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Preventing solvent evaporation for small scale synthesis
I have been conducting an organic synthesis with quite expensive starting material, so whenever I setup the reaction I only use 0.1 g of starting material. I use 4 mL of toluene to adjust the concentration of the starting material to about 0.1 M (although it is not essential option). I use 25 mL three-necked flask, sealed with a rubber septum, a glass stopper, and a reflux condensor. Top of the reflux condensor is connected with oil bubbler. So it is a closed system. I usually apply Teflon tape at the glass joint. Although this condition is generally okay, sometimes the solvent is completely evaporated during dawn. I searched for the alternatives, and I found someone that recommends glass thick wall pressure tubes or microwaves vial which don't require a condensor. Can I use these for alternatives, or anyone have recommendations? To be honest, I am hesitating to use grease because of the potential contamination.

additional question : Do the rubber septa assure more tight sealings than glass stoppers?

Your system is far from closed.  Are you adding reactants or sampling? Look up the composition of the rubber. You do not mention the other reactant[s]. Possibilities are better matching of the joints, reducing the number of joints, getting rid of the septum and the Teflon. You do not mention analysis. If it is spectrometric consider running the reactions in temperature controlled cells, uses less material and eliminates a transfer. If a solvent is needed try sealed tube reactions but practice before committing the expensive reactants. [In any case some practice on knowns is a good idea.]

Why do you need to reflux?  Why not use a higher-boiling solvent and better temperature control?

I have used a sealed microwave vial in a heating block successfully in similar circumstances


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Help with this mechanism?
I know that the bottom reaction is probably elimination, but I've never seen Na2CO3 being used as a reagent in this context. I've only taken Organic chemistry I and II, so that's the extent of my synthesis reaction knowledge. Any help would be appreciated. Sorry about the red squiggly box around Na2CO3; I'm not very familiar with ChemDraw.

Are you sure you have drawn the sodium benzenesulfonate salt correctly, or do you mean sodium benzenesulfinate?

I am going to assume that there is an error in the reaction diagram (either by the OP or in the original question) and the starting material is sodium benzenesulfinate (PhSO2Na) NOT sodium benzenesulfonate (PhSO3Na).
Sodium benzenesulfinate is nucleophilic through sulfur and has been shown to displace halogens from alkyl halides example.
Nucleophilic displacement of the primary bromide gives an alkyl sulfone with a beta leaving group. The methylene protons alpha to the sulfone are sufficiently acidic to be removed under basic conditions (though I have never seen carbonate used in this context) to drive the elimination giving the alkene.
If the reaction is genuinely run with sodium benzenesulfonate I have no idea what is going on.

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How heating of iron results in formation of new compound?
Question:

Which of the following processes results in the formation of the new compound?
A. Dissolving common salt in water
B. Heating water
C. Heating of the platinum rod
D. Heating iron rod

I've marked Option A but when all the questions and answers are officially published, it says Option D is correct.
I've studied that Heating of iron is a just physical change. But how does it forms a new compound? I tried to look it on google. I found this link
which says Option D is correct. How?

The question is incomplete in that it does not specify in what environment the iron rod is heated - in vacuum, in inert atmosphere, in oxygen? It can be argued that heating the iron rod in oxygen (or air) results in surface oxidation, creating oxides of iron.

Heating a element cannot lead to anything but that element. Even at T common in cores of massive exploding stars, it is difficult and energy consuming to form heavier elements. // A asking B why C provided a nonsense or implicitly conditioned answer does not make much sense.  Ask C or related resources instead. // BTW enough heating of water leads to H2 and O2.

@poutnik maybe heating water does not produce H2 and O2. The decomposition temperature could be high enough to dissociate one or the other into single atoms.

@Oscar Perhaps O, but there is still some way to H. Sure, all depends on considered eq. constants.

The question is underspecified and the complete conditions are not given. But, if we assume common lab conditions and not extreme ones, the first three options are obviously false. Only the heating of iron under normal conditions could result in a new compound (from oxidation in a normal atmosphere).

It forms iron oxides $\ce{2Fe + 1.5O2 -> Fe2O3}$

Hot iron is slowly oxidized in the presence of air, producing $\ce{Fe3O4}$
. It is mainly a surface effect, and the reaction is not stoichiometric. So the answer D is correct, if the hot iron rod is maintained a long time in air.

Presumably heating in air is meant, in which case iron does oxidize typically forming multiple oxides. At lower temperature we would get predominantly $\ce{Fe3O4}$ with a thinner layer of $\ce{Fe2O3}$ on top, the latter being the oxide that is actually equilibrated with air (but not simultaneously with the metal). Higher heating temperature results in the formation of $\ce{FeO}$ or more accurately $\ce{Fe_{1-x}O}$, which then becomes kinetically dominant.
It turns out that platinum can be oxidized to $\ce{PtO2}$ on heating in air or pure oxygen, up to a temperature of 500°C where this oxide decomposes again [1]. Thus technically, both (C) and (D) are correct if heating is done in air.
Reference
Chaston, J.C. (1964). "Reaction of Oxygen with the Platinum Metals" (PDF). Platinum Metals Rev. 8 (2): 50–54.

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Why is effective nuclear charge said to be decreased down the group, but the Zeff calculated is increased down the groups instead?
Some textbook write that effective nuclear charge is constant when gping down the group,while some write that the effective nuclear charge decrease down the group
So which statement is true? I try to dtermined it by using slater's rule.However,it came out a trend that $Z_{eff}$ is increasing,which mean effective nuclear charge is increase down the group instead of the 2 other trend i stated above.
I am very confuse about it

Consider the major feature for the Slater's rules: Weak mutual screening of neighbour electrons within the same ns or np orbital, versus strong screening by fully filled orbitals of the "lower" subshells. Applied to the chart, it should be clear.

It would probably be much easier to think about this question if you clearly defined what you mean by effective nuclear charge.
According to my understanding, the effective nuclear charge $Z_\mathrm{eff}$ is a fitting parameter appearing when you try to model your valence electrons by a hydrogen orbital.
In this sense, there's two effects in play: $Z_\mathrm{eff}$ will increase as the total charge of the nucleus increases, but $Z_\mathrm{eff}$ will decrease from the screening effects of the non-valence electrons (there's also some weak screening due to other valence electrons but that's a much smaller effect).
If you consider let's say the valence electrons of Na (well, the single one of them), then it will feel the positive charge of $+11$ from the nucleus, shielded by the negative charge $-10$ of the core electrons, leading to an effective nuclear charge somewhat larger than $+1$ because the screening is not perfect (in fact, $Z_\mathrm{eff}$ is about 2 for the valence shell of Na).
Then, consider Cl in the same row. The valence electrons will feel $+17$ charge from the nucleus, shielded by $-10$ valence electrons, meaning $Z_\mathrm{eff}$ will be in the order of magnitude of 7.

So now that we have established how to think about it, we can easily answer your question.
Compare Na to K. Both of these atoms will have one single valence electrons, so the effective nuclear charge is about $+1$, with the differences accounting for the efficiency of screening.
In general, the closer to the core an electron is, the more effectively it screens the nucleus. Meaning that a $2p$ electron will contribute more to screening than a $3p$ electron. This really all comes down to the radius of the orbital.
The core electrons in K are more diffuse than the ones in Na, so they don't screen as efficiently.
Therefore $Z_\mathrm{eff}$ is larger for the valence electron of K than for the valence electron of Na $\rightarrow$ $Z_\mathrm{eff}$ increases as you go down a group.

that means effective nuclear charge increase down the group, right ?If this the fact,then how would u explain the trend of atomic radius increasing going down the group as effective nuclear charge is increasing?

It is more or less the same explanation: the orbitals of the valence electrons are getting larger as you increase the principal quantum number. It is true that the effective nuclear charge is increasing, so the radius is smaller than it would be with a hypothetical constant effective nuclear charge, but it is not enough to offset the fact that the radius of the orbital is so much larger.

I am not sure which texts you saw which said that $Z_{eff}$ (effective nuclear charge) decreases down the group. Many will say that it is constant (an approximation, but it does simplify the explanation of trends) and a more in-depth analysis will show (using Slater's Rules and bearing in mind the fact that more diffuse inner electrons do a poorer job of shielding protons completely) that $Z_{eff}$ increases down the group - which is what your graph shows.
The simple explaination (constant $Z_{eff}$ down the group) assumes that all inner electrons exactly cancel the effect of protons in the nucleus. In a neutral atom, the number of protons not cancelled by inner electrons is equal to the number of outer shell electrons. As this is constant down a group, so is the $Z_{eff}$. Adding Slater's Rules simply shows that the inner electrons are not 100% effective at cancelling the charge of their corresponding protons, and the effectiveness of the inner electrons decreases as the orbitals get more diffuse, so $Z_{eff}$ increases down the group.
Where you may be getting confused is that although the $Z_{eff}$ is increasing down the group (or remaining constant in the simple model), the force of the attraction between the valence electrons and the nucleus is decreasing, as the distance between them is increasing. This is due to Coulomb's Law.
$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$
Many people confuse the meaning of $Z_{eff}$: it does not correspond to the force felt by the outer electrons, it is only the apparent charge of the nucleus, from the outer electrons' point of view. In Coulomb's Law, $Z_{eff}$ corresponds to one of the q charges on the numerator of the quotient.


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Calculating the net charge of DNA oligo
How can I calculate the net charge of a DNA oligo? I have a bunch of different sized oligos (vary from 9 to 50nt long) which I am using for microscopy (DNA-PAINT) and I need to know their charges depending on the sequence. How can I calculate that?
To top things up, the oligos often have modifications: azide on one end and a fluorophore on the other end. For example: 5' Cy5 TTATACATACTTA 3' azide.
Would appreciate any suggestions.

The charge is usually dependent on the pH of the solution. More information is required on this.

Negative charge at neutral pH is roughly equal to the number of nucleotides in the oligo. The length of single-stranded oligos is measured in nucleotides (nt), not base pairs (bp).

@BuckThorn pH is between 6.8 and 7.5


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What exactly happens in a Chemical Potential Energy Graph?
Along the x-axis, the progress of the reaction is observed, while along the y-axis, the potential energy of the reactants/products are observed. They keep rising until they attain the required activation energy for breaking of older bonds, and formation of new ones. But what exactly is this activation energy? Is it a form of potential energy (stored within the chemical bonds), or is it a form of kinetic energy (vibrations of the particles)? I don't really understand why, if the activation energy is just potential energy, it results in collisions (without any motion) and thus forms new compounds, given that, we all know, potential energy is the energy of a body at rest.
If it is not so, then what is the role of chemical potential energy into all of this, at all?
I am still a student, stuck within concepts, so please, keep the answer as brief as possible...

“In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors. “ This works at rest or in motion.

You'll find many similar posts on the site that might answer your question, for instance: https://chemistry.stackexchange.com/questions/115923/y-axis-of-the-reaction-co-ordinate-graph/ or https://chemistry.stackexchange.com/questions/119255/why-is-activation-energy-drawn-in-a-potential-energy-diagram-in-reactions

It is just potential energy: the potential energy needed for the reaction to happen.
This potential energy obviously needs to come from somewhere, and that's where kinetic energy comes in. If the molecules have high enough kinetic energy at the point of the collision for that reaction to happen, then part of that kinetic energy is converted to potential energy, and the reaction proceeds.

This is also why most reactions get faster as you increase the temperature. Higher temperature means more kinetic energy, meaning more successful reactive collisions.
Edit: I should add that there seems to be some sort of misconception in your question. There's nothing along a potential energy graph that would result in a collision.

But what exactly is the role of Potential Energy in a "collision". This kind of sounds weird, because potential energy is solely the energy at rest, whereas collisions come from Kinetic energy itself. Aren't both of them contradictory to each other??

I don't understand the question. The potential energy plays no role in a collision, apart from the fact that the total energy needs to be conserved. You start with a given amount of total energy, distributed between kinetic and potential energies, and those two can interchange into each other, as long as the sum is conserved. The activation energy tells you that how much of the total energy needs to be in potential energy at least for the reaction to happen.

The potential energy is the energy due to the positions of the atoms and their electrons have in a molecule that is a sort of combination of reactant and product, i.e. a transient structure from the path from reactants to products, some times called a 'transition state'. This barrier can be overcome by collisional (kinetic) energy between the reactant and product but also by thermal energy since by the Boltzmann distribution there is a small chance that a molecule may have, for a short period of time, many times the average energy expected at that temperature. Thus reactants may collide many millions of times before a reaction occurs, if the barrier is high, or only a few times if the barrier is low relative to average thermal energy. Consequently we observe rate constants in different types of reaction that vary from $10^{12}$/sec to less that 1 per hour.

Maybe this additional perspective will help.  Reactions involve bond breaking and bond making.  Breaking bonds results in an increase in potential energy (there is less attraction between atoms; you must add energy to break bonds, therefore, increase in PE).  Bond Making results in a decrease in potential energy (there is more attraction between atoms; less obvious, energy is released when bonds form).  So, in the first part of the reaction bond breaking dominates and the potential energy is increasing.  In the latter stages of the reaction bond making dominates and potential energy  is decreasing.  Thus, the graph goes up and then it goes down.  This is a very simplistic view, but it will give you a better perspective.


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Are carboxylic anhydrides in graphene oxides considered adsorption sites for metals? If so, are they more or less reactive than carboxylic acids?
The literature abundantly addresses the fact that carboxylic acids present in graphene oxides act as efficient adsorption sites for metals in aqueous solutions (e.g., HE et al., 2021; ZHAO et al., 2019; NOVÁCEK et al., 2017; WANG et al., 2018). Wang et al. (2018), for example, showed through DFT measurements that the adsorption energies of the complexes (–COOH)/Co, GO(–OH)/Co, and NGO/Co, where N corresponds to nitrogen-containing functional groups introduced into graphene oxide and Co to cobalt ions, were 44.05 kcal/mol, 16.44 kcal/mol, and 6.33 kcal/mol, in that order. This result was corroborated by the performance of the adsorbents in removing Co(II) from the water (GO (0.74 mmol/g) > NGO (0.46 mmol/g)). However, I have not found anything regarding whether carboxylic anhydrides, which can be identified via potentiometric titration, also capture metals through their complexation and/or via ion exchange. I would like to know if they can and if their reactivity is lower than, equal to, or higher than that of carboxylic acids in the water, present in graphene oxides, for the adsorption of metals.
Thanks in advance!
HEA, Lei; WANG, Lei; ZHU, Haomiao; WANG, Zhe; ZHANG, Luxia; YANG, Lutao; DAI, Yong; MO, Hong; ZHANG, Jun; SHEN, Jian. A reusable Fe3O4/GO-COOH nanoadsorbent for Ca2+ and Cu2+ removal from oilfield wastewater. Chemical Engineering Research and Design, v. 166, n. 171, p. 248-258, jan. 2021.
ZHAO, L.; CHEN, J.; XIONG, N.; BAI, Y.; YILIHAMU, A.; MA, Q.; YANG, S.; WU, D.; YANG, S. Carboxylation as an effective approach to improve the adsorption performance of graphene materials for Cu2+ removal. Science of the Total Environment, v. 682, p. 591-600, set. 2019. Elsevier BV. http://dx.doi.org/10.1016/j.scitotenv.2019.05.190.
NOVÁČEK, M.; JANKOVSKÝ, O.; LUXA, J.; SEDMIDUBSKÝ, D.; PUMERA, M.; FILA, V.; LHOTKA, M.; KLÍMOVÁ, K.; MATĚJKOVÁ, S.; SOFER, Z. Tuning of graphene oxide composition by multiple oxidations for carbon dioxide storage and capture of toxic metals. Journal of Materials Chemistry A, v. 5, n. 6, p. 2739-2748, 2017. Royal Society of Chemistry (RSC). http://dx.doi.org/10.1039/c6ta03631g.
WANG, X.; LIU, Y.; PANG, H.; YU, S.; AI, Y.; MA, X.; SONG, G.; HAYAT, T.; ALSAEDI, A.; WANG, X. Effect of graphene oxide surface modification on the elimination of Co(II) from aqueous solutions. Chemical Engineering Journal, v. 344, p. 380-390, jul. 2018. Elsevier BV. http://dx.doi.org/10.1016/j.cej.2018.03.107.

The key point which you have to clarify is the medium of adsorption. What is the solution? Are you interested in adsorbing metals from aqueous solutions with sorbents which have anhydride groups?

Two points I see easily: 1) there's probably not a significant difference between the chelation efficiency of the "free" molecule and the absorbed one, so I'd just use the unabsorbed pKa values as a first approximation (or, at least the ratio); 2) the medium really plays a huge role, I'm not sure there's a lot of anhydrate groups present when dispersed in water. In general, I'd not associate anhydrates with strong chelation, but then again it may just get hydrated in water.

@AChem, I want to adsorb metals from aqueous solutions with graphene oxides. However, potentiometric titrations showed that one of the materials that I'm testing present a large amount of carboxylic anhydrides groups. I just want to know if these groups also act as adsorbing sites, and if their reactivity is lower, equal, or greater than that of carboxyls. I added to my question some references for better contextualization.

There are no surprises that the metal adsorption literature is silent about carboxylic anhydride on carbon surfaces. These groups do exist on carbon surfaces but as you can guess from the name "anhydride" and "water" do not get along very well. This is why I had asked you to clarify your medium? These anhydride groups will enventually hydrolyze to carboxylic acid groups. There is a very nice paper "Formation and chemistry of carboxylic anhydrides at the graphene edge" [RSC Adv., 2015,5, 104198-104202, DOI https://doi.org/10.1039/C5RA23209K], that shows the hydrolysis of carboxylic anhydride in water and in humid air. This process is relatively fast as show in the Supporting File.

In the field of carbons and surfaces, I would advise to read at 12 papers before arriving at a conclusion. Surfaces are very very difficult to analyze chemically and a lot of speculation exists.


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Probability density and radial distribution function of finding the most probable distance of electron in 2p orbital in hydrogen atom
Referring to the answer by DSVA (Most probable point for finding an electron in the 1s orbital of a Hydrogen atom)

There's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance)

I face a problem in solving the maximum probability of finding electron in a 2p orbital. $$\psi=R_{2,1}Y_{1,0}=\sqrt\frac{1}{32\pi a_o^3} \left(\frac{r}{a_o}\right) \exp\left(\frac{-r}{2a_o}\right) \cos\theta $$
Using probability density function, differentiate and equate to zero$$ \frac{d\psi^2}{dr}=constants\left(2r-\frac{r^2}{a_o}\right)\exp\left(\frac{-r}{a_o}\right)=0$$
I obtain$$r=2a_o$$
Using radial probability distribution, differentiate and equate to zero$$P(r)=r^2|R(r)|^2 $$
Referring Atkins' Physical Chemistry (pg. 312), it is stated that spherical harmonics is normalised to 1.$$\frac{dP}{dr}=constants\left(4r^3-\frac{r^4}{a_o}\right)\exp\left(\frac{-r}{a_o}\right)=0 $$
I obtain$$r=4a_o$$
My question: When should we use radial probability or probability density to find maximum probability of finding electron and its most probable distance? What does the difference of values mean?

Related

What you are interested in is the maximum probability of finding the electron at a given position.
In general, the probability of finding an electron in a volume $\pmb{V}$ (not necessarily infinitesimal) is
$$ \int_{\pmb{V}} |\Psi(\pmb{r})|^2 d^3 \pmb{r}$$
This integral is, so far, abstract, because we haven't defined a coordinate system yet. If you want to use spherical coordinates - which is a common thing to do -, then you can translate this integral to
$$ \int_\pmb{V} |\Psi(\pmb{r})|^2 r^2 \sin \theta dr d\theta d\phi$$
So if you want the maximum probability, then what you are interested is in maximizing the integrand with respect to $r$
$$ \max_r |\Psi(\pmb{r})|^2 r^2 \sin \theta$$
If you then take the derivative of this quantity with respect to $r$ and set it to zero, you end up with
$$ -\frac{r^3 e^{-\frac{2 r}{a_0}} \left(r-2 a_0\right) \sin (\theta ) \cos ^2(\theta )}{16 \pi  a_0^6} =0$$
which has a solution indeed $r=2 a_0$.
A common mistake is finding the maximum of the wavefunction itself is the fact that it only takes into account the decay of the wavefunction with the distance from the nucleus, but it doesn't account for the volume of the infinitesimal disc at that range. Imagine if the wavefunction would be constant up to a certain critical radius $r_c$, and zero outside(I know it breaks quantum mechanics but don't think too hard about that for a second): the maximum probability of finding it would be right at $r_c$, because that radius has the largest geometrical probability to find the constant electron density in.
This is actually quite often misunderstood in the context of the definition of the Bohr radius: it is not the maximum of the electron density of the hydrogen atom, but of the probability density. I hope the distinction is clearer now.
(Two minor mistakes: I'm almost certain you need to include an angular factor in the probability density which you missed, and you shouldn't take the second derivative in the first case - which is irrelevant, anyways).

I'm really sorry and would like to apologise if you have referred to my hydrogen-like wave function for 2pz orbital in the question. I made a mistake for the exponential term. It should be (-r/2a) instead of (-r/a), so I edited just now. Therefore, I take the derivative of the integrand you mentioned with the corrected wave function. I get (r-4a) instead.

Yes, admittedly I just took your expression directly. This misprint will change the result probably, but not the explanation.


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Does iron rust because of impurities found in it (ex. carbon)?
If we, theoretically, get a piece of ideally 100% pure iron and it is left in moist air? Will it rust? My understanding of iron rust is that the Iron itself becomes the ANODE and carbon impurities (as an example of impurities) are CATHODE, so if this cathode is not there will iron itself become anode and cathode and rust?

Three substances are required to transform iron into rust : air (oxygen), water, and a impurity at the surface of the metal. If one of these substances is missing, iron will not rust. The whole operation occurs at the contact of the impurity and iron and it has an electric origin. The reduction occurs at the impurity working as a cathode. It is well described in N. N. Greenwood and E. Earnshaw, Chemistry of the Elements, Pergamon 1983, p.1250

Carbon dioxide?

Unless you are going to show some strong citations for the need for carbon dioxide to drive rusting, I'm going to assume you just made this up as it is not mentioned in most of the summary literature on rusting.

Matt Black is right. I have checked in my library, and especially in Greenwood and Earnshaw, where electrolytic rust formation is studied in detail. Nobody mentions the role of carbon dioxide. I have done it, because I took part once upon a time to a conference where the speaker had mentioned its role. So I will delete my whole message

Pure iron will rust but far, far more slowly than iron with impurities
The key point here is not whether iron will rust or not: it is about the speed of rusting.
Very pure iron will rust but very, very slowly. But even tiny impurities will promote much faster rusting. Many iron alloys (which means most iron you will ever see as very pure iron is rare) contain small amounts of carbon or other substances. On the surface of the iron these can promote tiny electrical potential that greatly speed the process of corrosion. The opposite is also true: some "stainless" alloys contain deliberately added ingredients that promote the formation of a strong, inert oxide layer on the surface that inhibits such reactions (making the substance a little like aluminium whose surface oxide layer is strong and inert, protecting a very reactive metal from oxidation reactions that would otherwise consume it quickly).
The issue with iron in general is that iron oxide is not strong and does not form a protective layer because it is flaky and friable. So, once started, rusting tends to get worse and happen faster over time.
Even pure iron will rust quickly in the wrong environment. Surface contamination will create impurity sites that can promote the electrochemical reactions that accelerate rusting. But those reactions need water, so keeping water away from the surface can slow any reactions very substantially. But, obviously, many varieties of impure iron can rust quickly when any water is present as the contaminants that promote the reactions are already present in the surface of the iron.
In summary, pure iron will rust. But slowly even when water is present. Impure iron will rust faster. All forms of iron will rust quickly if the environment adds other surface contaminants that can promote the electrochemical reactions that make rusting happen.

It depends; sulfide inclusions will enhance oxidation. Chloride surface contamination will promote oxidation. Phosphorus inclusions will slow oxidation . Copper and chrome  "impurities" will also slow oxidation. The latter three elements are the basis of "weathering" carbon steels like Corten. The carbon affect you describe causes "graphitic corrosion" of flake type cast irons ; oxidizing the iron ,leaving carbon flakes.

I am speaking of theoretically 100% pure iron!


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Why do atoms make bonds?
I have been studying chemistry for a long time in school till now, but, what I recently realised I do not understand is why do atoms really make bonds, why do they want to gain or lose electrons or why do they want to attain noble gas configurations when it results in putting a charge on them and comparing to the original condition, let's say oxygen has 8 electrons and 8 protons, why isn't this stable?
This question came to me when I was trying to understand why the ionisation enthalpy can never be negative and how an atom like Na only exhibits it's electropositivity when provided with some energy and is stable until then.
Thank you in advance for your answer
PS: I have realised in the past few days that I really understand very little chemistry:(

Why does anything happen in chemistry? Because it is energetically favourable.

Knowing you understand very little is natural consequence of actual learning. Unfortunately the more you find out, the more you see it's very little in grand scheme of things. That's the essence of Socrates' "I know that I know nothing"

Or this can be useful https://chemistry.stackexchange.com/questions/86918/the-impossibility-of-100-ionic-bond

I'd say for the same reason why an apple fells to the ground. Because there is an energy gain. Atoms and electrons follow the electromagnetic forces through which they interact, in the same way the apple and our planet do. Just in the case of the apple it is a two body problem with classical mechanics. In the case of bonds it is a few (many) body problem. This makes it much more difficult to visualize it and/or to have an intuitive picture. Luckily we have math and numerical simulations to help us with that.

why do atoms really make bonds, [...]

We never know why, but we can study when they do and when they don't make bonds. And mostly (on earth) they do make bonds.

[...] why do they want to gain or lose electrons or why do they want to attain noble gas configurations when it results in putting a charge on them and comparing to the original condition, let's say oxygen has 8 electrons and 8 protons, why isn't this stable?

A single (isolated) oxygen atom will keep is electrons, as will a single sodium atom. Making ionic compounds from isolated atoms (a thought experiment), there is usually a net cost of transferring electrons to make the anions and cations, and a large gain when these cations and anions combine to form ionic compounds. The cost is also offset when cations and anions are solvated in water. Mostly, the atoms in elements are already bonded in some way (metallic bonds for metals, covalent bonds for non-metals), so when compounds form from elements, you also have to consider breaking those bonds. Some elements react with many other elements, while others (like noble metals such as gold or platinum) don't.

Because some molecules have lower energy that the isolated atoms they are made of
Isolated atoms are usually stable if they don't interact with anything else. But, when things interact, you need to start thinking about what can happen and one way to do this is to understand the energy levels of the isolated atoms compared to the molecule.
In some cases when isolated atoms interact, nothing happens. Helium atoms, for example, just bounce off each other and no bonds are formed.
But two isolated hydrogen atoms, if they meet, will often produce a stable hydrogen molecule. When this happens, energy is released and we can measure how much. The hydrogen molecule, we say, has lower energy than two isolated hydrogen atoms. This we call a chemical bond.
But why is a molecule lower in energy than two isolated atoms? It is partially about electromagnetic forces (two positive protons attract two negative electrons in a hydrogen molecule). But knowing whether the net forces lead to a bond involves a lot of detailed calculations and you need a fair amount of quantum mechanics to even approximate the answer. The summary of the answer is that the amount of energy involved in the electromagnetic field of two isolated hydrogen atoms is a lot more than the energy involved in a hydrogen molecule. This isn't true for two helium atoms, which is why they keep to themselves and don't form bonds. But the details of those calculations are advanced.
A great deal of chemistry is about observing when and how bonds form and deriving general rules from observation that explain what happens without having to do those calculations (which are, in general, extremely hard). But the general rule is all about whether the amount of energy involved is lower with a bond than it is with isolated atoms.


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Why is hybridization different in a ring?
Why is nitrogen $sp^3$ hybridized in a compound such as NH3 but $sp^2$ hybridized in something like pyrrole, which has a ring structure? In both cases there are three bonds and one lone pair.

There are ring structures which do have sp3 nitrogen hybridisation. Being part of a ring is not the critical thing here.

Hybridisation is a mathematical description which usually supports our view on bonding and structure. It's is defined by structure. It's also not restricted to whole numbers. The orbitals in ammonia are close to sp³, but they are not exactly. However, in zeroth order approximation is still close enough. (In this scenario 3.14 would be a good approximation for pi.) So that arrangement of bonds is crucial for the description within the restrictions of hybridisation.

That second duplicate target has a terrible title, but anyway, the question of hybridisation in pyrrole/pyridine has been addressed before.

One might argue that the nitrogen in pyrrole does not really have a lone pair we can identify like that in ammonia or for that matter pyrrolidene ($\ce{C5H9N}$, adding enough hydrogen atoms to saturated the five-membered ring in pyrrole). In the pyrrole molecule, what "should be" the lone pair on nitrogen is instead coupled with the other pi electrons to form a continuous ring of conjugated pi bonding. Such a coupling is commonplace when it forms an aromatic ring, meaning in the most common cases $4n+2$ electrons in this ring (like the six pi electrons in pyrrole). There are also cases where what seems to be an empty orbital really isn't empty because of a similar coupling (like the cyclopropenyl cation). You should see more examples of this effect as you study aromatic compounds.


text:
Searching for electronic spectroscopy reference book
I'm looking for a reference book that will help me describe UV-VIS spectra of organic compounds in terms of Q band, MLCT band, Soret band, and so on. Is something like that available?
Thanks

Just a quick note: the terms you are using in your question are very specific, MLCT is used in multiple places in coordination chemistry whereas Q and Soret bands are (if I recall right) specific to porphyrin metal complexes. So you might be looking for introductory coordination chemistry textbooks here, not really organic chemistry.

@Szgoger Oh, great! Thanks for suggestion. Any recommended books on that?


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Does the charge on an atom impact the polarization of covalent bonds?
So far, I have generally seen polarization power only discussed in terms of ionic bonds: small, highly charged cations have more polarization power, and large anions are more polarizable. The more polarization power and polarizability, the greater the covalent character of an ionic bond.
Generally, one can estimate the polarity of a covalent bond by looking at the differences in the electronegativities of the two atoms involved in the bond. However, I am wondering whether one can apply the ideas of polarizing power to atoms in covalent bonds, specifically the idea that net charge impacts the polarity of the bond. Does a higher positive net charge on one atom cause electrons in covalent bonds to be pulled towards that atom? This is what I would conclude from a polarization power argument, because an atom with a positive net charge has more polarization power. Similarly, would an atom with a negative charge pull less electrons? These ideas also seem to coincide with a Coulomb's Law argument that a more positive charge would attract electrons better, and a more negative charge would not attract them well.
One piece of evidence supporting this idea is the inductive effect; polarization propagates down a chain of sigma bonds because partially positive carbons are able to polarize carbon-carbon bonds. This polarization occurs in spite of the fact that the difference in electronegativity between the two carbon atoms is 0. This seems to imply that one has to consider both electronegativity and the charge on the atoms when determining the polarity of bonds.
There are two particular cases that I am curious about: metals in coordination complexes and formal charges. Consider two complexes: one with a Mn (II) center and one with a Mn (VII) center. Will the Mn (VII) center withdraw the electrons of the M-L dative bonds better than an Mn (II) center (in turn inducing an electron withdrawal throughout the rest of the ligands)? For example, would an NH2 ligand be less basic when bonded with an Mn (VII) metal center than when bonded to an Mn (II) metal center, because the Mn (VII) center is able to withdraw electrons better (even though an Mn-N bond, with the same electronegativity difference between Mn and N, exists in both cases)?
As for formal charges, would an atom with a positive formal charge (such as the nitrogen in NH4+) withdraw electrons better than one with a negative formal charge (such as the nitrogen in NH (2-))? I know that formal charge does not always represent the true charge on an atom, but I am thinking that it could be used to compare the relative amounts of charge on related species - for example, I know most electron density is concentrated on the nitrogen in both NH3 and NH4+, but I assume that there would be a bit less electron density on the nitrogen in the NH4+ because the lone pair electrons of NH3 are now shared with a hydrogen.
I am also aware that there are other factors to consider when looking at the electron distributions of complexes and formal charge, such as pi donors/acceptor ligands and favored resonance states. However, I am wondering about the effects of charge on the polarization of bonds. Thank you for your help.

I am curious what Mn(VII)-NH2-R complex you could have in mind....(or rather what products you expect when the reaction between them is finished)

I didn’t have a particular complex in mind; I was using it as an example. Are those complexes not stable?

Basic chemistry knowledge is essential. Mn(VII)  does not form complexes, but MnO4- anion, often reacting explosively with easily oxidized compounds.

Everything seems to be correct here.

@Poutnik How about a Manganese (0) vs Manganese (II) complex?

@KanghunKim Did you mean everything in my question?

Yes, literally everything.

Can a charge not to have an impact on  other close charge? It changes the potential field and electron probability density.

Everything is correct here, including the explicit examples- the ammonium group is a very strongly inductively electron withdrawing group in electrophilic aromatic subtitution, far negating any hyperconjugative mesomeric electron donation effects, and "permanganamide" is likely to be strongly acidic, with the Mn-O and Mn-N bonds being highly covalent.

Do you have any links for further reading?

I would like to see "permanganamide", from a distance.

https://en.wikipedia.org/wiki/Electrophilic_aromatic_substitution

https://en.wikipedia.org/wiki/Polar_covalent_bond

"Permanganamide" would be definitely explosive, as it contains a nitride anion, which wants to be oxidised to dinitrogen gas, and a high-valent manganese, which wants to be reduced to something like Mn(II). But that doesn't mean that its properties cannot be extrapolated from those of the more stable permanganate ion.

2MnO3NH2 -> N2 + 2MnO2 + 2H2O (irrev.)

Is there a way to explain the higher polarizing power with molecular orbital theory? If I understand correctly, the Mn-O bond in (MnO)5+ would theoretically have more covalent character than the one in MnO. How would you explain this with molecular orbitals? Would the two species have the same molecular orbitals, just with fewer occupied in the (MnO)5+ species?

A higher charged species has a lower LUMO.


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Which is correct - molybdate or molybdenate?
Most websites (including Wikipedia) say that the anion MoO42- is called a molybdate ion, although some websites use molybdenate instead. Which one is correct, and if one of them is correct, is the other term still allowed?

Where did you find the term - molybdenate?

@OscarLanzi Molybdus is just the ancient greek for "lead" (see e.g. https://www.rsc.org/periodic-table/element/42/molybdenum , https://www.britannica.com/science/molybdenum ) - I'm not aware of any mythological aspect to this

Yes, and a Google search of "Molybdus" yields no results relating to a mythological character.

You may find my comment on this interesting. Yes, there is no mythological aspect to it. (cc @IanBush)

I have just checked Wikipedia. Note that the ion is called the molybdate ion, and the term "molybdenate" is just a redirect to it.

If you do a Google ngram of the terms 'molybdate' and 'molybdenate', you will find 'molybdate' standing out.

No one known why the 'en' was kept when you can simply shorten it to molybd-ate instead of molybden-ate (like phosph-ate and not phosphor-ate). However, the latter didn't got totally eliminated. Some research papers(here is a recent example) and textbooks still uses 'molybdenate'. The following image is from a physics textbook published in 2020.

Same thing with tungstate vs tungstenate: https://imgur.com/75IPtBq

I assume that for many non-native speakers (and possibly some natives too), the short right forms may not feel right by their guts and if they do not know they are actually right, they may use the longer form. E.g. I have learnt English - aside of very basics - after my formal education, therefore I had later often struggles with technical or scientific terminology.

According to Table X of the Red Book (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’), the anion name for molybdenum is molybdate.

Does the Table X of the Red Book say if molybdenate is allowed?

@Praseodymium-141 "molybdenate" is not mentioned in the Red Book.

I guess molybdenate may be in respective forms present in non EN language variants, that may occasionally by a sneaky way backport to English, if a chemist is unsure.  E.g., in Czech, "sodium molybdate" is "molybdenan sodný".


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Why isn't formyl chloride stable?
I am currently studying methods of synthesising ketones and aldehydes and one of the methods is Friedel-Crafts acylation. However, my teacher told our class that we can't use this to make benzaldehyde as formyl chloride, HCOCl, is unstable. Why is HCOCl unstable?

For formylation you'd need HCOCl - it's indeed unstable, but it can be done nontheless.

Oh,OK. Thank you. But why is it unstable though?

The problem of the stability is not yet perfectly understood in chemistry ?. Why is SO not really stable at  room temperature ? Why is NO so stable at room temperature ?  Why does HNO not exist long enough to be stored in a bottle ?


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Crystal Field Splitting in Octahedral Complex
So i was going through my 1st year engineering chemistry textbook as a i have test in about a few weeks. So my textbooks says:
"Repulsion by six ligands in an octahedral complex splits the d orbitals
on the central metal into t2g and eg levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d
electrons are symmetrically arranged, they will repel all six ligands equally.
Thus the structure will be a completely regular octahedron."
So i understood the first 2 lines. Basically what my understanding says is that in an octahedral complex the ligands usually approach the central metal along the x, y and z axis. So obviously the orbitals which lie along the axis (dx2-y2 and dz2) will rise in energy which is your eg level  than compared to the orbitals which lie in between the axis ( dxy, dyz, dxz) which your t2g levels. What i am not able to understand is the part where they say "If the d electrons are symmetrically arranged, they will repel all six ligands equally". The electrons in eg level rise in energy because they face more repulsion from the ligands right? so how is it possible that the d electrons will repel all ligands equally? Shouldn't the electrons in the eg level repel them more than the electron in the t2g level? I even tried to make my self understand by taking an example of a d5 configuration(which is symmetrically arranged) in the presence of a weak field ligand where two of your electrons would be in eg and the other 3 electrons would be in t2g level, but still I wasn't able to understand how is it that the d electrons are repelling the ligands equally?

Confusing wordings indeed in your textbook. For first year material, Chem LibreTexts are far more reliable. Read the section of CFT there. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Theory

And please capitalize "i's" in your post. I understand South Asian languages do not have capital letters but still "I" is "I".


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Stability comparison of europium(II) and ytterbium(II) ions
“Which one of the lanthanoids given below are most stable in divalent form?” This question was asked in JEE Main 2022 (Indian entrance test for engineering). Now the question had both $\ce{Eu^2+}$ and $\ce{Yb^2+}$ as the options.
My thought was that since $\ce{Yb^2+}$ has $f^{14}$ configuration it should be somewhat more stable than the half filled $\ce{Eu^2+}$. The answer key however says that $\ce{Eu^2+}$ is more stable. It would be helpful if someone could explain why that could be true. Relevant data of Reduction potentials would be helpful as well ( I tried to find it but without much success).

For Redox potentials, see https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29

Maybe in  $f^{7}$ there is less pairing interactions spin-spin than in $f^{14}$

Also Yb 2+ is smaller than Eu 2+ so the electrons are on average closer. But you can probably hand wave yourself to whatever conclusion you want.


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Can hydrogen fluoride form three or more hydrogen bonds?
As far as I know, each $\ce{HF}$ molecule has two hydrogen bonds, one formed by its hydrogen atom and one which the $\ce{F}$-atom forms with hydrogen atom of a third $\ce{HF}$ molecule.
In other molecules, like $\ce{H2O}$, where both of oxygen's lone pairs are used in hydrogen bonding each water molecule forms as many as four hydrogen bonds. Why does fluorine, which has three lone pairs, not use all its lone pairs up to form three hydrogen bonds instead?

Then which part of these very electronegative atoms are the hydrogen atoms attracted to?

They're just attracted to the atom, not any part as such

There are not enough H atoms to use all lone pairs.

@Infinite (cc ananta) Thank you for doing editorial work, this has been really helpful. One nuance: we prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. Also, it's a good practice to avoid chemical formulas and abbreviations in titles altogether whenever possible.

https://chemistry.stackexchange.com/questions/20149/how-can-the-hydrogen-bonds-in-solid-hf-be-best-represented

Hydrogen fluoride has only one hydrogen atom per fluorine atom, making a branched hydrogen-bond network difficult to form. (This is one reason hydrogen fluoride also forms hydrogen-bonded structures in the gas phase, unlike water where hydrogen-bond branching is more likely to occur and drive condensation.)
Ammonium fluoride, $\ce{NH4F}$, is not so constrained. In this compound each fluorine can use all four of electron pairs to form bonds that are apparently both hydrogen bonds and ionic bonds:

Ammonium fluoride adopts the wurtzite crystal structure, in which both the ammonium cations and the fluoride anions are stacked in ABABAB... layers, each being tetrahedrally surrounded by four of the other. There are N−H···F hydrogen bonds between the anions and cations.[1] This structure is very similar to ice, and ammonium fluoride is the only substance which can form mixed crystals with water.[2]

Cited References

A. F. Wells, Structural Inorganic Chemistry, 5th ed., Oxford University Press, Oxford, UK, 1984.

Brill, R.; Zaromb, S. "Mixed Crystals of Ice and Ammonium Fluoride". Nature. 173 (4398): 316–317. https://doi.org/10.1038/173316a0.

I guess in the context of hydrogen bonds, we should also consider $\ce{2HF <=> [H2F]+ + F-}$. (You might want to correct how you spell hydrogen.)

@martin I have a major problem typing on my device. Fingers are too fat for the tiny target offered by each letter. Thanks for the catch.

I do know exactly what you mean, I have somewhat similar issues sometimes. Anyway, I just noticed, in my last comment I've posted the wrong equilibrium. I meant of course: $\ce{2HF <=> [HF2]- + H+}$. Silly me...

Right, but note that $\ce{HF2^-}$ does not force more than one hydrogen bond from fluorine. Hence my choice.

No, exactly. I agree with what you wrote. The equilibrium is posted last still would also support the notion, why a network is hard to form though.

In a true intermolecular hydrogen bond, the X-H---Y bond angle is approximately 180$^\circ$. Thus, each H atom can only participate in one H-bond.
In pure HF, the ratio of H:F is necessarily 1:1. Since each H-bond requires one H and one F and each H can only participate in one H-bond, the F atoms necessarily participate (on average) in only one H-bond each as well.

So is the number of atoms participating in a hydrogen bond is proportional to the number of hydrogen atoms forming hydrogen bonds?

Yes, necessarily the number of H-bonds is determined by the number of H atoms that participate in H-bonds, since an H-bond must have an H.  But when you are counting H atoms involved in H-bonds, remember that sometimes the number of H atoms is not limiting. For example, in NH3 the N has only one lone pair, so only 1/3 of the H atoms can participate in H bonds.

Added "intermolecular". Our arguments apply in that case; intramolecular bonds may be constrained by overall molecular geometry. Then +1.

$\ce{HF}$ can not form two hydrogen bonds since the two positively charged $\ce{H}$ atoms would repel each other.

Also, unlike it has been said in the comments, $\ce{H}$-bonds do involve orbital overlap and are not a consequence of electrostatic attraction alone. Although the overlap does not result in very strong bonds, as is the case with overlapping in covalent bonds, $\ce{H}$-bonds can be surprisingly stable ($\Delta G \approx 3\text{kJ/mol}$).
Never thought I would find a possible explanation while comparing neighbor interactions. If we assume that the the zig-zag structure for $\ce{HF}$ is correct, we can not rule out the repulsion caused by the lone pairs of fluorine.

When there are sufficient H atoms available, F will happily bond with more than 2. For example, the solid crystal structure of NH4F is essentially identical to the ice-Ih form of solid H2O, with each F coordinated by 4 equivalent H atoms, just as O is coordinated by 4 H atoms in ice-Ih.

@Andrew even in questions comments, the number of hydrogen atoms vs. the number of fluorine atoms has been pointed out. However, I do not think the limitation is the number of hydrogen atoms, actually it is energetic considerations.

If you have a source for that claim, do cite it.

@Andrew You didn't cite a source for your claim either, don't you think it is unfair to ask me to do the same? An example of fluorine forming more than two bonds doesn't really answer the question if there can be more than three hydrogen bonds per $\ce{HF}$ molecule in an $\ce{HF}$ solid/liquid structure.

The claim for HF is based on simple arithmetic, and Oscar lanzi has provided citations for my statement regarding NH4F, so I'm not sure what other citation would be needed.


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What does the word "dynamics" mean in the context of proteins or biomolecules in general?
What does it mean by “dynamics” when we say “dynamics of protein” or “dynamics of biomolecules”? For instance, McCammon's paper has the title "Protein Dynamics" [1]. What does the word “dynamics” mean in this case?
Please, supply me with at least one reference.
Reference

McCammon, J. A. Protein Dynamics. Rep. Prog. Phys. 1984, 47 (1), 1–46. DOI: 10.1088/0034-4885/47/1/001.

It just means movements

And why not read the paper, or at least Wikipedia https://en.wikipedia.org/wiki/Protein_dynamics

@Mithoron, en.wikipedia.org/wiki/Protein_dynamics --- this is a total mess and utterly useless.

From Lottspeich' Bioanalytics, subsection Determination of Protein Dynamics [1, p. 463] (strong emphasis mine):

Determination of Protein Dynamics
Proteins are not rigid, static entities, but rather exist as ensembles of conformations.
The internal motions, which give rise to transitions between different structural states, are collectively referred to as protein dynamics and occur on a wide
range of time scales (ns–s).
Protein dynamics play crucial roles in protein functions particularly for the interaction with binding partners, in enzyme catalysis, and in allosteric regulation.
Luckily, many NMR spectroscopic parameters depend both on the mobility of the whole molecule (translational or rotational motions) and on internal motions (transitions between different conformations or bond rotations), providing insight into a wide variety of dynamic processes ranging from fast ﬂuctuations (lasting picoseconds) to slower conformational changes (lasting microseconds or more).

Protein dynamics studies have been used in biochemistry for studying genomics [2], enzymology [3] and many other areas, including GPCRs activity research which lead to the 2012 Nobel Prize in chemistry.
References

Bioanalytics: Analytical Methods and Concepts in Biochemistry and Molecular Biology, 1st ed.; Lottspeich, F., Engels, J. W., Eds.; Wiley: Weinheim, 2018. ISBN 978-3-527-33919-8.
Misteli, T. Protein Dynamics: Implications for Nuclear Architecture and Gene Expression. Science 2001, 291 (5505), 843–847. DOI: 10.1126/science.291.5505.843.
Klinman, J. P.; Kohen, A. Hydrogen Tunneling Links Protein Dynamics to Enzyme Catalysis. Annu Rev Biochem 2013, 82, 471–496. DOI: 10.1146/annurev-biochem-051710-133623.


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Can Thiamine Pyrophosphate be Prepared from Thiamine Hydrochloride by chemical means?
Is there a quick and easy way to prepare Thiamine Pyrophosphate in the lab (i.e. without using micro-organisms) starting by Thiamine Hydrochloride (by a simple displacement reaction maybe)?

Give some more details. Why do you need this?

Thiamine pyrophosphate is commercially available e.g. here This patent describes a preparation from thiamine hydrochloride.

Would you by any chance know how to increase the yield (7%) of Thiamine Pyrophosphate in the first step of the reaction?

Vary temperature and time if you are trying to improve the process, but as it is commercially available I would just buy it if it is for a one-off use..


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What is the standard method for activating molecular sieves with microwave radiation?
Typically, molecular sieves are dried under vacuum at elevated temperatures (about 200 to 300 °C depending on pore size) for several hours.
Alternatively, microwave heating has been used for preparation of mol sieves at least since 1980s and takes minutes instead of hours, but a standard protocol for this procedure appears to be something left to be desired.
Early technical report Microwave regeneration of molecular sieves [1] illustrates the efficiency of the method: over 90 % of water was removed under 10 min mark.
However, as convenient as it sounds, there is an issue of the microwave radiation being equally readily absorbed by both water and the sieves.
This causes overheating the sieves (up to 500 °C and higher) and, arguably more importantly, temperature non-uniformity causing sintering, change in porosity and cracking of the sieves.
The general recommendation was to use short regeneration cycles at full power and apply purge gas (air) flow with temperature control.
The report, however, only relies on the data obtained for microwave source operating at 2450 MHz and one of the recommendation for future studies was to test other frequencies.
Kappe's work from 2009 [1] draws a similar conclusion and calls for further optimization:

However, the use of these materials in conjunction with microwave heating is further complicated because molecular sieves are known to be very strong microwave absorbers.[10]
In fact, one of the best ways to dry/activate molecular sieves is to irradiate them with microwaves.[11]
[…]
[11] Although many chemists apparently use this method in the laboratory, we were not able to find an appropriate reference in the chemical literature.
According to our experience, the procedure given in the Experimental Section provides reproducible results but no efforts were made to quantitatively assess the water content of the sieves obtained using this protocol.
We believe that main reason to use microwaves is convenience due to rapid heating.
However, care must be taken not to overheat the sieves by prolonged microwave irradiation.
This will result in material with decreased efficiency for water adsorbance.

There are numerous websites with various recommendations for using microwave radiation for mol sieves activation (Reddit, Chemtips, Sciencemadness) and papers targeting narrowed-down applications [3], but it looks like a standardized procedure with recommended microwave frequency range and heating time strategy from an authoritative source is missing, and the recommendations are sometimes contradicting.
If the answer turns out to be dependent on morphology of the sieves, for the sake of simplicity it probably would make sense to post one for the most commonly used ones in the lab setting (3 Å beads, 8–12 mesh, I presume).
References

Singh, V. P. Microwave Regeneration of Molecular Sieves; F84023; Canada, 1984. (PDF)

Baghbanzadeh, M.; Kappe, C. O. Can Molecular Sieves Be Used as Water Scavengers in Microwave Chemistry? Aust. J. Chem. 2009, 62 (3), 244. DOI: 10.1071/CH08450.

Sârbu, E.; Călinescu, I. Microwave Assisted Regeneration of 3Å Molecular Sieves Used for Ethanol Dehydration. U.P.B. Sci. Bull., Series B 2014, 76 (1).


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Should change in enthalpy be zero in galvanic cells?
When an exothermic redox reaction happens by just mixing the reagents, the change in enthalpy at constant pressure equals heat which is consistent with the fact that:

For a closed system at constant pressure, change in enthalpy equals heat.

This can be derived from the definition of enthalpy and the first law of Thermodynamics.
When the same reaction takes place in a galvanic cell, there is no heat involved so the change in enthalpy must be zero. But as we know, the energy in galvanic cells is released in the form of electrical work. So $ΔH$ must be translated into electrical work which is inconsistent with the fact that $ΔH = Q_p$ (with $Q_p$ being zero in our case).
I want to understand where is the flaw in the above reasoning and how can we relate the change in enthalpy with the electrical work done by the galvanic cell.
I have looked here Link1 but the accepted answer is in contrast with the statement in the quotes (see also Link2 for the quoted statement).

$\pu{\Delta H = Q_p}$ can be positive or negative in a galvanic cell. It could in principle be equal to zero, although this result could only be due to a strange coincidence. But the enthalpy cannot be "translated" into electrical work, as you state. $\pu{\Delta H}$ cannot be directly related to the electrical work. Knowing the change of entropy is necessary to carry out this transformation.


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How do i know the order of the reaction?
$$\ce{2N2O5 ↔ 2N2O4 + O2}$$ is this reaction first order or second order ? And why ?
I know first order is when duplicating the concetration the velocity duplicates too, and secod order reaction is of order n if, when doubling the concentration, the speed increase by $\pu{2^n}$. For this cases i need the velocitys and concentrations but i only have the initial concetration and 5 hal life times

$$\ce{N2O5}= 2,4 mols_0$$
I dont know what to do, cuz to know, would have to have this data . Am i wrong ?

Temperature dependence is not useful here. You would need at least 2  timed data for the same temperature.

Order is an experimental data. It is never possible to derive the order of a reaction from its chemical equation. The order has to be determined by measuring the rate of a chemical reaction (in $\pu{mol/s}$, or in $\pu{mol L^{-1} s^{-1}}$) , and plotting a function of this rate vs. time. If the logarithm of the rate decreases linearly with time, the order is $1$. If the reciprocal of the rate increases linearly with time, the order is $2$. This is not dependent on the chemical equation of the reaction. In both cases, the slope of the line is the rate constant, $\pu{k_{1}}$ or $\pu{k_{2}}$.
This is what you should have to do with your experimental results about the decomposition of $\ce{N2O5}$. But here all you know about this reaction is its half-life. And this parameter (half-life) is a characteristics and a constant of the first-order reactions. So from this very property, you can state that the reaction is first order, without doing any calculations.
As the half-life is known at different temperatures, it probably means that, in a second time, you would have to calculate the activation energy of the reaction.

But have half life for second-order reactions. $$t_1/2\frac{1}{K*[A]_0}$$


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Zwitterionic Form in Amino Acid
In my laboratory workbook, it has been mentioned that the zwitterionic form of the amino acid is the dominant form of the amino acid at pI and no movement of the molecules takes place under an external electric field. I am a bit confused by this statement.
It is clearly not possible for all the molecues to exist in the zwitterionic form. Some of them must exist in any of the other charged forms, even at pI. This should lead to these few charged molecules, moving in an electric field.
My question is:

Is it true that all molecules are not zwitterions at pI?
When we say no migration of ions takes place at pI, do we actually mean no "NET" migration of ions occurs i.e. whatever number of ions move towards one side, causes an equal number of ions to move towards the other?

Try to think longer about it. It would lead to less questions. What happens to trace cation/anion forms if they migrate to regions with less than equilibrium concentration of the zwitterion?

@Poutnik They get converted to the neutral zwitterions. But still, in some other part of the solution, when zwitterions tend to be at an excess, the reverse reaction will dominate. Then charged molecules should form?

Check typical respective pKa and concentration ratios for used pH. Involve imagination of molecular dynamics It is quite analogical to chromatography and dynamic redistribution between both phases. (zwitterion like in stationary one and other forms like in mobile one). But here direction of "mobile phase" goes both ways.


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Ferrocene's band gap is supposed to make it colorless but it isn't. Why is it so?
Wang and Islam [1] state that the ferrocene molecule's band gap, i.e. the HOMO-LUMO gap at the Kohn-Sham level, is 5.03 eV, which corresponds to an absorption wavelength of at most 240 nm.
So, ferrocene should be colorless — except that it isn't. I found out that liquid water, which has an even higher band gap (7 eV) than ferrocene, is still colored because of vibronic, and not electronic, excitations.
Does the same hold for ferrocene as well, and, if it does, do the expected effects to the HOMO-LUMO gap of e.g. π-electron withdrawing/donating substituents (which would change the color of the parent molecule, were that color be dependent on the gap — again, except that it isn't) “change” the color in the “expected” ways?
P.S. Please read my long string of comments, involving explicit calculations done for the ionisation potential/negated Kohn-Sham HOMO energy of the hydride anion, on Oscar Lanzi's answer before adding more answers.
Reference

Wang, F.; Islam, S. Impact of ionization of ferrocene: EOES of α- and β-electrons and the fingerprint orbital 8a₁’ of ferrocenium. 2015. DOI: 10.48550/ARXIV.1509.05122.

Comparing the color of water with that of ferrocene is like comparing the atmosphere of the Moon with that of Earth. Technically the Moon has an atmosphere, but by standards we typically work with on Earth -- really?

Well, there's obviously some electron transition in there, other than what you mention. I don't think that's a right paper to tell the actual situation, better check out something about UV-Vis.

Please do not refer to any comments. Edit your question as a single concise post. This is not a discussion forum and comments are only here to improve on existing posts. Plus, they are gone now.

All of the traditional DFT is Kohn-Sham, that really does not specify any level of theory. That also does not provide any insight in which way they have confirmed that they are not using an outlier functional. DFT isn't known for this kind of accuracy and I expect a whole range of HOMO LUMO gap values for different functionals. A good paper would review these, and calibrate, and discuss.

They've used b3lyp with a split valence basis set because that reproduced the ionisation potential. And that was their only concern.  This level of theory is known to be right usually for the wrong reasons, error compensation and the lot. Also, they've used Koopmans' theorem, which in itself only works because of error compensation. Funny enough, they based parts of their argument on Hartree-Fock. They also don't discuss spin contamination (or I missed this). I have to be honest though, I didn't read the whole paper, it was too painful to follow.

Related: https://chemistry.stackexchange.com/q/16186/4945


text:
Why is oxygen an anomaly in this context?
In A Very Short Introduction: Earth System Science by Tim Lenton, the author discusses the composition of atmosphere:

the first observations from land-based telescopes showed that Mars had an atmosphere dominated by carbon dioxide, just as would be expected in the absence of life. So too does Venus. But the Earth has a remarkable atmosphere, containing a chemical cocktail of highly reactive gases, sustained by life.
Oxygen is the prime anomaly—at just over a fifth of Earth’s atmosphere it is essential for our existence as mobile, thinking animals, but without photosynthesis to create it, oxygen would be a very rare trace gas. Mixed in with oxygen are gases like methane that react eagerly with it—so much so that they are on the verge of combusting together.

Why is oxygen an anomaly in this context?

I’m voting to close this question because it's both hardly on topic and pretty much already answered in citation in the question itself.

Because oxygen is so reactive it won't be present in a planetary atmosphere unless there is a source continuously producing it
In the context of things found in planetary atmospheres, oxygen is a major anomaly (or, to put this another way, the earth has a very unusual atmosphere).
The reason is simple. Oxygen is so reactive that an atmosphere can only contain a significant amount if there is something continuously producing large amounts of it. On earth, that is life or more specifically living things that photosynthesise.
If photosynthesis stopped on earth it is estimated that there would be about enough oxygen to last only 20,000 years (based on the estimated rate of photosynthesis which is broadly in balance with the rate of consumption). Even if all the air-breathing animals died the oxygen would get used up pretty quickly by other processes like burning, rusting and geological weathering.
The atmospheres of planets typically depend on geological processes. But though some of these can produce some oxygen, there are many, faster, processes that use it up. So we don't expect to see free oxygen in a planetary atmosphere  where the planet has no life.

The conclusion may be true on "planets", but not on some other bodies. See my remark about Ganymede.

The OP was reading the citation literally and maybe thinking that oxygen was the anomaly. The real anomaly is the atmosphere of the Earth among those of other planets. Oxygen is simply the indicator of this anomaly. I don't get what the citation says about methane being in almost explosive proportions to oxygen in our atmosphere. In fact the proportion of methane in air is 1.7 ppm. Surely a much bigger proportion is needed for combustion ?

@OscarLanzi That is a fair point. OTOH the "atmosphere" of ganymede is so tenuous it was once thought to be nonexistent. So it is something of a edge case: a body with enough water and surface radiation to yield oxygen by photochemical splitting. But in very, very small amounts.

@Trunk An atmosphere with a lot of oxygen is the anomaly (so i'm not sure the phrasing of the quote is wrong). But notable amounts of methane is also an anomaly (doubly so in an oxygen rich atmosphere). Earth methane is nowhere near the explosive limit but there are plenty of reaction pathways that leave it with a very short atmospheric lifetime. Notably, methane on Mars is an anomaly (if measurements are accurate) as it is destroyed rapidly even in the absence of oxygen and most known processes making it are biological.

@matt_black Titan's atmosphere is made mostly of methane - it's not so unusual.

For the reasons mentioned in the quotation, oxygen -- unlike other major atmospheric components like nitrogen, argon or carbon dioxide -- cannot survive from primordial times and must be created by some active, ongoing process.
On Earth that process is, of course, life. But on some other worlds there are alternatives, namely spalling of oxygen atoms off rock or ice by ultraviolet radiation. Jupiter's moon Ganymede, with its water-ice surface, has a thin oxygen-bearing atmosphere.

Oxygen is not an anomaly when talking about "reactive gases" in the atmosphere as the author clearly mentions that oxygen is also highly reactive. In fact there are less reactive gases (nitrogen) present. I think the author means to say that oxygen is the anomaly when comparing atmospheric composition of planets. Other planets will contain all kinds of gases (but not oxygen). Here, it says that planetary atmospheres consist of many gases including carbon dioxide, methane, ammonia, and hydrogen and helium on distant planets. The Earth has a unique atmosphere of oxygen which helps sustain life.


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How can we define molar conductivity as the conductance of 1 ml of electrolytic solution?
I've been taught that conductance ($G$) of an electrolytic solution is the inverse of the resistance of the solution, and conductivity ($\kappa$) is the inverse of the resistivity. From this, we have
$$\frac 1 G = \frac 1 \kappa \frac l A$$
where $l$ is the distance between electrodes and $A$ is the area of the electrodes.
$$\kappa = G \frac l A$$
Therefore, when $l=1$ cm and $A=1$ cm$^2$, $\kappa = G$, i.e.

$\kappa$ can be defined as the conductance of $1$ ml of the electrolytic solution.

I don't think the above definition is a good definition. When $l=2$ cm and $A=0.5$ cm$^2$, the volume between the electrodes is still $1$ ml, but the conductivity is
$$\kappa=G \frac 2 {0.5} = 4G \neq G$$
However, in the class I attended, they proceeded to use this definition to derive the formulae for molar and equivalent conductivity by the unitary method (if the conductivity of $1$ ml is $\kappa$, then the conductivity of $1000$ ml is $1000 \kappa$ ...)
Is this a correct definition? Am I missing something?

$\ce{G}$ and $\ce{\kappa}$ can not be equal, because they don't have the same unit. Equalling these two parameters is similar to saying that $\pu{1 m = 1 m/s}$

Solution conductivity depends only on properties of solution itself (intensive property).
Solution conductance depends on solution conductivity AND geometry of the measurement.
Conductivity  has

the integral definition $G = \kappa  \cdot \frac Al$
the differential definition $\vec J = \kappa \cdot \vec E$, defined by the current density and potential gradient.(*)

Conductivity (in $\pu{S cm-1}$) is numerically equal to conductance of 1 mL if and only if the volume has the shape of a cube with 1 cm long edge.
Conductance for 10 cm cube is 10 times greater, but conductivity is the same, depending just on the solution itself.
Molar conductivity is not defined by conductance, but by conductivity, divided by molar concentration.

(*) For solid phases, conductivity is generally a tensor and vectors of potential gradient and current density need not to be parallel for anisotropic materials.


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Why are nitriles less basic than amides?
Like nitriles have a localised lone pair which amides do not have, as their lone pair is delocalised and  still amides are more basic. Why?
I could not find any satisfactory explanation so please explain in detail.

In amides, it's oxygen that is more basic.

Does not make sense, as nitrogen of cyanide being less electronegative and having a delocalised lone pair must be more basic.

Well, in protonated amide you get 2 full octet structures, in nitrilium, only one. Such analysis is hardly important, though, it's experimental data that matter. You didn't bring any reference...

related https://www.sciencedirect.com/science/article/abs/pii/S004040201100545X https://pubs.acs.org/doi/10.1021/jo800674h

From tags I assume you mean nitriles like acetonitrile CH3CN and amides of carboxylic acids like acetamide CH3CONH2 (and not anorganic counterparts like KCN and KNH2)

Yes I meant that


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Why does symmetrical distribution of electrons in d and f orbital lead to greate stability of these or orbitals in comparison to s orbital?
My high school textbook mentions the following as on of the reason for why Cr and Cu have thier 3d orbitals filled completely before their 4s orbital.

The extra stability of d and f orbital is due to:
Symmetrical distribution of electron :It is well known that symmetry leads to stability . The completely filled or half - filled subshell have symmetrical distribution of electron in them and are therefore more stable . This effect is more dominant in d and f - orbitals . This means three or six electrons in p - subshell , 5 or 10 electrons in d subshell and 7 or 14 in f - subshell forms a stable arrangement

I don't understand why symmetry should lead to greater stability. So I asked my teacher but I was unsatisfied by the explanation and looked over some books and websites but I couldn't understand why symmetrical distribution of electrons should lead to greater stability.
The completely filled s orbital also has Symmetrical Distribution of electrons(at least According to me ,correct me if am wrong) so why are the d orbitals filled completely before the s orbital?How does symmetry leads to greater stability?

I tried to add this as a comment, but I don't have 50 reputation points so I'm replying here.
I don't think symmetry of half-filled orbitals plays much of a role in the stability, look at the electronic configurations of Nb, Tc, Ru, Pt for example. They too have exceptional electronic configurations where the electron enters d orbital instead of s orbital, but in their cases it is not due to partial or full filling of the d orbital. Also, W (tungsten) has a configuration of d4s2, and not d5s1 as would be expected had symmetry of orbitals been the sole reason for stability.
I can't prove this, but it might be because s orbital being small and spherical in shape has less distribution of electrons, and consequently has greater electron-electron repulsion as compared to the d orbital, which might be resulting in these particular electron configurations with an electron in the d orbital instead of the s orbital to be more energetically favourable by having less e--e- repulsions.


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amorphous silica (diatomaceous earth) dissolution and precipitation using household bleach or ammonia
I'm a layman not a chemist. My question is whether it's possible to dissolve amorphous silica in room temperature water mixed with a commonly available alkaline household product and then have it precipitate naturally out of solution over a few days or weeks. To dissolve a kilo of amorphous silica, what would be the right amount of over-the-counter laundry bleach or household ammonia and room temperature water?
I'm looking for a way to prevent nests of tiny stinging red fire ants under patio pavers that will be laid on sand and jointed with sand. Diatomaceous earth (amorphous silica) is said to be fatal to ants when they come into contact with it, so I wondered if spraying the screeded sand underlayment in situ and also the sand in the joints until it was quite wet with the diatomaceous earth solution might do the trick.
I also thought it would pose a lower health risk and involve much less discomfort to mix up a batch while wearing a respirator and gloves than to work with the diatomaceous earth in powder form, mixing it dry with the sand, over the several weeks it will take me to lay the pavers.

Silica is not soluble in soda or any alkaline solution at rhom temperature. It can be  dissolved and transformed into a silicate solution in concentrated NaOH solutions but only at high temperature (above 100°C) and under pressure.

Once it's been wet, diatomaceous earth is less effective, so putting it into water, even as a suspension, defeats it's purpose. For that matter, it needs to be reapplied if wet by rain or damp earth. According to BugTech, "sharp edges make tiny cuts in an insect’s exoskeleton upon contact, and the diatomaceous earth then absorbs the oils and fats from its outer layer." If the silica is wet, it will hydrophilic, not oleophilic, and would not absorb oils.
Non-chemistry suggestion: disperse it with air. Use an old container from "talcum" (now starch, for safety) powder, or a container for boric acid insecticide. These squeeze containers are designed to spray out the powder, propelled by air. Diatomaceous earth is sold in air-spray bottles, too. It might be feasible to use a sandblasting gun on large areas, with proper respirator -- though at the risk of silicosis.

Is it because the rain washes the diatomaceous earth powder off the leaves that it becomes ineffective for gardeners? Or does the moisture interfere with its effect on the membranes of the bugs, possibly by "sheathing" the microscopic sharp edges?  In my case, I was thinking of placing the powder under pavers where it would not be washed away by rain but it would almost always be damp.

@Tim, it likely would be ineffective under pavers... see edited answer.

I can't tell from my reading whether the ant dies from desiccation that results solely from its exoskeleton being pierced by the sharp edges of the diatomaceous earth or as a result of the combined effects of that exoskeleton piercing and the desiccant qualities of the diatomaceous earth.  If the latter, then damp DE won't kill the ants, as you said, since it won't be drawing moisture from the ant.

Here's an article on recent developments in hydrophobic silica applications:  https://journals.ashs.org/hortsci/view/journals/hortsci/45/9/article-p1349.xml That type of amorphous silica might be better suited to my ant-killing requirements in a damp environment.  But the article speaks of greenhouse humidity, not outright wetness.

Found this patent relating to insecticidal use of hydrophobic silica, though it mentions "humidity" not wet soil: https://patents.google.com/patent/US3159536A/en

Patent on the use of silica gels in damp conditions to control insects:  https://patents.google.com/patent/US5192545A/en


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Why is orbital energy not the mean between ionization energy and electron affinity when the orbital has two electrons?
In this question it is asked why ferrocene is colored when the HOMO-LUMO gap seems to be beyond the visible light range. I tried to answer that orbital energies change with electronic transitions and therefore the transition energy need not match the difference in orbital energies. However, my answer crumbled under unexpected pressure and I wound up deleting it.
I tried to use a hydrogen atom as an example of changing orbital energy levels. When an electron is added to make a hydride ion, the combined energy of both electrons in the 1s orbital should be -13.6 eV (from the ionization energy of the neutral atom) plus -0.8 eV (from the affinity for the additional electron). With this total energy distributed between two electrons in the orbital the 1s orbital energy seemingly should br -7.2 eV (per electron) in the anion, versus -13.6 eV for the neutral atom when there is only one electron in the orbital. But I was barraged with comments that this analysis is wrong; the orbital energy in the hydride ion is only -0.8 eV from just the electron affinity. I do not understand why, and am baffled because this rendering seems inconsistent with conservation of energy.

Why does only the electron affinity figure into the orbital energy?

How does such a rendering square with conservation of energy?

Please don't tell me I am wrong, direct me to getting right.

I think the confusion stems from a misunderstanding of the meaning of the orbital energies.
Consider the hydrogen atom and its orbitals. You can obtain the energy of the electron in its groudn state using the Schrödinger equation
$$ \hat{H} \Psi = E \Psi  $$
For this atom you can write the hamitltonian (in atomic units) as
$$ \hat{H} = -\frac{1}{2}\nabla^2 -\frac{1}{R}$$
where $-\frac{1}{2}\nabla^2$ represents the electron kinetic energy (operator) and $\frac{1}{R}$ represents the Coulomb atraction between the electron and the hydrogen nucleus. Thus, the energy of the electron calculated corresponds to the binding energy of the electron to the nucleus or, if you prefer, the energy needed to ionize the atom.
Consider now the H$^-$ anion, the atomic system is different, you have an hydrogen nucleus and two electrons that interact amongst themselves. Thus, the hamiltonian is different too, and, accordingly, the orbitals and its energies are different from those of hydrogen.
The hamiltonian for the anion has the form:
$$ \hat{H} = -\frac{1}{2}\nabla^2(1) -\frac{1}{2}\nabla^2(1) -\frac{1}{R_1} -\frac{1}{R_2} + \frac{1}{r_{12}} $$
where there appears a new term in the hamiltonian $+\frac{1}{r_{12}}$ that represents the repulsion of the electrons.
As @Infinite has mentioned in his answer, you can describe the properties of the anion ignoring the electron-electron interaction.

With this method we would get the energy of the two electrons in the orbital. Would the orbital energy then become half that, per electron?

When you solve the Schröedinger equation, besides the total energy $E$, you also obtein both the orbitals $\phi_i$ and the energy of electrons in the orbitals ($\epsilon_i$). However the total energy, $E$, is not simply the sum of the energy of the electrons, $E \neq \epsilon_1 + \epsilon_2$. Why? Because  the energy of the electrons in the orbital includes the repulsion energy with the other(s) electron(s) and simply adding the energies of the electrons will overstimate the interlectronic repulsion energy. Hope this help.

I accepted this answer with one caveat. If you spell Schrödinger with the umlaut over the o, you do not also use the following e.

The orbital energy simply refers to the ionisation energy. So the orbital energy of the hydride ion should be numerically equal to the electron affinity value of the hydrogen atom.
And there is no violation of law of conversation of energy. For removing the first electron we should supply 0.8 eV and for removing second electron we should supply 13.6 eV for an atom. And I see that there is no loss of energy ( $\because$ the total energy of hydride ion is (13.6+0.8) eV in magnitude).
An explanation to the increased orbital energy from -13.6 eV to -0.8 eV is due to the electron-electron repulsions.

Please correct me if I did anything wrong because I don't have a good background in quantum mechanics.

Thank you, but what I needed to know is why the electronic energies do not add up as I expected they would.

If energy is distributed in such manner then 7.2 eV of energy is required for the removal electron but actually we need just 0.8 eV of energy for removing that electron. This energy is lesser than the energy expected by you, it is because electron-electron repulsions makes our task easy.

my interpretation would be 7.2 eV is required to remove either electron but 6.4 eV out of that is transferred from the remaining electron, this transfer being due to the disappearance of interelectronic repulsion.

Let us say we have supplied 1 eV of energy. If we don't observe any removal of electron then you interpretation could be correct. But I don't have enough sources for supporting this. May be someone can look into this with experimental results.

If the electron is removed, it could have taken energy from the other electron plus the supplied energy.

Yes this could be also a case...

Orbital energy but for H cannot be equal to ionization energy, as total energy of the  other electrons before and after ionization is not the same.

I think the confusion stems from a misunderstanding of the meaning of the orbital energies. If you consider the hydrogen atom, you can obtain the energy of the electron in its grouns state using the Schrörinder equation and the hamitltonia


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Regioselectivity of SN2 reaction on adenine
I found the following question in my examination:

I was able to make out from the options that the usage of a base is to remove the supposedly acidic hydrogen from either the Nitrogen $A$ or $B$. Then the Nitrogen with a negative charge would attack $\ce{CH3I}$ through an $\ce{SN2}$ path and hence overall, the Hydrogen is replaced by methyl.
Now I got stuck on deciding which Nitrogen would be involved here? Both A and B have reduced electron density as their lone pairs are delocalised.
A point of difference maybe that $A$ isn't directly responsible for aromaticity in the 6 membered ring while $B$ is directly involved in aromaticity of the 5 membered ring.
But I don't know if that creates a difference. It would be great if someone could highlight where I'm going wrong and how was I supposed to judge in the examination as I did not have access to $\ce{pK_a}$ values and other experimental data.

Some papers online suggest existence of different paths than this, but considering the level of the exam (JEE-main) it came in, real experimental reaction isn't much considered and more of theoretical approach is considered. I'll reframe my question.
Say that attack had to take place by either Nitrogen 'A' or, 'B' (Which is what the question hints us by using a base and so are the options framed in such a way to confuse)
Which Nitrogen would be preferred?
The correct answer given is (4) that is, Nitrogen $B$ is considered. How can we justify and explain this ?

1 is the major product according to this patent https://patents.google.com/patent/SU1100276A1/en

@Waylander thanks for that source, but that doesn't match with the answer given in my exam... Now obviously, a practical experiment is clearly what really occurs. But then I believe there is a difference in the patent and the question. The patent just uses DMF solvent while the question also has a base in it. While in the first case the SN2 reaction would directly occur, and hence most basic Nitrogen will attack $\ce{CH3I}$ but in our case, the base would first extract an acidic H from 'A' or 'B' Nitrogens and then the attack would take place. I'm not sure if that's whats going on here.

This is possibly solvent and base dependent i.e. whether you use a very strong base to deprotonate the starting adenine (and it would require a very strong base) or just use one to mop up the H+produced by the methylation - the question unfortunately does not specify. I have not be able to find further literature on this, but I do not have Scifinder access. I wonder if this question is based on a real experiment?

I've made edits considering inputs given by you. Rather than looking at what would happen through experimental approach, is it possible to explain the given answer by considering theoretical approaches?

Other posters may feel different about this, but I do not think I could say anything more than "you'll get a mixture" without knowing more details of the conditions.

If a strong base (not just the methyl iodide and the solvent) is present, then it will tend to deprotonate the $\ce{N-H}$ nitrogen on the five-membered ring, as this proton is about as acidic as one on an alcohol hydroxyl group; compare the $\mathrm{pK_a}$ values given for indole and ethanol. This $\ce{N-H}$ proton is exceptionally acidic because the negative charge left behind by removing the proton is effectively delocalized around the aromatic five-membered ring.
So the base removes that proton, and the deprotonated nitrogen then replaces the proton with the electrophile forming product (4).
Incidentally, once the five-membered ring is deprotonated nothing really prevents the electrophile from (alternatively) attacking the other nitrogen atom on that ring. That would give a different product from any of the choices given.


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Equivalence point in relation with law of equivalence
Why is that we apply law of equivalence only at equivalence point during a titration. Isn't the law valid always? In the sense, cant we use law equivalence even before we reach the equivalence point?

We can, but properly.

During titration, the molar amount of the used titrant is equivalent to the respective molar amount of the analyte it has already reacted with.
At the equivalence point, the molar amount of the used titrant is equivalent to the respective total molar amount of the analyte present in the sample.
After the equivalence point, the molar amount of the titrant that reacted is equivalent to the respective total molar amount of the analyte present in the sample.

So the amount of titrant that reacted is always equivalent to the amount of analyte that reacted. At the equivalence point, this is equal to the total amount of analyte and the amount of titrant added. At other stages, there will be excess of analyte or titrant, so the law is valid for the amount that reacted.


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How to neutralize bleach in a washing machine
i’ve seen a couple threads on this website about neutralizing bleach but none of them really say how to do it.
for some context, someone put 7 cups of bleach in our washing machine along with the water without any clothes inside because they wanted to disinfect the washing machine from a specific pathogen. They talked to Clorox on the phone and they said it was okay to use more bleach but now they’re double backing and saying that you weren’t supposed to use more than half a cup.
i’ve done multiple rinse cycles sense in order to wash away any residue but were nervous that because it was so much bleach that the cycles aren’t effective or that we need to do more and will never know exactly when it’s OK to use again.
So I just want to nip it in the bud and neutralize whatever residue that’s in there. I’ve seen people say to use Hydrogen peroxide but I  obviously don’t want to cause any accidents if there are residue in there. I’ve seen a different product but I don’t know if it’s safe to use in the washing machine or how much to use to neutralize 7 cups of bleach.
This experience has been extremely stressful, I don’t want clothing to have residue bleach on it and possibly put  pets and people in danger.
*** addition from og post*** I’ve read online that vinegar is not a good one i don’t know if that because of the possible reaction or because it’s not a neutralizer. Also does any one know of a way to test for residue?
Any advice would greatly be appreciated

Hydrogen peroxide does not neutralize bleach. Since the machine is fairly-well cleaned out by now, a half-cup of vinegar should eliminate any bleach left. If the vinegar smell lingers, give one more rinse, empty. BTW, that much bleach may have greatly shortened life of some parts of the washer -- expect repairs to be needed.

@DrMoishePippik everywhere i look people are saying to use peroxide to neutralize the bleach which i found weird because of the possible reaction of it. I’ve put it to do a heavy load four times almost everyday since so i figure it’s okay now but since it’s so much we’re terrified of it

putting vinegar in the washing machine also doesn’t make me feel good. Won’t it cause a reaction?

Hydrogen peroxide is another bleaching agent. It makes no sense at all to use it to destroy Clorox. Clorox is highly water soluble, the multiple rinse cycles you have run will have washed it out.

i don’t care about the machine. I’ll deal with the damage what i care about is to make sure there isn’t any residue inside so the people and pets that will sleep and use what we wash in there will be safe from bleach effects. Especially because we use HP in the wash and i don’t want to risk a reaction

@KarstenTheis I’m confused by the sentence  “following the rinse is recommended for disinfecting with zero. 5 cups of bleach”. What do you mean by that

@DrMoishePippik By neutralisation, they might not mean acid-base-wise, but in broader sense redox-wise, hypochlorite oxidizing peroxide to oxygen.

@DrMoishePippik Oh i see, i just want to make sure there isn’t any left over reside in there but are a little lost on how to do that

Easy way and test is - washing it away ( perhaps after rough/course elimination by chosen method) by empty water cycles, until smell of intentionally left residual chlorine is gone. This way you have a sensitive feedback. If you let it fully react with something, there would be possibly corrosive components and you would not have the feedback if you washed them away - what you have to do either way.

I agree with the idea of an empty wash cycle. Bleach (Sodium hypochlorite) is very soluble in water and fairly reactive, so it should be easy to wash away into the sewer with an empty cycle (or two if needed).

One issue I've had with bleach used in a cleaning cycle is that agitation is more vigorous without laundry, splashing bleach into areas not effectively rinsed. The next load sometimes has bleached spots on items. A bit of vinegar might be helpful after massive overuse of bleach... or not.

@Poutnik i’m a bit confused by your comment, are you saying the best way to test is to see if it smells like bleach? or are you saying i should use the elimination method and then do more empty cycles and see if it smells? When i said is there a test i was thinking like PH paper or idolize paper etc

@DrMoishePippik my washer does not have a agitation or a bleach dispenser. See my issue isnt left over bleach it’s any residue that left from applying it. Like how it leaves a residue when you use it on a table etc. that residue can then be harmful to you or others. I want to get rid of the residue not leftover

@JeffGustafson I’ve done several empty cycles. So im wondering you said if needed, how would you know if it’s needed? Since you can’t see the residue

Hydrogen peroxide does in fact neutralize chlorine bleaches, because it reacts with the active bleaching spcecies (hypochlorite) to make oxygen gas, water, and chloride ions.  https://www.sciencedirect.com/science/article/abs/pii/S0011916498000344

@CurtF. i thought so, clorox says to neutralize bleach with it but it’s seems to say that for clothing not to clean up on other material


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Is elongation a measure of plastic deformation?
I did a tensile test of a metallic alloy, and I want to calculate it's elongation. Would it simply be the (failure strain - yield strain)?
For reference, I got the following values from the test:
Yield strain: 0.015
Yield stress: 404 MPa
Failure strain: 0.22
Failure stress: 592 MPa
Would the elongation simply be 0.22-0.015 = 0.205
0.205 * 100 = 20.5%? I.e. simply the amount of strain that was done plastically?
Thanks in advance!

Wishing luck to meet someone familiar with material science of engineering.

Literally, elongation means "how much longer is the piece". If you know its initial length, and length (total length of the pieces) after failure, that's sufficient to calculate elongation. I can't stress enough this calculation should be no strain at all.

How is this a chemistry question?

I measured elongation and reduction of area on many bars, shapes, materials. Never calculated it, never heard on anyone calculating elongation. The elongation is the primary measure of ductility. The tensile machine graph puts out a strain  which is interesting , but we always measured the bars to be sure. That also give you a look at the fracture surface which can also be informative.  What are the units of your 0.015 strain; % ?, mm ? ( inches) of what gage length? Also various specifications have different definitions of yield.

"Mechanical Metallurgy" by  Dieter , McGraw  Hill; Is apparently in print after many decades. It has more than you want but must be worthwhile because it is still available.


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Full list of CAS numbers?
I'm trying to find a full list of CAS Registry Numbers, but it looks like it's not in open access.
Is it a proprietary database that can't be downloaded easily?

You may find CAS#s by typing individual chemicals you are interested in. Any vendor catalog such as Sigma-Aldrich provides long list of chemicals and their CAS#s.

CAS.org is the entity in charge of the registry. Various places to find numbers are shown on the FAQ page (https://www.cas.org/support/documentation/chemical-substances/faqs). They put time and effort into keeping the database and entering new compounds.