text:
Does more acidity = better electrolyte?
I am trying to figure out what makes an electrolyte better than another. I originally thought it was acidity, but a potato produces more voltage than a lemon. When I tried this same thing with acetic and muriatic acids, I found that acetic acid produced a constant voltage of a little under 1, while the muriatic (hydrochloric) acid started at about .60v and rapidly dropped into a negative voltage. So if acidity is not an indicator of a good electrolyte, what is?

The key terminology you want to look up is van't Hoff Factor.

Essentially, 'how good of an electrolyte' something is depends on how many particles the substance dissociates into. Let us discuss acidity. Now, from experimental evidence, it may seem that 'stronger' acids and mixtures, such as HCl, $\ce{H2SO4}$, Muriatic, are all good electrolytes. These are not because of their relative acidities. You will notice that in my last sentence, I put 'stronger' in quotes. This is because when talking about acids, 'stronger' has a different meaning than the one you are used to. A strong acid is one that dissociates virtually 100%. In other words, when you put a strong acid in water, all of it turns into ions, and you have none of the original acid left over. Similarly, a weak acid is one that does not completely dissociate into ions. By the definition of an electrolyte, stronger acids are better electrolytes (more ions in solution). This leads to our next point, ions in solution. As mentioned by Zhe above, the 'Van't Hoff Factor' is important. Essentially, it is just a measure of how many particles a substance dissociates into, for example NaCl has a VHF of 2, whereas $\ce{CaCl2}$ has a VHF of 3. The higher a substances VHF, the better of an electrolyte it is, the reasoning being that with an increase in number, there is an increase in dissolved particles, and therefore an increase in the state of being an electrolyte. Now, going back to acids for a moment, you may be wondering which acids are strong. There are 6, and they are HCl, HBr, HI, $\ce{HClO4},\ \ce{H2SO4},\ \ce{HNO3}$. Weak acids are everything else. These strong acids will have the best electrolytic results if you use them as such, much better than lemon juice (citric acid) or vinegar (acetic acid). In your example, you used muriatic acid (HCl) as one of your ion-producers, which explains the rapid voltage change, occurring from the dissociation of a strong acid.
$$\text{In essence, the answer to your question is:}$$ $${\text{How many particles a substance dissociates into determines how good of an electrolyte it is.}}$$

Which better electrolyte : $\ce{CaCl2}$ or $\ce{HCl}$ ?


text:
Is the conjugate base of pyrrole resonance stabilised?
This is something I am conceptually struggling with, if I deprotonate pyrrole on the nitrogen in the ring, is it true that the conjugate base is not resonance stabilised, because the orbital is orthogonal to the ring? However, with a pKa of 23, I feel like it must have some resonance stabilisation.

Yes - it isn't resonance stabilised, and 23 is sort of normal, what are you comparing it against? The pyrrole anion has the lone pair in an sp2 orbital which makes it slightly more acidic than a typical sp3-hybridised amine. For comparison you can look at the "amides and carbamates" section in the Evans pKa table. Also it would be helpful if you could use the full tag [tag:organic-chemistry] instead of just [tag:organic], thanks.

sorry I will do that in future! My organic tutor said that it was resonance stabilised and that concerns me greatly...

No problem. That's odd - well, technically the conjugate base is resonance stabilised, in the same way as how pyrrole itself is resonance stabilised (aromatic), because it does have that lone pair in the pi system. However, the additional lone pair created on N by deprotonation is orthogonal. Maybe it was a wording issue?

definitely a full conjugation was done. Same with imidazole-my tutor demonstrated full resonance, and argued that imidazole is more acidic because you can distribute the charge to the nitrogens in the ring (more electronegative than carbon so favourable)

I see. Well, in the conjugate base of pyrrole (and imidazole), the nitrogen has two lone pairs. One is delocalised into the pi system, the other isn't. The intent was probably to illustrate the delocalisation of the one that is delocalised (the one in the p orbital), which is perfectly legal.

i've added an image-so is this acceptable? It isn't too clear, but the negative charge resides in a p-orbital

There is like you assumed no reasonable amount of resonance stabilisation. However, the aromatic π-system contributes to the stability of σ-skeleton.
Below you can find the MO of pyrrole and its N-deprotonated anion (click it to enlarge). You can see that in MO 17 the electron density is shifted further into the centre of the ring, accounting for the larger negative charge on the nitrogen. It is a bit more difficult to see, but that is also true for MO 18 and 14 (pyrrole), which becomes 15 (anion). The most significant change is that the upon deprotonation resulting lone pair becomes much higher in energy (12 to 16), and it is as you note perpendicular to the π-system in a σ-type orbital. Note that the image only shows the trend and has no information of the actual orbital energies, apart from that they increase from bottom to top.

How do you get the MO diagram?

@Mockingbird I have calculated the MO on the DF-BP86/def2-SVP level of theory.

Is there any software?

@Mockingbird Yes, I used the quantum chemistry software package Gaussian 09, but there are other, openly accessible software packages, too. The graphics I have produced with ChemCraft.

Can you mention some of those softwaress?

@Mockingbird If you enter it into a search engine, you'll find plenty. Wikipedia even has a list https://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid-state_physics_software

As you pointed out, the lone pair is indeed orthogonal to the ring. Not that it would matter, since the aromatic ring already has 6 electrons, so you're not going to conjugate 2 more.
The lower pKa value is likely due to the face that $sp^{2}$ hybridized carbons are more electron withdrawing due to their higher $s$ character. That should be enough to help stabilize the conjugate base.

source: Stack Exchange

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sort: votes


text:
Chlorine Fluoride, why not Fluorine Chloride
Why is Chlorine Fluoride called "Chlorine Fluoride" rather than "Fluorine Chloride"? As I was taught in school that normally that when nonmetal combine with nonmetal then we write the compound with lower atomic number first.

Think about electronegativities.

Anionic part is written in the suffix.

Basically, with interhalogen compounds (and possibly most of binary inorganic compounds) are named like "element" "element'ide'", where no suffix is used for the electropositive part (eg: sodium, oxygen, nitrogen), and a suffix "ide" is used for the electronegative part (eg: chloride, oxide, nitride, phosphide, carbide).
Regarding $\ce{ClF}$, fluorine is clearly the electronegative atom here. Thus the name would be Chlorine fluoride.
Another example of this is from my personal favorite compound, $\ce{O2F2}$
Here, we have subscripts to indicate the number of atoms. In such cases, we add prefixes "di" or "tri" or "tetra" to the name.
$\ce{O2F2}$ would be correctly called as Dioxygen difluoride.

+1 for FOOF. Have a relevant blog link: http://blogs.sciencemag.org/pipeline/archives/2010/02/23/things_i_wont_work_with_dioxygen_difluoride

@DoktorJ I read that article too. FOOF is a pretty interesting compound indeed.

FOOF is tremendously amazing and terrifying

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text:
Is there a relationship between Free Radicals and Beta Particles?
This question is contingent on my understanding of the two terms in the title. As I was taught, a Free Radical is a reactive (high energy?) valence electron released for stability's sake because it is unpaired. I was taught that a Beta Particle is a high energy and fast moving radioactive positron or electron released to stabilize an atom. If both of these premises are true, is there a relation between the two terms, and if so, what is it?

Chemistry electrons are ~1eV, betas are ~1keV.

A free radical isn't an unpaired electron, it's a species containing one or more unpaired electrons.

Those sound like they should be common names for departmental intramural softball summer league teams - the Free Radicals and the Beta Particles. I'd bet the first is actually pretty common for Chemistry students, the second less so amongst Physics students.

not directly as most atoms are not radioactive, but indirectly as the electron is energetic enough to form many radicals (and ions) in other molecules.

There is high-energy in the sense of chemical reactivity, and then there is high energy in the sense of nuclear reactivity. The energy associated with the latter is many orders of magnitude greater than the energy associated with the former, so that there is generally no direct interaction between the two (where the free radical is chemically reactive, and the beta particle arises from nuclear reactivity).
A free radical is a chemical species, whose electron configuration is such that it is highly reactive towards other chemical species. However, this energy is far too low to affect any nuclear reactivity.
A beta particle is a high-energy electron or positron emitted by the radioactive decay of an atomic nucleus. As these particles are emitted from the nucleus of the atom, the electron configuration (i.e. a free radical) is irrelevant; only the stability/reactivity of the nucleus determines whether an atom will emit a beta particle, not the chemical species.
If any of this seems incorrect or unclear, please do not hesitate to ask for clarifications in the comments.


text:
Why would a solution of FeCl2 appear brown/yellow sort of like FeCl3?
If we prepare a solution of $\ce{FeCl2}$ by dissolving solid $\ce{FeCl2}$ powder in water, supposing the bottle of $\ce{FeCl2}$ was left open for sufficient time, is it possible for $\ce{FeCl2}$ to have oxidised to $\ce{FeCl3}$? I can't find a suitable reaction for this excecpt one which includes $\ce{HCl}$ which was definitely not present when preparing the mixture.

Yes, it is very common. You actually have to protect Fe(II) solution from air when you do experiments if oxidation is something you want to avoid.

The yellow/brown color you are seeing in the solution of $\ce{FeCl2}$ is due to formation of various types of hydrated iron oxide and not iron(III) chloride. They may include both the anhydrous($\ce{FeOOH}$)  or the monohydrate($\ce{FeOOH.H2O}$) often referred to as ferric hydroxide($\ce{Fe(OH)3}$). They are generally yellow to brown in color. The reaction is given here:
$$\ce{4FeCl2 + 6H2O + O2 → 4FeO(OH) + 8HCl}$$

Reaction between iron chloride (II), water and oxygen(moisture) to form
iron metahydroxide(Iron(III) oxide-hydroxide) and hydrogen chloride. The reaction takes place at
reflux.

Moreover, if you further heat iron chloride in oxygen at 450-480℃, iron oxide will form.
References(all pdf links)

http://www.jieas.com/fvolumes/vol081-5/3-5-11.pdf
http://www.gfredlee.com/SurfaceWQ/StummOxygenFerrous.pdf
https://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/77056/1/chd061_5-6_335.pdf

That's interesting... I guess what I've learnt in school is only for the Cambridge Advanced Levels, which is probably the simplified explanation. Your explanation seems credible and convincing as well.

But no solid is formed during the oxidation process... Wouldn't the hydrated iron (III) oxide-hydroxide give an insoluble precipitate. From my own observations in the laboratory, I have never seen any precipitate in a solution of yellow iron (III) chloride solution. How would you explain the absence of a ppt?

@TanYongBoon iron(III) hydroxide is indeed insoluble in neutral and basic pH and precipitates in the bottom of the solution(according to 1st link). However, the ferrous chloride solution present in the laboratory has somewhat an acidic pH for which the the hydrated iron oxide or iron(III) hydroxide solubilize in the solution and thus not get any precipitate.

$$
\begin{align}
\ce{Fe^2+ &-> Fe^3+ + e-} &\quad E^\circ &= \pu{+0.771 V} \tag{R1}\\
\ce{O2 + 2 H2O + 4 e- &-> 4OH-} &\quad E^\circ &= \pu{+0.40 V} \tag{R2}
\end{align}
$$
Thus, oxidation of iron(II) to iron(III) ions is feasible with atmospheric oxygen as the overall cell potential for the reaction is $\pu{+1.171 V}$ and it does occur at a reasonably observable rate.

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text:
Temperature change and gas evolution for vinegar and baking soda experiment
This is sort of 2 questions in 1, but I think they are related.
Could someone please explain why this endothermic reaction doesn't further decrease the temperature when the amount of baking soda is increased?
The amount of vinegar is constant, varying amounts of sodium bicarbonate is independent.
Attempt 1: 100ml vinegar and 1 teaspoon baking soda, temperature went from 16 degrees Celsius to 12 degrees to Celsius in <10 seconds.
Attempt 2: 100ml vinegar and 2 teaspoon baking soda, temperature went from 16 degrees Celsius to 12 degrees to Celsius in <10 seconds.
Attempt 3: 100ml vinegar and 3 teaspoon baking soda, temperature went from 16 degrees Celsius to 12 degrees to Celsius in <10 seconds.
Also, the volume of the effervescing gas collected into a balloon was the same for 1-3 teaspoons of baking soda. once again vinegar is 100ml for all 3 attempts.
I expected a further decrease in temperature as more baking soda is used.
I expected more gas when more baking soda is used.
What happened?

Hello and welcome to Chemistry.SE. If you have any questions about the site, feel free to visit the [help]. Regarding your question, my suspicion would be that you are using up all of the acetic acid from the vinegar with 1 tsp of baking soda, so adding more has nothing else to react with. If you knew the concentration of the acetic acid in the vinegar, and the mass (say, in grams) of the baking soda, you could calculate this. On the other hand, what else could the reason be?

thanks for reply. yes, i think its an exercise in reaction stoichiometry problem. amount of vinegar is constant at 100ml and each tablespoon is 6 grams of baking soda. i would love someone's take on this.

Plz chk this question, too. https://chemistry.stackexchange.com/questions/101586/vinegar-baking-soda-yields-far-smaller-volume-of-co2-gas-than-calculation

Although your observations make it pretty clear that you are using enough sodium bicarbonate to consume the acetic acid, even with $1$ teaspoon of sodium bicarbonate, we can also do a rough calculation to validate this hypothesis. It's a "rough" calculation because we don't know the concentration of acetic acid in your vinegar.
Vinegar is typically about $\pu{5\%}$ acetic acid by volume, but can be as high as $\pu{20\%}$. For our calculation, we will estimate that there is $\pu{5 g}$ of acetic acid in $\pu{100 ml}$ of vinegar, and use $\pu{6 g}$ as the mass for $1$ teaspoon of sodium bicarbonate, per your comment. Then, using the molecular weights of acetic acid ($\pu{60 g/mol}$) and sodium bicarbonate ($\pu{84 g/mol}$), we can calculate the following molar quantities of the reagents:
acetic acid: $\pu{0.083 mol}$ per $\pu{100 ml}$ vinegar
sodium bicarbonate: $\pu{0.071 mol}$ per teaspoon of baking soda
If our estimate of the acetic acid concentration is correct, it is close enough to a $1:1$ molar ratio with the sodium bicarbonate that you may not have noticed the difference in your measurements. In this case there is $\pu{17\%}$ more acetic acid than sodium bicarbonate. This means that the addition of excess sodium bicarbonate (i.e. $2$ teaspoons) should result in a $\pu{17\%}$ greater temperature difference, which would be $\pu{4.7^oC}$ rather than $\pu{4^oC}$. Since you are only reading your thermometer to $\pu{+/- 1^oC}$, these could be said to be the same and it is perfectly reasonable that you wouldn't have noticed the difference. It is not stated how the volume of evolved carbon dioxide was measured, but it would likely be even more difficult to notice this small change by the change in observed carbon dioxide volume.
Summary, TL;DR:
Your observations indicate that a single teaspoon of baking soda is consuming all of the acetic acid present in the vinegar, so when additional baking soda is added, there is no more acetic acid to react with and thus there is no indication of further reaction. Making some simple assumptions about the concentration of vinegar and mass of sodium bicarbonate in one teaspoon allows us to confirm that this is the case, within the detection limits of your method (measurement of temperature change and volume of carbon dioxide evolution).


text:
What is trans-addition?
A question from JEE-Advanced(2017) :

Bromination of both Z-but-2ene and E-but-2-ene involve syn and anti addition. So, there would be 4 products as suggested in the question. But, I don't understand the terminology of 'trans-addition', mentioned in the question.
What is actually trans addition?

Uh, I think that the correct term should be "anti" not "trans"

Why do you think there will be syn addition?

A screenshot or picture of written text is problematic as it cannot be indexed by our site search or other search engines. Chemistry.se is about the content, and your question is unlikely to be helpful in the future for other people. Please take a couple of minutes to transcribe the text, so that it can be found easily in the future.

@Martin Well, I am kinda underskilled at mhChem and with my mobile phone jt's hard to draw chemical structures. So, I am sorry. But I will edit the question.

Just like @QuantumAMERICCINO has pointed out, it's probably anti addition.
Furthermore, you seem to wonder why a syn compound doesn't form. It all lies in the mechanism of bromination:
The double bond in cis-2-butene polarises the $\ce{Br-Br}$ bond in a nearby $\ce{Br2}$ molecule. A $\ce{Br+}$ ion attacks the double bond and forms a bromonium ion.

The bromonium ion makes the adjacent carbon atoms electrophilic. The bromide may attack the most electrophilic out of them. Here, however, they are equivalent.

You should now be able to understand how the mechanism works.

So how are there 4 products?

What 4 products @Mockingbird360

M,N,O,P as suggested in the question.

@Mockingbird360 Oh that, they are just stereoisomers. I showed in my example, the bromine attacking the left carbon. If it attacked the right one instead, you would get the enantiomer of the above compound. For a trans-2-butene you would get a meso product.


text:
Why does strong Lewis acid-strong Lewis base interactions prevail over hard-soft acid-base interactions?
In Pearson's paper, he mentions that strong acid-strong base interactions are more favourable compared to hard-soft acid-base interactions. He cites the example of  $\ce{H+ + H- -> H2}$. In this example, the hydrogen cation is a strong and hard Lewis acid while the hydride ion is a strong yet soft Lewis base. By just considering hardness and softness of the acid and base, we would not expect a particularly favourable interaction between the two species. However, Pearson says it is because of a more favourable interaction which is the strong acid-strong base interaction which is at work in this case.
Some background information:
When the HOMO of the Lewis base and the LUMO of the Lewis acid are similar in energy, the bonding molecular orbitals produced are significantly lower in energy and thus, electron occupation of these bonding MOs give rise to significant stabilisation. This occurs when hard Lewis acids and hard Lewis bases react and also when soft Lewis acids and soft Lewis bases react. This is the molecular orbital basis of Pearson's Hard Soft Acid Base Principle.
Can we explain the favourable interaction of strong Lewis acids and strong Lewis bases using molecular orbitals as well?

Reference: Pearson, R. G., & Songstad, J. (1967). Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. Journal of the American Chemical Society, 89(8). Retrieved June 5, 2017 (link to paper)

The Klopman-Salem equation accounts for the energy gain (or loss) when the orbitals of one reactant overlap with those of another. The equation is
$$\Delta E=-\sum_{ab}(p_a+p_b)\beta_{ab}S_{ab}+\sum_{k<l}\frac{Q_kQ_l}{\varepsilon R_{kl}}+\sum_{r}^{\text{occ}}\sum_{s}^{\text{unocc}}-\sum_{s}^{\text{occ}}\sum_{r}^{\text{unocc}}\frac{2(\sum_{ab}c_{ra}c_{sb}\beta_{ab})^2}{E_r-E_s}$$
where $p_a$ and $p_b$ are the electron population in the atomic orbitals $a$ and $b$, $\beta$ and $S$ are the resonance and overlap integrals, $Q_k$ and $Q_l$ are the local charges on atoms $k$ and $l$, $c_{ra}$ is the coefficient of atomic orbital $a$ in molecular orbital $r$, where $r$ refers to the molecular orbitals on one molecule and $s$ refers to those on the other, $E_r$ is the energy of molecular orbital $r$ and $E_s$ is the energy of molecular orbital $s$.
The first term is a closed-shell repulsion term, the second term is the Coulombic repulsion or attraction, and the third term is the interaction of all the filled orbitals with all the unfilled of correct symmetry. This equation was derived using perturbation theory being the first term is a first-order perturbation term and the third the second-order perturbation term.
Now, the HSAB theory is affected by charge and orbital interactions. The h-h interactions is mainly electrostatic in nature, while in a s-s interaction the orbitals involved are close in energy. With the Klopman-Salem equation we can see that the second term is the hard contribution and the third is the soft contribution. There is a simplified form, which accounts only for frontier orbitals, that is
$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$
So, back to the question, a hard molecule or ion is characterized by a large HOMO-LUMO gap (definition of hardness) and the second term, in the simplified version, is very small, so the contribution is mainly electrostatic, i.e., the more charged the sites in the molecule/ion the greater will be the stabilization (or repulsion, in case of like charges). The electric field of the two parts in the molecules/ions would diminish $R$ and the term would be very large. In the case of soft-soft interactions, the orbital term is the dominant because the HOMO-LUMO gap is smaller.
But sure, the interaction has to be between to molecules or ions, not a single specie. You can apply this to nucleophiles and electrophiles as well. A hard nucleophile (negative charge localized and low-energy HOMO) would tend to react with a hard electrophile (positive charge localized and high-energy LUMO), leading to a great stabilization by the electrostatic term. Meanwhile a soft nucleophile (high-energy HOMO) would react with a soft electrophile (low-energy LUMO), stabilized by the orbital term, because $E_{\text{HOMO}(\text{Nu})}-E_{\text{LUMO}(\text{El})}$ is small.
If it is necessary, I can edit with a MO diagram to see this more visually.

Thanks for the insightful response. I now have a better understanding of Pearson's theory. Previously, I had the misconception that for hard acid-hard base, they would have similar high-energy HOMO and LUMO but that is not the case as you have pointed out.

Although u have explained very in a very detailed manner of the molecular orbital basis of Pearson's theory, u still have not answered my question as to why STRONG acid- STRONG base interactions would outweigh hard and soft interaction. Note that strength and hardness/softness are different.

Oh! you're right. Bad reading from me. For the moment I don't have the answer, but I think that this case is like comparing kinetic vs thermodynamic reactivity. The strength of an acid or base is determined by molecular structure, solvent effects, H-bonds, sterics, etc. Then, as far as I know, there is no a clear tendency based only on the electronic structure of the reactants.

Why does high positive charge result in a high-energy LUMO? Wouldn't a high positive charge stabilise the HOMOs and LUMOs and result in them being lower in energy?

Similarly, why does a localised, high negative charge result in a lower energy HOMO? Wouldn't the negative charge consequently destabilise the frontier MOs and raise their energies?


text:
How are the π bonds arranged in Osmium Tetroxide?
The structure of $\ce{OsO4}$ has puzzled me greatly for a while. I know that it has a tetrahedral geometry.

I am quite piqued at the thought of how the $π$ bonds were arranged in the molecule.
My knowledge is very limited, and I know that $π$ bonds are formed due to the sideways overlap of $p$ orbitals.

While, in a regular unhybridized atom, the 3 $p$ orbitals are arranged orthogonal to each other ($p_x$, $p_y$, $p_z$), I am unsure on how they are arranged in $\ce{OsO4}$. Is it something like this?

Or is it that some other orbitals may be involved?

Just a reminder not to play with OsO4 - it sublimates and is highly toxic. Nasty stuff.

I believe your confusion stems from the fact that you are trying to apply a basic bonding model to a situation where molecular orbital theory should be used. Yes, $d$ orbitals should be involved.

Your conception of $\pi$-bonds is a bit too restrictive. A $\pi$-bond is a bond in which there is one node along the internuclear axis. So, yes, two side-on $p$-orbitals do form a $\pi$ bond because there is one node along the internuclear axis (the one separating the bottom lobe from the top lobe). A $\sigma$-bond is a bond with no nodes (i.e. cylindrically symmetric), a $\delta$-bond has two nodes, etc. It's important to note that these symmetry labels work best in molecules where each bond axis either exactly lines up with or is exactly orthogonal to all the others (like diatomics and octahedral metal complexes) but we can work in other molecules with a little more effort.
Now, in osmium, the valence orbitals most involved in bonding are actually the $d$-orbitals. Unfortunately, because the molecule has tetrahedral symmetry and the bond axes are not all orthogonal or parallel, it's less easy to see how the oxygen and osmium orbitals come together to form $\pi$ bonds, and we need to resort to MO theory to get a qualitatively correct answer. If you don't know MO theory, the most basic way of approaching the following is to think of all the ligands and metal orbitals combining (much like the simple $p-p$ overlap in your question) and we label each combination by a label that has to do with the symmetries (rotations and reflections) of the combination of atomic orbitals.
Let's start with a qualitative MO diagram from this answer:

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs. Orbital energies here are qualitative (and in fact the order is wrong here for $\ce{OsO4}$ which has occupied orbitals energies $t_1>t_2>a_1>t_2>e$).
The oxygen $p$ orbitals that point toward the metal and will make up the $\sigma$ bonds form the $a+t_2$ orbitals. The oxygen $p$ orbitals that make up the $\pi$ bonds are the $e+t_1+t_2$ orbitals. The osmium $d$ orbitals are $e+t_2$. Since only orbitals of the same symmetry can mix, the $e$ orbitals are purely $\pi$-bonding while the $t_2$ orbitals are a mix of $\pi$ and $\sigma$ bonding.
So your picture is sort of correct for the oxygens but they have to be combined in specific ways in order to overlap with the osmium. Here is a picture of one of the $\pi$ bonds in this molecule:

You can see the $p$ orbitals on each oxygen, the node along the bond axis and the osmium $d$ orbital at the center taking part.
Here is a purely $\sigma$ orbital:

In this case, the Os $s$ orbital participates with the oxygen $p$ orbitals pointing towards the center.
And here are two views of one of those weird $t_2$ orbitals, which is a mix of both $\sigma$ and $\pi$ in tetrahedral symmetry:

You can see that the oxygen $p$ orbitals clearly have a node near the O, but as you get towards the metal center the node bends off-axis. Meanwhile, near the metal, the nodes are not oriented near the bond axes at all. Very confusing!
In tetrahedral symmetry, some orbitals are clearly $\pi$-type, some are clearly $\sigma$-type, and some are a mix and therefore not a well-defined $\sigma$ or $\pi$.
EDIT: The careful reader will note that I ignored the $t_1$ orbitals. These are $p$ orbitals on the the oxygens that have no osmium orbitals to bond with and so are non-bonding. Here is a picture of one of these non bonding orbitals.

You can clearly see all the oxygen $p$ orbitals, but note that the osmium center has no orbital.


text:
What would be the vapor pressure of water at above 100 degrees Celsius?
Water boils when it's vapor pressure equals atmospheric pressure (i.e., 760 mm of Hg) which is at 100 degrees Celsius. If the temperature is further increased, would the vapor pressure also increase or remain same?

Vapor pressure surely would increase, as long as you still have liquid water around.

Do you mean to ask what is the vapor pressure within the system during the phase transition? Or are you asking what happens as the temperature is increased after boiling is complete?

Steam tables are a forgotten art form. There is a reason for the development of super-heat for steam locomotives.

Your question is phrased inconsistently. As you noted under "normal conditions", water boils at 100 degrees Celsius when the pressure is 760 mm of Hg. At that point there is no liquid water left and hence, technically, there is no liquid water to have a "vapor pressure."
Now steam (gaseous water) can have a pressure above atmospheric pressure if heated above 100 degrees Celsius in a closed container. (as Jon Custer has commented...)


text:
Intuitively, why is it that when the HOMO-LUMO gap is small then the bonding molecular orbital produced is of significantly lower energy?
What is an intuitive explanation for why a small HOMO-LUMO gap between the electrophilic and nucleophilic species would produce bonding molecular orbitals much lower in  energy, giving rise to greater stabilisation upon electron population. I would like an intuitive explanation, not a mathematical one (such as by the Klopman-Salem equation).
Having done some thinking on my own, my explanation is that the closer the energy levels of the HOMO and LUMO, the greater the extent of orbital overlap. Is this logical? More importantly, is it accurate?


text:
What Would Hypochlorite Ions Prefer to React with, Hydrogen Ions or Ozone?
If I had a box that had inside it a UV light an equal quantity of ozone ($\ce{O3}$), and $\ce{H3O+}$, which would be broken down because of the UV light to $\ce{O}$, and $\ce{O2}$, as well as $\ce{H2O}$, and $\ce{H+}$ because that is what the $\ce{H3O+}$ would be broken down into. If I introduced one hypochlorite Ion ($\ce{ClO-}$) or a chloride Ion ($\ce{Cl-}$), what would the chloride Ion or hypochlorite Ion react with? Would it react with the $\ce{H+}$, or the $\ce{O}$ & $\ce{O2}$?

First of all, hydronium ion, $\ce{H3O+}$ is basically a proton in aqueous solution. A proton undergoes hydrogen bond in aqueous solution to forms various forms of hydrated proton. But for simplicity, we denote $\ce{H+}$  which is considered an acid. Now, UV light dissociates hydronium ion in ozone atmosphere into oxygen, singlet oxygen, proton and water. So, if hypochlorite  salt or its corresponding acid is introduced, it might react with these. Chloride ion in form of salts is very much stable and is unlikely to react with any of the dissociated products.
So, first of all, any hypochlorite salt or hypochloric acid added will dissociate into component ion in presence of UV light. Then will react with the following dissociated products:

Singlet oxygen

$$\ce{ClO- + 2O ⇄ ClO3-}$$
(Source, considered hypochloric acid)

Proton

$$\ce{2HClO + 2 H+ + 2  e{−} ⇌ Cl2(g) + 2H2O}$$
Water is present as moisture so it is unable to attack hypochlorite ion readily. Also oxygen is present in minute amount(mainly singlet oxygen is present), so it cannot oxidize hypochlorite ion.

So would the CLO- react with the H+ or the O family (O, O2)?

@ArjunJhaveri Both, but in the O family, only singlet oxygen will react.

What would it react with first? Also, Over time, would all of the the H+ ions precipitate back and not bond with the chlorine because they form water?

@ArjunJhaveri everything is possible. The reaction proceed in such a complex way that nothing can be said clearly. Indeed, hypochlorite ion is not that kind of thing you daily work with nor they can be found in labs so working with them require higher understanding of chemistry. Indeed, it is unstable and undergo numerous acid, it may form chloric acid, it may dehydrate to chlorine dioxide. So, nothing can be said clearly.


text:
To what microscopic quantity does Gibbs energy correspond to?
Is the Gibbs energy of a system equal to the potential energy of a system? The answers to this and this questions seem contradictory. Also, this does not seem to tell that Gibbs energy is equal to the potential energy. By potential energy, I mean the Electric potential energy between the atoms, electrons and nuclei.(I'm not considering other potentials due to nuclear forces or gravitational forces.)
If Gibbs energy is not potential energy, then what is the physical significance or physical interpretation of Gibbs energy at the molecular level? I know that Gibbs energy is equal to the maximum non-expansion work that can be extracted. I just want to know what does that mean at the molecular level.

Definitely not just the difference in potential energy. Free energy also accounts for entropic changes.

@Zhe I'm now quite convinced that Gibbs energy is not equal to potential energy. Now, I'm looking for a physical interpretation of Gibbs energy.

Part of the issue is how 'potential' is an overloaded term in physics and chemistry. Also, trying to take thermodynamic concepts down to the molecular level is fraught with problems, namely those problems solved by statistical mechanics.

Have you considered the possibility that Gibbs energy makes more sense in system level, not in (mono) molecular level?

@Greg Well, every bulk property arises from some microscopic quantities.

@ApoorvPotnis 1) Bulk properties often comes from ensemble behaviour. 2) Microscopic property =/= molecular interpretation

@Greg I hope I've changed the question accordingly.

As others have said, the Gibbs function (or Gibbs energy) makes sense at a macroscopic, thermodynamic level (instead of a microscopic level), but if you need to picture it, you can elaborate what it means from the deconstruction of its mathematical definition.
Internal energy
Let's start with internal energy, U. Internal energy is the mechanical energy (potential + kinetic) of all the particles in the system. Take into account that it isn't reducible to the energy of separate particles - these particles are interacting (for chemical systems, through intermolecular forces) and the potential and kinetic energy are intrinsically related (though quantum mechanics). The only energy that you aren't considering is the kinetic energy of the system as a whole - if your chemical system is moving through space or rotating around its centre of mass, that energy isn't relevant here (it's also trivially easy to calculate via Newtonian mechanics) - and potential energies that arise from external factors that affect the system uniformly - so if your whole system is subject to gravity, that potential energy isn't computed into the internal energy (also trivially easy to calculate via classical physics). In other words, the internal energy discounts anything that you could calculate if you picture your system as a point mass (as you do in high school mechanics); it deals with whatever is happening inside that point mass.
This mechanical energy is conserved, but the balance between its components will depend on the exact positions of all atoms; there would be constant transferences of energy from kinetic to potential, from inter- to intramolecular interactions, etc. It's easier if we start picturing the system in statistical terms - as a huge amount of molecules, moving in random directions at random speeds (following known statistical distributions), of which thermodynamic functions (such as Gibbs') capture an average.
So, in the internal energy you are taking into account: 1) the kinetic energy of molecules in your system, 2) the potential energy of intermolecular interactions, 3) the kinetic energy of intramolecular movements (rotations, vibrations), 4) intramolecular potentials (arising from the electronic structure of molecules).
Enthalpy
Since we are considering the movement of molecules relative to the centre of mass of the system, we know that the total movement of the system is zero - the centre of mass remains fixed. If you add all the momenta of the molecules, you get a zero.
However, there's a possible collective movement that, while averaging to zero, implies a macroscopic physical change: movements towards the system or away from it. It is possible, if you add all the momenta, to get a total momentum of zero and, at the same time, that more molecules are moving away from the centre of mass than towards it - if you picture it in macroscopic terms, that means that the system as a whole remains at the same position, but is expanding (think of a balloon). And, likewise, if more molecules move towards the centre than away from it, the system is immobile but contracting. Both these collective movements can, potentially, change the volume of the system - they are a microscopic interpretation of pressure.
Note that, ff the system actually changes volume, the intermolecular potential energy is affected - if the system expands, its molecules are further apart and its interactions are weaker; for a contracting system, the opposite is true. The energy involved in that process is purely a mechanical work - captured in the term $pV$. When we work with solids or liquids, changes in volume are usually negligible; when we work with gasses, we usually store them in rigid containers - so changes in volume are also usually negligible. Therefore, it's convenient for chemists to define a function that discounts this mechanical work from the internal energy - that's exactly what enthalpy, H, is.
So, in the enthalpy you are taking into account: 1) molecular movement (except expansion and contraction), 2) intermolecular potentials, 3) intramolecular movement, 4) intramolecular potentials.
Entropy
The molecular interpretation of entropy, S, is always contentious (disregard any explanation that includes the words "chaos" or "disorder"), but we can discuss it simplistically. Entropy is more complex to visualise because it's an intrinsically statistical magnitude, and it ultimately relates to how energy is transferred between the different forms (inter- and intramolecular, kinetic and potential). What entropy implies is that when more than one configuration produces equivalent states, that makes the state more likely - effectively stabilising it, lowering its energy.
If you took microscopic photos of your system at different points, you'd get random positions and states for its molecules - but some situations would be more likely, from a purely random viewpoint, than others. If you have a two-component system, it's very very unlikely that their molecules will be spontaneously segregated in different parts of the system - if you place molecules randomly, it's far more likely that they will be well mixed throughout the system, so, more entropy for the latter. Molecules with many internal movements (like large molecules with many vibrations) also produce more equivalent states - you could get many photos in which the only difference is what particular bond vibrates more strongly, while that's impossible for, say, diatomic compounds. An ideal crystal at T=0K, where atoms are fixed exactly in periodic positions, without defects, and without vibration because they don't have any kinetic energy, has only one way to be arranged - so it has minimum entropy.
The ability to populate these configurations is closely tied to temperature. Temperature provides the energy that produces these movements and allows the system to explore more configurations - through Boltzmann's distribution. Since entropy indicates how many configurations there are for a given state and temperature provides the energy to do so, the "entropic stabilisation" term depends on temperature, and, conversely, systems with higher entropy are more sensitive to entropic stabilisation. That's what the term $TS$ measures.
This is important because that "entropic stabilisation" means that a fraction of the energy of the system (precisely $TS$) is tied up in providing these alternative configurations, and cannot be extracted directly from the system. Unless chemical reactions or other structural changes take place, that energy is inaccessible.
Gibbs function
We can then describe a function that computes 1) the internal mechanical energy of the system, 2) discounts the energy involved in expansion/compression work, and 3) discounts the energy that is inaccessible due to entropy. This is what the Gibbs function, G, means at a microscopic level. Note that this doesn't make for a clear-cut distinction between what microscopic forms of energy are included in the Gibbs function - it isn't as simple as "kinetic out, potential in", because some things (mainly entropy) are statistically defined and only make sense for a collective of (interacting) particles.

The Gibbs Energy determines how favorable one reaction is over another. The Gibbs Energy tends to least action. According, to Gibbs whenever the equation:
Enthalpy - ((Temperature)(Entropy))
is negative the equation is favorable, the more negative, the more favorable. Just think what this means. Temperature and Entropy, both types of Enthalpies, measure the energy that can still be harnessed and reused from their current state of disorder. Enthalpy is the total energy of the system that is being used this includes pressure and volume. To answer your question, Gibbs Energy measures the amount of free energy in the isothermal and isobaric system available for work. In other words, Gibbs Energy is the thermodynamic potential to do reversible work.


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Reaction of fluorine with cold water
According to my textbook, when fluorine reacts with water, it produces oxygen and ozone.
\begin{align}
\ce{3F2 + 3H2O &-> 6HF + O3}\\
\ce{2F2 + 2H2O &-> 4HF + O2}
\end{align}
Is hydrogen peroxide ($\ce{H2O2}$) produced under any conditions on reaction of fluorine with water?

No, Hydrogen Peroxide is not produced under any conditions on reaction of fluorine with water. Fluorine is such a strong oxidizing agent that it even oxidize water to oxygen and ozone by ripping apart hydrogen bond from water and bonding with fluorine to form hydrofluoric acid. The reaction of water and fluorine can be found here and here. Yield of ozone is less than oxygen and can be considered as an impurity. Apart from oxygen and ozone, atomic oxygen is also produced.
$$\ce{F2 + H2O → 2HF + [O]}$$
If you use ice cold water, very minute amount of hypofluorous acid($\ce{HOF}$) is also formed which is present in solution in equilibrium.

$$\ce{F2 + H2O ⇄ HFO + HF}$$
Reaction between fluorine and water to form hypofluorous acid and
  hydrogen fluoride. The reaction proceeds at a temperature of -40 °
  C. The reaction is conducted M. Studer and E. Eplamenu in 1971, the
  first time obtaining hypofluorous acid (by fluorination of ice.(Source)

Hypofluorous is very unstable compound as it readily disproportionates into $\ce{HF, F2}$ or $\ce{H2O}$.  As such it has not been isolated yet as a solid compound.

Actually it was isolated https://en.wikipedia.org/wiki/Hypofluorous_acid

@Mithoron, yes it is in gaseous phase, not in liquid form as most acid exist in liquid phase.

I mean that solid HOF (and liquid) were obtained.

And on reading the Wikipedia article, one actually finds that $\ce{HOF}$ is the only hypohalous acid to be isolated in neat form.

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


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Chiral products from achiral reactants and 'principle of racemic mixture formation'
I have studied that if in a reaction system chiral reactants are absent but chiral products are produced; then all enantiomers of respective diastereomers are produced in equal concentration or a racemic mixture is produced.
But what is the reason behind this 'principle of racemic mixture formation'? Is it simply because as the enantiomers of respective diastereomers have no difference in physical or chemical properties, the reaction system shows no preference to any of them?

Because the starting material is achiral and the products are enantiomers, the reaction, by symmetry, must proceed via two enantiomeric transition states. Enantiomers have the same energy, so these two transition states have the same energy, so the rate of product formation is the same for both products. Therefore, they will be produced in equal amounts.
However, there is a complication that was not covered in your statement. There are examples of autocatalytic reactions like the Soai reaction where the product aids in catalysis and a tiny random preference for one enantiomer over the other leads to amplification of that preference. This is akin to how life uses preferentially one enantiomer instead of another, since presumably, we started with achiral materials in the primordial soup.

My next question would be what's so special in Soai Reaction?

It produces a chiral product that catalyzes the reaction that makes more of that enantiomer. So once you get any imbalance, the imbalance is amplified. It's a chemical version of the Kelvin water dropper except that the end state is reaction completion instead of discharge.


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Why a particular enantiomer of a diastereomer shows affinity towards another?
A particular enantiomer can show affinity towards another enantiomer of another diasteromer eg. body receptors with chiral drugs, chiral stationary phase of chromatography etc.
But what's the reason behind this affinity? A broad answer will be encouraged.

Hi @Mockingbird360. This question is very very very very very broad and emphasize the amount of "very"s I used. An answer to this question would need an entire dictionary of diastereomers to explain. It's wrong to expect such hard work from an answerer. Limit yourself to asking a particular question at a time. For example, you could point out a specific diastereomer and ask about it's selectivity. I hope you understand. Good luck and post better questions next time!

In a nutshell, try to shake someone's left hand with your right, and you'll see.

Let's set aside compounds for the moment.
Your hands are chiral objects, that is, they are non-superimposable mirror images of each other [see more here]. If you pick up something that is not chiral, like a cup, there is no difference whether you use the left hand or right hand. The "energy" of those two interactions (right hand + cup or left hand + cup) is the same.
Now, let's say you get a pair of gloves made to fit your hands. If you put your left hand into the right-hand glove, it feels different than if you put your left hand into the left-hand glove. You could think of it as those two interactions (left hand + left glove vs left hand + right glove) having different energies.
The reason for this is that enantiomers of chiral objects interact with non-chiral objects with the same energy, but interact with enantiomers of chiral objects with different energy (ignoring coincidental cases, as @Zhe suggests).
The way one enantiomer of a compound "fits" with enantiomers of other compounds (objects) like proteins or chromatography instruments is difficult to predict. Sometimes it is useful to actually model the interactions, but often, like in chromatography, it's enough to know that there should be a difference in energetics of the enantiomers of a compound with a single enantiomer of another chiral object.

Enantiomeric arrangements (molecules, complexes, etc) have identical energies. However, when you take two enantiomeric objects, their interactions are via diastereomeric arrangements. Barring the very unlikely case of accidental degeneracy, these diastereomeric arrangements have different energies. Different energies of interaction led to different chemical properties, including affinity.

Your answer would be better if you add some examples.

That's true, but "a broad answer will be encouraged." I'm just following directions. Ivan's example of hands is a good one.


text:
Why does increasing the temperature for a reaction at equillibrium cause the reaction to shift to the endorthermic side?
Increasing the temperature would also increase the rate of the exothermic reaction, however what drives the molecules present in the system to choose the endorthermic pathway?

Think about it this way:
There are two people. One guy likes chocolates and one guy hates them. They decide to transfer the chocolates. If you start throwing chocolates randomly to both of them, in which side will the reaction go? The guy who hates chocolates would keep giving it to the guy who likes them.
In a similar way, if you provide heat to a reaction, it would move in the direction of the where the heat is absorbed, i.e the endothermic pathway.


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Looking for a mechanism for the demercuration of organomercurial alcohol
I am teaching myself organic chemistry and am very frustrated that I cannot get a complete mechanism for the oxymercuration-reduction of an alkene to an alcohol. The reagents are an alkene, mercury(II)acetate, and water for the first step... the step I understand. It ends by forming organomercurial alcohol. This is then reacted in a second step with sodium borohydride in a solution of sodium hydroxide. I have searched for a mechanism for this step (specifically the reduction portion at the very end) but I cannot find one anywhere. The closest I have come to an answer is a source that says it probably had to do with radicals and inorganic chemistry. The products are an alcohol with the C-Hg(OAc) bond transplanted with a C-H bond, tetrahydroxyborate, reduced mercury (standard conditions), and OAc anion. Thanks any help is appreciated
Update:
I am attaching my attempt at a mechanism and would like someone to verify if it is the correct mechanism or tell me why it would not work. Thanks

The direct displacement of the mercury acetate by borohydride, as prooposed by @xavier_fakerat, looks attractive because it (1) makes some chemical sense and (2) gives the correct products. However, that mechanism can be disproven by looking at the stereochemical outcome of the reduction step.
Pasto and Gontarz did just that (Pasto JACS 1969). When cyclopentene was subjected to oxymercuration, followed by reduction with sodium borodeuteride, the product observed was the trans-configuration of the hydroxyl and deutero groups. The replacement of mercury for deuterium went with retention. This rules out direct displacement (SN2 like mechanism) which would lead to inversion of configuration.

When cis- and trans-butene were subjected to the same process in independent experiments, in both cases the result was a 50:50 mixture of syn- and anti-stereochemistries. This indicates that there is cleavage of the carbon-mercury bond and sufficient lifetime for the resulting intermediate to undergo bond rotation before carbon-hydrogen bond formation. For other substrates, C-C bond rearrangements were observed.

The authors propose that hydride displaces acetate at mercury. This is followed by homolytic C-Hg bond cleavage, leaving a carbon centered radical. Finally, hydrogen atom abstraction from the mercury-hydride gives the reduced product and elemental mercury.

Thank you for this. I wanted to make sure that in the last step, is the carbon radical abstracting the hydrogen from mercury so that the D-Hg bond is split homolytically and the D with its one electron then combines with the C radical? And in my example, the hydride is coming from the BH4+ but how is the B(OH)4 formed? Thanks

@Joe After borohydride delivers hydride, a trivalent boron species will be formed. This is Lewis acidic and can accept electrons from Lewis basic water. As boron continues to release hydrides, additional water molecules will add to boron, eventually getting to the maximum of four. If this is unclear, I can add it to my answer.

@Joe with respect to the final deuteration step, you are interpreting that correctly.

I managed to do a scribble for the reation mechanism.
This is an example of an alkene I took from my book, to explain:

The mechanism of the oxymercuration of an alkene to yield an alcohol comprises 3 main steps.
The first two steps you're familiar with, as you said in question but here are just for reference.

In (1), the electrophilic addition of $\ce{Hg2+}$ gives a mercurium ion.
In the second step, the $\ce{H2O}$ molecule attacks the more substituted carbon (bearing a greater share of the positive charge) forming an organomercury compound
In step (3), $\ce{NaBH4}$  cleaves the C-Hg bond. I believe this occurs through a hydride shift $\ce{:H-}$

Hope this helps

This may be close but I am certain the mechanism involves a radical intermediate. Thanks


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Parts Per Million
I am trying to calculate how many milliliters of a solution I cannot exceed to create $\pu{200ppm}$ in the air of isopropyl alcohol. The $\pu{26ml}$ bottle of solution is 70% isopropyl alcohol. Using the density ($\pu{0.786 g/ml}$) and molar mass ($\pu{60.10g/mol}$), I calculate a total of $\pu{0.24 mol}$ of isopropyl alcohol in $\pu{26ml}$ bottle of solution.
Now, I use the ideal gas law to determine the number of moles in the room. Using $\pu{P= 30 inHg}$, $\pu{V= 68.94m3}$, $\pu{T= 18.33°C}$, I get a number of moles of air in the room at $\pu{2889.79 mol}$.
First, am I doing the calculations right and second, where do I go from here? I am not factoring in A/C units, people walking in the room, or anything to change the air in the room.

I think you made an arithmetic error or typo on the moles of IPA you have in the question, I got .34 moles. Remember that "ppm" stands for parts per million, you can figure out the ppm level you have once you have the amount of IPA and amount of air in the desired units.

Here is my train of thought: 26ml x .7= 18.2. I am using the p=m/v where p is .786 (density), mass is unknown and v is 18.2. Once you work out the math, you have mass as 14.30 grams. Then, I am using molar mass (60.10 grams/mole) and canceling out grams to get .23 moles of isopropyl alcohol in 26ml of solution.

I whiffed that one, I skipped over the "70% IPA" in your description. So you have the right number of mols. Now that you have mols of IPA and mols of air molecules, you can setup the ratio and figure out what ppm of IPA you will have in the air (in terms of mols). Depending on how accurate you want this, you will need to find other ways to quantify the amount of air you have in the room. If you can work in terms of volumes, you can just use the room volume; if you can work in terms of mass, you can find charts for air density as a function of temperature.

Welcome to Chem SE! Take a [tour] of our site, and visit the [help] to learn how this site works. Good luck.


text:
Why does NO have a residence time of 4 days?
If $\ce{NO}$ is unstable, why does it have an atmospheric lifetime of 4 days? I know that it has a lone pair and one unpaired electron making it highly reactive. Therefore, shouldn't it react immediately in the atmosphere?

Before you ask "why", ask "if".

Four days seems long, and there are other variables involved, mainly ozone concentration as NO + O3 is a major sink process for NO. Could you give a source for your stated value? Thx.

It could be possible that you're thinking of the overall lifetime of $\ce{NO_x}$ ($\ce{NO2 + NO}$), which indeed does have a longer atmospheric lifetime than $\ce{NO}$ individually and which even, at around 5 km altitude, has a lifetime of ~4 days. The two interconvert during the day due to photo-chemistry, and the ratio of the two is dependent on the photolysis rate of $\ce{NO2}$ ($j_{\ce{NO2}}$) and on the rate coefficient of the reaction between $\ce{NO}$ and $\ce{O3}$.
As $j_{\ce{NO2}}$ is constant with altitude, the ratio between $\ce{NO}$ and $\ce{NO2}$ is dependent upon the reaction rate coefficient between $\ce{NO}$ and $\ce{O3}$. This, in turn, is governed by the temperature, and under lower temperatures, the reaction rate is slower. Combined with a decrease in the number concentration of ozone (due to the decrease in the number concentration of air with increasing altitude, if we assume a constant mixing ratio of ozone throughout the troposphere, of course), the ratio favours $\ce{NO}$ over $\ce{NO2}$.
At higher altitudes, then, you do see a situation where there is more $\ce{NO}$ than $\ce{NO2}$, and where the overall lifetime of $\ce{NO_x}$ can be higher than the 1 to 2 days found at the surface (reaching, sometimes, almost 12 days in the upper troposphere).


text:
Oxidation and reduction in organic chemistry
Why is it that in order to make an carboxylic acid from an alcohol you oxidise it twice...
ie alcohol + 2[O] ---> carboxylic acid
...but to get back to the alcohol you reduce it four times?
carboxylic acid + 4[H] ---> alcohol
My understanding was that oxidation is the opposite of reduction, and in organic chemistry [O] means addition of an oxygen atom or removal of a hydrogen atom, with [H] meaning the opposite.

One easy way to rationalize the relationship is to consider the $\ce{H2O}$ molecule where an oxygen atom can oxidize two hydrogen atoms.

Addition of oxygen and hydrogen can be useful to predict oxidation and reduction in most of the cases but this criteria fails to classify a reaction as oxidation or reduction when there are no oxygen or hydrogen atoms involved. This method is very primitive.
A better perspective :
Oxidation : Increase in oxidation state (by loss of electrons) of central atom.
Reduction : Decrease in oxidation state (by gain of electrons) of central atom.
If you already know how to find the oxidation state of the central atom (in your case carbon), then I hope you will be able to get your answer now.
The answer to the question why oxidation of alcohol to carboxylic acid takes 2 oxidation reactions, lies in the reaction mechanism. These mechanisms cannot be challenged, they are what they are. We can only observe and provide a acceptable explanation to why they happen like the way they do.

in reaction dont see how much O or H you are using.Just see how your derired substance is oxidized.During first oxidation to carbonyl alcohol loses hydrogen in case of methanol 1 hydroge.And nest on to convert to acid adds one oxygen.
Similar is the case whil reducing acid first adds remove one Oygen atom and then in carvonyl to alcohol adds one H atom.

This is kind of a mess and not very coherent.  Can you clean it up a bit?


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Equlibrium for a non-spontaneous reaction
Consider a reversible reaction $\ce{A -> B}$, at standard state wher $\ce{A}$ and $\ce{B}$ are at $\pu{1 atm}$. If the free energy of products is greater than the free energy of reactants, $\Delta_{\text{R}} G^\circ > 0$, and since $Q = 1$, the overall free energy for this reaction is $\Delta_{\text{R}} G^\circ > 0$ and this tells me the reaction is non-spontaneous. Yet, being at standard state does not mean that the reaction is at equilibrium, in fact $Q = 1$ could be greater or less than $K$, the products are not at equilibrium, and hence amount of $\ce{A}$ and $\ce{B}$ will adjust to a point where $Q = K$. Wouldn't that have just mean that the reaction is not at equilibrium and hence would spontaneously adjust to reach equilibrium. But yet $\Delta_{\text{R}} G^\circ$ tells me that it would not be spontaneous. Wouldn't this two concepts be conflicting against each other?

Nonspontaneous does not mean will not happen. Keep in mind that the macroscopic picture results from a very simple probabilistic model that governs how the individual molecules behave. Since the underlying model is probabilistic, even nonspontaneous processes can happen.

Related: https://chemistry.stackexchange.com/questions/43258/spontaneous-non-spontaneous-reactions-and-reversible-reactions

Yet, being at standard state does not mean that the reaction is at equilibrium, in fact Q=1 could be greater or less than K.

$\Delta_{\text{R}} G^\circ $ and K are linked. If $\Delta_{\text{R}} G^\circ $ is positive, K will be less than one. Consequently, Q > K and the reaction will go backwards.

But yet $\Delta_{\text{R}} G^\circ $ tells me that it would not be spontaneous. Wouldn't this two concepts be conflicting against each other?

No, there is no conflict. If the reaction is spontaneous, equilibrium is reached in the forward direction. If the reaction is non-spontaneous, equilibrium is reached in the reverse direction.
These rules only work for starting with concentrations at standard state. No matter what the value of the Gibbs energy of reaction, you can always choose an initial state such that the reaction will go forward or reverse. In some cases, however, it will very quickly reach equilibrium, so there is very little reaction.

If delta G is positive, that means if you start a reaction with both A and B in a container already and each has a partial pressure of 1 atm, the reaction is not at equilibrium. forward reaction is not spontaneous. Reverse reaction is. So it means the reaction will start with B changes to A to reach equilibrium.  At equilibrium, B partial pressure will be greater than 1 and A partial pressure will be less than 1 atm. This reaction has a k less than 1. It means this reaction favors the reactants.
Most kids think that a non spontaneous reaction means if I start with just A in the container, A won’t proceed to change to B. As you learn from equilibrium chapter, if you start with just A and no B, the forward reaction will proceed to change to B. The delta G (products free energy - reactants free energy) at this condition is negative since there is no products yet so The delta G at standard state is for when A and B both are 1 atm. Also remember if we say a reaction is nonspontaneous, we mean the forward reaction is non spontaneous, it also means the reverse reaction is automatically spontaneous.

This looks to be upside down: "At equilibrium, B partial pressure will be greater than 1 and A partial pressure will be less than 1 atm."

Note: This first part isn't actually that related to the specific scenario you are giving, but I think it's a misconception that you had that led you to ask this question. If you want a direct answer, go to the last paragraph.
Just because it is non-spontaneous, it doesn't mean that no reaction will occur. In fact, the reaction will always occur to some extent, even if only 0.001% of the particles react.
Consider this formula, which I'm sure you have seen before:  $$\Delta G=-RT\ce{ln}K$$
This makes sense because if the reaction is spontaneous ($\Delta G<0$), $K>0$, so the forward reaction, which is spontaneous is favored over the reverse reaction; vice versa.
If we, for example, say $\Delta G = +100$ kJ, at 298 K, $K$ would equal $3.4\cdot10^{-17}$. Thus, the forward reaction pretty much completely dominates the reverse reaction, so we say that the reaction is essentially irreversible. However, the reverse reaction is still happening, to a very small extent.
If we happened to make $\Delta G = +10$ kJ, $K$ would equal only about $56$, so the reverse reaction becomes visible. The terms spontaneous and non-spontaneous are only generalizations; it doesn't mean that a non-spontaneous reaction doesn't occur at all. There will always be some particles in that substance that will overcome the activation energy and react.
In terms of your question, since the reaction $\ce{A->B}$ is non-spontaneous, the reaction $\ce{B->A}$ is actually spontaneous (think about their equilibrium constants). Thus, the reverse reaction will be favored until the system reaches equilibrium, or when $Q=K$.

The expression $\Delta G=-RT\ce{ln}K$ is incorrect- it should read $\Delta G^\circ=-RT\ce{ln}K$.


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How to form an acetal from an enol ether?

I am having trouble understanding how the protecting group (ethylene glycol) is applied to the alkene next to the ether of highlighted step above. I thought maybe TsOH protonates the alkene as seen in the mechanism below so that it forms a carbonyl but it doesn’t seem right as a carbocation should be found on the tertiary carbon. I have no idea if I am going about this correctly, any suggestions?

Tosic acid just protonates to form an oxocarbenium. Go straight from 1 to 3 without intermediate 2 (that's just a resonance for 3). From here, you should add one of the alcohols of the ethylene glycol into the activated carbonyl. Proton transfer to the methoxy and eliminate. Then add the other end. Regenerate acid.

Steps (Source: from comment made by Zhe)

Tosic acid protonates alkene to form oxocarbenium through resonance.
One of the alcohols from ethylene glycol attacks the activated carbonyl and toluenesulfonate deprotonates the oxygen.
Methoxy group is protonated by the regenerated torsic acid.
Methanol is eliminated via SN2 with the other hydroxyl group of the ethylene glycol.
Tosic acid is regenerated by deprotonating the 1,3-dioxolane group.


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Island of Stability - Why not smash massive elements?
Why haven't scientists (or at least any reports of scientists) smash really heavy stable (or stable-enough) massive elements to produce heavier elements in the proposed island of stability?
Not being specific or accurate but the idea starts with something like this:

In late 1998, Polish physicist Robert Smolańczuk ... suggested that it
  might be possible to make oganesson by fusing lead with krypton

Why not smash lead with Thorium or Thorium with Fermium? Ignore the particle count, the main idea of the question is fairly layman.
There is also the obvious possibility that this has been done by the shadow society for their continued personal gain since there have been numerous leaks about the technology that they withhold from the rest of society (we're well beyond the 1950's technology we're "allowed" to toy with) such as free-energy-generation (not green tech, literal more energy output than input). So presuming that very high level elements are stable (even if under only certain conditions) what possible technological applications could they have that would clearly be intentionally withheld from public disclosure?

Before you ask "why", ask "if".

Is there a reason you put the conspiracy rant there, and then rolled it back when someone removed it?  All it really does is let people know that not only are they going to need a layman's argument, they're going to need one which is even simpler and harder to misinterpret than a typical layman's argument.  You might even be able to undo some of the downvotes.

Smashing two very heavy nuclei requires a lot of energy to overcome the electrostatic repulsion, think of trying to play golf where the cup is on top of a high hill.  Fortunately real golf holes are not designed that way.
And even if you get the ball in the hole, you still may not get what you want.  The probability of a nuclear collision making a specific product, measured as "cross-section", is in general very low for superheavy nuclei.  There is a lot of energy in the system (partly from what you have to put in to get atop the electrostatic hill) and there are many ways to redistribute this energy, of which your desired product is only one.
In superheavy element synthesis, a lot of failures per success is par for the course.  Physicists put a lot of work up front to identify the most probable path to success in the presence of all the hazards, and "smashing lead into thorium" is not among them.

Well, that's a lot of talking, but not a real answer. Simply we don't have nuclei with appropriate neutron count, that's it.

@Mithoron In short the bigger the nucleus the less stable the atom. The reason calcium 48 is used is because it is fairly light and so easier to accelerate to the required speeds, has lots of neutrons and reactions are more successful than when using other elements. The chances of synthesising super heavy elements would be too slim otherwise.


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Is it a coincidence that pKw = 14.00 at SATP?
High school chemistry student here. For our purposes in class for studying acids and bases, we are taught that $\text{p}K_\text{w}=14.00$, which is central for many of our calculations. I assume this is the value for room temperature and standard pressure.
However, it seems awfully suspicious that the negative $\log$ of an equilibrium constant should turn out to be a whole number. Is this just a useful coincidence? Or have scientists pulled some clever tricks, ex. by defining "room temperature" to be the temperature such that $\text{p}K_\text{w}=14.0000...$?

Pure coincidence. https://en.wikipedia.org/wiki/Self-ionization_of_water#Dependence_on_temperature.2C_pressure_and_ionic_strength

@Zhe I don't think it's exactly $1.0 \times 10^{-14}$

@PrittBalagopal That's exactly what the link indicates. I'm not claiming that it is. Though I can see my wording is not great here...


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Calculating internal energy of methane
I've recently found out my calculated value of internal energy of methane largely deviates from the ab initio output (at $\pu{1000 K}$ $0.058~\text{Hartree} = \pu{152.3 kJ mol-1}$). I used HF method, I know it's not highly-accurate, but I just need what internal energy should roughly be.
The way how I calculate:
$$H = U + PV = U + Nk_\mathrm{B}T$$
where $U$ is the internal energy, $H$ is the enthalpy.
In this case, $\Delta_\mathrm{f}H = \pu{-74.9 kJ mol-1}$ (I found the enthalpy value of methane from heat of formation table on Wikipedia).
So, I simply substitute the temperature $T = \pu{1000 K}$ into the equation and I got $\pu{-83.2 kJ mol-1}$. First, the internal energy of a system cannot be negative. Second, even it's positive, why it deviates that much from ab initio calculations? Can someone give me a hint?

I searched but I couldn't find about " ab initio output" what is it ?

It's from Gaussian09 calculation, by using HF/6-31G*.

I believe the answer to your question is summarized well by the documentation of Psi4, an open source quantum chemistry package which can do similar calculations. I am fairly certain this is the problem you're running into but I could be wrong.

It is important to know that PSI4, like any other quantum chemistry program, does not compute the usual enthalpies, entropies, or Gibbs free energies of formation provided by most reference books. Instead, quantum chemistry programs compute “absolute” thermodynamic properties relative to infinitely separated nuclei and electrons, not “formation” values relative to elements in their standard states. If you are computing thermodynamic differences, like a reaction enthalpy computed as the enthalpy of the products minus the enthalpy of the reactants, then these “absolute” enthalpies are perfectly valid and usable. However, they cannot be mixed and matched with enthalpies of formation from reference books, since the zero of energy is not the same. Additionally, the “thermal energies” reported in kcal/mol are the finite-temperature corrections to the electronic total energy, and not the overall thermal energies themselves. If in doubt, use the reported Total Energies in Hartree/particle.

Link to documentation
In other words, you're value may be perfectly reasonable once you use it to calculate a change in energy, but as an absolute energy both values are fairly meaningless because they both pick a reasonable, but arbitrary, zero of energy.
This is also why you're internal energy can be negative. This simply means it has a smaller internal energy than whatever the reference internal energy is. I didn't check your work, but it is possible to have a negative energy depending on the reference value.

Also, here is the very thorough documentation (with examples!) from Gaussian. They also note,

This is generally not the same as the output from Gaussian for a calculation on an isolated gas-phase atom. These values are referenced to the standard states of the elements. For example, the value for hydrogen atom is 1.01 kcal/mol. This is ($H_{H_2}^{\circ}(298K)- H_{H_2}^{\circ}(0K))/2$, and not ($H_{H}^{\circ}(298K)- H_{H}^{\circ}(0K))/2$ , which is what Gaussian calculates.

So you're going to have the same problem. Read through there and I think it tells you how to make the correction. I think you just need to inclue some zero-point energies.


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Mercaptan Detection
I have no chemical knowledge so please forgive me, I am working on a project to replicate the Ethanethiol smell that is in natural gas. With Ethanethiol liquid 99% how can this be presented to create the same smell as natural gas. I believe 0.5 on the sales scale.

Hello and Welcome to Chem SE. Take a [tour] of our site. Visit the [help] to learn more about how this site works. Good luck!

As natural gas (normally) is odourless, I understand your question as to how much of ethanethiol per unit of volume is necessary to replicate the alarming odour.  Be careful if you want to work with these materials, for example place a bowl underneath your equipment to limit spill, use a well ventilated fume hood, wear proper safety gear and be ready to rinse the equipment with bleach prior to further cleansing (glass ware) or waste disposal (lab gants).
According to this overview, ethanethiol is not the only mercaptan deployed in industry for this purpose, and sometimes these thioethers are blend with analogues or sulfur free compounds.  CFR 29 1910 (b) 1 mentioned here (page 2) requires 1 pound ($\pu{0.45 kg}$) of ethanethiol per 10000 gallons ($\pu{37.8 m^3}$) of liquified natural gas (LPG).
In NCBI's library, a dedicated entry for this compound provides some spread of "discomfort" if ethanethiol is present in air at a concentration of $4$ to $\pu{112 ppm}$ (or, $\pu{mg/m^3}$), and mentions as "level of distinct odor awareness (LOA) [...] $\pu{1.4E-4 ppm}$".


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What proof is there that the "Island of Stability" exists?
Often I see people talking about an "Island of Stability" existing for super-heavy elements on the periodic table, but I don't see any reason to thinks this exists? Surely the heavier you go the more unstable nuclei get? What, if any, proof or theory is there behind expecting an "Island of Stability" to exist?

The theory is that certain numbers of protons and neutrons are "better" than the others. As for the proof, there is none. The island may just as well not be there, or be "completely underwater".

People often misread "theory" as "proof"...

related https://chemistry.stackexchange.com/questions/21848/


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Reduction of a ketone to an alkane
What is the mechanism for the reduction of ketone to alkanes using $\ce{H2}$, $\ce{Pd/C}$ in ethanol? Searching on Google mostly yields the reduction of ketones to secondary alcohols using LAH.

Is there acid present?

For the reduction of ketone to alkane, no.

The ketone is reduced to an alcohol first.  I suggest looking up benzyl alcohol reduction to toluene.

Actually, you can convert ketones to alkanes, though not realistically with just palladium on carbon (It would probably only give the alcohol product). You could convert it to a dithiane using a dithiol and then reducing it with Raney Nickel.

The reduction of a ketone all the way down to an alkane using catalytic hydrogenation over palladium on charcoal is not a general/widely used process.
Quoting from Comprehensive Organic Functional Group Transformations III:

...while the reduction of benzylic ketones by hydrogenation over palladium is often an efficient procedure , it ... sees limited use. An isolated report of the reduction of aliphatic ketones by hydrogenation over platinum on Montmorillonite K-10 merits ...attention here and further study

The majority of examples of this reaction occur using benzylic ketones, carrying out the reaction in the presence of aqueous acid in order to facilitate the loss of the hydroxyl caused upon initial reduction of the ketone. The adjacent aromatic ring helps stabilise the developing (+) charge as the C-O bond cleaves, which is what makes benzylic ketones good substrates for this reaction.

Ref: J. Med. Chem. 2001, 44, 3424

In the context of lab scale chemistry (i.e. not industrial processes), several good methods exist including the Wolff–Kishner (or variants as Jan pointed out in the comments), the Clemmensen, and the Mozingo reduction. Some success is also noted via initial reduction, mesyaltion, and treatment with a hydride source.

Plus Wolff-Kishner variants such as Huang-Minlon.

Out of all the methods discussed, catalytic hydrogenation is the most efficient and practised at tonne scale. Acidic conditions, as mentioned, are best.

@Beerhunter, are they under conditions applicable to the lab? Or is it a case of high temperature, high high pressure, strong acid on simple feedstocks?

Pressure obviously but on tetracycline antibiotics with HCl or methanesulfonic acid,  less than 100degC.  Pressure depends on metal and substrate


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Van't Hoff factor for citric and phosphoric acids
I am trying to determine the van't Hoff Factor for citric and phosphoric acids.
When phosphoric acid is in aqueous solution, it dissociates as follows:
$$\ce{H3PO4 ->H2PO4- + H3O+}$$
Does this mean its van't Hoff factor is 2? For citric acid I cannot find what it dissociates into anywhere, so I am unable to determine the van't Hoff factor.

No. Phosphoric acid does not technically meet the arbitrary definition of a strong acid, and the dissociation is not as high as for something like hydrochloric acid. The van't Hoff factor will still be fairly high, but not 2. Ionization of the other species will be relatively low though they in principle will have an effect on the van't Hoff factor. The same idea applies to citric acid.

Thank you, but is there as way I can get the actual van't hoff factor or at least a number that is close to it?

You can compute from the pKa.

The Van't Hoff factor isn't an intrinsic property of any species - it depends on the dissociation degree of an electrolyte. Both of those acids are weak and triprotic, meaning they can lose up to three protons in solution (depending on the pH). Therefore, the Van't Hoff factor will likewise change depending on the pH.
1) If the acid is the only species in solution, and since it's a weak acid, you need to use the same formula as for any weak electrolyte:
$i=1+\alpha$
You can figure out the dissociation factor, $\alpha$, from the dissociation constant:
$K_1=\cfrac{[H_2A^-][H^+]}{[H_3A]}=\cfrac{(\alpha C_{H_3A})^2}{(1-\alpha)C_{H_3A}}=\cfrac{\alpha^2}{1-\alpha}C_{H_3A}$
where $C_{H_3A}$ is the formal concentration of the acid.
2) If you have more species in solution, it will depend on the pH of the solution. In general, you'll have this expression:
$i=\cfrac{[H_3A]+[H_2A^-]+[HA^{2-}]+[A^{3-}]+[H^+]}{C_{H_3A}}$
After some tedious substitutions you'll get
$i=1+\cfrac{K_1}{[H^+]+K_1}+\cfrac{K_2}{[H^+]+K_2}+\cfrac{K_3}{[H^+]+K_3}$
Some relevant values:
$pH=0 \rightarrow i \approx 1$
$pH=pK_1 \rightarrow i = 1.5$
$pH=pK_2 \rightarrow i = 2.5$
$pH=pK_3 \rightarrow i = 3.5$
$pH=14 \rightarrow i \approx 4$
and it changes logarithmically in between.


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Why rotate orbitals in a CASSCF calculation?
I'm trying to get the rate constant for the reaction of propene to give propenyl (radical) and H.
Using Molpro, I calculated the energy for the optimized geometry of the reactant using CASSCF.
It is required at one point to rotate two orbitals (the one with energy just below the HOMO and one in the valence shell).
I don't understand the point in doing this; what's the purpose?

Are you talking about orthogonalization of orbitals?

No, it's the rotate command, it's used to swap two orbitals at different energy and, if you are considering them in the active space they have different population too.

If I understand you correctly, you are swapping two occupied orbitals, is that right?
What is the symmetry of the rotated orbitals? Is that relevant at all?

I might guess that you are having issues during orbital optimization and the rotating of two orbitals is a convergence trick.

The reason for rotation of orbitals (really the reordering of orbitals) is that the starting orbitals are usually not in the correct order for any sort of MCSCF calculation (CASSCF, RASSCF, GASSCF, adding correlation, ...). Doing a single CASSCF calculation with no "preparation" beforehand should force most programs to start with Hartree-Fock-like orbitals; this is the reference wavefunction $\Psi_0$. Hartree-Fock includes almost no static correlation to speak of, which can manifest in an incorrect ordering of the orbitals when compared to MOs that do have static correlation accounted for.
From here:

Reorder your orbitals if necessary before running the MCSCF. Except for the most trivial cases, the orbitals that belong in the active space are not the orbitals that come straight out of your single-reference calculation. If they were, then you might not be running MCSCF to begin with.

Poor convergence of an MCSCF procedure is a good sign that the active space orbitals are not the correct ones. It is not a matter of the ordering within a space, since the energy is invariant to rotation/swapping within a space (inactive/frozen, active, or secondary/unoccupied). It is that an orbital Hartree-Fock predicts as a frozen core orbital should be a valence-type orbital in the active space, etc.
One of the reasons that MCSCF calculations are multi-step calculations and shouldn't be done in one shot is the generation of (hopefully) better orbitals than HF ones, inspection and any necessary rotation, then the MCSCF calculation. The best starting orbitals for MCSCF are other MCSCF orbitals, but you probably don't have those. Usual choices are MP2 or CI natural orbitals, which gives an indication of the total size of the active space and adds some dynamic correlation to "patch up" the lack of any real correlation in HF orbitals.
In my limited experience, HF orbitals are a fine starting point as long as the order is correct.

I'm aware of two reasons why this is done, and both are ways to go around a convention in the implementation of CASSCF that is shared by all the codes I know.
In general, the active space in CASSCF is comprised by a contiguous band of orbitals (in energy terms), which includes the $\mathrm{N}$ highest occupied orbitals and the $\mathrm{N'}$ lowest unoccupied orbitals. In ORCA (and, for what I could tell from the documentation, also in Molpro), there is no option to select which orbitals are included to generate the CAS - you can only select how many. In other words, if a relevant orbital is lower (if occupied) or higher (if empty) in energy than orbitals you are not interested in, you are stuck with either adding these uninteresting orbitals to the CAS as well (computationally expensive) or rotating orbitals so that you artificially swap their energies and "trick" the code into using them.
I've seen this used to build a symmetrically consistent CAS (if you are only interested in orbitals with a certain molecular symmetry) and in systems where small energy gaps mean that MO ordering can change throughout, say, a trajectory - to ensure that your CAS is always composed of the relevant orbitals. I can't offer deeper theoretical justification as I'm not a CASSCF expert.

Eventually I had an answer by the professor so, here's the full story:
As I said I had to study a reaction with formation of two radicals.
Performing a first calculation with two active orbitals I obtain an estimation of the energy of the molecule and occupation of HOMO and LUMO.
Having done this, I proceed in elongating the bond I want to break, until an imaginary frequency is detected, indicating the presence of a saddle point in my PES. Each calculation performed will have an inactive variable (to avoid optimization of geometry) and selecting the same active space, it will converge to the previous solution found.
The purpose of swapping two occupied orbitals is to highlight a delocalization of the sigma orbital between C and H that's the one that will pass from a double occupation to a single occupation.
This sigma orbital  usually has an energy much lower than the HOMO, so in the first CASSCF calculation is computed as a closed orbital.
Using the command rotate I perform again the calculation this time expecting that the sigma orbital will be the HOMO instead of the pi bond of propene.
This answer is not yet backed by theory, even if I guess pentavalentcarbon was pretty accurate in describing the process of multi-reference calculations.
Thank you all!


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How to clean a copper pan from black burned fat chemically?
I have several copper pans which I use daily. I don’t have the time to always clean them. Now they have a thick crust of black burned fat. When I wash it with dish soda, nothing changes. And I don’t want to use abrasive mechanics (steel sponge), because I am afraid this will scratch the copper surface.
Is there a chemical trick to dissolve the black crust? What is that black crust chemically and why is it so strong?

The crust is a giant mess of hydrocarbons that have been burned, caramelized, and congealed into the surface of the pan.
I would recommend that you try the steel sponge to get some of the crust off, you may find that as you start breaking into the top layers of crust that it will fall apart. Another strategy would be to put the pan in the oven to the highest heat that is safe for the cookware and this might cause the pan to expand enough to create cracks in the crust or release it from the micro-cracks that were already in the pan surface. As a last resort you will have to chisel it off slowly.

Aren’t hydrocarbons dissolved by any acid or the like?

You could use white vinegar (weak acid) and salt as an abrasive to help with the scrubbing. You could try a commercial copper cookware cleaner (typical composition is of weak acids and long chain hydrocarbons). I wouldn't break out the lab grade chemicals because the cookware probably couldn't handle it. If you used a dirty pan multiple times, you'll have to use a little elbow grease as your detergent!

Sprinkle a teaspoon or two of baking soda (bicarbonate of soda) on the surface, cover with water and boil to dryness.
Add fresh water and repeat a couple of times but do not heat after the last addition of water.
Then try to scrub off when still wet and warm.
Repeat for harder stains.
If this fails try washing soda (sodium carbonate) and with the same process.
If that fails try letting it soak with water and caustic soda (sodium hydroxide, drain cleaner, heating is not recommended however the water gets warm when you dissolve the caustic soda, use gloves and glasses), watch the progress as it will start to etch the copper.


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Will KOH work in place of NaOH as the electolyte in the electrolysis of water?
I am making a hydrogen and oxygen generator by electrolysis water. I found out that using salt as an electrolyte creates a compound of chlorine and oxygen with my electrodes clouding up the water and using all of the oxygen from the reaction. Bubbles are coming off of only one electrode. In my research I found that sodium hydroxide is a better electrolyte because it doesn't create chlorine gas and shouldn't react with my electrodes. Unfortunately I don't have any, but I am making lye ($\ce{KOH}$) from some hardwood ashes, will this work in place of $\ce{NaOH}$? I'm not sure if the potassium ions will conduct electricity in the same way. Thanks!

Yes, they should work the same.

There is a difference in conductivity between NaOH and KOH due to $\ce{Na+} $ having more spheres of hydration. I would look into conductivity values for the concentration solution you plan to use and see what the difference in conductivity is and determine if it will be an issue.

You will get a number of salts when you extract your ashes.  Fractional crystallisation could be used to isolate them if it becomes a problem but you need to accept that it is not just one salt in wood fire ash.

Using $\ce{KOH}$ rather than $\ce{NaOH}$ will work just fine.
Solutions of $\ce{KOH}$ are actually about $\pu{5\%}$ more conductive than solutions of $\ce{NaOH}$ at the same molar concentration. In terms of mass concentration, a given mass of $\ce{KOH}$ will have about $\pu{30\%}$ fewer equivalents of reactant than the same mass of $\ce{NaOH}$ because of the difference in molecular weights.
The combination of the difference in molar conductivity and molecular weights between $\ce{NaOH}$ and $\ce{KOH}$ just means that a given concentration of $\ce{KOH}$ should be about $\pu{35\%}$ less conductive than the same mass of $\ce{NaOH}$. Of course you can increase the concentration of $\ce{KOH}$ relative to the amount of $\ce{NaOH}$ you had planned to use if you like.
Please don't hesitate to ask for any clarifications in the comments below, and best of luck!

Do you know what happens to the hydroxide ions from KOH in the water? Do they break down during electolysis?

They will be oxidized to H+ and oxygen gas.

So would the % of H and O gas produced change to 3 to 2 instead of 2 to 1?

The ratio of H2 to O2 gas will still be 2:1. See this article.

Not sure where exactly it says that. I see the electrolyte section KOH is suitable.


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Complete Mechanism for Oxidation of an Alkene to a Glycol
My organic book does not write out the full mechanism for the reaction of an alkene with osmium tetroxide in a basic aqueous solution with sodium bisulfate. I have made an attempt at a mechanism and am looking for a confirmation or a correction to it. Thanks and I appreciate the help.
I am concerned because the mechanism I wrote creates hydrogen gas and I have not done the reaction so I don't know if a gas evolves.

We generally don't write detailed arrow pushing mechanisms for the details that occur on a transition metal because arrow pushing mechanisms can fail to describe what is going on.

In other words, don't try to do too much with the osmium. Focus on what happens to the organic component instead.

I agree with @Zhe. Whatever happens to the osmium compound doesnt matter, we are concerned with the organic molecules.


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Why does a smaller bandgap correlate to increased absorption of visible light?
I am currently involved in a project on metal organic frameworks in my AP chemistry class, and specifically their applications in $\ce{CO2}$ reduction into formate. All the explanations we can find on why a smaller bandgap correlates to higher absorption and quantum yields refers to PhD-level chemistry, e.g. the Density Functional Theory. Can someone provide a simplified version of this?

In your case, absorption takes an electron from a valence band and in to a conduction band. At energies below the narrowest gap between the valence and conduction band, the photon can't give enough energy to an electron to get it up from the valence to the conduction band - there is no absorption there.
Right at the band gap, you can now just get a small population of electrons from the valence to the conduction band, those right at the top of the valence band. Absorption starts to kick in.
As the photon energy increases, there is a larger and larger number of valence band states that can now be excited in to the conduction band.
Depending on details of the shapes of the conduction and valence bands (density of states of initial and final vs energy), and on the exact absorption process (if you need a phonon for momentum matching it is harder), you get various expressions relating the absorption to the band gap ranging from $\sqrt E$ to $E^{2}$ or higher powers.