Confusing stability order of free radical structures I have a question regarding the solution that has been given.
Question:

In the given compound, what would be the order of ease with which hydrogen atom can be abstracted from carbon atoms I to VI?:

The solution that has been given is:

In the solution they've posted above, how is the 2nd structure more stable than the third structure? The third one clearly has resonance with the pi bond, while the 2nd one has no resonance with the oxygen atom (it can't "accept" oxygen's lone pair as it is not a carbocation, and oxygen does not have an empty $\mathrm{d}$ orbital either).
Please can anyone shed light on this issue. Thank you!

Since this question was not answered considering it was posted for a long time, I'd try to give enough reasoning for the given stabilization order.
I'd start with mass spectroscopy. The most widely used technique in mass spectrometry up to date is electron ionization (EI), which produces molecular ions from gas-phase analytes. These molecular ions then fragment in a reproducible way. The energy sufficient for ionizing and fragmenting gas-phase analyte molecules is acquired by interaction with $\pu{70 eV}$ electrons. The range of the ionization energy of most organic compounds are fall between $\pu{10 eV}$  and $\pu{20 eV}$. The most reference spectra are obtained at $\pu{70 eV}$ because at that level the perturbations in electron energy have negligible effects on ion production (Ref.1).
A neutral analyte molecule absorbs energy during the interaction of the ionizing electron with its electron cloud. The analyte molecule interacts with an electron having $\pu{70 eV}$ of kinetic energy, which quickly causes the ejection of one of the electrons from the analyte (it needs to absorbs approximately $\pu{14 eV}$ as internal energy to make this ejection). Because the analyte molecule has many electrons, it is reasonable to ask, “Which electron leaves during the ionization process first?” Whereas sufficient energy may have been absorbed by the analyte molecule to eject any one of its electrons, the most probable ejection will be of an electron that is least tightly bound. Therefore, it is predictable that the most likely site of ionization will be where electrons are loosely bound (e.g., in the nonbonding orbital of a heteroatom) (Ref.1):
$$
\begin{array}{c|ccc}
\hline
\bf{\text{Compound}} & \bf{\text{Ionization Energy, }} \pu{eV}^\mathcal{a} \\
\hline
\ce{H3C-NH-CH3} & 8.23 \\
\ce{H3C-CH2-NH2} & 8.86 \\
\ce{H3C-O-CH3} & 10.03 \\
\ce{CH2=CH2} & 10.51 \\
\ce{H3C-CH3} & 11.52 \\
\hline
\end{array}\\
^a: \text{Reference 1}
$$
The above data shows abstracting an electron from $\ce{O}$ atom is easier than that from a carbon. That means, lone pair is loosely bound than $\sigma$- or $\pi$-bond. Therefore, it is safe to assume that resonance contribution of a lone pair is more than that of a double bond. Even though, this is not full a lone pair but one electron contribution for resonance, it is possible to contribute one electron from a lone pair and the other electron stays in one empty anti-bonding orbital as shown in below:

Again, after all, molecular ion of ketone stays as $\ce{R2-C=O^{.+}}$ in EI-mass spectrum of a ketone, usually with the intensity of ~40% of the base peak. That means $\ce{R-CH=O^.-R'}$ is pretty stable contributor (Recall both carbocations and carbon radicals have the same order of stabilization). Thus, it is safe to say that stabilization by oxygen atom is better than that by a double bond. This conclusion would be supported by user55119's comment (vide supra), which states that (homolytic) bond dissociation energy  (BDE) of $\ce{CH3O\color{red}{CH2-H}} = \pu{83 kcal/mol}$ while that of $\ce{CH2=CH-\color{red}{CH2-H}} = \pu{89 kcal/mol}$ (Ref.2). Keep in mind that the easiness of BDE is due to the stabilization of the resultant radical (either by oxygen or the double bond).

References:

J. Throck Watson, O. David Sparkman, In Introduction to Mass Spectrometry: Instrumentation, Applications, and Strategies for Data Interpretation; Fourth Edition, John Wiley & Sons, Ltd.: Chichester, West Sussex, England, 2007, “Chapter 6: Electron Ionization,” pp. 315-448 (ISBN: 978-0-470-51634-8).
Stephen J. Blanksby, G. Barney Ellison, "Bond Dissociation Energies of Organic Molecules," Acc. Chem. Res. 2003, 36(4), 255–263 (https://doi.org/10.1021/ar020230d).


Use of Spiegeleisen in manufacture of steel I recently read the following statements related to Manufacture of steel via Bessemer Process:

  When no more $\ce{CO}$ is produced (i.e. blue flame disappears) a calculated amount of Spiegeleisen (an alloy of $\ce{Fe, Mn, C}$) is added and distributed throughout the mass by blowing air for some time.

Why do we need to add Spiegeleisen at the end? Is it to reduce the oxides of iron (that might have been formed)? Or due to some other reasons, like, say $\ce{Mn}$ to make steel harder and increase its tensile strength? But latter doesn't convince me, since we could have added $\ce{Mn}$ directly on order to get the same results. Also what's the meaning of calculated amount, how do we calculate the amount to be added, and why?

I didn't think Bessemer was used anymore. Spiegeleisen is more likely to be called ferro-manganese today. I expect it is added at the end to minimize losses of Mn to the slag. Mn does contribute to hardenability  but the primary reason for its use is to combine with residual sulfur to make MnS . MnS stringers can be harmful but are an improvement because otherwise the S will mostly be in grain boundaries reducing ductility and toughness. ( Bessemer process adds nitrogen to the steel reducing the toughness).  Calculated amount would be based on the total weight of Mn that must be added to get the desired content in the steel , such as 0.5 weight %.The added amount would need to allow for expected loss to slag, etc.


Which is the currently accepted mechanism of a Wittig reaction? The Wittig reaction is one of the most significant advances in synthetic organic chemistry in the 20th century and rightfully won its discoverer, Georg Wittig, the Nobel Prize in Chemistry.

A Wittig reaction is the addition of a phosphorus ylide (previously thought to be an ylene with a $\ce{C=P}$ bond) to an aldehyde or a ketone resulting in a $\ce{C=C}$ double bond.

Mechanistically, there is little debate that the final step is a cycloreversion of an oxaphosphetane liberating a phosphane oxide. However, as far as I know there is no agreement on the preceeding step(s) to form said oxaphosphetane; with two different competing mechanistic proposals:

Nucleophilic attack of the ylide onto the carbonyl carbon to give a charge-separated betain structure. This rotates around the newly-formed $\ce{C-C}$ bond until oxide and phosphonium are in proximity whereupon a bond is formed to close the oxaphosphetane (stepwise mechanism)
[2+2] cycloaddition of the ylide and the carbonyl to immediately form the oxaphosphetane in a concerted manner.

Wikipedia just presents these two mechanisms side by side hence it is not really helpful.

Which of the two mechanisms is, at present, considered the most likely reaction pathway? Which principal pieces of evidence point towards it and disfavour the competing mechanism?

Please answer including references to the corresponding journal articles.

tl;dr: I don't think there is any mechanism that is 100% correct, and, in cases like this especially, I think it would completely depend upon what set of carbonyl/ylid/base/solvent etc. was used. But, of course, we like being able to generalise, and to my knowledge theres a lot more evidence to support a concerted type mechanism.
General Background
The question is hopefully summarised in the scheme below (no geometry, that, in itself, is a pretty lengthy discussion). Reaction of a carbonyl (in this case a ketone) with a phosphorus ylid is able to give rise to two species, either an oxaphosphetane directly or a betaine (an internal salt for the benefit of Loong) which then goes on to form the oxaphosphetane which itself is able to eliminate to afford the olefin with generation of triphenylphosphine oxide as a thermodynamic driving force.
This answer will present the case for a cycloaddition mechanism, and evidence against the betaine pathway. Importantly, only Wittig reactions of unstabilised ylids, as with stabilised ylids the rules of the game change due to reversibility of intermediate formation (see a Horner–Wadsworth–Emmons).

Evidence against the initial formation of a betaine
Betaines have never been observed during the course of a 'normal' Wittig reaction, that is, spectroscopically, we are unable to see it (this may, of course, just be due to the fact that the formation of the oxaphosphetane is so rapid relative to the timescale of the methods used).
Many reports of betaines have been from accidental (or otherwise) opening of oxaphosphetane intermediates, and indeed this seems to have been what led Wittig down the betaine pathway in the initial reports (though in the first ever paper he suggested that oxaphosphetane was the sole intermediate, but subsequently couldn't find any evidence for it back in the days before NMR etc).
Wittig[2] and others[3] reported the isolation of crystalline betaine salts. Phosphonium bromides were treated with PhLi to give a ylid, to which a carbonyl was added; subsequent addition of hydrogen bromide afforded crystalline solids which were able to be elucidated, proving the formation of a betaine.
Whilst this was convincing at the time, we now know that the betaine isn't observed, but rather the oxaphosphetane is. Vedejs (whom we'll come back to), has gone on to show that the earlier findings could be explained by quenching of the oxaphosphetane rather than as a result of directly trapping the betaine[4], which is more consistent with the other data available to us.

Scheme 1: Generation of the crystalline betaine derivative from the oxaphosphetane.

There are more recent reports of 'stable' betaines for very specific substrates where subsequent collapse isn't possible, again generated from oxaphosphetane intermediates.
Stefan Berger's group at Leipzig reported a betaine stabilised by a bipyridyl group, allowing NMR data to be reported for the first time[1].

Scheme 2 Reversible formation of a stabilised betaine

In this case, the lithium holds a chelated structure together, preventing it from forming the alkene (and also to a certain extent preventing reformation of the oxaphosphetane on steric grounds). Interesting, with addition of 18-crown-6 to sequester the lithium, the reaction proceeded normally in the forward direction with alkene signals observed.
The fact Berger was able to do this, does of course not imply that they are real intermediates along the Wittig pathway, but it was a nice piece of mechanistic work that I thought deserved mention.
In addition to this mechanistic work, there are some empirical issues with invoking a betaine intermediate that cannot be easily explained. Namely, all Wittig reactions using non-stabilised ylids should be under kinetic control and irreversible (hence highly (Z)-selective due to the inability of the initial intermediates to reverse and hence equilibrate to the thermodynamic product), however this is frequently observed not to be the case, and in certain cases, high (E)-selectivity can be achieved.
Evidence for direct oxaphosphetane formation
Vedejs was the one of the first to propose direct (irreversible) cycloaddition of the ylid and carbonyl to give rise to the oxaphosphetane, quickly followed by cycloreversion to form the desired alkene and a phosphine oxide.

Scheme 3: cycloaddition/cycloreversion mechanism for Wittig

In his early reports, direct observation of the oxaphosphetane by P-NMR was conducted[5], the paper reasons that the high-field chemical shift observed for the phosphorus containing intermediate ruled out charged, 4-valent species such as betaines, but was instead more consistent with formation of the 5-valent neutral species such as an oxaphosphetane.

Scheme 4: Vedejs' observation of oxaphosphetane.

This NMR analysis isn't really too convincing by itself, however in the scheme above, compound 4 had previously been characterised by X-ray crystallography[6] and various NMR methods and as such gave some weight to the species observed by Vedejs being similar to the known compound. More recent work[7] has also observed directly the cis/trans oxaphosphetanes formed via competing cycloadditions.
Summary
Overall, direct observation of the oxaphosphetane but not the betaine does seem to suggest that mechanistically the cyclic intermediate is formed directly rather than via the betaine closing.
As stated previously, we can't 100% rule out the possibility that the betaine closure is just sufficiently rapid to appear invisible to our currently spectroscopic methods however every report of isolated/observed betaines that I've came across thus far has either intentionally been formed from an oxaphosphetane or can be explained by this without the authors recognising it.
One thing that is hugely missing from the story is computational work, whilst some studies have calculated the relative energies of the two intermediates, to my knowledge, the entire reaction coordinate hasn't been fully explored.
In conclusion, I think the cycloaddition mechanism is safest, and indeed it was the one I was taught as an undergraduate (and the one I know several other leading universities use), so if in doubt i'd go for that explanation, but on the firm understanding that more mechanistic work is needed.

References
 Generally: Modern Carbonyl Olefination, Carey Advanced Organic A, and Comprehensive Organic Synthesis I.
[1] Neumann, R. A.; Berger, S. Observation of a Betaine Lithium Salt Adduct During the Course of a Wittig Reaction. Eur. J. Org. Chem. 1998, 6, 1085. Subramanyam had reported similar chemistry earlier, but the NMR data was incomplete.
[2] Wittig, G.; Haag, A. Über Phosphin-alkylene als olefinbildende Reagenzien, VIII. Allenderivate aus Ketenen. Chem. Ber. 1963, 96 , 1535. DOI: 10.1002/cber.19630960609. A very old paper in German.
[3] Schlosser, M.; Christmann, K. F. Olefinierungen mit Phosphor-Yliden, I. Mechanismus und Stereochemie der Wittig-Reaktion. Liebigs Ann. Chem. 1967, 708, 1. DOI: 10.1002/jlac.19677080102.
[4] Vedejs, E.; Meier, G. P.; Snoble, K. A. J. Low-temperature characterization of the intermediates in the Wittig reaction. J. Am. Chem. Soc. 1981, 103, 2823. DOI: 10.1021/ja00400a055.
[5] Vedejs, E.; Snoble, K. A. J. Direct observation of oxaphosphetanes from typical Wittig reactions. J. Am. Chem. Soc. 1973, 95, 5778. DOI: 10.1021/ja00798a066.
[6] Mazhar-Ul-Haque; Caughlan, C. N.; Ramirez, F.; Pilot, J. F.; Smith, C. P. Crystal and molecular structure of a four-membered cyclic oxyphosphorane with pentavalent phosphorus, PO2(C6H5)2(CF3)4C3H2. J. Am. Chem. Soc. 1971, 93, 5229. DOI: 10.1021/ja00749a044.
[7] Maryanoff, B. E.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108, 7664. DOI: 10.1021/ja00284a034.


How to produce dry pure oxygen I'm supposed to produce dry and pure oxygen using the decomposition of hydrogen peroxide with a manganese dioxide catalyst.

$$\ce{2H2O2 (aq) → 2H2O (l) + O2 (g)}$$

The gas will be collected in a bottle by downward displacement of water. How can I safely move the oxygen gas through the drying agent without mixing in atmospheric gas?

Commercial drying agents, such as Drierite (anhydrous calcium sulfate), or other desiccant, should do the job for small quantities of oxygen.

Most inorganic desiccants should not react with diatomic oxygen; as long as the oxygen is produced separately from the desiccant, there should be no issue. (Dropping 90% or stronger $\ce{H2O2}$ on many substances can cause dangerously rapid decomposition, but with 30% or lower concentration, that danger is minimal.)

As far as preventing atmospheric gases from entering, first bleed the apparatus of residual air by discarding the initial output. Seal the later containers underwater, e.g. by sliding a glass plate across the mouth. See Janice Crowleys' demo on YouTube.


Nitric acid vs. chloric acid strength Here the central chlorine atom is best described as $\ce{sp^3}$ hybridized. This makes sense; the molecule is pyramidal. Also, due to the molecule's pyramidal geometry, this rules out the possibility of effective resonance due to less overlap between the  $\ce{sp^3}$ lobes, which are angled, as opposed to the parallel pi lobes.

This explains quite well the difference in acidity between nitric acid and chloric acid; nitric acid is stronger than chloric acid. At first it seems difficult to differentiate between the acidities of both; the nitrate anion also has three peripheral oxygen atoms, which provide a similar (negative) inductive effect by delocalizing electron density.

With regard to electronegativity, the chlorate anion seems like it should be more slightly stable than the nitrate anion, implying that chloric acid is stronger than nitric acid; chlorine is more electronegative than nitrogen (if only slightly so - difference is 0.12 on the Pauling scale).

Oxidation state analysis suggests that the two should be similar in acidity; both the chlorine and the nitrogen have oxidation states of +5.

So it seems that the best rationalization of why nitric acid is stronger than chloric acid is the "better" (overlapping pi orbitals) resonance form of the nitrate anion.

With respect to the above rationalization,

1) How, exactly, do parallel pi orbitals enable "better" resonance? I get the principle of maximum overlap. If I had to guess, I would say that more overlap enables greater distribution of electrons among the relevant orbitals ... is there a more concise and scientific way of putting this? Is this even correct?

2) Are parallel pi orbitals necessary for resonance? I got the impression it is from organic class. Or are parallel pi orbitals only necessary for a special form of resonance - conjugation? Is the picture on the bottom referring to conjugation?

3) Are there any other ways to rationalize the difference in strength between these two acids?

The two ions you are comparing to each other are very different. In $\ce{ClO3-}$ there are two more electrons than in $\ce{NO3-}$. This is also the reason why the former pyramidalizes, while the latter does not.

In nitrate the negative charge can be delocalised over the whole molecule, since there is one molecular orbital with $\pi$ symmetry. One basic requirement for these orbitals is, that they are antisymmetric with respect to a mirror plane, that includes the bonding axis.

This MO scheme would also be thinkable for the chlorate ion, but here you have two more electrons, so you have to fill an antibonding orbital. The charge can no longer be delocalised as well in this molecule. The conclusion is, that the symmetry of the whole molecule is lowered and the $\pi$ orbitals vanish. A lone pair is always more stable in an orbital with high $\ce{s}$ contribution, which leaves only the $\ce{p}$ orbitals of chlorine for bonding with oxygen.

Resonance is a model which is necessary to describe bonding situations, that cannot be described by one Lewis structure alone. It is therefore more a rule of thumb than any evidence. $\pi$ bonds (or orbitals) are not necessary for valid resonance structures. They are however necessary for ([negative] hyper-)conjugation.


Why is WF6 stable whereas CrF6 is unknown? $\ce{F-}$ is a hard base since it is small and relatively polarizable.

Both $\ce{Cr^6+}$ and $\ce{W^6+}$ are hard acids, but shouldn't $\ce{CrF6}$ be favored since it would be a smaller hard acid?

The answer simply has to do with the accessibility of the high +6 oxidation state.

In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another; the only Cr(VI) compounds that we know of are paired with extremely hard bases like the oxide ion, viz. CrO3, CrO42−, Cr2O72−.

Apparently, fluoride isn't quite hard enough to stabilise Cr(VI). Another factor is sterics - squeezing six fluorines around Cr is relatively difficult as compared to having three or four oxygens.

In W, the 5d orbitals have two inner radial nodes. The two pockets of electron density close to the nucleus shield the bulk of the electron density, which is far away from the nucleus. So, when you continuously remove electrons from W, the remaining 5d electrons don't drop in energy so rapidly.

Here are Latimer diagrams for Cr, Mo and W (from Shriver et al., Inorganic Chemistry 6ed.); you can see that the stability of the higher oxidation states increases going from Cr to Mo to W (more obvious if you construct a Frost diagram, but I'm too lazy to do so):

This trend is hardly limited to the Group 6 elements. OsO4 features Os(VIII), whereas the highest oxidation state of iron is Fe(VI), which is already extremely difficult to achieve. Likewise Au(III) is pretty normal but Cu hardly goes above Cu(II).


Different chemoselectivity of transition metals in catalytic hydrogenation I am quoting my Clayden et al 2nd edition:

  Some catalysts are particularly selective towards certain classes of compound – for example, Pt, Rh and Ru will selectively hydrogenate aromatic rings in the presence of benzylic C-O bonds, while with Pd catalysts the benzylic C-O bonds are reduced faster.

My question is, why $\ce{Pt, Ru, Rh, Pd}$ all being transition metals still show different chemoselectivity in hydrogenation reactions?

1st edit: (After reading ringo's comment I think I should add something more) Well, I first thought that the number of $\ce{d}$ electrons is the cause. But, both $\ce{Pd}$ and $\ce{Pt}$ have same number of $\ce{d}$ electrons. Yet, they show different chemoselectivity. I think the difference is because $\ce{Pt}$ has an $\ce{f}$ orbital filled up.

And, I guess as $\ce{Ru, Rh, Pd}$ belong to same period, they show similar chemoselectivity.

Unfortunately, this is extremely complicated and the topic of active research in both computational and inorganic chemistry and chemical engineering. There is no simple answer to your question. We can discuss various forces that go to determining chemoselectivity though.

1) Crystal structure: You said that you are primarily focused on bare metal catalysis, so I will restrict myself to those systems. Not all surfaces of even the same material behave the same. These are commonly described by Miller indices (1), which describe the surface by how a plane cuts through the material's unit cell. The reactivity and selectivity are heavily determined by the exact surface structure (think dangling bond sites, binding sites, multicenter reactivity, all of which are influenced by the surface atomic structure both local and in its immediate environs). Different materials will have different crystal structures based on both electronic effects (like how many valence electrons there are) and size effects (how big atoms are, although there is relatively less variation on this in the 2nd and 3rd rows).

2) Chemistry of binding sites: Some materials are more likely to bind molecules than others. The Pt-H bond is extremely strong, much more than a Pd-H bond (these trends down a period are often explained in terms of the diffuseness of the 5d orbitals vs 4d which lead to better overlap with the main group orbitals of hydrogen and other light elements). This leads to greater coverage of a Pt surface with hydride equivalents than a Pd surface (generally, caveats apply). On the other hand, Pd tends to bind oxygen atoms more strongly than Pt does (one can think about hard-soft acid base theory here to explain that, but the situation is probably more subtle and complicated than that). One can easily imagine a benzylic C-O oxygen bind to the Pd surface which holds the site for hydrogenation close the the surface until a hydride equivalent can diffuse over to react (hydrides on surfaces are quite mobile and diffusion is often quite rapid, of course, hydride diffusion rates are also strongly determined by surface structure and material properties (because almost nothing can be simple)). On the other hand, $\pi$-system can interact strongly with metal surface (especially of soft (again in the HSAB sense) metal). So, for Pt, the aromatic system is brought close to the surface, allowing hydrides to migrate over to it.

So we can hand-wavily discuss why Pd is more likely to hydrogenate benzylic C-O bonds and Pt is more likely to hydrogenate aromatic systems as we did above, but the nitty-gritty detail is quite complicated. This is a massive area of research. Inorganic chemistry is its own field not because everything is so similar, but because metals have hugely varying behavior. The fact that these are "all being transition metals" should NOT imply that their reactivity is similar in the slightest. Being in the same period as a transition metal tells you as much about being in the same period in the main group. C and O and N have totally different chemistry, as do Rh and Ru and Pd.

P.S. f orbitals do not figure into this.


Calculating Commutator of Differential Angular Momentum I have tried for hours to calculate the commutator of angular momentum in the differential form, but I cannot get the correct answer. This is my first experience with actually checking if two operators commutes, so there may be some beginner's misunderstandings that causes the problem.

I know from my textbook the correct answer is:

$$
[\hat{L}_x,\hat{L}_y] = \frac{\hbar}{\mathrm{i}}\hat{L}_z = \frac{\hbar}{\mathrm{i}} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
$$

The angular momenta are defined as

$$
\hat{L}_x = y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \\
\hat{L}_y = z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \\
\hat{L}_z = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}
$$

Sets up the commutator expression, and writes it out:

$$
[\hat{L}_x,\hat{L}_y] = \left[ y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} , z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \right]
$$

From what I assume is a well-known algebraic rule for handling commutators, I get the following:

$$
\left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right] - \left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right]
- \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right]
$$

It was straight forward to find that the two middle terms commute, i.e. they equal zero. Hence, we are left with:

$$
\left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right] \tag{1}
$$

Upon evaluating these terms, I let them act on a function $ \psi $. Looking at the first term:
First I factored the complex fraction out. Then I wrote out the commutator and let each differential fraction act on $ \psi $:

$$
\frac{\hbar}{\mathrm{i}} \left(y \frac{\partial \psi z}{\partial z} \frac{\partial \psi}{\partial x} - yz  \frac{\partial \psi }{\partial x}  \frac{\partial \psi}{\partial z}     \right)
$$

The y in the second term above can be factored out from the differentials, because the differential is not done with respect to y. In the first term, however, the z must be differentiated along with psi by using the product rule. Doing this, and cancelling equal terms of opposite sign, yields:

$$
\frac{\hbar}{\mathrm{i}} y \psi \frac{\partial \psi}{\partial x}
$$

Repeating this calculation for the second term in equation $(1)$ yields

$$
-\frac{\hbar}{\mathrm{i}} x \psi \frac{\partial \psi}{\partial y}
$$

Substituting into $(1)$ yields

$$
[\hat{L}_x,\hat{L}_y] \psi = \frac{\hbar}{\mathrm{i}} \left(  y \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial y}  \right) \psi
$$

Now, comparing with the correct answer:

i) My signs are opposite of what they should be

ii) More importantly, I have a $ \psi $ in the differentials which should not be there. My question is, "why, and how do I rid my answer of it?".

I hope I have sufficiently showed what I have done. I have checked my notes several times, and I cannot get rid of those extra psi's in my answer.

Your problem with the sign stems from two things: Firstly, you use a wrong equation for the commutator $[\hat{L}_x,\hat{L}_y]$ - your prefactor is wrong by a minus sign: The correct equation (see here) is

\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \hat{L}_z
\end{equation}

Secondly, your equations for the angular momentum operators are slightly wrong because you leave out the prefactor $\frac{\hbar}{i}$ in them, although you use the correct versions in your equation [1].
So, substituting the differential operator expression for $\hat{L}_z$

\begin{equation}
 \hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}

into the commutator equation you should get:

\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) = \hbar^{2} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}

As for your problem with the additional $\psi$:
Everything is fine until you let equation [1] act on a wavefunction $\psi$.

\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \left( \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \right) \psi
\end{equation}

You seem to make the following mistake which then leads to the appearance of an additional $\psi$ in your equations:
When you let the commutator $ [\hat{A},\hat{B}]$ act on a wavefunction $\psi$ you did the following

\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \psi \hat{B} \psi - \hat{B}\psi \hat{A}\psi \qquad \qquad \text{(wrong)}
\end{equation}

which is incorrect. The correct way to do this is

\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A}\psi \qquad \qquad \text{(correct)}
\end{equation}

where $\hat{A} \hat{B} \psi$ means that first $\hat{B}$ acts on $\psi$ and only then $\hat{A}$ acts on the result.
So, in the case at hand this leads to

\begin{align}
  \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right] \psi &= \underbrace{\left(\frac{\hbar}{i}\right)^{2}}_{= \, -\hbar^{2}} \Biggl( \underbrace{ y \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial x} \biggr)}_{= \, y \frac{\partial \psi}{\partial x} + yz \frac{\partial^{2} \psi}{\partial z \partial x}} - z y \frac{\partial^{2} \psi}{\partial x \partial z} \Biggr) \\
  &= -\hbar^{2} y \frac{\partial \psi}{\partial x}
\end{align}

and

\begin{align}
  \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \psi &= -\hbar^{2} \Biggl( z x \frac{\partial^{2} \psi}{\partial y \partial z} - \underbrace{ x \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial y} \biggr)}_{= \, x \frac{\partial \psi}{\partial y} + xz \frac{\partial^{2} \psi}{\partial z \partial y}} \Biggr) \\
  &= \hbar^{2} x \frac{\partial \psi}{\partial y}
\end{align}

If you substitute these equations into the commutator equation you get

\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \hbar^{2} \left(x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right)
\end{equation}

which is exactly the result you should expect.


Redox reaction: Iron(III) and Iodide ion I'm a real beginner in chemistry and I'm currently reading about redox reactions. I'm quite confused about this simple redox reaction:

  ${\text{Iron(III) ion + Iodide ion} \rightleftharpoons \text {Iron(II) ion + Iodine}}$
  \begin{array}{|c|c|c|c|c|c|}
\hline
\text{Oxidation numbers}& +3 & -1 &\rightleftharpoons & -1 & 0\\ \hline
 & Fe^{3+}& I^-&\rightleftharpoons & Fe^{2+}&I_2\\ \hline
\end{array}

I understand the principles of transferring electrons. But how would I come from the upper formula to the result that the Iodine exists two times on the right side? After reading the reaction formula I would have written the Iodine just once (not as seen here twice).

Iodine exists as a diatomic molecule. This reaction indicates how the iodide ion is converted into diiodine. Your reaction just indicates the electron transfer: you still need to balance the reaction.
You need an $\ce{I_2}$ molecule on the right hand side, which would need two iodide ions on the left hand side:
$$\ce{2 I- -> I_2 + 2e-}$$
To accept two electrons, we would need two $\ce{Fe^3+}$ ions:
$$\ce{2 Fe^3+ + 2 e- -> 2 Fe^2+}$$
Now we have a balanced reaction. All the electrons donated by iodine are accepted by iron. Adding up the two reactions and cancelling the electrons gives us:
$$\ce{2 Fe^3+ + 2I- -> 2 Fe^2+ + I_2}$$


What is the absolute configuration of carbon 4 in glucopyranose? My colleague and I have been wrestling with the assignment of chirality to the 4th carbon atom in glucopyranose. In the linear form of glucose, the 4th carbon is definitely (R); therefore it will retain that chirality as it forms the hemiacetal. However, looking at the ring, we are having a hard time figuring out why the 3rd carbon gets priority over the 5th by CIP priority rules. Both the 3rd and 5th carbons are attached to $\ce{C,H,O}$. The question is where do you go after that to assign priority?

therefore it will retain that chirality as it forms the hemiacetal.

This statement is incorrect. Many chiral centres retain their CIP-descriptor upon reactions that don’t influence them directly, many others do not.

How do you assign CIP priorities? You basically had the same problem in the linear form of D-glucose. You move along the chain in order of priorities, finding the first priority difference and using that to assign the higher-priority side. On C4, you have (read this from left to right, i.e. everything with the same indention at the same time):

an oxygen (a)
the carbon C3

an oxygen (a)

a hydrogen.

the carbon C2 (b)

an oxygen (a)
a carbon C1 (b) <-- this one is higher priority than on the other side!
a hydrogen (c)

a hydrogen (c)

the carbon C5

an oxygen (a)

a hydrogen

the carbon C6 (b)

an oxygen (a)
a hydrogen <-- less priority than carbon
another hydrogen.

a hydrogen (c)

a hydrogen (d)

Following the same method in the pyranose form we get:

an oxygen (a)
the carbon C5

an oxygen (a)

a carbon <-- this one is higher priority than the other side!

the carbon C6 (b)
a hydrogen (c)

the carbon C3

an oxygen (a)

a hydrogen <-- this is lesser than carbon!

the carbon C2 (b)
a hydrogen (c)

a hydrogen (d)

Therefore, the absolute configuration of said atom changes from (R) to (S).

This is actually a pretty common thing that can happen in highly functionalised molecules with very similarly modified carbons. Because we cannot distinguish at any proximate position we need to go far down the chain and a single distal modification can change a lot.

Other examples that would have changed the absolute configuration of carbon C4 without changing its geometry would be:

methylation at C5
silyl protection at C5
oxidation at C5

All of these would make C5’s priority higher than C3’s.


Comparing formula for enthalpy change with bond dissociation energy and formation enthalpy I learnt that given a reaction: $$\ce{A -> B}$$ the enthalpy change is given by:

$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$

By convention, we keep the minus sign in the equation, and keep all bond enthalpies positive, regardless of whether bonds are made or broken. In this case, the bonds in the reactants are broken, and the bonds in the products are formed. I was also given the equation,

$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$

Are both $\Delta H$'s given by the above formulas equivalent?

Yes. The two methods are equivalent.

You can try this with the dimerization of $\ce{NO_2}$ into $\ce{N_2O_4}$. In this reaction we form one $\ce{N-N}$ bond. Every other bond stays the same.

$\ce{2NO_2 <=> N_2O_4}$

If we calculate the enthalpy of the reaction through bond energies, we get the negative of the $\ce{N-N}$ bond energy. Why? Because we only formed one bond in this entire reaction. Every other bond stays the same - we start with four resonance stabilized $\ce{N-O}$ bonds and end with four resonance stabilized $\ce{N-O}$ bonds. The net change in bonds is the formation of a $\ce{N-N}$ bond. Here is a picture of the process (note that this picture doesn't show any resonance stabilization but there is; the "pi" bonding electrons are delocalized despite what the picture implies).

Bond energy is a positive value; the negative of this value implies that the dimerization reaction is exothermic - which is obviously true! We are going from a radical species - nitrogen dioxide - into a species with no unpaired electrons. Radicals are reactive!

We can also calculate this using heats of formation for both nitrogen tetraoxide and nitrogen dioxide and we'd get the same enthalpy of reaction.


What does β stand for in β-potassium zirconate? What do the Greek letters (α, β, γ...) convey when they are prefixed to a crystalline substance?

For example: β-potassium zirconate

They signifies one polymorph (the beta polymorph) of barium zirconate.

Allotropy is the property of some chemical elements or compounds to exist in two or more different forms, in the same physical state at different temperature. Now, technically they are the same element or compounds but have different crystal structure at changing temperature. So, in order to name them, alpha, beta, gamma etc. are used in the prefixes to depict them as they are the allotropes formed at different temperature phases.. That helps in their differentiation.

For e.g. iron has 5 allotrope namely:

ferrite (α-iron): the ferromagnetic form at T < 770 °C (the Curie point, TC); BCC crystal structure
β-iron: T = 770 – 912 °C; the paramagnetic form of α-iron; same structure as α
austenite (γ-iron): 912 – 1,394 °C; FCC crystal structure
δ-iron: 1,394 – 1,538 °C; forms from cooling down molten iron; same structure as α
ε-iron (also called hexaferrum): forms at high pressures; HCP crystal structure

There are several more examples of such type of nomenclature of elements. You can find them in the wikipedia article of allotrope(see table).

Also, not only elements form allotropes. Compounds also form allotropes but have a different name i.e polymorphs. Generally, binary metal oxides are polymorphs. Polymorphs have different temperature phases in which one form of crystal structure exist. These phases are depicted by giving greek alphabet at the prefixes like α-phase, β-phase etc. Polymorphs or allotropes of compounds are given that greek alphabet prefixes according to that phase in which one form of that compound exist.

For e.g. iron oxide $\ce{Fe2O3}$ has 4 phases.

α-phase: at ambient conditions: has Corundum-type Rhombohedral (R3c) like structure
β-phase: below 773 K: has Body centered cubic (Ia3) like structure
γ-phase: up to 933 K: has Cubic, spinel structure (Fd3m) like structure
ε-phase: has Rhombic (Pna21) like structure

So, iron oxide $\ce{Fe2O3}$ has 4 polymorphs : $\ce{\alpha-Fe2O3, \beta-Fe2O3, \gamma-Fe2O3, \epsilon-Fe2O3}$.

For more information you can check this question(asked by myself) or this question that describes how polymorphs is formed.


Avogadro Conformer Properties with Linux Version My Windows version of Avogadro has the following menu point:

View > Properties > Conformer Properties

My vanilla Avogadro version under Ubuntu does not even have the menu. What do I need to do, to either activate the menu or to get to this option under Linux?

I can't reproduce the problem.

I'm using Ubuntu 16.04 and installed Avogadro from the repository. The version is 1.1.1, just as described on Launchpad.

This version does have the menu entry in question:
Ansicht > Eigenschaften > Konformereneigenschaften

After a conformation search, it does open a table with the found conformers and their energies.


When there's a photoelectric effect, do the electrons leave altogether or leave one by one? When a strong light, or even a laser, hits a metal surface or some molecules, and causes lots of electrons to fly away, do the electrons leave together, or do some of them leave first? If it's the latter, then do the valence electrons leave first and then inner layers? If so, how long are the intervals between their leavings?

In 1905, Einstein demonstrated that the photoelectric effect was a quantum phenomenon. That is, for each photon of light of a minimal specific wavenumber, one electron in a specific orbital is released.

For example, for cesium in its base electron configuration, $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 6s^1}$, about $\pu{3.43E-19 J}$ is needed to raise that electron to, effectively, an "infinite" orbital (where it is unlikely to be found near the atom). This corresponds to a wavelength of $\pu{\approx 580 nm}$ or wavenumber of $\approx \pu{17200 cm-1}$.

That said, very energetic photons may knock out more than one electron apiece, with some of the photon's surplus energy applied to the kinetic energy of the first electron, and some applied to ionizing the atom further. However, at the minimum photon level, for every photon absorbed, one electron is freed. A brighter light has more photons, so a brighter light releases more electrons (with limitations due to the build-up of free electrons near the $\ce{Cs}$ cathode, the space charge).


On the meaning of distinguishability, and wavefunctions for 3 electron atoms In a 2-electron atom at lowest energy, the $(1s)^2$ is occupied and the electronic wave-function must satisfy anti-symmetry requirements in the particle coordinates, as the spatial wave function is symmetric. How is the situation in a 3 electron system?

In a 3-electron atom (or in a nucleon with one excited quark) of lowest energy, say the $(1s)^2(1p)$ states are occupied, is the fermion in the $(1p)$ state distinguishable? I.e. must my wave-function still treat all 3 electrons indistinguishably, or are mixed symmetry wave-functions now physically viable?

The only case in which electrons can be considered distinguishable is when they are so well separated that their wave functions do not overlap. Any electrons on the same atom or within the same molecule are considered indistinguishable.

In chemistry we like to talk about electrons being in specific orbitals, but the indisinguishable nature of them means that they are all in all orbitals simultaneously. This is why a mathematically rigorous description of a molecule using hybrid orbitals is just as valid as one using canonical orbitals. Neither is true in the sense of describing where each electron is, and both are true in the sense of describing the collective electron density.


Why does a mixture of siloxene and cerium(IV) sulfate luminesce? I performed an experiment where siloxene and cerium(IV) sulphate were mixed together: when I did so  the mixture produced an orange-yellow glow. Why does it glow? What is it about the two chemicals in particular that makes this happen?

Method for synthesising siloxene:

Carefully measure out 50 cm3 of concentrated hydrochloric acid into a 100 cm3 measuring cylinder. Put the acid in a 1 dm3 beaker containing a magnetic stirrer bar and place the beaker on a heater/magnetic stirrer.
Switch on the stirrer and slowly add 5 g of calcium silicide, one spatula measure at a time, to the stirred acid. Continue stirring the mixture until the reaction subsides.
Measure out a further 25 cm3 of concentrated hydrochloric acid and carefully add this to the stirred mixture. Now turn the heater on and heat the mixture to boiling. Boil the mixture for five minutes. After boiling, slowly add 150 cm3 of distilled water then boil again for approximately five minutes.
Switch off the heater and leave the mixture stirring while it cools.
Set up a Buchner funnel and flask in a fume cupboard.
Carefully pour the yellow siloxene that you have made onto a filter paper in a Buchner funnel (try to leave the unreacted grey/black solid calcium silicide in the beaker). Now wash your sample of siloxene – first with 10 cm3 of ethanol and then with 10 cm3 of ethoxyethane. Drawing air through the sample will help to dry the yellow, solid siloxene.

Method for producing the glow:

Place 0.8 g of siloxene and 75 cm3 of 1M sulfuric acid in a 250 cm3 conical flask. In a second 250 cm3 conical flask, dissolve the 0.5 g of cerium(IV) sulphate in 75 cm3 of 1M sulphuric acid. Label and stopper both flasks.
Go into a dark room. Turn out the light, add the cerium(IV) sulphate to the siloxene and swirl the mixture. You should see a fairly bright orange-yellow glow.

The idea of chemiluminescence is pretty simple, you need an oxidizer and a substance that actually gives a glow.

According to Wikipedia cerium sulfate is rather an oxidizer especially under acidic condition, which actually explains why we need sulfuric acid, so basically, whatever is siloxene (I suspect it is a silicon oxide kind of structure) it is an oxidant that when oxidizes the difference in energy level causes a glow in the visible spectrum.

Unless more clarification about siloxene, this could be the only possible answer.

sources:

NurRage explanation about chemiluminescence
Wikipedia for properties about cerium sulfate

EDIT: what was meant by potential sources is that the method of using words was mine but the information is from the sources sorry for inconvenience


What is the difference between these equations for the Contact process? Some webpages describe the contact process as this (I left out $\ce{SO2}$, $\ce{SO3}$ production):

$\ce{SO3 + H2SO4 -> H2S2O7}$

After this, it is controllable to add water to the oleum therefore making liquid sulfuric acid. However, some other sites describe this:

$\ce{SO3 + H2O  -> H2SO4}$

I am confused about what really happens in the absorption tower. I've read about the wet acid, but it states that adding water causes gaseous $\ce{H2SO4}$ to form, which is condensed.

EDIT: In the absorption tower, trioxide sulfur enters as gas, and sulfuric acid comes down as liquid. As a result the liquid comes out as oleum (first eq.). The other contact process states different: the liquid entering is 98% sulfuric acid, and 2% water, this water reacts with the sulfur trioxide (this little water avoids the heat problem), giving strengthened sulfuric acid (let's say 99.5% sulfuric acid, and 0.5% water). Still confused.

In principle, your sources are correct and not contradictory.

In order to avoid the formation of problematic sulfuric acid mist, sufur trioxide is not directly absorbed into water. Instead, it is absorbed in sulfuric acid of 98 %–99 % concentration, which approximately corresponds to an azeotrope where the partial pressures of sufur trioxide, sulfuric acid, and water are relatively low.

Sulfuric acid reacts with the absorbed sulfur trioxide to form disulfuric acid:

$$\ce{H2SO4 + SO3 <=> H2S2O7}$$

The disulfuric acid can be converted to sulfuric acid by reaction with added water:

$$\ce{H2S2O7 + H2O <=> 2 H2SO4}$$

Thus, the overall net reaction is as follows:

$$\begin{align}
\ce{H2SO4 + SO3 \;&<=> H2S2O7}\\
\ce{H2S2O7 + H2O \;&<=> 2 H2SO4}\\
\hline\ce{SO3 + H2O \;&<=> H2SO4}
\end{align}$$


Hydride shift in carbocation I was practising some problems on hydride shift and got stuck in this particular question. I wanted to know if hydride shift would take place in this particular case and which carbon would get the positive charge. Will a hydrogen from left or right carbon would shift??

I strongly suspect that a hydride shift is not observed in this example.

The reason is that it seems much more likely that you will have an alkyl shift instead, accompanied by ring contraction and formation of a carbonyl. This is essentially a pinacol rearrangement.

EDIT:
Adding more details per @OscarLanzi comment...

I would think that this is the expected product after proton transfer.

It looks like the pinacol arrangement of 4,8-decalindiol has been known for quite some time. Hückel, Danneel, Schwartz, Gercke, Ann. 1929, 474, 121. and Meiser. Chem. Ber. 1899, 32, 2055.


Why do esters actually smell? We all know the very famous esterification reaction, where a carboxylic acid and an alcohol react to give an ester and water. But why do they actually smell? What is the reason behind its aromatic character?

I initially thought that it must be due to the conjugation between the two oxygens present. I do know that aromaticity is decided by the conjugated lone pairs present in the compound where the compound must follow Hückels rule. But it's only valid for a cyclic and a planar compound.

Why do esters have that fruity smell? What is the chemical explanation for a smelly non planar compound?

Dissenter is correct about aromatic compounds and that the basic requirement for a compound to be detected by smell is for it to be volatile enough to reach the nose. This is why things like dry salt don't really smell like anything. Once they reach the nose, things become more complex.

Why many esters smell fruity while most thiols smell rather unpleasant is a rather interesting question. Just as most sugars taste sweet, many molecules that have structural similarities also smell similar. Humans have about 350 different olfactory receptors. These are G protein-coupled receptors that change conformation when they bind a compound and activate a signal pathway that is eventually sent to the brain for processing. However, since we only have 350 different receptors and there are far more than 350 odoriferous compounds, there is not a one-to-one mapping of receptors to compounds—there isn't an "acetone receptor". Instead, olfactory receptors have partial and overlapping selectivity. Each receptor type responds to more than one compound, thiols with a certain general shape, for instance, and different receptor types can respond to some of the same compounds. A given compound then has a more or less unique pattern of receptor activation. All this information gets sent to the brain and is processed into the perception of smell. Compounds that have similar structural properties activate many of the same receptors, so they smell similar.


Could a lead-acid battery electrolyte be replaced by hydrochloric or nitric acid? Hydrochloric acid, as well as nitric acid, are also strong acids like sulfuric acid. So, why are not they used commercially in lead-acid batteries?

In the lead battery it is very important that the electrodes transform into insoluble lead sulfate when discharged, especially at the cathode where lead(IV) oxide is turning into lead(II) sulfate. Consider the two half reactions of the lead-acid battery:

$$\ce{Pb(s) + HSO4−(aq) → \color{red}{PbSO4(s)} + H+ (aq) + 2e−}$$
$$\ce{\color{red}{PbO2(s)} + HSO4−(aq) + 3H+ (aq) + 2e− → \color{red}{PbSO4(s)} + 2H2O(l)}$$

Lead nitrate is soluble, so metal ions would go into solution. The lead(IV) oxide would not form again from these ions on charging, metallic lead from the electrode would dissolve. The battery would no longer be rechargeable.

Lead chloride is insoluble, but lead(IV) oxide on the cathode would oxidize chloride into chlorine. Thus no more battery.


Name of this bond-line formula? I don’t know much about chemistry but I recently found the below bond-line structural formula painted on a wall near my place. I really can’t let go of trying to find the name of the compound. If anyone would be so inclined as to relieve me of my suffering, I sure would be grateful!
Thank you and greetings from Sweden!

The full name of the compound is 3,4-methylenedioxymethamphetamine, which is typically abbreviated as "MDMA". It is commonly used recreationally, and is casually referred to as "ecstasy", "molly", "E" and many other names.


Can dissolved salts (or any soluble chemicals, for that matter) 'bump' each other out of solubility? Will a more soluble substance dissociate, causing a less soluble substance to precipitate out?

in equation:

AB(s)+ C(aq)+ D(aq)--> A(aq)+ B(aq)+ CD(s)

Does that equation go through, under the assumptions that AB, as a whole, is more soluble than CD, and saturation point is achieved with both substances?

Thanks in advance! This question has been bothering me for a while...

Yes, this effect is termed salting out for non-ionic compounds and the common ion effect for ionic ones.

Note that in specific cases, the converse may hold true. For example, "the solubility of silver chloride in water is reduced if a solution of sodium chloride is added to a suspension of silver chloride in water (op cit)." However, adding more chloride re-dissolves the precipitate due to the formation of complex ions:

$\ce{AgCl(s) + Cl- (aq) -> AgCl2- (aq)}$


What's a good measure of the Eigen-ness of a wavefunction? Just wondering if someone can recommend an easy-to-compute measure for the Eigen-ness of a wave function, such as the wave functions computed using ab initio method?

For a wave function solved using a small basis set, the Eigen-ness should be smaller than that of wave function computed using larger basis set.

Naturally for arbitrary wavefunction $|\psi>$ one can compute :
$<\psi | (H/E-1)^2 | \psi>$,
where E is the expectation value, which is the dispersion of the energy. This, however, is too difficult to implement.

$\hat{F}\psi_i=\varepsilon_i \psi_i \implies \langle\phi_k|\hat{F}|\psi_i \rangle = \varepsilon_i\langle\phi_k|\psi_i \rangle$.  So the Fock matrix $F_{ki}$ can be approximated by the product of orbital energy $\varepsilon_i$ and the overlap matrix $S_{ki}$. The closer these two values are, the higher the "Eigen-ness".

Of course $\phi_k$ and $\psi_i$ have to be non-orthogonal, so $\phi_k$ could be, for example, a basis function or an MO for another molecule.  I've used this to approximate intermolecular Fock matrix elements here.


Is there a way to get hot water without altering plumbing or electrical wiring too much? Is there a way (other than collecting water and heating it on the stove ) to get hot/warm water from the tap without having to alter existing plumbing or electrical wiring ?

Perhaps use a chemical reaction to provide the heat to warm the water ?

Crystallization of a supersaturated solution of sodium acetate will release heat - once. But then it is just a solid. It works nicely for pocket warmers, but since you have to heat the flexible container in boiling water prior to re-use, this and other comparable processes are of little use for a water supply.

At the end of the day, you won't get around plumbing to tap an energy source.

Your options would be:

solarthermal
geothermal
heat released from the fermentation of organic waste


How to determine enzyme inhibitor efficacy? At what percentage reduction or increase in Km and Vmax is an enzyme inhibitor regarded as effective?- is any percentage increase or reduction regarded as being a clinically useful level of inhibition, or would it have to be over a certain limit?

If the inhibitor is a competitive inhibitor, to quantify inhibition the $K_m$  of the inhibitor ($K_I$) is determined.

$K_I$ is compared to the $K_m$ of the substrate.

Then, the degree of inhibition is a function the inhibitor and substrate concentrations.

$V_{max}$ is unaffected by a competitive inhibitor.

In the presence of an inhibitor the kinetics are modified such that the enzyme has an effect $K_m$ as:

$K_m(1 + [I]/K_I)$


Naming substituted bicyclo compounds I have to name the above compound. I am getting 5-bromo-2-chloro-7-fluorobicyclo[2.2.2]octane.
But my teacher is telling me that the answer should be 2-bromo-5-chloro-8-fluorobicyclo[2.2.2]octane.
His reasoning is that the bromo substituent should get a lower number as it comes first in alphabetical order. Which is correct?

You have already correctly identified the parent structure of the compound as bicyclo[2.2.2]octane.
The most important simplified criteria for the numbering for such structures are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have any principal characteristic group that is expressed as a suffix. Thus, Rule (c) is not relevant in this case.
The compound also doesn’t have any multiple bonds. Thus, also Rule (e) is not relevant in this case.
Next, low locants are assigned to the substituents that are expressed as prefixes according to Rule (f). Note that all substituents are considered together in a series of increasing numerical order. Therefore, the correct name is 5-bromo-2-chloro-7-fluorobicyclo[2.2.2]octane rather than ‘2-bromo-5-chloro-8-fluorobicyclo[2.2.2]octane’ since the locant set ‘2,5,7’ is lower than ‘2,5,8’.

Only if there is still a choice, a low locant would be given first to the bromo substituent in accordance with Rule (g).


Time required for a thin layer of ethanol to evaporate? If someone were to pour about 5 ml of 94% pure ethanol on a flat surface and at room temperature, approximately how much time would it take for it to evaporate?

EDIT: In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?

It be simpler and faster to do the experiment than to try to predict the time needed.

The kinetic molecular theory explains why liquids evaporate at temperatures lower than their boiling points. At any temperature, the molecules in a liquid have a range of kinetic energies described by a Boltzmann distribution. Some percentage of the molecules have enough kinetic energy to escape into the gaseous phase. These molecules contribute to the vapor pressure of the liquid. As the temperature increases, more molecules are above the gas threshold and the vapor pressure increases. As the vapor pressure increases.

In a closed system, equilibrium would be established with an unchanging ratio of liquid to vapor. Molecules would be exchanged between the liquid and the vapor, but their relative amounts would remain constant. The surface you describe is not closed - the vapor molecules can wander away by diffusion or convection. Equilibrium is disrupted, and Le Châtelier's Principle tells us that the equilibrium shifts to compensate. As more ethanol molecules escape, more vaporize to replace them until there are no molecules remaining in the liquid. In the following equation, $K$ is the equilibrium constant, $p$ is the partial pressure of ethanol vapor (the vapor pressure) and [$\ce{C2H6O}$] is the concentration of ethanol in the liquid.

$$\ce{C2H6O (l) <=> C2H6O (g)} $$
$$K=\frac{p_{\ce{C6H6O}}}{[\ce{C2H6O}]}$$

After establishment of equilibrium (fast), the rate-determining step of the evaporation is probably the diffusion of the gaseous ethanol molecules away. The average kinetic energy of a gas particle can be expressed as a function of mass ($m$, in kg) and root mean square velocity ($v^2_\text{rms}$) and separately as a function of temperature ($T$, in kelvin) times the Boltzmann constant ($k_\text{B} = 1.38 \times 10^{-23} \frac{\text{J}}{\text{K}}$). We can derive a formula for rms velocity.

$$\overline{E_\text{k}}=\frac{1}{2}mv^2_\text{rms}$$
$$\overline{E_\text{k}}=\frac{3}{2}k_\text{B} T$$
$$v^2_\text{rms}=\frac{3k_\text{B} T}{m}$$
$$v_\text{rms} = \sqrt{\frac{3k_\text{B} T}{m}}$$

You could calculate the velocity of ethanol particles escaping. If you set an arbitrary distance (one meter is probably fine), then you can calculate a time it takes one particle to travel that distance (on average). If we know the equilibrium constant, we can determine how much vapor is above the liquid and then calculate the amount of time it takes to move.

But how do we know the equilibrium constant? It varies with temperature! The value $\Delta G^\circ_\text{vap}$ in the equation below is the free energy change of vaporization of ethanol in the standard thermodynamic state. $R$ is the ideal gas constant.

$$K=\mathrm{e}^{-\frac{\Delta G^\circ_\text{vap}}{RT}}$$

The model above ignores complicating factors like mean free path, the fact that evaporation is endothermic (meaning the liquid cools as it evaporates and the vapor pressure decreases with time), the temperature and heat capacity of the surface govern how much kinetic energy is available to the liquid to begin with, and that any amount of air current in the vicinity will move the vapor away significantly faster than diffusion.

A complete model includes consideration of the following:

  Constants

the Boltzmann constant
The ideal gas constant
the free energy change of vaporization of ethanol (not actually constant, but it varies only slightly over the temperature range)
the vapor pressure of ethanol as a function of temperature
the mass of an ethanol molecule
the heat capacity of the surface

  variables

temperature of the air
volume of ethanol
temperature of the surface
atmospheric pressure (needed for mean free path corrections)
velocity of air currents

So, in principle, you could do it. Even so, the best answer cannot be had without some serious calculus. In practice, it would be faster to do the experiment (if often is). We often forget that science is empirical.


Does rearrangement take place in the reaction between ethenylcyclohexane and iodine chloride? I am stuck at the point if a non classical carbocation should be formed on the 2nd Carbon of the side chain.
One of my friend gave me the following suggestion:
I guess $\ce{ICl}$ will break in the form $\ce{I+/Cl-}$ so considering this if $\ce{I+}$ gets attached to the lower side of the chain then the carbocation formed on the 2nd element of the chain will not be a non-classical carbocation (because $\ce{I+}$ won't give a lone pair to the carbocation as it is electron deficient itself). A simple carbocation will be formed, and the rearrangement will be possible.
After rearranging it will get its most stable position on the carbon joining the ring and the chain so there the $\ce{Cl-}$ attacks and the requires molecule is formed.
I suspect this is wrong, What is the solution to this problem?

Yes, I believe the rearrangement can happen. However, the product from the rearrangement is probably not the major one.
The closest reaction to what you have drawn that I could find on SciFinder is this:

The paper is mentioned below in the reference. In the paper, the researchers performed similar reactions of alkenes with $\ce{I-Cl}$ and in cases where there is a possibility of a rearrangement to a tertiary carbocation, the by-product similar to $\text{(C)}$ is generated (for example in case of the cyclohexane ring being replaced with iso-propyl, the percentages are $\text{(A)=55%; (B)=2%; (C)=38%}$).
For the reactions drawn above, the researchers did not mention the A:B ratio, however, in most cases A is the major product, which is expected (Markovnikov's rule). Nevertheless, the product formed from the rearrangement is not the major one in any of the reactions.
The mechanism is not mentioned in the paper, but my guess is that it will not be much different from the usual carbocation rearrangement mechanism. Only in this case, the hydride shift will break the iodonium ring, while creating a tertiary carbocation. The Cl can then attack the carbocation and form the product.
This is my proposed mechanism:

To reiterate, this is likely not the major pathway. The usual addition to the double bond will be the major pathway and form the major product.
Reference:
T. Horibe, Y. Tsuji, and K. Ishihara, "Thiourea–I2 as Lewis Base–Lewis Acid Cooperative Catalysts for Iodochlorination of Alkene with In Situ-Generated I–Cl", ACS Catalysis, 2018, 8(7), 6362–6366


Cannizaro Reaction, Hydride Shift Reason? In Cannizaro reaction when concentration of OH- is less in the rds step hydride shift occurs, though it is very difficult for hydride ion to shift yet the reaction proceeds but what is the driving force here??

My teacher told me that this mechanism was proposed because the hydride shift occurs but there must be some reason as to why the hydride shift?

Hydride shift occurs because in the intermediate dianion there is no other suitable leaving group. It is either $\ce{H-}$, $\ce{R-}$ or $\ce{O^{2-}}$. Of those $\ce{H-}$ is the best leaving group. Note though that even under these conditions $\ce{H-}$ will of course not be solvated on its own (it is way too unstable for that). It is directly transferred from the dianion to the aldehyde.

This reaction will go "to the right" since the hydride shift is irreversible. There is simply no mechanism to form a dianion from the alcohol and carboxylic acid.


Do there exist aromatic compounds similar to azulene? So for this question I'm considering a chemical to be "similar to azulene" if it obeys the following conditions:

It is aromatic and obeys Hückel's rule,
It is a fused bicyclic compound,
One component ring is aromatic as an anion,
The other ring is aromatic as a cation,
It is not azulene, and
It has been synthesised before and the synthesis has been confirmed.

For instance, I believe bicyclo[3.1.0]hexa-1,3,5-triene should follow all of the previous conditions except possibly the last (but I don't know that it hasn't). Do any chemicals obey all of these conditions?

There is this compound which is I believe the closest to anything which you've asked for.
Except the fused part this satisfies all your conditions.
In a polar solvent the double bond breaks to form two highly stable aromatic rings(tropylium and dicyclopentene) which act as two separate poles. It also lowers the rotation barrier of this compound. Hence it even becomes soluble. Along with azulene this is as close as an organic compound gets to being ionic.


Why would Auger electron detection have potential for higher spatial resolution than EDX I understand that Auger electrons are emitted when an electron falls from a higher energy level to a open inner shell and EDX uses electrons that fall from outer shells when an inner shell is kicked off. I also have read that AED has a higher spatial resolution but I don't understand WHY this spatial resolution is higher. What is different about the Auger electrons that increases resolution?

I think this is a good question. The problem is more fundamental. In both cases, you would be using electron beams on the sample.
In EDX, you are shining a beam of electrons and detecting the X-rays. Here comes the main problem. It is hard to focus the X-rays. Mea culpa here. As Jon commented it is the depth of penetration of the electron beam and the corresponding X-ray emission that main affects spatial resolution. This picture roughly tells the story:

Auger electron electron spectroscopy, you are focusing a beam of electrons and you detect electrons. It is relatively easy to focus electrons using "lenses". Hence you can point point the exact origin where the electrons are being ejected.
Also, think about the depth profiling. EDX is not a surface technique unlike Auger. I used to have this misconception until I attended a workshop on surface analysis. The electrons can easily penetrate the sample and similarly originate from the some depth as well.


Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol I'm an electrical engineering and I hope you can give me some advice as I'm not much into chemistry.
I'm trying to start making my PCB (Printed Circuit Board) at home but I'd like to have some advice concerning how to handle and where to store these (potentially dangerous) chemicals.

I need NaOH as a developer, hydrochloric acid + hydrogen peroxide as etchant (they should be a better alternative than Ferric Chloride) and isopropyl alcohol (to make soldering flux).

As far as I understood I need:

Glasses (to protect against vapors and splashes)
Gloves (to protect when mixing the chemicals)
Some kind of protective clothes (not sure if I need to, but obviously I'm not wearing nice clothes when doing etching)

I have a few questions though:

Do I need kind of mask / filter (to put on my nose) - are these vapors dangerous ?
I read gloves should be stored in quite cold locations and not be exposed to sunlight. Does that apply to the chemicals listed above too ? I'd like to store them outside but maybe it's not safe ...
Hydrochloric acid comes in 36% concentration but I think I'll need to dilute it to 1%-5% max. After it has been diluted it should be quite safe, isn't it ?
Is it safe to store them inside a house (as long as it's well ventilated) ?

You probably don't need a mask - vapours from hydrochloric acid can be a little noxious, but if you're diluting it down and using it in a reasonably ventilated area, you shouldn't get much of that. Isopropyl alcohol vapour is toxic like most alcohols, but I don't think an ordinary mask will prevent that, and you'll be better off just making sure you have enough ventilation.

Standard white cotton labcoats are inexpensive and can be quickly removed if you spill a nontrivial quantity of something on them.

Hydrogen peroxide will probably come in an opaque container, but if not, it does break down in sunlight. Everything else is pretty stable to light, although you want to avoid storing the alcohol somewhere that's going to get too warm, because it evaporates readily and can pressurise the container more than is a good idea.

All these materials are pretty safe to store indoors in unspecialised storage, although for commercial purposes or quantities you may need to take regulatory precautions that will obviously vary by locale.

Edited to add: The copper chloride that comes out should be regenerated as many times as possible, but when you eventually end up with too much of it, you'll want to already have thought about what you're going to do with it - it's pretty toxic and you can't (or, well, shouldn't) just wash it down the sink.


Why are reactions with higher activation energy more sensitive to temperature? In the above picture it is stated clearly, with an example, that the more the Ea of a rxn the more it's sensitive to temperature, and hence even a small change in temperature in a reaction with high Ea, would have a drastic effect on the rate of rxn, as compared to a rxn with low Ea (about the probability factor and collision frequency, it's stated in the book as - "it must be understood that we are justified in doing this only when the reaction being compared are so closely related that differences in collision frequency and probability factor are comparatively insignificant" ).

My doubt is - If the energy-barrier (Ea) is high, doesn't it mean, (kind of) that, it's difficult to reach the intermediate stage, and hence a little change in temperature doesn't do good to the reaction as the barrier has still not been overcome and the reactants have a long way to go; and hence making the reaction more "temperature resistant" ? And on the other hand, if the energy-barrier is low, doesn't it mean that even a small change in temperature would be enough to overcome it and hence would increase the rate of reaction ?

After reading a bit more, I found that the statement is repeated again as - "the larger the activation energy of a reaction the larger the increase in the rate brought about by a given rise in temperature".    WHY ?

My appeal to the answerers is that - Can you plz answer in what-is-actually-happening-on-the-backstage way and without ,if necessary, equations ?

Thank you very much and sorry for poor English.

The first thing to note is that the rate constant with a large activation barrier will be far smaller than one with a small barrier at a given temperature. If the barrier is very small (e.g. 5 kJ/mol) at, say, room temperature, almost all collisions between the reactive molecules have sufficient energy to react and the barrier is crossed and this leads to a fast reaction. This means that increasing the temperature cannot lead to much of an increase, because the reaction is going almost as fast as it can at the lower temperature.

Now consider the case with a large barrier. In this case the rate constant is far smaller than when the barrier is small, and very few collisions have enough energy  to cross the barrier. Now increase the temperature and the number of filled energy levels the molecule has increased greatly, exponentially so by the Boltzmann distribution.  Now very many more collisions have sufficient energy to react and so the rate constant increases greatly, although it may still be small in value.

If you plot the Arrhenius equation (given in your text) vs temperature the slope of this gives the rate of change with temperature. With larger activation energies the slope gets larger. This is the effect also described in your text.

The figure shows rate constants calculated with the activation energies $E_a$ shown (in kJ/mol). You can see how rapidly the rate constant increases with temperature when the activation energy is large, i.e. the slope is greater when the activation energy is larger, than when the activation energy is small. (The rate constant $\displaystyle k=A_0e^{-E_a/(RT))}$ are with $A_0 =10,\; (E_a=10);\; A_0= 10^3,\;(E_a=20)$ and $A_0=10^6,\;(E_a=40)$ with $A_0$ chosen only to get them on the same scale).

The gradient of the rate constant is $\displaystyle \frac{dk}{dT}=E_a\frac{A_0e^{-E_a/(RT)}}{RT^2}$ which shows that the slope depends on the activation energy.


Is there an energy cost associated with flipping the spin of an electron? THE STORY:

A common example used to illustrate the limitations of restricted Hartree-Fock (RHF) theory is the H$_2$ dissociation energy ($D_e$) curves.  RHF enforces electrons to be paired into spin orbitals, $\chi$, or two spatial orbitals $\phi$ with the same set of spatial coordinates $\mathbf{r}$ but a different spin function (i.e. $\alpha$ or $\beta$ for spin-up $\uparrow$ and spin-down $\downarrow$, respectively).  Unrestricted Hartree-Fock theory gives each spatial orbital its own set of coordinates, allowing for $\phi_1(\alpha$) to have a different energy than $\phi_1(\beta)$, giving rise to a proper description of the H$_2$ dissociation energy (see the graph below).  I didn't include units in this graph but the energy is in kcal mol$^{-1}$.  The red curve gives the appropriate (and approximate) $D_e$.  The RHF formalism (blue curve) gives a $D_e$ that is three-times greater than the 'right answer'.

A corresponding electron configuration picture is shown below.  Here we clearly see RHF forcing the electrons to be paired at infinite distances, giving rise to an H$^-$H$^+$ description of the system which is clearly wrong.  (Note that this is a 'state' that included into the wave function by RHF).  UHF allows the electrons to reside in their own spatial orbitals giving rise to two doublets at infinite separation.

Now here is where things get interesting.  In terms of the actual computations implemented to generate the curves in Figure 1, the electrons adopt opposite spins.  However, let us imagine a theoretical scenario where two lone hydrogen atoms are floating around in space at infinite distance.  The corresponding electrons (which we will assume is not interacting with each other) can adopt any spin they want so both electrons can have a spin-up configuration.  Now imagine these two hydrogen atoms approaching each other.  For a covalent bond to form, the electrons will be paired into a spin-orbital which means that one electron must flip its spin according to the Pauli-Exclusion principle.  (We should note at this point that orbitals are merely mathematical functions whereas spin is an actual physical property).  This now leads us to...

THE QUESTION:

Is there an energy cost associated with flipping the spin of an electron such as in the scenario I've described?

The obvious follow up question (if the answer is yes to the first piece), would the H$_2$ dissociation energy curves be different for two H atoms each with spin up electrons and two H atoms with electrons of opposite spin (and how different would they be)?

I will try to describe what happens when two hydrogen atoms approach each other from infinity. At infinite separation the hydrogen atoms don't feel their mutual presence and each atom has one electron localized in its atomic 1s orbital. In the absence of magnetic fields it will not matter whether the spins of the electrons are parallel or antiparallel and you will basically have an ensemble of systems with an equal amount of triplet (parallel spins, $S=1$) and singlet (antiparallel spins, $S=0$) states, since $E_{\uparrow \uparrow} = E_{\uparrow \downarrow}$. But when hydrogen atoms are brought closer together the situation changes: For one thing each atom has a magnetic moment associated with its electronic spin and there will be a dipolar interaction between those magnetic moments. However this dipolar interaction will be very weak (in the order of $10^{-4} \, \mathrm{eV}$ at close interatomic distance) and won't have much influence. Another factor that comes into the game is spin-orbit coupling. Yet again, since $\ce{H}$ is such a light element spin-orbit coupling will be very, very weak and so it can be ignored too. But the predominant factor to be taken into acount is the exchange energy (its effect is in the order of $0.1$ to $1 \, \mathrm{eV}$). Here you can find a very nice treatment of the exchange energy for the hydrogen molecule which comes to the conclusion that

  For the hydrogen molecule the exchange energy is negative [...] so the state when the resultant spin is $S=0$ has the lower energy than the state when $S=1$.

This source also provides the potential energy curves for the hydrogen singlet and triplet states.

These curves amount to the following electron configuration diagrams:

${}$

So, the situation for the approaching hydrogen atoms is like this: If both atoms start out with electrons of opposite spin (i.e. the system is in a singlet state) you have the common case that the hydrogen atoms will bond together thus lowering the total energy of the system. But if both atoms start out with parallel spins (i.e. the system is in a triplet state) the total energy will rise when they get closer to one another and the atoms will not bond together but keep their distance. This can be understood from the electron configuration diagram of the triplet hydrogen molecule: The antibonding $\sigma^{*}$ MO is more destabilized than the bonding $\sigma$ orbital is stabilized by the interaction (this is generally true for 2-orbital interactions) and the energy splitting becomes larger the closer the hydrogen atoms get. But because of the Pauli principle the two electrons of like spin in the system can't both occupy the $\sigma$ MO but one is forced to occupy the $\sigma^{*}$ MO and thus the system's total energy rises when the hydrogen atoms approach each other. And because there is virtually no spin-orbit coupling for hydrogen the total angular momentum (represented by $L$) and the total spin (represented by $S$) of the electrons are essentially conserved quantities and there is no mechanism to facilitate intersystem crossing, i.e. the process of changing from the triplet to the singlet state has such a low probability that it virtually doesn't happen. The situation changes of course if you add an external bias to the system, e.g. irradiation with high-energetic light. Then a triplet-singlet transition might be possible but as I read it that was not your question.

So, the way MO-diagrams are often pictured with electrons of like spin pairing up in a bonding MO does not reflect the "real" situation of two atoms approaching each other from an infinite distance. But it is not meant that way although it is often pictured as such.

Update

So, in summary I established that if you have two hydrogen atoms in free space without spin-orbit coupling and relativistic effects they are stuck with the spin state they initially start out with and can't change it during their time evolution, i.e. they have to stay either on the triplet or the singlet Born-Oppenheimer surface and no switching between those is allowed. But I want to delve a little deeper into the reasons for it. If the hydrogen atoms wanted to change from triplet to singlet state, the atoms would have to transfer $1 \hbar$ of spin angular momentum somewhere else in order to comply with the law of conservation of momentum. In the absence of any particles to collide with the only possible mechanism by which this can be achieved is to emit a photon whose energy is equal to the energy difference between the triplet and the singlet state, i.e. $\Delta E = \hbar \omega = E_{\uparrow \uparrow} - E_{\uparrow \downarrow}$, and which would "carry away" the $1 \hbar$ of spin. But in order to emit a photon there must be a coupling between the electrons and the electromagnetic field. The magnitude of this coupling or in other words the probability for a transition from an initial to a final state under the emission of a photon is described by the transition dipole moment

\begin{equation}
  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \Psi_{\mathrm{i}} \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \Psi_{\mathrm{f}} \big\rangle
\end{equation}

where $q_{n}$ and $\mathbf{\hat{r}}_{n}$ are the charge and the position operator of the $n^{\mathrm{th}}$ particle in the system respectively, and the subscripts $\mathrm{i}$ and $\mathrm{f}$ signify the initial and final states of the transition. Under the conditions of the Born-Oppenheimer approximation the contribution of the nuclei to $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be zero for the situation at hand, so that we only need to be concerned with the electrons in the system. If there is a sizable spin-orbit coupling then the spin degrees of freedom and the spatial degrees of freedom are coupled via the total angular momentum operator $\mathbf{\hat{J}} = \mathbf{\hat{L}} + \mathbf{\hat{S}}$ because

\begin{align}
  \mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2} + \mathbf{\hat{L}} \cdot \mathbf{\hat{S}}
\end{align}

and $\mathbf{\hat{L}} = \mathbf{\hat{r}} \times \mathbf{\hat{p}}$ where $\mathbf{\hat{p}}$ is the momentum operator. However, if there is no spin-orbit coupling then $\mathbf{\hat{L}} \cdot \mathbf{\hat{S}} = 0$ and $\mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2}$, i.e.
the spin degrees of freedom and the spatial degrees of freedom are decoupled. That means you can make a product ansatz for the wave function where you seperate the spin-dependent part $| \psi(S) \rangle$ and the spatial part $| \psi(\mathbf{r}) \rangle$ such that $| \Psi \rangle = | \psi(\mathbf{r}) \rangle | \psi(S) \rangle$. So, for the transition dipole moment you get

\begin{equation}
  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \psi_{\mathrm{i}} (\mathbf{r}) \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \psi_{\mathrm{f}}(\mathbf{r}) \big\rangle \underbrace{\big\langle \psi_{\mathrm{i}} (S) \big| \psi_{\mathrm{f}}(S) \big\rangle}_{= \, 0} = 0
\end{equation}

where $\langle \psi_{\mathrm{i}} (S) | \psi_{\mathrm{f}}(S) \rangle = 0$ because the initial state would be the triplet state ($S =1$) and the final state would be the singlet state ($S=0$) and spin wave functions for different $S$ are orthogonal.
So, no spin-orbit coupling means that the probability for a transition from a triplet to a singlet state under the emission of a photon is zero.

In real atoms and molecules there is of course some spin-orbit coupling, so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be larger than zero. But the spin-orbit coupling in hydrogen is tiny and so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be tiny too and thus the probability of a triplet-singlet transition will be tiny as well and can be neglected for all practical purposes.


Producing sulphur dioxide for using in a reaction I am doing an investigatory project on studying the coloured reactions of d-block and f-block elements. I have included two reactions that require sulphur dioxide as a part of my study.

$$\ce{5SO2 + 2KMnSO4(purple) + 2H2O -> 2H2SO4 + 2MnSO4(pale pink) + K2SO4}$$
$$\ce{SO2 + K2Cr2O7(orange) + 3H2SO4 -> K2SO4 + Cr2(SO4)3(green) + 3H2O}$$

Now, my problem is I need an effective method to produce sulphur dioxide to be able to use for these in a school laboratory environment. I thought of using ferrous sulphate which on heating gives $\ce{SO2}$,
$$\ce{2FeSO4 ->[{$\Delta$}] Fe2O3 + SO2 + SO3}$$
But this reaction also produces $\ce{SO3}$. Will that screw with any of the above reactions? Also, are all the above reactions feasible in a school laboratory? Do suggest if there are better methods and how I can make it to react with the other reactants.

Note: Sulfur dioxide causes dyspnea. Prolonged inhalation of it may causes respiratory problems. So, reaction should be done under laboratory supervision.

The most common laboratory method of producing sulfur dioxide is the reaction of copper and sulfuric acid, both available in laboratories.

$$\ce{Cu(s) + 2H2SO4(l) → CuSO4(aq) + 2H2O(l) + SO2(g)}$$

Process: To a round bottom flask, 100 g of copper (small pieces, turnings, etc.) are placed. From a dropping funnel 100 ml of concentrated sulfuric acid is slowly added dropwise. The reaction flask is gently heated and obtained sulfur dioxide is dried by passing through a wash bottle with concentrated sulfuric acid. (Process taken from here. Read more to know how the reaction can be controlled and the precaution needed)

Alternatively, sodium sulfite can also be used.

$$\ce{Na2SO3 + H2SO4 -> SO2 + Na2SO4 + H2O}$$

Process: Sodium sulfite is placed in the bottom of the 300 ml Erlenmeyer flask of the apparatus. The dropping-funnel is filled with dilute sulfuric acid, made by pouring one volume of concentrated sulfuric acid into an equal volume of water and cooling the mixture. The glass elbow is connected with a gas washing-bottle containing concentrated sulfuric acid, and the dry gas is collected by displacement in two or three empty cylinders. On allowing the dilute sulfuric acid to drop slowly upon the sodium acid sulphite, a very regular evolution of sulfur dioxide is obtained.

For full setup, check here.


Dehydration of alcohol with Lewis acid aluminum trichloride I am told that this method yields ONE equivalent of water, but I seem to get two equivalents of water. Also does this mechanism actually work? I've never seen anything like it, but it seems to make sense. Would another Lewis acid such as boron trifluoride also work?

In the reaction $\ce{{}^{$i$}Pr-OH + H2O ->[\ce{AlCl3}] C3H6 + 2H2O}$ the $\ce{H2O}$ on the reactant side and one of the $\ce{H2O}$s on the product side cancel, so you get $\ce{{}^{$i$}Pr-OH ->[\ce{AlCl3}] C3H6 + H2O}$.

I think the reaction would also work with other Lewis acids. But it will rather proceed through the $\mathrm{E}1$ mechanism in any case. And it is likely that you'll get $\mathrm{S}_\mathrm{N}1$ side reactions.


How do you Synthesise a Fatty Acid Amide? How would you go about synthesising a fatty acid amide, preferably by means of a high-yield reaction. Please provide an explanation that I, as a grade 12 chemistry graduate, could understand.

I am familiar with:

Fatty acids
Carboxylic acid
Equilibrium reactions & constants
Amide groups
Amines
Alkanols
Esterification

If you need to use terms I may be unfamiliar with, it would be nice if you could explain them or provide a link to an explanation.

You are perfectly right on that you need to convert the carboxylic acid to a more reactive species to eventually obtain the amide!

Playing with the pH will not help:

under acidic conditions, the amine is protonated and therefore unreactive
under alkaline conditions, the acid is deprotonated and the carboxylate won't react at all

The Schotten-Baumann reaction of fatty acids with $\ce{SOCl2}$ will probably work fine:

reflux 1 mol of the acid with 1.5 mol of thionyl chloride
dIstill the remaining thionyl chloride off
purify the fatty acid chloride by distillation in vacuum. Note that the bp of stearoyl chloride is about 175 °C at 2 mm Hg!

This is quite some work to obtain a fatty acid derivative that eventually reacts with an amine!

Wouldn't it be much more convenient to form the amide in a one-pot reaction, possibly at room temperature, by activating the carboxylic acid  in situ?

(In situ means: at the same spot, without isolating the reactive intermediate.)

There's a large number of coupling reagents available, such as

carbodiimides like DCC (dicyclohexylcarbodiimide)
$1H$-benzotriazoles
chloro-substituted triazines

Eric Valeur and Mark Bradley published a critical review titled Amide bond formation: beyond the myth of coupling reagents, in Chem. Soc. Rev., 2009, 38, 606-631 (DOI) on the topic.


Why is oxygen absorbed in a colorless alkaline solution of pyrogallol and why does it turn dark brown? My book says:

  Oxygen is absorbed in colorless alkaline solution of pyrogallol and turns dark brown?

I'd like to know why and how.

Pyrogallol is the common name for 1,2,3-trihydroxybenzene.  Hydroxy groups are strong activavtors of aromatic systems.  By the time 3 hydroxyl groups have been added to a benzene nucleus you have a very reactive molecule, reactive enough to react with oxygen in the air.  The reaction is very complex and free radicals are thought to be involved.  Free radical reactions involving activated aromatic rings often lead to colored, high molecular weight products that are not easy to characterize.  Here's a link to a paper that suggests the initial products are probably the ortho-quinone and 2-pyrone pictured below.  Certainly the quinone will be colored, and as it reacts further to produce the higher molecular weight products, they will be colored too.  Here's another link to a very old paper suggesting that some of the higher molecular weight products contain at least 20 carbon atoms.

The balanced equation describing the initial radical oxidation of pyrogallol to the quinone can be written as
$$\ce{2~ pyrogallol + O_2 -> 2 ~o-quinone + 2H2O}$$


Isotope pairs in different states at room temperature? Does any pair of isotopes exist which are in different states at room temperature—either in isolation or as part of an otherwise-identical compound?

It’s not exactly room temperature but it’s probably the closest you can get to two isotopes having different states so that they can be separated by physical means:

At very low temperatures (below $1~\mathrm{K}$) liquefied helium will separate into a superfluidic helium-4 phase and a normal liquid phase containing mostly helium-3. The superfluid and the normal liquid can be considered different phases and are immiscible.


Memorizing polyatomic ions? Using Periodic Table In my Chemistry course, we must memorize a list of common polyatomic ions. Is their an easy way of memorizing ions such as Sulfate $\ce{SO4^2-}$ by looking at just the periodic table. I listed the ones we have to memorize bellow. I know that if they contain oxygen (which is most of them) they usually end in "ate" or "ite." But how do I know how many Oxygen an ion will contain and its charge. My teacher said the only reference table we can use on our test is the Periodic Table of Elements.

I have to memorize the name and the formulas of the ions. Any methods will be much appreciated. I guess the real question is their any trends in the periodic table that helps predict the charge and number of oxygens and polyatomic ions will have? So if I have Sulfate for example, Can I predict the formula by looking at Sulfur on the periodic table

Update: My question is not a duplicate of When to use -ate and -ite for naming oxyanions? because my question wants to **use the periodic table to identify and memorize polyatomic ion **. Voldemorts question had nothing to do with memorization what so ever!

  But, what I don't understand is how does the book know that NO−3 is Nitrate: how does the book know that NO−3 is "the most common oxyanion for the element". How does it know that a charge of −1 and 3 oxygen atoms create "the most common" Nitrogen oxyanion?

His question had nothing do with specifically looking for patterns in the periodic table to help him or her memorize the polyatomic ions. In fact he or she wasn't even asking for a technique or method to memorize polyatomic ions, but rather figure out which form of a polyatomic ion is "most common." For Example: He wanted to find out how the author of his book knows $\ce{NO3^-}$ is more common than  $\ce{NO2^-}$

Number of oxygens:

If your anion is in:

the 2nd period;
the VIIth main group/the 17th group/a halogen; or
is silicon:

then the -ate anion will have three oxygens.

All other -ate elemental anions will have four oxygens.

(Take note of ‘aluminate’ which I haven’t physically seen in that form yet but your teacher seems to insist be $\ce{AlO2-}$; disregarding this general trend.)

From the -ate anion, remove one oxygen to arrive at the -ite, remove two for hypo-ite. Add one for per-ate.

Charge:

Start at an anion with the most oxygens. (Note: Usually this is the -ate but because the halogens and permanganate are special they have the per-ate anion which is more oxygen-rich and important here.) Assume the maximum possible oxidation state for the non-oxygen atom. Assume $-\mathrm{II}$ oxidation state for every oxygen. Add up and take the negative value to arrive at the charge.

A hydrogen- will add $\ce{H}$ and lower the charge by one.

A thio- means one oxygen is replaced by sulphur.

A di- means take two of the (hydrogenated) anion and subtract water. Pyro means the same thing.

Memorise all the remaining! That is most importantly:

Chromite. Similar to aluminate I’  not sure if I ever saw it out in the real world.
Cyanide/cyanate. Thiocyanate is just like thiosulphate.
Oxalate, acetate and tartrate. Organic ions that will never fit into this scheme nicely.
Peroxide and superoxide. But those are -ides anyway.
Permanganate; which is almost like a perhalogenate. Luckily, molybdate and chromate fall into a category described above.

Periodic Table


Is a combustion necessarily a redox-reaction or can there be exceptions? I have a feeling that I have heard of an exception - either based on the material or the oxidizer.

Yes, every combustion reaction is a redox reaction. Combustion is an exothermic reaction of a fuel and an oxidant (that's a big clue). Most commonly, the oxidant is oxygen, which is reduced to water in the process. The other products will depend on the fuel. Organic compounds will have carbon atoms of various oxidation states that will be oxidized to the +4 state as carbon dioxide product.


Liquification of gas to liquid at critical temperature The gas that liquifies first, when cooled from 500 K to its critical temperature given in parenthesis is

  a) $\ce{CO2}$ (304.1 K)

  b) $\ce{NH3}$ (405.5 K)

  c) $\ce{O2}$ (154.3 K)

  d) $\ce{N2}$ (126.0 K)

Answer=b

I know that the critical temperature is the highest temperature at which the gas can be liquified and above which the gas can not be liquified whatever high pressures are used.

$\frac{p_\mathrm c V_\mathrm c}{T_\mathrm c} =nR =\frac{m}{M}R$

i.e.,  $T_\mathrm c$ is directly proportional to $M$, when all the other quantities are constant.

So, based on $M$ $\ce{NH3}\lt\ce{N2}\lt\ce{O2}\lt\ce{CO2}$
So, b should be the answer.

This is how I have solved this problem, please tell me if I am wrong and if so, then please tell me the correct way to answer this too.

Good question.  I think you got the right answer but by using some pretty suspect logic.

First, there isn't enough information to solve the problem in general because pressure isn't specified.  Is the cooling of the gas isobaric?  Or adiabatic?  Or ... ?  Let's assume its isobaric, i.e., it happens at constant pressure.  If so, what's the pressure?  If the pressure is below 5.1 atm, then $\ce{CO2}$ will never condense into a liquid, instead it would transition directly to a solid phase (dry ice).  Ammonia's triple point is roughly 0.06 atm.  But your question says "cooled to the critical temperature", which implies not below the critical temperature, which means the only way any of them will liquefy is if the pressure is at or above the critical pressure.  So let's assume the pressure is isobaric and is at least 11.8 MPa, the critical pressure of ammonia, the gas with the highest critical pressure among the possibilities.  But if we do that, the question is easy.  The gas that liquefies first is the gas with the highest critical temperature, which would be ammonia.

Relaxing this assumption makes the question much harder.  Complex calculations would be required.  The theory of corresponding states would likely be useful.

Your logic is OK, but in your equation, $w$, the weight, and $V_c$, the critical volume, also vary.   So the statement that

  $T_c$  is directly proportional to $M$

is not true, because the other factors are not constant.$%edit$