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Cation and anion bond
I'm new to chemistry still in a sense - I'm doing AP chemistry and the stuff we learnt in school left a lot of gaps that would have made understanding and grasping the concepts much easer if they had taught us some of the concepts you only learn if you wish to pursue chemistry further
My question is as follows and I know its such a stupid question from a AP chem student but yeah :
Is there a chemical bond or compound that can make a cation and anion of the element. I'm just going to use random elements to make my point without taking into consideration of valency and electrons for the sake of asking the question. Say we bond Ca with Br (that should be a easy bond with Br2 if I recall the properties) and now we have a cation of Ca and a anion of Br - now hypothetically we bonded another ion of Ca but the math's works out to show we have 1 Ca cation 1 Br anion and 1 Ca anion... I know that cant work out with these elements but hypothetically speaking my question is basically is there a way in a bond to have a cation and anion of the same element X as a result of the reaction with the other element Y - once again to simply what I'm asking to not appear that silly is basically 1X+ : 1X- : 1Y+ (X being the same element and Y being a different) is that possible?
Thanks - any feedback that educates me further will be appreciated ;)

Having both cations and anions of the same element is very rare.

Like the only way to form simple cations and anions with positive and negative oxidation state side by side are ionization processes in gaseous or plasmatic phase. But there is no mutual ionic bonding, like $\ce{O2+}$ + $\ce{O2-}$

The elements with positive oxidation state in both cases form complex anions with oxygen (  $\ce{Mn^2+}$ +  $\ce{MnO4-}$ or halogens ( $\ce{Cu^2+}$ + $\ce{CuCl4^2-}$ ) or negatively charged ligands ( $\ce{Fe^3+}$ + $\ce{[Fe(CN)6^3-}$ )

Alkali metals can form in very special environments alkalides. But it is not a simple ionic compound like e.g. $\ce{Na+ Na-}$. The cation must be stabilized by suitable complex agent, like 2,2,2-cryptand or 18-crown-6 ether.

We are not supposed to be mean to questioners but please bring that paragraph to your English teacher for a critique.
Your question: Can a compound  X[+2]Z[-1]X[-1] where X is an element and Z a different element exist.[the signed numbers in the brackets are ionic charges or possibly oxidation number]
You are certainly pushing the envelope for a beginner in chemistry; sort of like an infant trying to eat a tough steak without any teeth.  Advice! Talk to your teacher, learn the basics. The simple example you give Ca++Br-Ca- is a very unlikely candidate! That said, there are many ionic coordination compounds and mixed oxides that do have the metal ions in different oxidation states. Common example Fe3O4 probably involves Fe+2 and Fe+3 ions in a crystal lattice. Also the famous Prussian blue and Tunballs blue mixes of +2 and +3 iron cyanide complexes. Sorry I can not find a true + and - same element example. For Homework work out the electronic structure for ozone, O3, those O atoms are definitely not the same.  Organic chemistry is replete with carbon atoms in different oxidation states a simple example is carbon suboxide C3O2,  O=C=C=C=O  The C in the middle is definitely different.  The rules of chemistry are really relatively simple. How they are obeyed has always been beyond our wildest dreams. Dream away but learn some of the simple things too.


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I am using my phone as hotspot. What is the best charging strategy to maximize the lifespan of the battery?
This question could have been asked in Electrical Engineering SE, but I think it is more related with chemistry. It might be a Physics question too. If that is the case I will ask it over there.
Battery technology of my phone is Li-poly. I have read that Li-poly batteries are quite similar to Li-ion when it comes to maximizing the battery life.
For Li-ion batteries, it is widely recommended to keep the charge levels between 80% and 40%. I have also read that most electric car batteries do not exceed 80%. So, it seems not to be a good practice to keep the battery at 100%.
High temperature is also not good for Li-ion batteries. For that reason, fast charging is not recommended as it results in higher temperatures. I only have the stock charger and cannot adjust the charging rate. The application Ampere displays +1470mA (positive) while the phone is charging at 85%, and -300mA (negative) when unplugged at 85%. When the battery was at 99% the charging rate was around +380mA. So it might be that the charger adjusts the charging rate according to the charge level.
At 100%, while the phone is plugged in, the voltage is at 4.411V and the charging rate is around +300mA.
At 100%, while the phone is unplugged, the voltage is at 4.3V and discharging rate is around -300mA.
So, considering the battery chemistry and the information above, which charging approach is better to maximize the battery lifetime of a phone used as hotspot:

Always keep the phone plugged in at 100%?
Charge and discharge constantly, i.e charge up to 80%, then leave until 40%, then charge again and so on?  (Note that even shallow cycles cause some damage.)

What is the best method for the least amount of degeneration in battery?

Edit
When the phone is left plugged in at 100%, after some time, the app Ampere displays full. When this happens, the charging rate does not display on the screen and the voltage is stable around 4.35V.

Edit2
The links @Poutnik provided at the top of his answer do not involve more search effort (in terms of chemistry) than my question. Still they are in much better condition in terms of downvotes.
I specified that I use the phone as a hotspot to give information about the environmental conditions, e.g. how fast the power is being dissipated from the system, which would give an idea about the temperature and the number of chemical reactions happening inside. I also tried to provide the input power to a degree. Because these are what I thought would be needed for a chemist to make a decision. But the phone is a closed system to me and I cannot provide every detail a chemist would need.
If you want to say that "we are not interested in chemical reactions for specific environmental conditions, especially if it will make some benefit to people" or "we don't do science when we are not given every detail", then I think you should really question "how and for what reason" we do science.

Comments are not for extended discussion; this conversation has been moved to chat.

I'd prefer not to invoke the "effort" argument, to be honest. That's too easy to misapply, and I don't think it's an issue here at all. The real issue is that the relationship to chemistry is being stretched. There are plenty of things that chemistry can explain. There are plenty of other things that may require chemistry as the foundation upon which an explanation is constructed; however, not all of those would be valid questions here. Say, for example, explaining the symptoms of a medical disease. This is rooted in chemistry (inter-/intramolecular interactions between biomolecules) [...]

... and the beginning of an answer may start by talking about proteins, receptors, and ligands, which certainly is chemistry; but it would be a struggle for anybody to reasonably call that kind of question 'chemistry'. Simply adding "I want a chemistry explanation" to the question doesn't help, either, because it doesn't change the underlying question or the correct answer. Of course, it's not black and white: there are degrees of on- or off-topic-ness, and different people may have different opinions. But I strongly suspect the votes indicate that most people think that this is off-topic.

@orthocresol What field of science would be the most knowledgeable or credible when it comes to battery health in a specific usage pattern? I really thought only a chemist would give a through and reasonable answer to it by explaining the process and therefore clearing all ambiguous points. About medicine, it may well evolve to that in the future.

I'd suggest electrical engineering or materials science. It's not that chemistry isn't related; it's just one step removed, if you get what I mean. Just like how chemistry is one step away from physics or biology.

TLDR: The short answer would be not to use a device with a battery, as AC powered WifiTo4G hotspot, or a phone without battery or with bypassed battery charging (see below ). Or the phone kept being plugged in all the time.

There are closely related CH SE links

What is the optimal voltage for storing a 3.7V Lithium polymer battery? and
Keeping Li-Ion battery on charger vs plug the charger in and out regularly,

mentioning some chemical aspects of Li-Ion/Pol cell ageing.
But these effects are hard to quantify on basic principles, so evaluating Scylla of high charge and Charybdis of cell cycling is much easier to be based on empirical principles.
Additionally, there is A review on Lithium-ion batteries ageing mechanisms and estimations for automotive applications, that could be found by easy searching by widely used terms.
As near permanent usage of the phone as the hotspot is required ( mentioned in comments ), the activity would cause high charging/discharging rate with possible hot charging and high dynamic cell wearing. I would prefer static 100% charge here, unless there are electronic means to bypass the cell charging completely.
I recommend consideration the dedicated Wifi2Mobile AC powered hotspot devices for permanent-like usage, or some dedicated older phone.
Alternatively, there is android.stackexchange.com -
using-android-phone-without-battery or possibly similar interesting links.

Static aging at 100% ( even worse if hot and >85-90%) is reported 5x faster in some older tests than at 40%. This may not be fully true for newer cells due improved cell chemistry/arrangement, or due lovering the nominal "100%" charge level.
Dynamic aging by very shallow cycles is in context of total lifetime charge several times slower than for deep cycles. So lifetime-wise, for small cell loads, ideal would be shallow cycling around 40%. This does not apply for high cell loads if one cannot avoid continuous charging/discharging of partially charged cells.
What the cell would do with low voltage charging source depends on the charging control circuitry of the cell and possibly of the phone too. My guess is the too low voltage source would be rejected by the control circuitry/algorithms.
On my ThinkPad T470 Linux laptop, I have set charging start/stop levels, so both batteries are bypassed halfcharged on AC. I have meant keeping on charger and near end of hotspot session removing the charger so the charge - not the capacity - drops down. In some old tests, 100%/40% charged LiIon drops annually 20%/4% of capacity by static aging.

The thing is hotspot is almost always active and it is not easy to estimate when it won't be needed. Looks like I will leave it always in charge. Thanks a lot for your time.

Hm, So you use it like a nonstop device. I use frequently my phone as hotspot too, but usually only for working hours period. For like nonstop usage, consider rather some wifi2mobile standalone AC powered hotspot cell providers offer. ( I installed Linux Mint XCFE on 2 obsolete laptops for my friend )

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For a given packing (hcp, fcc, bcc), which ions occupy the corners and which occupy the faces or centers?
I recently came across the packing of atoms in unit cells. I wanted to know if there was a way to tell where the atoms in the unit cell are just by knowing the compound and its packing (hcp, fcc, etc.).
For example, if given the information that lithium iodide packs in fcc, how would I know where the lithium ions are and where the iodine ions are? Which would occupy the corners, or will both will share equally the corners and faces? Is there anything I could use to deduce where the atoms will be?
Another example, if NaCl exists in fcc where will the cations and anions be? Looking at its structure the chloride anions are on the edge centers. Can such forms be deduced or is memorizing the only way forward?

In my opinion, no techniques exist to explain why a given compound packs in fcc or bcc, or even hcp. It may even change from one form to another one if the temperature is changed

Many of the pretty pictures you find are not of primitive unit cells, so the full symmetries may not be obvious. Also, even primitive unit cells are not unique - they can be made into many different shapes and centered on different features of the atom basis. and still be primitive

The actual answer is that it doesn't matter. For many of the 1:1 solid-state structures, either the cations or the anions may be considered to be at the vertices (i.e. corners) of the unit cell. By symmetry, both representations are entirely equivalent.
To see why this is the case, it is helpful to look at a 2D analogue first. The following could be considered to be a 2D version of the cesium chloride (CsCl) structure.* The blue and orange dots represent cations and anions respectively (or the other way round; it doesn't matter, the point is that they're different things).

From the way I've drawn the unit cells (i.e. the black lines joining the dots), it appears that the blue dots should be at the vertices and the orange dot in the centre of the unit cell. It's obvious that this is a valid unit cell, as you can repeat it as many times as you wish to generate the full crystal structure.
However, I could just as easily have drawn this:

Again, this is a valid unit cell; but this time, the orange dots are at the vertices and the blue dot is in the centre.
This shows that there is no difference between the two representations of the unit cell. Both of them can be repeated (and translated) to form exactly the same crystal structure. That should already suggest to you the answer to the 3D case in your question.
Let's look at the 3D CsCl structure. You can draw it out in 3D and try to convince yourself again using the argument above. However, I think it's easier to use the 2D depiction on the left, where the numbers indicate the positions of the atoms (lattice points) in the third dimension, as a fraction of the unit cell length in that dimension. So, 0 and 1 indicates that there is a blue atom at the front and at the back, whereas 0.5 indicates the orange atom right in the middle of the unit cell.

What we need to do is to extend this unit cell a couple of times in each dimension. The first arrow shows us how to extend the unit cell in the third dimension, which has been collapsed: essentially, we need to add 1 to every number (which corresponds to adding a new unit cell behind the current one). The second arrow is just a tiling in the other two dimensions.

I'm now going to redraw the unit cell:

Now, all we need to do is to subtract 0.5 from every number (this is OK because it's just shifting the zero, i.e. moving our "point of view" backwards / forwards) and we get the same CsCl unit cell but with the atoms swapped round.
A similar exercise may convince you of the NaCl / rock salt case.†

Footnotes
* That's similar to the body-centred cubic structure, but is not the same: in the bcc structure the atom in the middle is the same as the one at the edges. Thanks to Karsten Theis for pointing that out.
† Similarly, the NaCl structure is similar to the face-centred cubic, but fcc has the same atom at all the lattice points.

This was great! Blew my mind.  I see how it works for the fcc case, depending on how it's viewed either can occupy the voids.

Technically, BCC implies the same atom on the vertices and the center of the unit cell. Statements like "lithium iodide packs in fcc" means that lithium ions form a FCC lattice and iodide atoms form a FCC lattice, shifted by half a cell edge. What you say about having a choice of where to place the origin is absolutely correct.

@KarstenTheis Indeed! I'm a bit rusty on this. I'll reword. It's CsCl and rock salt structure, IIRC. Thanks.

Yes, (I had to look it up too), the CsCl structure is primitive cubic but if you squint and tilt your head and imagine anions and cations to be the same, it reminds you of BCC. Beautiful figures, nice colors, hand drawn and still so neat.

@KarstenTheis thanks, the trick is to have an iPad, draw one unit cell, then copy-paste ;-)

Bravo for clear, concise, thorough, etc. Only slightly related (via surfaces exposed by the slicing of fcc) but anxious for even a lead towards an answer (in matter modeling SE): Straightforward formalism to get four sets of 2D hexagonal lattice vectors of fcc(111) planes that I can also cite?

Orthocresol’s answer is good at highlighting why it doesn’t make sense to distinguish between a rock-salt structure and an *anti-rock-salt structure: in the most basic cubic AB salt structures, the positions of cations and anions are symmetry-equivalent and thus either can be used to construct the unit cell and packing with the other sitting in the appropriate void.
This unfortunately changes rather quickly once you leave octahedral voids in the face and body-centred cubic structures. Trivially, this is the case with the fluorite structure ($\ce{CaF2}$ structure) in which calcium forms a face-centred cube and fluoride ions occupy all tetrahedral voids; as there are twice as many fluoride ions one cannot simply switch positions and arrive at the same structure. Instead, the opposite structure would be primitive-cubic with half of the cubes occupied. This is quite a mouthful which is why the shorthand anti-fluorite exists.
The same is true for hexagonal closest packing structures. While all closest packings have $n$ octahedral voids for $n$ atoms (and $2n$ tetrahedral voids), the octahedral voids of hcp do not form another hcp. For example in the nickel arsenide structure – hcp structure of the arsenic atoms with all octahedral voids occupied by nickel – if one were to take the nickel atoms as a lattice (as the image on Wikipedia does), the lattice turns out to be primitive hexagonal with every other central position occupied.
Things would get even more terrifying in a hexagonal structure with all tetrahedral voids occupied but gratifyingly such an $\ce{AB2}$ ionic structure does not exist in nature. The next-closest ist the wurtzit structure wherein half of the tetrahedral voids are occupied which, thankfully, forms interchangeable hcp structures. Unfortunately, other possibilities for occupying half of the tetrahedral voids exist in the patinum(II) sulfide and lead(II) oxide structures which feature square-planar platinum and square pyramidal lead atoms, respectively.
In these cases, where the different ions taken individually do not form the same crystal packing, one structure is the standard structure and the other is the anti structure. For example, in the fluorite structure of $\ce{CaF2}$, the calcium cations form an fcc substructure. Related structures in which the cation also forms the fcc substructure are known as fluorite structures. In compounds such as $\ce{Na2O}$ the positions of cations and anions are reversed: it is now the oxide anions that form the fcc substructure and the sodium cations occupying the tetrahedral voids. As this is opposite from the original fluorite, it is known as the anti-fluorite structure.
Similarly, anti-nickelarsenide, anti-platinumsulfide and anti-lead-oxide structures can be determined.
As can be seen when comparing the fluorite and the nickelarsenide structures, it is not always the same atom type that forms the lattice. In fluorite, the cations form the cubic lattice while the anions occupy the tetrahedral voids. In nickelarsenide, the more electronegative partner (although probably not strictly an anion) forms the hexagonal structure while the electropositive partner occupies the octahedral voids. So unfortunately, these have to be learnt by the student or researcher individually.
So far, this answer has barely scraped the surface. Take for example the spinel structure which contrasts with inverse spinels; notwithstanding various almost-spinels etc.

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Polarity of 1,2-dichloroethane
I don't understand how 1,2-dichloroethane is non-polar. Both the carbons have tetrahedral arrangements of atoms and I think there should be an overall dipole. Kindly explain why isn't there a net dipole?

Welcome to Chemistry SE site! When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit summary of partial answers/ideas/thoughts you have got until then. It is quite common requirement across all StackExchange sites.

I have provided my thoughts

Non-polar does not mean having zero dipole moment. Chloroform with significant dipole moment is frequently considered as non polar ( or little polar among non polar ) solvent in context of fuzzy sets of non polar and polar solvents..

If these are thorough thoughts after thorough search, I am curious how the shallow ones would look like.

Besides the comment above (non polar is more a classification of solvents etc, and might even include characteristics other than just electrical dipole moment...) consider that even a molecule with net overall zero dipole moment can locally be decomposed in fragments, and ultimately bonds, that do have electrical moment dipoles. In short, it is just a vectors sum. For the actual case, you can think of freezing the molecule in a conformation that indeed has a moment. But in reality, those are averaged out by rotation.

The dipole moment is a vector property, with arrow-like direction and absolute value.  On a local scale, within a chloromethyl group, there is a permanent dipole moment.*
You may define a global dipole moment about the molecule, too.  This then is the vector sum of all local dipole moments.
However, the two chloromethyl groups are bound together via a $\ce{C-C}$ $\sigma$ single bond.  Since the energetic barrier for a rotation is low, if the substance is in solution/melt/or gaseous state, there is continuous rotation between these two groups.  Thus, the time averaged global dipole moment of the molecule cancels out:

The situation from hot solution/melt may be well different from the one in the solid state and in the crystalline state (an ordered form of  the solid state).
*) Up to you which convention of direction of the vector you choose here.  For this question, it however is irrelevant.


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What causes plastic to become brittle and can it be softened again?
I have some old computer case parts made of ABS plastic. Two decades of storage has embrittled the plastic to the point where they easily crack or shatter with the slightest bend.
What is going on chemically that's causing the plastic to become brittle. Is there anything that can be done to reverse it?

This is an increasing problem of conservation problem, too.  Getty is just one of the collections and museums addressing this (brief overview, or spring 2014 edition about their conservation newsletters) to the non-specialists in the field.

I wonder if it's possible to impregnate plasticizers into plastic under high pressure and possible heat below the melting point of plastic.

I have had many Nalgene water bottles explode because of this brittleness.

Unfortunately, there is not too much to do. The original ABS substance is a copolymer made of acrylonitrile, butadiene and styrene. It is rather hard and brittle. To make it softer, some proportion of a plasticizer has to be added when it's hot and malleable. In the last century, the plasticizer was in general dinonylphthalate. It works well, but it is a bit volatile. So after a couple of years, the plasticizer gets evaporated, and the polymer becomes brittle again. This "illness" can only be cured by remelting the plastic and adding some new dinonylphthalate. Today better plasticizers have been discovered that are not so volatile.


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A better understanding of the screening effect and how it works?
What I am taught-
Inner shell electrons shield the outermost electrons from the attraction of nucleus. More closer the inner shell is to the nucleus, more its shielding effect. thus $n_s > n_p > n_d > n_f$ and such (in terms of screening effect as "s" orbitals are closest to the nucleus). $Z_{\text{eff}} = Z - \sigma$; where $\sigma$ is the screening constant and can be calculated via Slater's rule.
Why I think this happens -
Nucleus attracts the outermost electron just as well according to coulomb's formulae ; but the inner shell electrons provide an repulsive force towards the outermost electrons and thus "decrease" the attractive forces. But I don't get how $n_s > n_p > n_d > n_f$ is working; like s is closer to the nucleus thus is more far away from the outermost electrons thus it should provide less repulsive forces, right ? But what I think this order is due to is the shapes of these orbitals; like the p orbitals are like dumbells due to which some repulsive forces may get cancelled out (like derivation of electric dipoles and stuff); but I don't think that shows the full picture.
Adding to it; I know about the penetrating power of the orbitals ; but doesn't it just plot a graph of $4 \pi^2 \psi_r^2$ as a function of $r$ (distance from nucleus) of different orbitals and seeing the graph we can predict about the distance of the orbitals from nucleus; and we can deduct that "s" orbital stays the closest and the distance increases from there on. But it again pulls the question why then s orbital has highest screening effect, because if its closest to the nucleus it should be farthest from outer shell electrons; thus should have least screening.

As the $1s$  electrons are nearly against a nucleus with $Z$ protons, the outer electrons feelsthe attraction of $Z$ protons minus the $2$ electronic charges due to these $1s$ electrons. If these $1$s electrons were more cumbersome, and distributed around a large domain around the atom, the outer electrons will "feel" the attraction of the $Z$  protons minus something smaller than $2$

Welcome to Chemistry Stack Exchange! I've made some formatting edits to your post and added some MathJax. Please feel free to edit further.

Screening is direct application of classical electrostatics. Inner spherical charge distribution gives full screening like if the charge were in a nucleus. Deviations from spherical symmetry and reaching outer regions behind the screened electrons decrease the screening. Therefore, overlapping of 2 orbitals negatively affects screening by the more inner orbital.


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pH of PBS after adding acetic acid
Why is there a need for pH adjustment below ? Can pH go below 4.3 after adding acetic acid to PBS ?
“First prepare stock solution consisting of 50 ml of 10× PBS (pH 7.4), 12.5 ml of 80% glycerol, and 5 ml of glacial acetic acid. Adjust the pH to 4.3 with NaOH and then fill to 300 ml with distilled water. …”
Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5407559/

pH computations are not accurate. Authoritative is the fitting to the official practical pH scale defined by set of pH buffers. Unless particular solution is previously measured and precisely reproduced, it has to adjusted by measuring it pH..// Some pH buffers used for a given pH range are explicitly defined by adjusting with acid/base to the target pH. And yes, the pH is expected to go below 4.3, probably a lot.

Yes. Adding $5$ mL pure acetic acid to $50$ mL phosphate base saline (PBS) produces a pH value situated between $2$ and $3$, depending upon the exact concentrations of the phosphates of the PBS. Different mixtures are known. The buffer capacity of PBS is exceeded by the huge amount of added acetic acid.

The acetic acid/acetate concentration in the final solution will be about 290 mM. 10 x PBS could be about 100 mM phosphate, and it is diluted sixfold, so it will be less than 20 mM phosphate. At a pH of 4.3, the final solution is an acetate buffer rather than a hydrogen phosphate/dihydrogen phosphate buffer (PBS stands for phosphate buffered saline).

Thanks for the answers. The reason I am puzzled is actually if this passes the buffer capacity, why is the buffer used ? Does it have a benefit as a buffer other than providing saline at the end ?

@mete This solution is easy to make (because the stock solutions are typically already available in the lab) and it works. It is very possible that nobody checked whether there is any effect of leaving out one or the other ingredient.


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Does boron exhibit +1 oxidation state?
My teacher told that boron exhibits only a $+3$ oxidation state, but he also wrote the unbalanced reaction
$$\ce{B + C -> B4C}.$$
Carbon has greater electronegativity than boron leaving us with $\ce{\overset{+1}{B}_4\overset{-4}{C}},$ which contradicts what he taught.

B4C is a shorthand for boron carbide, which does not have a defined stoichiometry. https://en.wikipedia.org/wiki/Boron_carbide#Crystal_structure

B4Cl4 is a candidate, at least formally https://pubs.acs.org/doi/10.1021/ja01142a542  http://scripts.iucr.org/cgi-bin/paper?a00933 . If you want a -1 oxidation state there is MgB2

Wikipedia's template includes several known oxidation states for boron, including $+1$ and ranging down to $-5$.

Boron mostly favours the +3 oxidation state. However, boron does form some compounds in +1 O.S. but they are unstable and rearranges to stable molecule (such as boron monofluoride in @Oscar's  answer which polymerizes by itself to compounds containing 10-12 boron atoms).
Apart from the monofluoride, I found one example of boron(I) complex which is a radical intermediate.

The one-electron reduction of a cyclic (alkyl)(amino)carbene (CAAC)-stabilized arylborylene carbonyl complex yields a dimeric borylketyl radical anion, resulting from an intramolecular aryl migration to the CO carbon atom. Computational analyses support the existence of a $\ce{[(CAAC)B(CO)Ar]^{.-}}$ radical anion intermediate. Further reduction leads to a highly nucleophilic dianionic (boraneylidene)methanolate.

Ref.: Rang M, Fantuzzi F, Arrowsmith M, Krummenacher I, Beck E, Witte R, Matler A, Rempel A, Bischof T, Radacki K, Engels B, Braunschweig H. Reduction and Rearrangement of a Boron(I) Carbonyl Complex. Angew Chem Int Ed Engl. 2021 Feb 8;60(6):2963-2968. doi: 10.1002/anie.202014167. Epub 2020 Dec 11. PMID: 33191596; PMCID: PMC7898892. (link)

We may not have all boron atoms equivalently bonded to carbon in $\ce{B4C}$, so we might not be able to assign all boron atoms an equal oxidation state of +1. For that matter, assuming a carbon oxidation state of -4 might also be a reach.
A better candidate for boron(I) is boron monofluoride, $\ce{BF}$, which is unstable but isolable:

Boron monofluoride can be prepared by passing boron trifluoride gas at 2000 °C over a boron rod. It can be condensed at liquid nitrogen temperatures (−196 °C).1

Given the inherent instability of the $\ce{BF}$ molecule as such, the condensate would likely be a polymer.
Boron monofluoride is formally isoelectronic with molecular nitrogen and carbon monoxide, but the great polarity of the boron-fluorine bond and the different orbital sizes and energies of the two atoms prevents good pi overlap; therefore boron monofluoride is not triple-bonded:

Despite being isoelectronic to the triple-bonded species CO and N2, computational studies generally agree that the true bond order is much lower than 3. One reported computed bond order for the molecule is 1.4, compared with 2.6 for CO and 3.0 for N2.[2]

Thus boron monofluoride is much less stable than would be expected if the triple bond could have formed. For that matter, even carbon monoxide with its less decremented bond order thermodynamically favors dispropotionation (and sooting of industrial equipment with the resulting elemental carbon) upon cooling from high temperature.
Boron monofluoride is better known as a ligand in transition metal complexes, where it can indeed be considered a boron(I) species in a stable compound if it is considered a two-electron donor to a metal less electronegative than boron. For instance, the WP article mentions $\ce{(PF3)4Fe(BF)}$, citing [3]; counting the boron monofluoride (and the trifluorophosphine) as two-electron donors would be consistent with the 18-electron rule. A fully charactetized organometallic complex, with $\ce{BF}$ as a two-electron sigma donor and pi acceptor (again meeting the 18-electton ruke with two electrons from the boron), is given by Drance et al.[4], illustrated below from the abstract.

Cited References

Timms, P. L. (1972). "Low Temperature Condensation". Advances in Inorganic Chemistry and Radiochemistry. p. 143. ISBN 0-12-023614-1.

Martinie, R. J.; Bultema, J. J.; van der Wal, M. N.; Burkhart, B. J.; van der Griend, D. A. & de Kock, R. L. (2011). "Bond Order and Chemical Properties of BF, CO, and N2". Journal of Chemical Education. 88 (8): 1094–1097. Bibcode:2011JChEd..88.1094M. doi:10.1021/ed100758t.

Vidovic, Dragoslav; Aldridge, Simon (2011). "Coordination chemistry of group 13 monohalides". Chemical Science. 2 (4): 601. doi:10.1039/C0SC00508H

Drance, M. J.; Sears, J. D.; Mrse, A. M.; Moore, C. E.; Rheingold, A. L.; Neidig, M. L.; Figueroa, J. S. (2019). "Terminal Coordination of Diatomic Boron Monofluoride to Iron". Science. 363 (6432): 1203–1205. Bibcode:2019Sci...363.1203D. doi:10.1126/science.aaw6102. PMID 30872521. S2CID 78094683.


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What allows for colloidal solutions of polymers to act like solutions despite the large size of polymers?
I have been introduced to a paragraph in my textbook that I am trying to understand.

Macromolecular  colloids:  Macromolecules in  suitable
solvents form solutions in which the size of the macromolecules
may be in the colloidal range. Such systems are called
macromolecular  colloids.  These  colloids  are  quite  stable  and
resemble true solutions in many respects. Examples of naturally
occurring  macromolecules  are  starch,  cellulose,  proteins  and
enzymes; and those of man-made macromolecules are polythene,
nylon, polystyrene, synthetic rubber, etc

Upon some further research I did, 'macromolecule' is another term  used to refer to polymers as stated in the below paragraph from the same book.

The term polymer is defined as very large molecules having
high molecular mass (10^3-10^7u). These are also referred
to as macromolecules, which are formed by joining of
repeating structural units on a large scale. The repeating
structural  units  are  derived  from  some  simple  and
reactive molecules known as monomers and are linked
to  each  other  by  covalent  bonds.

As far as I have been taught, the principle factor that decides whether a particle in a dispersion medium (such as water) is a solution, colloid or suspension, is its particle size. Bigger the particle size, the less heterogeneous the mixture, and thus the more like suspensions they behave
However, for one, from a bit of research I did, polymer chain lengths range anywhere from 1 nm all the way to much beyond 1 micro meter
With that on hand, what I have seen about solutions is that the particle size ranges from 0.01 to 1 nm, usually consisting of ions and small molecules. According to my book, macromolecular colloids are relatively stable and behave like solutions. How do they remain stable and form psuedo-solutions despite their very large size?

That they aren't "very large". Chain length is only one dimension and polymers form random coils.

@Mithoron So what dimension should be considered when we have to see how soluble a particle is? Is it the area, or perhaps the volume?


text:
How does the rate equation account for solids and liquids?
The question is long because I wanted to include the whole thought process.
Given the hypothetical reaction:
$$\ce{ A(s) + B(aq) <=> C(aq) + D(aq)}$$
One would obtain the equilibrium constant:
$$K_c =\dfrac{[\ce{C}][\ce{D}]}{[\ce{B}]}$$
The "concentration" of $\ce{A}$ is excluded from the equilibrium constant because its activity is taken to be unity. As such, changing the amount of solid does not affect the position of equilibrium and does not lead to changes in the concentrations of $\ce{B}$, $\ce{C}$ and $\ce{D}$.  So far this makes sense to me.
However, I run into problems when I consider this from a kinetics perspective. Let us consider a water-filled container containing $\ce{B(aq)}$, $\ce{C(aq)}$, $\ce{D(aq)}$ and some excess $\ce{A(s)}$, and the system is at equilibrium.
The system would be in dynamic equilibrium, such that
$$\text{rate}_\text{forward}=k_\text{forward}\ce{[B]}$$
$$\text{rate}_\text{reverse}=k_\text{forward}\ce{[C][D]}$$
$$K_c=\frac{k_{f}}{k_{r}}=\frac{[\ce{C}][\ce{D}]}{[\ce{B}]}$$
Scenario (1):
If I were to now remove the excess $\ce{A(s)}$ from the container, the concentrations of $\ce{[B]}$, $\ce{[C]}$ and $\ce{[D]}$ wouldn't change, because the expression for $\text{rate}_\text{f}$ does not contain $\ce{A(s)}$ at all. Therefore, the rate of the forward reaction is not affected by the removal of $\ce{A(s)}$ and the system continues to remain in dynamic equilibrium.
Scenario (2):
However, if I set up a new water-filled container containing only $\ce{B(aq)}$, there would be no $\ce{A(s)}$ to react with, and the system would not approach the same equilibrium as above. This is despite: $\text{rate}_{f}=k_{f}\ce{[B]}$, and since $\ce{[B]}>0$, the forward reaction seems like it should proceed. Of course, this is impossible, so logically I would think that the participation of the solid should be accounted for by the pre-exponential factor.
If so, then going back to Scenario (1), would removal of $\ce{A(s)}$ not also affect the pre exponential factor for $\text{k}_f$, thereby shifting the position of equilibrium?
Essentially, how do I resolve this apparent paradox: when the system is at equilibrium, removing $\ce{A(s)}$ doesn't affect $\text{rate}_f$, but when an initial state devoid of $\ce{A(s)}$ is set up, the equation for $\text{rate}_f$ should equal 0.
In true stackexchange fashion, I have tried to resolve this issue myself. Perhaps the answer lies in some hidden way that $\ce{A(s)}$ is accounted for in the pre-exponential factor. Another way I tried to rationalise is by considering that removing $\ce{A(s)}$ from the system at equilibrium does not actually remove all the $\ce{A(s)}$ in the system, if one assumes that a small amount of dissolved $\ce{A(aq)}$ might have entered the solution. But then how would this explanation work for completely insoluble substances? And how would this explanation work if the system was gaseous instead, such that $\ce{ A(s) + B(g) <=> C(g) + D(g)}$, and $\ce{A(s)}$ is a completely non subliming substance like a chunk of metal?
*Important note: please don't answer using considerations of activities and chemical potentials. That would be missing the point. The issue I am having is when considering this problem from a kinetics perspective. Also, I would appreciate an answer considering both the mathematics and the possible conceptual errors.

Systems with more than one phase are called "heterogeneous". The kinetics that describe them can be much more complicated than the basic equations you have written, which assume homogeneous systems (though they can be applied to heterogeneous systems that meet certain criteria). Simply search for "kinetics of heterogeneous systems" and you will find more information than can be written in an answer here.

@Andrew thanks for the insightful comment. It seems I do not yet have the knowledge to sufficiently grasp the problem. However, (this may sound ignorant), it seems to me that heterogenous systems need not necessarily invoke such complications; a reaction A(s) + B(aq) -> C(aq) is adequately modeled by rate=k[B]. Using this same model equation, can it explain why rate=0 when the initial state of the system has no A(s) at all?

Why do you believe that "a reaction A(s) + B(aq) -> C(aq) is adequately modeled by rate=k[B]."? This is only true in a small subset of cases where other criteria are met, for example the surface area of the phase interface remains constant, and even in those cases, properties of A (such as surface area) are included in k such that k goes to zero when A is absent.

Thanks. I guess that is the essence of my question - yes, in that case k would go to zero, such that rate=0 when no initial A(s) is present. But this would contradict scenario (1) in my question, where removal of A(s) does not cause rate of forward rxn to drop to 0. I guess we then return to your first comment asking me to learn kinetics of heterogenous systems; I could be patient and slowly build up my learning to reach that stage. On the other hand I also see this as an opportunity for some expert here to provide an incisive answer to resolve this question's mathematics and concepts

It would not contradict scenario (1), since in scenario (1), the forward rate does indeed drop to 0. What you're missing is that such a scenario can only occur if the surface area of A also appears in kr, so both forward and reverse reactions go to 0 and the equilibrium concentrations do not change, which was explained in the answer you got to this question yesterday (see first sentence).

Thanks for the patience and pointing out my silly mistake @Andrew

Both the forward and the reverse rate are proportional to the surface area of A. Steelwool rust faster than a block of steel. Crystals grow faster with microcrystals instead of a single crystal in the mother liquor. Once you update your rate expressions to include the surface area of A in contact with the solution, it should all make more sense.

Thanks. But unless I am missing something, that would not fully resolve the issue, since in Scenario (1), after removal of A(s), the surface area should be 0 and it would seem rate(forward)=0<rate(backward), which contradicts the fact that dynamic equilibrium should not be upset. Perhaps it would help if it you could show me how surface area is accounted for in the rate equation mathematically. Thanks!

If you remove all of A, there is no longer an equilibrium, and both rates are zero.


text:
Why are MD simulations necessary for obtaining Boltzmann populations?
Given that MD simulations converge to the Boltzmann distribution $\rho \sim \exp(-\beta \epsilon)$ after sufficiently long times, and all the macroscopic quantities can be computed from the Boltzmann distribution itself, what is the need for MD simulations?
Specifically, I am asking in the context of short peptides (tripeptides or tetrapeptides).
For instance, in Beck, D. A. C.; Alonso, D. O. V.; Inoyama, D.; Daggett, V. Proc. Natl. Acad. Sci. U. S. A. 2008, 105 (34), 12259–12264, the authors use MD to generate Ramachandran distributions of pentapeptide conformations at a constant temperature. This distribution should obey statistical mechanics, and specifically the Boltzmann distribution. So in theory one should be able to write down the distributions using the Boltzmann weights as follows,
$$\rho(\{\phi_{i},\psi_{i}\}) \sim \exp(-\beta V(\{ \phi_{i},\psi_{i}\})).$$
Here, $\{ \phi_{i},\psi_{i}\}$ denotes the set of Ramachandran angle coordinates.
Why should I run MD to get the same distributions?

In the paper you reference, it is noted that the conformations do not converge and are dynamic in range. Most important and more broadly, experimentally-determined structures are static and do not represent full state space of the system.

I still have a question. Even if they get dynamic pentapeptide when they use MD to get the distributions. They can simply use boltzman weight to express that right? why should I use MD to get this result. Is boltzmann weight incorrect in predicting conformational distributions? @ToddMinehardt

In principle you can, assuming you are given $V(\{\bf{r_i}\})$ where $\{\bf {r_i} \}$ is the set of variables that define a configuration in the system - this is typically the coordinates of the atoms, but it might include things like external fields. The problem is in practice it will be at best horribly, horribly inefficient, and in fact almost certainly impossible.
To make progress let's try and define your problem at little more carefully. You have written the potential energy of the system as a function of just two variables, $\{ \phi_{i},\psi_{i}\}$. This is not the case - it is a function of the coordinates of all the atoms in the system, assuming no external fields, and so to get a function of just these two variables we are going to have to average over all configurations which contain the atoms in positions satisfying the constraint that $\{ \phi_{i},\psi_{i}\}$ are the values we desire. Thus in fact we should write
$$\rho(\{\phi_{i},\psi_{i}\}) \sim
\int d{\bf {r_1}}d{\bf {r_2}}... d{\bf {r_N}}\exp(-\beta V(\{ \phi_{i},\psi_{i}\};\{\bf{r_i}\}))$$
where the integral has limits such that the atomic configurations considered have the required angles, $\{\phi_{i},\psi_{i}\}$.
Now in principle this complicated multi-dimensional integral might be able to be solved analytically. If this is the case just calculate the value for all $\{\phi_{i},\psi_{i}\}$ you are interested in, normalise the probabilities appropriately, and you are done. In practice for any realistic forcefield you will never be able to perform this integral analytically. Thus you have to resort to numerical integration on the computer.
But now you have a secondary problem. To do this multidimensional integral on a computer naïvely is going to be impossible for anything but the smallest system. Let's say we just have 35 atoms, thus approximately 100 degrees of freedom, and we want to have just 10 points along each direction. That means we have approximately $10^{100}$ configurations to sample for each $\{\phi_{i},\psi_{i}\}$. This is impossible even on the most powerful computers we have today. And it will be impossible for a very long time.
But there's another problem, even if we could do this it will likely be extremely inaccurate to the point of uselessness. The problem is the vast, vast, vast majority of the atomic configurations will contribute almost nothing to the integral; their energy will be just too high - remember these are integrals over all space. As an example consider the case where the hydrogen bonded to a $\alpha$ carbon is stretched to almost infinity. We know this is very high energy, and therefore through the exponential will make a negligible contribution to the integral, but our above naïve sampling scheme makes no guarantee that we won't include such configurations and so pointlessly calculate their contribution to the probability, which will be zero. But worse is the flip side of this: We have absolutely no guarantee that we will actually include configurations that do have large contributions to the probability! Thus unless we have many more than 10 points along each axis most likely our result will be utterly, utterly wrong.
This is where Molecular Dynamics (or, alternatively, Monte Carlo) comes in. It is an "intelligent" way of sampling the above integral such that we only consider atomic configurations that have a sizeable contribution to the configuration integral. Thus we can calculate accurate estimates to the quantities of interest by using a relatively small sample of important configurations, and thus making the evaluation of those properties via the configuration integral viable on a computer in a reasonable amount of time and memory.
Finally note this is not specific to the case here - the same argument applies to virtually any quantity of interest, and the conclusion remains the same: In statistical physics the prime uses of Molecular Dynamics and Monte Carlo are as intelligent sampling techniques to overcome the Curse of Dimensionality


text:
Cathode vs Anode
Can someone please tell me a general, universally applicable definition for cathode and anode, that I can always use to recognize one or the other?
In a cathode ray tube, electrons are emitted from the cathode.   In a battery, electrons are absorbed into it, except when the battery is charging where it is the opposite. It gets very confusing.
What is the origin of these terms?

You don't have a universal definition. Anode and cathode are very old terms coined by Faraday and their etymology is not that helpful.
The negatively charged terminal of a battery or DC power supply is the cathode. This refers to the electrostatic sign. The particular electrode so connected to that power terminal is also the cathode e.g., in the cathode ray tube.
In electrochemisty, cathode is the always electrode where reduction is occuring.  Its electrostatic sign can be positive or negative. This applies to DC current only. Similarly, anode is the electrode where oxidation is occuring. I remember them as a(node) and o(xidation) are both vowels. C(athode) and r(eductio) don't start with vowels.


text:
The role of 2-nitrophenol in tin etchant (aqueous mixture of 2-nitrophenol and NaOH)
Aqueous mixture of 2-nitrophenol and NaOH is a time-honored etchant that dissolves tin-rich alloy and retain the intermetallic compounds from the tin. I wonder the role of 2-nitrophenol in the etching reaction and the purposes of adding 2-nitrophenol. Could anyone shed some light on this question?
Some research using such etchant
Silver, J., C. A. Mackay, and J. D. Donaldson. "A systematic investigation of the Mössbauer parameters of some intermetallic compounds and electroplated alloys of tin." Journal of Materials Science 11.5 (1976): 836-842.
Zeng, Guang, et al. "Kinetics of the polymorphic phase transformation of Cu6Sn5." Acta materialia 69 (2014): 135-148.

My intuition is 2-nitrophenol prevent intermetallic compounds from dissolution, but, how?


text:
How Does Differential Scanning Calorimetry (DSC) Differentiate Between Exothermic and Endothermic Changes?
In DSC, heat flux difference between sample and reference is measured as function of temperature while temperature of sample and reference are maintained the same. DSC is often used for polymer analysis.
What I am not clear is how does DSC differentiate between exothermic and endothermic changes like for example crystalline and melting point of polymer since in both state changes temperature of sample remains the same? In both cases, to keep both sample and reference at the same temperature, heat flux for reference is going to decrease because if not temperature of reference would become bigger than of sample since we are giving latent heat to the sample.

The sign of peak (negative vs. positive) tell us endothermic or exothermic changes. Se his graph https://www.researchgate.net/figure/Example-of-DSC-thermogram-heat-flow-against-temperature-exothermic-peak-endothermic_fig6_264589742

In the case of melting, the reference temperature goes up and up but the sample temperature remains the same.

Yes, sign of peak tells us if transition is exothermic or endothermic. However, how can we determine if process is endo or exothermic if in the case of both, reference temperature goes up?

The substance used as reference must absolutely have no exothermic or endothermic transformation when heated

@Dario, It cannot be both exothermic and endothermic with positive or negative peak if the reference temperature is going up.

@DarioMirić, If you are still confused, think about a specific example.

@M.Farooq I am not saying that it can be both exothermic and endothermic. I am saying that in either way TEMPERATURE OF REFERENCE would go up if heat flux of reference is not reduced regardless of process being exothermic or endothermic because at transition temperature, temperature of sample is constant regardless of process being exothermic or endothermic.

@M.Farooq Example is given in main question. Crystalline and melting point, first being exothermic and other endothermic. When polymer sample is heated it goes first through crystalline point and than through melting point. In both cases heat flux given to reference needs to reduce in order to keep the same temperature of sample and reference since polymer has constant temperature at phase shift and so keeping heat flux of the reference the same would heat it up above temperature of the sample regardless if process is exo or endothermic.

In DSC, differential power (heat) is provided to keep the sample and the reference at the same temperature. The DSC plot has differential power on the y-axis and temperature on the x-axis. Also, as reference is chosen in such a way that it will not undergo any phase change or state change in the chosen temperature range. The instrumentation is not as simple as one might think. There are two separate heaters which can be independently controlled. I quote from O'Reilly's Instrumental Analysis book (slightly outdated but still a pretty good reference for concepts). It describes DSC nicely,

There are two separate heating circuits, the average-heating
controller and the differential heating circuit. In the
average-temperature controller, the temperatures of the sample and
reference are measured and averaged and the heat output of the average
heater is automatically adjusted so that the average temperature of
the sample and reference increases at a linear rate. The
differential-temperature controller monitors the difference in
temperature between the sample and reference and automatically adjusts
the power to either the reference or sample chambers to keep the
temperatures equal. The temperature of the sample is put on the x-axis
(time) of a strip-chart (read "computer" today) recorder and the
difference in power supplied to the two differential heaters is
displayed on the y-axis. The power difference is calibrated in terms
of calories per unit time.

Now imagine that you are heating the polymer and the reference, their temperature is increasing but their difference is zero. You have a flat baseline, as shown in the figure (Taken from Google Images). Let us continue heating both, independently, a temperature is reached when the polymer is melting. There is a phase change and the sample temperature is not changing anymore, but it needs heat to stay at that temperature. The reference, when brought to the melting point, does not need further heating, so its temperature is at the melting point of the sample. The sample is consuming power, but the reference is not. You get a negative peak in the DSC due to an endothermic process.
Apply the same idea to an exothermic process, such as crystallization. The temperature of the sample is becoming higher than the reference. The reference needs power to catch up to the temperature of the sample. The sample heater is not consuming power but the reference's heater is, in order to be at the same temperature. This time you get a positive (exothermic) peak.

Buttonwood, I think the OP wanted clarification about the principal/instrumentation due to a misconception that power is continuously being supplied to the reference.

Thanks for detailed answer. I do agree with everything you said about melting. Reference is stopped heating while we need to heat the sample to keep the same temperature and to go through phase shift. I am not clear about crystallization since you've written that temperature of sample is increased during crystallization and that reference needs to catch up. However, what about latent heat of transition? Shouldn't temperature of the sample stay the same during phase change (crystallization) like in melting and any other phase change?

Since crystallization is exothermic we don't need to heat the sample until phase change happens, but if temperature stays the same during phase change we also don't need to and we shouldn't heat the reference to keep both at the same temperature.

Lets say, crystallization is exothermic at a temperature T. We need to heat the sample until it reaches the crystallization process. Now, the temperature of the sample cell is spontaneously increasing due to an exothermic process. It does not need heat at this moment from an external source. The reference cell must be heated now in order to bring the temperature at par with the sample.

As I said how can temperature of sample change during phase change which in this case is crystallization? When sample undergoes phase shift, its temperature stays the same until latent heat is provided. For example, When water is cooled, it turns to ice at 0°C. This phase change is also exothermic, but that doesn't mean that temperature increased. It is still 0°C until all water in the system is turned into ice (latent heat).  After that temperature can drop further.

Don't bring cooling at this moment. Let us stick to an exothermic process. The temperature must change due to an exothermic process. Don't you think so?

Well, actually I think you are correct. What confuses me is that usually in phase change, temperature is constant while in crystallization of polymers it is not as we discussed. What makes this exothermic phase change different from for example condensation which is also exothermic and in which temperature remains constant at phase change?

Temperature remains constant only for boiling and melting. Crystallization is a phase change and it is exothermic, leading to a temperature rise!

Okay, than everything has sense what you said. Thanks for clarification.


text:
Is BaSO4 (barium sulphate) soluble or insoluble in diluted HNO3 (nitric acid) and diluted HCl (hydrochloric acid)?
Here's the original question , I was asked in an exam :

Edit : the question in the picture is a single correct multiple choice question
I have concluded the the reaction of Barium chloride and Sodium sulphate will give BaSO4 as a white precipitate but I am having difficulty determining its solubility in HCl and HNO3

Have you tried searching with keywords baryum sulphate solubility dilute mineral acids ?

Yes I did see you answer to that question and your conclusion that " The BaSO4 solubility in concentrated HCl is limited, in diluted one is very low The solubility in concentrated sulphuric acid would be much higher " , but in this  scenario with HNO3 and HCl in frame , what could be the best answer to the question ( in the picture )

My answers are not all available resources... :-) // The question is a wrong question, having 2 correct answers.

I did research a lot before asking this question here , tried to find YouTube videos of the reactions  and  articles  but I was unable to find any such video on this reaction although there were some Quora threads which had contrasting answers

Rather ask yourself, why should be $\ce{BaSO4}$ soluble in dilute acids, if it is insoluble in water ? ( Be aware of rather fuzzy definition of insoluble.)

Barium+2 is toxic while BaSO4 has a low Ksp (10^-10). BaSO4 is used as a contrast agent in gastroenterology. What would be the consequence of BaSO4 entering the stomach? Look up the solubility of BaCl2 in water.

$\ce{BaSO4}$ is insoluble in dilute solutions of acids, because this dissolution would produce the following ions in solution : $\ce{Ba^{2+}, H^+}$ and $\ce{SO4^{2-}}$. And the preceding reaction has shown that $\ce{Ba^{2+}}$ and $\ce{SO4^{2-}}$ cannot exist simultaneously in solution : they must react to produce a precipitate of $\ce{BaSO4}$.
On the other hand, Baryum sulfate is known to be soluble in hot and concentrated sulfuric acid, because of the formation of $\ce{HSO4−}$ ions. But $\ce{BaSO4}$ is not even soluble in hot concentrated hydrochloric acid, according to F.-P. Treadwell, Analytical Chemistry, 1924.
As a consqquence, the answers $a)$ and $d)$ are to be chosen without any doubt.

Not quite. At pH<2 sulfate ion gets protonated. You need to develop a reaction equilibrium between solid BaSO4 and dissolved barium and sulfate ions using the Ksp of Barium sulfate and the Ka of bisulfate ion.

@Oscar Lanzi. You are right. But the problem mentions "dilute solutions" . Baryum sulfate is known to be soluble in hot and concentrated sulfuric acid, because of the formation of $\ce{HSO4^-}$ ions. But $\ce{BaSO4}$ is not even soluble in hot concentrated hydrochloric acid, according to F.-P. Treadwell, Analytical Chemistry, 1924.

Reference that in the answer. Thank you.


text:
Basic question about hybridization
I just started a basic course on chemistry and we are going really fast (at least for me).
During the last lesson our teacher left us with a question: "What is hybridization and how it relates to hund principle?"
Now I know from the slides what hybridization is and what the hund principle says, but for me they are not related. What I mean is that the hund principle state one thing and the hybridization states another thing, they are similar, but I wouldn't say that they are related.
A couple of details on what I've understood:
Hund principle: every orbital is first occupied by one electron before any one orbital is filled doubly occupied with an electron of opposite spin (like wikipedia say)
Hybridization is the mixing of a set of atomic orbital with sligtly different energy, that forms a new set of orbitals (also taken from internet/slides)
Are those definitions correct? can they be explained in a better way?
why and how they are related if they are?
PS remember that I'm a noob in this subject

Hybridization occurs in our heads.

Perhaps an example will help your understanding. Consider methane, $\ce{CH4}$, which is a common example when discussing hybridization. First, think about the electronic configuration of the central atom, carbon. It is in group 4A, which corresponds to four valence electrons. Carbon is a p-block element, which means its valence electrons are distributed between its S and P orbitals. You can write this out yourself by recalling the periodic trends you should be learning by now.

As you can see from the nifty graphic, Hund's rule tells us that despite carbon having four valence electrons, in its ground state (another way of saying its "normal state"), carbon appears to only have 2 valence electrons available for bonding. If carbon has only two electrons are available for bonding, such that only two bonds may form, how in the world do we have molecules like $\ce{CH4}$ floating around? This is where hybridization comes into play. According to wikipedia, Linus Pauling first suggested the idea that orbitals may hybridize (mix) into degenerate orbitals (orbitals with the same level of energy) so that bonding may occur. In other words, the S and P orbitals rearrange themselves in terms of equal energy (or at least near equal) so that carbon can create bonds in order to obtain a full octet of electrons.

Now it should be clear that we have the four electrons necessary to form covalent bonds like the ones we see in methane. Keep in mind, hybridization is a way to represent how atomic orbitals arrange themselves to promote bonding when the occasion rises. Hund's rule tells us how to organize electrons that end up in those orbitals, and shows us why the concept of hybridization is needed in the first place.

That second figure is just wrong. If you have atomic orbitals that are solutions to the time-independent Schrödinger equation with different eigenvalues (energies), then mixing those orbitals together does not any longer satisfy the time-independent SE. In other terms, hybrid orbitals do not have defined energies.

@Antimon I think you're missing the point of the explanation. If you have a problem with the image, you can argue with its creators (https://courses.lumenlearning.com/chemistryformajors/chapter/hybrid-atomic-orbitals/.) I don't think I made a statement about hybrid orbitals having exactly defined energies, other than that we can conceptually consider them degenerate, despite the idea of hybridization just being a tool to rationalize orbital mixing. I don't doubt your point, but it's not relevant to a student who is just beginning to learn chemistry, and only functions to obfuscate.

But why use an "explanation" if it is wrong anyway? Hybridization is about as correct as the Bohr model. We're not doing anyone a favour by leaning into outdated or oversimplified models only to have students unlearn them at a later stage (or, even worse, never). At the very least, it warrants a big "Don't actually believe this" disclaimer.

@Antimon  I suppose your issue is with many academic and educational institutions and what they teach, rather than the actual question here. I don't know how to help you. Write a letter to someone important I guess. Nevertheless, many outdated and unhelpful things are still taught in introductory chemistry. It's true, many things have to be rearranged and many ideas clarified later into a chemistry degree. I totally agree with you on this. But this is true in every field. Unfortunately we can't guarantee that every premed student taking intro chemistry class needs to learn quantum mechanics.


text:
Do Ions with less stability have less energy of ionization than those who are stable?
Problem. I've come up with the strange example of the third energy of ionization of both $\pu{Mg}$ and $\pu{Al}$, the standard logic that is to be applied on any problem of "which element has more energy of ionization", should be guided by how that element it's located on the periodic table, that way we can find it's periodic characteristics, like nuclear charge, and all of them.
My issue. The intuitive way to think about it, is that since $\pu{Al}$ is more to the right than $\pu{Mg}$ in the periodic table, it should be enough to prove that with all the periodic characteristics, Al has more energy of ionization than Mg. But what happens is all the way around actually, same thing happens with the first energy of ionization, which made me very confused.
What I tried to think to resolve this whole mess. I then thought that it has to happen due to another factor, and then I looked at the electronic configuration: $$\pu{Mg:[Ne]} \ 3s^2$$ $$\pu{Al:[Ne]} \ 3s^2\ 3p^1$$ and took a look at the ions $\pu{Mg+3 and Al+3}$, since I think that's where I have to look if we are investigating the 3rd energy of ionization, if this is wrong, please do let me know: $$\pu{Mg+3:[He]}\ 2s^2\ 2p^5$$ $$\pu{Al+3:[Ne]} $$ so then I thought that in these last ions, $\pu{Mg+3 is quite more inestable than Al+3, since Mg+3 has lost one full orbital while Al+3 has it full}$. And then that would kind of make sense intuitively? Since if one ion is more stable, it would be harder to take away one electron from it, hence the energy of ionization would be higher from the stable one, right?
Is this logic correct or is it flawed? And, if it's flawed, then how is it that Al has more energy of ionization than Mg? What should my general strategy to find which element has more energy of ionization be? Since just looking at the orientation on the periodic table doesn't seem to work always, like in this one.

Logic itself is misleading, if not based on domain specific data, facts, rules and laws. Vast majority of things is not deducible from first principles.

It would be higher because it is already charged. Your question confuses me a bit. But in principle you should be able to close thermodynamical cycles and put everything in the right order. That the latter is what you expect or something else is a different story.

"if this is wrong, please do let me know" - this is indeed wrong: to assess the third ionization energy you need to look at Al2+, not 3+. Apart from that, you are on the right track - see my answer below.

Rounded atomic ionization energy values:

Element
1st IE[eV]
2nd IE
3rd IE
4th IE

Mg
7.646
15.04
80.14
109.3

Al
5.986
18.83
28.45
120.0

Ionization_energies_of_the_elements
If energies are grouped according to orbitals:

Element
3p[eV]
3s
3s
2p
2p

Mg

7.646
15.04
80.14
109.3

Al
5.986
18.83
28.45
120.0

The respective orbital related ionization energies are greater for aluminium because of smaller respective ion sizes and greater effective ( screened ) nucleus charge.
Stability of ions has questionable meanings, as it is usually related to spontaneous changes, If one insists, it could be considered by 2 independent ways:

The energy released by capturing an electron, or all missing ones . ( A bigger one means thermodynamically less stable.
The energy needed for ionization of yet another electron.( A bigger means more TD stable.

While both values are related, the eventual stability order would depend on the chosen criteria.

When comparing the 3rd ionization energy of $\pu{Mg}$ and $\pu{Al}$, you are right to look at their electron configurations to assess their relative magnitudes. However, the ions you need to be comparing are $\pu{Mg^{2+}}$ and $\pu{Al^{2+}}$, not the three-plus ones (the first ionization energy relates to $\pu{Al}$, the second to $\pu{Al^{+}}$ and the third to $\pu{Al^{2+}}$).
When you write down the electron configurations
$$\pu{Mg^{2+}:[Ne]}$$
$$\pu{Al^{2+}:[Ne]} \ 3s^1$$
you find that you are comparing the ionization energy of a noble-gas-like ion with an ion that has noble-gas-plus-one-electron. It is well known that the noble gases are the most stable configurations possible, i.e. that they have the highest ionization energies among their neighbors (and are thus least reactive, whence the term "noble").
So, you may conclude that the third ionization energy of $\pu{Mg}$ is higher than that of $\pu{Al}$. With the same reasoning, you'll find that the fourth ionization energy of of $\pu{Mg}$ is lower than that of $\pu{Al}$, since in that case it is $\pu{Al^{3+}}$ that has the noble gas electron configuration.

The notion of relative stability of an ion is a non-sense when applied to comparing two different ions of the same element or two equally charged ions of two different elements. Production of $\ce{Mg^{3+}}$ requires a much higher energy ($7.73$ MJ/mol) than the $2.74$ MJ/mol necessary for producing $\ce{Al^{3+}}$. It does not imply that one of these ions is more or less stable than the other one.


text:
Photochemical cycloaddition of ethene
I've got a question regarding the (2pi+2pi)-photocycloaddition of ethene to form cyclobutane. If I understood correctly, the reaction doesn't take place under normal conditions, even if we increase the temperature because the entropy term of gibbs energy will favor more molecules at higher temperatures, so we need to use energy in form of UV light.
Additionally, as the molecular orbitals don't have matching symmetries, one of the ethene molecules needs to be excited by UV light, so one electron can enter the pi* orbital. So basically the HOMO of excited ethene A (pi*) and the LUMO of ethene B (also pi*)can form two sigma bonds needed for the cyclobutane, in addition to the already existing two sigma bonds, at the same time (concerted mechanism).
But why is it like this? Why can one single excited electron form (or basically induce) the formation of two sigma bonds only through orbitals (even though the LUMO is empty and the HOMO only has one electron)? And why is only one ethene excited and not two at the same time? Any tips would be of great help, thanks in advance.

Have you heard of the Woodward–Hoffmann rules? The explanation you suggest (about entropy / temperature) is a little bit superficial, to be honest; I'd be a bit skeptical of the source of that information.

Not yet, I just started taking organic chemistry 2, but I'm sure it will be covered sometime in the semester :D

OK, I'll try to answer in an appropriate way, and hopefully it will make more sense when you cover these things in the near future.

For any pericyclic reaction (including cycloadditions), the key set of rules are the Woodward–Hoffmann rules. The rules are actually quite general, but people often first learn about them in the specific context of cycloadditions*:

Cycloadditions involving $4n$ π-electrons ($n$ being an integer) proceed photochemically.
Cycloadditions involving $4n + 2$ π-electrons proceed thermally.

The [2+2] cycloaddition has four π-electrons and so falls into the first category, i.e. photochemical. On the other hand, the Diels–Alder reaction is a [4+2] cycloaddition and falls into the second category, i.e. thermal.
The full rules are way too long to explain here, so I'd suggest to look up a good textbook on it if you're interested. Clayden's Organic Chemistry is a good starting place, IMO, and if you want something more in-depth there's Fleming's Molecular Orbitals and Organic Chemistry Reactions.
Note that there is also Fukui's frontier molecular orbital theory, which in fact more closely matches the HOMO/LUMO-based analysis in your post. However, this generally sees less use than the W–H rules. You can already see that in the length of the respective Wikipedia articles. I think the Woodward–Hoffmann rules are perhaps a bit more complicated, but are more easily applied and generalised.

If I understood correctly, the reaction doesn't take place under normal conditions, even if we increase the temperature because the entropy term of gibbs energy will favor more molecules at higher temperatures, so we need to use energy in form of UV light.

I mentioned this in a comment, but this explanation is a little bit superficial. The key consideration isn't entropy, because if that were the case, then no cycloadditions should ever happen, not even thermal ones like the Diels–Alder reaction. (Basically, you can't selectively apply this argument to [2+2] cycloadditions and then suddenly forget it when moving to [4+2] cycloadditions.)

Why can one single excited electron form (or basically induce) the formation of two sigma bonds only through orbitals (even though the LUMO is empty and the HOMO only has one electron)?

The part of the system that reacts isn't just the single excited electron (that would just be a plain old radical reaction). Pericyclic reactions are quite special: they involve the cyclic movement of quite a number of electrons. So, in fact, it is not the single excited electron that reacts: it is the four electrons, three unexcited and one excited, which collectively react.
So, the followup question is: why does the "3 unexcited + 1 excited" configuration react (I'll abbreviate that to 3u1e), whereas the "4 unexcited + 0 excited" configuration (4u0e) doesn't react?
To truly understand this requires one to delve into MO theory. In particular, you need to understand molecular symmetry and electronic states: it's not the individual electron's states which matter (i.e. which electron is in which orbital†) but rather the total state of the molecule (loosely speaking, the product of all the orbitals). But I will try to give a flavour of the explanation here.
The first part of this explanation is given in the form of correlation diagrams, and you may likely find it in the textbooks mentioned above. Essentially, what it says is this: the electronic state of the reactants must "map to", or "transform into", the electronic state of the products.
There are many possible states for the reactants: for example, we have 4u0e, 3u1e, 2u2e, ..., and likewise, the electronic states for the product are similar: we have 4u0e, 3u1e, ...‡ It turns out that, based on symmetry arguments, the 4u0e reactant configuration maps to the 2u2e product configuration, which is very much uphill in terms of energy.
On the other hand, the 3u1e reactant configuration maps to the 3u1e product configuration. Both are high in energy compared to the ground state, but relatively speaking, they are fairly close, so the activation energy for the forward reaction is not too high, and the reaction can proceed. The tradeoff is that you need to start from an excited state of the reactant, and that's where the light comes in.
If you want to go even deeper, then there is this question from a while ago which I answered: Theoretical basis behind orbital correlation diagram for pericyclic reactions. It has diagrams, symmetry labels, etc. and also introduces the non-crossing rule, which doesn't change the conclusion but makes the discussion more technically accurate.

And why is only one ethene excited and not two at the same time?

The simple answer is that you don't need to excite two of them at the same time. Basically, the combination of 3 unexcited + 1 excited is sufficient for the reaction to proceed; you don't need to go to 2 unexcited + 2 excited. (Indeed, the reaction might not even be possible with that combination, for similar reasons as why the 4 unexcited + 0 excited combination doesn't work; I'm not sure, though.)
A secondary answer is that only a fraction of the molecules are excited at any one point in time. Getting two of the excited molecules to collide and react before they return to the ground state, is not likely.

Footnotes
* Technically speaking: cycloadditions where both components react in a suprafacial sense. But let's not go there (for now). Refer to the textbooks above for more info.
† Electrons are indistinguishable, so it's not possible to say which electron is in which orbital. But that's more of a technicality in the present context.
‡ These can be enumerated and labelled more precisely using symmetry labels.


text:
How do I find the volume of a gas when the pressure and temperature changes, given the original volume?
I am a sophomore in high school, just started honors chemistry this year, and I'm having trouble figuring out what formula to use for this question, and how exactly to do so:
If a gas has an original volume of 6 L, an original temperature of 298.15 K, and an original pressure of 1 atm, what would the volume be at a temperature of 288.15 K and a pressure of .92 atm?
So far I thought that it would probably be best to use the formula, pV=nRT , but I have not been able to figure out what exactly R is. I know that it is the constant, but what is the value and/or how is it calculated?  I have also been asked to use this formula in other questions, and I don't quite understand the value of R. If someone could help work me through this, that would be wonderful.

See this for more detail: https://chemistry.stackexchange.com/a/151451/79678.

$R$ is the universal gas constant, the value of which you can find e.g. here. For this problem, you don't need to know the value, though.
$$ p V = n R T \mathrm{\ \ \ or\ \ \ \ } R = \frac{p V}{n T}$$
is true for any ideal gas, and a pretty good approximation for most gases at room temperature or higher and atmospheric pressure or lower. So for two different samples 1 and 2 that behave like ideal gases, you can write:
$$ \frac{p_1 V_1}{n_1 T_1} = \frac{p_2 V_2}{n_2 T_2}$$
In your case, $n_1 = n_2$. You can solve for $V_2$ and plug in the numbers to get the answer.

Oh, okay! Thank you so much, I wasn't even aware I was using the incorrect formula. Appreciate the help!

Just to be clear - the formula is not incorrect, it is just not in a mouth-sized portion for consumption yet. @GraciousBean


text:
Naming of the chemical elements and original sentences by the discoverers
Does anyone recall a printed or an online compilation where the names of all the chemical elements are listed along with the original sentence, from a research paper, containing the name of the element? The unabridged Oxford English Dictionary is great resource for name origins but it obviously does not mention the very first non-English paper.
Take for example krypton, the etymology in the OED is "Etymology: < Greek κρυπτόν, neuter of κρυπτός hidden, concealed. Discovered by Ramsay in 1898." Such information is available everywhere. What was the original paper by discoverer when he wrote about it this new element? Many discoverers often gave some interesting logic behind their choice.
Another example is of Die Zerlegung des Didyms in seine Elemente by Carl Welsbach, in Monatshefte für Chemie 1885, 6 (1), 477–491.

Da sonach die exacte Zerlegung des Didyms in mehrere Elemente
realisirt ist, so schlage ich vor, die Bezeichnung Didym nunmehr ganz
zu streichen und beantrage, für das erste Element, entsprechendder
Grïnfärbung seinerSalze und seiner Abstammung die Benennung: Praseodym
mit dem Zeichen Pr und für das zweite, als das „neue Didym", die
Benennung: Neodym mit dem Zeichen Nd.
Since the exact decomposition of didymium into several
elements has been realized, I propose that the designation didymium
now be deleted altogether and that the first element, corresponding to
the green coloration of its salts and its origin, be designated:
Praseodymium with the symbol Pr and for the second, as the "new
didym", the name: Neodymium with the symbol Nd.

This website https://www.rsc.org/periodic-table/element/36/krypton#history, has a small history section for each element but I am looking for quotes (German French, English, etc.) like the above from the original papers that contain the proposed names of the elements. I tried searching but could not find a single compilation.

Does anyone recall a resource like that?

Where can one find the first official of modern elements (Z > 92), which state the name of the elements? Does IUPAC make that information public?

Side info: Note that there are very interesting podcasts chemistry in its element and Episodic table, with websites https://episodictable.com/ and chemistryworld - chemistry-in-its-element

I'll look into this actually because this is pretty interesting. The two resources I would go to would the the Merck Index and also the Chemical Rubber Company Handbook for Chemistry and Physics. It has some pretty cool explainations into symbolic representation orgins and the traceback to russian/greek.https://en.wikipedia.org/wiki/CRC_Handbook_of_Chemistry_and_Physics

The mere list of chemical element name etymologies probably will not satisfy you.

Reviewing the old elements up to #104, I recall «Discovery of the elements» by Mary Elvira Weeks/Henry W. Leicester published by the Journal of Chemical Education [sic!].  Pick up your library card on archive.org for a 1 h screening borrow here (7th edition, 1967).  While each element is presented in its own portrait, the chapters are grouped by topic, e.g. the elements known to the ancient world (Au, Ag, Cu, Fe, Pb, Sn, Hg, S), the halogens (except At), the noble gases, natural radioactive elements (Ra, Po, U, Rn, Pa, Ac, Th), discoveries by X-ray spectrum analysis (Hf, Re), or modern alchemy (Fr, Tc, Pm, ...).  The one hour offered to you there won't suffice for the 900+ pages compiled.  Includes cross references to primary literature, too.

If you don't have this much time, look out for The Merck Index.  Though the subtitle reads An Encyclopedia of Chemicals, Drugs and Biologicals, and contains many more organic than inorganic chemicals, you equally find a brief characterization of the elements including some of properties and (if not known for too long) some references about its discovery. For example $\ce{_{99}Es}$:

credit:  The Merck Index, 13th edition, Merck & Co Whitehouse Station NJ, 2001, p. 3562, e.g. on archive.org.  Available for a borrow for up to 14 days, link to the page after using your library card.
Because of the handy (sub)structure search, some groups/institutions favour the electronic access of this resource on https://www.rsc.org/Merck-Index/reference (currently 15th edition, 2013).

Thanks Buttonwood. I was not aware of Merck Index listing the origins, I will certainly check. I had a copy of Elvira's book. Pretty nice. I had actually used for my Journal of Chemical Education article on re-discovering iodine.


text:
Mixing hydrogen peroxide and sodium hydroxide doesn't lead to sodium peroxide anymore
I've been using same bottle of hydrogen peroxide and same bottle of 50 % of sodium hydroxide to make sodium peroxide for a month now. You could see the presence of sodium peroxide in the solution because it turned yellow, but it won't turn yellow now and I don't know why, but hydrogen peroxide is not efficient oxidizer for my current use. Where is the problem?

If the result is suddenly different the conditions are suddenly different. We should ask you what is different and not vice versa. Perhaps slow decay of H2O2, or absorption of CO2 by hydroxide.

Poutnik is right, what likely happened is one or both of the reagents lost its potency through chemical reactions to which they are susceptible.
In the case of the sodium hydroxide, if it reacted with carbon dioxide from the air (it is best stored and handled under a CO2-free atmosphere), the resulting carbonate is no longer a strong enough base to react with the hydrogen peroxide (this is an acid-base reaction, not a redox one). If the sodium hydroxide is/was as strong as 50%, you may be able to see a sign of this contamination: sodium carbonate is not as water-soluble and might be seen as cloudiness or a white precipitate.
If the hydrogen peroxide decomposed, that is that as far as forming sodium peroxide goes. Decomposition of hydrogen peroxide, unlike carbonation of sodium hydroxide, cannot be prevented by using an inert atmosphere, and it has no visible effect on the solution until, as you describe, it's called on to perform. If there was no apparent outward change in your reagents, I would therefore suspect hydrogen peroxide decomposition first.


text:
Where can I find any assay of common baking soda (for example Arm and Hammer or generic grocery store brands)?
I am looking for any assay of common baking soda, listing approximate percentages of impurities or trace elements for example potassium, boron, chloride, fluoride, etc.
For example my naive guess would be it's 98 percent sodium bicarbonate, with the remaining 2 percent consisting of potassium, chloride, aluminum, calcium, and so forth.  What are the percentages of each of these.
I just want some order of magnitude assay.
Not requiring super accuracy nor large sample size.
I didn't find results on the web.
Please direct me to any assay that would give me such estimates.

Sodium bicarbonate (also called sodium hydrogenocarbonate) is produced with a rather high degree of purity. Whatever its quality, the purity is always greater than $99.5$ %. The main impurities are calcium and chloride, which are both smaller than $0.01$%, after Merck catalog.

The only relevant source of such info is what the vendor ( or in some cases independent agency ) provides for the product. Baking purpose may generally involve improving additives.


text:
How strongly affected is the buffering capacity of sodium carbonate - sodium bicarbonate by temperature changes?
I would like to do a lab regarding the affects of temperature on the buffering capacity of sodium carbonate - sodium bicarbonate in particular for sulfuric acid. Currently I have two competing theories,
a) according the la chatliers principle the increasing heat should increase the reactivity of the acid thus reducing the effective buffering capacity,
b) according to Van 't Hoff equation the acid dissociation constant should be reduced, increasing the effective buffering capacity.
can anyone offer further insight or resources regarding this, theoretical or experimental?
A lot of my information especially point B came from this post What's the effect of temperature on buffering capacity? I don't believe this is a duplicate but rather an extension which wouldn't be suited for that stream.

Temperature has effect on the $\ce{HCO3-}$ dissociation contant and therefore the pH of buffer. The buffer capacity temperature dependence is negligible and indirect.


text:
What is the ∆H_vap when temperature of a given liquid is equal to the boiling point of the liquid?
What is the $∆H_{vap}$ when temperature of a given liquid is equal to the boiling point of the liquid?
For the other cases ie (i) T>BP, $∆H_{vap}$ > 0 (ii) T<BP, $∆H_{vap}$ < 0
So what will be the $∆H_{vap}$ when T (Temperature) is exactly equal to BP of the liquid?
Because we know that heat has to be an input to convert from liquid phase to gaseous phase so can the the $∆H_{vap}$ = 0 ? If so, how is it possible since we are inputting heat into the container.

$\Delta H_\mathrm{vap} \gt 0$ is valid regardless of the relation $T$ vs $T_\mathrm{vap}$, as it is an endothermic process, requiring energy to break intermolecular bonds. Its value drifts with T, depending on the relation of the specific/molar heat capacity of the liquid and the vapour, following the Hess's Law.
You probably confuse $\Delta H_\mathrm{vap}$ which is always positive and $\Delta G_\mathrm{vap} = \Delta H_\mathrm{vap} - T \cdot \Delta S_\mathrm{vap}$, which is zero for the phase equilibrium for given $T$ or $p$ ( sitting on the (l)-(g) line of the phase diagram.)
If $T \lt T_\mathrm{vap}$ for the given $p$ then $\Delta G_\mathrm{vap} \gt 0$
If $T \gt T_\mathrm{vap}$ for the given $p$ then $\Delta G_\mathrm{vap} \lt 0$


text:
Why total stereoisomers is 1 even if no geometrical isomers and optical-isomers are present
Let's for example take the complex  structure type $\ce{MA_6}$, What will be the Total number of Stereo-isomers (=[Geometrical-isomers]+[optical-isomers]).
My effort was that There will be no G.I and O.I present in this octahedral structure. So there should be 0 (zero) stereoisomers.
But the correct answer is that total stereoisomers are 1 (one), but the total G.I and O.I are same (i.e 0).
I want to ask the reason for this.

The equations you use assume $\ce{MA6}$ as the first form to join the atoms in a molecule/a complex as the first isomer to count.  Consider this first form of spatial arrangement as a reference.
Depending on the atoms at disposition, their number, their chemistry (in case of ligands, for example, their denticity) there may be additional possibilities to arrange all these atoms in a molecule/a complex.  In a pairwise comparison of the first form, these may be e.g., an enantiomer, or a constitutional isomer.  So you add them up and report the total.


text:
Calculating the electrode potential from thermochemical data (part 2)
This question is in continuation of this question.
I am taking an example of $\ce{Cu^2+}$ in aqueous medium.
$$\begin{align}
\ce{Cu (s)    &-> Cu (g)}         & \Delta_\text{a}H^\circ    &= \pu{339 kJ mol-1}  \\
\ce{Cu (g)    &-> Cu+ (g) + e-}   & \Delta_\text{i}H_1^\circ  &= \pu{745 kJ mol-1}  \\
\ce{Cu+ (g)   &-> Cu^2+ (g) + e-} & \Delta_\text{i}H_2^\circ  &= \pu{1960 kJ mol-1} \\
\ce{Cu^2+ (g) &-> Cu^2+ (aq)}     & \Delta_\text{hyd}H^\circ  &= \pu{-2121 kJ mol-1}
\end{align}$$
Net reaction:
$$\begin{align}
\ce{Cu (s)    &-> Cu^2+ (aq) + 2e-} & \Delta_\text{r}H^\circ    &= \pu{923 kJ mol-1}
\end{align}$$
From Lange's Handbook of Chemistry- 15E, at page 6.91:
$$\begin{align}
\Delta_\text{r}S^\circ &= S^\circ(\ce{Cu^2+ (aq)}) - S^\circ(\ce{Cu (s))} \\
                       &= (\pu{-98.4 J K-1 mol-1}) - (\pu{33.15 J K-1 mol-1}) \\
                       &= \pu{-131.55 J K-1 mol-1}
\end{align}$$
and hence
$$\begin{align}
\Delta_\text{r}G^\circ &= \Delta_\text{r}H^\circ - T\Delta_\text{r}S^\circ \\
                       &= \pu{923 kJ mol-1} - (\pu{298.15 K})(\pu{-131.55 J K-1 mol-1}) \\
                       &= \pu{962.2 kJ mol-1} \\
E^\circ &= -\frac{\Delta_\text{r}G^\circ}{nF} \\
        &= -\frac{\pu{962.2 kJ mol-1}}{(2)(\pu{96485 C mol-1})} \\
        &= \pu{-4.986 V}
\end{align}$$
But the value of $E^\circ(\ce{Cu|Cu^2+})$ is $\pu{-0.34 V}$.
Where did the calculations go wrong?

I'd strongly suggest looking at my edit for typography. In particular, it's far easier to enclose the number with the unit inside \pu{...} i.e. use \pu{2 mol} not 2~\pu{mol} (the latter is correct but a bit more tedious). And you should really make use of the align environment for this sort of thing, it makes things look far better. See this or this. Lastly, don't leave out units!!

Ok... But why plimsoll replaced by circle?

No particular reason, to be honest; I guess I just retyped the equations and I'm more used to using the circle so ended up using that "by default". Both symbols are acceptable according to the IUPAC Green Book. Feel free to change that back if you wish.

I'm going to drop the standard state symbol throughout this answer for ease of notation.
Your value for $\Delta_\mathrm{r} H$ is way too large. Specifically, if you want to do it this way (by counting ionisation energies, etc.), then you must do the same for the reference electrode i.e. the standard hydrogen electrode (SHE), because $E^\circ$ is not an absolute value but is rather measured against the SHE. So, you should calculate the full $\Delta_\mathrm{r}H$ for the reaction
$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$
and in this, you'll find that there are extra terms corresponding to the solvation of protons, ionisation energy of hydrogen, and the $\ce{H-H}$ bond formation which you've neglected in your analysis.
Instead, a more straightforward way to calculate $\Delta_\mathrm{r}H$ is to use formation enthalpies:
$$\Delta_\mathrm{r}H = \Delta_\mathrm{f}H(\ce{Cu^2+(aq)}) - \Delta_\mathrm{f}H(\ce{Cu(s)})$$
The second of these terms is obviously zero. Looking up the first term in Table 2.8 of the "Resource Section" in Atkins's Physical Chemistry (9th ed.) I get a value of $\pu{+64.77 kJ/mol}$. Plugging this in:
$$\begin{align}
E^\circ &= -\frac{\Delta_\text{r}H - T\Delta_\text{r}S}{nF} \\
        &= -\frac{\pu{64.77 kJ mol-1} - (\pu{298.15 K})(\pu{-131.55 J K-1 mol-1})}{(2)(\pu{96485 C mol-1})} \\
        &= \pu{-0.539 V}
\end{align}$$
which is not quite perfect, but at least it's much closer.
The reason why this value works is because the enthalpies of formation for ions are already normalised against $\ce{H+}$: that is to say, by definition, $\Delta_\mathrm{f}H(\ce{H+ (aq)}) = 0$, and so the enthalpy for the SHE half-reaction is zero. (The electrons can be safely ignored, because they would cancel out on both sides of the equation.) In Atkins's own words (p 71, Section 2.8):

The standard enthalpy of formation of ions in solution poses a special problem because it is impossible to prepare a solution of cations alone or of anions alone. This problem is solved by defining one ion, conventionally the hydrogen ion, to have zero standard enthalpy of formation at all temperatures:
$$\Delta_\mathrm{f}H^\circ(\ce{H+, aq}) = 0$$
Thus, if the enthalpy of formation of $\ce{HBr (aq)}$ is found to be $\pu{-122 kJ mol−1}$, then the whole of that value is ascribed to the formation of $\ce{Br-(aq)}$, and we write $\Delta_\mathrm{f}H^\circ(\ce{Br-, aq}) = \pu{-122 kJ mol-1}$. That value may then be combined with, for instance, the enthalpy formation of $\ce{AgBr(aq)}$ to determine the value of $\Delta_\mathrm{f}H^\circ(\ce{Ag+, aq})$, and so on. In essence, this definition adjusts the actual values of the enthalpies of formation of ions by a fixed amount, which is chosen so that the standard value for one of them, $\ce{H+(aq)}$, has the value zero.

It's worth pointing out that the standard molar entropies that you quoted are also already adjusted in this way. That is to say, the standard molar entropy of $\ce{H+}$ is already set to $0$, by definition.

Suppose, I want to calculate it by first method: (i) Enthalpy for reverse of solvation of proton= +1150, multiply by 2 gives= +2300 kJ/mol; (ii) reverse of ionisation energy of H atom= -1312, multiply by 2 gives= -2624 kJ/mol; (iii) reverse of bond enthalpy of H2 molecule= -436 kJ/mol; (iv) addition of these gives= -760 kJ/mol; (v) Now adding my DH=+923 kJ/mol, I get= +163 kJ/mol against your +64.77 kJ/mol?

I'm not really sure how much agreement we should expect to get in the final number, to be honest. When you add up many terms on the order of 1000 kJ/mol, even if each of them have an error of 1% i.e. 10 kJ/mol, the final result may easily end up being not perfect. Generally, I'm thinking more in terms of broad strokes here: your original calculation and the correct one differs by a factor of 9 or so. I'd say that even +163 is a much better improvement over the original value of +923.

I'd expect the solvation enthalpies to be the most inaccurate of all the terms in the sum.

I am getting -1.05 V, which is not at all in agreement with accepted value of -0.34 V. I think, I missed something.

Well, how do you judge what "in agreement" is? How close must it be before you decide it agrees enough?

I don't think that there is any accepted value of %error which works for every scientific calculation. But consider this; the error is more than 200%, which is huge!

Hence my point about the numbers that you're subtracting, in my previous comment. If $X$ and $Y$ are both large numbers of similar magnitude, then $X - Y$ is a small number. Even if the relative error in $X$ and $Y$ is small, the absolute error is still gonna be large compared to their difference, which translates into a large relative error for $X - Y$.

I think, you missed the $S^o (\ce{H2 (g)})=\pu{130.68 J K-1 mol-1}$; you will get exact -0.34 V. In my case, now I am getting -0.85 V.

Aha, indeed! Good catch.


text:
Why doesn't NaCl work for electrolysis?
So I found out online that to increase the conductivity of water for electrolysis, NaCl or NaOH is used.
However, upon using a low amount of NaCl in water and passing 12V through the solution, I noticed something weird, the cathode was bubbling a lot more than the anode (as expected in electrolysis - Hydrogen)  but after 20 mins, there was no collection of gas on the anode. Why could this be?
When repeating the experiment with NaOH, the desired result was obtained with hydrogen and oxygen on the cathode and anode respectively.

What was it that you expected, really?

Important info is missing: Electrode materials!

@Ivan Neretin I expect to see oxygen on the anode, which was not happening

@M. Farooq I used stainless steel

Steel could be reacting with chlorine. Did you see it corroded?

Yes the anode was a bit corroded

After running the setup for sometime with NaCl, the solution had turned a slight yellow

Oxygen? Why not chlorine?

At the cathode, $\ce{H2O}$ is decomposed according to $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ So some $\ce{H2}$ gas is produced, and the solution becomes basic. At the anode, two reactions occur simultaneously and independently : $$\ce{2 Cl- -> Cl2 + 2e-}$$ $$\ce{Fe -> Fe^{2+} + 2 e-}$$ So the anode is corroded. Now the chlorine $\ce{Cl2}$ produced by the first reaction is a gas relatively soluble in water, which reacts with the ions $\ce{Fe^{2+}}$ produced by the second reaction, according to :$$\ce{2 Fe^{2+} + Cl2 -> 2 Fe^{3+} + 2 Cl^-}$$ The ion $\ce{Fe^{3+}}$ is usually hydrated in aqueous solution taking a orange-brown color due to the ion $\ce{[Fe(OH)]^{2+}}$ produced by the reaction $$\ce{Fe^{3+} + 2 H2O -> [Fe(OH)]^{2+} + H3O^+}$$ And the solution becomes brownish and acidic around the anode