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Determination of metal state from protein X-ray structure
Many proteins or enzymes contain metal ions within their active sites.
These metals are often directly involved in the reaction catalysed bu the enzyme. In certain cases the function is well known and understood, including the identity of the metal and the changes in oxidation state throughout the reaction.
I wonder how the metal and oxidation state is determined a priori when a new structure is obtained.  For examples if I solved a structure of a previously unknown enzyme by crystallography I may be able to determine that it contains Zinc (iron a better example), but knowing the density map tells me nothing about the oxidation state which may be important to function.
I think it must be possible to determine this, however I do not find much reliable information (inorganic complexes typically have other techniques in addition to X-ray). The situation is made harder as many protin structures in the pdb appear to be assigned without any real study.

No way there's Zn other then (II) in a protein - Fe would be a better example.

Well, if your crystal is good enough, you know that the cell should be neutral and then you can basically count the charges. If you have multiple metal ions in there it becomes more difficult.

X-ray diffraction experiments, regardless if about small molecules or proteins / RNA / DNA, are usually one element characterizing a sample.  On occasion, you may infer the oxidation state from the transition metals environment: how many closest neighbors, their distance, their number, their spatial arrangement.
You are free to complement findings by any other technique either using the same crystal sample as mounted on the diffractometer (e.g., UV-Vis), or with sample material from the same batch characterized at the same temperature as the crystal mounted on the diffractometer: spin characterization by EPR, Moessbauer, NMR (your are not constrained to CHN) and SQUID  to mention a selection. Because PubMed includes records about the structure characterization of large biomolecules, this database may serve as a source of inspiration how this was done, e.g., for heme


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What is the definition of thermodynamic stability?
The term thermodynamic stability is used on this site, but I can't find a good definition. Is is a quantitative or a qualitative concept? Does it apply to a single compound, or a pair of compounds? Does it a relative or an absolute value?
An answer to another question, Thermodynamic stability of benzene derivatives , states

Thermodynamic stability of compounds can be determined by obviously enthalpy of formation ($\Delta H_{_\mathrm f}$) of individual compounds. The enthalpy of formation will be lesser if the compound is formed from its constituent elements enjoys some greater stability.

The enthalpy of formation considers the synthesis of a given compound from the elements. If the enthalpy of formation of two substances like bromomethane and chloromethane are different, it might be due to differences in bond strengths (or electronic states) in the two compounds, or differences in bond strengths (or electronic states) comparing elemental bromine and elemental chlorine.
I do understand the concept of thermodynamically favored product. In this case, there are competing reactions leading to different products. Because they start with the same reactants, there is no issue about the reference state. However,  if I compare two substance with different composition (i.e. not isomers), I don't know what it means to compare their thermodynamic stability. I also understand local and global minima, as discussed in this answer which references a single set of atoms that undergoes transformation. The problem arises when comparing compounds with different sets of atoms.
Thermodynamics usually is pretty well defined. Thermodynamic stability, however, is a bit of a mystery to me.

The lack of a clear cut definition (i.e. proposed by those well-known bodies) clearly shows that you are not alone. I personally think that 'stability' in any form only makes sense as a relative concept and that is why I am always very aggravated by the homework assignments to sort carbocations (et.al.) by their stability. As an absolute property: any well in a energy surface is thermodynamically stable, like a molecule that survives plenty of vibrations... but well, that's just my opinion.

Oh well, that linked question and answer really gut my blood pumping... who is teaching this stuff?!

Take it as an energy content. Of course in ideal case it would be better to have a "normalised stability", which in some cases is possible. For instance benzene is more stable than cyclohexane as per pi electrons. I agree that comparing different things isn't really necessary, but I have no examples in mind in which we do this comparison other than in a vague way. When it comes to quantities, we analyse reactions or transformations in which elemental composition is globally the same.

@Martin-マーチン at least the first answer isn't that bad. What one can do with that is a different story, but it is not the fault of the answer. If it would have been me, I would have answered all three questions with something having more didactic value. Perhaps I will do it.

Bumps on energy surface are the very example for kinetic stability (opposed to thermodynamic stability).

I think that it should be abandoned as an idea unless it is restricted to a system at equilibrium where there can be no further change. Thermodynamically speaking diamond is unstable in air,  but this is clearly nonsense in any practical sense. There are undoubtably many species with larger $\Delta G$ change i.e. in thermodynamical terms more stable that are fast in comparison so would appear less stable.

@porphyrin but this is unrelated to the question, it just bring in the issue of kinetics. But OP doesn't seem concerned by this aspect, but about the quantity H of formation itself.

Especially to OP: https://chemistry.stackexchange.com/questions/18898/why-is-the-standard-enthalpy-of-formation-of-elements-in-their-native-forms-zero. It might be what you are looking for.

@Alchimista I don't know what answer you mean. At the same time I do not wish to fall further down that rabbit hole; I see absolutely zero value in discussing anything in relation to these types of exercises. In the best case scenario they are just playful oversimplifications, in a more serious way they mask actual chemistry, in the worst way they teach some rule of thumb that works only for these systems and would be completely incorrect for others. I see absolutely no way in making anything of didactic value out of this, but I am looking forward to being proven wrong.

@Martin-マーチン Don't mix things. You are right for that the way exercises are proposed (or understood) is often wrong but this does not mean that any series of carbocations has an order of stability, and for that there is quantity call enthalpy of formation. It is simple as that.

Answer and comments you got seems to negate that enthalpy is a state function.

@Alchimista And there we are: falling down the rabbit hole. I really have no idea where I am mixing things, but that is of no concern here. If you want to make the argument that enthalpy of formation is akin to stability, then you are well within your right of interpreting data. You might even side with the majority or 'common sense' here. For me there is just no evidence that this interpretation is valid or even sound. If it is 'simple as that', then I really do not understand why IUPAC hasn't formalised it as that. But again, I'm digging myself out of that rabbit hole now…

@Martin-マーチン Don't mix things: I do have the impression that commonly administered exercises (or the way they are intended by students) are indeed non sense, and that this deviate your judgement in this specific discussion. I only agree that IUPAC isn't exceptionally clear perhaps, but as far I can read it does the same but just in the correct free energy space. Point is that independently of reactions, one can compare molecules for their energy content. All is needed is a "scale", and H, as it is constructed to be a state function, is. I will really try to write an answer to the above....

.... question body. The title somehow ask for something else and the IUPAC definition actually suffice. The body need a more articulated one. We will see. But one of my main motivation is provided by answers and comments here that seems to swap the question proposing something that it is true but only shifts the problem. Nice to have such discussions (although comments isn't the best place. I apologise).

The full phrase should be thermodynamic stability with respect to ____, where the dash indicates a process, or a chemical reaction.
A mixture of hydrogen and oxygen is thermodynamically unstable with respect to water formation.
Similarly, a diamond is not forever (which may not please De Beers and ladies). It is thermodynamically unstable with respect to conversion to graphite.
Also, thermodynamic stability is a relative term which is often contrasted with reactivity or kinetic stability.
Diamond is kinetically stable at room temperature for the same process (lucky ladies can smile again).

This is as to say "with respect to its formation". I upvoted but it does not say anything that is not already contemplated. Basically OP questions about the meaning of assigning 0 enthalpy of formation to elements in their standard state, and specifically to the fact that Cl2 and Br2 (or anything which is not a single atom) can have different energy stored in. Also, it is nice to recall the kinetic aspect, but that isn't really related to the question. Personally I cannot answer if not that enthalpy of formation must be used for what it is. This answer is correct but doesn't really answers.

We shouldn't restrict to formation only. As I said, it could be any chemical reaction or process. We can say Mn(III) is thermodynamically unstable with respect to disproportionation to Mn(II) and Mn(IV).

Yes fine. But it does not answer the question, for which I also do not have a clear and coincise answer different than what I said in my first comment. That is, we don't really (or should not) assign any meaning to thermodynamical stability different than what it is: standard enthalpy (or G) of formation of the given compound. That can be used of course to close other cycles, as you said.

The modern definition of thermodynamic stability is the state of maximum entropy.
Some background information is necessary to make sense of this.  I hope you will find the following helpful!
Phenomenologically, thermodynamic stability is the absence of visible change. This is the 'original' definition, employed by experimentalists during the 18th and 19th centuries.  If repeated observations of your system - such as measurements of its temperature, pressure, density, colour, etc - don't indicate any change, you can tentatively regard it as stable.
Why tentatively?  Because as you've alluded to, some changes can be tortuously slow, so unstable systems can appear stable because the intervals between each observation are fleeting by comparison with the system's rate of change.  Such states are called 'metastable'.
The existence of metastable states severely limits the scope of this observational-based definition of stability.  A more fundamental definition, that can distinguish between truly stable and merely metastable states is clearly desirable.  This alternative, quantitative approach, involves measuring the energy changes that accompany different chemical reactions.  This is a tricky process, because the differnet forms of energy transfer accompanying any reaction can be are numerous: heat (thermal conduction); work (exertion of a force or pressure); current (transfer of charge across an electrical potential); to name the most common ones.  The basis of this approach is that chemical compounds store energy in their bonds, so by tabulating the energy changes associated with many different reactions, their capacities for storing energy can be calculated.
But remember!  The defining property of energy is that it is conserved!  A table of bond energies such as that described above cannot by itself function as an indicator of stability.  A final step is needed, which is to identify a particular form of energy that is minimized by all chemical reactions, and which will therefore be amenable to the kind of 'potential well' analysis described in the  linked answer by Thomij.  This form, commonly called 'Gibbs energy', is the energy associated with a change in entropy.  It's the increase of entropy that is the true driver of spontaneous processes.  Accordingly, the condition of maximum stability for a chemical system is defined by the maximization of its entropy.
What is entropy?
Entropy is a measure of how the energy in a system is distributed among it's constituent particles.  More statistically probable distributions have higher entropy.  The most probable distribution has the highest entropy.  Entropy is often described as a measure of disorder, although I personally find this exposition misleading. Entropy is a subtle and unnerving concept, which whole books have been written about, and which I've taken literally years to make peace with.  The best text I can recommend is 'The Second Law' by Henry Bent.
What do you mean by 'the energy associated with an increase of entropy'?
Transfers of energy always accompany some other change, such as an increase in volume, or a flow of current, or a transfer of mass.  Indeed, the 'forms' of energy familiar from high school are defined by the nature of their accompanying change (e.g. 'work of expansion' energy is that accompanying the exertion of a pressure, and thereby an increase in volume).  An increase in entropy is simply another (admittedly more obscure) example of this.  But you can think of entropy as a property somewhat analogous to volume, in the sense of being a feature of a system that can be changed by the application of a particular form of energy.  In the case of entropy, the corresponding 'form' of energy is heat, rather than work.  As alluded to here, there is a very close connection between temperature and entropy; in fact, a definition of temperature is the limiting ratio between the heat supplied to a system and the change in entropy that results.

“The modern definition of thermodynamic stability is the state of maximum entropy” Modern? You mean 100+ years old? Thermodynamics as a concept didn’t exist in 18th century, so it makes absolutely zero sense to talk about “old definition”

This is not a constructive criticism now is it Greg.  Firstly, it is perfectly reasonable to use the term modern to describe something that occurred more recently than something else, even if it is itself 100+ years old.  By nature, 'modern' is a relative term.  Secondly, while it's true that the laws of thermodynamics hadn't been discovered in the 18th century, the fundamental notions like heat, work and equilibrium had been recognized, and even put to use in the construction of steam engines.  The concept of stability (the subject of the question) is certainly at least this old.

People knew about materials from the beginning of time, so does heat, yet talking about fire as an element that converts metals to gold or phlogiston theory doesn't say much about chemical processes or their stability points. The understanding of energy-like statefunctions, their role in stability criteria is the very beginning of what you call "modern thermodynamics" (nonstandard use). Also, the distinction between thermodynamic and kinetic stability, which is the core of the question can only be interpreted with "modern thermodynamics". I am sorry if it came out harsh.

Who said anything about fire as an element or phlogiston theory?  I don't see what purpose your continued sniping serves.  As for modern thermodynamics' distinction between thermodynamic and kinetic stability, I made this the core of my answer; the bit you completely ignored.

@Greg not to enter in your discussion but the distinction between thrrnodynamic and kinetic stability is not at the core of the question. It has just been inadvertently introduced into the discussion confusing the thread. I am thinking of writing an answer not really for the question itself but to fix some wrong concepts brought in the thread. Most seems to negate the existence of a state function they speak about, and its use.

Yes a lot of the concepts introduced in the question-comments section are besides the point.  Stability is the state of maximum entropy (equivalent to minimum Gibbs energy).  That's the quantitative definition.  I'm really surprised that neither of those variables even appear in the comments.

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Removing Heavy Rust from Very Old Carbon Steel Carving Knife on the Cheap
I am trying to remove heavy rust from a carbon steel 10-inch carving knife. Mothers Polish did not get rid of much of the rust. I turned to vinegar soaking and scrubbing: At first it took off a good bit of the rust leaving visible residue in the bottom of a glass carafe. With a lot rust left I turned to vinegar and salt (see below). I have been soaking the knife using this solution in the 1-quart glass carafe for three or four days. Every day I pull out the knife and scrub it vigorously with a wire brush. Lots of black stuff comes off on a paper towel but there is still a lot of stubborn rust. When I take the knife out of the carafe the knife is blacker then when I put it in. After a vigorous wiping an alligator-like pattern remains on both sides.
I am using the same solution daily without changing it save for topping off the carafe. Also there seems to more chemical activity going on at the top of carafe than at the bottom, for some reason: is there an explanation of the phenomena?
If the rust is indeed being dissolved slowly I am wondering if the solution is loosing potency? Is the stoichiometry changing as the rust is being removed? Would changing the solution daily/ frequently help speed this process? Would frequent stirring help uniform the gradient? Is there a better method using household products?
I do not want to resort to expensive OTC cleaners or high-end sandpapers. Oh! FYI: pun intended-I am many decades rusty in Organic and Physical Chemistry.
Thanks for any thoughts.
Source of the idea: Hunker....Fill a large container made of a non-reactive material, such as plastic or glass, with 1 gallon of distilled white vinegar. Add 1 cup of table salt and stir the mixture to combine. The salt will increase the potency of the acid in the vinegar, making it even more effective. Use 1/2 gallon of vinegar with 1/2 cup of salt for smaller objects

Here's an idea: just ditch the rusty thing and buy new. BTW if you'd try too hard you could dissolve also metal - all in all, not worth the effort.

You can't reasonably remove rust by dissolving it; mechanically remove it with wire brush , sand blast, etc. Then "covert" the small remaining amount with phosphoric acid containing solution such as Navel Jelly. Presumable converting it to iron phosphate. Coat the surface with something like oil as it can oxidize again fairly quickly.

You are better off to use a bucket (or any other container which does not conduct electricity, like you already do), and to fill it with some of luke warm water, add kitchen salt (needn't be saturated).  Vinegar alone can not reverse the reaction of rust in iron, but if you use a charge carrier plus a metal less noble than iron, you may remove quite some of the rust.
Find some Al foil which will serve as galvanic/sacrifical anode, scratch it with a metal brush, then wrap the iron in question with this foil and put both into the water such that both metals are in contact with each other and both Al and iron equally are in nice contact with the water/salt solution, too.  It may take hours, it may take more than one «bath»; while there is no guarantee that all rust is removed this way, at least a significant portion is going to be removed with much less manual labor. During this process, slowly, the foil is eaten away.

First, read my answer here where I cite the electrochemistry background. The latter include source links to galvanic corrosion and, in particular, in the current instance, a classic case of pit corrosion.
Stop long-term soaking, as this in the presence of salts and an H+ together with dissolved oxygen will just further promote the slowly evolving electrochemistry.
I, personally, ignored stake-knives left in a sink (water and oxygen exposure) that were used to cut a mango. The sticky mango juice, rich in Ascorbic acid (Vitamin C) is apparently particularly notorious for promoting pit corrosion.
Never have completely removed the stains, so do not have high expectation here,

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Does formal charge affect bond polarity?
Bond polarity, as far as I understand, is a measure of the degree  to which shared electron density is distorted, and thus solely depends on the electronegativity difference.
Up until now, I had learnt that the dipole moment (charge on each atom x separation) is used as a measure of the polarity. However, for molecules with a formal charge like CO, even though the shared electron density is distorted towards the more electronegative atom, the dipole moment points in the opposite direction. For other molecules like Ozone, even though there is approximately no distortion of bonding electron density, the formal charges alone result in a dipole moment.
For a general molecule, am I supposed to take into account the formal charges for bond and thus molecule polarity, and it it even a measure of degree of distortion/ionic character at that point?

it all depends on whether you are talking about the polarity of a bond or the molecule

At the end of the day, my major use of bond polarity is to find the overall molecular polarity. Could you elaborate on why there’s a difference?

polarity is a vector quantity: it is not a simple addition but a vector addition.
For example, for the molecule of $ \ce {CO_2} $, each bond $ \ce {C-O} $ is polarized but as the molecule is linear, the moments of the 2 bonds are compensated: the bonds are polarized but the molecule does not 'is not

I am aware that polyatomic molecule polarity depends on bond polarity and molecular geometry. my question is regarding formal charge and has nothing to do with thus

If there is formal charge, there is a corresponding vector. Just sum it to all the other el. dip. moments due to elemental electronegativity etc. Of course it does. The extent of the effect depends on the weight of the limiting form with charge separation, in a resonance frane.

https://chemistry.stackexchange.com/questions/30797/how-can-the-dipole-moment-of-carbon-monoxide-be-rationalised-by-molecular-orbita https://chemistry.stackexchange.com/questions/35603/why-is-co-practically-nonpolar

Formal charges are a simple model of bonding used to count electrons. They don't affect anything. They are also a poor model for the character of real bonds or real electron distributions which need more complex models.

@matt_black I disagree. A formal charge represents the charge on each atom if the bonds were perfectly covalent. The electronegativity difference can then be used to get a rough idea of the actual charges on each atom. The negative formal charge of C in CO causes it to be partially negative despite the lower electronegativity. They clearly affect the charge on the atoms

@OVERWOOTCH You just described a model that helps us to explain the real electron distribution, which is basically what I said. What do you disagree about?

@matt_black You said that "they don't affect anything" while in this model, they DO appear to affect the charges on each atom. Maybe I misunderstood you.

@OVERWOOTCH I think my (subtle) language is about the difference between models (they explain things) and reality (where some complicated quantum stuff causes the real electron distribution).

When you compute formal charges you split bonding electrons evenly between bonded atoms. This does not account for differences in electronegativity. It is similar to the difference between formal charge and oxidation state, as well explained in the Wikipedia.
A better picture (closer to the real electronic distribution) is, as Pauling would probably suggest, intermediate to the two, assuming a hybrid between possible Lewis structures (covalent and ionic).
In any case, when considering bond polarity, electronegativity overrules formal charge.    On the other hand inductive effects should not be dismissed.
The OP presented an excellent counterexample: CO. Here the carbon has a very slight negative charge relative to oxygen, as predicted based on the computation of formal charges.
I am not an expert on the development and application of electronegativity scales, but suffice it to say that these scales are (1) empirical and (2) provide useful guidelines. The original Pauling formulation of electronegativity cannot account for all possible bonding schemes. Case in point is the modification of the electronegativity scale to account for orbital hybridization, the following data via the Wikipedia:

Hybridization
χ (Pauling)1

C(sp3)
2.3

C(sp2)
2.6

C(sp)
3.1

'generic' C
2.5

Based on this modified scale the electronegativity of oxygen, 3.44, is no longer much larger than that of the carbon atom in carbon monoxide.
References

Data found in Wikipedia, original reference: Fleming, Ian (2009). Molecular orbitals and organic chemical reactions (Student ed.). Chichester, West Sussex, U.K.: Wiley.

What I’m trying to ask if I should take into account the formal charges in addition to the electronegativity when finding the bond polarity. Is the triple bond in CO polarised towards carbon or oxygen? The carbon appears to have a partial negative charge despite the lower electronegativity

See eg https://en.wikipedia.org/wiki/Inductive_effect

@OVERWOOTCH You're right, CO is a strange beast.

I do not see why CO should be called counterexample. It is consistent with your answer, at least it seems so to me. OP you appear to dissect bonds. You can of course think about how the bond are polarized in a structure with formal charges. But the resulting polarity will be determined also by the separated charges.

@Alchimista the question as I understand it is whether you should incorporate information from formal charge computations into an estimate of bond polarity, as opposed to relying just on the relative electronegativities of the atoms involved in the bond. In this case electronegativity would seem to predict $C^{\delta+}O^{\delta-}$ whereas formal charges predict $\delta-\ce{:C#O:} \delta+$

@BuckThorn yes that is clear. But your answer seems to me 1) correct 2) it does include right the CO case. One note to both of you, electronegativity won't be that as tabulated in a periodic table. This is as to say that =O and --O^- aren't the same.

Thanks for your encouraging comments @Alchimista. With it in mind I made a modification to my answer.

@BuckThorn right. In my comment to OP under the question I wanted to mentioned right C as faced from different directions. I skipped because I was afraid to makes thing unclear, in a comment. The main point is that whatever analized the el   dip moment would be the sum or the various components we - correctly or not - search for.


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Comparison of boiling points of para-derivatives of phenol
The boiling points of four para-derivatives of phenol are:

How can we compare the boiling points of p-nitrophenol and p-aminophenol?
Apart from the hydroxyl group, p-nitrophenol has two oxygen atoms to form inter-molecular H-bonding. Whereas, p-aminophenol has a nitrogen atom and two hydrogen atoms to form inter-molecular H-bonding. Also, the molecular mass of p-nitrophenol is more than that of p-aminophenol.

what kind of bonds break during the change of state?

@Nicolas As far as boiling point is concerned, our interest is limited to various inter-molecular forces between molecules, which will reduce in phase conversion from liquid to gas.

And what kind of intermolecular bonding are possible in p-aminophenol but which is not possible in p-nitrophenol?

Whatever consideration, note that the difference is just 5 °C.

@Nicolas I can't think of any. Out of these two, which one has higher dipole moment?

Such comparison hardly makes sense - apples vs oranges.


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How do we know orbitals of atoms and elements exist?
How would we know if we have never experimentally seen them?

They don't. They are mere figments of our imagination which just happen to come in handy.

Also, we've seen them experimentally (for example: https://chemistry.stackexchange.com/questions/38605/why-is-it-possible-to-image-lumo-if-these-orbitals-are-by-definition-unoccupie). How so, would you ask? A contradiction? Well, quantum stuff is weird.

@Ivan Neretin I don't understand whether you are bung sarcastic or serious

Then you still underappreciate the weirdness of quantum mechanics. Who said I can't be both?

Nobody says that orbitals exist. But calculations made with them explain correctly the real chemistry like the shapes of molecules.

@Maurice At least (although I am not authoritative at all :) I am not saying that. Perhaps that applies to orbital as reserved to wavefunctions, certainly not to the quantities can be derived from. The linked paper on pentacene speaks for all.

Its just for our understanding ..its a tool


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Reaction pathway - Extra molecule used in one step only?
I am currently building a reaction mechanism through Gaussian calculations. I successfully found each TS and intermediates through the different jobs (opt, freq, irc) for my 3-step mechanism from reactants to my products.
However, I was questioning myself about one step in particular which could involve one extra molecule. Here's a scheme of my system:
$$\ce{A + B -> C <=> D -> E}$$
Where A and B are in same equivalence experimentally.
All steps are characterized by stable intermediates and reasonable energy barriers confirming experimental data, except one step. The step from C to D is a simple tautomerism from enol to ketone. However, this proton exchange needs to be driven by a proton acceptor like B to have a energy barrier that is closer to the reality. Then, for my system, an extra molecule B could be implied in the mechanism only from this C to D step.
Practically, you perform optimization of your TS and complex of products and reactants involving all molecules for the reaction step. However, I was wondering if I can draw a reaction pathway from A and B to E without the extra molecule B and involving it only for the step where it is required (as a sort of catalyst). Let's make it more clear:

I take the step A and B to C as already calculated
I perform C to D with an extra molecule B acting as proton acceptor for that step
I take the step from D to E

In this way, the extra B is used only in the second step of the mechanism and I can get from there the relative energies. However, in that case, how can I draw the mechanism plot with all relative energies ? C or C + B have totally different energies. Am I forced to re optimize all my system with an extra B molecule ?

Are these A, B, C, D and E substances corresponding to real molecules ? Or are they here simply for the fun of writing kinetics formula ? Is it possible to imagine that for example$\ce{A = C2HR, B = H2O, C = RCHCHOH, D = RCH2-CHO}$ and $\ce{E}$ is something else ? So, why thinking that a second B is necessary ?

It is as example instead of providing a real system. The other B would act as a sort of catalyst for only one step of the mechanism and would not be involved in the other steps (except in the first one but as reactant).


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Combined Effect of Steric Inhibition of Resonance and Inductive Effect
While qualitatively comparing the acidic nature of o-methyl benzoic acid and benzoic acid, we consider the steric effect due to which the $\ce{-COOH}$ group loses its planarity and the +M effect of $\ce{-CH3}$ isn't considered any longer. But still, the +I of $\ce{CH3}$ is functioning, which increases the electron density of the compound and reduces its acidity. Thus o-methyl benzoic acid should be less acidic than benzoic acid. But while confirming my results with the experimental data, I found that $\ce{pK_a}$ of o-methyl benzoic acid is 3.90 while that of benzoic acid is 4.20. How can this be explained using the basic concepts of electronic displacement effects?

+I is a very weak effect compared to loss of coplanarity due to sterics.

the grouping $ \ce {-CH_3} $ is too far for its + I effect to be felt on the acidity


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Interatomic spacing vs electric conductivity
I have a theory which may be wrong but I will write it down.If 2 elements at the same group (in solids)have the same crystal structure then the one with the biggest interatomic spacing has better electric conductivity.I don't know if anyone else has found this but I will give you 2 examples:
The crystal structure of diamond,silicon and germanium is face-centered cubic.We know that the order in conductivity of Ge>Si>Diam.
The order in interatomic spacing is Diam<Si<Ge.
Is this a coincidence or not?

What about the next member of the series: $\alpha$ - tin? It is a semiconductor with the same structure as diamond, silicon and germanium.

What group are you talking about? A family of chemical elements from the periodic table, a space group of a crystal or something else?

If they have the same structure, then (generally speaking) the one with the biggest interatomic spacing is also the one with the greatest atomic number, and hence behaves more metal-like. So yes, it is no coincidence; it is a genuine relationship, albeit trivial.

@andselisk same group of the periodic element


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Why does dissolved oxygen concentration decrease with increasing atomic number of metals?
I am looking at how metal ions ($\ce{Fe^2+}$, $\ce{Ni^2+}$, $\ce{Cu^2+}$, $\ce{Zn^2+}$,  and $\ce{Ag+}$) affect the biochemical oxygen demand in water and used the Winkler's method to find out the oxygen concentration of each solution. I graphed this with oxygen concentration ($y$ axis) and increasing atomic number ($x$ axis) and found that the order: $\ce{Fe^2+}$, $\ce{Ni^2+}$, $\ce{Cu^2+}$, $\ce{Zn^2+}$,  and $\ce{Ag+}$ had decreasing value of oxygen concentration (almost a decreasing exponential graph like an "l" shape rather than a straight negative linear line) but I do not really know the specific chemistry behind it.
I thought with an increase in atomic number, there will be an decrease in dissolved oxygen concentration because as the atomic number increases, the proton number increases. The greater nuclear charge attracts the water molecules more and a ligand can be formed. Thus, as the atomic number increases, the metal ion will form a ligand with water molecules more easily, and the dissolved oxygen concentration will decrease as more water molecules will form ligand with the metal ion.
But I am confused because with increasing atomic number, the atomic size gets smaller and wont that make it harder to form a ligand bond with water?

What is your sample matrix i.e., what else do you have in the sample besides metal ions?

It is Winkler's method not Wrinkler of oxygen determination.

@M. Farooq : I think OP should elaborate his method. This seems a modification to the Winkler's method.

@M.Farooq Thank you so much for your reply i only have distilled water in the sample besides metal ions as a controlled variable - first data point in the graph (atomic number = 8). For the method, the metal ions were in solution as a metal nitrate and distilled water was added. the wrinkler's method was then used and sample was titrated w sodium thiosulfate. The ratio of the steps in wrinkler's method was used to find the oxygen concentration. I am a bit confused here. How is this a modification of the wrinkler's method? and how would that affect my data?

Does your comparison make sense ? You present a curve going from $x = 13$ to $x = 47$. But this curve is based on the only points where $x = 26, 28, 29, 30$. The point at x = 47 (Silver $\ce{Ag+}$) may not be part of the same line as the corresponding ion is singly charged, and the four previous points are doubly charged ions. Si to be sure that the dependance does exist as you imagine, I would recommend to repeat this measurements with maybe Magnesium ($ x = 12$), Calcium ($x = 40$), Cadmium ($x = 48$), Baryum ($ x = 56$), and forget about silver.

@Maurice Thank you, I will try with Magnesium. How does a singly charged ion compared to a doubly charged ion affect its ability to form a ligand with water molecules?


text:
Rigorous but terse introduction to molecular orbital theory
Is there a terse but rigorous introduction to molecular orbital theory (MOT)? I'm looking for the 'Rudin's Real Analysis' of MOT, if that analogy resonates with anyone. If not, I'm looking for a book that I can use as a reference rather than a gentle tutorial.

What is your level in math, physical chemistry  and group theory? I can explain e. g. crystallography one a note card if you are familiar with the 3D Fourier transform and the convolution theorem.

If you want to throw Schrodinger's equation, symmetry groups, or Fourier analysis at me, I'm cool with that. What do you have in mind for a book?

A good answer would depend on how deep you want to go. Maybe Szabo and Ostlund would work?

In my opinion, it is impossible to be more terse than Szabo and Ostlund. I'm not too sure whether it covers MOT to a rigorous extend. I think it's more for the nitty gritty solving that thing part of it.

Rigorous and terse are often mutually exclusive. The first few chapters of Albright, Burdet and Whangbo give a nice quick intro that's semi-quantitative, but hardly rigorous. Maybe that in combination with Szabo and Ostlund will give you a good start?

Is there even a clear cut definition of what is considered "canonical" molecular orbital theory ? I mean, there exists qualitative Hückel MO theory and then quantitative Hartree Fock MO theory. Should the reference cover both or only quantitative methods ? And then there is stuff "in-between" like frontier mo theory, ligand field mo theory etc. I don't think there exists a book that covers all of those in a terse manner.

From commenters' interpretation of "rigorous but terse" it would appear that the current answer would be "no". Perhaps you want to edit your answer. Also, in case you've not seen it there is a guide to textbooks in the meta section: https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry


text:
HCN, or my nose says I'm dead
As you see, I still post...
I studied chemisty decades ago (and when I was through with all the compulsory practica, I was sent to a computer chemistry workgroup and politely suggested to never touch a beaker again, which I happily followed and made my PhD there, otherwise probably the chemistry dept. of the university of Hamburg would no longer exist :-)
Here's one of my specialities as a walking disaster area. Hmmm, "Trennungsgang" seems to be nonexistent in English, make that Qualitative Inorganic Analysis. Don't ask me to remember the details, but Fe3+ is detected with NCS-. For another cation later you need H2O2. I never learnt to use a different test tube. The result were always the same - a geysir shooting out of the test tube and over my fingers (now that you say it, gloves exist) accompanied with some almond stink that suggests I just got a deadly dose of cyanide.
I'm remarkably hard to kill but methinks I rather misinterpreted the perceptions. So, maybe it only was less toxic HSCN that blew off? (I think the H2O2 solution was acidic.) BTW, Wiki says "odor pungent".

HCN as a poison is greatly overrated.

Related: How do people know HCN smells like almonds?

You don't get HCN from oxidation of SCN. Your reaction was mostly catalytic decomp. of H2O2.

You can get also the cyanide odor from wilting leaves of stone fruits like peaches and cherries.

From the Toxic Substance Portal of ATSDR, "The [almond] odor of hydrogen cyanide is detectable at 2-10 ppm (OSHA PEL = 10 ppm), but does not provide adequate warning of hazardous concentrations. Perception of the odor is a genetic trait (20% to 40% of the general population cannot detect hydrogen cyanide)." However, from that same source, the NIOSH IDLH (immediately dangerous to life or health) is 50 ppm, so if your among the ~70% who can smell $\ce{HCN}$, you could note an odor and have a safety factor of ~10 before immediate death occurs. Of course, the smell is less noticeable as the sense of smell fatigues, and long term effects are another question.
As to whether you smelled thiocyanic acid, $\ce{HSCN}$, its odor is described as "pungent", not at all like bitter almonds, so likely that was not what you smelled. Thiocyanates are about 100 times less toxic than similar cyanides, and are even used internally in medicine.
The toxicity of $\ce{HCN}$ is roughly comparable to that of $\ce{H2S}$, which is frequently encountered in the lab and in nature, such as at geothermal features, but since most people can sense $\ce{H2S}$ at 0.03 ppm, it is treated far  more casually.
But $\ce{HCN}$ is toxic enough that the old joke still applies. Question: "What's the difference between cyanide and sayonara?" Answer: "Sayonara means goodbye in Japanese, and cyanide means goodbye in any language."

In the main body of the query, the OP is asking about HSCN. The title is a typo.

@M.Farooq, reread the question: he is unsure if it was HCN or the relatively nontoxic thiocyanate "So, maybe it only was less toxic HSCN that blew off?". However, I'll add that to the answer.


text:
Is there a specific historical figure or era that is most responsible for our current definitions of "physical change" and "chemical change"?
I understand that there is a very blurred distinction between the idea of a physical and a chemical change at advanced levels of chemistry, so I'm referring mainly to the definitions provided in most introductory texts and websites:

physical change - processes that change the form of a substance and not its chemical composition.
chemical change - processes in which new substances are created through changes to a substance's chemical composition.

The closest thing I found to an answer is the formation of Dalton's atomic theory in the early 19th century, which defines a chemical reaction as a rearrangement of atoms in the reactant and product compounds. I am hoping to confirm that this is the earliest definition closest to our modern perception of physical and chemical change, and that I am not missing anybody important.

Why "blurred"? The distinction is very clear also in advanced modern natural science, and based on distinct changes in the electronic structure of materials (excluding phase transitions). Physicists can also handle a test tube, chemists do a lot of physics.

@Karl Shame on you that new user gets it better ;)  @ OP. It's not "blurred at advanced levels of chemistry" but non-existent. Also there's hardly any such "current definition". Write the source of what you cite and the one who wrote the source has made the "definition".

Consider also the SE site History of Science and Mathematics.


text:
Does partially/completely removing solid shift the equilibrium?
For the reaction: $$\ce{P4(s) + 5O2(g) <=> P4O10(s)}$$
Let's say the reaction is at equilibrium, and SOME $\ce{P4O10(s)}$ is removed from the reaction. Will that affect equilibrium? My assumption is no, since $\ce{P4O10(s)}$ is not included in the equilibrium expression.
However, for me, it gets confusing when let's say $\ce{P4O10(s)}$ is COMPLETELY removed. In this case, I would assume that the reaction shifts to the products (since no products are present). I'm not sure if this is right.
Help is appreciated. Thank you!

There isn't much of an equilibrium here; the reaction is shifted pretty much all the way to the right anyway. That being said, your reasoning is correct.

@IvanNeretin That makes sense. Thank you. I'm just confused because I thought solids/pure liquids do not effect the equilibrium?

If there's no equilibrium then there's no effect on equilibrium. Find a valid example.

Hmm but in my example, there was an equilibrium. It's just that I'm saying what happens in  P4O10 is completely removed?

I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate.
If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer dissolving because there aren't any. Less obviously, there is no crystallization process anymore (there are no crystals that solvated ions could attach to). So this is not a dynamic equilibrium, i.e. the reaction or process no longer is observed at all.
This is different from a system out of equilibrium where one direction of the process is faster than the other direction. In that case, the system will approach equilibrium, and the process (in either direction) will continue to happen when equilibrium is reached (which is why it is called a dynamic equilibrium).
You might ask how we can ever get solid sodium chloride again. If the solution is sufficiently supersaturated, nuclei will form (in solution, or on the container or on some impurity), and then crystal growth will be observed again.

What do you mean by "it is not longer an equilibrium?" If it isn't, wouldn't a change follow?

@BuckThorn Sorry, sloppy language. Either we have a reaction or process where both forward and reverse reactions are happening. This can either be at equilibrium or approaching equilibrium. Or, like in this case, neither forward nor reverse reactions are happening.

Ok, I see your point. @theorist made the additional good point that you have to keep things constant, e.g. pressure, to speak of equilibtrium when you remove the salt.

Nothing would happen. The reaction would not shift to produce more product.
Let's suppose you allow this reaction to equilibrate in a sealed container.   Equilibrate means you've reached the equilibrium value for $p_{\text{O}_2}$ at that temperature.  Now further suppose you were able to remove all the product without changing the free volume (say you slightly decreased the volume in the container by an amount equal to the volume of the solid product you removed).  The system is still at equilibrium ($p_{\text{O}_2}$ is unchanged).  Thus nothing will happen—no new product would be formed.
If you're not convinced, imagine that you had some fluctuation in the equilibrium such that a bit of product was produced.  In that case, $p_{\text{O}_2}$ would increase above its equilibrium value, and thus the product would need to disappear (the reaction would need to shift back to the left) for the system to go back to equilibrium.

I agree with "nothing would happen" but think it is important to say that something was happening before (dynamic equilibrium), and now the underlying molecular processes (forward and reverse reaction) are no longer in place.

@KarstenTheis  That's a v. good point—that while nothing changes macroscopically, there are changes microscopically. But while you wouldn't have fwd and rev rxns as they would normally occur in an equil sys, I wouldn't go so far as to say that the fwd and rev rxns no longer take place, since there would still be fwd and rev rxns due to fluctuations about equil (as describred in my last paragraph).  So this raises a very interesting ques: How would the nature and magnitude of the fluctuations about equil differ between the original equil sys (where product was present) and this "one-sided" sys?

I would agree with your last paragraph for a homogeneous equilibrium. If the reaction involves forming a solid (as in a precipitation reaction) and the solution is not saturated, no solid is formed. On the microscopic level, there might be some ion pairs formed, or even some transient nuclei but I don't think they rise to the concept of solid.

It might help to think of the system as containing three phases: $\ce{P4(s)}$, $\ce{O2(g)}$ and  $\ce{P4O10(s)}$. As long as the pressure, temperature and compositions of each phase remain constant (and P,T and chemical potentials across phases are in equilibrium i.e. equal), we can alter their extents arbitrarily without altering the condition for equilibrium to be sustained.
An example is freezing of water. When water is cooled to $\pu{0^\circ C}$ at 1 atm pressure it can remain at equilibrium without forming ice (it will not freeze), provided no energy is removed (as heat) from the liquid phase. Otherwise ice will begin to form. If you remove an ice cube from a glass of water at  $\pu{0^\circ C}$ more ice does not spontaneously form, provided T and p are not altered (say by providing a means for the water in the glass to lose heat to the surroundings, which would of course describe a nonequilibrium situation).
This is admittedly somewhat strange and not the end of the story because we have considered the system to be an abstract body in which the interfaces between phases are not important. In fact they can be, and we might describe a system in which substituting the interface between two phases with another type of interface might cause a disequilibrium.


text:
Why do we use nCv∆T=∆U almost universally(for ideal gases)
In my understanding $C_V$ which is defined when the volume is constant is written as
$$C_V=\frac{q_V}{\Delta T}$$
Thus,
$$q_V={nC_V\,\Delta T}$$
since for an isochoric process work done is zero
$$\Delta U=q+w$$
(which is zero)
$$\Delta U=nC_V\,\Delta T$$

Related question from Physics.SE: https://physics.stackexchange.com/questions/336945/when-is-delta-u-nc-v-delta-t-true. Also, Work done by adiabatic expansion of gas.

https://chemistry.stackexchange.com/a/145585/95340

What is the question? This is practically the definition

Because $\Delta U$ is a state function!
It does not depends on how you progress.
For $$V=\text{const}\Rightarrow w=0$$
$$U=q+w=q=nC_V\,\Delta T$$
For $$p=\text{const}\Rightarrow w=-p\,\Delta V=-nR\,\Delta T$$
Since $nR=\text{const}$, only $T$ changes which makes $V$ changes
$$\Delta U=q+w\\
\Leftrightarrow nC_p\,\Delta T-p\,\Delta V\\
\Leftrightarrow nC_p\,\Delta T-nR\,\Delta T\\
\Leftrightarrow n(C_p-R)\,\Delta T\\
\Leftrightarrow nC_V\,\Delta T$$
So $\Delta U=nC_V\,\Delta T$ when $p=\text{const}$.
Only for $T=\text{const}$
$$\Delta U=0$$
For $p$ or $V=\text{const}$
$$\Delta U=nC_V\,\Delta T$$

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial

The below equation is definition of the molar heat capacity.
$$C_V=\frac 1n \left(\frac{\partial U}{\partial T}\right)_V$$
The condition $\mathrm{d}V=0$ is not required for ideal gases, as their internal energy depends only on temperature. It should be therefore obvious why the defining equation is used so often.
Note that the increase of internal energy and temperature during adiabatic processes is achieved even at $q=0$, so  $$C_V=\frac 1n \left(\frac{\delta q}{\text{d}T}\right)$$ cannot be used for definition of $C_V$ of real gases.


text:
What is the change in enthalpy for constant volume?
Which of the following is the correct equation of enthalpy?
$$
\begin{align}
\Delta H &= \Delta U + \Delta (pV) \tag{1} \\
\Delta H &= \Delta U + p\,\Delta V \tag{2} \\
\Delta H &= \Delta U + W \tag{3}
\end{align}
$$
I am facing issues understanding them. Also, when am I supposed to use them? Finally, does $\Delta H = \Delta U$ when only the volume is constant, or both the volume and the pressure are?

We define a state function enthalpy as $$H=U+PV$$
Since we cannot measure the absolute value of enthalpy, but the change in it, we modify the equation to $$\Delta H= \Delta U +\Delta(PV)$$
Which is the correct relation. Now, if you assume pressure is constant, then you can take the $P$ out to get $$\Delta H=\Delta U+P\Delta V$$
Now $PV$ work (or the expansion work) is defined as $W=-P\Delta V$, so we have from the first law of theromodynamics ($\Delta U= q+W$), $$\Delta U= q-P\Delta V$$
Therefore at constant pressure , we have $$\Delta H= q-P\Delta V+P\Delta V=q$$
Which can be stated in words as "At constant pressure, the change in enthalpy is equal to the heat flow".

and finally when is $$\Delta H=\Delta U$$

This happens whenever $\Delta(PV)=0$, which implies $P_1V_1=P_2V_2$ must hold.

For an ideal gas $\Delta (PV) = 0 \Rightarrow \Delta (nRT) = 0 \Rightarrow \Delta U = \Delta (\alpha nRT) = 0$ so $\Delta H = \Delta U \Rightarrow \Delta H = \Delta U = 0$

Enthalpy is the heat released during a chemical process under conditions of constant pressure.  Therefore all the equations you have presented are correct.  Expression two is the most specific definition of enthalpy.
There's no reason P can't be a constant in the first equation (although it's appearance inside the bracket admittedly gives the impression that it could be a variable, making the expression unnecessarily general).  If P is a constant, expression one reduces to expression two.
As for expression three, W (the 'work of expansion') is merely another designation for the second term in expression two.  Expression three is in fact the first law of thermodynamics; the statement that an overall energy change (for a closed system) comprises two terms, one for heat and the other for work.  If no work is done (there is no volume change), then all the change in energy takes the form of heat, which gives you your final expression.  This expression is valid if the product of P and V is constant (i.e. no work is done).  I don't think it's necessary that both pressure and volume be constant here, just that their product be constant (i.e. the two variables are codependent), though I stand corrected by anyone who knows better.


text:
Can endergonic reactions be driven by heat?
I am having trouble conceptualizing endergonic reactions and their difference from endothermic reactions. An endergonic reaction is one that has a positive Gibbs free energy change, such that energy is taken in for the reaction to take place; in biological systems the very exergonic breakdown of ATP is a major source. The expression for Gibbs free energy (for a reaction at constant temperature and pressure) is given by:
$$\Delta G = \Delta H - T\Delta S$$
It is known to me that some endothermic reactions can in fact be exergonic (at certain temperature thresholds) as the entropic increase will be able to counteract the enthalpy change. However, if the reaction is endothermic and associated with a decrease in entropy, the reaction is endergonic at all temperatures. I have a hard time understanding this; isn't heat energy a source of energy as well? If we raise the temperature high enough, surely there would be enough energy to drive the reaction. Intuitively, I imagine that heat energy can't flow into the system due to a thermodynamic law, but I can't seem to figure it out. Does it have to do with entropy or anything related to transfer of heat?

if the reaction is endothermic ($ \Delta H> 0 $) and there is a decrease in entropy (I suppose you mean $ \Delta S <0 $) then we necessarily have $ \Delta G> 0 $ so the reaction is endergonic

I think you can look at this entirely from the perspective of entropy (forget about the Gibbs free energy for a moment). You can divide the universe into two parts, the system and its surroundings.
When a change happens in your system, and some heat is exchanged with the surroundings then the change in entropy of the surroundings is: $$\Delta S_\mathrm{surr}=-\frac{\Delta H}{T}\tag{1}$$
(minus sign because if $\Delta H$ is positive, that means heat is taken in from the surroundings so the entropy of the surroundings must go down).
The change in entropy of the system itself is: ${\Delta S_\mathrm{sys}}$
(which you are referring to as $\Delta S$ in your question).
Now the total entropy change of the universe is:
$$\Delta S_\mathrm{univ}=\Delta S_\mathrm{surr}+ \Delta S_\mathrm{sys}=-\frac{\Delta H}{T}+\Delta S.\tag{2}$$
If you have a reaction where $\Delta H$ is positive, then the first term (i.e. $\Delta S_\mathrm{surr}$) is negative. If the entropy change in the system is also negative (i.e. $\Delta S_\mathrm{sys} < 0)$ then the total entropy change of the universe becomes negative. Of course, the universe is an isolated system, so the second law forbids any change that reduces the entropy of the universe.
Now what happens when you increase $T$? Well… nothing. Assuming that the signs of enthalpy and entropy change of the system stay negative regardless of rising temperature, the total entropy change is going to stay negative. All you do by raising $T$ is decrease the magnitude of the first term.
If you look at the third equation, and multiply both sides by $T$ and change signs, you will basically get the expression for Gibbs free energy. In the end, the sign of Gibbs free energy basically tells you if the total entropy of the universe is going down or not when you do that change in the system.
So even if you raise the temperature, an endergonic reaction will not go, and heat won't flow into the system because that would mean reducing the total entropy of universe.

As an aside, I should mention here that all of this applies to one system. In real world, there are many things you can do to drive an endergonic reaction. For example you can couple it with another reaction/change that makes more entropy than it consumes, and that can drive the endergonic reaction. You can supply electricity (think of it as the entropy that was produced during the generation of electricity feeding the endergonic reaction).
Lastly, even if a reaction has $\Delta_\mathrm{r} G < 0$ that does not mean that you won't get any product. It just means that the equilibrium lies to the side of the reactants. The values of $\Delta_\mathrm{r} G,$ $\Delta_\mathrm{r} H,$ $\Delta_\mathrm{r} S$ that we write consider only the reaction happening, it does not include the entropy of mixing that comes from mixing of the products and reactants. So, even though the reaction is disfavoured, you might get a tiny amount of the product.

You may find IUPAC's brief guide on using italic and roman letters (PDF) useful.


text:
Relationship between Quantum Numbers and the Wave-function
I recently started learning about quantum mechanics and its applciations in atomic structure in chemistry.
In this inorganic textbook Inorganic Chemistry, it describes
"Each of the wavefunctions obtained by solving the Schrödinger equation for a hydrogenic
atom is uniquely labelled by a set of three integers called quantum numbers."
What is the relationship between the quantum numbers and the wave function? If the above is saying that each wavefunction is uniquely labelled by 3 quantum numbers, then it sounds to me that the 3 quantum numbers can describe the wavefunction on their own. But how can just 3 numbers describe a function?
On this website, it says the following
"A complete solution to the Schrödinger equation, both the three-dimensional wavefunction and energy, includes a set of three quantum numbers ( n,l,ml )."
Is the complete solution to the Schrodinger's Equation the wavefunction in 3D? what is the "energy" mentioned in the last quote to do with solution of the equation?
Thank you.

Have you looked at the general solution of the S.E for a hydrogen atom ? Have you seen the quantum numbers there ?

Imagine you have a differential equation leading to a general solution y=k+l.x+m.x^2, where k,l,m are integers. Does the  knowledge of k,l,m completely describe the function y=f(x) as the particular solution of that implied equation ? In both cases, here and wave function case,  the general  function is already known but the parameter values.

A number can not define a function. It is just part of the function, like the constant $k$ in trigonometric problems, where you find a solution with an angle $\alpha + 2k\pi$

@Maurice A number can define a function, if the general form of function is a priori known. If y=n.x is given, than giving n defines the particular function. By other words the wave function is partially predefined, waiting just for particular  values of allowed quantum numbers.( as allowed numbers are predefined as well ).

The hydrogen atom
Indeed, three numbers cannot define a function. There are several more criteria, one of which you mentioned already: it must satisfy the time-independent Schrödinger equation. The Schrödinger equation is a nasty differential equation, which looks something like the following:
$$-\frac{\hbar^2}{2\mu}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) - \frac{e^2}{4\pi\varepsilon_0r}\psi = E\psi$$
On top of this, it must also satisfy the usual criteria for a wavefunction ("boundary conditions"): for example it must be square integrable (i.e. its integral over all space must be finite, meaning that it can be normalised):
$$\int |\psi|^2\,\mathrm{d}\tau < \infty$$
Once you put all of these conditions together, it turns out that not every plain old function $\psi(x,y,z)$ is a permissible atomic orbital. There is a family of functions $\psi$ which do, and they all have a certain form which is completely determined by the three quantum numbers $(n, l, m_l)$.

A simpler example
This is pretty abstract, so why don't we make it more concrete with a simpler example. (The hydrogen atom is not easy to solve explicitly.) Consider the one-dimensional function $f = f(\phi)$, where $\phi$ is an angle. You should recognise at this point that, given no other information, one quantum number is not sufficient to specify what $f$ is, as there is an infinite possible different $f$'s.*
However, assume now that $f$ obeys the differential equation
$$\frac{\mathrm{d}f}{\mathrm{d}\phi} = \mathrm{i}mf$$
The general solution of this differential equation is (after dropping the normalisation factor)
$$f = \exp(\mathrm{i}m\phi)$$
But note also that since $\phi$ is an angle, we have that for all $\phi$
$$f(\phi) = f(\phi + 2\pi)$$
meaning that
$$\begin{align}
\exp(\mathrm{i}m\phi) &= \exp[\mathrm{i}m(\phi + 2\pi)] \\
&= \exp(\mathrm{i}m\phi) \exp(\mathrm{i}m2\pi) \\
1 &= \exp(\mathrm{i}m2\pi)
\end{align}$$
which means that $m$ must be an integer. Consequently, the allowed $f$'s are
$$f = \exp(\mathrm{i}m\phi) \quad\quad m \in \mathbb{Z}$$
and so we can now label the $f$'s by the value of $m$. The quantum numbers $m$ do not represent actual values of $f$ at any point in space, nor do they represent actual values of $\phi$: they instead serve to parameterise the $f$'s. That is to say, the value of $f$ at any value of $\phi$ is completely determined by the value of $m$. We do not write this as $f(m)$, but rather $f_m$, and thus we have:
$$\begin{align}
&\,\,\,\vdots \\
f_{-2} &= \exp(-2\mathrm{i}\phi) \\
f_{-1} &= \exp(-\mathrm{i}\phi) \\
f_{0} &= 1 \\
f_{1} &= \exp(\mathrm{i}\phi) \\
f_{2} &= \exp(2\mathrm{i}\phi) \\
&\,\,\,\vdots
\end{align}$$
Indeed, $m$ can be said to be a quantum number here. Notice how we've used a simplified problem: it has one instead of three dimensions, and it has a much simpler differential equation. However, it illustrates that by enforcing additional conditions on the wavefunction — namely that it obey a differential equation and boundary conditions — we can get a family of solutions which can be labelled with quantum numbers.†

What is the energy?
When you find the allowed $\psi$'s that solve the Schrödinger equation, you can plug it back into the differential equation and evaluate the value of $E$ inside it. That will give you the energy of the atomic orbital.

Footnotes
* Technically, there is an uncountably infinite number of $f$'s. If it were countably infinite, then we could establish a one-to-one mapping between the different $f$'s and the natural numbers, which would allow us to label the $f$'s as $f_1, f_2, f_3, \cdots$ which is exactly what a quantum number does.
† With regards to the previous footnote, this set of permissible $f$'s is still infinite, but it is countably infinite this time. Also, this actually happens to be the exact way in which the quantum number $m_l$ is derived, so my choice of $m$ as the quantum number was not arbitrary.

You need to add two quantum numbers for the electron: s and ms to take in account its Fermionic nature.
The concept of quantum number is quite general and can be applied both to monoelectronic wavefunctions both poliectronic ones. Given a physical system, a quantum number is an integer or a semi integer associated to an invariant. An invariant is a quantity that does not change during the dynamics. For an isolated system, the most important invariant is the total energy.
Why quantum numbers are so useful? Because they can label in a unique way the wavefunctions. That's it.
I would suggest you to read the masterpiece of Eyring, Walter and Kimball "Quantum Chemistry". It uses a very old notation, but if you want to understand the root of QM it is the best choice.


text:
How long can iodine bottle left open? Can it go bad sooner than expiration date?
I have bought iodine about a year ago but the bottle cap got damaged during the transportation and it leaked all over the package, but I have decided to keep it anyway. Unfortunately I did not use it to anything and it was just sitting on my shelf with cracked cap with about a one millimetre sized hole.
I wonder if it could go wrong and now it's useless. As far the expiration date, it's still not expired, but that's probably when the bottle is fully sealed.
So is it still safe to use if needed or I can return it to the drug store so they can safely dispose it?
Content for 100ml: iodopolyvidonum 7,5g with 10% content of Povidone-iodine (M.h. 40 000)
Thank you

Iodine itself has a comparatively high vapor pressure, ~40 Pa at room temperature, so iodine solutions, e.g. tincture of iodine (iodine in ethanol/water), lose iodine by evaporation (well, by sublimation, from the solid). This just means that the strength of the antiseptic is reduced if left open, not that it has not "gone bad,"  in the sense of being dangerous.
For povidone iodine, $\ce{C6H9I2NO}$, the vapor pressure is far lower, ~1.3 Pa at room temperature, so the rate of loss of iodine would be far slower. Again, though the antiseptic will slowly lose potency, it does not "go bad". If the color looks noticeably lighter, I'd discard it to get a new bottle.
That said, iodine is not a perfect bactericidal or fungicide. for example, Pseudomonas bacteria have been found in production systems for commercial iodine antiseptic. Not to worry, though, as long as povidone iodine is used according to directions. Just cover the bottle with material that does not react with iodine. Note that plastic food wrap and polyethylene bags allow iodine vapor to go through slowly, so use a few layers if that's your choice for a cap.

Thank you, I did not expect to get such a detailed answer. I have learned something new.  Thank you

A solution of iodine antiseptic is a medical item that is used for improving health. There are two ways to treat your cracked cap.
#1. If you inspected the bottle in the drug store before you bought it, you would not buy this bottle. Almost all over-the-counter medicines have warnings: Do not use if the safety seal is broken. So, out of utmost caution, do not use this iodine, because it might be contaminated (even tho you are the only one who could have contaminated it). Look at it this way: you bought a bottle of iodine solution, you had its potential usefulness for a year to keep your skin abrasions antiseptic (even tho you did not have the need to use it). It can be considered to have served its purpose usefully; get a new bottle and be more careful.
#2. Being the cheapskate that I am, I would examine the bottle: has the liquid level dropped and dried out? Is it gooey so as to be difficult to apply? If not, it might be OK to keep and use. On the other hand, a cracked lid is an invitation to disaster - if the bottle gets knocked, it will leak - probably onto the item you just laid down, which will, of course, be very important and/or delicate. To avoid catastrophes of this sort, dispose of the iodine solution. You will feel more secure. Get a new bottle.
Or, get a tube of antiseptic ointment, which will have less tendency to spill, unless you step on it.

Thank you for the easy to understand answer even for someone who does not have much in common with chemistry. As far the #1 goes, I have bought it online, the delivery man messed up and I got my money back and I was thinking it might be useful someday. For the #2 Nope, it did not dry out. Everything looks perfectly fine but I am not a chemist nor know anything about medicine, so that was the reason I came to seek more professional insight. Considering DrMoishe gave more professional answer and this is stackexchange, I have accepted his answer, but I am glad for yours too. Thanks again :)

BTW, please don't consider this medical advice in any way... we're just commenting on the state of the chemical, not it's usefulness for any purpose.


text:
Why electrons do not distribute evenly among the atoms in a molecule?
I was wondering why the state where electrons are evenly (or the closest to being evenly) distributed among the atoms in a molecule is not the lowest energy state? For example, in a water molecule it is known that oxygen attracts electrons from hydrogen due to its high electronegativity (oxygen wants to fill its shell). But if we instead took 4 electrons off the oxygen outer shell and just filled hydrogen states with them we would have 2 hydrogen atoms with 1s$^2$ 2s$^1$ configuration and our oxygen with 1s$^2$ 2s$^2$ configuration. This way we eliminate supposedly high-energy p-states of oxygen and populate lower-energy s-states of hydrogen and lower the total energy.
I know that my example is extremely naive if not brutally wrong. I thought that maybe energy levels of different atom types have different energies which results in the correct electronic distribution (not like in my example above) in a water molecule to be truly the lowest energy state. Or maybe my logic violates some quantum mechanical postulate...I couldn't immediately find why my reasoning is wrong so I hope someone here would help me out.
Thank you!

What do mean by "distribute evenly"? They've distributed in a way that minimizes the energy. I'd call that pretty even.

@Zhe as I tried to suggest by my example, if you take high energy electrons from the p orbitals of oxygen and put them on energetically lower orbitals of hydrogen the total energy would be lower. "Even" means to take all the electrons available and put them on the lowest levels of all atoms in the system. My model is clearly wrong, but I believe I introduced it clearly.

Why don't the protons all distribute evenly across the atoms in the molecule?

@JonCuster I don't know how proton energy states/levels look like. If you wanted to make an analogy, could you please expand on this?

But the system has accounted for the all factors and the energy-optimized system is what's observed. The primary issue may simply be that your view of the system (atomic orbitals in a molecule) is not accurate enough to produce the meaningful result you want.

Atomic orbitals are not the same as molecular orbitals
You seem to be assuming that the electron orbitals of isolated atoms are the same as the orbitals that exist in stable molecules. So the issue is which orbitals get filled to create the lowest energy configuration.
But the quantum mechanics of orbitals is a lot more complex than that. The orbitals that exist in atoms are the possible configurations of electrons for an isolated molecule. But if you solve the relevant equations for a molecule you get different orbitals and they depend on the properties of all the nuclei in the molecule.
Getting that full solution is usually a computational nightmare and chemists tend to think and describe the solutions in terms of combinations of atomic orbitals. But those combinations need to recognise the nuclear properties of the component atoms. So an oxygen atom has more charge than a hydrogen and the molecular orbitals will show a higher electron density near oxygen (or, higher than you would get for an isolated oxygen filling its atomic orbitals).
The reason we don't get an even distribution is because the nuclei are not the same and have different charges. The stable molecular orbitals recognise this giving uneven distributions of the electron density.

Thank you for your answer. Certainly the complete physics is complex, but there are simple rules like the "octet rule" that I am trying to make sense out of. The fact that atoms want to make their shell complete just isn't so clear for me, and that is why I started to think about possible states in terms of their energies.

@CHILLQQ consider that independently of the frame, just external electrons are involved in bond formation.

@CHILLQQ The octet rule is a very simplistic rule of thumb to explain how bonding works and it totally fails to capture anything useful about the details of bonding or electron distributions. It is certainly not suitable for describing anything about electron distributions in bonds, for that you need some molecular orbital theory.

I found an answer online that satisfied me.
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Miessler_Fischer_Tarr)/05%3A_Molecular_Orbitals/5.03%3A_Heteronuclear_Diatomic_Molecules
It shows that 1s orbital of Hydrogen has energy of -13.64 eV whereas 2s and 2p orbitals of oxygen have -32.36 eV and -15.87 eV respectively. This suggests that it is indeed energetically more favorable to keep oxygen electronic states occupied instead of transferring electrons to hydrogen orbitals, which are apparently higher in energy. I suggested in my question that "energy levels of different atom types have different energies " but I couldn't find the confirmation, and now I found it.

We do operate this energy distinction. And it is even not our choice. While is drastic in some approximation, it is natural. It is even not a choice. It it would be for electrons deeper in energy there won't be chemistry. Very much as when you touch a body you encounter its surface first (macroscopically, else I am afraid someone may show up with "no, touching is electron repulsion". Which albeit a bit hand waving serves to the purpose as well :)


text:
In reaction equations, why are j and k sometimes capitalized to J and K?
From 'Leighton relationship' on Wikipedia:
through the following mechanism:[2]
($J_1$) $$\ce{NO2 + h\nu (\lambda < 420 nm) → NO + O (^3P)}$$
($k_2$) $$\ce{O (^3P) + O2 + M → O3 + M}$$
($k_3$) $$\ce{NO + O3 → NO2 + O2}$$
This series of reactions creates a null cycle, in which there is no net production or loss of any species involved.
And here:
Since $\ce{O (3P)}$ is very reactive and $\ce{O2}$ is abundant, $\ce{O (3P)}$ can be assumed to be in steady state, and thus an equation linking the concentrations of the species involved can be derived:
$$\ce{[O_3]} = {J_1\ce{[NO2]} \over k_3\ce{[NO]}}$$
$${\displaystyle [{\ce {O3}}]={\frac {J_{1}[{\ce {NO2}}]}{k_{3}[{\ce {NO}}]}}}$$
$${\displaystyle [{\ce {O3}}]={\frac {J_{1}[{\ce {NO2}}]}{k_{3}[{\ce {NO}}]}}}$$
The ${J}$ is capitalized, but the $k$ (or $k$s) are not....
Does this signify anything?

For eventual writing and formatting of chemical formulas or equations, see how to use MathJax with mhchem extension

Typically lower case k is used for rate constants and upper case K is for equilibrium constants, but I don’t know what j is there

The entry in Wikipedia misses to define the entries in the formula.  But note equations 1 and 2 in a paper by Griffin et al. (doi 10.1029/2006JD007604, open access) in the Journal of Geophysical Research, Atmospheres. To establish context of the paper by Griffin et al., the authors introduce the reader to the topic citing earlier work.

«In the troposphere, a molecule of $\ce{NO2}$ is degraded photochemically to form a molecule of $\ce{NO}$ and an oxygen atom ($\ce{O}$). It should be noted that this process depends on the sufficient flux of photons associated with light of the appropriate wavelength to photolyze $\ce{NO2}$. The $\ce{O}$ atom quickly reacts with molecular oxygen ($\ce{O2}$) in the presence of a third body to form $\ce{O3}$. However, the $\ce{O3}$ formed can react rapidly with the $\ce{NO}$ to reform $\ce{NO2}$ and $\ce{O2}$. The net result of these reactions is a null cycle, and when steady state is assumed, the $\ce{O3}$ concentration can be predicted as [Seinfeld and Pandis, 1998]
$$ [\ce{O3}] = \frac{j_\ce{NO2} [\ce{NO2}]} {k_1 [\ce{NO}]}$$
where the bracket notation represents a concentration or mixing ratio (here, parts per billion by volume, ppbv), $j_\ce{NO2}$ ($\pu{s^{-1}}$) is the photolysis rate of $\ce{NO2}$, and $k_1$ ($\pu{ppbv^{-1} s^{-1}}$) is the temperature-dependent rate constant for the reaction between $\ce{NO}$ and $\ce{O3}$.»

The article continues by

«This is referred to as the photostationary state for $\ce{O3}$. If each side of equation (1) is divided by $[\ce{O3}]$, the right hand side is termed the Leighton ratio, $\Phi$ [Leighton, 1961]:
$$\Phi = \frac{j_\ce{NO2} [\ce{NO2}]} {k_1 [\ce{NO}] [\ce{O3}]} $$
[...]»

Griffin, R. J.; Beckman, P. J.; Talbot, R. W.; Sive, B. C.; Varner, R. K. Deviations from ozone photostationary state during the International Consortium for Atmospheric Research on Transport and Transformation 2004 campaign: Use of measurements and photochemical modeling to assess potential causes. J. Geophys. Res. 2007, 112, D10S07. DOI 10.1029/2006JD007604.


text:
Derivation of Enthalpy Equation

Why are they taking delta H in equation (iv) to be at constant pressure while in the same equation they are considering delta E to be at constant volume to arrive at equation (v)? If they are arriving at equation (iv) from equation (i) then shouldn't P and V be variable?

I do not understand where the problem is. for a reaction at constant pressure, the use of the ideal gas law allows to go back to the relation (iv)

Equation (v) is describing the relative heat flows from two different processes, one at constant pressure (where the heat flow at constant pressure, q_p, is given by Delta_H), and one at constant volume (where the heat flow at constant volume, q_V, is given by Delta_E). So it's OK (and, indeed, necessary) to have two different constraints (const. p vs const. V), beause they are comparing heat flows from processes with those two different constraints.

See https://chemistry.stackexchange.com/questions/84207/work-done-by-adiabatic-expansion-of-gas/146588#146588

You determined the $\Delta H$ in going from products to reactants at constant temperature and volume.  But, you are dealing with an ideal gas mixture, so the enthalpy of the product mixture is a function only of temperature, and not pressure.  So the enthalpy change of the product mixture in going from the final pressure in the reactor to the initial pressure is zero.  So, even though it is a constant volume reactor, the enthalpy change you determined also applies to carrying out the reaction at constant temperature and pressure.


text:
Steady State: Exclusion of Term
Synopsis
We are given a proposed mechanism for the condensation of $\ce{(CH_3)_2CO}$ as follows
First Step: $$\ce{(CH_3)_2CO + OH^- <=> CH_3C(O)CH_2^- + H_2O}$$
Second Step: $$\ce{CH_3C(O)CH_2^- + (CH_3)_2CO -> Product}$$
Attempt and Question
I assume $\ce{CH_3C(O)CH_2^-}$ is in steady-state, and obtain subsequently that $$ k_1 \ce{[(CH_3)_2CO][OH^-]} = k_{-1}\ce{[CH_3C(O)CH_2^-][H_2O]} + k_2[\ce{CH_3C(O)CH_2^-][(CH_3)_2CO]}$$ since I'm assuming the rate of change of $\ce{[CH_3C(O)CH_2^-]}$ is $0$.
From here, in my original attempt, I factored out the $\ce{[CH_3C(O)CH_2^-]}$, solved for it, and plugged it in to my rate equation $$\ce{Rate = $k_2$ [CH_3C(O)CH_2^-][(CH_3)_2CO]}$$ to obtain $$\ce{Rate = \frac{$k_1k_2$ [(CH_3)_2CO]^2[OH^-]}{$k_{-1}$[H_2O] + $k_2$[(CH_3)_2CO]}}.$$ But looking at my solution, I realize I didn't eliminate the $\ce{H_2O}$ term. This is where my question lies. In the steady state approximation, do I assume $\ce{H_2O}$ is in steady state too and follow the same steps, substituting it into my equation for the concentration of $\ce{CH_3C(O)CH_2^-}$? I tried that, but in doing so, I obtain $$\ce{$k_1$[(CH_3)_2CO][OH] = $k_{-1}$[CH_3C(O)CH_2^-][H_2O]}$$ and that $$\ce{[H_2O] = \frac{$k_1$[(CH_3)_2CO][OH^-]}{$k_{-1}$[CH_3C(O)CH_2^-]}}.$$Plugging that in to my equation for $\ce{CH_3C(O)CH_2^-}$, I obtain something that seems absurd, mainly that $$k_1 \ce{[(CH_3)_2CO][OH^-]} = \ce{$k_1$[(CH_3)_2CO][OH^-]} + k_2[\ce{CH_3C(O)CH_2^-][(CH_3)_2CO]}.$$ This, being obviously wrong, led to great confusion. How do I deal with the $\ce{H_2O}$ term? I recognize it has no role in forming the products, but it does have a role in the rate of disappearance of $\ce{CH_3C(O)CH_2^-}$, so I can't just throw it away right? If I can, why is that? What am I missing here? This is my first chemistry course in college, so if this is a stupid question or if it's already been asked, please just kindly point me to a helpful resource.

if $\ce{CH3C(O)CH_2^−}$ is in steady-state so we can write that $\ce{\frac{d[CH3C(O)CH_2^−]}{dt}=0}$

Yes. I'm aware. I know what our assumption is, but I'm asking about the $\ce{H_2O}$ term, and how it relates to our steady-state assumption.

I use the assumption in my development of the equation $ k_1 \ce{[(CH_3)2CO][OH^-]} = k{-1}\ce{[CH_3C(O)CH_2^-][H_2O]} + k_2[\ce{CH_3C(O)CH_2^-][(CH_3)_2CO]}$.

what data do you have in your exercise? I think you can reasonably consider the low water concentration if the reaction takes place in an organic medium

There is no data. The entire question is in the synopsis.


text:
What is the electrophile in Perkin's condensation reaction?
In Perkin's reaction, an aromatic aldehyde reacts with an acid anhydride to give an alpha-beta unsaturated aromatic acid. Which reactant is the electrophile and which one is the nucleophile?

Google is your friend! If you search "Perkin Reaction Mechanism" you will find your answer

@Waylander I did but I couldn't really get the answer to my question

The reaction mechanism shown on the Wikipedia page for the Perkin Reaction is perfectly clear.

The first step is deprotonation of acetic anhydride by acetate anion.
The second step is nucleophilic attack by the anion of acetic anhydride on the aromatic aldehyde; thus the aldehyde is the electrophile.


text:
How is Gibbs free energy related to stability?
I have been learning thermodynamics for quite some time and I fail to relate Gibbs free energy and stability.
For example, how do I compare stability of metal oxides using the following data?
Also, do stable compounds decompose easily or hard?
$$
\begin{array}{clrr}
\hline
\text{#}  & \text{Compound} & \Delta_\mathrm{f}G^\circ/\pu{kJ mol^-1} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} \\
\hline
1 & \ce{ZnO(s)} & -318.4 & -348.3 \\
2 & \ce{Cu2O(s)} & -146.0 & -168.8 \\
3 & \ce{HgO(s)} & -58.5 & -90.8 \\
4 & \ce{PbO(s)} & -187.9 & -217.3 \\
\hline
\end{array}
$$

Generally the compound with more negative Gibbs free energy of formation is the most stable one.

can you elaborate???

Gibbs free energy is basically an indicator of how far and in which direction the equilibrium is pushed. So if the Gibbs free energy is lower (i.e. more negative) that means the equilibrium is pushed towards the product side. Here the free energy of formation means the formation of the metal oxide from its elements in standard state. So the lower the Gibbs free energy, the metal oxide is more stable w.r.t the elements.

The thermodynamic data refers to formation of the given compounds from the elements in their standard states. For instance for $ \ce{ ZnO(s)}$  it refers to the reaction
$ \ce{ Zn(s) +1/2O2(g) ->ZnO(s)} \tag{1}$
with each element or compound present at a pressure of 1 bar and (unless noted otherwise) a temperature of 298.15 K.
A more negative value of $\Delta G^\circ$ means greater product stability. This follows from the relationship
$$\Delta G^\circ = -RT\log K \tag{2}$$
where K is the equilibrium constant for the reaction at that temperature. An increased value of K implies an increased amount of products relative to reagents. By the relation in equation (2) a larger value of K also means a more negative value of $\Delta G^\circ$.
By definition a more stable compound decomposes less readily. Decomposition here refers to the reverse of reaction (1).
When comparing the tabulated oxides, a more stable oxide has a more negative free energy of formation. Therefore $ \ce{ZnO(s)}$ is the most stable of the oxides relative to oxygen and the reduced metal. Of the metals the mercury(II) oxide would require the least energy to reduce.
Note that for all of the reactions here
$$K = p^\circ/p_\ce{O2}$$
where $p^\circ=\pu{1 bar}$ (assuming the pressure is low). When starting from metal oxides, allowed to decompose until at equilibrium, a less stable oxide is at equilibrium with a higher partial pressure of oxygen (smaller K) due to more of the oxide decomposing. Therefore imagine you have an evacuated box containing the oxide of one metal and an amount of another metal in pure (reduced) form. If the equilibrium partial pressure $p_\ce{O2}(\text{eq})$ for the oxide is higher than $p_\ce{O2}(\text{eq})$ for the oxide of the second metal, then (if you wait long enough) all of the oxide will be reduced and the second metal oxidized until either the original oxide or pure metal is exhausted.
The enthalpies of formation refer to how much heat evolves when  1 mole of the oxide is formed from the reagents at constant T and p, with all compounds in their standard states. The more negative the value the more heat is released (the reaction is more exothermic). The oxidation of zinc is the most exothermic reaction (per mole of product formed) of those shown in the table.


text:
Why these compounds don't show and show Geometrical isomerism?
I was recently studying organic chemistry related to Isomerism. I have doubt based on an example that is,

Why Cumulenes show GI?
Why 1-chloro cyclohexene don't show GI?

The basic idea of geometrical isomerism is the difference between the spatial arrangement
of atoms/groups between two compounds which have same molecular formulae and connectivity.
One way of judging GI is analysing the difference between the distances of groups.
Cumulenes having an odd number of double bonds are capable of showing GI (granted that the two atoms/groups attached to the end carbons aren't same), however cumulenes with an even number of double bonds cannot show GI since the groups connected at the terminal carbons aren't in the same plane and are equally distant.
Don't treat double bond in a ring like a double bond in a chain. In case of a ring, for trans isomer the ring (sort of) wraps around the double bond. This is not possible in small rings due to strain and tendency to remain planar. GI around a double bond in a ring is seen from cyclooctene onwards. And it doesn't depend on the groups attached.

This is my first answer on ChemSE, so please feel free to point out mistakes.

sorry it's cyclohexene

@RitujithManoj I appreciate your contribution but we have a policy of not answering homework questions. Since you are a new user, you might want to take a tour of this site.

Can you elaborate how distances are same ? If they have four terminal groups and each group is distinct, then each bond of terminal group will have different bond length , and geometrically different distances as well by cosine rule?


text:
What happens when solution containing two or more cations are electrolysed at an excess potential?

$\pu{250 mL}$ of a waste solution obtained from the workshop of a goldsmith contains $\pu{0.1 M}$ $\ce{AgNO3}$ and $\pu{0.1 M}$ $\ce{AuCl}$. The solution was electrolyzed at $\pu{2 V}$ by passing a current of $\pu{1 A}$ for 15 minutes. The metal/metals electrodeposited will be:
(1) only silver
(2) only gold
(3) silver and gold in equal mass proportion
(4) silver and gold in proportion to their atomic weights

I came across this question whose answer was given as (4). But I doubt the answer. I think that the molar ratio should depend on the ionic mobilities of the cations and shouldn't equal 1.
Am I thinking right? A qualitative answer explaining this would be appreciated.

It is highly inprobable you could make 0.1 M AuCl. And AuCl3 forms with water $\ce{[AuCl3(OH)]-}$ anion. The task seems to me as made up, being detached from reality.

This problem is impossible for two reasons. First $\ce{AuCl}$ cannot form $\pu{0.1 M}$ solutions, as it is decomposed in water without $\ce{HCl}$ . And second $\ce{Ag+}$ will react with chlorine compounds like $\ce{AuCl}$ to produce silver chloride $\ce{AgCl}$ which is insoluble in water.

Choice 4 is roughly correct, instead of atomic weights, they should write equivalent weights.

@Maurice I thought of that also. But consider this as a hypothetical question or, replace AuCl with any other compound that doesn't react and is stable under the conditions

@Govind. I like impossible questions, like : How long will it take for a sparrow to fly up to the Sun ? This question can be solved, by dividing the distance by the speed.


text:
Why does the electron withdrawing group increase acidic nature of carboxylic acids?
The doubt in my mind is that the electron withdrawing group is attached to the carbon atom of the carbonyl group and that carbon doesn't have any negative charge..the negative charge lies on oxygen atom so how does this electron withdrawing group pulls the electron from oxygen.?

The inductive effect decays very rapidly but it is not limited to a bond. Let quantitative assessment aside, it does not take much intuition to see that if the C atoms see less electron density in one side, it will attract some electron density from the other. If it would be a localized effect, then we would not even speak of inductive effects.

The carbon atom attached to oxygen and electron withdrawing group(suppose halide) in an Acyl halide is strongly polarized and that is why they are very reactive and unstable.


text:
3-propyl-penta-1,3-diene vs 3-ethylidene-hex-1-ene
From the rules I know I would name the following compound as 3-propyl-penta-1,3-diene because maximum number of functional groups (double bonds in this case) are in the five numbered carbon chain:

However, ChemDraw names the above compound as 3-ethylidene-hex-1-ene. Why? Am I missing a rule?

This compound also has stereochemistry ($E/Z$).

If we ignore E/Z nomenclature at the moment @Mathew Mahindaratne , what about the main name?

Naming rules include identifying the longest carbon chain.

I know that, but there is another rule that if there are functional groups present, we choose the longest chain which has maximum number of functional groups, and since it contains two double bonds IMO 5 numbered chain should be considered as parent chain. Can I get a reference? @z1273

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

@Loong Can you provide a source from where I can refer nomencluature rules, I don't have access to the blue book.

@LightYagami Start here: Brief guide to the nomenclature of organic chemistry (IUPAC Technical Report) 2020


text:
How can DI water inhibit peroxidase activity?
I'm troubleshooting an IHC staining issue, and according to Thermo-Fischer's website, "Deionized water can sometimes contain peroxidase inhibitors that can significantly impair enzyme activity." What exactly could deionized water contain that would inhibit peroxidase?

Would be nice to have a link. Found this: https://www.thermofisher.com/se/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/ihc-troubleshooting-guide.html

I'll hazard a guess: minute amounts of heavy metals maybe? Oxygen (although that sounds farfetched)?

@BuckThorn, I think it must be leachates from the ion exchange resins and perhaps bacterial contamination. I don't think deionized water has traces of metals.

@M.Farooq Yes, good point. The presence of metals would violate the designation DI. I'll leave my comment up in case someone else has this odd thought.


text:
pH probe reading greater than 14
Today I was using my pH meter on wastewater samples but the meter was reading values near 14 and sometimes greater, I changed the probe with the spare one and it read well (sure I calibrated before reading) My question is if the probe must be damaged or is it only need cleaning?

Try testing it with standard buffer solutions, e.g., pH values of 4, 7, and 10.

You did not mention what kind of waste water are you talking about. Is it from a textile industry, paper mill, leather factory? Waste water is not a "compound", it can be any junk and its pH can vary over several orders of magnitude!
As I have stated before, there is nothing fundamental about pH. It is a scale of convenience. It can be negative and it can be greater than 14 depending upon the activity of the hydrogen or hydroxyl ion in water respectively.
In case the wastewater is in fact highly alkaline:
The point to ponder is whether a glass electrode is suitable or not for measuring really alkaline solutions. The answer is no. Glass electrode have something called as alkaline error at high pH. So you may have to dilute your samples quantitatively and quote the pH of dilute samples. If the samples are too alkaline, simply express the titre with 0.1 M HCl since pH measurements will not be reliable.

Just to add on, the pH can be greater than 14 if the temperature of the solution is low, in which case the water dissociation constant will decrease. If a given solution had a really high pH that is close to 14 at room temperature then it would definitely increase to a value beyond 14 if the temperature of the solution were to decrease.


text:
What chemical could be used as carbon monoxide scrubber?
I am trying to produce nitrogen by passing air over a heated tube of coals. This leaves a mixture of $\ce{CO},$ $\ce{H2O}$ and $\ce{N2}.$ Only $\ce{CO}$ is unwanted in this case.
Could I use calcium hydroxide to scrub carbon monoxide out of the gas? What other substances could be effective?
edit: this is part of challenge to make other chemicals from natural resources, therefore i have to obtain nitrogen myself instead of buying it

If you just want pure nitrogen, locate your nearest specialty gas company. Cylinders of nitrogen are danged cheap. And any welding shop has pure argon if you just want something pretty inert.

Whole this is particularity bad and convoluted idea

CO could be absorbed in solution of copper(I) chloride  in ammonia or hydrochloric acid. ( almost insoluble in water ). But I am not sure about residual CO left and using the reagent may affect the intended nitrogen  usage, unless cleaned up.

@joncuster It is definitely the easiest and preferred way, but the OP may want just small volume. The whole cylinder may be "a cannon for a sparrow". It would help if the purpose was revealed, if he really needs nitrogen, or just a gas without oxygen. For the later, CO2 would be easier way.

It is an experiment done for the pleasure of doing it? OP please explain. Otherwise buy N2. And what about CO2? There should be some, although is easy to be scrubbed away.

it is. I was thinking of ways to make chemicals from sources found in the wild. Nitrogen could be used to turn charcoal mixed with potash into potassium cyanide at 1200°C, which hydrolyzes to give ammonia, as artifficial fertilizer (bucher process). carbon monoxide would shift the equillibrium the wrong way

Specify a) the quantity (volume at pressure x and temperature y) you need, b) check if you refer to CO as carbon monoxyde, because $\ce{Ca(OH)2}$ normally is used to catch CO2 / carbon dioxyde.  c) Offer a realistic estimate about the highest concentration of CO acceptable for you.  d) An indication of the intended use of this purified gas could offer a second perspective if your estimate c) is reasonable or suggest an alternative more suitable for your needs.

Air reacting with charcoal will produce $\ce{N2},$ $\ce{CO}$ and $\ce{CO2}.$ $\ce{CO2}$ will be absorbed by calcium hydroxide or sodium hydroxide.
However, $\ce{CO}$ cannot be absorbed at ordinary temperatures and pressures. Only at $\pu{200 °C}$ and $\pu{10 atm},$ it reacts with $\ce{NaOH}$ to produce sodium formate $\ce{HCOONa}.$
Finally, $\ce{CO}$ can be oxidized into $\ce{CO2}$ by passing over hot $\ce{MnO2}$ or $\ce{Fe2O3},$ and then be absorbed by a hydroxide.


text:
Does anybody have any ideas about the trend that I found for luminol chemiluminescence reaction? Or can anybody help me analyze them?
The reaction that I did is shown below. I'm testing the duration and luminosity of the light the reaction produces against different pH. The way I used to control the pH is by adding more or less $\ce{Na2CO3}$ since it is basic with a pH of 11.
Solution A: In $\pu{200 ml}$ water, add in $\pu{0.16 g}$ $\ce{Na2CO3}$, $\pu{0.04 g}$ luminol, $\pu{5 g}$ $\ce{NaHCO3}$, $\pu{4.6 g}$ $\ce{(NH4)2CO3}$, and $\pu{0.08 g}$ $\ce{CuSO4}$.
Solution B: In $\pu{200 ml}$ water, add in $\pu{10 ml}$ $\ce{H2O2}$
Then, pouring $\pu{50 ml}$ of solution B into $\pu{50 ml}$ of solution A, a bright light will be produced. The light will gradually fade away but the light will be produced again after some time, a guaranteed second reaction for every trial. (Could someone also tell me how could this happen?)
The trend that I found is shown below:

https://pubs.acs.org/doi/10.1021/acs.jpcb.0c06301


text:
Why does boron have a lower affinity than oxygen?
Electron affinity of oxygen is 141 kJ/mol and electron affinity of boron is 27 kJ/mol. I suspect it is because boron nucleus is much better shielded than oxygen nucleus (in the p-shell), but I am not sure if this is the only factor since the difference is so large.

Oxygen's nucleus consists of more protons than boron an thus its Effective nuclear charge is stronger than in case of boron. This means that oxygen will be "more electronegative" than boron, because his nucleus has bigger + charge and attracts electrons more than boron.
Boron's nucleus is actually less shielded than oxygen's as we can determine from Slater's rule, but it is simply not as strong as oxygen's nucleus and because of that, boron has smaller electron affinity.

While I understand that oxygen should have larger effective charge, the difference between affinities just seems too large. Using Slater's rule, effective charge of Boron should be 2.25 and effective charge of Oxygen should be 4.2, i.e. less than a factor of 2 difference. At the same time, electron affinities of the actual elements differ by a factor of 5.
Slater's rule does not seem to work well for affinities quantitively at all.