text:
Which combinations of orbitals lead to pi bonds according to Valence Bond Theory?

How many of the following combination of the orbitals will lead to formation of $\pi$-bonds with $z$ axis being the internuclear axis:
$$p_x+p_x,\,p_z+p_z,\,p_y+p_y,\,d_{zx}+p_x,\,d_{zy}+p_y,\,s+p_y,\,d_{yz}+d_{zy},\,d_{zx}+d_{zx},\,d_{z^2}+s$$

My attempt/Understanding:
According to  Valence Bond Theory, any orbital having any relations with $z$-axis or $z$ related planes should form $\sigma$-bonds. This is because $z$ is internuclear axis and these orbitals will have direct overlapping, forming $\sigma$-bonds.
Rest orbitals will form $\pi$-bonds by lateral overlapping, so I counted them as answer $(4)$:
$$p_x+p_x,\,p_y+p_y,\,s+p_y,\,d_{xy}+d_{xy}$$
But answer given is $(6)$, and it is not mentioned which are orbitals are counted. So, please tell which orbitals are to be counted for, and tell any flaws in my understandings, if any.

Here are the requirements for a $\pi$ bond:

Each contributing orbital has to have exactly one nodal plane containing the bond axis.  There can be other nodal planes that might cut through the axis, but don't contain it so those planes don't count.  Two nodal planes containing the axis, as with a $d_{xy}$ orbital, also fail because that's too many.

The nodal planes identified in (1) that contain the bond axis must line up, so they end up as a single nodal plane going through the bond.  Draw, let us say, the $p_x-p_x$ pair, which is easy to see it's a $\pi$ bond, to see what I mean.

If you sketch the orbital combinations for each of the nine given cases you see immediately that three fail test (1) (the $s$ orbital is no-go for a nodal plane, and $p_z$ has its nodal plane oriented wrong), but the others pass both tests (1) and (2) and they are the six combinations identified by the book.
The hidden lesson here is that $d$ orbitals can form $\pi$ bonds too.  You will see them do so when you study transition-metal complexes.

So can you clearly say which orbitals are to be counted for? Sorry but your some spelling mistakes are hampering  me to understand the answer.

Dammit, I can't type for crap on this device !  Can someone clean up?  Thanks!

No problem, but do you mean that orbitals combinations containing s, and $P_z+P_z$ don't form $π$ bonds?

Exactly, because the nodal planes of $p_z$ don't contain the bond axis, just cut through a point, so they do not satisfy test (1).

I have corrected/edited most of the spelling mistakes, by the way thanks for your answer.

I think it should be mentioned that $d_{xy}+d_{xy}$, which is not mentioned by the question, but by the OP's attempt, would form a $\delta$-bond.


text:
ChemDraw: how to change the default aromatic ring style for drawing from SMILES
I'm a beginner at ChemDraw.
I'd like to draw a molecule from SMILES and the ring bond should be drawn like the left image:

In my case, my image is drawn like the right image.
Is there any method to change the default style setting?

One circle counts 6 electrons if I remember that correctly; if that is the case, the rightmost image would even be wrong.

@Martin-マーチン I am not sure if there is a rule. I don't think so. Both representation requires underlying chemical knowledge, and in principle for new and exotic structures that isn't even available, strictly speaking. In other words left structures can be taken as having double bonds, one has to know what happens even in benzene. What is sure is that inner rings do not always represent 6 electrons. If there is a rule, it does not make other "localized" structures right, at least at first glance.

It's worth noting that there is generally an "equivalence" between SMILES and resonance forms in most chemical drawing software, including ChemDraw.  Specifically, c1ccccc1 which I view as a better SMILES representation of benzene, will be rendered with the circle, as you've shown on the right, and I think this is a worse visual representation of benzene.  Conversely, if you want most drawing programs to show explicit double bonds (which I think is generally better), if you were pasting from SMILES you'd have to do C1=CC=CC=C1 (which is a worse SMILES representation).

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the rightmost image is wrong. "It is generally not acceptable to use curves in two adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings."

From pp. 91–92 ChemDraw 20.0 User Guide, PDF version shipped with the ChemOffice suite:

Toggle Aromatic Display
An aromatic structure can be displayed either as an alternating double and single bond or as an aromatic circle. Toggle Aromatic Display option lets you change an aromatic structure from Alternating double/single bond to Aromatic circle
and vice versa. To do so:

Select the structure.
Go to Structure → Toggle Aromatic Display.

When opening a file of non native chemical format or pasting as SMILES, you can now define the way aromatic structures are displayed.
To do so:

Go to File → Preferences. The Preferences Dialog appears.

Click Open/Save tab.
Click the Normalize Aromatic Display checkbox and select either Alternating Double/Single Bond or Aromatic Cirlces as desired.
Click Apply and then click OK. All aromatic structures in the file loaded or the SMILES string pasted will be of the defined representation.

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text:
Any examples of reactions where simpler molecules are combined into more complex ones and are endothermic at the same time?
Can you name examples of reactions that are endothermic and simple molecules are combined into more complex ones at the same time? Like what plants do they take heat and sunlight and they synthetize air and water into organic fuels such as glucose and ethanol. I found similar experiments made for carbon capture some call it reverse combustion, they take air and water, and with help of heat or electricity they turn it into the ethanol or glucose. So can you name similar reactions, where something similar happens? I really can’t find an answer anywhere. Please help . Thank you for taking your time to read this.

Is there a serious reason for not writing chemical formulas properly ?

What are you really looking for ? In the beginning, you are looking for the  "synthesis of $\ce{CO2 + H2O}$". Well ! this operation is carried out every day in the combustion of any fuel like gasoline, oil, petroleum, alcohol, wood, etc. It is not a synthesis and it is not endothermic. In your second sentence, you speak of a company that makes vodka out of air and water. This is non-sense, for lack of Carbon atoms. In your third sentence you speak of the synthesis of glucose out of CO2 in the lab. No labs are able to do it. Only plants can do it. I repeat : What are you looking for ?

Its not non-sense, look at this for example https://phys.org/news/2017-09-alcohol-thin-air.html

Also synthetis is different from combustion

"There is never time to write questions properly, but there is always much more time for their explanation."

I thought this platform is for unexperienced person to ask a question and get answer from someone experienced. Like that’s the whole point of asking questions. So in order to get anyone to at least try to answer my question it has to be perfect? Like how do I write perfect question if I don’t speak “professional chemist” or “experienced entitled stack exchange user” language?

Do kids ask perfect questions? Would you act like this if your kid asked something?

Apparently you have entirely rewritten your question. Now you are looking for examples of reaction producing organic compounds out of CO2 and H2O. Well you can name all the reactions carried out by plants, like the synthesis of cellulose, of starch, of sugars, of oils, of proteins, of acids like citric acid or ascorbic acids, of dyes like carotene, of perfumes like menthol, thymol, limonene, etc. Well I stop this list here. These substances are made by plants without heating, without distilling, and in aqueous solution. Presently chemists are not able to synthesize them under these conditions.

I think this question should not have been close or it might be reopened. OP should clarify if s/he put special emphasis in the word synthesis. If it is meant for reaction, then the answer is to look for the Free Energy of the process. If it is meant for a reaction organising matter and placing energy into it (at least one part of the system sees lowering of entropy), it becomes the topic of why reactions within the living kingdom are special. Not that I know the answer or I can debate it with lucidity, but the issue is interesting and opens, like in the photosynthesis chemistry community.

@Alchimista thank you for your comment. Could you please help me rephrasimg the question to make it better? When I researched word synthetis, I understood that it is combining smaller, simpler molecules into more complex ones. Thats where I am coming from. So I wanted to know about more processes where more complex molecules are formed and at the same time chemical reaction takes hear or energy from the environment. Also I think it is interesting topic and even more important nowadays because of environmental problems

You are right that synthesis can be taken as to build up complex things from smaller. Perhaps it was even the original meaning. However synthesis can be read as whatever reaction is conducted. So dismantling a complicated molecule into pieces of interest can be called somehow the synthesis of the latter. This said, on the light of your question as for the latter comment, I am afraid only living organisms do that. There is a question that I can't find right now, here on in physics SE,  to which I have contributed full of wish but I am crystal saying that at the end the issue isn't that clear.

.... clear to me. If I have understood correctly what are you referring to, it is indeed a difficult thermodynamics issue linked to the origin, or at least the prerogatives, of life. Try to look for that thread. Same I'll do.

Related. If you are not concerned with the entropic aspect then forget the link and look for spontaneous reaction - free energy. Decreasing of Free Energy gives you the criterion for spontaneous reaction at given conditions. https://physics.stackexchange.com/questions/453853/is-it-true-that-only-living-things-and-man-made-objects-can-make-an-endothermic/454208#454208

@Alchimista oh that explains a lot and nicely connect to this topic. Just another proof for me that chemistry, biology or physics - all of these don’t exist in vacuum. It’s just how people categorised the nature and it’s processes. But everything is connected and sometimes we need to see bigger picture to understand things better.

Here are three examples of endothermic synthesis reactions.  Heats of formation are given at 1 bar, 298.15 K.  The first of these, the production of ozone, is naturally-occurring.
$$
\begin{align}
\ce{3/2 O2(g) &-> O3(g)} &\quad \Delta_\mathrm{f}H^\circ &= \pu{142 kJ mol^-1}\\
\ce{N2(g) + 1/2 O2(g) &-> N2O(g)} &\quad \Delta_\mathrm{f}H^\circ &= \pu{82 kJ mol^-1}\\
\ce{1/2 N2(g) + O2(g) &-> NO2(g)}&\quad \Delta_\mathrm{f}H^\circ &= \pu{33 kJ mol^-1}
\end{align}
$$

Does this classic example count?

@EdV and this one: https://chemistry.stackexchange.com/questions/86855/the-coldest-the-most-endothermic-reaction

@NilayGhosh Yes! And I was kind of hoping that entropy and free energy could get an oblique nod (nothing more than that: simply name-checking them), just to indicate that enthalpies, while cool characters here, are not the entire cast of the play!

@theorist thank you very much for the answer. But after some research I found out that both n2o and no2 are naturally occurring too.

https://chemistry.stackexchange.com/questions/101691/why-does-ozone-have-higher-entropy-than-oxygen

@EdV  I don't think so, since that seems to be a replacement reaction rather than a synthesis reaction.

@Joe839949  Yes, those compounds are naturally-occurring, but I don't know if those reactions are. E.g., most NO2 in the atmosphere is not made combining N2 and O2, but rather by the burning of nitrogen-containing fuels. And that is not an endothermic process.

I guess I take a pedestrian view of synthesis. For example, I used to synthesize white fuming nitric acid (in my basement chemistry lab when I was in high school) by reacting concentrated sulfuric acid with strontium nitrate. I used a retort and heat. So I got the WFNA and strontium sulfate was the byproduct. But I take your point!

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text:
Determine how much ethylacetate is formed from reaction equation
I have the reaction:
$$\ce{C2H5OH(l) + CH3COOH(l) <=> CH3COOC2H5(l) + H2O(l)}$$
and want to calculate how much ethylacetate is formed if I have $\pu{3 mol}$ of ethanol and $\pu{1 mol}$ of acetic acid. $K = 4.12$
This is how I calculated this:
$$
\begin{array}{c|c|c|c}
\ce{C2H5OH} & \ce{CH3COOH} & \ce{<=>} & \ce{CH3COOC2H5} & \ce{H2O}\\\hline
 \pu{3 mol} & \pu{1 mol}  &   & - & -\\\hline
 3-x & 1-x  & & x &x\\
\end{array}
$$
$$\begin{align}K &= \frac{x^2}{(3-x)(1-x)} = 4.12\\[1.5em]
4.12 &= \frac{x^2}{x^2 - 4x+3}\\[1.5em]
x^2 - 5.28x + 3.96 &=0\end{align}$$
solving for x:
$$x = \frac{5.28 \pm \sqrt{(-5.28)^2 - 4(3.96)}}{2}$$
why do I have to choose the negative sign here?
$$x = \frac{5.28 - \sqrt{(-5.28)^2 - 4(3.96)}}{2} = 0.905$$
How do I know that I have to choose the negative sign? Is it because if I take the positive sign then $x=\pu{4.375mol}$ which is more than the mole of ethanol? So I should always choose the $x$ that is less than greatest amount of reactant?

Not all the roots of polynomial equations have physical/chemical meaning. You cannot get more of acetate than there was the acetic acid, can you ? You have already restrictions in place 0<=x<=1

Please take a look at this posting to learn more about the preferred use of MathJax on this site.

Mathematically, there are two possible options: the one where you add the square root and the one where you subtract it.
However, chemically only one of the two makes sense. If you were to add the square root, you would receive:
$$x = \frac{5.28 + \sqrt{(5.28)^2 - 4 \times 3.96}}{2} = 4.37$$
As this dimensionless value is – according to your initial table – supposed to be a quantity in moles which you subtract from the $\pu{3 mol}$ of ethanol and $\pu{1mol}$ of acetic acid, this would give you $\pu{-1.37mol}$ of ethanol and $\pu{-3.37mol}$ of acetic acid at equilibrium. This is impossible according to the rules of science and the universe as we know it. So while this answer is mathematically sound it is chemically meaningless.


text:
How to efficiently weight DCC (N,N'-Dicyclohexylcarbodiimide)?
I have to run a reaction using N,N'-Dicyclohexylcarbodiimide and at room temperature, it is solid, but it's very hard to weigh. I know its melting point is around 40 degrees. Is it better to melt it and measure it as a liquid or is it no worth the struggle? I just don't want to have it in my hands anymore after reading it is very toxic in contact with skin and I feel like I had a few grams in contact with my hands everytime I try to break the chunks and weight it. Any tips on dealing with this reagent?

Well, you obviously shouldn't touch it with bare hands. I don't think melting is good idea though, for eg. would lead to a bit more reaction with water vapor, I guess

Good god, don't touch that stuff. Use gloves and a heavy duty metal spatula. The stoichiometry doesn't need to be perfect, so who cares if you are 5% off.

Why is it coming into contact with your hands? Do you not routinely use gloves....?

I usually break it up wih a heavy duty spatula then weigh it out - using gloves!

I do use gloves lol. But it's so messy breaking that extremely well packed solid that a lot of it goes in the gloves and of course some of it will reach my skin eventually (and I think in my first reaction I did all the grinding without gloves on and I'm still alive, so maybe I'm a survivor idk)

Try using a drill (~7mm wood drill, using a small electrical screwdriver or simply by hand).


text:
Best way to handle phosphorus tribromide
I need to do an alcohol bromination using PBr3 and I've never done this reaction before. Reading about PBr3 I found it is water/air-sensitive and the bottle I bought doesn't have a seal. So how am I supposed to manipulate it if I can't expose it to air? Just open the bottle and insert an Ar flow rapidly? Is there a more refined procedure to deal with this without a glove-box? I tried reading procedures on papers but they usually just say "PBr3 was added to the flask", nothing else.

Hmm, I don't think it's very sensitive. More like too much contact with air decreases yield a few percent sensible, probably.

PBr3 is a high boiling liquid that can be measured gravimetrically or volumetrically in a vented hood.

Do you have a nitrogen gas supply?


text:
Why does acid hydrolysis of glucose-1,6-bisphosphate produce glucose-6-phosphate and not glucose-1-phosphate?
A paper I found details treating glucose-1,6-bisphosphate with 6M HCl to produce glucose-6-phosphate. I'm curious why glucose-1-phosphate (and then even plain glucose) isn't formed instead?

You should follow up your reasoning in the question further - think what's the difference between positions 1 and 6?


text:
Solution of the Roothaan Equations of H2 by Symmetry Arguments
FWIW my background is in physics and maths, but I am just starting a chemistry PhD (the last time I took a chemistry class was high school). I have only some background in representation theory, and only an abstract taste, as it was in a maths class (bonus: suggest a reasonable representation/group theory reference with applications to chemistry, preferably a concise treatment).
In Szabo and Ostlund's Modern Quantum Chemistry, section 3.5.2 they solve the ${\rm H}_2$ molecule (in Hartree-Fock theory) using a minimal STO-3G basis, $\phi_1$ and $\phi_2$ (centred on each proton). I believe I understand how to set up the Roothaan equations, guessing the density matrix, and solving self-consistently. However, they go on to say "The canonical molecular orbitals will form a representation of the point group of the molecule." Um, okay. Why? And what is meant by canonical here? They go on to say they can be labelled by their symmetry $\sigma_u, \sigma_g, \pi_u, \pi_g$, etc. (I know these terms from wikipedia, and it is discussed briefly in another chapter) I don't fully understand why they (can) do this.
Now, I understand we have two molecular orbitals so it makes sense that they choose only the 2 lowest energies. However, from the set of options above, they state without proof that the $\sigma_g$ is the lowest energy, followed by $\sigma_u$ (I also don't really understand where the wave function comes from). How do we conclude this? Is it because the hydrogen atoms are 1.4 Bohr radii away, hence close and hence we expect bonding (not antibonding?), which I understand to be represented by $\sigma_g$? If this is correct, then it's still not rigorous so I would appreciate any more explanation.

If I'm interpreting this correctly, the main thing confusing you is how the MOs are associated with the symmetry labels...? So you understand that you can solve the SCF equations to get a bonding MO $\propto \phi_1 + \phi_2$ and an antibonding MO $\propto \phi_1 - \phi_2$, but don't know why one is $\sigma_g$ and the other $\sigma_u$? And the second question is why the bonding MO is lower in energy than the antibonding MO?

These things are usually covered in undergrad, and there are quite a lot of books for this: most undergrad inorganic chemistry books will have a chapter on "molecular symmetry", and there are some specialised books on symmetry. But they probably do not have the rigour which you might prefer (though you may well be able to fill it in yourself). If you want rigour, my recommendation is probably Albright et al. "Orbital Interactions in Chemistry": https://www.wiley.com/en-gb/Orbital+Interactions+in+Chemistry,+2nd+Edition-p-9780471080398 if you can get your hands on it, Chpt 4 is very good.

Cotton's book on Group Theory would probably help with this.

@orthocresol Yes, I am wondering about the two questions you list. However, my main question is actually how they write $MO\propto\phi_1\pm\phi_2$ without solving the Roothaan equations (I understand how to get this numerically though, and purportedly it will converge to their symmetry argument). Also, thanks for the reference!  I figured it is taught in undergrad and that's why they skim over it, and I'll need to do some catch up.

I'd also recommend Levine's Quantum Chemistry text as a better place to start than Szabo (which is still a very good text, just doesn't include the introductory material that you appear to want). Read Albright for the conceptual understanding of MOs, then Levine for the deeper math understanding, then come back to Szabo

@Andrew I think it's only really the symmetry arguments that have me confused. I am actually using Szabo as an easy supplement to Helgaker et al Molecular Electronic-Structure Theory. Most of Levine seems familiar from undergrad/master's quantum mechanics and condensed matter classes.

well, Levine has a whole chapter (13 in my fourth edition) devoted to diatomic molecules that explains the source of all of the electronic term symbols of $\ce{H2+}$ (not just $\sigma_g$ and $\sigma_u$) and the calculation of their associated energies. Perhaps worth reviewing.

After speaking to a coworker and reading some more I think I have a better understanding.
The symmetry of the wave function needs to follow the symmetry of the molecule (unless we are studying a broken symmetry solution, e.g. charge transfer, but that doesn't apply in this case). We are using a minimal basis, and hence we have only two orbitals. From atomic theory, these are s functions. Hence, they cannot form $\pi$ orbitals, and we are left with $\sigma$ orbitals. Between $\sigma_g$ and $\sigma_u$, $\sigma_g$ has the lower energy because it has no nodes, whereas $\sigma_u$ contains a node. From QM we know that this generally implies greater energy, but there's a nice explanation on this here (at the end he also specifically discusses a homonuclear diatomic molecule).
The specifics of how to get the symmetry representation for the molecule and match those with wave functions is still not entirely clear to me, but for this I will read a book on group theory and its chemical applications. It is, however, clear that $\phi_1+\phi_2$ and $\phi_1-\phi_2$ respect $\sigma_g$ and $\sigma_u$ symmetry respectively.


text:
What does per-O-acetylated glucopyranoside mean?
What does per-O-acetylated glucopyranoside mean?
Can you please break down the roots?

Per-O-acetylated sugars are sugars where each (thus, per) hydroxyl group (thus O) is converted into an ester of HOAc. E.g., https://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532015000400755 For glucose, for example, to get you glucose pentaacetate.

Per-O-acetyl-glucopyranoside chemical structure is shown below. It is a fully acetylated glucose saccharide unit in the pyranose ring form.
It is mentioned in Molbase and
Chemsrc. Pubchem prefers calling it B-L-Glucose pentaacetate and Chemspider 1,2,3,4,6-penta-O-acetyl-B-L-glucopyranose.
A search for CAS 66966-07-2 should return an assortment of suppliers and database details for the β configuration of acetylated carbohydrate.

I hope you are familiar with the concept of a glucopyranoside.
Acetylated means that at least one hydrogen atom has been replaced by an acetyl group $\ce{-C(=O)-CH3}$, typically abbreviated as $\ce{Ac}$. It is connected to the rest of the structure by the bond I have drawn at the front.
O-acetylation means that the substitution occurred on an oxygen atom. Thus, instead of an alcohol group (or a hemiacetal) we are dealing with an acetate (or an acetal ester) group: $\ce{-O-C(=O)-CH3}$ or $\ce{OAc}$.
Per-acetylation means that the substitution as described above has been performed on all possible atoms. This leads to the structure z1273 drew.


text:
Will the amount of solute change if the solution evaporates? Will the solvent carry the solute with it if it evaporates?
If a solution of dissolved metals in aqua regia is heated in an open container and left to dry overnight, will the dissolved metal ions be carried by the evaporated solution or will it stay in the container?
I was trying to get a solution of dissolved metals in aqua regia for ICP measurement but my sample got dried up so I was wondering if this would affect the ICP measurement.

Solvent shouldn't carry the solute. Common observations prove that unless a solute is volatile all by itself, it will not evaporate along with the volatile solvent.

Depending on way of drying and solution behaviour, it may get partially away mechanically. Another issue may be quantitative redissolving of residues.

My experience was that when heated to drying the solidifying sluge at the bottom of a beaker tended to splatter. You can easily tell if there was splattering by solids splattered on the sides of the beaker. Typically the beakers were covered by a watch glass and the reside on the watch glass was rinsed back into the beaker at the next step.

It depends on the elements. If your element of interest were arsenic or mercury losses can be significant. You should not have left them open to dryness without supervision. The second important issue related to drying them again was that nitrates/ chlorides can decompose or hydrolyze easily, and your metal of interest may not dissolve easily again. The analysis should be repeated without heating to dryness in an open vessel.

Hard to think of a solution of As or Hg. But of course others solids can decompose or even be volatile. So OP should just retain the message that if dryness is not desired then it must be avoided.


text:
Oxidation State of Nitrogen in N2O
If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.
I tried to do it with the structure and that's where I got confused:
$$\ce{\overset{-}{N}=\overset{+}{N}=O}$$
The $\ce{N=N}$ bond is a coordinate bond, so that should give a $-1$ oxidation state for the left $\ce{N}$ and $+1$ for the middle one. Since the middle one has a double bond with oxygen, it gets additional $+2$ for a total of $+3.$
But if I consider this resonance structure:
$$\ce{N#\overset{+}{N}-\overset{-}{O}}$$
The left $\ce{N}$ is getting an oxidation state of $0$ and the middle one gets $+2.$ So I am confused as to what is actually correct.
Is it possible for atoms to exist in superpositions of oxidation states? Or maybe I am overlooking some basic concept?

It is possible for molecule to be in a superposition of different quantum states involving different electronic structure. Why should not be possible to be in superposition of oxidation states, what are, in contrary to the former, just a thought conceptual construct ? See also wikipedia: Nitrous oxide

Some jurisdictions set the age of consent to 16 years, some to 17, and some to 18. Which is actually correct? Er, well... Same thing here. Oxidation states are human concept; atoms don't care about them.

related: https://chemistry.stackexchange.com/a/31272/102629

If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.

Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases (e.g. hydrazine) they are and the result you get algebraicly is as you would expect from a Lewis depiction. In other cases (e.g. here) this is not the case as the nitrogens are not equivalent (only one is bound to oxygen). Thus, it is obvious that a simple algebraic approach should fail.
But what about the ‘real’ result? What about the resonance structures? Well, this is where things get really difficult. You essentially have two π systems orthogonal to each other and each occupied by four electrons which can manifest as a lone pair on either end and a multiple bond to the other atom. If you really wanted to play the game, you could include another resonance structure as shown below, where all π lone pairs are centred on the terminal nitrogen:
$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O <-> \overset{2-}{N}-\overset{+}{N}#\overset{+}{O}}$$
(It is clear that this third resonance structure contributes least to the overall picture as it has a greater charge separation and charges are separated opposite to what electronegativity would predict.)
The experimental structure shows that the $\ce{N-N}$ distance is slightly shorter than the $\ce{N-O}$ distance which one could use to assume that the $\ce{N-N}$ bond have a slightly larger bond order than the $\ce{N-O}$ bond. But ultimately, they are still very similar (the difference is merely $\pu{4pm}$) so equal bond orders might also be an option. Long story short: short of calculating the electronic distribution (i.e. solving the Schrödinger equation) you won’t be able to arrive at a definite answer for the ‘real’ oxidation states.
So what can you do on paper? Comparing the resonance structures, the leftmost as I have ordered them is slightly better than the central one as the formal charges are distributed according to the different electronegativities. Thus, I would be inclined to give it a slightly higher weight and – in a classroom setting – use it to determine oxidation states. That said, it is too much of an ambiguous example to be seriously used in any examination unless the goal of said examination is to develop the chain of arguments that formed this answer.

Actually this was a question from an exam and the given answer was +1 but what you said makes sense. I could challenge the answer key, I suppose.

@Physicsa Well … did they ask or imply using the algebraic method? Or did they strongly suggest using the structure? Because if the exam question intended to test the algebraic method, that’s okay imho.

They didn't imply using any one method over another.


text:
What would be the product of this reaction?

A 3 membered carbon ring with a double bond to oxygen from one carbon and a double bond between the remaining carbon with a methyl group attached to one of the carbon is made to undergo acid catalyzed hydrolysis. What will be the product?

The compound is 2-methylcyclopropenone. What do you think of the 4 options offered?

I was thinking compound A could be more stable due to aromatic stability and would not further react with H2O to form B. However the answer is B.

A is a charged species and that is not going to be stable in the presence of water. There are lit references to acetals and hydrates of cyclopropenones but they're all behind paywalls.

According to the official answer key, the answer is given as (A). But I still don't understand how a charged specie is more stable than a neutral specie?


text:
Finding equilibrium constant for the reaction between non-stoichiometric mix of gaseous components

$$\ce{CH3OH(g) + NOCl(g) <=> CH3ONO(g) + HCl(g)}$$
The volume of the container is $\pu{433 cm^3},$ $T = \pu{50 °C}.$ Methanol was added until the pressure was $\pu{50.1 mbar},$ then $\pu{0.059 g}$ of $\ce{NOCl}$ was added. At equilibrium the partial pressure of $\ce{NOCl}$ is $\pu{27.6 mbar}.$
Calculate $K.$

I assumed that $\pu{50.1 mbar}$ was the partial pressure of methanol (I do not know if this is correct) and calculated the initial partial pressure for $\ce{NOCl}$ from the ideal gas law:
$$p = \frac{nRT}{V} = \frac {(\pu{0.059 g}/\pu{65 g/mol})\times 8.314\times 323}{\pu{4.33E-1}} = \pu{5.63 Pa} = \pu{5.63E-5 bar}$$
Then
$$
\begin{array}{cccc}
&\ce{CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)} \\
&\pu{0.0501 bar} && \pu{5.63E-3 bar} && 0 && 0 \\
&\pu{0.0501 bar} - x && \pu{0.0276 bar} && x && x\\
\end{array}
$$
When I then solve for $x$ I get a negative value which isn't correct. Is $\pu{50.1 mbar}$ supposed to be the total pressure and if yes, how do I use that then? The correct answer for $K$ is $1.33.$

Upon looking at your table I immediately noticed that you initially start out with 0.00563 bar of NOCl and end up with 0.0276. How is possible to end up with more than you started with?

Your calculation is wrong. You report a volume of $0.433 L$. Here the volume should be in cubic meter. So the volume should be $\ce{4.33·10^{-4} m^3}$ and the initial  pressure of $\ce{NOCl}$ is $5630 Pa = 0.056 bar $. With this value, you can obtain a reasonable value for $x$..

As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with $p_0$ and the change in partial pressure with $x$:
$$
\begin{array}{lccccc}
&\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\
&\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) && 0 && 0 \\
&\text{C} && -x && -x && x && x \\
&\text{E} && p_0(\ce{CH3OH}) - x && p_0(\ce{NOCl}) - x && x && x
\end{array}
$$
You are interpreting given conditions wrong.
The problem explicitly provides you with the initial partial pressure of methanol $p_0(\ce{CH3OH}) = \pu{50.1 mbar}.$
As for the initial partial pressure of nitrosyl chloride, it can be found applying the ideal gas law:
$$
\begin{align}
p_0(\ce{NOCl}) &= \frac{m(\ce{NOCl})}{M(\ce{NOCl})}\frac{RT}{V}\\
  &= \frac{\pu{0.059 g}}{\pu{65.46 g mol^-1}}\frac{\pu{8.314 J mol^-1 K^-1}\times\pu{323 K}}{\pu{4.33E-4 m^3}} \\
  &\approx \pu{5590 Pa} \\
  &\approx \pu{55.9 mbar} \tag{1}
\end{align}
$$
Note that methanol and nitrosyl chloride reaction mix is apparently not equimolar.
Also, the total pressure is not relevant for obtaining the solution.
Do keep in mind though it is not the initial partial pressure of the first component because the second gaseous reagent has been added afterwards.
The unknown $x$ can be found trivially since you are given the equilibrium partial pressure of nitrosyl chloride $p(\ce{NOCl}) = \pu{27.6 mbar}$:
$$
\begin{align}
p_0(\ce{NOCl}) - x = p(\ce{NOCl}) \quad\implies\quad x &= p_0(\ce{NOCl}) - p(\ce{NOCl}) \\
  &= \pu{55.9 mbar} - \pu{27.6 mbar} \\
  &= \pu{28.3 mbar} \tag{2}
\end{align}
$$
Finally we can write an expression for $K_p$ (by the way, here $K_p = K_c$ since $\Delta n = 0)$:
$$
\begin{align}
K_p &= \frac{p(\ce{CH3ONO})\cdot p(\ce{HCl})}{p(\ce{CH3OH})\cdot p(\ce{NOCl})}\\
  &= \frac{x^2}{(p_0(\ce{CH3OH}) - x)\cdot p(\ce{NOCl})}\\
  &= \frac{(\pu{28.3 mbar})^2}{(\pu{50.1 mbar} - \pu{28.3 mbar})\times\pu{27.6 mbar}} \\
  &\approx 1.33 \tag{3}
\end{align}
$$
Don't forget to do dimensional analysis to check yourself and never omit units in your calculation.
Also, keep an eye on significant figures: the number for the molar mass you used was too sloppy; instead of two figures $(\pu{65 g mol^-1})$ you had to go with four $(\pu{65.46 g mol^-1}).$


text:
Why are Hydrogen bonds directional?
While reading about Hydrogen bonding in a book (O.P. Tandon's G.R.B. Organic Chemistry), I came across a sentence that was mentioned in a rather unstructured way relative to how comprehensive I was finding it until then.
Here is the excerpt of that.

And here I quote two of the sentences from the excerpt again which will bring me to my question.

(i) Hydrogen bond is merely an electrostatic force rather
than a chemical bond.
(v) All the three atoms in X-H- - -X lie in a straight line.

Question:
If Hydrogen bonds are electrostatic, why are they directional?

So I did some searches and found a pretty good publication by a professor of Indian Institute of Science about the exact same thing.
Here you can find a PDF of the same.
Initially I wondered why any electrostatic bond would be directional against the common understanding of Physics that I have.
Here there is a really good rationale about it.
But majority of the explanation given in it is really alien to me. Atomic basins and the computational methods are something that I've never even heard about in my senior year.

So the problem here is actually I think I don't understand the terminologies. I'm afraid my question might get closed but I really need someone to explain about it in a rather simple way. And also, could I get a suggestion about where I can read more about things like atomic basin, etc.?

Get a better book. Only place where first statement you quote would be OK is wrong option in a test question. The second one is serious idealisation.

https://chemistry.stackexchange.com/a/5506/9961

Yeah the second link directs to something that I would accept as an answer. @Mithoron


text:
Which compound releases more energy upon combustion?

I know that the last two on the right are stable so they are not the answer, but can someone give me a hint on how to decide between the two on the left? what is the way of solving such question, I don't think it's only dependent on stability because there is more ring strain (two cyclobutanes) in the molecule at far left.

How do you decide stability, to begin with?

well usually comparing different rings, the smaller the ring, the higher the strain and less stable it is. but here I'm comparing cyclooctane which is less stable than cyclohexane, but comparing the last two confused me two cyclobutanes connected vs cyclooctane.@IvanNeretin

At a glance I assumed they were all isomers, but they are not. I'm not sure whether this makes the answer harder or easier.

The very left one has 2 rings and therefore 2 less hydrogens. The rest 3 are isomers. So the answer depends on if the molar or specific combustion enthalpy is to be considered, wrt to the C-C and C-H bond energies.

First of all, great notice matt, but it makes it harder for me, there no specific enthalpies given in this question, (it's a question from an exam in 2017 btw, I'm not a student anymore). so i am not sure if it is related to these 2 hydrogens or not.

There is practically no strain in the cyclooctane, as the molecule is not in a plane. The Carbon atoms are all tetrahedra.

so why is it answer? what is the logic?


text:
Calculating the zero-point energy for the C2N2 molecule
I shall calculate the zero-point energy for the $\ce{C2N2}$ molecule in the $^{1}\Sigma_{g}^{+}$ electron ground state with the following vibrational wavenumbers (symmetries in parentheses) which were found in IR and Raman spectra:

$v_1 = 2330~\mathrm{cm}^{-1}$ ($\sigma_g$)
$v_2 = 854~\mathrm{cm}^{-1}$ ($\sigma_g$)
$v_3 = 2158~\mathrm{cm}^{-1}$ ($\sigma_u$)
$v_4 = 507~\mathrm{cm}^{-1}$ ($\pi_g$)
$v_5 = 233~\mathrm{cm}^{-1}$ ($\pi_u$)

I have no clue about how to do it - maybe somebody can help me.

Do you know what a zero-point energy is?

The adiabatic electronic energy only contains the energy of the electrons. You are missing the energy that the molecule has in form of the vibrational groundstate of the nuclei. This energy can be calculated trivially in the harmonic approximation. Within the harmonic approximation we can define normal modes and each mode corresponds to a 1D harmonic oscillator with its respective normal mode frequency. The eigenenergy of a harmonic oscillator is given by
$$
E_{n} = \hbar \omega(n+1/2), \ n\geq0
$$
Each normal mode of your molecule contributes thus such an energy when you include vibrational energy. The zero point energy refers to the system where each mode is in the groundstate i.e. $n_i=0 $, where $i$ is an index for the normalmode. The zero-point energy  correction is then simply
$$
E_{ZP} = \sum_i^{3N-5} \hbar\omega_i 1/2
$$
where $N$ is the number of atoms and $3N-5$ is the number of normal modes in a linear molecule. In a nonlinear molecule you would have $3N-6$ normal modes. $n$ is the quantum number of the oscillator state.
In your case, you have some degenerate eigenvalues due to symmetry. The $\pi_u$ and $\pi_g$ modes are actually sets of two degenerate modes, which means that their frequency would appear twice in the sum that i denoted. This also works out when we consider your example, we have the following number of modes
$$3\cdot 4 -5=7$$. Since we have to sets of doubly degenerate modes we see only 5 distinct frequencies although there are 7 normal modes.
Plugin all thatt into the formula yields,
$$
E_{ZP} = 1/2(\hbar \omega_1+\hbar \omega_2+\hbar\omega_3 + 2\hbar \omega_4 + 2\hbar \omega_5)
$$
with $\omega_i = 2\pi \nu_i.
$
It's sometimes a bit vague what people mean with zero point energy. Some people use the term for the correction that i defined above, but technically it should include also the electronic adiabatic energy. Which means that you have to add the electronic state energy to the $E_{ZP}$ value.
One might also want to include the contribution due to the rotational states of the molecule, but we can't do that without more information.


text:
How to verify that ethanol is infinitely soluble in water?
My textbook [Chemistry Live, Declan Kennedy] states that ''Ethanol is said to be infinitely soluble in water''. It then says ''verify this for yourself by mixing some ethanol with water in a test tube''.
Both water and ethanol are colourless. How can I observe that they are infinitely soluble ?

I think what they mean by infinitely soluble is they are miscible. And when they are miscible, any concentration will be mixed to form a homogeneous solution. See here the first paragraph is explaining it.

That you don't see a difference is the proof. Add some naphtha and difference is visible.

@Mithoron What's wrong with this question? I think it is a very natural one. How would you know the difference, if you never mixed two liquids before?

@IvanNeretin Well, not asking yourself what would happen otherwise, lack of research in homework-type question... And even observing a slowly dissolving sugar in tea would show that lack of difference in color is no problem.

@Mithoron There is a huge wall between any chemistry class and the life outside. The knowledge of sugar dissolving in tea can't be brought inside. Ditto for any other common knowledge. Likewise, the chemistry knowledge is of no use outside. At least that's what seems to be universally taught.

https://chemistry.stackexchange.com/a/41592/102629

OP your question makes some sense given the book wording. The statement infinitely soluble in water is ridicolous. Sounds like "let's make things more fancy physics". You do not need any infinite amount of ethanol to get a mixture spanning all the range from 0 to 100% ethanol and behaving as a solution. The proper statement is miscibility at any ratio. It might be time consuming if you want to check for bizzarre holes instead of a continuous behaviour, but the way to prove it experimentally is what the book suggests.

I comment again because I really wonder how an author of a chemistry book can invent such a bizarre sentence. It is wrong from every point of view. Practically, philosophically, mathematically.

"It then says ''verify this for yourself by mixing some ethanol with water in a test tube''. Does this textbook give any hint on how you should get 100%, water-free ethanol? Burn your textbook trying to distill 100% (or at least 99%) pure ethanol.

The better term is miscible in all ratios. The idea behind the term is that you can have a mixture from 99.9999… % water and 0.0000…1 % ethanol to a mixture of 0.0000…1 % water and 99.9999…% ethanol. No infinity necessary.
The obvious opposite to miscible is immiscible (in some or all ratios). A common pair of household liquids that are immiscible are fatty oils and water. In the chemistry lab, hexane and water might be a better example.
To verify the miscibility of water and ethanol, set up a parallel experiment with water and hexane/a fatty oil such as sunflower oil. You will notice that a fully immiscible pair will never mix, there will always be two different bits of liquid visible. You can try a little oil/hexane and lots of water and vice-versa. On the other hand, a miscible pair will mix and you get a solution where you cannot tell one from the other.
For completeness, partially miscible: if you repeat the experiment with ether and water you will observe that the first couple of drops of water will dissolve in sufficient volumes of ether (and vice-versa) but if more is added you will again get two phases.

Good answer. "All ratios" raises a (stupid?) question in my mind. Is a single 50-50 ratio experiment adequate? Or is it possible that some ratio like 1:10 or 1:100 would show immiscibility? In solids, different ratios can give different phases.

@JamesGaidis This phase diagram (hexane, nitrobenzene) suggests that at high temperatures, there is one phase at 50:50 but two at 60:40. http://people.cst.cmich.edu/teckl1mm/pchemi/chm351ch8af01_files/image024.gif

@Karsten: Thank you. That suggests there may be many others if the two components are similarly disparate.

Both water and ethanol are colourless. How can I observe that they are infinitely soluble ?

They have different refractive index. You will see Schlieren as you mix them, which will eventually disappear. If two solutions don’t mix, there will be a visible interface. If you have trouble seeing it, mix it and let it settle.

"How to verify that ethanol is infinitely soluble in water?".
You cannot do it because it is WRONG. And not because it is false for ethanol, it is simply plain wrong conceptually and mathematically.
Solubility is the ratio between the amount of the specified substance and the volume of the phase solution.
As for a substance occupies its own volume, solubility cannot tend to infinity but at most reach the value dictated by its molar volume in the given P, T conditions.
For instance, let me take water - for which I remember the value - instead of ethanol. Its concentration as pure substance is about 55.5 mol/l.
This is aimed to correct a disturbing error in the text book rather than provide a practical answer.
For finding out if water and ethanol are miscible in any ratio - this should have been the question - the book is at least correct that a simple test tube suffices to get at least a reasonable idea. How and why is already in the other answers and comments and I assume it is clear to the OP as well, at this point.

I would like to know why this has been downvoted.

It is sad to see how somebody uses the downvote function. My answer does correct a terrible mistake of a chemistry book. Infinity soluble is understood in day life perhaps, but to even think of infinite solubility is completely wrong from too many viewpoints.

It's interesting to see your viewpoint. The term "infinitely soluble" can indeed be misunderstood to mean that an infinite amount of solute could be dissolved in a finite amount of solvent. However, that is so farfetched that it would be glossed over.

@JamesGaidis the point is the it does not matter how farfetched it is. Is simply wrong and stupid. The very point is not even abstraction to infinity. It is not even the point. At most solubility tends to a finite amount (whatever depending on how S it is expressed) and such a wording isn't acceptable on a chemistry book or chemistry forum. I am not saying it can be misunderstood in practice, but conceptually yes it can be. Your comment is in that direction. Farfetched is something about mass or amount of substance, not about solubility or more generally conc.

@Alchemista: Well, how much water is soluble in a drop of ethanol? Quite a lot, and if not infinity, then mighty close to it, because the amount is boundless.

I upvoted to say that I agree with that of course, except the term is miscibility not solubility. Solubility is never unbound. If we want to be even more pedantic that would be possible when expressing solubility as molality, but then one as to keep to call solvent what is finally present in traces. In short, solubility is not mass nor volume nor amount of substance.

@Alchemista: I agree with you; that is a better word. It's interesting to think of the situation with extremely soluble salts, like Ca(NO3)2.4H2O, 2710 g/L at 40C, density 2.5, mp 43C. There is more weight and more volume of solid than the water solvent, so it could be said that the solid has dissolved the liquid. I would just call it an interesting quirk of nature - doesn't happen that often, so not to worry about it!


text:
How do I go about calculating the enthalpy of a solution with an unknown salt?
I have the $\Delta T$, I have the specific heat of the calorimeter, I have known weights/mass. I know the salt is lithium-based. What steps do I take to calculate the enthalpy of the solution so I can compare my answer to the specific molar enthalpies of possible lithium salts?
If the equation I use is $Q = m \times c \times \Delta T$, what mass do I use and do I use the specific heat of the calorimeter?
I just don't understand this. I've been hitting my head off the problem with research and trying to attack it from different directions and I'm just stuck.

https://chemistry.stackexchange.com/a/1105/102629

You determine Q from your equation using the mass and specific heat capacity of the calorimeter (you ignore the heat capacity of the salt, as it's small relative to that of the calorimeter).   Then you divide Q by the number of moles of salt you added to get the enthalpy of solution.  If you want further details, most general chemistry textbooks should have worked calorimetry examples.


text:
Cleavage of 2-Methyltetrahydropyran with HBr

After this cyclic ether is activated with H+, will the bromide attack the more substituted site on the left or the less substituted side on the right? I think it will attack the right because SN2 favors less steric hindrance, but what is confusing me is that for acid activated epoxides, nucleotides attack the more substituted site. Can anyone clarify this for me? Does it have to do with the size of the ring? The charge of the nucleophile? Please help.

Which is the more stable cation after breaking of the C-O bond?

You will likely obtain 1,5-dibromohexane. The question is which intermediate bromo alcohol forms faster and by what mechanism.?

related: https://chemistry.stackexchange.com/q/138998/102629

@cngzz1: The reaction you cited is not related here. This is ether opening. The reaction you mentioned is acetal opening.

@MathewMahindaratne How about https://chemistry.stackexchange.com/q/29488/102629

@cngzz1: Yes, that could be related.


text:
Is N--O--N known to exist?
Commonly, the empirical formula $\ce{N2O}$ refers to the molecule with structure $\ce{N\bond{~--}N\bond{~-}O}$.  Is the isomer $\ce{N-O-N}$ known to exist? I'm guessing it is significantly less stable than the NNO isomer, but that doesn't mean it's completely unknown.

What's your definition of existing? For that, being in minimum on PES would be already something. It's existence is fleeting, at best.

I don't know what PES is, but I'd take any one paper claiming to have synthesized and characterized it as a good start on evidence.

@zwol PES is a potential energy surface.

That you won't get. "synthesized and characterized" are wrong words. It's rather detect and that's optimistic.

Synthesized and characterized are the things that it was not.

Unfortunately N-O-N does not exist. At least, it has never been observed or synthesized. Maybe in some future ?

@Maurice Can you write an answer explaining how you know that?

The nitrogen atom never makes stable ions or molecules with only one covalent bond. N-H or N-Cl or N-OH do not exist. In this situation, N would have been surrounded by six electrons, and would not look like a noble gas. It would not be stable.

@Maurice Your last comment may be a useful "lie for children", but again isn't really correct. Such molecules may be very reactive, but may be present in interstellar space. https://en.wikipedia.org/wiki/Imidogen

See this papers where cyclic and linear NON isomers of N2O are calculated: https://pubmed.ncbi.nlm.nih.gov/11209870/ ... https://www.researchgate.net/publication/225137278_Valence_bond_and_molecular_orbital_studies_of_three_N2O_isomers_and_valence_bond_representations_for_some_azide_decompositions


text:
Degassing HI from Reaction (Via Ultrasound Maybe?)
I'm just going to put the questions first and the explanation after in case y'all don't have the patience.
I want to know if it's reasonable to think that :

Using an ultrasonic bath on something like a rotovap would prevent bumping, with the cavitation effects serving as basically the ultimate boiling chip.

Whether ultrasound could be used to degas an organic solution while a reaction is occurring, in order to definitely confirm that a gas is not supersaturating the reaction mixture.

Does it make any sense whatsoever that hydrogen iodide could stay in solution in toluene?

I'm considering strapping an ultrasonic transducer onto a metal pan we use as our oil bath. Is there a potential reason that this wouldn't be effective for 1) and 2)?
Context: working on a mild cannabinoid conversion (at a Canadian university, definitely not sketchy) via the following method:
https://pubs.acs.org/doi/10.1021/acs.jnatprod.7b00946
Long story short, we're investigating why side products are being produced and halving the yields when any larger amounts are used in the reaction. I'm seeing brown residue (immiscible with the reaction solvent of Toluene) collect and return down the condenser into the mixture, formed via this reaction:
4 HI + O2 → 2 H2O + 2 I2
Of which the HI and H2O then form concentrated hydroiodic acid droplets which return to the mixture.  I think this means that purging with nitrogen and otherwise degassing the solution is necessary. I think HI should basically boil out as it is formed due to very low boiling point and negligible solubility in toluene, so just doing it under nitrogen flow should be fine. Supervisor thinks some is sticking around in solution (which is why larger volumes at identical reagent concentrations also lower yields, as the smaller volumes means higher surface:volume and takes less time for the entire volume to boil/recondense/cycle compared to larger volumes), so basically what I'm trying to do is go so ham on degassing this solution that we can diagnose whether this is accurate.

Could you react the HI somehow to remove it? Long time since organic chemistry but would cleaving an ether work?

I can't view the full reaction procedure you are using as it is behind a paywall but if your system is generating water due to the presence of oxygen then running under nitrogen/argon should help. Can you remove the HI by running in the presence of a basic resin?

In short, to answer your question "Why side products are being produced",  you appear to have ignored all possible radical activity!
Per this source: Free radical formation from sonolysis of water in the presence of different gases, for example, to quote:

$\ce{H2O + Sonolysis -> .H + .OH}$

Here, replacing H2O with HI likely results similarly in:
$\ce{HI + Sonolysis -> .H + .I}$
Some possible further reactions:
$\ce{.H + .H -> H2 (g)}$
$\ce{.I + .I -> I2 (s)}$
And, in the presence of an organic (RH):
$\ce{.H + RH -> .R + H2 (g)}$
$\ce{.OH + RH -> .R + H2O (l)}$
$\ce{.R + .I -> RI }$
$\ce{.R + .R -> R2 (g)}$
Also, with any dissolved oxygen (or even nitrogen, see cited reference), expect even more, as for example:
$\ce{.H + O2 -> .OOH}$
$\ce{.HO2 + .HO2 -> H2O2 + O2}$
$\ce{.R + .OOH -> ROOH}$
....
As such, expect a range of products, as would occur, by the way, if you performed photolysis in lieu of sonolysis. However, there are related reactions, like in the presence of ammonia, where the final photolysis products can display a high degree of product selectivity.


text:
Origin of the NMR shielding
In a paramgnetic material the induced dipole moment of atoms or (molecules) of the material aligns with the external field. I can understand this by Zeeman splitting. A term symbol splits into states with different energy. According to Boltzmann distribution the lower energy states should have more population. These states would correspond to dipole moments aligned with the external field therefore the field inside the material should be enhanced. Why then the nuclei in NMR "feel" a lower magnetic field? I read that it was according to Lenz law. Is that because we take into account only the field in the region near to the nucleus? I mean if we had a paramagnetic material then the field should be greater.

According to my understanding, resonant frequencies are effectively a measure of the effective (felt) magnetic field on the magnetic dipole moment of a nucleus. Therefore you don't include the nuclear magnetic dipole in your calculation of the effective field. As for shielding, it is caused by the electron density surrounding the nucleus. The applied magnetic field causes electrons to precess, generating an opposing magnetic field and lowering the effective magnetic field on the nucleus.

Lenz' law is about electrons, specifically about the direction of the field that moving electrons induce!

@Karl I can't understand though why in paramagnetic materials the induced field is in same direction with the applied. Are these two phenomena (NMR shielding and paramagnetism) related at all?

They are not. The polarisation of spins in a field gives a very small static net magnetisation in the direction of the field. Chemical shift is caused by induction of moving electrons, and creates a field against $B_0$.


text:
salt hydrolysis and pH calculations
I am a high school student and I am very confused in salt hydrolysis
My confusion is that: when we talk of hydrolysis of salt of weak base and weak acid, we internally assume the degree of hydrolysis of both cation and anion of the salt is assumed to be same but I think it is a very wrong assumption, why don't we simply do the pH calculation like I have done in this example? what is the problem with this?,if my approach is also correct then why doesn't it matches with the standard answer? what is the mistake in it? I have found some similar questions but no one has given the exact answer to those that's why I am asking it here please, explain it in a simplified way because I am only at high school level.

The way you write the letter H is very unique as a mirror image of N! It is hard to follow the picture, can you clarify your question by typing? "We internally assume the degree of hydrolysis of both cation and anion of the salt is assumed to be same". No this is not correct nor it is a requirement.

No, it is a assumption,,, this is even my problem that if we are assuming the hydrolysis of both to be same  then concentration of OH- as well H+ will be same due to hydrolysis and then we consider the dissociation of the formed acid and base,, to find the concentration of H+ and OH- ,but then why we dont'' subtract their concentrations here to find net concentration of H+ or Oh-?,,,In this image I have mentioned the results if we don't assume it,,,which is very different from the result which we drive by using the direct formula,,,,please try to understand the image

How does your textbook solve this or a similar problem like this?

I have added one more image  you can see that when we derive the standard formula we internally take an assumption that the cation and anion are hydrolysed   to the same extent,,,why do we cannot solve it without taking assumption like I have done in above problem??what is the mistake in it?

Please practice writing H properly. People will have trouble reading your hand writing in general. It looks like a Russian letter. https://en.wikipedia.org/wiki/I_(Cyrillic)

Using photos/screenshots of ( even handwritten) text instead of typed text itself is strongly discouraged. It is impossible to index/search for it, or reuse it in answers referring to it. In a  case  of handwritten text, it puts extra burden on responders to properly decipher it.
That all may lead to the question being ignored or even closed.
Consider copy/paste or retyping and using eventually  MathJax for expressions and formulas.

sorry, but I am a high school student and doesn't know how to type these equations so that all people can understand the last time I tried typing ,my question was get closed,, so here I prefer sending the handwritten image ,,,I will gradually learn how to type these equations properly ,sorry for all the stress.

@Poutnik I have searched it ,,,but don't get any detailed answer,,,,,,my question here,,simply is why don't we do the problem simply like I have done,,,what is the mistake here??why do we go for approximations which are not needed

We can assume the same degree of hydrolysis for both cation and anion in this specific case, as pKa of acetic acid is to 2 decimal points equal to pKb of NH3. What happens to acetate, producing acid and OH- is mirrored by NH4+, producing NH3 + H+. The result is neutral pH 7.00 ( +- 0 01, depending on exact constant values ).

Your simplified formula [OH-]=sqrt(c0 . kB) requires  the assumption that auto-dissociation of water can be neglected. Reality is the very opposite. Practically all the ions H+ and OH- would be from water auto-dissociation.

The general approach to solve acido-basic equilibrium problems is:

respecting charge balance and charge conservation
respecting mass conservation
respecting dissociation equilibrium constants for acids, bases and water itself.

Generally, it leads to a set of nonlinear equations for a set of independent variables. The application of a substitution method usually results to a polynomial equation for $\ce{[H+]}$ of the second up to the forth order. The latter is hard to solve analytically.
Fortunately, in many scenarios, there can be made justified simplifications, like

neglecting water autodissociation
neglecting concentration of one substance form wrt the other one or their sum.

That largely simplifies an equation to be solved.
E.g. to calculate $\mathrm{pH}$ of solution of acetic acid, ammonia, or sodium carbonate,
we can consider none of $\ce{H+}$, resp. $\ce{OH-}$ are created by the water autoionization and just negligible part of the compound underwent the acido-basic reaction.
E.g. for sodium acetate hydrolysis
$$K_\mathrm{b}=\frac{\ce{[HA][OH-]}}{\ce{[A-]}}\approx
\frac{{\ce{[OH-]}}^2}{\ce{[A-]}}\approx \frac{{\ce{[OH-]}}^2}{c_0}
 $$

but,,,I want to know what is the problem with my solution of above problem,,why  it is incorrect,,,or if isn't incorrect, why don't we simply do it like this and not go for the formula?

I kind of refuse to read your manuscript, sorry.

@ArunBhardwaj, Can you type the solution in Word? It is hard to follow that picture. Convert that Word typed solution in pdf and then you can use Mathpix Snip to convert your equations into LaTeX which be pasted here. https://mathpix.com/

Regarding your degree of hydrolysis expression in the textbook, we have to keep charge balance in mind. It is not an assumption, it is a charge balance fundamental requirement because a solution can never have a uncharged specie without its counter ion. A solution is always electrically neutral. For example, $\ce{Na+}$ cannot exist alone in a solution, it also needs one singly charge anion like $\ce{Cl-}$ or $\ce{NO3-}$ to balance the charge. Same is the case of hydrolysis, if you lose one ammonium ion on the left hand side you must also lose a single formate ion.

but I think that the net charge is balanced even in my answer that I have written above,,,,as CH3COO-  ion gives ch concetration of OH^- ion ,,,and NH4+ gives c*alpha concentration of H+ ions if you write the concentration of all ions and balance then you will notice the charge is balanced

LOL - I spent so much time trying to decipher your handwriting I missed what you question was.
You actually asked a very pertinent question. It is very very wise to wonder about the boundary conditions of some derived formula.
First let me point out that chemistry problems aren't like pure math problems. In math pi has been calculated to millions of decimal places. In chemistry 2-4 significant figures is typically the best that can be done.
You are absolutely right. For ammonium formate there are two hydrolysis reactions.
$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$
$$\ce{HCOO- + H2O <=> HCOOH + OH-}\tag{2}$$
Now the solution must be electrically neutral so for the charge balance we have:
$$\ce{[NH4+] + [H+] = [HCOO-] + [OH-]}\tag{3}$$
However let's consider making a strong ammonium formate solution say 0.1 molar. The ammonium cation is a weak acid, and the formate anion is a weak base. Thus little of either will hydrolyze. So we can assume:
$$\ce{[NH4+] \gg [H+]}\tag{4}$$
and
$$\ce{[HCOO-] \gg [OH-]}\tag{5}$$
thus for charge balance we can assume:
$$\ce{[NH4+] = [HCOO-] }\tag{6}$$
This in turn implies that the majority of ammonia and formic acid in solution are not formed by the reactions (1) and (2) but rather by the reaction:
$$\ce{[NH4+] + [HCOO-] <=> [NH3] + [HCOOH]}\tag{7}$$
Thus the two ions from the salt are hydrolyzed by same amount.
Now if we have instead made a weak solution of ammonium formate, say $1.0\cdot 10^{-5}$ molar, then equations (4) and (5) wouldn't be valid and the individual hydrolysis reactions would have to be considered making the formula to calculate the pH much more complicated.

I have one more question that, by assuming the degree of hydrolysis of both cations and anions to be same,, don't we indirectly assuming the concentrations of H+ and OH- to be same? as they will be produced in smae amount and will react in same amount to form water


text:
Why are s orbitals more “penetrating” than p orbitals?
In polyelectronic atoms, the reduction in the net central force due to electron-electron repulsion is accounted for through an effective nuclear charge that depends on a “shielding effect” of inner electrons.
The reason for the splitting of energies of orbitals in the same quantum shell (s<p<d<f) is usually given that s orbitals are more "penetrating" i.e have a greater probability density near the nucleus, so experience less shielding and thus a greater effective nuclear charge.
However, I can't understand why a comparatively larger p. density is used to measure the penetrating power. the mean orbital radius actually decreases from s to f orbitals. Sure, s orbitals have a relatively higher p.d near the nucleus but the rest of the p.density is also relatively more further away than that in p,d or f orbitals. Using the mean distance as the appropriate measure, shouldn't the energy be in the order f<d<p<s?
the only explanantion that comes to mind is that the decrease in energy from being unusually close some of the time outweighs the decrease in energy due to being more farther away for the rest of the time (and thus on avg), but this is just a guess and a rather non-intuituve explanation

The point is, the mean distance is not an appropriate measure, due nonlinear distance dependence of electrostatic potential and also due the distance dependent effective nucleus charge.

related: https://chemistry.stackexchange.com/q/152/102629

Also do not see it as "most of the time". Is as it is.

I found an old slide from when I was teaching.
These are plots of the 2s and 2p orbitals for hydrogen (assuming 1 electron). The radial wave function is on the left and the probability density as a function of distance from the nucleus is on the right ($4\pi r^{2}R(r)dr$ if you understand the calculus).
You can see that there's a blob of density that comes closer to the nucleus for the s orbital. There's your penetration.


text:
Melting points of benzoic and stearic acids
I wonder, why does benzoic acid have a higher melting point than stearic acid?
Here is the data:
Benzoic acid melting point = 122.3 °C
Stearic acid melting point = 69.3 °C
I do understand that organic acids can bond with both London dispersion forces (van der Waals) and hydrogen bonds (due to the -COOH group). I have also learned that the longer carbon chains, the higher melting point due to the increase of the London dispersion forces.
However, in the case with benzoic acid vs stearic acid, I can't figure out an explanation as to why benzoic acid has a higher melting point. I suspect it is because of the resonance structure in the benzene ring in benzoic acid?

Consider the dimer $$\ce{Ph-C(OHO)2C-Ph}.$$ The respective dimer for stearic acid is less likely to be formed, because of entropy. The BA dimer would be nicely planar, with great intermolecular cohesive forces.

@Poutnik - Thank you! Please excuse me if I am very wrong in my following guess: So I can say that benzoic acid has a higher melting point because larger contact area gives stronger intermolecular forces?

Well the resonance within the aromatic ring can be seen as a reason in the sense that results in a regular flat structure that can orderly pack; perhaps a source of electromagnetic interaction between   molecules, ecc. At the end you are comparing very different things. The "longer the molecule..." rule of thumb is OK for comparing terms of the same class of compounds that differ, eg, by a few carbon atoms. Roughly, think of a tile versus a flexible spaghetti. The first il likely to pack and interact with the neighbours tightly. Cyclohexanecarboxylic acid is a liquid, to let you finish....

@Alchimista - Thanks, but I still find it difficult to understand. Is it the shape of the molecule that makes benzoic acid more prone to arrange itself more tightly?

@Csharpyikes more tightly than what? All this comes from you comparing benzoic acid to stearic acid. They are different enough to render the more than reasonable rule of thumb not applicable here. I should write an answer? It would be tedious for me and at the end convey the message that everything should be taken with a grain of salt. Even aromatics are not lining down on each other because flat, normally a kind of tilted arrangement called herring-bone is found. This point to interactions - more specific than generic dispersive/London/VdW forces - associate with the aromatic rings. Still

... benzene or toluene are liquids. So the fact of dealing with the carboxylic acid no doubt has a major role. Note also the dimer mentioned in another comment. If you really want a on point answer between stearic acid and benzoic acid then consider my first comment again, or that of Poutnik. Long chains are inherently flexible. At a certain T an entropic term kicks in, while a closed ring or something anchored will be rather insensitive to this aspect. Let me know if this helps

@Alchimista - Im sorry but I still don't understand that much. (I am just a 16 yr old reading high school chemistry, trying to understand what my book doesn't tell me) :(

I mean, I do understand that the structures are affecting the melting point, where benzoic acid is exemplified as a tile and stearic acid as a spaghetti. But I still don't get it 100%, is it only that that affects the melting point? That it is easier for a "tile" to structure itself into a solid form and therefore needs more energy to break it down?

@Csharpyikes no it is not only that otherwise benzene would be a solid too. You cannot reduced everithing to a single and primary cause. But given a series of compounds you can reasonably order them for mp or or a number of properties. Even with no calculation (in many cases they will be difficult indeed) but in the way we proceed in this thread. Fortunately James Gaidis wrote a detailed answer that at first glance shall answer your question.

Puzzles are more puzzling when the pieces almost fit together, but for some unexplained reason, just don't fit. In other words, a comparison should be made between two almost  (but not quite) similar things.
In this case, stearic and benzoic acids are quite different. They have some carbons (quite different, and not even the same number!), and one similar group of four atoms ($-COOH$) - not very much similarity.
How about we compare the melting points of octadecane and stearic acid - and their structures? There's a lot of similarity there, and a little dissimilarity. Octadecane melts at 28-30 C (Ref. 1) vs. stearic acid at 69.3 C (Ref. 2). Obviously the difference is caused by the added stickiness of the carboxyl groups for one another, because they form dimers in the crystal, a structural feature (Ref. 3) lacking in the paraffin.

That establishes that a carboxyl group raises the melting point of a paraffin chain. An x-ray structure for $C_{24}H_{50}$ (octadecane not available) looks just like the one for stearic acid, but without the carboxyls (Ref. 4).

Now let's compare benzoic acid with toluene. The numbers of carbons are the same, and we just convert one $CH_3$ group to a $COOH$, as in the previous comparison. Toluene stacks in the solid (Ref. 5) with energies about 3-4 kcal/mol, which isn't much, considering that toluene melts at -95 C. But benzoic acid stacks nicely (Ref. 6) and melts at 122 C.

Now there's a puzzle: why does a carboxyl group raise the mp of a paraffin by 40 degrees, but raises the mp of an aromatic ring by 217 degrees? Looking at the formulas, looking at the drawings of a molecule (or two) doesn't help. Looking at actual data (because those little molecules have a mind of their own, and we have a hard time thinking like they do) like x-ray crystal data, suggests that putting a little planar group ($COOH)$ onto a long chain might strengthen intermolecular bonding a little, but extending a ring's planarity and adding in some hydrogen bonding, which amplifies the extent of planarity could do some powerful intermolecular stickiness.
I guess the wiggly straight chains don't have such good contact with their neighbors as the benzoic acid dimers do. Who would have guessed, without the crystal structure data?
Will puzzles never cease? If octadecane melts at 28-30 C, and its mp is raised by turning a methyl into a carboxyl, you might wonder what a phenyl group would do. Apparently, not much. The mp of 1-phenyloctadecane is 29 C (Ref. 7).
Ref.1 https://en.wikipedia.org/wiki/Octadecane
Ref.2 https://en.wikipedia.org/wiki/Stearic_acid
Ref. 3 https://www.researchgate.net/figure/Crystal-packing-of-left-the-A-super-14-and-right-the-A-1-13-two_fig1_233992144
Ref. 4 https://www.researchgate.net/figure/Crystal-structure-of-C-24-H-50_fig1_270344720
Ref. 5 https://www.x-mol.com/paper/6024417
Ref. 6 https://www.researchgate.net/figure/Selected-intermolecular-contacts-in-the-benzoic-acid-crystal-viewing-the-dimers-from-the_fig2_51049984
Ref. 7 https://pubchem.ncbi.nlm.nih.gov/compound/1-Phenyloctadecane#section=Experimental-Properties

+1 I particularly liked your first paragraph. Many budding chemistry students don't get that they are asking "Why is an apple different than an orange?. They are both fruit."


text:
Common Ion Effect explanations
From my textbook, it is stated that:
e.g. If we add a basic salt $\ce{NaF}$, it will fully dissociate to give $\ce{F-}$. Then if we add $\ce{HF}$, since it is a weak acid, the high concentration of $\ce{F-}$ already present in solution shifts the equilibrium to the left, hence fewer $\ce{H+}$  forms than expected so higher $\mathrm{pH}$.
But can I explain like this: after dissociation of $\ce{NaF}$, the formed $\ce{F-}$ reacts with water to give $\ce{OH-}$ as the following equation:
$$\ce{F- + H2O <=> HF + OH-}$$. Due to the presence of $\ce{OH-}$, upon addition of $\ce{HF}$, the $\mathrm{pH}$ will be higher than expected. Are there any flaws in this explanation?

You picked a really bad example. For example for KCN both paragraphs would be OK. For HF thing look very differently https://chemistry.stackexchange.com/a/34829/9961

I agree, with forming HF2- and rapid increasing of HF acidity for concentrated solutions. ( $\ce{3 HF <=> H2F+ + HF2- }$ )

Both your paragraphs are valid, as all present components are balanced by 3, mutually linked equilibrium reactions, with the respective equilibrium constants:
$\ce{HF + H2O <=> F- + H3O+}$, $K_\mathrm{a}$
$\ce{F- + H2O <=> HF + OH-}$, $K_\mathrm{b}$
$\ce{2 H2O <=> H2O+ + OH-}$, $K_\mathrm{w}$
with the relation $K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}$
$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[F-]}}{\ce{[HF]}}}$

thanks. To clarify, so both can explain the effect, and are there any causal relationship between the 2 explanations since you said they are 'linked'?

Rather 2 complemental sides of the same thing. Like 2 pieces of torn up paper sheet. What is the causal relation between them ?


text:
Are electron orbital orientations filled in any particular order?
This question is about the magnetic quantum numbers and their corresponding orbital orientations, e.g., px, dyz, etc. Are the electrons in any given subshell distributed across the orbitals in any particular order as the number of electrons increases?  Such as m = -1 first, then m = 1, etc.?

No. Also, there is no one-to-one correspondence between $p_x, p_y, p_z$ and $m=-1,0,1$.

a little more detail would be super helpful.  How ARE they distributed then?  Randomly? Why is there no correspondence?  What difference do the notations represent?

Let me explain with the example of p-subshells. The first electron entering the p-subshell could occupy anything - x or y or z. Suppose it occupied x, then the next will occupy either y or z but not x again yet. Suppose z is occupied now, so then the electron occupies y. Now after all three are occupied by one each, pairing occurs again randomly. From my memory, all -x,y,z are of same energy but only their orientation in space is different.

Consider that x, y and z are completely arbitrary (other than being orthogonal to each other), and m_l was arbitrarily chosen to align with the arbitrarily chosen z (ie it could have been the y component just as easily). Also recall that electrons aren't connected to individual orbitals until we measure them. An atom with a single electron in p orbitals is better thought of as a superposition of the states with px occupied, py occupied and pz occupied, and even that is a greater degree of localization than is "true".

No, you cannot say that there is a particular order in which the shells are filled.
Relation between $\mathrm{p_x,p_y,p_z}$ with $\mathbb{m_l}$ values
The magentic quantum number ${m_l}$ represents the projection of the angular momentum $\vec{l}$ of an electron in an orbital, on an axis, usually taken to be the z-axis. The orbitals, and their $\mathrm{m_l}$ values, come from the solution of the Schroedinger's equation for a Hydrogen atom.
Now, the problem is, the solution for the orbitals with ${l=1}$ (i.e. p orbitals) produces one real wavefunction (corresponding to $m_l=0$), and two other complex wavefunctions (corresponding to $m_l=1$ and $m_l=-1$). If the main axis is taken to be z, then the $m_l=0$ orbital has the usual dumb-bell type shape and points in the z-direction, so it is the $\mathrm{p_z}$ orbital. However, the complex orbitals are conjugates of each other, and if you plot the real part, they both look like donuts. (You can find more on this here and here). So the other two orbitals, let's call them $\mathrm{p_1}$ and $\mathrm{p_{-1}}$, do not really correspond to anything we usually see in the chemistry textbooks.
However, wavefunctions themselves don't hold any physical meaning, because we cannot observe them. We can only observe things like electron energy, electron density etc. So, we can take linear combinations of the two complex orbitals to get two new real p orbitals, as long as it conserves the total energy. However, doing this means that the new $\mathrm{p_x}$ and $\mathrm{p_y}$ orbitals are no longer eigenfunctions of the $\hat{L_z}$ operator (i.e. they have no well defined angular momentum along the z-axis, so no well defined $m_l$ value). In most cases angular momentum is not really important, so we can safely use $\mathrm{p_x}$ and $\mathrm{p_y}$.
I suspect that there is a similar case with the d orbitals as well.
Which orbital is filled in first?
Inside one shell (e.g. 2p) the electrons are indistinguishable, so it is not possible to say which orbital is occupied by which electron. However, if a shell is not completely filled, i.e. there are unpaired electrons, then there would be an total orbital angular momentum and total spin angular momentum coming from all of those electrons combined. It is possible to identify states which have different angular momenta, as they have different energies. This forms the basis of the coupling schemes (LS coupling or jj coupling). However, you still cannot say exactly which electrons are residing in orbitals with which $m_l$ values, because multiple arrangements (microstates) can contribute to the same energy level (term).


text:
Clarification of term symbol explanation of laser crystal doping from textbook
I am currently studying the textbook Diode Lasers and Photonic Integrated Circuits, second edition, by Coldren, Corzine, and Mashanovitch. Chapter 1.2 ENERGY LEVELS AND BANDS IN SOLIDS says the following:

In gas and solid-state lasers, the energy levels of the active atomic species are only perturbed slightly by the surrounding gas or solid host atoms, and they remain effectively as sharp as the original levels in the isolated atom. For example, lasers operating at the $1.06 \ \text{$\mu$m}$ wavelength transition in Nd-doped YAG, use the $^4\text{F}_{3/2}$ level of the Nd atom for the upper laser state #2 and the $^4\text{I}_{11/2}$ level for the lower laser state #1.

I don't understand what is meant in this part:

For example, lasers operating at the $1.06 \ \text{$\mu$m}$ wavelength transition in Nd-doped YAG, use the $^4\text{F}_{3/2}$ level of the Nd atom for the upper laser state #2 and the $^4\text{I}_{11/2}$ level for the lower laser state #1.

I understand that these are term symbols. However, these are usually taught in the quantum mechanics part of physical chemistry textbooks, rather than physics-focused quantum mechanics textbooks. I have only just begun slowly studying physical chemistry, so I do not yet have any exposure to term symbols. Furthermore, after skimming through the textbook, this looks like it might be the only section (or at least one of the only few sections) that use term symbols. Therefore, I do not want to get hung-up on this part, and would rather keep progressing and then learn the details of term symbols when I get to them in my physical chemistry textbook. So I was wondering if people here with physical chemistry understanding would please take the time to explain what this is saying in a way that does not require understanding of the details of term symbols, so that I may understand "the gist" of this part of the textbook and move on.

All this is really saying is that the Nd ions in the Nd-doped YAG matrix have extremely similar environments in the crystal. So their interactions with the pump beam are almost the same and a narrow laser emission profile occurs. This is what happens in standard green laser pointers: the Nd-doped YAG emits at 1064 nm and is frequency doubled to 532 nm. In a Nd-doped glass laser medium, the Nd ions are in slightly different environments, because glass is not ordered, so the energy levels are perturbed and emission from the ensemble of perturbed Nd ions results in broader emission profile.

@EdV Oh, so it's referring to the environments of the Nd ions themselves relative to each other? And, in the glass case, all of the Nd ions are not in extremely similar environments, so the energy levels are perturbed?

Minor fix: in green laser pointers, Nd-doped yttrium orthovanadate is used rather than Nd-doped YAG.

Correct! Glass is just not the nice ordered crystal environment and this has consequences.

@EdV Ok, this makes sense to me. Thanks for taking the time to clarify.

The answer provided by orthocresol provides a very nice explanation of term symbols, so my answer will focus on the laser specifics aspect of the question. To begin, figure 1 shows two laser rods: ruby crystal on the left and Nd-doped phosphate glass on the right. Both laser rods can be optically pumped to achieve the necessary population inversions. The neodymium lasers, i.e., Nd-doped glass or YAG (yttrium aluminum garnet) are 4 level systems, while the ruby laser is a three level system and is not discussed further.

Source: my photo of two of my laser rods.
Side note:  The Nd-glass rod is polished on the ends and on bands around the ends. That is where the o-rings would seal the rod in the flashlamp cavity. The ruby rod is only polished on the ends.
Verdeyen 1 shows the usual emission manifolds for the Nd-YAG laser and a typical glass Nd-glass laser.

Source: 1, Fig. 10-5(a) and 10-5(b).
This is what Verdeyen 1 has to say about the emission manifolds:

Only the lowest level in each manifold is shown in Fig. 10-5 (a) and only a few of the important ones are shown in Fig. 10-5(b). For instance, the $\mathrm{^{4}I_{9/2}}$ manifold consists of five states extending from $\pu{0}$ to $\pu{857 cm^{-1}}$, the $\mathrm{^{4}I_{11/2}}$ manifold consists of six states extending from $\pu{2001}$ to $\pu{2518 cm^{-1}}$, and so on. (See Ref. 6 for more details.) One consequence of this large number of states in a given manifold is that the fluorescence line shape is inhomogeneously broadened (even in YAG) and that one would be hard pressed to describe the fluorescent line shape functions, shown in Fig. 10-5 (e) and (f) by a simple mathematical expression.

Source: 1, p. 261. Reference 6 in the quotation is 3 in the references below.
Yariv 2 provides a slightly more detailed emission manifold figure:

Source: 2, Fig. 10.11. Reference 8 in the figure caption is 4 in the references below.
The emission profiles mentioned in the first quotation are shown below:

Source: 1, Fig. 10-5(e) and 10-5(f).
Verdeyen 1 says

The line width of the transitions around $\pu{1.06 \mu m}$ for YAG is much smaller than that for glass $\mathrm{(\Delta \lambda_{YAG} \sim 7 \overset{\lower.5em\circ}{\mathrm{A}} \ll \Delta \lambda_{glass} \sim 300 \overset{\lower.5em\circ}{\mathrm{A}})}$ but is still asymmetric. The asymmetry is due to the fact that there are two different transitions present in the smooth curve of Fig. 10-5(e) - a strong one at $\pu{1.0641 \mu m}$ and a weaker one at $\pu{1.0645 \mu m}$ (in about a 3:1 mix). Nevertheless, the huge difference in the line shapes points out the role of the atoms’ local environment. YAG, being a crystalline substance with a well-defined structure, presents a nearly identical environment for each Nd atom and its line width is narrow. Glass, being an amorphous material, presents a different site at each location; consequently, its line shape is much broader.

Source: 1, p. 261.
The transmittance spectrum of a Nd-silicate glass is shown below:

Source: 2, Fig. 10.13. Reference 8 in the figure caption is 4 in the references below.
The strong absorbance just past $\pu{800 nm}$ is especially useful for laser diode pumping at $\pu{808 nm}$. Inexpensive green laser pointers use yttrium orthovanadate crystals, similar to YAG, excited by IR laser diodes with emission wavelength near $\pu{800 nm}$. The resulting IR lasing is near $\pu{1064 nm}$, which then gets frequency-doubled to the green $\pu{532 nm}$ output. It is important to filter out residual IR, to avoid inadvertent eye damage. Not every inexpensive green laser pointer does this properly.
Also note the strong absorbance just below $\pu{600 nm}$: this absorbs the famous sodium D lines extremely well, making “didymium” glass useful for blocking the over-whelming sodium emission that occurs in glass blowing.
One more thing: the OP’s question looks reasonably easy on the surface, but it is easy to find contradictory information. For example, Lakshman and Kumar 5 provide the following branching ratios table:

So they claim that the major lasing line is to the $\mathrm{^{4}I_{9/2}}$ manifold, instead of the the $\mathrm{^{4}I_{11/2}}$ manifold. N.B. SASP glass is sodium acetate and sodium hexametaphosphate glass, fused and then quick quenched.
References:

Joseph T. Verdeyen, Laser Electronics, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1981.

Amnon Yariv, Quantum Electronics, 2nd Ed., John Wiley & Sons, New York, 1975.

H.G. Danielmey, “Progress in YAG Lasers”, in Lasers, Vol. 4, ed. A.K. Levine and A.J. DeMaria, New York: Marcel Dekker, Inc., 1976, p. 8.

E. Snitzer and C.G. Young, “Glass Lasers”, in Lasers, Vol. 2, ed. A.K. Levine, New York: Marcel Dekker, Inc., 1968, p. 191.

S.V.J. Lakshman and A. Suresh Kumar, “Lifetimes of laser lines of $\ce{Pr^{3+}}$, $\ce{Nd^{3+}}$, $\ce{Er^{3+}}$ and $\ce{Tm^{3+}}$ in acetophosphate glass”, J. Phys. Chem. Solids 1988, 49, 807-811; doi 10.1016/0022-3697(88)90032-7.

Since I never dared to open the Surelite NdYAG's this might be a naïve question: why does the rod next to «Nd-glass» looks like a polished cap, while the lower end of the rod next to «phosphate glass» still is less glossy (like the ruby rod along all of its mantle)?

The ruby rod is from a 1970s ruby laser. The ends are polished, but the rest is unpolished. It was in a silver-plated water-cooled brass elliptical housing, at one foci and a Xe flashlamp was at the other foci. The ends of the ruby rod has o-rings on them, so the design was not good for several reasons: the ends after the o-rings were not optically pumped, so they were actually detrimental. And the rod was really heavily doped: I have three other ruby rods that are lighter doped: one is basically pink rather than pigeon blood red. The Nd-glass rod is polished on the ends and on bands around

the ends. That is where the o-rings would seal the rod in the flashlamp cavity. I never used it: it was a gift from a colleague at Corning. The ruby rod in my photo was used, but not by me. It is beautifully dichroic: rotating it about its axis is quite neat. I used to pass it around class (a grad class), telling students to be extremely careful not to drop it. Looking through it is wonderful: the most intense red I will ever see, most likely. By the way, my photography skills are poor, so maybe your question was caused by my bad photo?

@Buttonwood At CSI XXV in Toronto on June, 1987, I saw 1981 Physics co-Nobelist Arthur Schawlow give a plenary lecture. He brought along his ruby laser in a toy pistol housing and fired it at the projection screen. The red was so intense it appeared to me to be white in the center and it visually seemed to ‘peel off’ the screen. Unforgettable!

I think it might be worth to add this as a side note to the answer (because comments are prone to be removed at some point) that the upper part of the middle glass rod was polished intentionally on this face and the small portion of the mantle.  Otherwise on might get (erroneously) get the impression this were a cap added to the rod.  The observation about the «laser pistol» reminds me that some researchers are good to explain their point convincingly to the point, accessible even to the non specialist.

(Continuation) It is similar to Kurt Wuethrich, once asked in a lecture how they are able to assign so well flexible conformations of large proteins by solution NMR. He put his belt on the table to demonstrate some of the problems they faced.

I notice that Ed V has addressed the chemistry aspects in the comments. To address the title question about the term symbols themselves: if you have a physics / quantum mechanics background, then actually term symbols are not all that complicated.
When two sources of angular momentum (with quantum numbers $j_1$ and $j_2$) are coupled, the resulting total angular momentum quantum number has the permitted values
$$j \in \{ j_1 + j_2, j_1 + j_2 - 1, \cdots |j_1 - j_2| \}$$
and the corresponding projection quantum number is
$$m \in \{j, j-1, \cdots, -j\}$$
There are two ways of representing the eigenstates, either the "coupled" representation $|j m j_1j_2\rangle$ which uses eigenstates of $J^2, J_z, J_1^2, J_2^2$; or the "uncoupled" representation $|j_1m_1j_2m_2\rangle$ with eigenstates of $J_1^2, J_{1z}, J_2^2, J_{2z}$. All of this is described in any physics textbook (indeed, IMO, it is explained far better in physics textbooks than in chemistry ones).
The term symbols are just shorthand ways of writing the coupled eigenstates $|jmj_1j_2\rangle$, where we don't care about the value of $m$ (because the $2j+1$ states of different $m$ are degenerate in the absence of a magnetic field):
$${}^{2j_2 + 1}(j_1)_j$$
In the case of atoms / molecules $j_2$ refers to the spin angular momentum and $j_1$ to the orbital angular momentum, so it's common to label them as $s$ and $l$ respectively. Also, chemists often prefer to use capital letters, because the small letters usually refer to single electrons and atoms have many electrons:
$${}^{2S + 1}(L)_J$$
In particular, letters are used in the middle instead of numbers: $l=0$ is labelled as S, $l=1$ is P, and so on. The full sequence is SPDFGHIK....
So $^4\mathrm{F}_{3/2}$ is just refers to the "state" $S = 3/2$, $L = 3$, and $J = 3/2$. Actually, this is not a state: it is a collection of four states with $m = 3/2, 1/2, -1/2, -3/2$. Hence, "term" symbol, instead of "state" symbol.
Likewise, $^4\mathrm{I}_{11/2}$ means a collection of 12 states with $S = 3/2$, $L = 7$, and $J = 11/2$.


text:
Why is benzoic acid less soluble in water than acetanilide?
Why should acetanilide be more soluble in water than benzoic acid when benzoic acid has the same number of h-bonding sites and has an ionizable site, allowing it to form ion-dipole forces while acetanilide can't be ionized? I don't see how the dipoles would be significantly different as to make acetanilide more soluble. They are of the approximately same size so dispersion can't be a factor either.

The acid dimer is much less polar than the monomer.

You compare what is in solution, but not what happens in solid.

@poutnik are you sure that benzoic acid dimerizes in water?

@sweetandtangy I have not said that. What I have meant is, it could have been more soluble, if it had not formed dimers in solid.


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Why does rate of chemisorption decrease with increase in pressure?
My textbook says that rate of chemisorption decrease with increase in pressure.

I do found a similar statement on the web but couldn't understand this.

The rate of chemisorption depends on the flux of chemical species incident on the substrate surface. High incident flux results in high chemisorption rate. As we keep on increase the pressure the incident flux gets reduced due to low diffusion length of species and hence a decrease in the rate of chemisorption. Low pressures result in larger diffusion length of chemical species towards the substrate. As a result the flux incident on substrate surface increases which eventually results in high chemisorption rate.

Source- https://www.quora.com/Why-does-chemisorption-decrease-with-an-increase-in-pressure
Any explanation for this Statement would be highly appreciated.

And it does favour it.
Decreasing of rate of chemisorption is not decreasing of chemisorption.
Decreasing of this rate is another instance of application of the principle. A system approaching saturation reacts by a way of decreasing the rate this saturation is approached.

Can you please elaborate the last line of your answer

I understood the thermodynamics part, that the equilibrium would shift in forward direction in accordance to the principle

But can you explain how the rate is decreasing, since the pressure is increasing so number of molecules striking the surface should increase and hence the rate should also increase

There is possible confusion, if you / your textbook mean kinetic rate like dn/dA/dt, or chemisorption curve rate dn/dA/dp. A context could help.

Yes , I do think that the statement itself is not clear.

The former increases with pressure, assuming the surface has not reached equilibrium saturation yet . The latter decreases with pressure.