e2 don't individually conform to the E symmetry species, but together they do. Is that correct?

So just like you wrote $a_1$, $e_1$, and $e_2$ in terms of the AOs, you can write further functions in terms of $a_1$, $e_1$, and $e_2$. So you can write any function you would have written with the AOs in terms of your functions instead. So, for $e_1'$, I solve for a combination of your original functions that will produce the vector.


text:
How fast will oxygen be produced through electrolysis with a 9V battery?
What kind of rate of oxygen production could be achieved with a simple 9V electrolysis setup using water?

Adding to what @Poutnik said, a 9 V battery is literally a battery of 6 AAAA cells. If you dissect a 9 V battery, you will see this. Better to use a surplus DC “wall wart” from some dead (or gone) electrical gizmo. And 3 or 4 volts is usually fine, if the electrolyte concentration is not too low.

By the way, this is a trivially simple experiment to try at home. Put water in a drinking glass, dissolve Epsom salt (magnesium sulfate heptahydrate, from the grocery store) in the water and completely submerge a new 9V battery in, standing upright. You can collect either gas using an inverted test tube or substitute. You can figure out which battery terminal has oxygen bubbles rising from it. The battery will have to be tossed later. Anyway, don’t expect anything but a slow gas evolution rate.

If you use pure water, you will not get any H2 or O2 at all, because no electrolysis can be done in pure water, whatever the voltage. You have to add some solute, like sulfuric acid ou sodium sulfate. And the yield in H2 and O2 will be proportional to the concentration of this added substance. Anyway, the yield in H2 and O2 is proportional to the surface of the electrode and inversely proportional to the distance between them.
So there are three parameters that should be known to be able to answer your question : the concentration of the solute, the surface of the electrode, and the distance between them. The voltage is important, but the three mentioned parameters are much more important.

There is too many unknown variables, like composition and conductivity of electrolyte plus internal resistance of battery.
Generally, these batteries do not provide high current and most of the too high voltage is wasted on heat production.


text:
Does an orbital have its own energy, separate from the electron? Can I picture them as a entity independent from the hosted particle?
Does an orbital have its own energy, separate from the electron? If so, is it possible for an electron to have a different energy than the orbital it occupies?

Nope, energy is not stored in the energy levels themselves but the states which occupy them. The different orbitals and energy levels are just possible states that the electron can be in, some high energy and some low energy.  therefore, the idea of an electron and orbital are not separable when talking about atoms and molecules.

Does your passport have any money or property of its own, separate from you?

Very interesting question. It doensn't make sense to separate the energy, it would be like dismissing a fine because your car couldn't be above the speed limit, just its velocity. But the closest to that, I belive, would be treat the particle and its wavefunction as separate physical entities. That approach was tried by De Broglie and David Bohm early in quantum theory, to solve the duality problem. See pilot-wave theory, and this very nice macroscopic analogue discovered in 2010.

An orbital of current quantum models of atoms and molecules has 3 major interpretations:
orbital(1) as  a wave function, being a solution of the quantum wave equation describing behaviour of electrons around an atomic kernels. Functions do not have energy, even if they describe it.
orbital(2) is the 3D region describing probability of occurance of an electron, related to orbital(1) . Space regions do not have energy either.
(Not exactly true, considering radiation energy and quantum vacuum energy, but nothing comparable to the electron energy at usual conditions.)
orbital(3) is a quantum state of an electron, described by orbital (1) and orbital (2), consisting of set of discrete values of physical quantities as functions of quantum numbers, including energy. In the case this quantum state is not occupied, the orbital energy means "an electron would have this energy".

orbital (4) from the obsolete Bohr model means a path of an electron, which is a crude approximation of the quantum model of an atom.

Last, but not least: Orbitals are not physical things, they are part of the quantum theoretical model of atoms and molecules. Within the given quantum model, an electron cannot have different energy than energy of it's quantum state.
As models of multielectron cases are imperfect, real electron energy usually more or less differ from its energy predicted for the given quantum state.

I'll add a fourth // orbital (4) - path of an electron in the Bohr model which is a crude approximation of the quantum model of an atom.

Hmm, I am not sure, it is not interpretation of post Bohr quantum model orbitals.

Sorry, that was the point that I was trying to make. The Bohr model isn't the best model but it is a useful step into understanding atoms. It takes considerable math skills to understand/appreciate the quantum model.

Sure. I have added it.

Thank you for your responses. If orbitals don't really exist by themselves as separate entities, why do we talk about them that way? Example: when reading about the hydrogen atom, my textbook comments that 2s and 2p are actually degenerate for hydrogen... but those are unoccupied and hence shouldn't exist.

They do not exist as a physical object, but they do exist as objects of the quantum mechanical model of atoms/molecules. The 2s, sp, and even higher orbital CAN be occupied, if the hydrogen atom has the electron in an excited state. See Lhydrogen spactral series


text:
Why is the Standard Entropy of a Proton Zero?
It was my assumption that the standard entropy was the absolute entropy of a substance at standard state; however, my textbook states various standard entropies ranging from $-144.77\ \mathrm{J/(mol\ K)}$ for $\ce{Al^3+}$ to $0\ \mathrm{J/(mol\ K)}$ for $\ce{H+}$ to $213\ \mathrm{J/(mol\ K)}$ for $\ce{CO2}$. It is clear that these values are not absolute entropies, as the only substance that can have an absolute entropy of $0$ is a perfect crystal at $0\ \mathrm K$ according to the third law of thermodynamics. Why are standard entropies notated like this instead of just stating their absolute form? My assumption is that this is just a way to make calculations involving entropy easier, since they merely involve changes in standard entropies; however, why is the standard entropy still notated $S^\circ$ and not instead $\Delta S^\circ$ (since it is not absolute)?

Indeed, it is clear that these values are not... Wait! What "these values"?
You are putting two entirely different things in the same basket. (If your textbook does so, then it does a poor job.) It is much like treating dogs and cupboards similarly, on the basis of both having four legs. Then, however, many generalizations will fail quite miserably. You feed them meat, but the results are not always great. You call in the carpenter to fix them, but...
Same thing here. See, ions and molecules are very, very different things. You totally can have a mole of $\ce{CO2}$. It is a thing. You can work with it, can measure its heat capacity and stuff, and eventually work out the entropy. Indeed, the textbook value of $\ce{CO2}$ entropy is absolute.
On the other hand, you totally can't have a mole of $\ce{Al^3+}$. It is not a thing. Unless you have some counterions, that is, but in that case you can't tell how much of the entropy comes from the counterion and how much from $\ce{Al^3+}$.
How can we assign entropies to ions, then? Why, it is as simple as that: we assign one ion arbitrarily, then work out the entropies of all its counterions from there, and then turn to their counterions. Just what one ion could that be? Why, let's say it is $\ce{H+}$, because so much of our chemistry is dominated by the acid/base equilibria. It is kind of natural to consider its entropy $0$, much like the absolute elevation of the sea level is $0$. Now there you have them, the textbook entropies of ions, which are not quite absolute.
So it goes.

So basically, it is impossible to determine the absolute entropy of an ion, so instead we determine its entropy relative to a H+. In the case of performing calculations (such as Gibbs free energy) it would not matter whether or not the value is absolute, just if there is a change in entropy values. This makes sense; however, if we ran a reaction in which ions were produced, such as the reaction between sodium and chlorine to form NaCl, these values would no longer be relative to each other (with one set absolute and the other relative to H+), so how do we compare these changes in entropy?

Well, the values for cations can all be increased by the same arbitrary constant, but then the values for anions will be decreased by the same constant. See, these things are not quite absolute, but not quite arbitrary either.


text:
How to calculate the amount of water required to increase the pH of a solution from 2.1 to 4?
After working out the problem with certain approximation of considering the initial pH=2, my answer came out to be around 1 litre of water. But the correct answer is 79 litres of water. Can anyone kindly help me with this problem and if possible tell me what I probably missed out on?

Unless we know what's in there, the answer can be anything.

Exactly sir, I don't understand the same. But I think we can assume it to be a litre of strong acid in the initial case since the question came under calculation of pH of strong acids.

Good. Now what is the initial volume of the solution?

One litre , sir.

Then the correct answer makes sense, and your one doesn't. To tell more, I need to know more. How did you come to that answer of yours?

If the acid is $\ce{HCl}$, the problem is how to dilute HCl with pH $2.1$ up to pH $4$. But pH $2.1$ means that $\ce{[H^+] = 10^{-2.1} = 7.94 10^{-3} M = c_1}$. Applying the expression $\ce{c_1V_1 = c_2V_2}$, with $\ce{V_1 = 1 }$ L, and $\ce{c_2 = 10^{-4}}$ M, the final volume $\ce{V_2}$ will be $79.4$ liters. It also means that the volume of water to be added to $1$ liter is $79.4 - 1 = 78.4$ liters


text:
Calculating pH of reaction mixture of silver nitrate and ascorbic acid
I am designing an experiment but before I do, I wanted to understand if the following method is right to calculate the expected $\mathrm{pH}$ when I mix $\pu{0.736 M}$ $\ce{AgNO3}$ and $\ce{C6H8O6}$.
I want the $\ce{AgNO3}$ to be the limiting reactant for this reaction. I think chemical equation will be:
$$\ce{2AgNO3 + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2HNO3},$$
which would mean that there would be 2 hydrogen ions.
$$\ce{2Ag+ + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2H+}$$
If I start with $\pu{0.736M}$ $\ce{AgNO3}$ and I plan to have that in $\pu{1L}$ water solution (also intend the $\ce{C6H8O6}$ to be in $\pu{1L}$ water solution), then the moles of hydrogen generated will be:
$$\pu{0.736 mol} \ \ce{H+}$$
This would mean the concentration of $\ce{H+}$ will be
$$\frac{\pu{0.736 mol}}{\pu{2L}} = \pu{0.368M}$$
So the $\mathrm{pH}$ of the solution would be $-\log 0.368  = 0.434$? This number does not seem right since ascorbic is a weak acid, is my logic correct?

If you start equimolar and the reaction goes to completion, there is no more ascorbic acid. What pH do you expect at the beginning of the reaction?

To rephrase Karsten's comment - The acid build up in the solution is the result of a redox reaction where ascorbic acid is oxidized to dehydroascorbic acid, not the dissociation of ascorbic acid.

That is quite a lot of silver. Is that experiment worthy of it ?

Would the pH of ascorbic acid when added to water be the pH in the reaction with silver nitrate? Is that what you mean @MaxW

user510 - If you don't add any other acid, then yes the initial pH would be due to the dissociation of ascorbic acid.

Ok so the pH initial pH will be the pH with silver nitrate? Wouldn't the extra volume increase the pH? Because you initially put x mol ascorbic acid in 1L water solution but then you add it to another 1L solution, wouldn't this impact pH?

Poutnik has a good point. I'd sure use smaller quantities of solution until I knew the reaction was going to work like I wanted. 10 ml of each would reduce the cost of the experiment to 1% of using 1 liter quantities.

@user510 yes diluting would increase the pH not lower it. // pH increase indicates the solution is less acidic, pH decrease means more acidic.

Yes sorry that is what i meant. So that is what I was trying to calculate, how much the pH will increase by

Whether it is worked or not, you should make it more diluted. As Poutnik pointed out, this silver nitrate would cost you an upside of $250.00.

Ok point considered but could you help me in understanding how the pH will change when ascorbic acid solution will be added to the silver nitrate solution? How can I calculate that?

With high concentration of reagents, there would possible the created nitric acid would be slowly dissolving silver back to the nitrate.

For solutions of weak acids and with simplifying assumptions [H+]=[A-], [A-] << [HA ], there is formula:
$$\mathrm{pH}=\frac12(\mathrm{p}K_\mathrm{a} - \log {c})\tag{1}$$
For our case, if the stoichiometric ascorbic acid concentration $0.368/2=\pu{0.184 mol/L}$ is used:
$$\mathrm{pH}=\frac12(4.1 - \log {0.184}) \simeq 2.42\tag{2}$$
If reaction is completed, $\mathrm{pH}$ would be given by $\ce{HNO3}$ only.
Situation would get complicated, if nitric acid starts to react with metallic silver or residual ascorbic acid. These side reactions would decrease mineral and eventually also organic acidity.
$$\ce{3 Ag + 4 HNO3 -> 3 AgNO3 + 2 H2O + NO}$$
$$\ce{3 C6H8O6 + 2 HNO3 -> 3 C6H6O6 + 4 H2O + 2 NO}$$

If more of the ascorbic acid is used, initial $\mathrm{pH}$ is again given by the above equation.
The ending $\mathrm{pH}$ is given by equation:
$$[\ce{H+}]= c_1 + [\ce{A-}]\simeq K_\mathrm{a}\frac{c_2}{[\ce{A-}]}\tag{3}$$
where $c_1$ is the nitric acid concentration, $c_2$ is the residual ascorbic acid concentration. and the right equation side is taken from $(1)$
Concentration of $\ce{H+}$ is equal to sum of $\ce{HNO3}$ concentration $c_1$ + $\ce{H+} from weak acid, that is equal to A- concentration ( HA -> H+ + A-). The latter is then approx equal to simplified extression derived from the acid dissociation constant.
$$c_1 \cdot  [\ce{A-}]  + {[\ce{A-}]}^2 - K_\mathrm{a}\cdot c_2  \simeq 0 \tag{4} $$
$$[\ce{A-}]=\frac{-c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{5}   $$
From (3):
$$[\ce{H+}]=\frac{c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{6}  $$
$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ \frac{4\cdot c_2 \cdot K_\mathrm{a}}{c_1}}}{2} \tag{7}  $$
In the case by we use acid concentration twice as much as stoichiometric one, then $c_1 = 2c_2$. As $K_\mathrm{a}=\pu{7.94e-5}$:
$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ 2 \cdot K_\mathrm{a}}}{2} \simeq c_1 \cdot  1.00004 \tag{8}  $$
You can see the dissociation of the weak acid is almost suppressed by $\ce{H+}$ from nitric acid, so we can neglect the weak acid contribution to $\mathrm{pH}$.

Could you explain how eq (3) was derived

Ok so 0.1 M $\ce{AgNO3}$ will have pH then of essentially -log(0.1) = 1?

But if it reacts with ascorbic acid the pH will still be close to 1 which you described above because the H+ concentration will be dominated by nitric acid

Ok I see. I am going to add 1.25 times stoichiometry concentration (to make sure silver nitrate is limiting) so I do not think that will drive up the pH of the initially $\ce{AgNO3}$ solution

is c1 = $\frac{\sqrt {0.736 x K_b}}{Volume}$?

Also, if I start with 0.736 moles (0.736 M) $\ce{AgNO3}$ and 0.368 moles (0.368M) ascorbic acid, by stoichiometry, I should get 0.736 moles hydrogen ions. However, by using $\ce{[H+] = c1 + [A-]}$ I would get total hydrogen concentration of 0.736+0.368 = 1.104 x 2 L = 2.208 moles hydrogen ions and not 0.736 moles that I should get by stoichiometry?

How is this accounting for the change in volume that will occur from mixing the two solutions?

If c1 = 0.368 mol / L then the pH = 0.434. Isn’t this very low? Because it is a salt reacting with a weak acid I feel like the pH cannot be below 1.

Is my logic correct?

Right, my concern was if the pH is so low I probably cannot use stainless steel equipment to mix the reaction (which is what I have been using but I have not seen any eating away of the steel that would happen at such low pH causing my doubts of if the pH could get this low). It is for reduction purposes, therefore, I am using ascorbic acid (which should give the ascorbate ion). If you could please clarify what you meant by "better is using ascorbate, or hydrogenascorbate" that would be very helpful.

Ya sorry I got it. But why would research paper just talk about Ascorbic acid rather than using the ion? Could it be they use dispersants which convert the Ascorbic acid to its ionic nature before mixing it with the silver?

Yes I was looking into that but the reactions become complex adding mix of all three with all the intermediates.

Wait I am confused again, if the H+ is coming from the ascorbic acid to begin with which is a weak acid, how can the product get so acidic. Sure, nitric acid is the product but the hydrogen ion is coming from the ascorbic acid dissociation so there should not be as many hydrogen ions to begin with

Right but only a small fraction of ascorbic acid will dissociate into hydrogen ions? So I guess the 0.184 mol/L of ascorbic acid you begin with, only a small fraction of that, maybe 0.01 moles hydrogen ions are  produced, so I am confused how then in the product side (by stoichiometry), there would be 0.368 mol/L

What happens to the equilibrium reaction when something is eating the product ?

to regain equilibrium it will produce more of the product? But how does it get more hydrogen ion from?

Write all equations and think.

Well the only places it can get it from is ascorbic acid or water. I do not think the acid will dissolve more under these conditions so then I am guessing it is coming from $\ce{H2O}$

Think longer. H2Asc <=> 2 H+ + Asc^2-. Asc^2- + 2 Ag+ -> Asc + 2 Ag. What will remain, if there is stoichiometric ratio AgNO3 and H2Asc ? H+ and NO3-.


text:
Concerning Solubility Product Equations and Phases?
NaOH(?) <---> $Na^+$(aq) + $OH^-$(aq)
What is the phase of the NaOH, I originally thought it was Aqueous due to it being readily soluble in water but I also heard that it was common practice to always write Solubility Product Eq'ns putting the substance on the Left hand side as solid as Solubility is a continuum and every substance is soluble to a certain degree.
Thank you very much.

Even in $\ce{NaOH(s)}$, there are no $\ce{NaOH}$ molecules, but ions $\ce{Na+}$ and $\ce{OH-}$, as sodium hydroxide is a ionic substance.
When it is being dissolved, water molecules pull ions from the solid ionic lattice and wrap them into their hydrated forms.
So it is written
$$\ce{NaOH(s) ->[\ce{H2O}]Na+(aq) + OH-(aq)}$$
If it were e.g. a benzoic acid, its crystals contain molecules, that are dissolved and then it dissociates:
$$\ce{HA(s) <=>[\ce{H2O}]HA(aq)<=>H+(aq) + A-(aq)}$$


text:
Acidic strength
How is ammonia more acidic than methane? My teacher told me it is so as I am confused as ammonia is basic in nature and methane is neutral so how can ammonia be more acidic than methane?

Does this answer your question? Strength of weak acids and bases: Is methane better base than ammonia?

The question is a very good one. I upvoted it because it shows how a student can see quite clearly what the teacher is either too unfamiliar with or too familiar with.
The basicity of ammonia and the neutrality of methane is the situation in the world of the student. These characteristics are readily observable. But the teacher is describing a quite different world.
Chemists are never really satisfied with the status quo, so they do reactions where extra energy is added. While methane is neutral with respect to water, because it is unreactive toward water, I could rip a proton off if I tried hard enough, and if I did that, I would get CH3-, which is extraordinarily basic and would itself rip a proton off almost everything else to go back to CH4. Compounds like methyl sodium or sodium methide (NaCH3) are so air-sensitive that they are handled in special equipment. Some alkali salts of hydrocarbons can be dissolved in solvents, like butyl lithium in hexane, but they are quite air-sensitive too.
It's easier to rip a proton off NH3 to form NH2-. Amides are easier to handle: NaNH2 is a solid that melts at 200C, and can be handled (if quick) in dry air, but it will react with water vapor, pulling off a proton to give NH3 and NaOH. If you thought NaOH was a strong base, think how much stronger a base NaNH2 is! But NaNH2 won't pull a proton off of CH4, or hydrocarbons in general, so NH2- is not a strong enough base, or CH4 can be said to be a weaker acid.

Just want to ask one more thing is there any other nature of a molecule except acidic,basic and neutral as if it is not so how can the total nature of a molecule can exceed 100%?

Not sure what you are asking. We don't say that a molecular nature has 100% of anything; we describe certain properties - they can go from zero to some large number (like polarity, or dipole moment), or from diamagnetic to paramagnetic (repelled by or attracted to a magnetic field), or colored (photon absorption energy). Perhaps buoyancy could go from - to + (something sinks or floats), but that would be characteristic of density of a quantity greater than a single molecule. Do you want to clarify?

For ammonia, there are 2 different acido-basic reactions:
$$\ce{NH3 + H+ <=> NH4+}$$
$$\ce{NH3  <=> NH2- + H+}$$
In the former, ammonia is relatively strong base with weak conjugated acid $\ce{NH4+}$.
In the latter, ammonia is very weak acid with very strong conjugated base $\ce{NH2-}$.
It is related to the liquid ammonia  autodissociation:
$$\ce{2 NH3(l) <=> NH4+ + NH2-}$$
The base $\ce{NH2-}$ cannot exist in water, as it reacts with it as with an acid:
$\ce{NH2- + H2O -> NH3 + OH-}$
Methane is an extremely weak acid, weaker than ammonia,
$$\ce{CH4  <=> CH3- + H+}$$
making $\ce{CH3-}$ much stronger base than $\ce{NH2-}$.


text:
Which site would NaCN attacks (SN2)?
Question

Attempt
I thought that the bromine atom farther from the ring would be substituted because of less hindrance, but the book says opposite, why?

The reaction proceeds by SN2, which is highly accelerated at the benzylic position, due to stabilisation of the transition state by conjugation with the aromatic ring.

So you'll have to replace the benzylic bromine.
References: Benzyl chloride: chemtube3d


text:
Reaction of metal carbonate with acids
It is said that acids react with metal carbonate to form salt carbon dioxide and water.
Sodium carbonate reacts with hydrochloric acid to form sodium chloride , carbon dioxide and water
But if we ionise them then sodium carbonate dissociates to form  two sodium ions and carbonate ion and hydrochloric acid disassociates to form two hydrogen ion and two chlorine Ion so the sodium Ion will combine with the chloride Ion to form NaCl and the hydrogen ions will combine with carbonate ion and form carbonic acid but this doesn't happen and they form carbon dioxide and water.
WHY?

Hydrochloric acid is already dissociated to $\ce{H+}$ and $\ce{Cl-}$
Sodium carbonate as ionic compound dissolves as
$$\ce{Na2CO3(s) <=> 2 Na+(aq) + CO3^2-(aq)}$$
Chloride and sodium ions are in this reaction called the spectator/bystander ions, not being involved in the reaction.
The ongoing reactions are these :
$$\ce{ 2 H+ + CO3^2- <=>> H2CO3 <=>> H2O + CO2(aq)}$$
followed by the dissolution equilibrium:
$$\ce{ CO2(aq) <=> CO2(g)}$$
If solution is lead to crystallization, the spectator ions then meet to form ionic crystals, keeping their charges.
$$\ce{Na+(aq) +  Cl-(aq) <=> NaCl(s)}$$

In solution, Na+ and Cl- will stay hydrated, they won't recombine to NaCl (which actually is the representation of the ions organized in a crystal lattice).
The CO32- is in an equilibrium with HCO3- and CO2, depending on pH. Adding acid is shifting the equilibrium towards CO2, that's why you are observing bubbles when you are adding the acid.
With solid carbonates, essentially the same is happening, as the acid you are adding (in case of HCl) brings water with it and dissolves the carbonate (or may react with it, as with liquid pure acetic acid). You may mix carbonates with solid acids (e.g. citric acid) and not much will happen until you add water, as you know from fizzy tablets.


text:
Can Cyanide ion substitute methoxide?
Question

Attempt
Since CN is better nucleophile than both OMe and Cl, I believe that it should replace both. But none of the options match my answer. Where am I wrong? Please help.

MeO- is a really poor leaving group

But compared to CN?

CN- is a better leaving group. The product here is b

Nucleophilicity is an entirely different concept from leaving group ability. One does not imply the other.

You need to reckon with how strong a base the nucleophile is compared with what you are trying to displace.  Only if the nucleophile is the stronger base can you expect the reaction to go based on thermodynamics.
Thus, both of the following reactions can go:
$\ce{RCl + OCH3^- -> ROCH3  +Cl^-}$
$\ce{RCl + CN^- -> RCN  +Cl^-}$
because both nucleophiles are stronger bases towards carbon than the chloride ion that would be displaced.  You can then say that the second reaction will go faster with a better (kinetic) nucleophile.  But if you have
$\ce{ROCH3 + CN^- -> RCN  +OCH3^-}$
where you are trying to displace the strongly basic methoxide ion rather than adding it as a nucleophile like the first reaction above, then cyanide ion is too weakly basic to push through.  Cyanide ion can be a good nucleophile only when displacing a weaker base or adding to an appropriate unsaturated compound (such as a carbonyl compound) without displacement.
Try it now.


text:
Comparing between nucleophilic and basic strength of acetylene ion
Question

In this question I understand that 2 reactions- Substitution and elimination- will occur. But which will be major?

I think you can immediately eliminate answer a and d. The conjuated alkene wil be preferentially formed.

The answer is (b); majority elimination by an E2 mechanism.
Quoting from chem.libretext.org:

Acetylide anions are strong bases and strong
  nucleophiles[...] Secondary, tertiary or even bulky primary substrates
  will give elimination by the E2 mechanism.

But how do u compare between its basicity and nycleophilicity ?

That is really rather difficult to compare because solvent effects, counter-ion and electrophile differences make very reaction different. Chemistry is complex!


text:
How is exact racemization possible?
A racemate consists of 50 % $d$ and 50 % $l$ forms of an optically active compound. But how can someone ensure exact 50% quantity comparing molecule by molecule? There will always be some difference in number of molecules in both forms of optically active compounds. Is this sufficient for the mixture to rotate the plane polarised light in clockwise or anticlockwise direction?
According to wikipedia "Racemization can be achieved by simply mixing equal quantities of two pure enantiomers."

Racemization isn't "exact," but rather very very close to equality. It is just simple probability.
Think of flipping a coin, p=probability for heads, and q=probability of tails. Now for a fair flip p=q=0.5. From binomial theory  the standard deviation is $\sqrt{n\cdot p \cdot q}$ where n is the number of flips. Now let's assume 2 standard deviations difference, which is roughly at the 95% confidence interval.
If you flip 10 pennies then a two standard deviation difference is $2\times \sqrt{0.5^2\times 10} \approx 3$ in the number of heads.
Now flip $6.022\times10^{23}$ dimes then a two standard deviation difference is $2\times \sqrt{0.5^2\times 6.022\times10^{23}} \approx 7.8\times10^{11} $ in the number of heads.
But now think of the % difference.

$3$ heads in 10 tries for the pennies is $30\%$.
$7.8\times10^{11}$ more heads when flipping $6.022\times10^{23}$ dimes is only a difference of $1.3\times 10^{-10}\%$ which is an insignificantly small difference.

Nice way to interpret! I believe this explains the racemization occurring during a reaction. What are your thoughts about manually mixing equal quantities of two pure enantiomers (as stated in last line of question)?

@Apurvium - mixing enantiomers would be much less precise. It would easy to create a solution that was measurably optically inactive, But about 4 digit equality would be a very precise manual mixture. So 0.01% difference would be very very good work.

It's just theory vs. real life. When mixing components, you always have limitations with the purity of chemicals and the accuracy of the balance available.
When you look at chemical reactions which yield chiral compounds, starting from achirals and there is no bias towards d or l, then you will end up with a true racemate.

It is pretty much possible in asimptotic way - if you enable l and d form to change into each other. For example, there are enzymes called "something-racemase".
Even if you start with pure L or D form, you will get a mixture approaching equal concentrations of L and D.

Stereoselectivity is a very important factor in organic synthesis. Most of the natural chemistry in the world, at least the world that involves biology and chemistry, is chiral, favoring one enantiomer over another. Speaking from the perspective of pharmaceuticals, chirality is critical to biological activity with respect to therapeutic benefit or toxicity. It is often desirable to produce one enantiomer in large excess to increase the yield of the desired "useful" product and reduce expense, waste, or extra steps required to isolate or purify the product, especially when multiple steps are required in synthesis and even small losses accumulate with each step.
In the case of many reactions, the ratio of stereoisomers produced can be random, leading to a near 50:50 split. In other cases, factors such as the intrinsic chirality or sterics of one or more reactants may influence the enantiomeric ratio. In some cases, one enantiomer is preferred almost exclusively because a reactant cannot approach the reaction site because of steric hindrance or factors including molecular conformations, such as hydrogen bonding, solvent, or temperature.
Books have been written about this topic, as well as numerous individual publications and review articles. A few references that might be helpful are listed below.
https://en.wikipedia.org/wiki/Stereoselectivity
https://www.researchgate.net/publication/316322015_Principles_of_Asymmetric_Synthesis
https://www.researchgate.net/publication/262607017_Editorial_Thematic_Issue_Enantioselective_Synthesis_in_Organic_and_Medicinal_Chemistry
http://www.carborad.com/Volume%20I/Chapter%2011/Stereoselectivity.pdf


text:
Reducing power of ammoniated electrons
I came across a question asking which of the following conversion/reductions can be accomplished using ammoniated electrons.

$\ce{O2}$ to $\ce{O2^{2-}}$
$\ce{K2[Ni(CN)4]}$ to $\ce{K4[Ni(CN)4]}$
Aromatic ring
Non-terminal alkyne

I have been able to find on Wikipedia that $\ce{O2}$ is reduced to $\ce{O^{2-}}$. So I believe 1 should be incorrect. I am unsure of 2 and could not find any literature regarding the same. About 3, I know about Birch Reduction. About 4 I don't know.
The answer is 1, 2, 3, 4.

Question 1 is probably the first step for producing $\ce{O^{2-}}$ later on, as stated by Wikipedia. Question 2 makes sense, because ammoniated electrons are produced when dissolving metallic Na or K in liquid ammonia at low temperature. So, if $\ce{K}$ is used, this electron has no problem reducing $\ce{[Ni(CN)_4]^{2-}}$ to $\ce{[Ni(CN)_4]^{4-}}$ and producing  $\ce{K_4[Ni(CN)_4]}$

The reducing power of alkali metal such as $\ce{Na}$ and $\ce{K}$ in liquid $\ce{NH3}$ is duee to the solvated electrons ($e^-$), which can reduced certain compounds into unusual oxidation states. One such example is reduction of oxygen ($\ce{O2}$). The the solvated electrons can be reduced $\ce{O2}$ to $\ce{O2^{.-}}$ (superoxide ion) first and then to $\ce{O2^2-}$ (peroxide ion) (Ref.1). However, a solution of $\ce{Na}$ in liquid $\ce{NH3}$ is treated with oxygen, the peroxide ion is generally formed (Ref.2). When a solution of $\ce{Li}$ in liquid $\ce{NH3}$ is rapidly oxidized with $\ce{O2}$ at $\pu{-78 ^\circ C}$, a bright lemon-yellow solution is formed (which is believed to be a solution of $\ce{LiO2}$). When the solution is warmed towards $\pu{-33 ^\circ C}$, the yellow color faded and white suspension is formed (which is believed to be solid $\ce{Li2O2}$). The color change is not reversible (pp. 131-132, Ref.2).
Another example is conversion of $\ce{Ni^{II}}$ to $\ce{Ni^{I}}$ and $\ce{Ni^{II}}$ to $\ce{Ni^{0}}$. Potassium tetracyanonickelate(II) ($\ce{K2[Ni(CN)4]}$) has been reduced by $\ce{K}$ in liquid $\ce{NH3}$ at $\pu{-33 ^\circ C}$ to the red potassium tetracyanonickelate(I) $(\ce{K3[Ni(CN)4]})$ (Ref.3), which is slowly reduced further to yellow potassium tetracyanonickelate(0) $(\ce{K4[Ni(CN)4]})$ at $\pu{0 ^\circ C}$ (Ref.4).
It is known fact that solutions of alkali metal in liquid $\ce{NH3}$ reduce several organic compounds including alkynes to trans-alkenes and aromatics to 1,4-cyclodienes (Birch reduction; e.g., reduction of benzene to cyclohexa-1,4-diene).
Therefore, correct answer to the question is all of (1)-(4).
References:

William H. Schechter, Harry H. Sisler, Jacob Kleinberg, “The Absorption of Oxygen by Sodium in Liquid Ammonia: Evidence for the Existence of Sodium Superoxide,” J. Am. Chem. Soc. 1948, 70(1), 267-269 (https://doi.org/10.1021/ja01181a083).
Nils-Gösta Vannerberg, “Chapter 3: Peroxides, Superoxides, and Ozonides of the metals of groups  Ia, IIa, and IIb,” In Progress in Inorganic Chemistry; Volume 8; F. Albert Cotton, Ed.; John Wiley & Sons, Inc.: Easton, PA, 1962, pp. 125-198 (ISBN:978-04-7017-6733).
John W. Eastes, Wayland M. Burgess, “A Study of the Products Obtained by the Reducing Action of Metals upon Salts in Liquid Ammonia Solutions. VII. The Reduction of Complex Nickel Cyanides: Mono-valent Nickel,” J. Am. Chem. Soc. 1942, 64(5), 1187-1189 (https://doi.org/10.1021/ja01257a053).
George W. Watt, James L. Hall, Gregory R. Choppin, Philip S. Gentile, “Mechanism of the Reduction of Potassium Tetracyanonickelate(II) and Potassium Hexacyanocobaltate(III) with Potassium in Liquid Ammonia,” J. Am. Chem. Soc. 1954, 76(2), 373-375 (https://doi.org/10.1021/ja01631a016).
Additional reference: V. Gutmann, "Chapter III: Coordination Chemistry in Proton-containing Donor Solvents," In Coordination Chemistry in Non-Aqueous Solutions; Springer-Verlag: Wien, Austria, 1968, pp.35-58 (ISBN-13: 978-3-7091-8196-6).


text:
Major product of the reaction of m-hydroxyphenol in boiling aq NaHCO3

The reaction: (a) m-hydroxyphenol + aq. $\ce{NaHCO3}$/boil; (b) workup with $\ce{H3O+/H2O}$

I know that $\ce{CO2}$ will be formed by boiling aq. $\ce{NaHCO3}$, and so the reaction will be like Kolbe-Schmitt type, i.e. $\ce{COOH}$ group will be attached on the benzene ring.
But I have a doubt that where will the addition of $\ce{COOH}$ group take place,

(1) ortho-position w.r.t. both hydroxy groups
  $$\text{or}$$
(2) ortho-position w.r.t. one hydroxy group and para-position w.r.t. the other group.

So, Which will be the major product and why?

The major product (proportions dependent on conditions) is option 2 (beta-resorcylic acid) according to these papers$\ce{^{[1]} ^{[2]}}$. The relatively low proportion of option 1 (gamma-resorcylic acid) is attributed to steric hindrance$\ce{^{[3]}}$.

References

The carboxylation of resorcinol and the separation of β-and γ-resorcylic acid by ion-exchange chromatography by
D. K. Hale,  Audrey R. Hawdon, J. Idris Jones  and  D. I. Packham. DOI: https://pubs.rsc.org/en/content/articlelanding/1952/JR/JR9520003503#!divCitation
The chemistry of resorcinol carboxylation and its possible application to the CO2 removal from exhaust gases by Vincenzo Barbarossa, Francesco Barzagli, Fabrizio Mani, Sarah LaicGiuseppina Vangaa
DOI:https://doi.org/10.1016/j.jcou.2015.04.004 (picture source)
https://www.studocu.com/en-us/document/university-of-north-carolina-at-charlotte/organic-chemistry-2/assignments/lab-report-4-carboxylation-of-resorcinol/2141490/view

Isn't OH a small group? Can it contribute to steric hinderence so much so as to change the major product?

Apparently so. This is the accepted explanation

These are a few points I'd like to add in addition to Waylander's answer for readers' benefits:
It is true that $\beta$-resorcylic acid (2,4-dihydroxybenzoic acid) is the major product of this reaction (option 2),a significant amount of $\gamma$-resorcylic acid (2,6-dihydroxybenzoic acid) would also form (option 1). For example, heating resocinol (1,3-dihydroxybenzene) with excess $\ce{KHCO3}$ or $\ce{NaHCO3}$ in water had afforded $\beta$-resorcylic acid exclusively, and $\ce{KHCO3}$ had given higher yield (Ref.1). As Waylander pointed out, the proportions are dependent on conditions, specifically by excessive heating (Ref.1). According to the authors, about 80% $\gamma$-resorcylic acid tends to be decarboxylated. And also, with 20% sodium hydroxide solution, it afforded some $\gamma$-resorcylic acid but considerable decarboxylation occurred simultaneously. This effect on base to give more $\gamma$-resorcylic acid is evident in Ref.2, which speculate formation of dianion would be the reason.
Further, $\beta$-resorcylic acid also tends to be decarboxylated in organic solvents above $\pu{150 ^\circ C}$ to give resocinol (First order kinetics; Ref.3 & Ref.4).
References:

D. K. Hale, Audrey R. Hawdon, J. Idris Jones, D. I. Packham, "The carboxylation of resorcinol and the separation of β-and γ-resorcylic acid by ion-exchange chromatography," J. Chem. Soc. 1952, 3503-3509 (https://doi.org/10.1039/JR9520003503).
Vincenzo Barbarossa, Francesco Barzagli, Fabrizio Mani, Sarah Lai, Giuseppina Vanga, "The chemistry of resorcinol carboxylation and its possible application to the $\ce{CO2}$ removal from exhaust gases," Journal of $\ce{CO2}$ Utilization 2015, 10, 50-59 (https://doi.org/10.1016/j.jcou.2015.04.004).
Louis Watts Clark, "The kinetics of the decarboxylation of β-resorcylic acid in amines and glycols," J. Phys. Chem. 1963, 67(12),2831-2833 (https://doi.org/10.1021/j100806a078).
M. A. Haleem, M. A. Hakeem, "The kinetics of the decarboxylation of β-resorcylic acid in catechol," Australian Journal of Chemistry 1976, 29(2),443-446 (https://doi.org/10.1071/CH9760443).


text:
How to predict what metals form when they react with water?
How to predict what metals form when they react with water? For instance sometimes they form a metal oxide, but sometimes they form a metal hydroxide. How does this vary and why?

Some metal oxides e.g. sodium, potassium react with water to give hydroxides

In general, the reaction of metals with water could produce either the oxide or the hydroxide.  Roughly speaking, most metals are capable thermodynamically of displacing hydrogen from water to form the hydroxide, but in many cases you need such a high temperature to run the reaction that the hydroxide is not stable and thus we go over to the oxide.
To show the variety of possible reaction routes let's look at a few cases:
Iron
Iron does not react with water alone under ambient conditions; a simple reaction between iron and water requires steam at a high enough temperature to destabilize any hydroxides.  You therefore get only oxides, generally $\ce{FeO}$ or $\ce{Fe3O4}$.
But if you add the supplemental oxidizing power of air, then iron can react with both the water and the air to generate rust -- a mixture of hydroxide and oxide compounds, or even  oxide-hydroxides like $\ce{FeO(OH)}$ with iron(III).  The hydroxides become possible precisely because the combination of air and water lowers the reaction temperature to make it so.  See Wikipedia for a fuller explanation of the many reactions involved in the rusting of iron.
Magnesium and Calcium
Magnesium is something of an intermediate case between metals that clearly react with water alone at ambient temperatures and those that do not.  Magnesium reacts slowly at ambient temperature, possibly generating enough hydrogen bubbles to temporarily float pieces of the light metal in the water.  Reaction is faster in boiling water.  In both of these cases the hydroxide $\ce{Mg(OH)2}$ is a stable product; but this decomposes at $350 \, \pu{^\circ C}$   From that temperature upwards, including burning the magnesium in steam, you will get $\ce{MgO}$.
Calcium more clearly reacts with water under ambient conditions to give $\ce{Ca(OH)2}$, but this also transitions to forming $\ce{CaO}$ as the hydroxide decomposes at $580 \, \pu{^\circ C}$.
Alkali metals
In general, alkali metals react so rapidly with water and the hydroxides are so stable that at least for all practical purposes, we expect exclusively the hydroxides.

The reaction of metals with water generally proceeds, for example, with aluminum, which has been slowly cleaned of its protective $\ce{Al2O3}$ coating via aqueous NaCl, as follows:
\begin{align}
  \ce{Al-NaCl &-> Al^3+ + 3 e-}\\
  \ce{ H2O &<=> H+ + OH- }\\
  \ce{H+ + e- &-> .H }\\
  \ce{.H + .H &-> H2 (g)}
\end{align}
Net reaction:
$$\ce{ 2 Al + 6 H2O -NaCl-> 2 Al(OH)3 + 3 H2 (g)}$$
Now, burning Magnesium, as another example, reacts with steam to create $\ce{MgO}$, but this due to the thermal decomposition of $\ce{Mg(OH)2}$:
$$\ce{ Mg(OH)2 + Heat -> MgO + H2O }$$
Find a source at bbc.co.uk.


text:
Angular Momentum of an orbital and orbit
While pursuing forth in Atomic Structure, I encountered the following concepts...
According to Bohr, the angular momentum of an orbit remains quantized, i.e $\displaystyle mvr = n\hbar .$
But he wasn't able to expalin the reason that why is it so.

But later Louis de Broglie gave his hypothesis:
Every particle in the universe has dual nature, one wave nature an other particle nature.
Focusing on wave nature of particle he gave an equation, known as de Broglie equation or de Broglie wavelength as
$$\displaystyle \lambda = \frac {h}{mv}.$$
This was proved years later by Davisson and Germer's Diffraction Experiment.

Now, we can also prove the quantization of angular momentum of an orbit as:
For any orbit the number of waves produced by an electron in one complete revolution is $\displaystyle \frac {2\pi r}{\lambda}.$
Substituting the values of $r,v$ given by Bohr, then
$$\displaystyle \frac {2\pi r}{\lambda} = \frac {2\pi r \cdot mv}{h} = \frac { n^2 \hbar^2 \cdot mkZe^2}{\hbar \cdot mkZe^2 \cdot n\hbar} = n$$
As $\displaystyle r =\frac { n^2 \hbar^2 }{ mkZe^2}  $ and $\displaystyle v = \frac {kZe^2}{ n\hbar}.$
Now , as $ \displaystyle \frac {2\pi r}{\lambda}= n ,$ From this we can prove that angualr momentum of an orbit is quantized.

Now according to Quantum wave mechanical model, the angular orbital momentum of an orbital is : $ \displaystyle \sqrt {l(l+1)} \hbar .$
Now which one is actually valid, because in many of the books both of the momentums are refered only by the term angular momentum.  But both are not same.
My question might sound silly, for such topic... But I would like to know which one is to be refered when we encounter angular momentum.

Seriously, erase the Bohr model from your brain. It is scientifically wrong. Its only use is giving us a glimpse into how quantum mechanics developed historically.

Ok, means we have to consider Quantum wave mechanical model whenever angular momentum is refered... Thank you.

Yep. The Bohr model did manage to get some things right, but those were essentially by coincidence. Electrons simply don't move in a circular "orbit" around the nucleus.

@orthocresol what is the significance of angular momentum in quantum model when electron does not got in circular motion?

@Vishalprabhulawande, even classically, angular momentum is simply the cross product of position and momentum: $\mathbf{l} = \mathbf{r} \times \mathbf{p}$. It is not something that is exclusive to uniform circular motion. The same is true of quantum angular momentum, except that $\mathbf{r}$ and $\mathbf{p}$ are operators.

@orthocresol  got it. thank you so much! Any good reading matter you can suggest in understanding quantum model of an atom?... Would be a great help

@Vishalprabhulawande My suggestion is to read an actual QM book, and by that I mean a physics one. In my opinion, (notwithstanding the famous Feynman quote) one needs to "understand" QM itself before they can "understand" QM as applied to atoms. I learnt from Griffiths' Introduction to Quantum Mechanics, but there are many texts out there. You don't need to follow it through to the end with all the advanced topics, but getting the basics right is really important.

Bohr Model came before the quantum theory and it was a success for explaining the non- continuous emission spectrum of excited hydrogen atoms. However, Bohr model was not sufficient for explaining many other phenomena related to the interaction of matter with light, such as the non-continuous  emission spectrum of multi-electronic atoms,  Zeeman Effect,... Therefore, the quantum theory is considered as a more complete theory.
The quantification of the angular orbital momentum should be linked to the second quantum number $l$ not $n$.
Up till now, we use Bohr model in chemistry textbooks in spite of its limitation. This is because it's simple and doesn't require complex mathematical tools as quantum theory.

Ok, I agree to that. But my question is that when de Broglie hypothesis can prove consevation of angular momentum and the Schrodinger's equation also satisfies the  de Broglie hypothesis, then why we do not consider $\displaystyle mvr = n\hbar $ ?

The Bohr Model does not account for the fact that accelerating electrons do not emit electromagnetic radiation.

Thank you for the help, I understood the situtation..

Ok, I didn't knew that...

The circular trajectory of  the orbit is also an error. This is only true for Bohr model.

I am happy that I helped you.

Yes, I agree to that, there are no orbits, rather orbitals...


text:
does stirring honey make it less viscous?
After making some research, I found that crystallized honey does get less viscous as we stir it, so is it the same thing with pure honey or not? and also if not could we use the Zahn cup experiment to test for the viscosity of crystallized honey or no?

A very good reference to the rheological properties of honey, both in a liquid and a crystallized state, is https://www.intechopen.com/books/honey-analysis/rheological-properties-of-honey-in-a-liquid-and-crystallized-state.
The summary is that some varieties of honey may be slightly thixotropic (shear thinning), but that honey can be generally considered as newtonian. My impression is that the concentration of sugars, or the amount of water, is the chief cause of viscosity type. After all, a solution of sugars (glucose, fructose and even some sucrose) will have a viscosity which is quite newtonian (linear shear force vs shear rate), and it will increase as the concentration increases, until you can just visualize the honey plowing up in front of the blade as the honey gets more concentrated and viscous.
Minor constituents in the bee barf are probably also thickening, just like minor constituents in saliva.
Another possibility is that stirring a viscous material like honey develops some heat with a small temperature rise which causes a drop in viscosity.


text:
Calculate third and fourth nearest neighbours in fcc packing
My textbook has given data for third and fourth nearest neighbours to be 6 and 8 with distances  a and √(3/2)a.
I have been able to calculate for the first and second nearest neighbour but it has become difficult to visualise for the other two to calculate. Can you help me with hints on how to proceed preferably with a diagram.

Does crystal has FCC structure or NaCl like structure?

Crystal has fcc structure surrounded by n number of identical atoms around it.

I don't understand "n atoms surrounded by it". Is there any description of n given in exact question.

There must be a mistake in your textbook, when it states that there is a neighbor at a distance $a√(3/2)$. It should have been written $a(√3)/2$. I explain and develop this point now.
Let's start from any apex of the elementary cubic cell. The nearest neighbors of any apex in FCC are the atoms in the middle of a face. And there are $8$ such atoms, at a distance $(a√2)/2 = 0.707 a$. The next neighbors are in the center of the cube, and there are $8$ such atoms, at a distance $(a√3)/2 = 0.866 a$. The third next neighbors are the $6$ next apexes, with a distance $a$. And the fourth next neighbors are the $8$ atoms at the opposite apexes on each face, with a distance $a√2 = 1.414 a$.

How second neighboura are in centre of cube as in FCC atoms are present on middle of faces and vertices of cube not in centre of cune?


text:
Oscillation of particle in harmonic oscillator
In harmonic oscillator, as vibration quantum number increases then does the distance of oscillation of particle from the mean position where there is minimum potential energy increases?
If yes, then the expression of potential energy $(kx^2/2)$ which we derived from Taylor expansion of potential energy at mean position should not hold as we neglect high powers terms $(x-x_0)^3$ etc.but as x increases so we can't neglect high power terms. Then the expression of quantized energy $(n+1/2)\hbar\omega$ which we get by solving Schrödinger equation also doesn't hold.
The wave function diagram also shows that the distance from mean position increases as vibration quantum number increases.
Please tell me where is flaw in my argument.

Classically speaking, the (idealised) harmonic oscillator is defined by Hooke's law. Quoting from Wikipedia:

In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force $F$ proportional to the displacement $x$:
$$\vec{F} = -k\vec{x}$$
where $k$ is a positive constant. If $F$ is the only force acting on the system, the system is called a simple harmonic oscillator.

In one dimension we can drop the vectors and write this as $F = -kx$. But the force is also the (negative) derivative of the potential, i.e. $F = -\mathrm dV/\mathrm dx$, and integrating this to get $V = kx^2/2 + C$ is trivial. The constant of integration $C$ is arbitrary, but it doesn't fundamentally change the physics (it just moves the energy scale), so we can ditch it by setting it to zero.
If you had a potential energy with higher-order terms in $x$ (i.e. $x^3$ and above), then the derivative with respect to $x$ would contain terms that are $x^2$ and above. This directly contradicts the definition of the harmonic oscillator. It could be some other kind of oscillator, but its motion would not be called simple harmonic.
Consequently, for this particular oscillator, trying to perform a Taylor expansion is a little pointless: there are no terms in $x^3$ or above, so all the higher-order derivatives $\mathrm d^3V/\mathrm dx^3, \cdots$ are all zero, and all those terms in the Taylor expansion are zero. Of course, in the general case, Taylor analysis of a different oscillator could be very useful. Just not here.
The potential energy of the quantum harmonic oscillator is directly inherited from the classical simple harmonic oscillator, by replacing the classical quantities $x$ and $p~ (= mv)$ with their quantum analogues $\hat{x}$ and $\hat{p} ~(= -\mathrm i\hbar(\partial/\partial x))$. So given that the classical potential energy is $V = kx^2/2$, it's not too much of a stretch to get the quantum potential energy $V = kx^2/2$.

In a strictly harmonic oscillator then the energy and potential are exactly as you state. This is not the case in any real molecule because all real molecule dissociate at some energy, and so higher terms in $x$ should be included. The energy levels are changed in accordance with how the potential energy in the Schroedinger equation is changed. A common potential is the Morse potential of the form $(1-\exp(-bx)^2$ as this allows the Schroedinger equation to be solved exactly, see Wikipedia for details and graphs. (Very often, in practice,  the potential is derived from experiment as it is very hard to calculate the potential accurately using MO methods. )
In your second paragraph, for the harmonic potential the average internuclear distance is the same for all energy levels. The average value is $\langle x \rangle =\int_{-\infty}^\infty \psi(x)_n x\psi(x)_n dx$. This is an 'odd function' which means that the integral over all space (or any symmetrical limits) is exactly zero. The wavefunction multiplied by itself must be symmetrical $\psi(x)^2 = \psi(-x)^2$ and multiplying by $x$ makes the integral odd meaning that if the function is $F$ then $F(x)=-F(-x)$.
The square of the average value is not zero, $\langle x^2 \rangle =\int_{-\infty}^\infty \psi(x)_n x^2\psi(x)_n dx$ as this is an 'even function'.


text:
Intutive understanding of internal energy
Our high school teacher told us that once we calculate the formula of internal energy, it is valid everywhere. I cannot understand how? I asked him, he simply answered because it is state function. It made me even more confused.

From first law, $\mathrm dU=\mathrm dq+\mathrm dW$
In constant volume process, volume differntial $\mathrm dV=0$, hence $\mathrm dW=0$, so
$(\mathrm dU)_V=(\mathrm dq)_V$ and $(\mathrm dq)_V=C_V\mathrm dT$

So, $(\mathrm dU)_V=C_V\mathrm dT$

Why we use this formula everywhere even when volume is changing (since we derived this relation for a constant volume process), at least for ideal gas?

This formula is used only when the process occurs at constant volume like a reaction occurs in closed vessel in which gaseous reactants converts to gaseous products. As volume is fixed so there is no expansion or contraction work.

Energy is a state function, but that only matters if you're going from the same point A to the same point B every time. In the formula where you have derived, the process is constant-volume, so the system at points A and B must have the same volume. Now if you have a process in which the volume changes, like point A to point C where the system at C has some different volume, then the energy change for A->C won't be the same as A->B, because the initial and final states aren't the same, and your formula becomes invalid. Unfortunately your teacher is not explaining this very well.

Because, for an ideal gas, internal energy is a function only of temperature.  So it doesn't matter what the volume change is.  For a non-ideal gas, it can also be a function of volume.

@ChetMiller Is this because of ideal gas' molecules do not have attractive forces?

For an arbitrary fluid (not necessarily an ideal gas), a combination of the first- and second laws of thermodynamics tells us that, $$\left(\frac{\partial u}{\partial v}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_v\right]$$where U is the molar internal energy and v is the molar volume.  For a fluid which obeys the ideal gas law Pv=RT, the right hand side of this equation is zero.  So the internal energy of a substance which obeys the ideal gas law is independent of volume.  An ideal gas is a substance which obeys the ideal gas law because the interactive forces between its molecules are negligible.  So, yes, it is because ideal gas molecules do not have attractive forces.

I think you misswrote the partial derivative on the rhs.

@BuckThorn Thanx.  Ive corrected it.

I might say, even for a HS student, something like: "ideal gas molecules do not have any attractive or repulsive forces". But that's a judgement call.


text:
nucleophilic aromatic substitution reaction
I am in a project in organic synthesis writing about nucleophilic aromatic substitustion reactions. I cant´t figure out why the nucleophile (benzylmethylamine) attacks the chloro-binded carbon when bromide is a better leaving groupe? (See picture)

A halogen in the 2 (and 4) position to N is activated towards nucleophilic attack because of the polarity of the C-N bond and the capacity of the heterocycle to stabilise -ve charge as you have drawn. There is no such polarity with the C-S bond and it cannot stabilise -ve charge

That's the main reason, but also "Br is a better leaving group" isn't true: for SNAr reactions it's F > Cl > Br > I, the reasons are covered in all organic chemistry texts and (for example) here: https://chemistry.stackexchange.com/questions/5690/why-are-fluorides-more-reactive-in-nucleophilic-aromatic-substitutions-than-brom?rq=1


text:
Algae Cultivation Using Polyethylene Film Reactor Vessel Floating In Pool with Algicidal ClO2 Concentration
I'm attempting to grow chlorella algae in a polyethylene film bag (which contains the growth medium) submerged in a temperature-moderating, above-ground swimming pool.  This presents a challenge in that I would like to kill algae in the pool with  $\ce{ClO2}$ while cultivating it in the bag despite diffusion of $\ce{ClO2}$ across the polyethylene film.
How often must I replace the growth medium?
The constraining quantities:

7.62e-16kg/m/m^2 Pa $\ce{ClO2}$ diffusion across polyethylene
Polyethylene film thickness 0.5mm
Bag volume to surface area ratio 5cm
5mg/L average $\ce{ClO2}$ concentration in the pool
0.1mg/L maximum $\ce{ClO2}$ concentration in chlorella growth medium

Justifications for my choice of constraints:

Chlorella 30 minute kill rates:

0% @ 3ppm (and below)
80.7% @ 4mg/L
87.2% @ 5mg/L
99.0% @ 6mg/L

High concentrations of $\ce{ClO2}$ are desirable for disinfectant power.  I'd like to keep the pool at an average of 5mg/L (swinging between 6mg/L and 4mg/L) to suppress infection and keep the chlorella growth medium at 0.1mg/L since according to the below table, the "Experimental phenomenon", shows "ball in shape" morphology at 0mg/L which is lost at some concentration below the sublethal 2mg/L.  Such loss of morphology implies a potential impact on growth rate.
Table 5 of "Disinfection effect of chlorine dioxide on viruses, algae and animal planktons in water":


text:
I'd like to model some compounds with DFTB, but some elements don't have parameter files in the official site. Is there a workaround?
I was reading a bit about this technique, Density Functional based Tight binding (DFTB). It's claimed it can provide a speed improvement of several fold when compared to plain DFT, still giving acceptable results. I downloaded one of its free implementations, the DFTB+ package, and started to work on their tutorial. The problem is, like a semiempirical method, this technique demands parameters. In DFTB the parameters are downloaded separately, as a series of files for each bonded element pair. The problem is, in the page linked as the repository for such parameter files, the options listed don't cover every possible element pair in the periodic table. I my case, I'm interested in heavier members of the alkaline earth group, Sr and Ba. But the only elements in this group with parameter files available are Mg and Ca.
So, I wonder if there is a workaround, short of doing the parametrization on my own, what I imagine would probably be a whole research project on itself, and far beyond my league. Perhaps that repository page just doesn't list every Slater-Koster file in existence, ond others may be found somewhere.

I recommend you try out xtb by the Grimme group. It is a state-of-the-art Tight Binding program parameterized for elements up to Radon (already built-in). There is a small caveat: the only precompiled binary they distribute is for Linux (but the program is open source, so you can compile it yourself if you want.)
And yes, it is really fast.

Thank you, Raphaël. The fact it runs on Linux is a plus for me. Ditched Windows for Ubuntu years ago, never looked back.


text:
Unit cell structure of ionic crystal
I have a question about the structure of an LiH unit cell, and while this is related to a homework problem, it isn't the problem itself, I'm just looking for conceptual understanding. I've already found that there should be 4 Li+ ions and 4 H- ions in one unit cell. However, I am struggling with how they will be arranged. Because the number of LiH molecules is 4, I am fairly certain that the structure intended for this problem is face-centered cubic lattice. This is where my conflict begins. In my class we have had the face-centered cubic unit cell described as (for one element) an atom at each corner and an atom in each side. From this arrangement we get (1/8)*8 + (1/2)*6 = 4 atoms. Now, I've seen a description online of the face-centered cubic structure for NaCl as "We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl" (Chemistry LibreTexts). For me this makes a certain amount of sense, because this would give 4 cations when you use the same counting principle as before. However, I've seen multiple diagrams online that contradict this, such as: https://images.app.goo.gl/CGAsc4xAJM39tMgi8.
This seems to claim that the center image is the unit cell, and while it does have 4 of each ion "on" it, it does not seem to fulfill the same structure as the aforementioned unit cell suggestion. Is the problem here just that they are showing 1/8th of the full unit cell for simplicity? If not, I am even more confused.
Thank you for reading this extraordinarily long question for what could be a very simple answer! And please don't roast me too hard, I've never posted here before.

I looked and there are a lot of questions on the site about units cells, many of which are pertinent. However I couldn't find one that answers your question directly.
First there is no reason to guess that $\ce{LiH}$ has a FCC structuree. You can look that up, for instance in Wikipedia.
Second $\ce{LiH}$ doesn't really have molecules in the solid. Think of $\ce{LiH}$ as  an empirical formula rather than a molecular formula.

UNIT CRYSTAL
I lifted the following from website https://opentextbc.ca/chemistry/chapter/10-6-lattice-structures-in-crystalline-solids/

Now the text says:

The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula NaCl.

That is correct. The unit cell must have the correct stoichiometric ratio. However the section seems to imply that the middle drawing, as show below, is the proper unit cell. It isn't!

First realize that when looking at proper unit cells is that that the lattice points between which the edges are drawn are the nuclei, not points in space that would contain whole atoms. So consider the drawing below. For the cell as shown in the illustration, there are 8 corners with 1/8 atom each. Four corners are Na, and four are Cl. Thus the cell shown only has $\frac{1}{2}$ of a Na cation and $\frac{1}{2}$ of a Cl anion.

Now down the rabbit hole!
The proper unit cell for NaCl is shown below:

From Wikipedia article on Cubic crystal system:

The face-centered cubic system (cF) has lattice points on the faces of the cube, that each gives exactly one half contribution, in addition to the corner lattice points, giving a total of 4 lattice points per unit cell (1⁄8 × 8 from the corners plus 1⁄2 × 6 from the faces). Each sphere in a cF lattice has coordination number 12. Coordination number is the number of nearest neighbors of a central atom in the structure.

So the unit cell contains $\ce{Na4Cl4}$ !?!
The reason is that the proper unit cell must also contain all the symmetry elements of the crystal. Look at the drawing below:

I won't get into all of how group theory applies to crystals and molecules, but the  red line shows a four-fold rotation axis. In other words all the atoms are identical. You can rotate the crystal 90 degrees about the red axis and the sodium and chlorine atoms are in identical spots. The axis is a four-fold axis because you have to rotate 90 degrees four times to get the identical atoms back to their starting position.

Thank you so much for your response! This was related to a homework problem where I had to solve for the number of atoms, which is why I said "I am fairly certain" that the structure is FCC, and why I referred to LiH molecules (I combined the molar masses when solving). You perfectly addressed my confusion about this unit cell. The insufficient number of ions in the proposed unit cell that you showed above was why I was conflicted, and the FCC unit cell you showed below was what I was assuming based on the Chem LibreText source. Also, thank you for your comment on the symmetry requirement.


text:
Deriving fluorescence intensity equations
I've been having trouble with deriving the equations in the following problem.

The interaction between DNA and AO to form the AO–DNA complex can be expressed by the following reaction:
$$\ce{AO + DNA <=> AO-DNA},$$
whose equilibrium constant is
  $$K = \frac{\ce{[AO-DNA]}}{\ce{[AO][DNA]}}  \tag{1}$$
Consider that initially there is only AO in the measuring cell displaying an emission at $\lambda_{em} = \pu{520 nm}$, and finally at equilibrium both AO and AO–DNA complex have emission at the same wavelength.
Additionally, the binding equilibrium constant for AO intercalation to DNA (ignore AO self-aggregation and dimerization) can be determined from the equation:
$$\frac{C_\ce{AO}}{ΔF} = \frac{1}{Δϕ}+\frac{1}{ΔϕK}\frac{1}{\ce{[DNA]}} \tag{2}$$
$F - ϕ_\ce{AO} \cdot C_\ce{AO} = ΔF$, $F = ϕ_i \cdot C_i$, and $ϕ_\ce{AO} - ϕ_\ce{AO-DNA} = Δϕ$ is given.
$F$ is the overall intensity, $ϕ$ is the fluorescence constant and $i$ denotes a particular component.
1) Show that $ΔF = [\ce{AO-DNA}]Δϕ$
2) Derive equation (2) starting from equation (1)

In terms of the first part, I tried working out formulas for both initial and final fluorescence intensities $F$ using $F = ϕ_i \cdot C_i$ but didn't really get anywhere after that. In terms of the second part, I tried substituting $ΔF = [\ce{AO-DNA}]Δϕ$ into the equilibrium equation but didn't really get further from that either. I would really like to get some tips on where to get started.

It would help if you would edit to define all the terms. I can guess what the concentrations and quantum efficiencies are, but you can eliminate the guessing.

I'm guessing that $\phi$ = quantum efficiency, F = final fluorescence intensity. Something that disturbs me is that the molar attenuation coefficient isn't considered which makes this more complicated.

I must admit being totally baffled as to what the subscript $i$ means in  $\ce{F = ϕ_iC_i}$. Maybe initial? If so then why isn't it  $\ce{F_i = ϕ_iC_i}$ ?

dah... I think I figured it out. F is the total intensity and $i$ denotes a particular component. So it should be $$\ce{}F = \sum\limits_{i = 1}^n ϕ_i*C_i$$

Sorry about the unclarity. $F$ is in fact the total intensity, $ϕ$ is the fluorescence constant and $i$ denotes a particular component.

Concentrating on fluorescence intensity for the moment the equation becomes $I_O = I_D $ where $I_O$ is that for the free dye and $I_D$ for that bound to the DNA. If $\alpha$ dissociates the equilibrium is then
$$\begin{align}  &I_O \quad \rightleftharpoons  &I_D \\ &1-\alpha &\alpha \end{align}$$
and the equilibrium constant $K=I_D/I_O=\alpha/(1-\alpha)$.
The total emission is $\displaystyle F=I_O(1-\alpha)+I_D\alpha=I_O\frac{1}{1+K}+I_D\frac{K}{1+K}$
Rearranging this gives $\displaystyle \frac{1}{I_O+KI_D}+\frac{1}{I_D}=\frac{1}{F}$
Replacing intensity with product of fluorescence yield and concentration and using the rather odd definitions given  should give the equation you seek. Note that $F-\phi_{AO}C_{AO}$ is the intensity of the bound dye.
The first question states that the intensity due to the bound dye $F-\phi_{AO}C_{AO}$ is equal to the amount bound times the difference in yields, this because the dye fluoresces at the same wavelength whether bound or not.

I don't quite understand how $I_O$ and $I_D$ are defined using the definitions given. In other words, I'm not sure with what I should replace the intensities $I_O$ and $I_D$. If I understood correctly $I_D = {F - ϕ_{AO}C_{AO}}$. If, so how should the $I_O$ be replaced?

$\phi_{A0}C_{A0}$ the measured intensity being proportional to yield and how much is there and similarly for the complex.

Thanks! Would it also be possible to provide help on how to derive the equation shown in the first question using the given definitions?

see the last paragraph which explains this in words


text:
Proposed mechanism for the catalyzed oxidation of CO by O3: Are NO and NO2 catalysts or intermediates?

The steps below represent a proposed mechanism for the catalyzed oxidation of $\ce{CO}$ by $\ce{O3}$:
$$\begin{align}
&\text{Step 1: } & \ce{NO2(g) + CO(g) &-> NO(g) + CO2(g)} \\
&\text{Step 2: } & \ce{NO(g) + O3(g) &-> NO2(g) + O2(g)} \\
\end{align}$$
Which of the following statements is true based on this mechanism?

The overall products are $\ce{NO2}$ and $\ce{O2}$.
The overall products are $\ce{NO}$ and $\ce{CO2}$.
The overall products are $\ce{NO}$ and $\ce{O2}$.
Either $\ce{NO2}$ or $\ce{NO}$ can be viewed as a catalyst. (correct answer)
Both $\ce{NO2}$ and $\ce{NO}$ are intermediates.

Based on the given proposed mechanism for the catalyzed oxidation of $\ce{CO}$ by $\ce{O3}$, the answer for above question suggests that either $\ce{NO}$ or $\ce{NO2}$ can be considered as a catalyst.
Thus, I'm very confused as to why the answer is 4. I would think it is 5) because $\ce{NO}$  is produced first and than consumed later on. I understand $\ce{NO2}$  does not have that same pattern, however I thought that it cannot be a catalyst because $\ce{O3}$  is already a catalyst.

Ozone is what is oxidizing the carbon monoxide and the two nitrogen oxides are catalysts, just as answer 4 says.

Why is NO not an intermediate though? I learned it that an intermediate shows up in the products first and then shows up in the reactants

I do not know what you learned about intermediates, but what matters is that ozone cannot (in this problem anyway) directly oxidize CO to $\ce{CO_2}$, so $\ce{NO_2}$ does the task and would end as NO, except that ozone then oxidizes NO back to $\ce{NO_2}$. The nitrogen oxides just cycle around. I do not see any harm in thinking of NO as an intermediate, but it is the cycling pair that comprise the catalyst.

@Ed V: I think $\ce{NO2}$ acts as a catalyst but $\ce{NO}$  is intermediate (it does not regenerate).

@MathewMahindaratne Ah, but suppose you start with a gas mixture of $\ce{O3}$, CO and NO. Then the $\ce{O3}$ will oxidize the NO to $\ce{NO2}$, then the first reaction takes place and the net result is the same. So, for the order in the problem, I tend to agree: $\ce{NO2}$ gets the ball rolling and NO is intermediate and essential in regeneration of $\ce{NO2}$. But if the second reaction goes first, then the nitrogen oxides change roles. Maybe this is a distinction without a difference!

@Ed V: You are absolutely correct. Reaction could start either way. Would you put this as an answer. I guaranteed you got my vote. :-)

@MathewMahindaratne Done! I am still woefully slow at doing the formatting.

@Ed V: You are not alone on that matter.

My take on this problem is as follows. Suppose a flask contains an equimolar mixture of $\ce{CO}$ and $\ce{O_3}$. The net reaction between the two gases will be
$$\ce{CO (g) + O3 (g) -> CO2 (g) +  O2 (g)} \tag{1}$$
But it needs a catalyst. If $\ce{NO_2}$ gas is introduced, then the reaction proceeds via the two reactions given in the OP's question:
$$\ce{NO_2 (g) + CO (g) -> CO2 (g) +  NO (g)} \tag{2}$$
and $$\ce{NO (g) + O3 (g) -> NO2 (g) +  O2 (g)} \tag{3}$$
So the nitrogen oxides cycle around. On the other hand, if $\ce{NO}$ gas is introduced, then the reaction proceeds via 'flipped' equations:
$$\ce{NO (g) + O3 (g) -> NO2 (g) +  O2 (g)} \tag{4}$$
and $$\ce{NO_2 (g) + CO (g) -> CO2 (g) +  NO (g)} \tag{5}$$
and again the nitrogen oxides cycle around. So it does not really matter whether $\ce{NO_2}$ gas or $\ce{NO}$ gas is introduced first: the two nitrogen oxides are the overall catalytst and either could be considered to be an intermediate. Basically, this is a distinction without a difference.

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text:
Change in the number of ions due to exchanging ions
If the position of Na+ and Cl- are interchanged in NaCl having FCC arrangement of Cl- ions then what change we can observe in the unit cell of NaCl.
I am confused as how the contribution will change is it due to the difference in the size of the ions or something else.
Please give answer in terms of the reduction or addition of number of ions of sodium and chloride ions
Thanks
The correct answer is Na+ ions will increase by 1 and Cl- ions will decrease by 1.

Basically anions are present in Crystal arrangement and cations in voids. But we can change their position for visualization purposes. It still remain FCC.

With the change in the unit cell I am talking about the effect on the number of ions of sodium and chloride ion.

@Manu - I think the notion would be more generally  expressed as the smaller ion is considered to be in the voids.

No. of ions per unit cell remains same. As now $Na^+$ forms FCC arrangement so their effective number is 4. And $Cl^-$ is present in octahedral voids, so their effective number is 1(at centre of unit cell)+ 3(at edges)=4.  So their effective number  remains same, as arrangement remain same, only point of view changes.

One can visualize this way, consider two unit cells having $Na^+$ at voids and $Cl^-$ in FCC. Now join these two unit cells and see only 1st half and second half which are joined together (not that half which are free). Now this arrangement has $Na^+$ in FCC arrangement and $Cl^-$ in voids. This good only for visualization point of view but generally the cations are present in voids due to there small size.

Can you tell me what will be the effect individually on the number of ions of Na and Cl because the answer is given in that term?

Your correct answer isn't.

@Ivan Neretin you means to say that the answer of this question is incorrect.

Why, the answer given by James Gaidis is quite correct. Your answer isn't.

@Ivan Neretin this is the question of IIT-JEE question paper without any correction so the answer which I gave is correct otherwise NTA(which organise all of them) would have changed it.

Reverse engineering
I googled the answer given by the OP to find out what the question might be

Source: 42 Years (1978-2019) JEE Advanced (IIT-JEE) + 18 yrs JEE Main (2002-2019) Topic-wise Solved Paper Chemistry 15th Edition, Disha Experts
It makes note of two chloride ions and one sodium ion on each edge of the unit cell.
The question that would have the correct answer
How do the number of chloride and sodium ions on each edge of the unit cell change when you move the origin from the chloride ion (conventional choice) to the sodium ion (unconventional choice)?

So in this question you means to say that at a particular edge we are shifting the edge from the chlorine to the sodium atom which is causing the increase of one sodium atom and decrease of one chlorine atom.

Yes but just along the edge. The total number per cell is unchanged. @BitthalMaheshwari

NaCl has a cubic unit cell. It is best thought of as a face-centered cubic array of anions with an interpenetrating fcc cation lattice (or vice-versa). The cell looks the same whether you start with anions or cations on the corners. Each ion is 6-coordinate and has a local octahedral geometry. http://www.ilpi.com/inorganic/structures/nacl/index.html
So you have two interpenetrating FCC lattices. If you exchange the cations with the anions, you get: two interpenetrating FCC lattices! The unit cell will have to have the same number of ions, just exchanged, one for the other.
Maybe this was a trick question to get you thinking about FCC lattices.

The answer is not the same as you are telling. Actually I know the answer but I don't know the reason. If you want I can tell u the answer just tell me in the comments.

@Bitthal Maheshwari: Please do. I just copied mine from the reference. Maybe there is a better answer.

Sorry for delay, the answer is Na+ ions will increase by 1 while Cl- ions will decrease by 1.