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How can compounds be called reduced if nothing is known about the reaction they formed in?
I'm reading a research article about a model trying to explain the elemental compositions of meteorites.[1]
What confuses me, are statements of the following type (here in chapter 2.1, second paragraph about enstatite chondrites, a type of meteor):

[...] Enstatite condrites contain minerals that are remarkably reduced chemically, including abundant sulfides like oldhamite (CaS).

In my understanding reduction or oxidization only refers to a specific element in a specific reaction, but here they talk about a compound  with unknown reaction path to form it.

How can CaS be considered to be (remarkably) reduced, if it is not an element in a reaction?
How can CaS be considered to be (remarkably) reduced, if we don't know the reaction path that it formed in?

Steven J. Desch, Anusha Kalyaan, Conel M. O'D. Alexander. The Effect of Jupiter's Formation on the Distribution of Refractory Elements and Inclusions in Meteorites. arXiv:1710.03809 [astro-ph.EP]

CaS can be considered reduced compared to Earth conditions (where it is likely to get oxidized). The reaction path that created it is irrelevant.

@Ivan Neretin: You mean because both components, Ca and S would end up in compounds which would give them positive oxidization numbers? Else one would have to refer to either Ca or S as being reduced, but not CaS as molecule itself, no?

@AtmosphericPrisonEscape You actually do not need to know the reaction details to consider that a given atom is in a more/less reduced form than another one. "Reduced" is also used to describe a state (i.e. the oxidation state) rather than a process or an action.

While we don’t know exactly how minerals on meteorites were formed, we can make very educated guesses on the conditions they were formed in. Most notably, outer space only contains very low concentrations of the compound most associated with oxidation on Earth: gaseous oxygen.
But this is not the main reason why the label ‘remarkably reduced’ is accurate here. We will immediately recognise the mineral in question as an ionic compound and given that it is made up of a metal and a nonmetal, we can equally immediately assign oxidation states as shown below.
$$\ce{\overset{+II}{Ca}\overset{-II}{S}}$$
There is no surprise about calcium which generally is found in its $\mathrm{+II}$ oxidation state. The surprise here is that sulphide is the anion which is a rare occurance. On Earth, most mineralic sulphur is oxidised and mainly to sulphate ($\mathrm{+VI}$). This is basically the extreme opposite and caused among others by the high atmospheric oxygen concentration.
Of course, the label ‘remarkably reduced’ can only be applied to some of the elements of the compound. Language-wise, we are looking at the oxidation state and merely stating that a state this low is uncommon.

The claim about no oxygen in space is highly debatable. Oxygen is one of the most abundant heavy elements floating around in the universe, but I guess it will be outcompeted by the reducing, still way more abundant hydrogen in planet-forming environments. Anyway, so by having this remarkable statement, we're still comparing with Earthly conditions. For chemists that might be ok, but I'm worried that this biases our view on minerals in space. Thoughts?

@AtmosphericPrisonEscape First, I never said ‘no oxygen’. Second, just because oxygen atoms are very abundant doesn’t mean $\ce{O2}$ molecules are. Third, I specifically said concentration which — I hope you will agree — is extremely low in space and still very low on meteorites and other small rocky space objects with little gravity.

On the topic of biased views, that could very well be the case; I don’t know what I think about it though.

Sure, but the concentration of anything is low in space. The relative abundance still favors hydrogen  reactions, just on longer timescales. Still minerals form, and reaction times can be rescaled with the proper physical parameters, so no point arguing about that. Thus the aim of my question was to view the space environment similar to a reducing environment, only on longer timescales for reactions.


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Magnetic moment μ approximation
I've been reading a bit about the magnetic moment (spin-only) $\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons
$$\mu_{s.o}=\sqrt{n(n+2)}$$
where $n$ is the number of unpaired electrons.
However in our lecture today we were using the approximation $\mu_{s.o} \approx n+1$. Is this  an acceptable approximation for the magnetic moment or should I stick to using the previous one.
Obviously using $\mu_{s.o} \approx n+1 $ is easier to use for calculations but I would like someone's opinion on this.

I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\sqrt{x(x+2)}$ about $x=\infty$.

This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\sqrt{n(n+2)}=\sqrt{n^2+2n}\approx\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1$$
By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.

In fact, using algebra, one can show that the error of approximation is bounded by ${1 \over 2n}$.

I would hesitate to call this a great approximation because it's really quite bad for small n which is the only part of the plot that makes physical sense since you're rarely going to have close to 10 unpaired electrons. Good answer though mathematically.

@jheindel true though I can see it be useful for back of the envelope/order of magnitude type calculations. And sure 0 and 1 aren't great, but for those its hardly worth making the approximation in the first place because you could probably just memorize those. By n=2, you only have about 6% error.

Ya true. I suspect this is not very useful beyond a coarse approximation anyways so approximating it more can't be too bad.

Using $\sqrt{x}-\sqrt{y} = { x -y \over \sqrt{x} + \sqrt{y}} $ we have
$\sqrt{n(n+2)} - (n+1) = -{ 1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )} $, so
$|\sqrt{n(n+2)} - (n+1)| \le {1 \over 2n}$.
The approximation is reasonable for large $n$.
Addendum: To see where the approximation comes from, note that
$\sqrt{1+ {2 \over n}}+1 + {1\over n}  \ge 2$, hence
$ {1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )}  \le {1 \over 2n}$.

@Tyberius: Good suggestion. Done.

To me this seems like a blind usage of the Taylor expansion.
Let's say we want to get a Taylor approximation of $\mu (n) = \sqrt{n(n+2)}$. Then we know we can expand any function $f(x) = f(x_0 +h)$ if $h/x_0 \ll 1$ into $f(x_0 + h) \approx f(x_0) + \frac{\partial f}{\partial x}(x_0) \cdot h $.
If I apply this to the function $\mu(n) = \mu(0 + n)$ then I get the approximate result $\mu(n) \approx \frac{n+1}{\sqrt 2}$. But here of course it is not true that $n/0 \ll 1$, so the assumption needed for using Taylor is violated.
However, if I plot both those functions in Wolframalpha they seem to agree OK enough for large n. So I guess a different approximation technique might have been used deriving this result, and it depends if you're working at large n or small n.

It's not really a Taylor expansion, so much as using the arithmetic mean of two numbers ($n$ and $n+1$) as an approximation to their geometric mean.

@psmears: You mean $n$ and $n + 2$, right?

Why do you assume that it's a blind usage of anything?

@DavidRicherby: The approximation apparently transforms a nonlinear function into a linear one. The simplest and most common way to do this is to do a Taylor approximation.

My (literal) emphasis was on "blind". Saying it's "a blind usage of Taylor expansions" means that you think that the teacher used Taylor expansions without giving any thought to whether they were suitable. Why are you accusing them of that?

@DavidRicherby: Math is usually not the strong side in a chemistry department.

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two questions about character tables
I have learnt about group theory before I learnt MO theory, so I have a mathematical understanding of those orthogonality theorems and character tables.
I will use the character table of $D_{4h}$ to ask my questions:

About basis functions
In the character table, we can see that the basis functions of $A_{1u}$ are not listed. Does it mean that any function can be a basis function of the corresponding invariant subspace?
For $E_{u}$, we know that the dimension of the invariant subspace is two, then does it mean that any pair of functions listed can be regarded as a set of basis functions?
About labels of irreducible representations
I learnt that those labels for irreducible representations are called Mulliken symmetry labels. But I don't understand what are the symmetries that we check? For instance, $A_{1g}$, is that we check the symmetries of $x^2 + y^2$ or $z^2$ (which are basis functions) that we get such a label?

For 1, yes, though it's really a matrix in $GL_{n}(\mathbb{C})$. For 2, the notation for Mulliken symbols is well documented, e.g., http://mathworld.wolfram.com/MullikenSymbols.html

@Zhe thank you for the answers! But for 1, I just can't build the intuition of any function can serve as a basis function, are there some materials that can help me make sense of this? For 2, I read the explanation,  but I think it doesn't answer my question, whose symmetry that we are checking? basis functions?

I'm confused by these questions. We're trying to describe matrices and their symmetries under the group operations. The functions just correspond to matrices that have the correct symmetry under the operations of the group.
As an aside, why can't functions serve as a basis? For well defined functions, you can Taylor expand the function, and $x^{n}$ for non-negative $n$ is your basis set. For periodic functions, you can expand into a Fourier series.

@Zhe ... I just realized that the choice of basis is arbitrary, thank you! But there's a quick question, if a function can be represented by the linear combination of all the basis functions of a point group, then are there some properties that the function should satisfy?

Yes, the functions should have the same symmetries as the matrices in the irreducible representation.

@Zhe which means the functions have to be invariant under all the symmetry operations of the group?

Let us continue this discussion in chat.

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Effect of the pH on the synthesized nanoparticles
I have synthesized gold nanoparticles in water through the seed-mediated thermal method by cationic additive $\ce{Ag+}$ as an assisted growth procedure [article]. The pH of the nanoparticles solution is about 2. I would like to reach the pH of the solution to 5.I found that when I use NaCl solution or buffer (such as sodium acetate) for adjasting pH, it cause the aggregation of nanoparticles.
How to increase the pH without any aggregation in the nanoparticles solution?

It should have, because with pH you change the electostatic forces between particles.
I guess you would increase pH adding more electrolytes and this might decrease double-layer thickness of the NPs, so repulsive forces would decrease and the NPs might agglomerate. It strongly depends on electrolyte conc.
Therefore I think you should rather try to synthesize them in pH5 then increase pH from 2 to 5.


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Cooling down and heating of solution unexpectedly during cation identification test
While conducting qualitative tests for identification of cations in the given salt, I encountered this unexpected temperature change. Following are the steps of tests:
Step 1: We made an aqueous solution of the salt and added dilute hydrochloric acid. There was no precipitate, which showed that Group 1 cations ($\ce{Pb^2+}$) are absent.
Step 2: We then divided the solution into two parts, and through one part, passed $\ce{H2S}$ gas. There was no precipitate, which showed that Group 2 cations ($\ce{Cu^2+}$) were absent.
Here, the temperature of the solution through which we passed the gas, rose above the room temperature. It could be easily felt while holding the test tube.
Step 3: To the other part in the test tube, we added $\ce{NH4Cl}$ to make the solution completely alkaline, and then added $\ce{NH4OH}$ till the solution was ammoniacal. We did not get any precipitate, which showed that Group 3 cations (viz. $\ce{Al^3+}$, $\ce{Fe^2+}$ and $\ce{Fe^3+}$) were absent.
It is here that we encountered the temperature change. Before adding $\ce{NH4Cl}$, the solution was at room temperature. However, as we kept on adding $\ce{NH4Cl}$ to make the solution saturated and alkaline, the temperature of the solution went down considerably. After some time, it was so cold that water molecules in the atmosphere had started to condense on the outer side of the test tube. The temperature still went down when we added $\ce{NH4OH}$.
Step 4: Now we divided the solution into two parts and through one part, passed $\ce{H2S}$. We got a white precipitate, indicating the presence of Group 4 cation ($\ce{Zn^2+}$).
It must be noted here that as soon as we passed the $\ce{H2S}$ gas, the temperature of the solution again increased to room temperature.
Step 5: To the other part of the solution, we added $\ce{(NH4)2CO3}$. We got a white precipitate, which indicated that Group 5 cations (viz. $\ce{Ba^2+}$, $\ce{Sr^2+}$ or $\ce{Ca^2+}$) was present.
It must be noted that the temperature remained almost same now.
Later, by confirmatory tests, we determined the salts to be $\ce{Ca(CH3COO)2}$ and $\ce{ZnSO4}$.
Can anyone explain what chemical reactions had occurred which were exothermic and endothermic, thus resulting in a change in temperature of the solution?

Just so you know, dissolution of ammonium chloride in water is endothermic; so adding it to an aqueous solution shouldn't be a surprise ;)

After the help by @paracetamol, I found that the process of dissolution of $\ce{NH4Cl}$ is an endothermic reaction, which results in the decrease in the temperature of the solution.
An experimental demonstration can be found here.
Greater information may be found here.
Important parts of the second reference:

When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.
$$\Delta G = \Delta H - T \Delta S\tag{1}$$
There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):
$$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$
Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.
$$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$
Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).
$$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$
All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table.
$$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline
\text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline
\ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$
So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.
Thus:

The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).
This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)

I suppose you didn't get the answer for the second bit of your question (why passing $H_2S$ gas through some solution caused a rise in temperature). If you wish, consider posting that as a separate question (since you've already found the answer to the first part); try to scrape together some thought on that and be sure to include it in your question (reduces chance of closure). Great find on the ammonium chloride dissolution bit! :)


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Is dissolution of H2S in water exothermic?
While doing certain laboratory tests for qualitative identification of cations, I found that on passing $\ce{H2S}$ in the solution, the temperature of the solution increased considerably. When the temperature decreased due to addition of $\ce{NH4Cl}$, the solution again attained room temperature in passing $\ce{H2S}$.
This rouses the question: is the dissolution of $\ce{H2S}$ in water exothermic? Or does it react with other components of solution to make an exothermic reaction?
The Wikipedia article stated only this:

Hydrogen sulphide is slightly soluble in water and acts as a weak acid (pKa = 6.9 in 0.01–0.1 mol/litre solutions at 18 °C), giving the hydrosulfide ion $\ce{HS^−}$. Hydrogen sulphide and its solutions are colourless.

The full experiment can be found here.

Koschel et al. (2013, p. 14486) report that

dissolution of gaseous hydrogen sulfide in aqueous phase is exothermic at conditions where $\ce{H_2S}$ is gaseous, i.e., at a temperature of 323 K and pressure 2 MPa, the saturation pressure $p_\text{sat}(\mathrm{H_2S})$ being 3.5 MPa. The dissolution process is endothermic at the medium and highest temperature where hydrogen sulfide is liquid and supercritical, respectively, the critical parameters of $\ce{H2S}$ being $T_c = 373.4$ and $p_c = 8.94$ MPa.

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IUPAC nomenclature of molecules containing alkoxide and carboxylate groups
I am studying for my A levels and recently came across the above-mentioned reaction.

I tried to think up a name like disodium 4-oxidebutanoate, but this is not based on any IUPAC rule. Internet searches have revealed no name for this, at least to me.
This compound is both an alkoxide ion and an alkanoate ion. How are such compounds named?

The principal characteristic group is expressed as a suffix. The name of the carboxylate anion alone (without the alcoholate function) would be ‘butanoate’. The name of the anion derived from the hydroxy compound (without the carboxylate function) would be ‘butan-1-olate’.
For the compound that is given in the question, one anion is chosen as the parent anion and the other one expressed as an anionic substitutent group. Since the carboxylate anion is the senior characteristic group, the parent structure is butanoate.
According to Subsection P-72.6.2 of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefix for the anionic group $\ce{-O^-}$ is ‘oxido’. (The prefixes is derived from the name oxide by changing the final letter ‘e’ to ‘o’.) Therefore, the name of the dianion that is shown in the question is 4-oxidobutanoate. The complete name of the salt is disodium 4-oxidobutanoate.


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Would alpha particles react to form compounds?
Alpha particles, to my understanding, are just $\ce{He}^{2+}$ ions. Therefore, if one were to emit alpha particles in oxygen, wouldn't the oxygen, ($\ce{O^{2-}}$), react to form $\ce{HeO}$?
If so, would the bond be strong or weak?
If not, why not?

There is no $\rm O^{2-}$ in oxygen. Then again, some particles will form, but not for long. Related: https://chemistry.stackexchange.com/questions/37260/why-cant-helium-nuclei-alpha-particles-react-with-fluorine/37262#37262

This is ionising "radiation".  Strips molecules of electrons and gives helium.

@IvanNeretin Oxygen has a valence of (-2, 6)

I don't know what does that mean (especially the "6" part) and how is it related to the question at hand, but I still stand by my statement.

Oxygen is $\ce{O2}$, there is no $\ce{O^2-}$ (oxide) in oxygen.

Do you want to radiate $^4_2He$ INTO oxygen, or do you say oxygen and helium should be involved in a nuclear decay? If the latter, please clarify if oxygen is the product or the educt of the nuclear decay.

@AtmosphericPrisonEscape Radiate into

There are two ways of looking at your question, from a chemical perspective or from a nucleosynthesis perspective. As stated in the comments and in StephenG's answer, helium is chemically inert and cannot easily form bonds. Instead He$^{2+}$ will strip electrons from any compound it encounters. This is because the ionization energy of He$^+$ is extremely high, more than 54 eV, so that a lot of energy is released when He$^{2+}$ recombines with an electron from another atom or molecule.
However, collisions between alpha particles can produce new elements under conditions that are encountered in stars like our sun. Stars consist mostly of hydrogen and helium at high denisty and temperature. If you consider the potential energy between two alpha particles you'll see that for the most part it is dominated by the repulsive Coulomb potential. However at very short distances, on the order of 10$^{-15}$ m, the Strong Force creates a very steep bounding potential and the two alpha particles form a $^8$Be$^{4+}$ ion. Under normal conditions, the Coulomb potential prevents nuclear reactions to happen, but under the conditions in stars, the kinetic energy of some of the alpha particles is equal to the ground state energy of the (instable) $^8$Be nucleus and quantum mechanical tunneling allows for the reaction to proceed (this is basically the opposite process of alpha decay). Although $^8$Be is radioactive and lives only very short, the density in stars is so high that some of the Be nuclei collide with another alpha particle. Again, by chance, the collision energy equals an excited state of a carbon nucleus (this is a nuclear excited state and involves the ordering of the protons and neutrons in the nucleus and has nothing to do with electrons) that can decay to a Be and He nucleus or to the ground state of the carbon nucleus. In this way carbon is formed in stars and this process is called the triple-alpha process. Further collisions with helium nuclei will produce heavier elements like oxygen.
It is interesting to note that the existence of the excited state of the carbon nucleus was postulated by the astronomer Fred Hoyle because it needed to exists to explain the large amount of carbon in the universe.

Alpha particles, to my understanding, are just He2+ ions. Therefore, if one were to emit alpha particles in oxygen, wouldn't the oxygen, (O2−), react to form HeO

If Oxygen itself were emitting alpha particles it would on longer be Oxygen - it be be Carbon (it would have radioactively decayed).
If you mean that an external source of alpha particles were directed at oxygen, could they interact, then yes.  But not to form stable compounds.
He is chemically inert.  There are few known Helium compounds and I don't recall ever hearing of any Helium-Oxygen compound (i.e. just Oxygen not other elements as well).
Any interaction would be scattering - basically a collision.  This can ionize the oxygen, but it won't form a stable compound with Helium so the oxygen will simply interact with "more willing" partners.  I think @ivan-neretin added a comment that linked to a related answer.

I meant in regard to alpha radiation hitting oxygen


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How can copper(II) sulfate crystallize outside a glass vial?
I'm making copper sulfate crystals in glasses. Within a few weeks, the copper sulfate seems to crystallize both inside and outside the glass:
(A few weeks after making a new solution)

(Yet another few weeks later)

How is that possible? The glasses haven't moved for all those weeks. Is that pure capillary action, or is there another phenomenon?

Indeed, capillary effect is enough. New portions of solution get higher and higher up, and deposit crystals to build more capillaries, much like a horde of guys with ladders storming some medieval fortress.


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t-butyl VS s-butyl
In the IUPAC system the common names for t-butyl and s-butyl are still accepted. Usually the "t" from tert- and the "s" of sec- is not included in the alphabetic ordering of substituents. If you have both these substituents on a hydrocarbon, which will get alphabetic preference in the ordering of the name? Does it matter?

sec- has priority over tert-, but the use of sec- should be avoided.
The 2013 IUPAC Blue Book has this to say:

P-14.5.3 When an alphanumerical ordering is required and Roman letters do not permit a decision for the order of citation, italicized letters are considered.

The text and examples are essentially unchanged from those of the 2004 Provisional Recommendations, which are freely available. Three examples are provided, one of which happens to be 1-sec-butyl-3-tert-butylbenzene:

However, the preferred IUPAC name (PIN) of this molecule is given as 1-(butan-2-yl)-3-tert-butylbenzene. An alternative name is 3-tert-butyl-1-(1-methylpropyl)benzene.
(Actually, the 2013 Recommendations do explicitly say that this molecule is not to be called 1-sec-butyl-3-tert-butylbenzene, but by one of the other two names instead. This is because the name sec-butyl is no longer retained. The 2004 draft, though, does explicitly place the sec- alphabetically before the tert-.)


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How is the alpha carbon defined in alkyl halides?
The α carbon refers to the first carbon that attaches to a functional group. According to this information, the alpha carbon in 2-bromobutane should have been the carbon next to the one with bromine. But in my book it says that the alpha carbon in that compound is the one with bromine atom. How is this possible?

In the case of simple substituents like halides, I think you're likely to have unambiguous notation if you use geminal and vicinal.

The notation using Greek letters, albeit very popular, is not unambiguous and therefore not recommended by IUPAC. Butan-2-one is point in case, having tow alpha carbons (C1 and C3). Enumeration is preferable. I doubt there are any official rules for using Greek letters.

The α carbon is the one directly attached to the functional group. There is no contradiction to your book’s claim that the α carbon of 2-bromobutane is the one directly bound to bromine.
This is due to the different definitions of functional groups and what they include. The keto functional group is defined as $\ce{-C(=O)-{}}$, i.e. it includes the carbon atom doubly bonded to oxygen. Halogen functional groups, for example the bromo functional group are defined as only the heteroatom, so in this case $\ce{-Br}$. Note that because the carbon is not included in the definition of this functional group, it is instead the α carbon.
This means that reducing the ester functionality of a β-halo ester will result in a γ-halo alcohol even though no group migrated.


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Difference in bond orders between formaldehyde and carbonate
So I am asked to find the bond order for both of them. I drew the correct Lewis structure, and there's one C-O double bond (and 2 other single bonds). I determined the bond order in CO3 (2-) to be 1.33, but why is it not the same for H2CO? Any hint would be greatly appreciated guys!!
Thanks in advance

Because the $\ce{C=O}$ and $\ce{C-H}$ bonds in formaldehyde are not identical.

All three oxygens in carbonate are equal and these three share a charge of -2. That means you can draw these three resonance structures, where each bond in 2 out of 3 has an order of 1, and in the third there's a double bond. The average of that gives you exactly 1.33.
But you can't draw any reasonable resonance structure for formaldehyde, which would share the partial charge of oxygen to hydrogens. There's only one other, which has a positive charge on carbon and negative on oxygen, but that won't make the greatest contribution, as carbon would have only 6 electrons


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Explaining color differences using bond conjugation
I'd like to ask about how the color differences between the dyes methyl violet and malachite green can be explained. Looking at the conjugated bonds of each, methyl violet seems to be more conjugated than malachite green with two $\ce{NMe_2}$ auxochromes, which causes a bathochromic shift, suggesting that methyl violet's absorption peak to have a longer wavelength compared to malachite green's absorption peak.
However, after having collected data using a spectrophotometer, methyl violet appears to absorb at a shorter wavelength ($\pu{593 nm}$) compared to malachite green ($\pu{623 nm}$), with both dyes solvated in $50~\%$ ethanol.
How can this color difference be explained?
Methyl violet:

Malachite green:

The auxochromes in MG are $\ce{NMe2}$ groups (as in MY). Conjugation is not the only factor, the HOMO-LUMO gap is also affected by the presence of lone pairs that can take part in the conjugation. Keep in mind that the structures presented are only one of many resonance structures.

What does MY stand for? Also, you say that lone pairs can also affect the HOMO-LUMO gap because lone pairs "can take part in the conjugation", but if you take them into account, wouldn't that also point to Methyl Violet having an absorption peak at a longer wavelength?

I meant to write MV (as in methyl violet).  I am not sure which way the N lone pair would influence the gap.

Oh, I see. I think amines produce a bathochromic shift, reducing HOMO-LUMO gap by extending the length of the conjugated system.

https://chemistry.stackexchange.com/questions/84720/explaining-color-differences-using-bond-conjugation


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How do we calculate a character table from its basis functions
This question is a actually a homework question, I'm given the basis functions of all the irreducible representations of $C_{2v}$ group, then I'm asked to complete the character table of $C_{2v}$ point group.
For example, for $A_{1}$ irreducible representation, I'm given the basis functions {$z$, $x^2$, $y^2$, $z^2$, $z^3$, $z(x^2-y^2)$ } which I can choose from one of them.
My questions are:

Why we need to complete the character table based on those basis functions? Simply by using the two orthogonality theorems of characters we can readily fill the character table.
How can I use the basis functions given to complete the character table? Should I represent those symmetry operations by using the basis (x, y, z), then transform the basis to those given to calculate the characters? I think it is too tedious.

The basis functions are not required. It just helps understand the irreducible representations.

@ I agree that the basis functions are not required. But if we have to use basis functions, should we do what I asked in the second question?

I can't think of another way to do it if you're not just using orthogonality...

I think now I answer my question.
Although in principle, I can calculate the full character table totally from the orthogonality theorems of characters. Namely, I need to use the two relations:
$$\sum_{k} N_k\chi^{(\Gamma_i)}(C_{k})\chi^{*(\Gamma_j)}(C_k) =\delta_{ij}$$
$$\sum_{\Gamma_j}N_k\chi^{(\Gamma_j)}(C_{k})\chi^{*(\Gamma_j)}(C_{k'})= \delta_{kk'}$$
But in practice, it is not always easy to use these two relations to calculate the characters. Even for $D_3$ group which has only two three irreducible representations, we need proper initial guess if we don't want to solve the set of equations.
On the other hand, if we know the basis functions in the first place, then we can effectively get the characters.
For example, if we want to calculate the characters of $A_1$ of $C_{2v}$ group. First, I need to choose a basis function. Let's say I choose $z$ as my basis function (I have listed the basis functions of $A_1$ in my question), since the irreducible representation is of one dimension, I simple need to see how the operations will change the sign of $z$.
There are four classes for $C_{2v}$ group, and the representatives are $E$, $C_2$, $\sigma_v(xz)$ and $\sigma_{v}^{'}(yz)$

$E$
It is evident that $E$ won't change the sign of $z$, so the character is 1.
$C_2$
The principle axis is $z$, so the character is also 1.
$\sigma_v(xz)$
It is a reflection across $xz$ plane, but it doesn't change the sign of $z$, so the character is still 1.
$\sigma_v^{'}(yz)$
It is a reflection across $yz$ plane, the character is also 1.

So, based on the above discussion of using basis function to calculate characters, we see that basis functions can be readily used for calculating characters. Although the example is really trivial (we can write down the characters of $A_1$ even without thinking), but the example indicates that if we need to calculate the characters of $A_2$, we just don't need to solve the set of equations.


text:
Most probable point for finding an electron in the 1s orbital of a Hydrogen atom
My professor says that the most probable point for finding an electron
in a 1s orbital of a hydrogen atom is at its origin.
He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.
At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

I can understand the reasoning he gave about the origin being the highest probable point for finding an electron, but then why are the graphs of the radial distribution function so?

It will depend on the hidden context of the question. If the question is "what is the most probable radius for finding the electron" then you will get a different answer as there is zero probability of the electron being at zero radius - the elemental ring (2d) or elemental shell (3d) has no space at zero radius -the wonders of mathematical imprecision. Further, it gets much worse as the dimensionality increase - everything tends to be at the 'one sigma' radius in high dimensional spaces.

The alternate context is that the professor wants to be clear that the electron is 'tightly' bound to the atom. It is always (on average) right there, aligned with the atom. They are together  as one. A comparative statistic is say WWII bomb (c.f. electron) aiming - was the bomb cluster on target, and what was the spread of the bomb cluster. The electron is on target, and has spread - they are distinct measures.

The most probable point vs most probable distance are two different things

I provide here a few supplementary comments; the major part of your question has been answered by DSVA.

For the electron's position in a 1s orbital, the probability density $\rho(r)$ is maximal at the origin, whereas the radial probability density $p(r)$ is maximal at the Bohr radius.
probability versus probability density: The probability density integrated over a volume yields a probability. We cannot talk about probabilities at a point for a continuous system because a point has measure zero---that is, if we make a measurement of a random variable for a continuous system, we will never get exactly the value represented by the point. Instead, we need to consider a higher-dimensional object, which has non-negligible measure. A natural choice is a shell centered around the origin with radius $R$ and thickness $\Delta R$, assumed small. We can find the probability that the electron is located within the shell as $$P(R < r < R+\Delta R) = \int_\text{shell}\mathrm{d}\mathbf{r}\,\rho(r) = \int_R^{R+\Delta R}\mathrm{d}r\,4\pi r^2\rho(r)\approx 4\pi R^2 \rho(R)\,\Delta R.$$ Motivated by this calculation, we define the radial probability density $p(r) := 4\pi r^2\rho(r)$, which is a more important measure of where the electron actually is, because of the radial symmetry in the system---we are more interested in the distance between the electron and the origin than in the position vector of the electron.
probability density versus radial probability density: The probability density takes as its fundamental unit a point---imagine a cube of infinitesimal volume, as your professor uses. The radial probability density takes as its fundamental unit a sphere---imagine a shell of infinitesimal width, as before. We will restate the contents of the first bullet point in this context: The probability of finding the electron in a given infinitesimal cube is maximized for a cube centered about the origin. The probability of finding the electron in a given infinitesimal shell is maximized for a shell at the Bohr radius. In the latter case, although the probability density is lowered for large $r$, the increased surface area means that there is more space we can hope to find the electron in, and the competition of these two factors result in a maximum for the radial probability density at the Bohr radius.

+1! The claim from the title of the question simply makes no sense at all: the probability of finding an electron at a certaint point is effectively zero. The position of a particle is described by a 3-tuple of real numbers, the set of reals is uncountable, and the probability that a random variable takes a certain value from an uncountable set is 0.

It makes a difference if you look in a point or in a volume element.
For the radial distribution we sum up the probabilities within (very thin) shells at different distances from the core.
If the radius get's bigger the volume in this shell gets bigger.
Remember that the surface of a sphere goes by $4 \cdot \pi \cdot r^2$ so the volume of the shell is essentially for all practical purposes $4 \cdot \pi \cdot r^2 \cdot \Delta r$.
At the core the radius is zero which means the volume is zero and thus the probability is zero too.
At the same time the probability of finding the electron in a point at a certain distance goes down, but the volume goes up.
Because of the scaling of the volume and the "decay" of the probability to find it in a point further away it turns out that there's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance).

So whats the significance of the graph of the radial distribution function? And isn't it true that when r tends to zero, we get the Origin?
And what's the physical significance of the different conclusions we get with probability density and probability?

The radial distribution is usually the interesting part since it actually describes the probability of finding the electron somewhere at a certain distance around the nucleus, which is a useful information for the chemist. But from a physics view there's actually no difference, it's just two ways of representing the same thing. And yes, r=0 is the origin which we usually set right at the position o the core.

So, can we a take a volume element of radius dr and consider the probability of the electron at the origin?

He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.

Not exactly.
The answer by @dsva explains why this is wrong and I'll just expand that in a minute, but first note that it's easy to see why this is wrong.
If the electron was most likely to be at the origin it would be likely to interact with the nucleus, something we expect not to happen.  In fact there is a small chance of this because the nucleus is a non-zero size.
Essentially he's missing the need to sum over the shell, and the shell's volume is zero at the origin because of the volume element $dV = 4\pi r^2dr$ with $r=0$.
The probability of finding the electron in a region of zero radial size is zero, but we can evaluate the relative probability for two radii.  For 1-s orbitals that is :
$$\frac {r_1^2 e^{-{r_1}/a}}{r_1^2 e^{-{r_1}/a}+r_2^2 e^{-{r_2}/a}}$$
and
$$\frac {r_2^2 e^{-{r_2}/a}}{r_1^2 e^{-{r_1}/a}+r_2^2 e^{-{r_2}/a}}$$
For $r_1=0$ we clearly get a relative probability of zero compared to any $r_2\neq 0$.

At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

This is because of what an expectation value is and how it is calculated in QM.
An expectation value is an average over the entire space of the effect of an operator on a wavefunction.  It is not simply the operator multiplied by the probability density.
In short an expectation value can be considered an average over all space, whereas the probability density is not an average over all space.

The professor takes a small cubical volume for the origin. I don't know why?

That's because to get a probability from the wavefunction you have to use a non-zero volume.  The probability of the electron being in a volume of zero size is, naturally, zero.  Are you sure the prof is using a cubical volume ?  Do you mean spherical, which would be more natural, as the wavefunction is in spherical coordinates ?

Just to explain his statement regarding the most probable point at origin he takes a small cubical volume element at the origin.


text:
How do you count sp3 hybridized configuration electrons for methane?
I've been reviewing my organic chemistry and upon reviewing sp3 hybridization have become confused.
For methane CH4, the electron clouds rearrange into sp3 hybridization configurations of 1s2 2sp3 2sp3 2sp3 2sp3; with 1 electron in each of the 2sp3 orbitals to equal 6 electrons all up.
When bonded with hydrogen, do we count each of the 4x hydrogen electrons as belonging to the carbon like we do for the hydrogen (hydrogen wants to reach 2 electrons, so we count the carbon's electron as belonging to the hydrogen)? If so this would result in carbon having 10 electrons (1s2 = 2) + (2sp3 = 2) + (2sp3 = 2) + (2sp3 = 2) + (2sp3 = 2) = 10 all up despite the fact that it wants to have 8 electrons.
Could someone please point out where I'm going wrong here?
I have been following Khan Academy for revision and here is where I got the information that has confused me: https://youtu.be/u1eGSL6J6Fo?t=10m35s.

You don't count 1s2. These are deep down below and play no role.

The basic way to think about it that all those s- and p-elements want to have as much electrons in total, as noble gases do. 10 electrons in total is exactly non-bonded neon.

Thanks for that, exactly what I needed and suddenly it all makes sense again :).

When we want to specify the type of hybridization (sp3, sp2, sp) we count the number of regions with high electron density (bonds or lone pairs) and attribute it to the valence sub-shells such as 2s and 2p in the case of carbon. we can call s a sub-shell or orbital,  1s lies outside the valence shell so it is not considered in bonding.
So if we have 4 regions (bonds) as in methane they are made by the contribution of the four orbitals (2s, 2px, 2py, 2pz), while in case of ethene and ethyne (acetylene) the density regions are lesser than methane due to condensation of electron densities between carbons as double and triple bonds.
But still we have to mix from the 2 sub-shell types of the valence shell, S and P, so we have sp2 and sp hybridizations, respectively.
The electron density concentrates between the carbons in ethene and acetylene.

Please use bolding more sparingly. See here


text:
Are there any guidelines for what constitutes an “organic acid” regarding storage requirements?
I am moving a large amount of chemicals to a new lab and I’m curious about any guidelines for chemical separation. I’ve seen suggestions that organic acids and non-organic acid be stored separate from one another, and away from general chemicals. However, where would a long chain fatty acid lie, general chemical, or organic acid? When they get long enough, they become a solid at room temperature. Does storage requirements change for organic acids which are solid?

@Jan is correct.
Organic vs. inorganic is not the main concern for acid storage.  The strength and oxidizing ability of the acids is a much more serious issue.
Your institution/company may already have internal safety guidelines that you should follow.  If not, your country/province/state will have occupational health and safety regulations that you will need to respect.
You can find the American Chemical Society's safe storage guidelines here.

In general, I would not pay too much attention on the difference between organic acids and inorganic ones in principle. Rather, the deciding factor should be how strong the acid is (i.e. how bad things are if it is spilt or the bottle breaks) and whether or not it produces acidic fumes.
For a general rule, see if the acid has been rated corrosive. If so, store in a separate acid cabinet. This will include inorganic ones like hydrochloric or sulphuric acid but also organic ones like acetic acid. It will ignore weak inorganic acids and weak organic acids alike because they can easily be stored with the general chemicals.

$\ce{HNO3}$ and $\ce{HClO4}$ and their salts should be stored as far as possible from any organic compounds, because oxidizers should be stored away from reducers.
That said, some organic compounds might produce toxic compounds in strongly acidic or basic solutions. Formic acid, in particular, decomposes in concentrated sulfuric acid, producing toxic and flammable carbon monooxide.
===
As for storage suggestion.
Concentrated volatile acids ($\ce{HHal}$, $\ce{HNO3}$, formic and acetic acids) will leak through most containers, causing rapid corrosion of the surroundings. Same for bromine, ammonia and to lesser extent for some organic solvents. Storing them in separate location with very good ventilation is common sense. Furthermore, many concentrated acids might produce toxic fumes on reaction with rather innocent salts. Mixing nitric acid and table salt without reason is not advised. Mixing strong acids with each other isn't advised either, because again toxic fumes might develop.
===
In general, you should either ensure that any possible mix in the batch you move or store does not explode or produce large amount of toxic fumes (especially if they are not irritating) or take special cautions to ensure that the containers will not break even under high stress.


text:
Do you know book's name under the topic of anti corrosion materials and binders?
I need book's name talking about anti-corrosion materials or binders to complete my graduation project

I'm voting to close this question as off-topic because we are not a reference request service.

Although closed ; I will recommend Corrosion Handbook , Ulhig, pub. John Wiley & sons , 1948. Old , but if one understood everything in this one book you would be an expert today.


text:
Why do electrodes need to share an electrolyte in a lemon battery?
From what I understand of a lemon battery, there are two electrodes one zinc, and one copper. The electrodes are then both placed separated in the same electrolyte, in this case a lemon. The Zinc dissolves into the electrolyte oxidizes and loses 2 electrons. Those 2 electrons stay on the zinc electrode. On the copper electrode 2 h+ combine with 2 electrons from the copper and then form h2. the h2 bubbles up and leaves the electrolyte. The extra electrons from the zinc then move to the copper when there is a connection between the anode and cathode. This reaction continues until the zinc is too dissolved or if there is not enough h+ in the electrolyte to take electrons from the copper.
If my understanding is correct (please correct me if I'm wrong), then why can't each electrode have is own electrolyte, completely separated. For example the anode in half a lemon and the cathode in the other half. Both half's separated by air.

Electrons don't flow through air easily. You can't say they don't at all because there is lightning.

The circuit won't be complete.

I don't understand how this post was closed for being off topic. I thought I demonstrated the underlying chemistry quite well. I always knew circuits needed to be closed, but wondered why in this particular case it needed closing as I felt the hydrogen gas at the time bubbled balancing out the two sides, allowing the reaction to continue. I answered my own question eventually.

Like @Pritt said, the circle has to be closed, 2 electrons are traveling from one half-cell to the other, means 2 negative charges are missing, the half-cell needs to replace it by other 2 negative ions or send 2 positive charges, since charges cannot exist separately.

Thank you guys for the help. So from what I understand, if there is no way for the positively charged zinc ions to leave the zinc half cell then that half cell gets too positively charged. This means the electrons will not want to go the other half cell, stopping the reaction.


text:
Formula of the compound based on occupancy of lattice voids

Atom of element $\ce{B}$ forms HCP lattice and those of the element $\ce{A}$ occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements $\ce{A}$ and $\ce{B}$?

My doubt is that since the element $\ce{B}$ is in HCP, the number of octahedral and tetrahedral voids will be 6 and 12, respectively, which means that $2/3 \times 12 = 8$ tetrahedral voids occupied by $\ce{A}$.
Since $\ce{B}$ is in HCP lattice, there must be 12 voids, i.e. the ratio will be $8 : 12 \implies  2 : 3$, so $\ce{A2B3}$ must be the answer.
But according to NCERT textbook (example 1.2) the answer is $\ce{A4B3}$.

Since B forms an HCP lattice, it's z=6 (z is the number of atoms power unit cell of the lattice) then the ratio comes out to be 8:6 that gives us A4B3.

Let no. of atoms of B used in hcp lattice =n
Number of tetrahedralvoids=2n
Number of atoms of A=(2/3)2n=(4/3)n
A:B=(4/3)n(n)
formulaof the compound is A4B3

NCERT  answer is correct


text:
Getting Molecular Orbital contributions to the ESP fitted charges
For a single determinant wavefunction, the electron density is the sum of the densities arising from each molecular orbital. The molecular Electrostatic potential $\phi_{mol}$ is calculated from the total density and therefore, at a given point $\vec{R}$, its value is given by the sum of all the contributions arising from each MO.
\begin{equation}
\phi_{mol}(\vec{R})=\sum_{i=1}^{N_{MO}} \phi_{i}(\vec{R})
\end{equation}
Atomic partial charges can be obtained by fitting the total molecular ESP in different flavors and with different constraints, however I am interested in knowing for a given atom $I$, the charge contribution of each molecular orbital to the total ESP-fitted charge, namely the elements in the summation below
\begin{equation}
q^{_I}=\sum_{i=1}^{N_{MO}} q_{i}^{_I}
\end{equation}
Is there anyone knowing a software for wavefunction-analysis that can do this from a wfn file or a way of getting the solution without writing a piece of code from sctratch?


text:
Isolation of salt from solution
I just want to clarify my understanding of whether metathesis can be used to form other products by combining an acid and a base (do special rules apply whether its strong/weak acid/base)?
Say I want to form some $\ce{Na2CrO4}$. There's many ways to do this but the most obvious one I can think of is:
$$\ce{2NaOH(aq) + H2CrO4(aq) ->Na2CrO4(aq) + 2H2O}$$
Can you then just filter out the $\ce{H2O}$ to get your desired product? How would you do so? I'm used to looking at these in terms of net ionic equations where two aqueous reactants for a solid precipitate but since everything here would cancel out (except water) is this even a viable possibility?

That's going to be difficult since $\ce{CrO4^{2-}}$ does not have a charge of -1. Other than that, you're basically talking about evaporating the solvent, which is definitely a thing you can do. Ideally, it's not water, because evaporating water is a pain. Though lyophilization might work.

As Zhe pointed out, you'd need to evaporate not filter the water out. But I'd agree that this is a metathesis reaction since you could isolate the two different compounds by physical means. // Evaporation with heat is a messy process since as the liquid surface disappears the solid starts to splatter.

This is not what I thought when I read ‘metathesis’ …

You obviously cannot ‘filter water out’ as water is a liquid at STP and the solvent in which your reaction occurs in. However, you can evaporate your solution whereby the solvent gets removed and the lower vapour pressure contents (mainly solids or oils) will remain. While water can be more difficult to evaporate at a standard rotavap, it is by no means impossible.


text:
What is the physical state of the gas adsorbed on a solid adsorbent?
I am studying surface chemistry in high school I have this doubt and I am unable to find a answer your help is appreciated
 I searched for the answer on google I asked my chemistry teacher he said the question is beyond high school's curriculum and I even asked some of my friends but all in vain.

If you could edit the question such that the question includes how you tried to solve it, please do so. In its current state, the question is likely to be closed.

I have edited the question is it ok now

Who said that everything must have a physical state?

The old standard characterization was solid, liquid, gas, then plasma was discovered, and so on. So there all sorts of add characterizations. A "gas adsorbed on a solid adsorbent" is redundant. A chemist would normally just say an adsorbed gas. If the gas is in a liquid then it is dissolved. If the gas is in a gas, then it is a mixture. So the only way a gas can be absorbed is onto a solid. There isn't any better way to characterize an adsorbed gas than that specific phrase.

Also see wikipedia article https://en.wikipedia.org/wiki/List_of_states_of_matter

@IvanNeretin Every kind of particular matter is in some physical state. Otherwise it'd have to be in some unphysical state, which sounds unnatural to me. ;-)

@Jan edited the question.

Well, it usually depends on the temperature: if temperature is too low, molecules cannot move so freely so they are localized on top of the solid without much movement.
However, in general cases the adsorbed gas behaves like a liquid in two dimensions. Near the surface, the gas molecules are 'trapped' in the force field of the solid, so they don't escape so easily. Then the freedom of movement occurs only in two dimensions (the molecules travel on the surface).
I made some time ago a movie calculating (using molecular dynamics) movement nitrogen molecules adsorbed on a graphene surface at 77 K. This can give you a clear image of what happens on the surface. As you can see, the molecules can travel on top of the graphene in two dimensions (we can call it a 2D phase). This movie is the case when still a monolayer of gas does not complete. There can be cases with less molecules, say, less pressure of nitrogen in the gas phase; or more molecules.
The physical states can be observed more clearly in phase diagrams of the adsorbed surfaces. Typical examples shows a plot of the adsorbed phase transitions in a P vs. 1/T plot. This topic is extensively covered in the nice monograph on Physical Adsorption by Bruch, Cole and Zaremba, for example.

Is this movie an animation or a simulation?

It is a molecular dynamics simulation. Simulation was done in HOOMD and visualization with VMD.

Does the adsorbed gas behave like a 2d gas.(See Wikipedia for what's 2d gas)

@Shub
Yes! This is correct! Actually, it can behave as a 2D gas (molecules move and diffuse a lot in the surface) or like a 2D solid (just localized molecules). It generally depends on the temperature. You can even find phase diagrams for these 2D/3D transitions. (I will edit the answer to reflect this). Thanks!

Since solid means ordered structure and when gas gets adsorbed orders itself on the surface (at least partially) and is not free in volume, I would consider it a solid.

Gas molecules adsorbed on a surface are typically highly mobile in the plane parallel to the surface.


text:
How can you inhibit crystallisation of hot superconcentrated magnesium sulphate
I’m trying to create a gel like solution of primarily magnesium sulphate heptahydrate or Epsom salt and water.  It’s simple to create a hot super concentrated solution but when it cools it’s obviously begins to crystallise. Continuous stirring as it cools and cooling quickly creates a slush of tiny granules which eventually settle to the bottom of the solution like sand in water if too thin.
How can a thick concentrated smooth liquid be created?
I’ve looked into ways of inhibiting crystal formation but no good examples.  Maybe an impurity if some kind and then possibly a thickener if the mix is still watery.

First, your question involves inorganic chemistry, not organic, unless you're talking about the unmentioned gelling agent.
Second: Impurities serve as a source of nucleation, and stirring increases crystallization. If you leave the supersaturated solution in a clean container and allow it to cool slowly, e.g. in a Styrofoam cooler chest, it may take a long time to form crystals. If you use some thickener, such as PEG or gelatin, it might take longer yet. Left undisturbed, though, supersaturated $\ce{MgSO4}$ can persist for years. There are videos of it crystallizing after a seed is added.

Thank you for your assistance. I’m obviously not a chemist so your help with this is more than useful. You’ve given a direction so thank you again. Much appreciated.

Just a quick update. I found that cooling this super saturated solution slowly did leave some solution as liquid but it actually caused much larger crystals to form also. It would be fantastic if it could all be kept as a liquid solution. I read somewhere that adding cellulose can do this. I don’t need a good safe solution in particular, anything within reason could be added.

Amendment for above “Food safe”

If the supersaturated solution contains no seed crystals, i.e. it is heated until all the Epsom salt dissolves, it should not recrystallize until "provoked" (e.g. shaking, stirring, scratching the side of the container, dropping in a seed crystal).


text:
How does hemoglobin adhere to cotton?
Blood stain on cotton t shirt.How exactly does hemoglobin connect to the fabric? The blood stain if washed in warm water can become very long lasting,being able to remain after many washing cycles.
From chemical perspective,how does it stick to the cellulose based fabric as found in cotton t shirts?

Denaturated proteins act as glue.

Building on what @aventurin said: There's a load of other (real) proteins in blood, which exist in amounts far greater than hemoglobin. Large sized molecules give rise to a whole host of reactive fragments that probably end up reacting with some portion of cellulose (not sure, but the three $-OH$ groups on each cellulose monomer sounds like a good place to start). Good question though!


text:
Can the experimental mole fraction equal one?
I recently did an experiment on the temperature dependence of phase equilibrium acetone $\ce{C3H6O_{(l)}} \to \ce{C3H6O_{(g)}}$ I'm at the stage where I need to find the experimental mole fraction $\chi_e$ which I did by plotting a graph (see below)

From this graph I can simply find $\chi_e$ by finding the slope. I took two points $(56,50)$ and $(66,60)$ and found $$\chi_e = 1$$
This doesn't look right to me, so I'm just asking if someone can let me know if what I've calculated is correct or way off the mark
EDIT: I just calculated the theoretical mole fraction there and got $$\chi_T = 0.94$$
The moles of the non-volatile solute and pure acetone were known.

Hmmm... neither of the two points used in your calculation are on the line you've shown.

$(66,60)$  looks like it is, no? I've changed my other point and I'm getting $\chi_e > 1 $ which is defintely bad.


text:
Has Crabtree's photochemically-activated mercury-catalysed hydrocarbon activation process ever been used commercially?
In the 1980s, while searching for new hydrocarbon activation catalysts, Crabtree's research group accidentally found that mercury vapour activated photochemically could perform some interesting and unusual alkane activation reactions yielding unusual products.
Some of the work is summarised in this Journal of Chemical Education paper from 1988. The reaction seems to be able to produce some unusual (and hard to get) dimers of hydrocarbons and is also tolerant of other functional groups (eg alcohols, amines) and can also generate cross coupled products (eg it will couple alcohols with hydrocarbons).
At the time it was thought to be an interesting curiosity but with some prospect of yielding easy routes to some hard-to-get targets.
Has it ever been used in a significant synthesis? Or for any industrial processes?

You will find that in general,  use of photochemistry in commercial processes is rare as so few companies offer it as a service (as a CMO) and photochemistry is a pain in batch operations (due to efficiency and fouling).  Also, due to trade secrets, certain companies might only release this technology information under confidentiality agreements. For these reasons, it would be hard to know for sure, but the probability is very small.


text:
Intramolecular lactonisation in an unsaturated carboxylic acid
I am having some trouble figuring out how this reaction proceeds:

Attempt 1: The most obvious thing to do is deprotonate the acid group. I tried using the carboxylate group to attack the alkene, leaving the molecule negatively charged and not really knowing what to do next! If this happened, water would certainly protonate it and then it wouldn't be unsaturated, so it can't be that simple.
Attempt 2: Using the alkene to deprotonate the acid, there will be a positive charge at (one of) the alkene carbons. Then I tried using hydroxide to add in an alcohol group but to no avail.
Is there something special going on because of the cyclic structure?

This doesn't make any sense. How did you lose an equivalent of $\ce{H2}$  from the reactant?

The reaction as you are given turns up no results in SciFinder, meaning it has not been reported in any scientific journal. It is thus the imagination of whoever set the test.
However, closely related reactions are reported and these may come as no surprise to the advanced chemist. They mainly differ in that iodine $\ce{I2}$ is added to the reaction mixture. The mechanism is called iodolactonisation.
Basically, iodine is attacked by the double bond to form an iodonium ion (whether or not this is actually a three-membered ring shall be deemed irrelevant in the context of this answer). This is then rear-side attacked in an intramolecular $\mathrm{S_N2}$ reaction by the deprotonated carboxyl group to form the five-membered lactone kinetically. In a subsequent step, the iodo substituent is removed in an $\mathrm{E2}$ type reaction with a base to regenerate the double bond; due to steric constraints relating to the required anti-periplanar transition state, this newly-formed double bond is shifted along the six-membered ring.

This is described for your compound multiple times in the literature using different bases, for example by Khanjin et al.[1] They use DBU as a milder alternative.
Reference
[1]: N. A. Khanjin, J.P. Snyder, F. M. Menger, J. Am. Chem. Soc. 1999, 121, 11831–11846. DOI: 10.1021/ja992453d.

Makes a lot more sense. I wonder if steps 1 and 2 are under Curtin-Hammett conditions or if there is good selectivity in the first step to put the iodinium anti from the carboxylate...

my question sheet had an error! thank god, otherwise I would have been very confused! thanks for clarifying the mechanism with halogen.

@Zhe I guess we can argue with steric hindrance of the carboxy group to direct the attack of iodine to the back side. Or, we get equilibrium conditions.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Solubility of cholesterol in water - data not matching up?
I found two papers with solubility data for chol in water.
One says 1.8 mg/L at 30 deg C
Other says 10^-9.3 (mol fraction basis) at 25 deg C
However when I tried to convert the latter to mg/L I get:
So for 25 deg C = 10^-9.3 = 5E-10 mol/mol
5E-10 mol chol/ mol total = 5E-10 mol chol/ (1-(5E-10)) mol water = 0.99
(5E-10 mol chol/ (1-(5E-10)) mol water) *(386.65 g/mol chol / 18 g/mol water) =1.074E-8 g chol/g water
1.074E-8 g/g * (1000 g/L water) * (1000 mg/g chol) = 0.0107 mg chol/ L water
Is there a mistake in my conversion? They seem way off for only being 5 deg difference.
Thanks!

A quick lookup tells me that mol fraction=moles of solute/total mole.
Let us consider the mole of cholesterol to be x.
In 1 litre of water, mass of water is 1000 grams.
Assume Mr of water to be 18 (rounded up). So, moles of water=1000/18=500/9
=> 10^(-9.3)=x/{x+(500/9)}
Solving this find x gave me x = 2.784...E-8
So, mass of cholesterol in mg is (x)(386)(1000).
This is 0.01074... mg.
As you can see my calculations matched up with yours.
I was off to Google and https://pubchem.ncbi.nlm.nih.gov/compound/cholesterol#section=Solubility
 states the solubility at 30 degrees Celsius is 0.095mg/L. So, I'd say one of those papers might have stated a value under different conditions or a wrong value.


text:
Python command line tool to get chemical structure
In computational chemistry, we need to provide an starting molecular structure to start our calculations. Is there any way to get that from a chemical database? Say, I want to run H2, it would call something like
url="www.chemicaldatabase.com"
H2=(url,H2)

or it can use smile/other standard symbols for chemical representation.

Would something like http://blog.matt-swain.com/post/16893587098/chemspipy-a-python-wrapper-for-the-chemspider work?

For most computations you need 3D structures, so you either need to pull 3D structures directly (example: PyMol and the protein database) or convert identifiers like SMILES into 3D structures, not sure if there's a way to do that. Maybe ObenBabel can do it?

@chipbuster: Thanks for the link, but it doesn't give you any coordinates.

@DSVA: Yeah, it would be very hard to list 3D molecular geometry into a database as there are tons of isomers and rotamers involved. I'm running some well known molecular structure using DFT. I don't want to spend time making those initial structures. In ASE, they have a molecular library for some structures, but I am looking for a more detailed database.

Have a look at molget https://github.com/jensengroup/molget

There are multiple approaches in Python.
My suggestion would be to use a cheminformatics library like Open Babel or RDKit to convert from SMILES (for example) to 3D coordinates.
If you want to grab from chemical databases, I can suggest two approaches:

CIRPy - Uses the NIH chemical resolver to convert from names, SMILES, etc. into 3D structures.
Webel - This is a web-based cheminformatics tool. I've used it in the past, but I'm not sure if it's still maintained.

There are other databases, including my PQR and PubChemQC that offer QM-optimized geometries.
The catch with databases is that you might want the geometry of a molecule that's not in the database. In that case, Open Babel or RDKit is a better solution.
One other caveat. Nothing I've indicated above does a very good job with metal-containing species. If you want ferrocene, that's a trickier problem with current solutions.

And as Jan noted above, if you don't care about Python, molget is great.

If you are after quantum chemistry Psi4 offers an all-in-one solution:
import psi4

mol = psi4.geometry("""
    pubchem:Water
""")

mol.print_out()

scf_e = psi4.energy("SCF", molecule=mol)

I’m curious - since this searches PubChem, what happens if it finds multiple matches (eg ibuprofen) - does this return a list?

It plays the "I am feeling lucky" google card.

One other related question - some PubChem searches return molecules with undefined or unknown stereochemistry. Same thing? Would you get a 2D molecule?

AFAIK an error should be thrown. Psi does not have the right tech to do 2D->3D guesses. We use this a lot in educational settings, but it probably does not pass muster in a professional one.

Can psi4 uses SMILES as input?


text:
Keeping an open, partially empty soda bottle bubbly longer
I remember as a kid being able to get a little pump which you could screw onto a half empty soda bottle (I believe the bottles were glass at the time) to pump air into it to keep it fizzy for longer.  I surmised or was told that this worked by increasing the density of the air, thus making it harder for gas to leave the liquid and enter the air.
I now find myself with partially empty soda bottles.  Can I keep the fizz longer by squeezing the air out of the bottles and screwing on the lid?    (My guess would be that this would work if the compressed bottle is altered such that it does not try and reform back to its original round shape, but if it would pop back save for the lid being screwed on, I'm actually going to make it loose its fizz slightly faster ?)

The fizz is dissolved carbon dioxide (or carbonic acid with which it is in equilibrium) which is put into the bottle under quite some pressure (around $\pu{120 kPa}$ at $\pu{4 ^\circ C}$ or $\pu{250 kPa}$ at $\pu{20 ^\circ C}$; from a first glance on the google results).
$$\ce{CO2 (g) + H2O (aq) <=> H2CO3 (aq)}$$
Because the bottle (in first approximation) is a closed system, there is an established equilibrium between the "air" in the bottle (gaseous $\ce{CO2}$) and the dissolved $\ce{CO2}$. Once the bottle is opened the equilibrium is disrupted and the dissolved carbon dioxide is no longer as soluble as before.
If you close the bottle again and use the little pump you essentially add pressure again, shifting the equilibrium back to where more dissolved carbon dioxide is stable. It won't be as effective as the factory produced bottle, but it will slow down loosing the fizz.
The volumes also play some role here, ideally the aqueous phase should be much larger than the gaseous phase, as you need more molecules producing the pressure if you have more space. An half empty bottle therefore is not ideal to start with.
Squeezing the air out of the bottle would in the worst case result in the opposite effect that you want, in the best case it won't change much. As you correctly remarked, the bottle reforms which means that you would have created lower pressure, essentially dragging out more carbon dioxide molecules. If that was not the case, you would reduce the volume of the gaseous phase, which would probably help a little. Without adding high pressure though it will still loose most of the fizz over time, until equilibrium is reestablished.
In principle there is not really very much you can do. Keep the bottles refrigerated (as carbon dioxide is more soluble at lower temperatures) and try to finish them quickly. Alternatively, get yourself a soda maker and enjoy all the fizzy drinks in all the customised quantities whenever you want. (I guess, this is not a plug for soda maker machines, I personally prefer non fizzy drinks.)


text:
Calculating the heat of formation of propane given its enthalpy of combustion
Given the equation $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$ and that enthalpies of formation for $\ce{H2O (l)}$ is $\pu{-285.3 kJ/mol}$ and $\ce{CO2 (g)}$ is $\pu{-393.5 kJ/mol}$, and the enthalpy of combustion for the reaction is $\pu{-2220.1 kJ/mol}$, I need to find the heat of formation of propane.
My initial idea was to use Hess's law, and I got
$[3(-393.5) + 4(-285.3)] - [-2220.1] = \pu{-101.6 kJ}$
I then doubted myself because Hess's law gives the $\Delta H^o_\text{rxn}$ which is different from the $\Delta H^o_\mathrm{f}$ for  $\ce{C3H8}$ that we are trying to find.
So I used another approach using the individual equations for the formation of $\ce{H2O, CO2}$, and $\ce{C3H8 + 5H2O -> 3CO2 + 4H2O}$ and combining the equations to get the equation for the formation of $\ce{C3H8}$ (which is $\ce{3C + 4H2 -> C3H8}$). After manipulating the equations, I got $+2220.1 - 1141.2 - 1180.5 = \pu{-101.6 kJ}$.
I got the same answer, and I want to understand why Hess's law still works even though the equation I used in Hess's law had nothing to do with the formation of $\pu{1 mol}$ of $\ce{C3H8}$. I understand my second method, but why does the first method work?

That's the very gist of the Hess's law, or more broadly, of the fact that enthalpy is a function of state: the path is not important, only the destination is. (Fix your equation, BTW.)

Yes, fix your equations. Sums of numbers do not equal kilojoules!

The equation I had earlier in my first method was wrong. Using Hess's law, we know the change in enthalpy of combustion to be $\pu{-2201.1 kJ/mol}$. Thus:
$$\Delta H^\circ_\text{rxn}=-22201.1 = [3(−393.5) + 4(−285.3)] − [5(0) + x]$$
where $x$ is the heat of formation of propane.
Solving the equation, we get $x = \pu{-101.6 kJ/mol}$.

by combining 3 equations $$3CO_2 + 4H_2O -> C_3H_8 + 5O_2 \Delta H = 2220 kJ$$
$$3C + 3O_2 => 3CO_2 \Delta H = 3(394) kJ$$
$$4H_2 = 2O_2 => 4H_2O \Delta H = 4(286) kJ$$
$$3C + 4H_2O = C_3H_8 \Delta H = 2220 + 3(394) + 4(286)$$ why i was wrong?


text:
How to obtain the distance of maximum probability of finding an electron on x axis
I have the radial component of the wavefunction
$R_{nl}(r)=r^2e^{-2r/3a_{0}}$ so to obtain the maximum distance on x axis I do not know if I have to compute the second derivative of $R_{nl}^2(r)$ or $R_{nl}$ or of the radial distribution function ($r^2R_{nl}^2(r)$).
Thanks in advance.

Just compute the first derivative of $R^{2}$ and set to zero.

Thanks for you answer Zhe. I really appreciate if you could give me some insight about why this is the correct option.

Probability is proportional to $|\Psi|^{2}$, so maximum probability must correspond to $\frac{d(R^{2})}{dr}=0$.

Sorry, your radial component doesn't already include $r^{2}$, so you do need $r^{2}R^{2}$ because probability is proportional to $r^{2}R^{2}$.


text:
Brown ring from the anion detection of NO3-
Today in our practicals I observed that when I did the wet test for nitrate (Making a 1 ml solution of salt solution+$\ce{FeSO4}$ solution+hot concentrated $\ce{H2SO4}$);
The brown ring appeared for a very short time and then disappeared.
But later on I came to know that it was nitrate only. What may be the possible reason for this temporary appearance of a tinge of a brown colour and then its sudden disappearance?

Probably transient formation of $\ce{NO2}$.

See this question and links therein.

The brown ring formed in the confirmatory test for the nitrate ion is due to the complex called nitroso ferrous sulphate, or pentaaqua nitrosyl iron (II) sulfate with the formula $\ce{[Fe(H2O)5NO]SO4}$.
The concentrated sulfuric acid decomposes this complex and brown ring disappers in a few seconds when $\ce{H2SO4}$ moves to the upper layer. This is why slow addition of the acid (through the sides of the test tube) is advised.

So you mean that excess of concentrated H2SO4 makes brown ring disappear.

But if we add excess then it turns black...

I tried it out too


text:
Salt bridge in galvanic cell
When we make a galvanic cell using zinc rod (in zinc sulphate solution) and copper rod (in copper sulphate solution), there is a flow of sulphate ions from the copper compartment to the zinc compartment through the salt bridge. Once the sulphate ions reach the zinc compartment, what happens to them?

Nothing. They are no different from those sulphate ions which were there from the beginning.

According to my opinion i think you are wrong because the sulphate ion does not enter the other compartment. This is because the function of the salt bridge is to maintain electrical neutrality by either allowing the sulphate ion to reside in it or by allowing the cations inside the salt bridge to enter the compartment so as to balance the ions.

Can you please confirm that no sulphate ions are flowing through the salt bridge. And that, the salt in the salt bridge supplies the required ions. Thanks.


text:
HOMO Odd, LUMO Even: Why does the affected orbital change depending on the carbon? How are carbons numbered?
While looking at how substituents affect the absorption spectra of molecules, I came across this:

Substituents alter the emitted wavelengths of these molecules by affecting the electron densities. Electron Withdrawing Groups (or EWGs) withdraw electron density, thereby making the arrangement more stable. The energy level of the orbital affected subsequently lowers, causing a typically-noticeable red shift in the maximum wavelength. Alternately, Electron Donating Groups (EDGs, for short), add to the electron densities of the molecule. This causes the affected orbital to be less stable, and requires more energy to obtain. The orbital affected depends on the placement of the substituent. A useful mnemonic device in this matter is HOLE: HOMO Odd, LUMO Even. This explains that the affected orbital alternates between odd and even placements.

Why does the orbital affected depend on the placement of the substituent? I thought substituents that caused bathochromic shifts just extended conjugation and hence reduced the HOMO-LUMO gap? Is this odd-even rule only applicable to azulene, or can it be applied to other chromophores?
Also, why are the carbons numbered the way they are? Here's a picture with the numbering.

Shouldn't the carbons with the question mark also be counted?

This is more general than extending conjugation. It is a kind of rule that applies always. Then is matter of the extent of the bathochromic ecc effect. Eg compare benzene and chlorobenzene

The numbering question is unrelated. It could make for a question of its own. Therefore, I am voting to close as too broad (asking too many distinct questions at once)

I intended for the numbering question to add on to the even-odd question, since if it's numbered differently, then an odd carbon might become an even carbon etc. From Tyberius' answer, I now understand that the even-odd behaviour is due to the shape of the HOMO and LUMO rather than an intrinsic property of odd/even carbons. I'm sorry that the question was phrased too broadly.

It's not too broad in my opinion (and others, too), don't worry about this. If the answer by Tyberius helped you, please  consider accepting it.

To start with, I'll address the frontier orbitals. Lets look at what the HOMO and LUMO look like.

Notice the LUMO has no density on carbons $1$ or $3$ and (while it is a little difficult to see in this image) little density on carbons $5$ and $7$. Similarly the HOMO has little to no density on any of the even carbons. So, if we want to affect the energy of these orbitals, we have to add a substituent to a carbon that has electron density from that orbital. The odd-trend is likely not a general rule, but if you know the HOMO and LUMO reside mainly on different carbons, you can assume that selective functionalization of those carbons will affect the orbital of interest. I say "assume" because you should always keep in mind that the orbital picture is an approximation and that once you make a substitution you have to check to make sure you haven't perturbed the system enough where your new compound doesn't have frontier orbitals that look like this anymore.
In regards to the numbering, this is really just a matter of convenience for a fused ring system like this. We don't number these bridging carbons because they can't be functionalized; that is, we can't add a group to these carbons without turning the molecule into something other than azulene. So the numbering starts from next to these bridging carbons and skips over them as you move around the rings.

This numbering is the one used for azulene alone (see http://www.acdlabs.com/iupac/nomenclature/79/r79_72.htm ). Carbon atoms which are part of a "bridge" (as are the carbons denoted by "???") are actually numbered "3a" (between atoms 3 and 4) and "8a" (between atoms 8 and 1). This is the standard numbering for those compounds, aka "fused polycyclic hydrocarbons".


text:
what are some books and research papers that an aspiring theoretical chemist must read?
Can anybody suggest books that must be read by someone who wants to work in the field of theoretical and computational chemistry?

That is a very good question and I am quite interested in the answers even if my field (organic chemistry) is a bit far from it.

Related: https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry. This should help, though it kind of depends what you are looking to do. The computational chemists on the site tend to skew towards electronic structure theory rather than molecular mechanics, so you will likely find more resources for electronic stuff. (@SteffX so you see this as well)

@Tyberius Thank you very much. This looks really helpful to me because I am looking for pursuing electronic structure theory as well.

@ArchyTripathi not a problem, though it is looking like the site is trying to condense reference requests to that single post (see here), so this question will potentially be closed a duplicate.

The physical basis of biochemistry and the physical basis for chemistry are outstanding books.


text:
I'm pretty sure this doesn't follow Le Chatelier's Principle
The likelihood is that I'm misunderstanding what's going on here.
Consider the reaction $\ce{A <=> B}$, where $K_\mathrm{c}=1$.

Initially, the system at equilibrium, where $[\ce A]=\pu{1M}$ and $[\ce B]=\pu{1M}$.
The system is then disturbed to increase $[\ce A]$ by $\pu{1M}$. Instantaneously, $[\ce A]=\pu{2M}$ and $[\ce{B}]=\pu{1M}$. Le Chatelier's principle suggests that the equilibrium position moves to the right to minimise the increase in $[\ce A]$.
The system then returns to equilibrium, such that $[\ce A]=\pu{1.5M}$ and $[\ce{B}]=\pu{1.5M}$.

Clearly, the relative amounts of $[\ce A]$ and $[\ce B]$ have not changed, so the equilibrium position has not changed, which is in direct contradiction with Le Chatelier's principle.
I think my misunderstanding may lie in one of the following:

My definition of equilibrium position (the relative amounts of reactants and products at equilibrium).
Between which numbers the equilibrium position is considered to move, let me elaborate on this: the relative amount of $[\ce A]$ and $[\ce B]$ have not changed between 1 and 3 ($1/1=1.5/1.5$), but they have changed between 2 and 3 ($1/2\ne 1.5/1.5$), so I see how the equilibrium position could be said to have moved right from step 2 to step 3 as the amount of B increases and the amount of A decreases.

In order for you to get maximum returns on any answer, I think you should tell us the definition of Le Châtelier's principle that you are using.

"If a change is made to a system in equilibrium, the position of the equilibrium will move to minimise that change" @Zhe

That description is qualitative and not quantitative. But certainly, you increased the concentration of $\ce{A}$ by $1\ \mathrm{M}$, but the final increase was significantly less than that.

You say "the relative amount of [A] and [B] have not changed between 1 and 2 (1/1=1.5/1.5)". Do you mean between 1 and 3?

You misunderstand the principle, It doesn't say K will change, it says if A changes B will change so and so in order to keep K constant (applied to the given situation).

@greg $K$ does not change. I think you meant that the equilibrium changes so that $Q$ well remain constant and equal to $K$...

@Zhe What are you talking about? I didn't say K changes. I said B changes in order to keep K constant. What is Q anyway?

@Greg "in order to keep $K$ constant"

$Q$ is the standard notation for the reaction quotient which describes the composition of the system relative to equilibrium.

Yes, K is constant, AB changes and LeC principle is about that. OP assumed LeC said otherwise. What is your problem?

@Greg, I think Zhe means that by saying "in order to keep K constant" one implies that K could change.  After step 2 above, K = 1 but Q = 0.5 = [B]/[A].  It's more rigorous to say that [B] changes so that Q = K than to say a change in [B] keeps K the same, since K stays the same what ever [B] might be.

Le Chatelier's Principle does not directly say what happens to concentration ratios.  Nor does it directly compare conditions after a disturbance with those before.  It says what happens only after a disturbance has been applied.
Here, the disturbance is adding one mole of A per liter.  When equilibrium is re-established you have only an additional one-half mole of A per liter, the rest having reacted to form B.  The initial addition of A has been partially offset, which is exactly what Le Chatelier's Principle says.

So Le Chatelier's Principle applies between the instant a change is made to a system in equilibrium until the moment it reaches equilibrium once again, not from an initial state of equilibrium to reaching equilibrium once again? @Oscar Lanzi

Exactly.  Hope this helps.

And in terms of the equilibrium position in my example: the equilibrium position moves to the right between the instant [A] is increased and the moment equilibrium is reached with [A]=[B]=1.5M. Between the initial and final states of equilibrium, the equilibrium position has not moved, since the chemicals are present in the same relative amounts? In general, equilibrium position does not move between states of equilibrium, it moves between states of instantaneous disturbance and return to equilibrium? (Sorry if this is unclear) @Oscar Lanzi

Basically that is how a reaction works.  You have to start out of equilibrium, or disturb the system out of equilibrium, then the reaction will move back to equilibrium.  It cannot move directly from one equilibrium state to another without something to set it off.

You are simply slightly misapplying Le Chatelier. The principle is used to predict the reaction of a system to changes in the external conditions, i.e. temperature or pressure, not to you wilfully changing the internal composition of the system.
The general idea (system minimises the change) is of course right, but at constant conditions, the law of mass action allows a direct quantitative predicition, so why bother with Le Chatelier?
Anyway when the system reaches equillibrium after your addition of compound A, the relative change in the system is already zero again, so, following Le Chatelier, the change in the equillibrium constant would also be nil. No contradiction at all.


text:
Defining equilibrium position vs equilbrium constant
My chemistry textbook says that $K_{p}$ indicates the equilibrium position. It then says that $kp$ will not change if pressure is increased, but it also says that the position of equilibrium does change if pressure is increased.
Let me restate that: $K_{p}$ tells you the equilibrium position. Increasing pressure does not change $K_p$ (indicating the equilibrium position does not change) but does change the equilibrium position. This is clearly a contradiction, so how should I properly define equilibrium position and equilibrium constants? I've never really had a clear understanding of what equilibrium position means.

Mind your Ps and Qs -- or rather, mind your ks and Ks. A lowercase 'k' is usually used to denote a rate contstant, while an uppercase K is an equilibrium constant.

You're using the word "pressure" too casually. There are partial pressures of species and total pressure (which is the sum of all partial pressures of all species, regardless of if they're participating in the equilibrium or not).
Partial pressures of reactants and products factor into the equilibrium constant $K_{\mathrm{p}}$. If you increase the partial pressure of a gas that is not participating in the equilibrium, none of the other partial pressures change, so the equilibrium is unaffected, but the total pressure will change. Thus, depending on the circumstances, changing the total pressure may or may not change the position of equilibrium.
The key is you should focus only on the partial pressures of interest. All other pressures are irrelevant, unless they somehow tell you about a change in a partial pressure of interest.

Kp tells you the value of the partial pressure term at equilibrium. The partial pressure term itself may vary but at equilibrium, it is always the value of Kp.
The partial pressure term mentioned is the expression for Kp used to calculate Kp but the concentrations are not at equilibrium but at the point in time when you made the measurements.


text:
Is there a direct relationship between the adsorption energy and the catalytic activty
I have seen in several papers, where people correlate adsorption energy with the catalytic activity; the greater the adsorption energy, the greater would be the catalytic activity.
I understand that in order for a catalytic reaction to take place, the adsorbates should get adsorbed on the surface of the catalyst. But what if the adsorbates are too strongly adsorbed on the catalyst. This is unfavorable for the reaction. Isn't it?
Therefore, I don't understand how one can correlate higher adsorption energies with greater catalytic activity.

You think right. But then again, maybe the system in question just never adsorbs too strongly?

Adsorption energy is indeed related to catalytic activity, but not "greater energy means greater activity".
As you were guessing, there is an optimum window in adsorption energies for which catalytic activity is maximum, this is explained by the Sabatier principle:

The Sabatier principle is a qualitative concept in chemical catalysis named after the French chemist Paul Sabatier. It states that the interactions between the catalyst and the substrate should be "just right"; that is, neither too strong nor too weak. If the interaction is too weak, the substrate will fail to bind to the catalyst and no reaction will take place. On the other hand, if the interaction is too strong, the product fails to dissociate.

There is a nice recent paper in which you can find much more mechanistic details.

The energetics of adsorption and the hydrophobicity of a surface play an important role in catlytic activity at that surface. Catalysed pathways are complex, and actually quite poorly understood mechanistically. The range of catalysed reactions is incredibly broad, and each reaction tends to be the subject of intense discussion regarding mechanistic pathways. Needless to say, to answer your question there is an indirect relationship between adsorption energy and catalytic activity. Generally you want good adsorption of reactants. However adsorption, for example of water, can be associative (molecular) or dissociative adsorption. This is of great consequence in catalytic activity. Different surfaces, owing to their surface lattice, show differing affinities for adsorbed species and thus when examining catlyst activities, it is important to consider the level at which individual planes are expressed.


text:
Which has a higher priority while naming: Double or a triple bond?
Which has higher priority in IUPAC naming: A double or a triple bond? given that both are located at the terminal/equivalent positions in the molecule.

My chemistry textbook says a double bond has a higher priority than a triple bond but I'm skeptical over it.

If someone can provide some more authentic source citing this rule, I'll be highly grateful.
EDIT: Drawing the hydrocarbon linear

Here is a wikipedia link https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry that refers to the IUPAC publication Nomenclature of Organic Chemistry (the Blue Book).

Please, draw that alkyne linear.

@orthocresol Edit Made. Thanks for notifying!

@MaxW The priority table of functional grops given on wikpedia does not have alkene/alkyne in it.

@orthocresol I didn't knew this has been previously answered. Thanks!

@orthocresol is it always necessary to make alkynes linear? Are zig zag formulas not allowed for them?

@Abcd This should be a new question, really. See: https://doi.org/10.1351/pac200880020277 (p 345)


text:
In QSAR, what is the simplest (but most powerful) way to create an applicability domain model?
QSAR means Quantitative Structure Activity Relationship.
An applicability domain model is a model that tells you how far you are from the domain where your model was trained. I.e. it is a kind of map where you can place molecules. And on this map, there is some delimited territory that you have explored while you were training your model.
Let's say I have created a QSAR model on some training dataset, and validated it on a test set.
Next, I'd like to create a model that gives me a 1.0 (or nearby) if the molecule I present to the QSAR model looks similar to the training set, but a 0.0 (or nearby) if it looks dissimilar.
Currently, I was thinking about training a one class Support Vector Machine on the training set.
Broad picture: I'd like to get a warning in case the QSAR model is presented with something that looks too far from anything it was trained on. The warning would mean: "be careful with this prediction".

Welcome to the site! It might be a good idea to identify the acronyms (QSAR, SVM) you use and elaborate on your question further (what is an applicability domain model?).

I believe the paper you want look at is: Netzeva, T. I., Worth, A. P., Aldenberg, T., Benigni, R., Cronin, M. T., Gramatica, P., ... & Myatt, G. (2005). Current status of methods for defining the applicability domain of (quantitative) structure-activity relationships. ATLA, 33, 155-173. It describes several different methods of choosing the AD. All of have pros and cons, so you will have to choose the one which best fits your problem.


text:
What is the relation between ionization energy and electropositivity?
Electropositivity of an element is defined as a measure of element's ability to donate electrons while Ionization energy or Ionization potential is defined as the energy required to dispel a loosely held electron from the valence shell. These two terms sound closely related to me or they are just measure and energy?

Electropositivity is rarely used by itself. It is much more commonly seen as just the opposite of electronegativity. While electronegative is more or less defined to the point where numeric values can be found for each element in a given system (sometimes excluding noble gases), electropositivity has no such scale I am aware of.
Ionisation energy or potential are clearly defined energies that can be experimentally directly determined and are tabulated across physical chemistry textbooks with little to no variance (the main variance being experimental error).
The two terms are hardly related; neither by definition/physical meaning nor by value trend.


text:
Relation between oxoanion acidity and number of oxygen double bonds
https://en.m.wikipedia.org/wiki/Oxyanion
Was reading the above Wiki page when I came across the 'heuristic for acidity' section which mentions that the oxoanion's acidity can be predicted by the number of oxygen double bonds, with more double bonds making the oxoanion more acidic. My question is, why is this so? Can someone explain it to me by the definition of an acid, and is this prediction basis only limited to oxoanions or can it be extended to other molecules?

This rule is doomed to fail. The only oxyacids with element-oxygen double bonds are carbonic acid and nitric acid, both counting 1. Unless you want to include boronic acid’s resonance structure containing a double bond.

related https://chemistry.stackexchange.com/questions/35284/how-to-determine-which-substance-is-the-strongest-base-acid

The strength of an acid is determined by how far the reaction
$$
\ce{HA} \rightarrow \ce{H^+} + \ce{A^-}
$$
proceeds. One of the products will always be the same for all acids ($\ce{H+}$), so no influence there. The other product is the anion ($\ce{A-}$) and the basic question is: "How stable is that anion?", or equivalently, "how well can that anion stabilize its negative charge?"
Many factors go into answering that question, the one under consideration in the rule alluded to in the question is: Are there atoms surrounding the central atom that can stabilize the charge effectively? Oxygen atoms bound only to the central atom (in the protonated form) are a prime example, because oxygen is a very electronegative element. (Another example not involving oxyanions/oxyacids is trifluoroacetic acid, which is more acidic than acetic acid.) The more such oxygens are in your acid, the stronger it will be - $\ce{H2SO3}$ is weaker than $\ce{H2SO4}$.
Jan commented on the question that double bonds between the central atom and oxygens only exist in the first row of main group elements (i.e. $\ce{C},\ce{N}$). This is true, however, the rather basic rule cited in the question does not go into this level of detail and simply assumes that the outdated picture of double bonds in say, $\ce{ClO4-}$, is correct.


text:
How to remove pyridine from Reaction mixture containing chiral molecule with diacetal protection group?
While working with chiral molecule, we have pyridine in our reaction mixture. Molecule has four chiral centre and protection group-diacetal and acetyl groups which may break in harsh conditions. Please suggest work up methods to remove pyridine.

Please, provide name/structure of the compound. As a general idea, I think pyridine might be evaporated given enough time, but it definitely is not fast or easy.

Aquaeous extraction with a saturated ammonium chloride solution is usually the way to go.

Extract the organic soln of the product with dilute aq. Cu(II) until the characteristic blue of the Cu-pyridine complex is absent from the aq. phase.