text:
Are nucleophilic aromatic substitutions stepwise or concerted?
Recently, there has been a lot of discussion about nucleophilic aromatic substitution (SNAr) reactions apparently being concerted instead of stepwise (e.g. Krämer, K. Textbook aromatic substitution mechanism overthrown. Chemistry World, 23 July 2018):

This is in contrast to the usual mechanism taught at an undergraduate level, where the anionic Meisenheimer complex is considered to be an intermediate:

What is the current consensus on the mechanism, and what evidence is there for it?

TL;DR Most SNAr reactions proceed via concerted mechanisms without a discrete anionic intermediate. This is typically the case when there is a good leaving group (e.g. chloride or bromide) and if the ring is not extremely electron-poor.

Over the years, there has been increasing evidence of concerted nucleophilic substitutions at sp2-hybridised carbons. One notable example was published by Neumann et al. in 2016, who studied the mechanism by which PhenoFluor (2) converted phenols to aryl fluorides:[1]

For the conversion of [1,1′-biphenyl]-4-ol to 4-fluoro-1,1′-biphenyl (i.e. the above reaction with a phenyl substituent at the para position), a 16O/18O kinetic isotope effect of 1.08 ± 0.02 was observed. This implies that Ar–O bond breaking takes place in the rate-determining step, which is inconsistent with the traditional stepwise mechanism. The concerted mechanism shown above was further supported by DFT studies which revealed a transition state 4 with partial C–O and C–F bonds.

Building on this, Kwan et al. went one step further to show that many SNAr reactions were indeed concerted.[2] In particular, for three examples, the authors used 13C/12C kinetic isotope effects to evaluate the mechanism. These KIEs were determined by integrating the 13C satellites in 19F NMR spectra (an adaptation of Singleton's method[3] which improves sensitivity).
The presence of a (primary) 13C/12C KIE suggests that the C–X bond is being broken in the transition state. However, the KIE itself does not a priori differentiate between stepwise and concerted mechanisms, since in both pathways the respective rate-determining steps involve weakening of bonds in the transition state. In order to identify the mechanism, the authors compared the experimental KIE to a calculated "maximum" KIE. Compounds with stronger bonds (such as C–F in 6) have more vibrational zero-point energy to lose in the TS, and hence have larger maximum KIEs (1.070 for reaction 1).

For a concerted mechanism which "alters both bonds simultaneously", the ratio $\mathrm{{KIE}_{exp}/{KIE}_{max}}$ is expected to be large; and for a stepwise mechanism which "alters one bond at a time", $\mathrm{{KIE}_{exp}/{KIE}_{max}}$ is small. [The paper does not go into detail about exactly why this is so; I suggest reading pp 225 onwards of an article by Westaway[4] which explains this in the context of SN2 reactions.]
This allows the three reactions above to be classified into three different categories:

a stepwise reaction (6 to 7) with a discrete anionic intermediate. This intermediate is heavily stabilised by the electron-withdrawing nitro groups, and also because fluoride ion is a poor leaving group;
a concerted reaction (8 to 9) with the Meisenheimer complex as the transition state. Here, the presence of a good leaving group (chloride) means that the Meisenheimer complex is not sufficiently stable to be an intermediate;
and a borderline case (10 to 11) where there is formally no intermediate but a shallow region in the potential energy surface near the TS.

For the concerted pathway, DFT calculations revealed that the transition state was essentially a Meisenheimer complex: the ring bears a significant amount of negative charge, is no longer aromatic (as indicated by NICS values), and NBO analysis shows that the electron density in the TS is mostly described by the Lewis structure of the Meisenheimer complex. Consequently, even though many "textbook" SNAr reactions are in fact concerted and not stepwise, the same factors that stabilise the previously purported intermediate (electron-withdrawing groups at 2- and 4-positions, and heteroatoms in the ring) also stabilise the transition state and facilitate the reaction.
Finally, a range of SNAr reactions on different carbocyclic and heterocyclic rings were then surveyed by DFT. It turns out that, at least within the scope of the reactions investigated:

All SNAr reactions on nitrogen-containing heterocycles (2-halopyridines, 2-halopyrazines, and 2-halopyrimidines) proceed in a concerted fashion, regardless of leaving group or nucleophile.
Likewise, the theoretical SNAr reactions on unsubstituted benzene are concerted.
For electron-deficient benzene rings (with 4-nitro, 2,4-dinitro, 2,4,6-trinitro, or 4-acetyl groups), SNAr reactions with fluoride as the leaving group are stepwise, but chlorides and bromides mostly react in a concerted manner.

References

Neumann, C. N.; Hooker, J. M.; Ritter, T. Concerted nucleophilic aromatic substitution with 19F− and 18F−. Nature 2016, 534 (7607), 369–373. DOI: 10.1038/nature17667.
Kwan, E. E.; Zeng, Y.; Besser, H. A.; Jacobsen, E. N. Concerted nucleophilic aromatic substitutions. Nat. Chem. 2018, 10 (9), 917–923. DOI: 10.1038/s41557-018-0079-7.
Singleton, D. A.; Thomas, A. A. High-Precision Simultaneous Determination of Multiple Small Kinetic Isotope Effects at Natural Abundance. J. Am. Chem. Soc. 1995, 117 (36), 9357–9358. DOI: 10.1021/ja00141a030.
Westaway, K. C. Using Kinetic Isotope Effects to Determine the Structure of the Transition States Of SN2 Reactions. Adv. Phys. Org. Chem. 2006, 41, 217–273. DOI: 10.1016/S0065-3160(06)41004-2.


text:
Work function and ionization energy
I've seen that the first ionization energy of a particular metal is always greater than its work function. This implies that the valence electrons in an isolated atom are more strongly bound than the de-localized conduction electrons in the solid metal.

For example, the first ionization energy of sodium is $5.14\mathrm {eV}$/atom.
(source: here)
while its work function is just $2.28 \mathrm {eV}$.
(source: here)

I'm thinking that in a metal lattice, the de-localized electrons are attracted by many positive kernels at the same time. Also, they suffer repulsive forces from the electrons around. So the concerned electron is shielded from the attractive force from the positive centers  by the other nearby de-localized electrons and the the case isn't much different from that of an isolated atom. But that doesn't seem to be the case because the work function is (very roughly), just half the value of the ionization energy.
In metallic solids, after the formation of bands, are the de-localized electrons (which now exist in bands) attracted less strongly by the positive cores which gives rise to such a difference?
Is it the case always? If so, why?
Thank you very much for your valuable suggestions.

They differ by definition, which states that work function is difference in energy between Fermi level (work required to add one electron to the body) and vacuum level (energy of an electron that drifts in vacuum, free from any material), while ionization energy is difference in energy between HOMO (highest occupied molecular orbital) and our old friend, vacuum level.
It's clear that electrons that occupy orbitals (HOMO energy level) are lower in energy than Fermi level (the boundary between occupied and unoccupied states in a continuum of states). Below is the picture which you might find useful. Also check for the source of picture: it has good and detailed explanation about your next possible questions on topic.

[source]: Antoine Kahn. Fermi level, work function and vacuum level. DOI: 10.1039/C5MH00160A (Focus) Mater. Horiz., 2016, 3, 7-10.

source: Stack Exchange

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

sort: votes


text:
Calculating the standard reduction potential for the oxidation of water
I was working with Latimer and Frost diagrams for oxygen when I came across what seems to me a contradiction.
From the Latimer diagram for oxygen below, we know the standard reduction potential for the reduction of $\ce{O2}$ to $\ce{H2O2}$ and the standard reduction potential for the reduction of $\ce{H2O2}$ to $\ce{H2O}$, as represented in the following chemical equations:
$$\ce{O2} \overset{+0.70}{\longrightarrow} \ce{H2O2} \overset{+1.76}{\longrightarrow} \ce{H2O}$$
$$\begin{align}
\ce{2H+ (aq) + 2e- + H2O2 (aq) &-> 2H2O (l)} & E^\circ &= \pu{+1.76 V} \\
\ce{O2 (g) + 2H+ (aq) + 2e- &-> H2O2 (aq)} & E^\circ &= \pu{+0.70 V} \\
\end{align}$$
If we add these two reduction chemical equations, we can find the standard reduction potential for the reduction of $\ce{O2}$ to $\ce{H2O2}$, which would be $\pu{+2.46 V}$ according to the aformentioned reduction potentials and chemical equations.
The reduction of $\ce{O2}$ to $\ce{H2O}$ is the reverse reaction of the oxidation of water:
$$\ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}\qquad E^\circ = \pu{-1.23 V}$$
Why does the standard reduction potential for the oxidation of water calculated from the Latimer diagram differ from the value reported in the table of standard reduction potentials? More specifically, why is the value calculated from the Latimer diagram double of that reported in the table? I checked my work multiple times and do not understand why this occurs, but I might have still made a silly mistake.
References
Shriver, D. F., Weller, M. T., Overton, T., Rourke, J., & Armstrong, F. A. (2014). Inorganic chemistry 6th Edition.

Related (pretty much the same thing, tbh): Deriving a reduction potential from two other reduction potentials

According to Hess’ law, we expect the Gibbs energy change $\Delta G$ of the two partial reactions to add up to the total Gibbs energy change:
$$\Delta G_1+\Delta G_2=\Delta G_3$$
For an electrochemical reaction, the change in Gibbs energy corresponds to an electrochemical potential $E$ according to
$$\Delta G = -zFE$$
where
$F$ is the Faraday constant and
$z$ is the number of electrons transferred.
Since for the two partial reactions the number of electrons is $z_1=z_2=2$, but for the total reaction $z_3=4$, we expect:
$$\begin{align}\Delta G_1+\Delta G_2&=\Delta G_3\\
z_1E_1+z_2E_2&=z_3E_3\\
2\times0.695\ \mathrm V+2\times1.763\ \mathrm V&=4\times1.229\ \mathrm V
\end{align}$$

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text:
Why does standard reduction potential get multiplied by the change in oxidation state and not by the number of electrons transferred?
The whole basis of Frost diagrams is to have a graphical representation of the Gibbs free energy of formation ($\Delta{G}_{f}$) of different oxidation states of an element using the proportional relation between Gibbs free energy and standard reduction potential:

$\Delta$$G$ = $-nFE^{o}$

From which we can derive:

$\frac{{-}\Delta{G}}{F}$ = $nE^{o}$

Thus, if we graph $nE^{o}$ against the oxidation number (N), we can accurately describe the relationship between oxidation state and ($\Delta{G}_{f}$).

I encountered a problem in Shriver's Inorganic Chemistry that asks to draw the Frost diagram of oxygen from its Latimer diagram.

Construct a Frost diagram for oxygen from the Latimer diagram below.

I initially though that we would use the standard reduction equations below, which give $nE^{o}$ as $-1.36V$ ($-2$ times $0.68V$) for $\ce{H2O2}$ and $-4.92V$ ($-4$ times $1.23V$) for $\ce{H2O}$.

$\ce{O2(g) + 2H+ (aq) + 2e- -> H2O2 (aq)}$ $E^{o}$=$+0.68V$
$\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}$ $E^{o}$=$+1.23V$

However, the book seems to be using the change in the oxidation number of oxygen to calculate $nE^{o}$ for $\ce{H2O2}$ and $\ce{H2O}$.

Online images of Frost diagrams corroborate this calculation.

Why is the change in oxidation state used as the standard to calculate $nE^{o}$ instead of the number of electrons transferred given that $n$ in $nE^{o}$ is the number of electrons transferred?
References
Shriver, D. F., Weller, M. T., Overton, T., Rourke, J., & Armstrong, F. A. (2014). Inorganic chemistry 6th Edition.

You have just multiplied the given equation by 2. In your case $\Delta G^0$ will change and yes it is multiplied by no. of electrons transferred only. Think properly.

I've made what I'm asking more clear with a specific example.

I have to agree with you, it's confusing. To calculate $\Delta G^0$ you'd do as you suggest. Presumably the point here is to provide a measure of the potential on a per atom basis. It seems a matter of convention. In real life, reporting how you calculated a result is often enough, if a general convention is not evident.

The reaction used to calculate the nE values is the formation of the zero oxidations state (designates as nE=0) from the oxidation state in question, correct? If so, then measuring the potential on a per atom basis would be flawed and would not provide a plot that reflect accurately upon the formation of the oxidation state from the zero oxidation state (opposite of reaction used to calculate nE)?


text:
Which is more activating among Phenol, Sodium Phenoxide & Anisole? Also arrange them in increasing order of their activating effect
This is in reference to Kolbe's Reaction, actually I think that Sodium Phenoxide is the most activating because, in Kolbe's Reaction we don't make react $\ce{CO2}$ directly with phenol for the electrophilic substitution(Maybe because of it's comparatively less activating effect), but we do make Sodium Phenoxide react with $\ce{CO2}$ (Maybe because of it's comparatively more activating effect) so that it will get attach to the Ortho position of Benzene Ring.
Along with this, please also tell me why in Kolbe's Reaction, the main product is Ortho directing & in most of other reactions, the main product is para directing?

Sodium Phenoxide is much more activating than phenol itself. Because here directly one negative charge comes and bond with the C below and the pi bond shifts to the ortho position as negative charge. And it also answers your doubt that why ortho product is major.
You can refer to this mechanism. After reacting with OH- then the tautomerization takes place but before it no effect where as in case of phenoxide it can perform the same reaction even without OH-
So activation order
Anisole

What about anisole? In the title of the question, anisole was also mentioned...

Out of $\ce{O-}$, $\ce{-OH}$ and $\ce{-NH2}$, I would say that O- is the hardest nucleophilic centre,so by HSAB, phenoxide should be more activated.
For Kolbe's reaction, since it is carried out in basic medium, so ortho position gets more activated as
$\ce{O-}$ has a positive mesomeric as well as a positive inductive effect on the molecule
Also, it may depend on the stabilization of the transition state by the base. Take a look at this (During electrophilic substitution by carbon dioxide)-


text:
E1cb reaction conditions
I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.
I thought the conjugate base of the leaving group must be stable enough not to de-protonate one of the newly formed alkene's H's. But if this is the case, why can't a good leaving group like Br be used in E1cb which would neither de-protonate the alkene's H's.

May be considered as a duplicate of: https://chemistry.stackexchange.com/questions/24145/why-must-the-leaving-group-in-e1cb-be-poor?noredirect=1&lq=1

However, 2nd part regarding whether Elcb can involve good leaving groups is new, thus technically not a duplicate.

I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.

In an E1cb reaction, the rate-determining step (RDS) is the $\ce {C-H}$ bond cleavage. According to Carey & Sundberg (2007), p. 552, for the irreversible E1cb reaction (when the hydrogen abstracted is very acidic), the rate of the reaction may increase in the following order of leaving group: $\ce {I^- < Br^- < Cl^- < F^-}$. Why does the reaction rate increase as the leaving group becomes "lousier"?
From kinetics, we known that the rate of a reaction is determined by the slowest step, the RDS. In the E1cb, the RDS involves removal of a proton to form a carbanion. This step would be accelerated if the $\ce {C-H}$ bond is weakened and when the carbanion experiences greater stabilisation of the negative charge. Poor leaving groups, such as $\ce {OH^-}$ and $\ce {F^-}$, are often able to provide inductive stabilisation of the negative charge, e.g. $\ce {-OH}$ and $\ce {-F}$ substituents are electron-withdrawing in nature. This would thus help to speed up the RDS and thus, the overall reaction rate would be faster with these poor leaving groups.
Note: The explanation for this is similar to that for this question: Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?

Paraphrased: If this is the case, can E1cb reactions still involve good leaving groups, such as $\ce {Br^-}$?

On the same page, the authors also conclude that:

E1cb reactions generally require both carbanion stabilization and a relatively poor leaving group

If the leaving group becomes a better one, e.g. if the $\ce {F^-}$ becomes $\ce {I^-}$, then the $\ce {C-X}$ cleavage may be more advanced than the $\ce {C-H}$ bond cleavage, resulting in the reaction becoming E2, E1-like or even E1. Thus, if good leaving groups are present, likely the elimination will not proceed by E1cb but by the other pathways.
Update
I just realised someone had a similar question answered already: Why must the leaving group in E1cb be poor? You can refer to that answer for a different perspective. But ultimately, the essence of the correct explanation is well-contained in my answer above.
Reference
Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


text:
Why is it important in the iodine clock reaction that there is only a small quantity of sodium thiosulfate?
These are 2 questions .
Why is thiosulphate needed ?
And why In small quantities ?
I know that the thiosulphate reacts with the I2 to form I ions
Why is that important though ?

To look at why Thiosulfate is needed, it is helpful to first break down the Iodine clock reaction into two parts:
$$\ce{H2O2 + 2I- + 2H+ → I2 + 2H2O}$$
This is the slow part of the reaction and why there is a clock component to it.
$$\ce{2S2O3^2− + I2 → S4O6^2- + 2I-}$$
The conversion back into Iodide ions happens much more quickly than the first part of the reaction, so you don't see the solution change color as almost all of the Iodine is in its ionic form.
The Thiosulfate is required in this reaction to convert the Iodine back into its ionic form; if there was no Thiosulfate present, the solution would just very quickly turn black and it would lose its surprise value. Since the Thiosulfate is used to convert Iodine back into Iodide, if you added more, it would take longer for the quantity of Thiosulfate to react in the reaction. It wouldn't affect the mechanism of the reaction, only the duration until the solution would change color. This is why a small amount is typically used.


text:
What is symmetry of p- and d-orbitals in ${D_{3h}}$?
To identify IR and Raman active, symmetry of p- and d-orbitals are used.
If we start since making ${D_{3h}}$ character table, how to match between symmetry of p- and d-orbitals and Mulliken symbol? because when I tried to find out, symmetry after doing symmetry operation isn't similar with any symmetry (Mulliken symbols).
For example, considering ${AsCl_5}$

compared with ${D_{3h}}$ character table

The p orbitals can be considered to be unit vectors along their respective axes. Thus they transform as displacements in $x,; y$ and $z$, hence transform as $E'$ for $x$ and $y$ and $A_2''$ for $z$.


text:
Sugar solutions have a neutral pH in themselves, but it makes your body acidic. Why?
I'm not an expert, but as far as I understood a sugar solution is completely neutral since sugar can't take hydrogen ions out of the water or donate them in. Sugar is a non ionic compound, so it does not release H and OH ions in the water so it will not make the solution acidic or alkaline.
I keep on reading and seeing charts of how sugars make your body acidic, like this one:

What process makes a neutral pH solution into an acidic one?
I'm not into chemistry at all and therefore the simpler the answer the better.

Charts like these are unscientific nonsense, ignore them.

@Waylander ok but is it proven sugars acidifies the body? If so, how?

@Mr.Web the opposite is proven - these foods (and any other foods) do not acidify the body. Science formed the hypothesis in the 19th century and disprove it in the early 20th century. People keep making these charts, but they are simply wrong.

You've got to love the sheer gumption of putting citrus fruits in a column headed 'Alkaline'...

@AakashM Oh, but you see the citrus fruits make your body produce bases to neutralise them and when you stop eating them your body doesn't react fast enough to stop making them and makes too much base (/s, just in case)

@Mr.Web In general, if you see a chart alleging anything about food without citing at least one scientific study in a journal which, when Googled, doesn't turn up on any predatory journal lists, it's a steaming pile of bullshit. Especially about which foods are "healthy". (That's not always the case, of course -- for example, the charts found in those studies -- but if people are stealing those charts without citation they're plagiarizing and probably leaving out important context.)

Can we tag wherever this came from "troll" because that chart wasn't made to actually help people.

@AakashM the claim is not that the food itself is alkaline. The claim is that it makes your body fluids alkaline. The hypothesis comes from burning food in the late 19th and early 20th century. The foods in the right column produced alkaline ash (usually due to high levels of potassium, magnesium, etc., which reacted into alkaline salts) and the foods in the left column produced acidic ash (usually due to sulphur levels). It took some testing to find out that 1) burning food is not a good model of human metabolism, and 2) homeostasis renders the whole hypothesis moot.

Okay everyone, why in the name of the chemistry gods has this been closed and of all reasons why was unclear chosen? The question is crystal clear. As the comments point out, it is based on an outdated, inherently flawed concept but one that was originally considered scientific and it can be answered easily by this site. It should remain open.

Is even the concept of acidic or alkaline body that is nonsense. Blood has been mentioned. Your fluid excretion can indeed vary in pH depending on diet. But indeed is excretion!

I definitely agree with Jan on this one. This is exactly the sort of question that should be on the site because it helps to remove old misconceptions about chemistry.

It is not proven that "sugar makes your body acidic"!
Your body's pH is very tightly regulated by the body's internal systems; it is also different in different parts of the body - the stomach is acidic (1.0-2.5), the intestine are mildly basic (jejunem 7-9) terminal ileum 7.5 reference here. Blood pH is 7.35, and any deviation from this is indicative of serious illness.

It is almost impossible to "make the body" acidic while keeping it alive and functioning. Diets that advertise "acidic" or "alkaline" do little (nothing) to change the body PH. Instead they focus on other properties. ref

It is perfectly true that the "alkaline" foods recommended above will do you little harm, and some good if you replace the "acid forming foods" with them - just not for the reasons advanced.

in the end they are just (wrongly used) marketing buzzwords (lies).

@Mindwin While it is correct that there is a lot of quackery around using diet to alter "the body's pH", you will find that your counter is completely false. The existence of a physiological buffer does not mean the pH doesn't change in disease states. It  often does. There is an entire chapter in Cecil Medicine on the topic (Ch 120). Yes, alkaline diets are quackery. But no, it's not because your physiologic buffer is inviolable. Various pathophysiologic processes of varying severity can cause disturbances in the buffer system.

@Mindwin don't counter quackery with false simplifications. Counter it with facts. Absent some pre-existing disease process, alkaline diets don't substantially change the pH of any body compartment (other than urine), and they aren't shown to be any more effective than any other kind of diet. (vs. "it is almost impossible to make the body acidic while keeping it alive and functioning")

I thought the quacky explanation was more about making your body not have to regulate to counteract acidity, than about making it more alkaline/acid than it wants to be.

@immibis That's how you get so many people believing in any quackery - you invent a hundred explanations that are pure nonsense, and let people choose which they like the best. This means it needs a lot of effort to convince people it's a quackery - you debunk one explanation, and then someone says "Oh, but that's not really what X is about - the real mechanism is Y." Sooner or later, you give up trying to convince people, and the "quackers" win.

@denovo  be nice.

One place where sugar does cause acidity is your mouth. The bacteria living in your mouth feed on sugar and excrete lactic acid. It's part of their metabolism. Wikipedia has more details about dietary sugars and acidity in mouth.
It's not clear from the chart what part of the body it refers to, or what it claims the acidity to cause. But it does sound similar to the alkaline diet advice. It was originally related to acid ash hypothesis, which claimed acidic diets to cause osteoporosis. However, scientific research has not found evidence for the hypothesis. There is also a significant difference: the original acid ash hypothesis classified foods based on the pH of the ash left when burned - not the sugar content (all sugar will burn away).
In conclusion, there is no general link between sugar and acidity in body, but in some very specific cases such as caries bacteria in mouth, there is a connection.

On the point of ‘makes your body acidic’ you already have an answer to which I have nothing to add.
However, you also claim that sugar be completely neutral—at which point I must intervene.
A lot more compounds can act as acids and bases than the shortlist of common acids you probably had in chemistry class. Most things with a lone pair can be a base and practically any $\ce{X-H}$ bond where $\ce{X}$ has a higher electronegativity than hydrogen can be an acid. In $95~\%$ of cases, these compounds are probably still not acidic or basic though.
And then there is the (much smaller, but still vast) class of compounds that can act as acids or bases in standard aquaeous solutions but usually do not. In sugar’s case, it has a lot of $\ce{O-H}$ bonds and a lot of oxygen atoms. All oxygen atoms can be protonated if a strong acid is introduced to the mixture so sugar is a base about as strong as water. On the other hand, the hydrogen of the $\ce{O-H}$ bonds can also depart as a proton if a strong enough base is in solution. This means sugar’s hydroxy groups are also weak acids—again, about as strong as water.
Then there is a final convoluting factor. One of these hydroxy groups in glucose is attached to a carbon that is bound to a second oxygen ($\ce{R-O-CHR'-O-H}$). Here, the electronegative oxygen exercises a negative inductive effect, i.e. draws electron density away from the hydroxy group. Therefore, this one proton is slightly more acidic than all others. It is also ever so slightly more acidic than water so it acidifies the solution ever so slightly.
For all practical intents and purposes, this effect can be fully neglected, however, as it is so minor.

Ok so, to keep it easy, pouring pure sugar into a stomach, doesn't it make it more alkaline? Or does it change the ph in any way?

@Mr.Web: The initial effect will be to dilute your natural stomach acids, making it less acid. But the acidity of the stomach is regulated by the body, producing more acid to digest consumed food.

@MSalters and, as the stomach produces more acid, the blood becomes more alkaline for a period of time (an effect known as the alkaline tide). There is no "keeping it easy" when it comes to acid base physiology.

@Mr.Web All of my answer should be titled ‘what happens in water’. As for the body itself, I refer to Waylander’s answer (as in my first paragraph): the pH range of all parts of the body is tightly regulated. While adding anything will briefly disrupt the stable state, counter-processes immediately kick in and re-regulate the pH to where it should be.


text:
Balancing the Redox reaction by Half reactions
I'm having trouble balancing this reaction using half reactions:
$$\ce{I2 + KOH -> KI + KIO3 + H2O}$$
I so far know that iodine is reduced when it forms $\ce{KI}$ (oxidation noumber decreases from 0 to -1) and oxidised in $\ce{KIO3}$ (the Oxidation number increases from 0 to 5).
Oxidation half: $\ce{I2 -> KIO3}$
Reduction half: $\ce{I2 -> KI}$
The potassium atoms on both halves seem to be the problem because they are only on the product side. When I balance the iodine atoms the $\ce{K}$ atoms aren't.
I've tried this several times and can't work it out.

To make your problem simpler(equivalent of your problem) just balance this reaction:
$\ce{I2  -> I^{-} + IO3^{-}}$ in basic medium.
While balancing, the $\ce{OH-}$ on the LHS correspond to the number of moles of $\ce{KOH}$ and the $\ce{IO3-}$ on the RHS correspond to the number of moles of $\ce{KIO3}$ (the $\ce{K}$ is just a distraction).

1- Determining   oxidation number, Oxidation and Reduction:
$$
\ce{
\overset{\color{red}{(0)}}{I2}+
\overset{\color{black}{(+1)}}{K}
\overset{\color{ black }{(-2)}}{O}
\overset{\color{ black }{(+1)}}{H}
->
\overset{\color{black}{(+1)}}{K}
\overset{\color{green}{(-1)}}{I}
+\overset{\color{black}{(+1)}}{K} \overset{\color{ blue }{(+5)}}{I}\overset{\color{ black }{(-2)}}{O3}+\overset{\color{black}{(+1)}}{H2}
\overset{\color{black}{(-2)}}{O}
}$$
When you look at the reaction: it is an auto-redox reaction, in which:
The atom$\ce{I}$: went from $\color{red}{(0)}$ to $\color{green}
{(-1)}$, it gains electrons, so it is reduced.
The atom$\ce{I}$: went from $\color{red}{(0)}$ to $\color{blue}{(+5)}$, it lost electrons, so it is oxidized.
2- Separate the redox reaction into two half-reactions: one for the oxidation, and one for the reduction
$$\text {Reduction half: } \ce{
\overset{\color{red}{(0)}}{I2}+
{KOH}
->
{K}\overset{\color{green}{(-1)}}{I}
}$$
$$\text {Oxidation half: }
\ce{
\overset{\color{red}{(0)}}{I2}+
{KOH}
->
{K}\overset{\color{ blue }{(+5)}}{I}{O_3}
}$$
3- Balance all elements except oxygen and hydrogen in both half equations:
$$\text {Reduction half: } \ce{
\overset{\color{red}{(0)}}{I2}+
2{KOH}
->
2{K}\overset{\color{green}{(-1)}}{I}
}$$
$$\text {Oxidation half: }
\ce{
\overset{\color{red}{(0)}}{I2}+
2{KOH}
->
2{K}\overset{\color{ blue }{(+5)}}{I}{O_3}
}$$
4- Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms:
$$\text {Reduction half: } \ce{
\overset{\color{red}{(0)}}{I2}+
2{KOH}
->
2{K}\overset{\color{green}{(-1)}}{I}
+2\ce{H2O}} $$
$$\text {Oxidation half: }
\ce{
\overset{\color{red}{(0)}}{I2}  +
2\ce{KOH}  + 4\ce{H2O} ->
2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}
}$$
5- Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms:
$$\text {Reduction half: } \ce{
\overset{\color{red}{(0)}}{I2} +
2\ce{KOH} + 2\ce{H+}
->
2\ce{K}\overset{\color{green}{(-1)}}{I}
+2\ce{H2O}} $$
$$\text {Oxidation half: }
\ce{
\overset{\color{red}{(0)}}{I2}  +
2\ce{KOH}  +4\ce{H2O} ->
2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}+ 10\ce{H+}
}$$
6-  Use electrons ($\ce{e^-}$) to equalize the net charge on both sides of the equation:
$$\text {Reduction half: } \ce{
\overset{\color{red}{(0)}}{I2}+
2\ce{KOH} +2\ce{H+} + 2\ce{e-}
->
2{K}\overset{\color{green}{(-1)}}{I}
+ 2\ce{H2O}  } $$
$$\text {Oxidation half: }
\ce{
\overset{\color{red}{(0)}}{I2}  +
2\ce{KOH}  +4\ce{H2O} ->
2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} + 10\ce{H+} +10\ce{e-}
}$$
7- Make the electrons gained equal to the electrons lost:
$$\text{Reduction half: } \ce{
5\overset{\color{red}{(0)}}{I2}+
10\ce{KOH} +10\ce{H+} +10\ce{e-}
->
10\ce{K}\overset{\color{green}{(-1)}}{I}
+ 10\ce{H2O}  } $$
$$\text {Oxidation half: }
\ce{
\overset{\color{red}{(0)}}{I2}  +
2\ce{KOH}  +4\ce{H2O} ->
2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} + 10\ce{H+} +10\ce{e-}
}$$
8- Combine the two half-reactions:
$ \ce{5\overset{\color{red}{(0)}}{I2} +
10\ce{KOH} +10\ce{H+} +10\ce{e-} + \overset{\color{red}{(0)}}{I2}  +
2\ce{KOH}  + 4\ce{H2O}
->
10\ce{K}\overset{\color{green}{(-1)}}{I}
+ 10\ce{H2O} + 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}+ 10\ce{H+} + 10\ce{e-}  } $
9- Cancel out the electrons and anything that is the same on both sides then collect the things that is the same on the same side to form one balanced redox equation:
$$ \ce{6\overset{\color{red}{(0)}}{I2}+
12\ce{KOH}
->
10\ce{K}\overset{\color{green}{(-1)}}{I}
+ 6\ce{H2O} + 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}  }$$
10- Shorten the equation in a simplified form:
$$ \ce{3\overset{\color{red}{(0)}}{I2}+
6\ce{KOH}
->
5\ce{K}\overset{\color{green}{(-1)}}{I}
+ 3\ce{H2O} + \ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}  } $$

Here's a little guide.

Start with determining "relevant" species (those that reduced or oxidized):
$$\ce{I2^0 + K+ + OH- -> K+ + I- + K+ + IO3- + H2O}$$
Notice that you have $\ce{OH-}$ here, thus, you have basic conditions.
Half-reactions
$\ce{I^0 -> I^-}$
$\ce{I^0 -> IO3-}$

So, you should "compensate" oxygens in $2^{nd}$ reaction (either using water or $\ce{OH-}$, since your conditions are basic, you should use $\ce{OH-}$ and "add" water to the right to compensate for "new" hydrogens).

$\ce{I^0 + 6OH- -> IO3- + 6H2O}$

Usually you simply use twice as many $\ce{OH-}$ as "compensated oxygens". Here you have 3 of 'em, so you should use $\ce{6OH-}$.

New half-reactions, but with electrons so net charge is 0:

$\ce{I^0 + e- -> I^-}$
$\ce{I^0 + 6OH- - 5e- -> IO3- + 6H2O}$

Balance electrons so number of electons for each half-reaction is equal (by multiplication)

$\ce{5I^0 + 5e- -> 5I^-}$
$\ce{I^0 + 6OH- - 5e- -> IO3- + 6H2O}$
then add halves:$$\ce{3I2^0 + 6OH- -> IO3- + 5I^- + 6H2O}$$

Almost there! Don't forget about cations:

$$\ce{3I2 + 6KOH -> KIO3 + 5KI + 6H2O}$$
Equation is balanced :)

Quibbling I'd write the half cell as $\ce{I^0 + 6OH- -> IO3- + 6H2O + 5e- }$ to better conform how it will be written in standard reduction tables.

convert it into ionic form = I2 + OH⁻ ⟶ I⁻ + IO3⁻
oxidizing number of I went from 0 ⟶ -1(I⁻) and +5 (IO3⁻)
There's an "OH⁻" in the equation, which means the reaction is carried out in a basic medium.
Separating the equation into oxidizing and reducing halves,
↪ oxidizing half: I2 ⟶ IO3⁻
↪ Reducing half: I2 ⟶ I⁻
Now balancing both halves,
↪ O.H :  6H2O + I2 ⟶ 2IO3⁻ + 12H⁺ + 10e⁻
(1. Balance the number of Iodines.  2. Balance oxygens by adding 6H2O because there's 6 Oxygens on RHS, then balance the Hydrogens on LHS by adding 12H⁺ on RHS, then balance out the charges by adding electrons.)
↪ R.H :  2e⁻ + I2 ⟶ 2I⁻
(1. Balance the number of Iodines.  2. since there's no oxygen, balance out the charges by adding electrons.)
Multiply the reduction equation by 5 so that the number of electrons in both halves gets cancelled out.
↪ The new equation for R.H will be: 10e⁻ + 5I2 ⟶ 10I⁻
combining both halves together,
↪ 6H2O + I2 + 10e⁻ + 5I2 ⟶ 2IO3⁻ + 12H⁺ + 10e⁻ + 10I⁻
Since the reaction is carried out in basic medium, add the appropriate number of OH⁻ to both sides, which will neutralize all H⁺ and convert them into water molecules.
↪ 12OH⁻ + 6H2O + 6I2 ⟶ 2IO3⁻ + 12H⁺ + 10I⁻ + 12OH⁻
↪ 12OH⁻ + 6I2 ⟶ 2IO3⁻ + 6H2O + 10I⁻  ..... ×½
↪ 6OH⁻ + 3I2 ⟶ IO3⁻ + 3H2O + 5I⁻
adding K to balance the final equation,
↪ 6KOH + 3KI2 ⟶ KIO3 + 3H2O + 5KI
( hope that helps. )

Some listed procedures are so complicated I would be in danger of unlearning to solve such trivial tasks just in my head within few seconds. Enough is ability to operate with small integers and using "magical" equation n.m=m.n. // ON change by 1 and by 5, therefore the reverted ratio 5:1:  i.e. 5 KI + 1 KIO3. We then need 6 KOH and 3 I2.  // 3 I2 + 6 KOH -> 5 KI + KIO3 + 3 H2O // Finished.

I solved it here using redox rules. You can do it if that is comfortable for you. you gotta solve more equations to get used to it.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.


text:
Is the acetyl protecting group cleavable with ammonia?
I read in several sources that one can cleave an acetyl protecting group attached to an amine (thus an amide!) with gaseous ammonia. I can't imagine that a nucleophilic attack by ammonia at the carbonyl group would cleave the acetyl group and release the amine. Is this possible? I find no evidence for this during my research. I would be very grateful for any advice.

In general N-acetyl takes a bit of shifting and NH3 is not strong enough (it will readily cleave O-Acetyl). Standard conditions are refluxing 1N HCl, heating with hydrazine, KOH in refluxing EtOH/ethylene glycol.
If you are struggling to cleave RNHAc then forming RN(Ac)BOC with BOC2O/DMAP/MeCN followed by reaction with hydrazine can be effective.


text:
How to derive the molar fundamental relation
We have the fundamental relation:
$$dS = \frac{1}{T}dU+\frac{P}{T}dV-\frac{\mu}{T}dN$$
how do we get now the following:
$$ds = \frac{1}{T}du+\frac{P}{T}dv$$ where $s = \frac{S}{N}, u = \frac{U}{N}, v = \frac{V}{N}$.
I tried substitution but it got me nowhere..

Did you use the product rule for differentiation?

@ChesterMiller yes

Let's see what that gave you.

$Nds+sdN = \frac{1}{T}(Ndu+udN)+\frac{P}{T}(Ndv+vdN)-\frac{\mu}{T}dN$

but why should $\frac{s}{N}dN+\frac{u}{NT}dN+\frac{Pv}{NT}dN-\frac{\mu}{TN}dN$ be zero

You're missing a factor of v in the term involving P.  Factor out the dN/N.  What does that give you?  What is the definition of $\mu$?

we did define it as $\mu = \frac{\partial U}{\partial N}$ but if we use the Gibbs-Duhem-relation it equals zero

You have  $$ds=\frac{du}{T}+\frac{P}{T}dv+\left(\frac{u}{T}-s+\frac{Pv}{T}-\frac{\mu}{T}\right)\frac{dN}{N}$$
But, $$\mu=u+Pv-Ts$$
That is, $\mu$ is the Gibbs free energy per mole.  So the term in parenthesis drops out.


text:
How to understand MO diagrams for cationic octahedral transition metal complexes
I am having problems understanding the construction of MO diagrams in octahedral transition metal complexes within ligand field theory (LFT) when the metal center is asumed as an cation:
I am completely fine understanding the MO diagram when the metal is neutral and thus the (say) 3d, 4s, 4p orbital energies are above the ligand orbitals as shown e.g. here:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/ligand.php
However, when the metal becomes cationic, wouldn't the metal orbitals be significantly lower in energy?
In textbooks, you find quite a few pictures like these:
However, when I perform some quantum chemical calculations (it does not really matter if based on Density Functional Theory or Hartree-Fock), the metal orbitals drop significantly below the ligand orbitals (as you would expect).
With such an energy difference, the metal and ligand orbital should not be able to interact properly as they don't match energetically.
Consider for example orbital energies of [Fe(H$_2$O)$_6$]$^{2+}$ on BP86/SVP level of theory:

HOMO [Fe(II,LS)(H$_2$O)$_6$]$^{2+}$ : -12.0997 eV
HOMO [(H$_2$O)$_6$]            :  -4.7910 eV
HOMO [Fe(II,LS)]$^{2+}$       : -25.9057 eV

Now, if Fe is assumed neutral, the problem more or less "vanishes" as the metal orbitals rise in energy and the diagram makes way more sense:

HOMO [Fe(0)]             :  -5.1303 eV

However, this is not what the diagram shows.
Can you help in explaining the discrepancy?
Is the MO diagram making things way to easy?
Is the assumption of a cationic metal center non sensible?
Are orbital energies overrated and cannot be used for problems like these?


text:
XRD-Amorphous materials
I am interested in sealing some air-sensitive samples inside of xrd-amorphous material to allow for XRD and hopefully XRD-texturing analysis. Of course, Kapton is the common material used for this purpose, but I am unaware of a good way to seal two pieces of kapton together. Impulse sealers (ie here) are commonly used for creating plastic seals in food processing. However, they don't reach a high enough temperature to melt Kapton. I also tried to just heat the kapton with a soddering iron, to no avail. Are there other materials that maybe suitable for sample storage during xrd ?

More classically, glass was the choice.

Really, what type of glass?

See this link for an example.

Not really sure since I never did that kind of a thing, and glass departments associated with chemistry departments aren’t a thing either. But in literature of the 50s and 60s there are lots of papers with distilling or otherwise getting samples into small tubes and crimping the glass shut. Great way to seal it. Best guess would be a standard boro-silicate glass with a reasonably low softening point.

In our department, we have glass capillaries for this. There are actually some suggestions for this from the Argonne beam-line here: https://wiki-ext.aps.anl.gov/ug11bm/index.php/Air_Sensitive_Samples

The other trick that I recently learned is that apparently thin packing tape also works, wherein the tape is 1-2 mil thick and the signal is either negligible or that it can be subtracted out.

PMMA (acrylic) may also be a good choice. It's used in the air-sensitive holder we bought from Bruker.


text:
Is it possible to turn food into fuel for a combustion engine?
According to McDonald's Nutrition Calculator page, the following order would contain 2480 kilocalories (or 10376 kilojoules):

1 x Double Bacon Smokehouse Burger
1 x Large Chocolate Shake
1 x Large World Famous Fries®

Would it be possible to somehow extract/convert that energy and use it to power a car engine? If so, what would be the involved chemical processes, equations and byproducts?
Note that this question is entirely motivated by curiosity and by an ingenuous notion of Lavoisier's principle of mass conservation.

Could you please explain the downvote?

http://en.citizendium.org/wiki/Energy_consumption_of_cars after a bit of number crunching I found out it takes $2953.3 kj$ to go one km.

You can combust all you have and obtain 10376 kJ (human will obtain less energy, because we are not campfires to burn everything indiscriminately). But, instead, you can be a little bit more smart and combust fats of the same mass, which give 38 kJ per gram (proteins and carbohydrates give less than half of it). This idea allows you to use less fuel for same productivity. Actually, the idea is already widely used - check for biodiesel. Source below gives detailed basic-level insight.

Biodiesel are produced through a simple technology called trans-esterification reaction. Degummed oil free of all forms of impurities is reacted with a reasonable alcohol (ethanol, methanol, butanol etc.).

Citation source:
Biodiesel: Fuel for the Future (A Brief Review). Owolabi R. U., Adejumo A. L., Aderibigbe A. F. International Journal of Energy Engineering 2012;  2(5): 223-231.
DOI: 10.5923/j.ijee.20120205.06


text:
How did J. J. Thomson prove that the cathode rays were made of negative particles and not negative rays?
In Thomson's experiment, he used a discharge tube to prove that the cathode rays that emanate from the cathode were made of "a stream of negatively charged particles" because they were repelled by an applied negatively charged electrical field. Why did Thomson conclude from this experiment that the negatively-charged cathode rays were made of particles, and hence the discovery of the electron? Why did this force him to conclude that the rays were really particles? What am I missing here?

You are pursuing a distinction which is not there.

Check for wave–particle duality :)

I apologize. What I am asking is that I understand how Thomson discovered that the rays were NEGATIVE, but how did the experiment prove that the rays were made of PARTICLES? ( wave-particle duality was not discovered yet.....)

There is evidence that they have a Kinetic Energy that has the ability to accomplish work

I edited my post, so if you could please take off the hold....

@IvanNeretin Wave particle duality does not imply what you think here unless nothing in the entire universe can ever meaningfully be described as a "particle". Electrons in cathode ray tubes behave far more like particles than waves and the relevant experiment was a major breakthrough in our understanding of matter well before quantum stuff became important.

The take home message is that a particle-stream is a ray. For example, how do you distinguish between a light ray and a stream of photons?

Wikipedia gives the outline of the experiment(s). They were called 'rays' beforehand, but acted like particles (taking all experiments into account). Perhaps you could be clearer on what you think the distinction between 'ray' and 'particle' was at the time, and how you think Thompson did (or did not) resolve the difference.

@Zhe That is not the take home message here unless you want to forget one of the most important breakthroughs in the development of modern physics and chemistry. Quantum wave/particle duality is totally irrelevant here and obfuscates the central point of the Thompson experiment which showed that a wave didn't explain the cathode ray.

@matt_black I'm just saying that the classical ray is just a stream of QM particles. It just so happens that the stream in this case is charged, so you can manipulate the ray with an electric or magnetic field. When I said "take home message," I didn't mean for the OP's question, just with respect to the the comments before.

Thompson was able to measure the ratio of their mass to their speed
Cathode "rays" had been known for some time before Thomson. They were first observed as experiments in gas discharge tubes started to exploit better and better vacuums (the early experiments observed the varying forms of discharge in low pressure gases; cathode rays only become significant when there is very little gas left in the tubes).
At very high vacuums, people observed that the glass at the end of the tubes would glow even though the contents of the tubes did not (as there were too few gas particles left to cause a discharge). So the idea that some sort of ray was emitted by the cathode in the tube was developed. Initially the rays were seen to travel in straight lines.
Later, scientists such as Crookes observed that the rays could be deflected by magnetic fields (though he thought they were just negatively charged gas particles). Hertz, though, claimed electric fields would not deflect them (he was wrong).
The critical experiments done by Thompson relied on having an excellent glassblower who was able to construct an apparatus where electrodes inside the discharge tube which could be charged with an electric field (while retaining a very high vacuum). This enabled Thompson to refute the claim by Hertz that electric fields didn't cause deflection (maybe Hertz's glassblower wasn't as good). Once he had this apparatus further experiments with electric and magnetic fields could reveal a great deal more about the particles.
Thompson describes the critical experiments like this:

As the cathode rays carry a charge of negative electricity, they are deflected by an electrostatic force as if they were negatively electrified, and are acted on by a magnetic force in just the way in which this force would act on a negatively electrified body moving along the path of these rays, I can see no escape from the conclusion that they are charges of negative electricity carried by particles of matter. The question next arises, What are these particles? Are they atoms, or molecules, or matter in a still finer state of subdivision? To throw some light on this point, I have made a series of measurements of the ratio of the mass of these particles to the charge carried by it (“Cathode rays,” Philosophical magazine, 44 (1897), 293-316, pdf of original available here)

[A good review and description of modern recreations of this experiment is given here]

The point of the experiments is that the deflection of the particle in a uniform electric field depends on its mass, velocity and charge. And its speed can be measured by carefully balancing the electric and magnetic fields to give zero net deflection (why this is so involves equations, but this is Chemistry.SE not Math.SE and you can read them in the paper).
So Thompson observed that cathode rays were best explained by a charged particle (a classical wave would not have been deflected) with an observable charge to mass ratio (which implied either a very large charge or a very small mass). And he observed that the particles were always the same regardless of the cathode material or the original gas in the tube.
Later experiments clarified the mass and charge. Which correspond to what we now call electrons.
Note that this happened well before any of the ideas of quantum mechanics and wave/particle duality appeared. Thompson was concluding, in contradiction to some theorists at the time, that the properties of cathode rays were better explained by particles obeying the known laws for electricity and magnetism than by a wave. And he was right. His discovery was the first time a key component of the atom had been identified.

Good answer. Do you it's helpful to explicitly say here that classical electromagnetic waves are not affected by electric/magnetic fields?

@Zhe good point. That was why Thompson was convinced it was particles not waves.


text:
How can f-f transitions happen?
The question might seem quite easy to answer but still I am unable to find any useful information. I am aware of laporte- or symmetry-forbidden transitions and for f-f also spin-forbidden transitions in lanthanides but still I can't seem to find any good solution how they can happen.
For a d-d transition there are many exceptions depending on the arguments you come up with. From inverse symmetrical polyhedra that use ligand vibrations to become distorted to s- and p-mixing with the d-orbitals.
But f-orbitals, at least for the lanthanides, are said to not contribute to chemical bonds. Then there is often the argument about spin-orbit coupling but that would probably solve the spin-forbidden transitions. I'm not sure how it would affect the actual f-f transition being forbidden. In an old paper I also read about df- and fg-mixing. That would explain it but I am not sure if this is really the case.
So the question remains can the f-f transition in lanthanide complexes be somehow influenced to become at least a tiny bit more probable and therefore cause the transition to happen?

So I am currently reading the book: Physical Inorganic Chemistry A Coordination Chemistry Approach, S. F. A. Kettle
One of the chapters deals with Lanthanides and Actinides and their spectra. After a long introduction on different energetic levels and spin-orbit coupling there are a few remarks on $\ce{f->f}$ transitions although, as the author chooses to explain different ideas and phenomena during his explanation I am not sure if I fully understood his final conclusion. I guess it's probably a sum of many contributions. To summarize it quickly the aspects he is discussing are:

We can see that the intensities for $\ce{f->f}$ transitions in coordination polyhedra with high symmetry tend to be weaker than in low symmetries. That means that there are probably mechanisms such as vibrations and mixing with other orbitals, presumably d-orbitals (like in $\ce{d->d}$ transitions).
For the vibronic coupling (in order to compensate for a parity forbidden transition) he gives $\ce{NaCs2[YCl6]:M^3+}$ as an example, where at low temperatures both electronic and vibrational transitions seem to be observable for the trivalent lanthanide dopant.
Then there are also 'bands' under high-resolution conditions that cannot be explained by vibronic coupling. This is where he introduces magnetic dipole allowed transitions. In his explanation he compares how an electric dipole will invert at an inversion center while, much like in the detectors for mass spectrometers, electrons will be forced to move in a spiral along the magnetic vector and a spiral or circle does not invert here. According to his explanation the magnetic dipole is an even function and as f-orbitals are odd an $\ce{f->f}$ transition is $$u * u = g$$
and therefore magnetic dipole allowed.
Then he also refers to the fact that even though light may not be absorbed in matter it will still travel slower through a medium than through vacuum since it causes oscillations that have to propargate through the medium first.
I'll just quote the book at this part:

It has been suggested that the light wave used in absorption studies
  polarizes the ligands and that the lanthanide ion feels the dipoles created.
  That is, it is suggested that the ligands provide a non-centric field at the
  lanthanide ion not by virtue of any vibrational movement of their nuclei but
  because of the instantaneous polarization of their electron density caused by
  the light wave itself.

Then there was a remark on the environment as well. Since the f-orbitals don't really take place in any bonding for the lanthanides, do they actually feel their environment? And for this they considered weak nephelauxetic effects.
And then there are so called hypersentisitve-transitions (transitions with environment sensitive intensities). And those seem to be related to quadrupole-allowed transitions. The two dipoles in a quadrupole may arise from a polarization by the electric vector as mentioned above. So one dipole is the light vector itself (with the energy corresponding to the hypersentisitve-transitions and the other one being the induced dipole).
The difference in environments would arise from different polarizabilities of the ligands in different environments.
The other mechanism seems a bit more complicated. It considers low-symmetry crystal fields with an inherent dipole. So high-symmetry crystal fields distort via a vibration to a low symmetry. The ligand environment is distorted to a point where it become dipolar.
And then the light itself is the other dipole.
The final remark was on the bonding in f-orbitals. Here the complex $\ce{[UF8]^3-}$ was considered. By repulsion a square antiprism would be favored if there was no interaction at all with the ligands. What we find however is a cubic arrangemenent.
And the $\ce{f_{xyz}}$ orbital can contribute to a bond only in the cubic case.

As far as I understand this is explained by Judd–Ofelt theory. When lanthanides are doped in the crystals, the crystal perturbation renders the Laporte selection rule inadequate to explain these transitions. The accumulated influence of the electrons and nucleus of the crystal host creates an electric field called crystal field. This crystal field generates a Stark effect (similar to Zeeman effect, only with an electric field instead of a magnetic one). The perturbation has to be added to the Hamiltonian of the free ion leading to new selection rules, called, Judd–Ofelt selection rules.
These selection rules are:
For electric dipole: $\Delta \textbf{S}=0$, $\Delta \textbf{L}\leq 6$, $\Delta \textbf{J}\leq 6$, $\Delta J=2,4,6$, opposite parity.
For magnetic dipole:  $\Delta \textbf{S}=0$, $\Delta \textbf{L}=0$, $\Delta \textbf{J}=0,\pm1$ , same parity.
For electric quadrupole: $\Delta \textbf{S}=0$, $\Delta \textbf{L}=0,\pm1,\pm2$, $\Delta \textbf{J}=0,\pm1,\pm2$ , same parity.
You can check about the Judd–Ofelt theory in Advances in Spectroscopy for Lasers and Sensing [1, pp. 403–433].
References

Advances in Spectroscopy for Lasers and Sensing; Di Bartolo, B., Forte, O., Eds.; NATO science series II; Springer: Dordrecht, 2006. ISBN 978-1-4020-4787-9.


text:
How to rationalise the resonance structures of nitrogen dioxide?
The following image depicts resonance in $\ce{NO2}$ molecule:

I don't know if it is obvious or not, but I'm puzzled by this structure. I've read about the structure of the $\ce{NO2}$ compound, but it wasn't related.
In the first structure, nitrogen is having a single electron with a positive charge, and oxygen is having a negative charge. So, that means nitrogen has donated a single electron to Oxygen, without forming a bond. I know it can be possible, but I couldn't understand how it happens.
Also, I thought that nitrogen donates a lone pair to oxygen, as it can have a maximum co-valency of 4. As in the following image:

But, this isn't the case here. Why?

Possible duplicate of Structure of NO2 compound?

But, @NilayGhosh that post asks about whether the single electron will stay on Nitrogen or on Oxygen. Whereas I want some insight on how a single electron stays on Nitrogen.

I don't understand the question. There is an odd number of electrons, they have to be distributed as best as they can, preferably so that all elements get as close to 8 as possible (but not more).

@Martin-マーチン I was saying that why did N donated only one electron (I've read that electrons are always donated in pair) to O, instead N could also make a bond with O, gaining a co-valency of four.

Where ever you read that electrons are donated in pairs, you either have missed some context, or the book is grossly oversimplifying, or even completely wrong. $\ce{NO2}$ is an odd electron species, or a radical. It is also a very complicated bonding situation. The way you are trying to rationalise is insufficient for it. If there were a double bond between oxygen and nitrogen, then one of the elements would exceed the 8 valence electrons, which is not possible for these elements. Please [edit] your question with all the conceptual issues you have to make it clearer.


text:
If the heat exchanged is different for reversible and irreversible processes, how can enthalpy be a state function?
From the definition of Gibbs free energy, we can write for an irreversible process:
$dG=dH-TdS<0$
With $TdS=\pu{q_{rev}//T}$ this becomes:
$dH-dq_{rev}<0$
Since G is defined for constant (p,T) $dH=dq$, in the case of an irreversible process we can write $dq=dq_{irrev}$, so:
$dq_{irrev}-dq_{rev}<0$
Or, by default:
$dq_{rev}>dq_{irrev}$
BUT: since $dH=dq$, this would mean:
$dH_{rev}>dH_{irrev}$
Which is in contradiction with the fact that H is a state function (the path between 2 states shouldn't matter)! What am I missing here?

When you write dG=dH-TdS, you are implying that T is constant.  Did you mean to imply this?  Can you give a specific process example to focus on to illustrate what you are asking?

The definition of Gibbs free energy always implies constant T. It is valid for all processes assuming constant T and p.

This is incorrect.  The definition of Gibbs free energy is G=H-TS, so dG=dH-TdS-SdT=VdP-SdT

Can whoever downvoted this question tell me why? Or answer the question at least? I asked this to quite some people with knowledge of thermodynamics and they weren't able to answer me so it must be a valid question.

It wasn't me.  I will try to help you.

From Wikipedia:
In thermodynamics, the Gibbs free energy (IUPAC recommended name: Gibbs energy or Gibbs function; also known as free enthalpy[1] to distinguish it from Helmholtz free energy) is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric).

Let us continue this discussion in chat.

I was hoping we could continue our discussion.  I know I can help you if only you are willing to participate.

Even though the OP does not seem to be active in this tread any longer, I'm going to provide an answer anyway, since it seems like such an interesting question.
First of all, even if a closed system undergoes a process in which it is in contact with a constant pressure surroundings throughout the entire process, and this pressure is the same as that of the system in its initial and final states, the heat absorbed Q is still not necessarily equal to the change in enthalpy unless PV work is the only form of work being done. For example, if there is a stirrer doing work to agitate the system (and cause irreversible viscous heating), and no PV work occurs, and the system is in contact with a constant temperature bath throughout, the change in enthalpy will be equal to zero, but Q will be equal to minus the amount of work that the stirrer does on the system.
Secondly, assume that throughout the irreversible process, the system undergoes a change in which it is in contact with a single constant temperature reservoir at the same temperature as the initial and final temperatures of the system, and is also in contact with a constant pressure surroundings at the same pressure as that of the system in its initial and final states.  But consider that a reversible process between the same initial and final end states does not have to resemble the irreversible path in any way whatsoever, as long as it matches the initial and final temperatures and pressures (and species concentrations if chemical reaction is involved).  Even if the boundary temperature for the reversible path is held constant at the same value as the irreversible path, the pressure of the surroundings during the reversible path definitely does not have to be (and will not be) constant throughout the reversible path.  And, given that this is the case, the heat Q for the reversible path would not be required to be the same as that for the irreversible path, even if the enthalpy changes for both paths is the same.  In virtually all cases, it will actually not be possible to devise a reversible path at a constant surrounding pressure between the same pair of end states.  So the heat Q for the reversible path will not be equal to the change in enthalpy for the reversible path (and thus the change in enthalpy for the irreversible path).

Thanks and sorry for my inactivity first of all. Regarding the first point, ok true but let's assume only pV work. In the second point you make the case that qrev is not equal to q(irrev), which I agree with. But since H=pV+U=q-pdV+Vdp+pdV=q+Vdp, H equals q at constant pressure in absence of pV work. So saying q(rev) is not equal to q(irrev) is saying H(rev) is not H(irrev). And H is a state function (unlike q). That was the origin of my confusion.

Please don't use differentials for finite path lengths in irreversible processes.  It will lead you astray.  Now, let's focus on a specific irreversible process, and see how this plays out.  I am proposing we look at the irreversible expansion of an ideal gas, in which the gas is held in contact with a constant temperature reservoir at its initial temperature and in which we suddenly (at time zero) drop the external pressure to a new constant value and then hold it at that value until the gas re-equilibrates.  Is this an OK case to consider?

The problem is that H is defined for constant pressure, usually H is used for processes involving a change in chemical potential. I guess an isobaric expansion would fit the definition of H (although for G as stated in the original question both T and p are constant)

H is not defined for constant pressure.  The definition of H is simply H = U+PV.  Are you saying you don't like the specific focal problem I proposed?  If not, it's your turn to propose a focal problem.

Here is an alternate irreversible process for you to consider.  I have a system that is going to react chemically.  There are two ideal gas reactants in separate chambers at the same pressure and temperature separated by a barrier.  The system is in contact with a constant temperature reservoir.  At time zero, the barrier is broken and the reaction is allowed to proceed.  The imposed pressure on the system is maintained constant at the initial value.  Compare the values of the heat transferred and the change in enthalpy for this irreversible with a reversible for the same end states.

Yeah you're right, but I started from the definition of G which does assume constant p,T. And for constant p dH=dq. Ok that's an interesting process, but I honestly don't know how to start with it. No pV work will be done here right? So the change in internal energy will be down to heat and chemical potential (making abstraction of other kinds of work). How do you make this reversible/ irreversible?

Let us continue this discussion in chat.

The simple answer to your question is that, for an irreversible process, $\text{d}H \neq \text{d}q_\text{rev}$ under the given constraints (i.e., constant $T$ and $P$, $P_\text{sys} = P_\text{surr}$, only $PV$ work).  On the other hand, $\text{d}H = \text{d}q_\text{irrev}$ is valid.  As Chet Miller suggests in his answer, there is no reversible path for a change of composition at constant $T$ and $P$ (when there is only $PV$ work). Thus, while $\text{d}H = \text{d}q$ at constant $T$ and $P$, $\text{d}H$ nevertheless cannot equal $\text{d}q_\text{rev}$, simply because $\text{d}q_\text{rev}$ does not exist at constant $T$ and $P$.
Note that at equilibrium, there is no change of composition, $\text{d}G=0$ and $\text{d}H = \text{d}q = \text{d}q_\text{rev}$ as it should.
One way to have a reversible change of composition at constant $T$ and $P$ is to allow non-$PV$ work, such as that in an electrochemical cell.  By carefully and slowly adjusting the current, we can allow the reaction to proceed reversibly.  However, in that case $\text{d}H \neq \text{d}q$.  In the presence of non-$PV$ work, $\text{d}H = \text{d}q + \text{d}w_o$. If the non-$PV$ work allows the reaction to go reversibly, then:
\begin{align}
\text{d}H = \text{d}q_\text{rev} + \text{d}w_o
\end{align}
and
\begin{align}
\text{d}G &= \text{d}H - T \text{d}S \\
          &= \text{d}q_\text{rev} + \text{d}w_o - T \text{d}S \\
          &= \text{d}w_o
\end{align}

In your last line, you claim that $\text{d}H_\text{irrev} = \text{d}q_\text{irrev}$. But this cannot be true, because $\text{d}q_\text{irrev}$ is an inexact differential, and it does not then make sense to talk about $\text{d}H_\text{irrev}$ because the system’s thermodynamic state is ill-defined along irreversible paths. $\text{d}H = \text{d}q|_\text{const. P}$ only for reversible isobaric processes.

Suppose I have an ideal gas in a cylinder at temperature $T_0$ and pressure P.  The external pressure on the piston is held at P throughout the process.  I suddenly place the cylinder in contact with a bath at temperature $T_1$ and hold it in contact with the bath until the system re-equilibrates.  In this irreversible process, the work is $P\Delta V$ and, the heat added is equal to $Q=\Delta H$.  Now, same setup, except here I add heat to the system very gradually using a sequence of reservoirs and temperatures running from $T_0$ and $T_1$.  In this reversible path, $Q=\Delta H$ again.  So???

@ChesterMiller, your example involves states and finite differences $\Delta$. My answer concerns paths and infinitesimal differentials $d$.

How can you consider differential  changes in a state function like H along irreversible paths if the states along the path are not thermodynamic equilibrium states?

@ChesterMiller, that’s a good point, and I have updated my answer accordingly. The rest of my answer and my response to your proposed counterexample should still be valid, however.

I do not understand this. The derivation I've seen is exactly this: H=U+pV, so dH=dq-pdV+pdV+Vdp. So at constant pressure and temperature this becomes dH=dq. Enthalpy is used a lot in chemistry, and is always treated as a state function (also Wikipedia confirms btw :-)), depending only on initial and final state. Just like U is a state function and Delta(U) is the same for reversible and irreversible processes H should be as well. You are then basically saying that H is not a state function.

See also https://en.wikipedia.org/wiki/Enthalpy#Relationship_to_heat

Also, Delta(H) is defined for processes such as combustion, which are not reversible.

@Stikke, read my answer carefully, and in particular my first response to ChesterMiller’s comment. $H$ is certainly a state function. Nowhere does $\Delta$ appear in my answer. The standard derivation that you’ve seen is imprecise and requires care to make it work.

@a-cyclohexane-molecule, the OP is correct that $\text{d}H_\text{irrev} = \text{d}q_\text{irrev}$ at constant $T$ and $P$, when $P$ of the system equals $P$ of the surroundings throughout the change of state, and no non-PV work is done.  There are many cases in thermodynamics when an inexact differential is equivalent to an exact differential.  For instance, when no work is done and only heat is exchanged, $\text{d}U =\text{d}q$. Since $\text{d}U$ is path independent, in this case $\text{d}q$ is also path independent.

@ratsalad, at constant T and p, I’m not sure how you plan on perturbing your system. Even if you remove the constraint of constant T, how do you intend on interpreting $dH_\text{irrev}$? Your subsequent example is correct, but it misses the gist of my argument. Reversible or not, the equalities $\Delta H = q|\text{const. p}$ and $\Delta U = q|\text{const. V}$ hold. I am stating that their differential analogues do not in an irreversible situation.

@a-cyclohexane-molecule Given the conventional constraints for using H and G (i.e., constant $T$ and $P$, $P_\text{sys} = P_\text{surr}$, only $PV$ work), it is straightforward to derive $\text{d}H = \text{d}q$ (see OP comment a few above).  The derivation does not assume reversibility, so the differential equality holds for irreversible processes.  This equality $\text{d}H = \text{d}q$ is the fundamental basis of constant pressure thermochemistry and calorimetry.

At constant $T$ and $P$, the system is typically perturbed by a change of composition (e.g., a chemical reaction) that releases or absorbs heat (exo/endothermic reaction).  Any out-of-equilibrium chemical reaction is by definition irreversible.

@ratsalad, the OP’s derivation most certainly assumes reversibility upon taking dw = -pdV. The equality $\Delta H = q|_\text{const. P}$ forms the basis of calorimetry, not its differential analogue. You are correct in stating that changes in N can perturb the system, but I don’t understand why you have chosen to fix T and introduce N, rather than allow T to vary and not worry about N altogether.

@a-cyclohexane-molecule $\text{d}w = -P \text{d}V$ does not assume reversibility, it only assumes that $P_\text{sys} = P_\text{surr}$.  Which could be occurring quasistatically.   Note that $P_\text{sys} = P_\text{surr}$ can be true during an irreversible change of composition.  The OP fixed $T$ and $P$, and that is also the conventional constraint when using Gibbs free energy $G$.  $G$ is typically used by chemists and biochemists since the reactions they are interested in (changing $N$) frequently  occur at constant room temp or body temp and constant atmospheric pressure.

@ratsalad, I must admit that I have never found a fully satisfactory distinction between quasistatic and reversible. My understanding is that quasistatic implies a well-mixed system, so that it can be said to have a thermodynamic state at any point, whereas reversible implies the same internal and external intensive parameters. From my perspective, therefore, quasistaticity is insufficient. If you interpret quasistatic a different way, please clarify this in the context of this problem.

Regarding your other points, an irreversible change of composition—-and indeed any change of composition—requires the addition of a chemical work term to the OP’s derivation, rendering it insufficient. I propose that we restrict the conversation to fixed N and varying T for simplicity. There is no reason to consider G, and hence require fixed T and p, because G is not the subject of our disagreement or indeed of the OP’s post.

Finally, while these points are worth discussing, I believe the crux of the issue can be resolved by considering what you mean by $dH_\text{irrev}$. To me, this is not a meaningful quantity, because the system state during an irreversible process is not well-defined.

Quasi-static just means that the change is slow/controlled enough for internal equilibrium of the system, which allows the internal state functions of the system to be defined, even for an irreversible process.  It is applicable here because we are assuming that the system has defined constant T and P during the change of state.

The derivation of $\text{d}H = \text{d}q$ does not require anything about a chemical work term or chemical potential.  The derivation only assumes the definition of enthalpy $H=U+PV$, the first law $\text{d}U = \text{d}q + \text{d}w$, and the constraints constant $T$ and $P$, $P_\text{sys} = P_\text{surr}$, only $PV$ work.  These constraints in fact allow a well-defined system state during an irreversible process (at least enough to conclude that $\text{d}H = \text{d}q$).

@ratsalad, in that case, our definitions of quasistatic agree, and reversibility is imposed by the condition that the internal and external pressures are equal. The derivation thus holds solely for reversible processes. There is no need for the constraint of constant temperature. I agree that you need not consider chemical work (at least when dN = 0), but if you insist on constant T and p, then the only irreversible perturbation of the system you have access to requires changing N, and hence a chemical work term.

First, reversible implies quasistatic, but not the converse.  Reversibility implies $P_\text{sys} = P_\text{surr}$, but constraining $P_\text{sys} = P_\text{surr}$ does not imply reversibility.  Second, a chemical reaction can occur in a closed system spontaneously and irreversibly without doing any work (and without requiring work to be done on it from the surroundings).  Hence there is no chemical work term required for a change of composition.

@ratsalad, the combination of (i) internal and external parameters being the same, (ii) only heat and PV work (in particular, no frictional or dissipative heating) being produced as a result of the reaction, and (iii) quasistaticity should imply reversibility. Regarding your second point, that's incorrect. There will be a term for the chemical work in the expression for energy because the number of moles of chemical species is changing.

Let us continue this discussion in chat.

@a-cyclohexane-molecule Both of the points in your last comment are incorrect, and the last one violates the first law of thermo for a closed system.  We should move to a chat at this point if we wish to continue.


text:
Is paper aging(oxidation) and paper burning(rapid oxidation) the same thing?
I woke up today with a question of what is fire; looked up the definition and the first response was that fire was rapid oxidation.
I was curious what the difference was between rapid oxidation and regular oxidation that we have all around us as seen with metals rusting, book pages yellowing, artificial aging, paint drying, apples browning due to air exposure.
I cannot find an answer to what the difference between rapid and regular oxidation is, other than the speed of the reaction, and the visibility of the reaction when it becomes fast enough.
I am curious whether there is any difference at all between metals "rusting", paper "burning"/"aging", paint "drying", or if these are all different names for the same process at different rates.
Is it reasonable to say that there is fire everywhere there is oxidation occuring, but it's invisible until the reaction is fast enough?

well fire isn't really an actual thing, its the result of reaction giving out energy.

From what I understand, fire is an actual thing in that if we have two universes, one with oxidation occuring, and one without oxidation occuring, and get temperature of the two universes at every point in space, then subtracting the gradient of the universe without oxidation occuring from the one with oxidation occuring will result in a gradient where the differences may be hot enough to suggest that the point/cluster of points in space is emitting visible light via thermal radiation, in which case we would call fire. Is this a valid model?

With regards to what you are looking at i am unsure. But the traditional yellow flame that we could see when you burn paper is simply the excitation and then de-excitation of carbon atoms.

Oxidation of paper, apples, polymers, etc. goes selectively. You have
chemical substances or regions of molecules which tend to oxidize and those that not.
Combustion goes indiscriminately. You have a bond - it will be broken with suitable fire.

Paint drying is a different process. It is literal drying.

@Jan several types of paint such as oil paint dry through oxidation.

@Kelly Shepphard what do you mean by "selectively"?

@Dmitry, I mean that only "parts" of material oxidize. It can be chemical compound, repetitive unit of copolymer, etc.

Rapid oxidation- burning - leads to two main products carbon dioxide and water. You cannot keep any records if paper is burned.
In case of slow oxidation some of paper components are oxidized which changes the colour of paper, but just that. Old manuscripts can be stored for hundred of years if no none burns them.

But wouldn't that be the same case if the fire under which the paper was uniformly burnt at a rate so slow as to not be able to burn the paper in hundreds of years? Wouldn't the paper still turn black/burnt given a significantly longer period of time?

No. as the point of burning it is that you are supplying the paper with enough energy for complete oxidation. whist when just leave it, there isn't enough energy for complete oxidation to take place.

At the room temperature molecules doesn't have enough energy to pass the activation barrier for the process of turning everything into CO2 and water. This is why combustion never happens without initial heating
Hope it helps.

are you claiming that slow oxidation, given any amount of time, will never be able to accumulate the effects equivalent to rapid oxidation? And if so, is there still technically a "fire" at the point in space where oxidation is occuring, since oxidation emits heat, would there technically still be a visible light emitted from this point in space, no matter how dim, that would look like fire if amplified?


text:
How do catalysts provide alternative routes for reactions?
According to collision theory, a reaction will take place between, say, two molecules, if the collision between the atoms has a sufficiently high kinetic energy (to meet the activation energy threshold) and have the correct orientation. If these two conditions are not met, the collision will not result in a reaction.
However, there is one way to make a reaction easier, and that's with the use of a catalyst. Catalysts do not lower the activation energy or increase the kinetic energy, as I've just learned trying to figure out this question, but they provide an alternative activation energy, lower than the initial one, that, if met, will allow the reaction to take place.
My question is this: what is the actual chemistry behind how catalysts provide this alternative route? What's happening with the molecules or atoms? How does it work?
For example, if I wrap zinc in oxidized copper wire and mix it with some hydrochloric acid, the oxidized copper will act as a catalyst for the reaction between zinc and hydrochloric acid. I'm not sure how this takes place, though.
Thanks.

So catalysts provide an alternative pathway by allowing a new intermediate to be formed.
For example consider the reaction of $A$ and $B$ to form $C$. this means that
$$\ce{A + B -> C}$$
Say this reaction has  an activation energy of $x$ $kJmol^{-1}$.
This reaction could also be possible with a catalyst $X$. This means that a new possible route for the reaction could be (there a multiple routes the new reaction could take this is just a simple example):
$$\ce{A + X->AX}$$
$$\ce{AX + B->C +X}$$
Now you have ended up in the same position using a catalyst and when not using a catalyst, seen as you add both reaction and cancel out common species then you get back to the equation of:
$$\ce{A + B -> C}$$
These two reactions with the catalyst will have their own activation energies. Say these two reactions have activation energies $\pu{p}$ and $\pu{q}$ $\pu{kJmol^{-1}}$ and that $\pu{p>q}$ (doesn't matter if $\pu{p}$ applies to the first or second reaction) then if $x>p$ a particle needs less energy to proceed via the route with the catalyst compared to the route without the catalyst.
As a particle now needs less energy for the product $\pu{C}$ to be formed, more particles will now be able to take part in the reaction thus increasing the amount of successful collisions per second which is the same in an increase in rate. This can be seen nicely in a general Maxwell-Boltzmann distribution of molecular energies as the new route will mean there's a lower activation energy overall meaning a lot more particles can react.

Thanks a ton. Could you replace the - > in your comment for arrows with the symbol →? Just copy and paste it in, it makes the answer more visually clear.


text:
Why aren't primidone's lactam groups acidic?
I understand that 5,5-disubstituated derivates of barbituric acid only show lactam-lactim tautomerism opposed to the lactim-lactam and keto-enol tautomerism of barbituric acid, hence the 5-mono and 5,5-disubstituated derivates are less acidic.
Primidone differs from phenobarbital in a C2 oxo group, so it's lactam groups are not acidic and it is not soluble even in alkali-bases. Why aren't the lactam groups acidic exactly? The oxo groups should still present quite a bit of electron withdrawing effect, shouldn't they?

Amides are not acidic and that is all these are

There's a carbonyl group missing (i.e., a barbituate is made from urea and this is made from diaminomethane). Compare the pKas of a ketone versus a $\beta-$keto ketone.


text:
Half-reaction for ammonium nitrate comproportionation into dinitrogen monoxide.
Ammonium nitrate comproportionates into dinitrogen monoxide in the following reaction:

$\ce{NH4+ (aq) + NO3- (aq) -> N2O(g) + 2H2O (l)}$

I attempted to write the half-reactions by calculating the oxidation state for nitrogen in the ammonium and nitrate ions, and then finding the electrons needed for each half-reaction.
For example, for the ammonium half-reaction, I noticed that $\ce{N}$ is in a -3 oxidation state in $\ce{NH4+}$ and in a +1 oxidation state in $\ce{N2O}$. I concluded that four electrons must reduce ammonia and wrote the following preliminary equation:

$\ce{NH4+ (aq) + 4e- -> N2O (g)}$

Which I balanced the elements using standard acidic redox into:

$\ce{2NH4+ (aq) + H2O (l) + 4e- -> N2O(g) + 6H+ (aq)}$

However, I realized that my chemical equation is not balanced with respect to charge, indicating I have a flawed half-reaction.
What would be the correct half-reactions for this chemical reaction?

It is half-reaction(oxidation)$$\ce{2NH4+ (aq) + H2O (l)  -> N2O(g) + 10H+ (aq) + 8e-}$$

Ahhh. I had two mistakes: recognizing which species undergoes oxidation vs reduction and using the appropriate number of e-.


text:
Can LCAO be seen as superposition?
In wave physics, when waves interact, the principle of superposition applies. It states that the displacement of the resultant wave is made up of the sum of the individual displacements of the interacting waves. This appears to be very similar to the expression of a molecular orbital $\psi$ as a sum of the different atomic orbitals $\psi_i$ with different weighted coefficients. Is there any overlap between the quantum chemical basis of LCAO and superposition of waves?

Wouldn't be just superposition implies weighing coff to be unity? Otherwise yes, interference of waves is surely at the core.


text:
Why is the combustion of Nitrogen endothermic?
Other combustion reactions are exothermic but this combustion reaction endothermic. Why ? Please explain..

consider the strength of the Nitrogen-Nitrogen triple bond

And, not all reactions with oxygen are exothermic - the ones that are tend to be called combustion reactions since, well, they actually combust.

As said in the comments a reaction with oxygen doesn't have to be exothermic it can be endothermic - as of this we only really say the reaction with oxygen is combustion if the reaction is exothermic.
In the example of the reaction of $N_2$ with $O_2$ several nitrogen oxides can be formed $N₂O$, $NO$, $N₂O₃$, $NO₂$, and $N₂O₅$. But well do an example of $NO$ formation as this one of most common products.
For a reaction the enthalpy change of the reaction using bond enthalpies is given by $$\Delta H=\sum{ broken}-\sum{ formed}$$
So are reaction for $NO$ formation is
$$N_2 + O_2\rightarrow 2NO$$ and the bond enthapy of $N\equiv N$ is 891 kJ/mol, $O=O$ is 495kJ/mol, and $NO$ is around 90.2 kJ/mol
Therefore $$\sum{ broken}=1386 kJ/mol$$
$$\sum{ formed}=180.4 kJ/mol$$
so $$\Delta H = 1389-180.4=1205.6 kJ$$
This means that $\Delta _f H[NO]=602.8 kJmol^{-1}$ this shows that the reaction is endothermic as $\Delta H$ is positive despite being a combustion reaction.

See https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here


text:
Why it is very rare to find a compound that lacks carbon to be liquid at standard temperatures and pressures?
It is the first time I heard such a generalization, I had no idea. Why being liquid in standard conditions is special to compounds which contain carbon atom? Which property of carbon cause this?

It is a property of human mind, and not that of carbon. If you hear about a remote island inhabited by two-headed people, will you seek an explanation, or try to verify whether the thing is true in the first place? Water, mercury, H2SO4, Br2, POCl3 all are liquid, despite containing no carbon. Methane, CO2, naphthalene, adamantane, diamond all contain carbon, but...

@IvanNeretin - while I totally agree with you, the fact remains that organic chemistry sure makes a lot of things that are liquid at STP. Far more than, say, Osmium...

@JonCuster True, but then again, organic chemistry also makes far more gases and far more solids than osmium.

@IvanNeretin - OK, I'm getting a little silly, but carbon should get a lot of credit simply for ethanol... Cheers!

@IvanNeretin Hello there, asking here is my way to verify actually :) I searched and couldn't find anything satisfying that's why I asked here. I already know there are liquid compounds that's why I wrote "rare" not "impossible". Because organic compounds form a majority. Thank you anyway.

The property of carbon that causes this is the ability to form rings and chains that are stable under atmospheric conditions.
The only reason why ‘so many liquids contain carbon’ is basically confirmation bias. There are orders of magnitude more compounds known that contain carbon than those that do not. For the longest time in the history of chemistry, inorganic chemistry was restricted to very simple molecules of only a handful of atoms while polymerisation was already a thing in organic chemistry. Furthermore, a large proportion of ionic compounds is inorganic—or, in the context of this question better: carbon-free—and most of these are solids at standard temperature and pressure adding to the confirmation bias.
Thus, by sheer economy of scale the number of known carbon-containing liquids is far higher than the number of known carbon-free liquids.


text:
Experimental measurement on the ring-strain
for my thesis i calculated the ring-strain of a molecule by MP2 on the computer. I was wondering if there is an easy possibility to get the experimental value of the ring strain. I guess I have to combust the compound and somehow calculate it from the combustion-enthalpy. My Problem is, that there is no access to the open form of the molecule to compare it. Is there still a way to calculate it?
I thought of the Hess-law maybe there is a way with some standard values?
It is an organic molecule only containing C, H and O.

I am looking for an experimental setup such as calorimetrie oder TGA ore something to determine the ring strain of my specific molecule. The links are not helping, because my molecule is not a cyclo alkane.
You're answer seemes quite rude regarding that it is not helpful or targeting the problem at all.

You need a measure of ring strain, ie comparison of combusion energy wrt non-strained compound. Wikipedia is helpful: https://en.wikipedia.org/wiki/Ring_strain

I edited my question. There is no "open form" of my molecule, well there ist but it is no accesable because I open the ring and polymerize the molecule via ring-opening metathesis polymerization.

In any case to determine something like ring strain you need an unstrained reference molecule to compare to. Then you can use Hess's law for instance, if data is available.

There has to be another way. There is a lot of experimental data in publications, but no one is explaining how it is done. There is no open form of the molecule. In Metathesis, you break the doublebond and attach it to another broken doublebond.
https://en.wikipedia.org/wiki/Olefin_metathesis
Therefore, the open form with double bonds has at leaast two more methylene groups.

If I understand correctly the problem is that the initial strained molecule is a monomer and the unstrained molecule is a polymer? If so off the top of my head two ideas to start with: 1) "trap" the open form of the monomer by reacting it with something that will quench polymerization; 2) perform the combustion on the polymer but normalize by the degree of polymerization to determine individual monomeric contributions.

Basicly that is the problem yes. I can not trap the open form. The ring opens via metathesis, so the doublebonds breaks and forms 2 new doublebond with 2 new carbon atoms. So even if i stop the reaction, it wont be the correct form as it contains more carbon. If I trap it without adding carbon, then I loose my doublebond which also has an impact on the energie of the molecule.
It is hard to determine the exact chain length, because I' measuring it via GPC which is an relativ measurement no absolut. As the molecule is new, there is no standard for it to get an exakt amount of repeating units.

In the absence of more info I'd say you want to study the transition state during the reaction, and there are definitely ample ways of doing this at the ab initio QM level.

I don't understand what you mean with that. Did you look at the Metathesis mechanism, take cylcopentene for example. The open form would be penten. That already has 2H more. Now there is no way to just open cyclopentene. When you do ring-opening metathesis, you end up with Hept-1,6-en. When you do the ring-closing of the Hept-1,6-en you end up with cylcopentene and ethene. So it would be the correct linear form for this reaction type but include 2 CH2 more than the original one.
How do you want to trap the transition state as a stable product which you can combust in a calorimeter afterwards?

I mean ab initio (eg DFT or other) modeling of different structures. You start with monomer and transform into transition state, finally product. There are people well versed in how to do this. See eg references associated with https://cordis.europa.eu/project/rcn/99638/factsheet/en , or search for "ab initio modeling olefin metathesis"

For an older ref: https://pubs.acs.org/doi/10.1021/om00002a024

I don't mean at all experimental determination, unless you have some spectroscopic way of catching signatures of the intermediate. This may be possible in the gas phase, not sure if in solution. And when I say "in gas phase" I mean, if that is at all possible.

But now I'm really out of my league: https://pubs.acs.org/doi/abs/10.1021/ja078149z?journalCode=jacsat  etc ...

Perhaps also this: https://s3.amazonaws.com/academia.edu.documents/46972552/om020581w20160703-24993-bp1p.pdf?AWSAccessKeyId=AKIAIWOWYYGZ2Y53UL3A&Expires=1545146847&Signature=EyC1D%2FNMf6RUg5BSg4tU7U5jDZU%3D&response-content-disposition=inline%3B%20filename%3DMolecular_Modeling_of_Ruthenium_Alkylide.pdf


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How to identify whether a metal carbonate is basic or not?
Magnesium carbonate is basic, but why isn't lithium carbonate also basic? And how does one determine whether a carbonate is basic or not?

Who says lithium carbonate is not basic?

lithium carbonate and magnesium carbonate behave near enough the same. lithium carbonate is basic just not as basic as other group 1 carbonates. It also has low solubility in water and is the only group 1 carbonate that isn't thermally stable.
It is still is basic but it just doesn't fit to the general trends.


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Preparation of Saturated Salt Solutions - W/O Being Wasteful
Question
What is the most cost effective way to prepare saturated solutions of each of these compounds? LiCl, K2CO3, and NaCl.
Background
I am trying to create a rudimentary controlled humidity device within my lab. I am essentially taking a desiccator, removing the desiccant, cleaning thoroughly, and inserting various saturated salt solutions in order to have a controlled vapor pressure of water above the surface of the saturated solution; assuming there is a constant temperature.
I have included a temporary link to the two journal articles I am using to try and accomplish this goal, but I will try and give all the relevant information here as well so people don't have to go and check out my shadylinks.Dropbox w/ journal articles
I have the salts needed to make up solutions of LiCl, K2CO3, and NaCl. They were quite expensive for our small research group though and I don't want to waste any materials. The volume in the bottom of the desiccator is about 200mL and I have 100g of each salt to use for the course of the short experiment.
I know that a saturated solution in general chemistry is simply prepared by dissolving as much salt as is possible into a heated solution and then cooling it. However, as I have been reading some of these papers there are many reports of non-diffusing barrier layers on the surface of many saturated salt solutions. This means that the relative humidity of these solutions changes a significant amount over time. Some papers report the use of "slurries" or "sludges" in which just enough water is added to a pile of salt in order to make it appear damp.
I have also read that the preparation of a saturated salt solution is very dependent on the relative humidity which is established by that salt solution. More specifically, that low rel. humidity saturated solutions might need the slurry approach and higher rel. humidity solutions favor the standard saturation techniques. However, I have no idea what the cuttoff for "low" and "high" is and I am left to assume it's whatever the atmospheric humidity is.
I am more than happy to do the legwork and read up on as much as need be, and I am still doing so as this post sits. I just don't know where to go or how to approach the question of how much saturated solution, or how much slurry, will be enough to control the humidity inside of about a 1L container. I can't afford really to go about preparing the solutions in order to find the best method experimentally and I was really hoping there were some experienced people out there who could point me in the right direction.
Thank you for all of your time reading this regardless of response! It is much appreciated!

I would go by simply adding while stirring until a solid remains. Then a little more. The temperature in the lab should be stable enough (presumably) not to alter Kps. Just check that no more salt dissolve, I.e. take a periodic look at the solid deposit at the bottom. That is.

"Some papers report the use of "slurries" or "sludges" in which just enough water is added to a pile of salt in order to JUST make it appear CLEARLY damp," adding water drop by drop uniformily (personal experience).

I went through and tried to calculate how much water vapor I would need to fill the chamber, then added that much water to a slurry pile which was large enough to not dissolve when that much water was added. However, I am getting some pretty erroneous results with my K2CO3 samples. I am currently still trying to find out what could be going on here.

Thanks for all the help everyone. Sorry for the delay in response.


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Ionization enthalpies of Transition elements
“There is generally expected increasing trend in the values as the effective nuclear charge increases. However the value[First ionisation enthalpy] of Cr is lower because of the absence of regarding any change in d configuration and the value of Zn higher because it represents an ionisation from 4s level”
For, Cr it is 653 kJ/mol and for Zn it is 906 kJ/mol
It seems that the statement regarding Zn undoes what is pointed out by the statement regarding Cr, as for Zn it changes from 4s2 3d10 to 4s1 3d10 and for transition from Cr to Cr +1 the electronic configuration changes from 4s1 3d5 to 3d5, clearly both have change in their 4s configuration and not in 3d configuration.
My first question is what does this mean?
Problem in understanding this is that the book has not mentioned anything about filled 4s2 configuration of Zn, it seems that it is supposed to be understood from the statement alone.
Even after ignoring that, if I think in terms of effective nuclear charge, which is higher for Zn and as a result higher energy required for ionisation of electron from 4s2, but the problem in that that it does not seem to be the explanation of that because of that However , which seems to point to some exception.
I have checked other questions on this site regarding the irregular trend of ionisation enthalpy for transition element but they don’t seem to me to address this.

Strongly related: https://chemistry.stackexchange.com/q/151

in the latest version of ncert, there are many changes in this part, it's better than the earlier version

there were mistakes/inacccuracies in the earlier version, like the one you have quoted ... ionization from 4s level happens in all 3d metals, not just Zn

Latest version of NCERT can be found here.

I don't think I need the answer of the question anymore, though you can post your answer here, If it is reasonable I can accept it.