text:
Is my bottle of "RooR"-brand glass cleaning liquid pure isopropyl alcohol or something else?
I have this exact bottle: https://www.grasscity.eu/media/catalog/product/cache/2b55fa971a29a7844993ad77b24950ac/i/m/img_3633_1_1.jpg
I originally purchased it to clean my vaporizer (I've since stopped doing that and thrown all the equipment away), and now I'm wondering if this liquid is "isopropyl alcohol", because I want to use it to clean video game cartridges' connectors/internals with cotton swabs.
The text is 100% in German, and does not mention "isopropyl" or "alcohol" anywhere. It has no ingredients listed anywhere. It's described as "biological glass cleaning". The store from which I bought it described it as "pure isopropyl alcohol", but who knows if they are lying/confused?
So, is this actually pure isopropyl alcohol? And can it thus safely be used for my cartridge/hardware cleaning purposes? Or is it some sort of weird proprietary mix which is only to be used for cleaning glass and causes horrible damage to anything else?
I'm surprised that they would not be legally required to have some sort of "list of ingredients". I thought that was definitely a law in Germany, like every other country I know of. And even if not a law, it seems reasonable to mention "isopropyl" or at least "alcohol" somewhere on the bottle. If it indeed is 100% pure isopropyl alcohol, you'd think they would want to brag about this and make a big deal of it.
It was very expensive.

It's impossible to tell from the image, why don't you edit your question to add the text ?

@porphyrin What do you mean? The text doesn't say anything useful beyond what I've already described. The "model number" thing doesn't yield any search results.

One thing that they can't (hopefully) skip is a clear sign showing flammability. If there is no warning about flammability then the product isn't just alcohol (in the sense that the alcohol conc is so low to compare with, say, vodka, and assuming the product is alcohol based). Perhaps there is not alcohol at all. Or there is just ethanol and much water. Avoid biological cleaners at least to replace alcohol because it is a relatively expensive non sense.


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A contradiction in the nucleophilic attack of hydride ion based on charge densities
On page 130 of Organic Chemistry by Clayden, it is stated that

Nucleophilic attack by the hydride ion, $\ce{H-}$, is an almost unknown reaction. This species, which is present in the salt sodium hydride, NaH, has such a high charge density that it only ever reacts as a base. The reason is that its filled 1s orbital is of an ideal size to interact with the hydrogen l atom’s contribution to the $\sigma ^*$orbital of an H–X bond (X can be any atom), but much too small to interact easily with carbon’s more diffuse 2p orbital contribution to the LUMO
(π* of the C=O group.)"

We may draw a direct proportionality between molecular radius and charge density.
But according to radii values, Inorganic Chemistry: Principles of structure and Reactivity states,

it has a radius of 208 pm compared to 216 pm for the iodide ion

Calculating the charge density, as
$Q_\mathrm{density} = \frac{[\mathrm{Charge}]}{\mathrm{[Volume]}} = \frac{[1.6\times 10^{-19}]}{[4/3π  \times 208^3]} $ (nearly equal to iodide which has low charge density.)
But in the next few lines it is written that the charge density is sufficient to interact with the σ* orbital of H in H-X molecule, but is too small to interact with 2p orbitals of carbon.
The two appear contradictory as on one hand size of hydride is big and soft, while on the other it cannot react with C as "as carbon has too diffused 2p orbitals."
Please explain the contradiction.

See this question for why the ionic radius of the hydride ion has been historically overestimated: https://chemistry.stackexchange.com/questions/125814/radius-of-hydride-h%e2%81%bb-ion?rq=1

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Proton removal during benzyne formation in substituted haloarenes
I encountered the following question in an organic chemistry paper -

The reaction follows benzyne mechanism. My answer was B.
However, the answer given in the test solutions was D. The mechanism given was as follows:

My question: The tert-butoxide being a bulky base is more likely to abstract the least hindered hydrogen, which is clearly present at the para-position to $\ce{-OCH3}$. Then the benzyne triple bond would form between the meta and para carbons, following which product (B) would be obtained. However, the given mechanism involves abstraction of the hydrogen from the ortho-position, which has two large groups on either side, and hence shouldn't occur.
Is my reasoning right? Are there any other factors affecting the reaction that decides which hydrogen will be removed?
I am also open to the possibility of the test solution's mechanism being wrong.

I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (like $\ce{OCH3}$ here), or accept a lone pair from the aromatic nucleus (like $\ce{NO2}$ would do) is not relevant in this case. However, inductive effects of the substituents in aromatic nucleus are still relevant (i.e. electron-withdrawing effects that occur through single $\sigma$-bonds).
However, $\ce{t-BuO-}$ is not as strong a base as $\ce{H2N-}$ in usual cases of benzyne reactions. Thus, it would abstract relatively most acidic available proton from to aromatic nucleus. Compared to para-position, ortho-position is more acidic due to proximity (ortho is one $\ce{C-C}$ bond away compared to para which is three $\ce{C-C}$ bonds away). Following is an example you can find in literature (Ref.1):

References:

José-Antonio García-López and  Michael F. Greaney, "Synthesis of biaryls using aryne intermediates," Chem. Soc. Rev. 2016, 45(24), 6766-6798 (DOI: https://doi.org/10.1039/C6CS00220J).

Would $\ce{NH2-}$ deprotonate the para-position then?

To some extend yes, but ortho is preferred. If you look at experimental values with $\ce{PhLi}$, it preferably gives ortho intermediate.

Short answer:  The $\ce{OMe}$ stabilizes the negative charge in the ortho position
From Master Organic Chemistry.com here    see Case#2 below

So how do substituents on the ring affect addition to the triple bond?

Since the aromatic pi-system is at right angles to the triple bond, what’s NOT relevant is the ability to donate a lone pair to the ring (like $\ce{OCH3}$), or accept a lone pair from it (like $\ce{NO2}$).
However, inductive effects are still relevant (i.e. electron-withdrawing effects that occur through single bonds).
Why?  Addition to the triple bond creates a negative charge on carbon, and electron-withdrawing groups stabilize negative charge: the closer, the better.
So a key principle in the addition of nucleophiles to arynes is that  addition tends to happen so as to place the negative charge closer to an electron-withdrawing substituent.

What’s interesting is that $\ce{OCH3}$ behaves like an electron-withdrawing group in these examples, since the oxygen lone pairs can’t interact with the triple bond.

Your answer is most informative, but it mainly explains the point where nucleophile (NH2 in your case, OD in my question) attack after the benzyne triple bond is formed. Which will explain why the answer can be (A) and not (B), also can be (D) but not (C).

However, I am ignorant as to whether a negative charge is formed during the benzyne formation. If it is, then of course for stability it will be at the ortho position, and your answer will be helpful in that regard also. Could you please elucidate if a negative charge temporarily appears during benzyne formation, on that carbon from where hydrogen is taken? Also, will the negative charge stability by -I of OCH3 group be sufficient to counter the steric hindrance experienced by the bulky tert-butoxide while abstracting the ortho hydrogen?

A transient -ve charge is formed by deprotonation before elimination of the chloride ion.

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Practical meaning of omega potential
In statistical thermodynamics, the quantity omega potential is defined and is different for fermions, photons, phonons and other particles .
I have been understanding the derivations of omega potential for varied particles but I feel I am not really able to physically interpret what it actually is.
Is it similar to Gibbs energy? Since there are many energy parameters like internal energy, Gibbs free energy, energy of states - I am unable to understand what exactly omega potential signifies.
I tried to search over internet but I mostly found literature which was hard to decipher and get a quick answer.
Below is the omega potential for a fermion. What exactly does this equation provide us with?
\begin{align}
\Omega^* &= \sum_i\Omega_i^* \\
&= -KT\sum_i \ln\left[1+ \exp(\frac{\mu -\epsilon_i}{\mu T})\right]
\end{align}


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How to go about deciding the acidic strength of a functional group in an aromatic compound?
The question requires me to identify the decreasing order of the acidic strength of the following functional group. But I'm afraid I don't know how to distinguish one from the other since all the hydroxyl groups would be in resonance. What would be the next step in this case? I felt like the answer should start with (i) as there is a carbonyl carbon near and the delocalisation would be stabilised.

The (ii) OH has a seperate delocalisation and is not connected to the others so it will  probably be the most acidic. Delocalisation hinders the acidity of (i). So, according to me, the order should be ii>iii>i. I may be incorrect though.

I don't understand what you mean by separate delocalization. Wouldn't (iii) OH exhibit this as well?

There are only two resonance structures for (ii) a keto and an enol. It would not be possible to draw extended delocalisation structures for ii. But the resonance of i and iii is connected. In other words two seperate electron clouds exist. An extended one for i and iii and another cloud in ii.

@Aditya Roychowdhury: The order is $iii \gt i \gt ii$. More the resonance structure of conjugaste base, more the acidity.

@MathewMahindaratne Yes I missed that.

Well, as far as I'm concerned, only important thing about structure you present is that it's a non aromatic tautomer of polyphenol, which makes all this a pointless exercise.

@Mithoron: Agreed!

This question is, at best, flawed. The most acidic protons on the structure provided are not any of the labeled OH protons. Rather, the most acidic position is the one I have drawn a rectangle around below.

Deprotonation at this position produces an anion with a resonance contributor in which both rings are aromatic. The circa 150 kJ/mol resonance stabilization for each ring will vastly outweigh any delocalization that might occur for the oxyanions formed by deprotonating any of the OH groups.


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Can chlorine show negative mesomeric effect due to presence of d orbital?
I was wondering in this problem that if chlorine will show positive mesomeric effect as it is at para-position and strength of inductive effect decreases with distance from carbon atom and chlorine can expand its octet due to presence of $\mathrm{d}$ orbitals so will this structure be more stable than toluene carbocation I mean if we remove $\ce{Cl}$ from the para-position of same carbocation?

Cloro substituent in an aryl ring has two effects combined:

I- effect, owing to its electronegative character and tendency to pull electrons towards itself from the ring and thus deactivating the ring.

M+ effect to donate electron to the ring by resonance.

Generally the I- effect is stronger than the M+ effect, leading to an overall decrease in carbocation stability and deactivation of the ring towards attack by an electrophile.
The M+ effect occurs owing to the formation of strong pπ-pπ lateral bond due to relatively similar sized, 2p orbital of carbon of ring and 3p orbital of chlorine. Since a number of lone pares occur in chlorine these may be donated to the ring by resonance.
However due to the large size and highly diffused nature of the 3d orbitals of chlorine the overlap with 2p of carbon in ring is not very efficient and therefore back bonding or reverse conjugation is not observed.
The above 4-chlorocarbocation is less stable than benzyl carbocation (formed by removing a proton from toluene).

"large size and highly diffused nature of the 3d orbitals" The biggest factor is simply that the 3d orbitals are just very high in energy. Backbonding into the 3d orbitals is just not energetically feasible.


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What is the crystal structure of ammonium hydrogen sulfate? It's beautifully birefringent
I make art using a microscope and camera. I have heated ammonium sulphate that decomposes to ammonium hydrogen sulfate, melts, and then recrystalises to produce exceptionally beautiful birefiringent colours.
I am interested in its typical crystaline structure as perhaps other chemicals with similar structure may also be interesting artistically.
I have tried looking it up in ChemSpider, but no luck. I took some university courses in chemistry, so I have some (out of date) chemistry knowledge.

I looked up “polarized light microscopy art” and got hits, like this one: https://fineartamerica.com/art/polarized+light+microscopy.

Crystal structure won't tell you much. Compounds with similar structure are not necessarily as good-looking, and moreover, the chemically related compounds do not necessarily have similar structure.

Thanks for your answers/time. I occationally sell my images although I certainly need my day job. Good to know that maybe just structure might not best guide to crystals aesthetic qualities in polarised light...I would love to be able have a more systematic way of hunting for "good" crystals and thought this might have been a way forward.

Birefrigence is a result of structural anisotropy observed for crystals other than of the cubic Bravais class. Then, interaction of vectorial properties (e.g., light) with said material is anisotropic, too. It is a phenomenon seen both for inorganic, as well as organic materials (examples listed e.g., here). As seen e.g., by cooling films of molten polymers on a heat stage in polarization microscopy, local deviations from isotropy suffice to observe birefrigence, too (example). Thus, there is no «special functional group» which, if present, suddenly would switch on / switch off a sample to be birefrigent. It's observation and measurement by polarization microscopy allows to characterize the microstructure of e.g., minerals, and thus one of the  elder tools in geology (still used today, e.g., Stoiber's «Crystal Identification with the Polarizing Microscope» by 1994 lists some in its freely available back matter).  It may be used on organic samples, too (e.g., this tutorial).
The models to describe the crystal structure about ammonium sulfate, $\ce{(NH4)HSO4}$, depend on the temperature of the sample. This might increase or decrease birefrigence (I did not look out for a temperature-dependent recording of this about ammonium sulfate).  The explored variation of sample temperature in the work below (at ambient pressure) however do not change the unit cell symmetry this much that birefrigence suddenly would be cancelled out for symmetry alone.
At room temperature ($\pu{293 K}$), the best description of this material that there two hydrogensulfate groups per unit cell which are described independently from each other:

(credit 2012JChemPhysChemA223, Fig. 1a / $\pu{293 K}$)
In the illustration, these are marked by O (ordered) and D (disordered).  This is to say that the model describes the reality best assuming positions with $\ce{HSO^-4}$ with more than one orientation and local geometry.  (Similar to sitting either forward, or backward on a chair.)  This is, taking a different perspective on the sample (the representations omit hydrogen atoms):

(modified representation, based on 1972ActaCrystA445)
with ammonium ions, ordered and disordered $\ce{HSO^-4}$, the sites with the disordered would be either one of the following:

(credit 1972ActaCrystA445)
At lower temperatures, e.g. at $\pu{200 K}$ and at $\pu{125 K}$, it is energetically more favourable if both $\ce{NH^+4}$ and $\ce{HSO^-4}$ are in ordered position and orientation only, e.g.:

(credit 2012JChemPhysChemA223, Fig. 1b / $\pu{200 K}$)
This phase transitions and changes of unit cell symmetries (space group $P2_1/c$ (at $\pu{293 K}$) $\rightarrow$ $Pc$ (at $\pu{200 K}$) $\rightarrow$ $P1$ (at $\pu{125 K}$)) however will not cause a cancellation of  birefrigence because of symmetry reasons.

You do not state (this much) why knowing the the spatial arrangement of the atoms, and hence a model of the crystal structure is important for you.  However, because you mention to possess some command of chemistry and interest art, be aware among the crystallographers there are some equally with a sense for beauty.  One of the low-level entries is IUCr's Crystal growing competition for school children (note the their brochure with examples in the later section) considerably newer than their teaching pamphlet.
On the more artistically appealing side, though, selectively dyeing crystals may be of interest for you.   Work in this field was much popularized again by e.g. Bart Kahr:

(credit, includes protocol)

(credit 2001ChemRev893)
with protocols for chemicals easier to access (e.g., potassium dihydrogen phosphate [KDP] and amaranth dye, cf. here).  Beside beauty, the selective incorporation of dyes provides insight into the growth mechanism of crystals. You may find additional inspiration in other works by him, e.g., in 1996AdvMat941, 2001ChemRev893 including a 20 page long table of examples («reported
during the past 147 years»), or 2002CrystEngComm514 about «painting crystals» by an easy-to-follow protocol.

Nice answer. But I'd like to see an explanation of the birefringence given the crystal structure.

Thanks for the time you put into your answer including the actual structure and helpful guidance for me when I want to look up further structures. The crystals of interest to me have to be interesting under a microscope through crossed polaroids. I was interested in the structure as I had hoped that structure is guide to birefrigence which it appears it is only loosely. Thanks also commenters. What would be useful (to me and other microscopists) would be catalogue of crystals that are strongly birefrigent. Minerals are interesting but petrographic sildes are hard to make.

From considering crystal symmetry, we can only say if birefringence must be zero, not how strong it may be. Consider a wave propagating along an axis about which the crystal structure has a four-fold rotation symmetry. It should be clear that there can be no birefringence, since two waves that differ only by a 90-degree rotation of polarization will each see exactly the same crystal structure. This kind of thinking shows, for example, that a cubic crystal cannot be birefringent for any direction of propagation, while an orthorhombic crystal will have one direction with zero birefringence.

Thanks for this. Certainly I have found that sodium chloride has no birefringent properties....


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Why does a potential on the reference electrode create an offset on the measure electrode of a pH probe?
I am making an analog front end for a pH probe and i encountered a design where creating an offset on the reference electrode, creates an offset on the measure electrode.
How does that work ? Does the charge go through water ?
Experience based on TI application notes : https://www.ti.com/lit/an/snoa529a/snoa529a.pdf?ts=1619613280868&ref_url=https%253A%252F%252Fwww.qwant.com%252F
(I don't want to details electronics design because it's not the point)
So here is a pH probe and my mesurement setup:
The voltmeter measure the potential described by the nernst equation.
What happens when I shift the polarity of the reference should be like that:

Now the voltmeter measure U= Nernst eq + 300 mV
But how having this polarised metal rod in the solution change the measure of the pH electrode ?
My understanding is that the charge on the electrode mooves to the water and to the other electrode ?
But the glass membrane should only let hydrons go through so it "should" be isolated from other change of ions. But it maybe let the current flow from the reference electrode to the measure electrode ?

How does this differ from what you asked over at the EE stack exchange: https://electronics.stackexchange.com/q/562549/223146? There was an answer by an extremely high rep user and you accepted that answer, which also had multiple upvotes.

I asked the question on an electronics context, and i had an unstatisfying answer: "it behaves like a battery". It has to be expected from an electronic points of view.But I want to understand the process behind it. Why does it behave like a battery ? How the potential of the ref electrode interact with the potential of the pH electrode ?

That EE answer is correct: the pH electrode and reference electrode are simply an extremely high impedance voltaic cell. It can supply only extremely small current, i.e., pA or so. The smaller the better. So offsetting the reference electrode with respect to circuit zero reference (ground) does not change the potential difference between the pH and reference electrodes. But, it allows for use of a unipolar power supply instead of a bipolar power supply, just as that answer said. You might want to look up three and four wire potentiostats for more insight.

"the pH electrode and reference electrode are simply an extremely high impedance voltaic cell." @EdV So how does that work ? My hypothesis is the AgCl is polarised positively (300mV) so it should have less electrons. So I suppose oxydation is made on the AgCl wire. Those new negative ions should go to water through the porous junction ? Therefore changing the potential of the water. Water being negatively charged, I don't see how it goes through the glass membrane tho...

This is just a level shifting issue: the pH electrode and reference electrode work as usual. The electrical circuit is referenced to electrical ground, so its output voltage is level shifted by the deliberately added 512 mV. This is not a chemistry issue and there is nothing more. Sorry, but your guesses are just off track.

@Damezumari Please don't cross-post, it is against policy. It appears to me that you have a correct answer on EE and that Ed V has contributed quite a bit to validate that.


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How is kinetic energy accounted for when describing diatomic bonds with a Morse potential?
The Morse potential can be used to describe the potential energy of the a diatomic bond:

This curve suggests that the lowest state $v = 0$ is the most stable, since it has the lowest potential energy. However, the energy of a system comprises both potential as well as kinetic energy, and stability should depend on the total energy, not just the potential energy. How does kinetic energy come into play here? Is it being neglected for some reason, or is this description of a diatomic molecule insufficient?

This is the potential energy plot; kinetic energy has to be accounted for in the usual manner, i.e., as given by the kinetic energy term in the {Hamiltonian,Lagrangian}.

The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation values of the total energy and potential energy.
Because the math is simpler I will limit the discussion to the harmonic oscillator (ie quadratic potential). For the QM harmonic oscillator the expectation value of the potential energy turns out to be
$$\left< E_\text{pot} \right> =\frac12 \left(v+\frac12\right)\hbar \omega=\frac12 E_\text{tot}$$
The expectation value of the kinetic energy is then
$$\left< E_\text{KE}\right>=E_\text{tot}-\left< E_\text{pot} \right>= \left< E_\text{pot} \right>$$
So it turns out that the kinetic energy and potential values have identical expectation values, consistent with the prediction of the virial theorem for a quadratic potential. This  is all outlined very nicely for instance in Atkins' physical chemistry textbook.
It is difficult to tell what energy levels are displayed on the diagram but they might very well correspond to the total energy given by $ E_\text{tot}  = \left(v+\frac12\right)\hbar \omega$.

As $\langle T\rangle =-E-RdE/dR$ and $\langle V\rangle =2E+RdE/dR$ the sum is still the total energy  but kinetic and potential energy have opposite signs with harmonic potential as $E=kR^2/2$

@porphyrin For the harmonic oscillator $\left<V\right>=\left<T\right>$. The relation between kinetic and potential expectation values is straight from the virial theorem. See eg https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/02%3A_Review_of_Newtonian_Mechanics/2.11%3A_Virial_Theorem The link is to a classical treatment but the result holds for the QM case.

Yes I am wrong, I picked up the equation for the electronic energy of a diatomic not the vibrational energy.

The curves labelled Harmonic/Morse show how the potential energy changes with internuclear separation, and the horizontal lines are the allowed total energies, potential plus kinetic. The potential energy is a maximum at the turning points (end of the horizontal lines in your picture) and zero at the $r_e$ value but because the total energy is constant the kinetic energy is the opposite and zero at the turning points and maximal at $r_e$. Its shape is just the inverse of that of the potential's at the energy for each quantum number.

For a QM system (which the OPs diagram presumably describes) there are no discrete turning points, that is, the wavefunction "leaks" into classically forbidden regions of the potential.

@Buck Thorn yes of course thats what you get from the wavefunction via expectation values but for the OP I have simplified things a little.   However, the wavefunctions are for a time independent model you really need to use wavepackets to get any actual motion (bond vibration) but that is a whole other story.


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Forces operating in molten ionic solids
What type of forces exist between ions  in molten ionic solids?
If this force is electrostatic attraction only, then how is it different from electrostatic interaction when ionic compound exist as solid?

The forces operating in liquids and solids are the same, it is just the kinetic energy of the ions or molecules that differ.
A variety of forces operate between ions or molecules in solids and liquids. In ionic solids, the electrostatic forces are often strong and dominant, but that doesn't mean that weaker dispersion forces are not present (or that the strong forces preventing atoms close to each other from getting even closer are missing).
The reason why things become liquids is simply that, when the temperature is high enough, the average molecule or ion has enough kinetic energy to break free from the crystal lattice of the solid. At some level of kinetic energy the attractive forces are no longer sufficiently strong to prevent the vibration of the surface molecules from shaking them from the lattice.
But even in ionic solids or liquids, the other forces also exist and make some contribution. The strength of the forces don't change with temperature, just the amount of vibration or the speed of the individual components. Ionic forces are particularly strong for small, compact ions (think Cl- and Na+ in table salt. After all the electrostatic force strength decays with the square of the distance between the ions.
But, if you make a salt using large floppy ions like 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) (see this question), the irregular shape and large size of the ion can weaken the strength of the ionic bond enough that the compound is a liquid at room temperature. in these molecules the forces are no longer strong enough to make the compound crystalline even at low temperatures. The specific balance of forces and the kinetic energy of the ion movement in these liquids has been altered from the more typical situation in more familiar ionic compounds so they can be liquids at lower temperatures.
In short, there is nothing different about the forces acting in an ionic liquid
and an ionic solid. The difference in the liquid is just that the kinetic energy of the molecules are large enough to overcome the forces.

that means bounded but moving

@kanchantolani All ions and molecules move all the time in solids and liquids. The major difference in solids is that the movement, as you say, is bounded: the components vibrate around a common point defined by the various forces they experience. I liquids they wander around with no fixed point and more freedom.

can you please look for my other question also

https://chemistry.stackexchange.com/questions/150844/how-is-kinetic-energy-accounted-for-when-describing-diatomic-bonds-with-a-morse

i didn't understant why large size of ion can weaken strength of ionic bond

CsF will be more ionic than LiF (fajans rule) but by above point it will be weaker than LiF

@kanchantolani A large ion with the same charge as a smaller one will see the overall electrostatic force between it and another ions as weaker as the force (approximately) declines with the square of the distance between the charges. Large ions can't get so close to each other so the forces will be much smaller.

therefore my CsF and LiF example is correct


text:
Calculating specific heat capacity from entropy
I'm currently struggling with a specific thermodynamics problem.
I'm given the entropy of a system (where A is a constant with fitting physical units): $$S(U,V,N)=A(UVN)^{1/3}$$
I'm asked to calculate the specific heat capacity at constant pressure $C_p$ and at constant volume $C_V$.
I know that the two are given by the following equation:
$$C_P=T \left(\frac{\partial S}{\partial T}\right)_{N,P}$$
$$C_V=T \left(\frac{\partial S}{\partial T}\right)_{N,V}$$
I've tried to calculate a expression for the entropy in terms of the temperature $T$ by calculating $$\left (\frac{\partial S}{\partial U}\right)=\frac{1}{T}=\frac{1}{3}(NV)^{1/3}U^{-2/3}=\frac{1}{3}\frac{S}{U}$$
but I think i'm going in a circle with this approach.
I guess I'm also having problems with the notation $\left(\frac{\partial S}{\partial T}\right)_{N,P}$ vs $\left(\frac{\partial S}{\partial T}\right)_{N,V}$. I know that this should mean "partial derivative of $S$ with respect to $T$ with constant $N$ and $P$ or $V$". But I'm not quite sure on how to evaluate that mathematically. For me personally, if I were to evaluate this two partial derivatives, they would be equal to eachother. But this can't be really true because of the relationship $$C_P-C_V=\frac{TV\alpha_P^2}{\kappa_T}$$
It would be awesome if anyone could give me starting point for this kind of problem because I'm kinda stuck with my method right now. Any help would be very much appreciated


text:
Why are we not getting the most stable product in Aldol condensation?

The picture above is from my class notes. I am familiar with the mechanism of both acid and base catalysed Aldol condensation. I also understand the reason why KCP was told as the major product in base catalysed Aldol condensation. But we know that Aldol condensation is reversible throughout so shouldn't TCP product be major all the time?

See https://en.wikipedia.org/wiki/Aldol_condensation. Are both steps reversible (addition AND dehydration)? The other observation is that in 'real' synthesis one usually tends to look for conditions that favour the desired product, rather than allowing equilibration and conversion of the products to undesired side products. Think of Michael additions: those are indeed reversible on paper. But then no Michael addition product should exist or be stable. I am trying to make you reflect on the fact that the reaction conditions are very important.

Yes your last line is important. What I have understood is that dehydration would become reversible if enough temperature is provided. So, you can keep the temperature a bit low and get the desired result. Am I correct?

Not sure about this specific case, but there is a general principle: 'kinetic' products tend to be favoured by low temperatures, and 'thermodynamic' products tend to be favoured by higher temperatures. E.g. in reactions involving organometallic species, low temperatures and the use of a strong base may favour the abstraction of the kinetically preferred proton, whereas higher temperatures and a weaker base may lead to the formation of a mixture, with a prevalence of the most stable anion.
But for your case I think it's more important to check if some reactions are reversible at all.


text:
How do electrostatic effects effect facial selectivity in addition to cyclic oxo-carbeniums?

I was reviewing Dave Evans' notes for stereochemistry of C=X electrophile additions, and found this slide detailing differences in additions to cyclic oxo-carbenium electrophiles, but I don't understand why this selectivity is seen. With the OBn group, I can see that adding on the same face as the OBn is favored, compared with the methyl substituent where adding trans is favored, but I cannot figure out why.
In looking at the mechanism for allylsilane addition I can see that a positive charge builds up on the nucleophile's central carbon. Is this what is stabilized by the neighboring OBn group that effects selectivity?

The direction of attack of the nucleophile is governed by the geometry of the oxocarbenium ion, not directly by the substitution. The nucleophile comes in from the direction that will cause it to become axial in the product because the transition state is chair-like and lower in energy than whatever happens attacking the other side.

After 15 years, I can't quite remember how the electrostatic effects come into play here. Is it to control the reactive conformation (since attack is axial) or is it to help direct the nucleophile? Maybe look up the Woerpel articles...


text:
When is it realistic to model diffusion without advection?
Brownian motion abounds in nature, but so does drift current.  In theoretical models of solutions and colloidal suspensions it is sometimes assumed that advection is negligible.  I'm wondering if this is a natural assumption to make, or if it is generally better (e.g. more informative, whether in a positive or negative sense) in settings where accuracy is desired to account for correlated motion owing to mesoscopic currents from the beginning.
This question is partly motivated by the following paradox (or non-paradox, as the case may be): correlated motion of solute and suspended particles from bulk currents is only irrelevant when the particles are near enough to a solid surface, like a porous material or substrate, for those currents to be negligible.  But no surface is perfectly rigid, and to some extent each exposed surface may act as an additional source of random current: could vibrations transmitted by the substrate introduce additional correlations in particle diffusion that depend on the substrate-solvent interface?
Finally, is it known to what extent hydrodynamic interactions between suspended particles account for drift induced by boundary fluctuations (e.g. deformations or perturbations to the container walls), or would these be better modeled as an independent source of randomness?
A related question would be, is it possible to reproduce Brownian dynamics with hydrodynamic correlations through a randomly fluctuating flow field? (and if so, to what extent should these fluctuations involve independent activity or spontaneous motion of the particles, i.e. as though they were pulled along by invisible forces, and to what extent should the particles be driven by currents that are induced by fluctuating boundary conditions?)
References:
Hydrodynamic coupling to a wall (i.e. effect of random currents on the diffusion tensor, both away from and near a perfectly rigid wall)
https://physics.nyu.edu/grierlab/leshouches2/node20.html
"Currents" induced by fluctuating surfaces:
https://www.scientificamerican.com/article/making-sound-waves/

Could you add a reference or two in support of the proposed arguments?

Or references in support of the claims and choice of terminology?  (e.g. that 'bulk currents' produce 'correlated motion' only when particles are far away from surfaces, and that surfaces are capable of generating random currents?)

I suppose it depends on whether you want the system to be homogeneous from the start or not. I guess engineers want well stirred reactors in their study of reactions but if you are studying reactions in the  seas, for example, then advection is going to be important. How do your surfaces get vibrations? Do you mean waves?

@porphyrin: I'm not entirely sure what you mean in your first sentence (it should be clear that this question is more about understanding how things are than trying to force anything), but vibrations in solids have several origins, including thermal excitation, which can become more pronounced at large wavelengths depending on the shape of the material at large scales.  I could imagine vibrations playing a non-insignificant role in the diffusion of colloids near a graphene sheet, for example, or near a thin wire.


text:
Why does Co(I) have a 3d8 configuration?
Why is the electron configuration of $\ce{Co^+}$ $[\ce{Ar}](\mathrm{3d})^8$?
Since neutral $\ce{Co}$ itself has a $[\ce{Ar}](\mathrm{4s})^2(\mathrm{3d})^7$ configuration, wouldn't the ionised electron be lost from the $\mathrm{4s}$ orbital, leading to an electron configuration of $[\ce{Ar}](\mathrm{4s})^1(\mathrm{3d})^7$?

For lighter elements, the shells fill in order. Starting at the transition metals, an outer s orbital may fill before an inner d orbital, so the electron configuration of unioninzed cobalt is written $\ce{[Ar]}4\mathrm s^1\,3\mathrm d^7$, rather than $\ce{[Ar]}3\mathrm d^7\,4\mathrm s^1$.
There is a video diagramming the electron configuration of $\ce{Co}$, $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$, thought it does not explain the reasoning, nor does it cover the less common $\ce{Co^{+}}$ ion, produced by photoionization or as found in some esoteric metal-organic compounds, or in the theoretical $\ce{CoCl}$.
That said, the situation becomes murky for transition elements, and downright turbid for lanthanides and actinides, where f orbitals are added. For example the outermost shells of $\ce{La, Ce and Pr}$ are $\ce5\mathrm d^1\,$, then $\ce4\mathrm f^1\,5\mathrm d^1$, and then $\ce4\mathrm f3$. What happened to the Pr d electron? My understanding is that energy levels are quite close for the larger outer shells, and it is not intuitive which orbital fills first.

I have always been under the impression that for transition metals, 3d < 4s in energy. The $3d^8$ configuration of Co(I) ion is thus the norm, and it's the $4s^2 3d^{n-2)}$ configuration of neutral atoms that is the exception. Also, what exactly is the energetic drawback of having an incomplete orbital? Exchange energy only really plays a crucial role in the d5 and d10 case; it's always there for d6/d7/d8/d9, of course, but isn't large enough to affect the electron configuration.

I'm probably wrong about the 3d/4s thing, though, sorry. cf. references in https://chemistry.stackexchange.com/a/33310/16683. I think I should probably do some research before commenting on that.

What is it meant by 'has no incomplete orbital'. Are there not 2 half-filled orbitals in a $\ce{d^8}$ configuration?

Thanks, @M.L, my mind must have been wandering. Max numbers are, of course, s2, p6, and d10. The answer will be corrected.

So there is now way once can predict the configuration of transition metals, lanthanides, actinides ions?

So there is no way to predict the configuration for lanthanides, actinides or transition metal ions? I thought 4s was prefered to 3d and 3d9 or 3d1p prefferes to 3d9 e.g. Ni, Pd?

There are rules, but calculation can be very difficult. Consider the effort to calculate wave functions for H https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.10%3A_The_Schr%C3%B6dinger_Wave_Equation_for_the_Hydrogen_Atom


text:
Acidic nature comparison; benzoic acid and phenylacetic acid
The $\ce{pK_a}$ values of benzoic acid and phenylacetic acid are around 4.2 and 4.31 respectively.
In benzoic acid, you have the resonance being the dominating effect, destabilizing a conjugate base anion. In phenylacetic acid, resonance doesn't work anymore, and only the -I inductive effect remains, which should stabilise the anion.
Why, then, is benzoic acid the stronger acid?

You considered the resonance, but neglected that phenyl rings are electron-withdrawing via the inductive effect. (An often underappreciated fact!)

@orthocresol: But isn't the resonance considered to be slightly stronger than the inductive effect here? Isn't that how you explain benzoic acid being weaker than formic acid?

Usually resonance effects are more important, but the keyword is usually. It's not always true: think for example about aromatic substitution of halobenzenes, there are competing effects which give them a "special" place as o/p-directing but deactivating substituents. I think with benzoic/formic acid it gets very messy because there are significant effects arising from solvation and it might not be easiest to compare these solely based on electronic effects.

@orthocresol: so just... Knowing how it is is more important in chemistry, I guess. And you just make something up for whatever the situation's like. Is that it?

If you want a fully theoretical, fully rigorous answer, then you need to solve the Schrodinger equation for these molecules and reactions. That’s extremely complicated, and doesn’t give you any real insight into the problem anyway. If you want to have simplified and understandable models, then you need to sacrifice something: and what you lose is the applicability or generality to every situation. The difficulty lies in striking a balance between these two.

In some sense, resonance and inductive effects are huge successes in this regard. They are easy to understand, and can be applied to almost every organic molecule. But you cannot make overgeneralised comments like resonance is always more important than inductive effects (how do you even quantify it? There isn’t a SI unit for resonance effects.), or if you do, you must bear in mind that there will be exceptions that don’t fit. Part of the problem is that these acidities are very close to one another, so the explanation needs a bit more subtlety than just some broad, sweeping statement.

TL;DR: You’re looking for an exact, accurate answer to a quantitative phenomenon (pKa’s), but we only have inexact, qualitative tools to use in the explanation. On top of that, we have multiple inexact tools which point in opposite directions, which makes it even less clear which one is more important. The way I see it, we have two choices: either use more exact tools (QM and computational chemistry) at the cost of insight, or fudge it with the inexact tools and accept that this is a bit of a limitation.

The acidity of benzoic acid $(\ce{C6H5-COOH})$ and its substituted derivatives, $\ce{R-C6H4-COOH}$ $(e.g., \ \ce{4-NO2-C6H4-COOH})$ have been studied extensively using the Hammett Equation. Similarly, acetic acid $(\ce{CH3-COOH})$ and its substituted derivatives, $\ce{R-CH2-COOH}$ $(e.g., \ \ce{Cl-CH2-COOH})$ have been studied extensively as well using similar Hammett Equation. You may read some brief discussion of Hammett Equation here.
Actually, you are comparing benzoic acid and a $\ce{R-CH2-COOH}$ where $\ce{R = Ph}$. Their is no relationship between these two acids except for both having $\ce{Ph}$ functionality. Benzoic acid has $\ce{Ph}$ directly connected to $\ce{COOH}$ group while acetic acid derivative has it connected through $\ce{-CH2\!-}$ group. Because of that, acidity of $\ce{Ph-CH2-COOH}$ is depend solely on inductive effect. It isworth noting that I cannot find any comparative studies of these two type of carboxylic aid in one single research.
The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20$ $(K_\mathrm{a} = 6.3 \times 10^{-5})$ while the $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.75$ $(K_\mathrm{a} = 1.77 \times 10^{-5})$. Thus, benzoic acid is about 3.6 times acidic than acetic acid. That is because when you consider $\ce{R-COOH}$ acid, it is more acidic if $\ce{R}$ group is electron withdrawing group (EWG) such as $\ce{Ph}$ group, compared to $\ce{R}$ group being electron donating group such as $\ce{CH3}$ group (EDG) like in the case of acetic acid.
Now, let's increase the electron withdrawing ability of $\ce{R-CH2-COOH}$ acid compared to acetic acid (where $\ce{R = H}$): The $\mathrm{p}K_\mathrm{a}$ value of 2-chloroacetic acid (where $\ce{R = Cl}$) is $2.87$ $(K_\mathrm{a} = 1.36 \times 10^{-3})$, which is about 76.8 times acidic than acetic acid. Thus, we can conclude that $\ce{ClCH2}$ group has larger inductive effect than that of $\ce{Ph}$ group. Thus, if we change $\ce{Cl}$ in $\ce{ClCH2}$ group to $\ce{Ph}$ group, then resulting carboxylic acid should be less acidic than that of 2-chloroacetic acid.
As given in OP's question, this carboxylic acid, 2-phenylacetic acid $(\ce{Ph-CH2-COOH})$ has a $\mathrm{p}K_\mathrm{a}$ value of $4.31$. It is more acidic than acetic acid, but not quite acidic as benzoic acid because of the extra $\ce{CH2}$ group. Remember, inductive effect is weaken by additional $\ce{C-C}$ bonds (c.f., ortho-, meta- and para-inductive effects).
Source of  $\mathrm{p}K_\mathrm{a}$ values


text:
How to apply the Fajans' rules?
Outside of bonds between metal atoms=metallic bonds, non-metals=covalent bonds (molecules), it is difficult for me to determine what kind of bond it is.
I know that in the rule metals and non-metals form ionic bonds and that the ionic part of such bonds increases with increasing electronegativity difference and that Fajans' rules states that large cations with small positive charge and small anions characteristically form ionic bonds. But to me all these measures seem very fuzzy. I can say that SnCl2 is more ionic than SnCl4 because of the smaller positive charge of Sn (when considered as a cation) in SnCl2 (Sn2+) compared to SnCl4 (Sn4+). But how do I know that SnCl4 is a molecule?
Are there clear rules for determining the type of bond? How to proceed if criteria of the Fajans' rule contradict each other and at what point is it no longer an ionic bond with covalent character but a covalent bond (for metal-non-metal bonds like SnCl4, would SnO be a molecule or salt?)?

These classifications are made by us and not by nature and whenever we humans do so there is always ambiguity regarding the borderline from which you can say that type of bind has changed. Rules like Fajan's rule help us the covalent character in different ionic bonds. Rules like this cannot tell you if something is too covalent to not be called ionic.

Note: Salt ≠ Ionic and Molecule ≠ Covalent.

See: https://chemistry.stackexchange.com/questions/9038/how-would-one-compare-the-magnitude-of-covalent-character-between-sncl4-and-snf2/9968#9968

Thank you for your help, so one can't be sure if a compound is covalent or ionic only looking at the chemical formula/geometry?


text:
Yield of alkyl chloride in the Hunsdiecker reaction
Apparently the yield of alkyl chlorides is low as compared to that of alkyl bromides when preparing alkyl halides using the Hunsdiecker reaction.
Why does this occur? The only factor I can see is the lower stability of the free radical of chlorine as compared to that of bromine, as the reaction proceeds by the free-radical mechanism.
Is that what causes the lower yield?

The Hunsdiecker reaction is a halogenation-decarboxylation reaction of silver salts of carboxylic acids.

To understand what causes the yield to change, first we have to consider the mechanism of the reaction.

As seen from the mechanism the homolytic cleavage of the $\ce{O-X}$ bond is the slowest step in the reaction and thus is the rate determining step. So as you predicted it all comes upon the stability of halogen radical. It's commonly known that the bromine radical is fairly stable than the chlorine radical and thus the Hunsdiecker reaction gives better yield with bromine rather than with chlorine.


text:
Finding mass percent through molality of potassium nitrate solution
Question

An aqueous $\ce{KNO3}$ solution has a molality of $\pu{4.16 m}$ and a density of $\pu{1.08 g/mL}$.  Calculate the percentage by mass $\ce{KNO3}$ of the solution.

My Approach
I know that molality ($b$) is defined as:
$$b = \frac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$
In order to the mass of solution I multiplied $\pu{(1.08 g/mL)}\times\pu{1 mL}$, since, $m_\mathrm{solution} = m_\mathrm{solvent} + m_\mathrm{solute}$
However, I don't know where to use this information. I am also not sure what I to do with the $\pu{4.16 m}$ since I don't have moles of solute or mass of solvent. I am thinking that I have to assume some number here, but I might be wrong about this. Any help will be appreciated.

Density is excessive input, not needed to calculate conversion. All you need is molar mass of KNO3. Density would be needed for calculation from molarity.

I am not sure you even need the density in this case.
Say you have $\ce{1kg}$ or $\ce{1000g}$ water (the solvent). That would mean that there is $\ce{4.16 mol}$ of $\ce{KNO_3}$.
Since we can figure out the molar mass of $\ce{KNO_3}$ = $\pu{101.11g mol-1}$, the mass of $\ce{KNO_3}$ in the sample must be $(4.16)(101.11)= \pu{420.62 g}$.
We already assumed that we had $\ce{1000g}$ of water, the mass percent of $\ce{KNO_3}$ is then equal to: $$\frac{420.62}{420.62+1000
} \approx 29.6\%$$

You can now upgrade your formatting skills and go beyond using just $\ce{}$, try using $\pu{}$ for physical units.


text:
Naming a nitrile in the middle of an aliphatic chain
What class of compounds does this belong to, if it exists; $$\ce{CH3-CH2-C\equiv N^{+}- CH3}$$
I honestly don't know if it exists; it popped up in my head while I was reading up on nitrile compounds, and I haven't heard of this type of molecule, thus the question.
If it doesn't exist, why not?

You appear to have a carbon with 5 bonds in there

@Waylander: Wouldn't it be $-C\equiv N-$?

Then you have a N with 4 bonds and no charge

The charge on the nitrogen was implied, sorry. I didn't know how to do that with MathJax. Does it exist, then?

Yes. Nitriles can be methylated by a variety of reagents. The N-methyl nitrilium species generated are highly reactive: example https://pubs.rsc.org/en/content/articlelanding/1980/c3/c39800001151/unauth#!divAbstract

Just so you know, charges are never implied.

Compounds like this are named based on the nitrile from which they are derived. So, in your case, the nitrile is propanenitrile:
$$\ce{CH3-CH2-C#N}$$
A protonated nitrile would be called a nitrilium cation, following the general form:

amine --> ammonium
imine --> iminium
nitrile --> nitrilium

Thus, the following cation is named propanenitrilium:
$$\ce{CH3-CH2-C#N^+-H}$$
If we swap that proton with an alkyl group, we can call it an N-alkyl substituent. Thus, your cation is the N-methylpropanenitrilium cation.
$$\ce{CH3-CH2-C#N^+-CH3}$$


text:
Cyanide hydrolysis; could it yield an oxime?
The mechanisms for both acid and base catalyzed hydrolysis of nitriles are clear;

(With the subsequent deprotonation from the hydroxyl) and

(Source)
In both cases, the hydroxyl adds to the carbon while the proton adds to the nitrogen.
I thought up an alternate mechanism, with the proton attaching to the carbon, and the hydroxyl going to the thereby positively-charged nitrogen, yielding an oxime.
Why doesn't this occur? I thought of the fact that the nitrogen atom, with a lone pair, would cause any nearby protons to come straight to it, but I'd like clarification on that.

Well, nitrogen is the electronegative atom here and it has the lone pair as you mentioned, so its very unlikely that the proton will go to the carbon

Yes, since nitrogen is more electronegative than carbon, and also since it has the lone pair available for taking, the proton is likelier to go to it.


text:
pH of water after getting heated
When water is heated, the molecules of the water gain energy which leads to dissociation of water molecules. Does this means that pH of water has changed?

Yes, but not by much.

Dissociation equilibrium constants are temperature dependent. You can find tables of pKw and water neutral pH versus temperature on many places on internet. It has been supposed to be found  before asking.

http://www.iapws.org/relguide/Ionization.pdf

It's an intriguing question though. One might think that a sample of water that has equal concentrations of $\ce{H3O+}$ and $\ce{OH-}$ always has $pH = 7$. But if $pK_w \ne 14$... hmm...

The neutral pH is equal to 7 just accidentally, as pKw=13.99 for 25 Deg C. People are too fixed on this "magical" value.

Ionic product of $\ce{H2O}$ is changing with temperature. It is $14.94$ at $0$°C, $14.17$ at $20$°C, $13.83$ at $30$°C and $12.26$ at $100$°C.
So the pH of pure water is  $7.47$ at $0$°C, $7.08$ at $20$°C, $6.92$ at $30$°C, and $6.13$ at $100$°C


text:
If a metal's solvent extraction process is exothermic, then the stripping is necessarily endothermic?
I checked some articles, including Hydrometallurgy 2012, 133-144, 185-191 and Hydrometallurgy 2000, 57, 269-275. In all research papers I checked, an exothermic stripping process followed an endothermic solvent extraction process and vise versa. First, what reasons can be behind such phenomena? Second, is such a claim necessarily right in all cases?


text:
Why is Enthalpy of Compound at Some Temperature Equal to its Enthalpy of Formation at the Same Temperature?
This is a question I don't understand in thermo. Enthalpy of compound would be sum of U + PV at some temperature, but since we can't calculate absolute value of H, we calculate it relatively to some reference. Enthalpy of formation is heat needed to be exchanged between system and surroundings at some pressure to keep the system at the same temperature when compound is formed from elements in the most stable state. Why are these two quantities equal?

They are not equal.  The enthalpy of a compound at temperature T is relative to the pure elements at the reference temperature $T^0$.  So it is equal to the standard heat of formation at the reference temperature plus $C_p(T-T^0)$.

So, you are saying that enthalpy of elements in their most stable form is taken as zero only at reference temperature and not the other temperatures? If so, it makes sense to me.

In practice the enthalpies of the elements are always going to cancel from any practical calculation, whether you do it this way or whether you use heats of formation at temperature T, since you are dealing with. balanced chemical reaction equations, and the elements contribute both to the products and the reactants in equal. amounts.  However, it is easier not to have to make use of the heat capacities of the elements to calculate their enthalpies at T if it is not going to matter.  Try a sample reaction both ways to satisfy yourself that this is correct.

Yes. So, do we agree that enthalpy of compound at temperature T is equal to its enthalpy of formation at the same temperature if we regard enthalpies of elements to be zero at that temperature?

I would never think of it that way.  I would always regard it as relative to a fixed specified state.

If you do it the way you are suggesting, then the simple issue of getting the change in enthalpy for heating an ideal gas at constant pressure would have to take into account the changes in enthalpy of the elements.  Otherwise, the change in enthalpy would not equal the amount of heat added.

Well, yes. Choosing enthalpies of elements to be zero is a reference since than enthalpy of formation becomes equal to enthalpy of compound because enthalpy of any reaction is stochiometric difference of enthalpies of products and reactants at constant P and T. In formation reaction reactants are elements in their most stable for which we chosen that their enthalpy is zero, so enthalpy of formation is equal to enthalpy of component at this temperature. Since formation happens at specified state (as you noted) this enthalpy is one at that specified state.

So you are in agreement now that the enthalpy at temperature T should be represented as the heat of formation at the reference temperature plus the heat capacity times the difference between T and the reference temperature?

Yes, I do agree with that. But, what I am not quite sure is that can we say that enthalpy of compound at some is T equal to its enthalpy of formation at the same T so that cp(T-T0) term becomes zero. I would say we can define reference however we want, but we should know according to what reference we calculated our enthalpy of compound.

This may be more practical if we know enthalpy of formation at some temperature closest to our working temperature since if temperature difference is small we may not need to know/take into account how cp varies with T for our compound.

Let us continue this discussion in chat.


text:
What is the reasoning behind the assumptions of Transition State Theory?
We were told in a first year introductory Thermodynamics and Kinetics lecture, and also from Wikipedia, that there are 3 main assumptions for the derivations used in Transition State Theory, but I'm not sure how or why they have been assumed?

Reactants are in constant equilibrium with the transition state structure.
The energy of the particles follow a Boltzmann distribution.
Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

I'm concerned with 1. and 3. particularly, as it seems as though they contradict each other, because how can the reactants be in equilibrium with the TS, if the TS structure cannot revert back into reactants? How do we then utilise these assumptions, is it so that we can write
$$\ce{A + B <=> AB^‡ -> P}?$$

This is deeper discussed in Rate Constant Units and Eyring Equation, but I am going to post a very short take-home message here.
First, the three assumptions you have cited are not complete. They completely gloss over the most important one, i.e. nuclear and electron motion can be separated, energies (translation, vibration, rotation) can be treated additively.
Having gotten that out of the way, reactants are in constant thermal equilibrium with the transition state structure. This actually means that the reactant structure(s) is in equilibrium with all possible structures that can be reached from it. It is hence more correct to say that the reactants are in thermal equilibrium with all possible transition state structures.
This also basically ensures, that all these structures can be treated within a single, combined Boltzmann statistic, where the energy is distributed according to all the states of the assembly/molecule.
The next assumption is that the transition state will be crossed only once (point of no return), which means that when an assembly/molecule reaches this (high) energy state, it will form the product. This basically sets the limit to the Boltzmann statistic as all states following the transition state are not included anymore.
Putting both together we arrive at
\begin{align}
  \ce{A + B <=> &AB^\ddagger -> P}\\
  \ce{P <=> &AB^\ddagger -> A + B}.
\end{align}
However, you see that you can look at it from both channels, but that'll become to involved here. Please read the other post first.
All these assumption basically lead up to that the transition vector can be treated classical as translation.
In conclusion, the equilibrium part is necessary to write the combined Boltzmann statistic; the transition state collapses into products part is necessary to only include everything up to and including the TS, but nothing more into the statistic.

Also related: References for Transition State Theory and Quantum Transtion State Theory

Thank you for your reply and point in the right direction, I have my final exams right now, so I'll take some time to look into this a bit more and perhaps post some follow up.


text:
Detection of sulphide in solution also containing sulphate
Question:

The reagent(s) that can selectively precipitate $\ce{S^2-}$ from a mixture of $\ce{S^2-}$ and $\ce{SO4^2-}$ in aqueous solution is(are):

(A) $\ce{CuCl2}$
(B) $\ce{BaCl2}$
(C) $\ce{Pb(OOCCH3)2}$
(D) $\ce{Na2[Fe(CN)5NO]}$

Answer:

 (A) or (A) and (C)

Why (C) is considered as correct option even though lead sulphide forms black precipitate and lead sulphate forms white precipitate?
Question source: JEE Advance 2016

This is an ill-posed question. I think A is a better answer. Your concern is right both sulfate and sulfide should ppt in excess of the lead salt.

Please write the Q in text. Pictures are not recommended.

You are correct about (C). Both ions form precipitate with lead acetate. However, the question can also be reworded as "What reagent can help distinguish between sulphide salt and sulphate salt?". In such case, sulphate salt does react with BaCl2 to form white BaSO4. In case of sulphide, no ppt. forms. Additionally, the sodium nitroprusside test is actually a selective test for sulphide only,

No doubt question can be formatted in better way (words) but it is the real question I am concerned of. I just want to know if I am missing something special knowledge.

I will add one thing that may be the reason behind why (C) is an acceptable answer choice. Yes, both $\ce{PbS}$ and $\ce{PbSO_4}$ are insoluble. However, if you look at the $\ce{K_{sp}}$ for both compounds, $\ce{PbS}$ is about $\pu{3.2 x10^-28}$ while the $\ce{K_{sp}}$ for $\ce{PbSO_4}$ is $\pu{1.3 x 10^-8}$. When you have a mixture of ions, the resulting molecule with the lowest $\ce{K_{sp}}$ will precipitate first, given that the same stoichiometric ratio is the same. In this case, since $\ce{PbS}$ has a lower  $\ce{K_{sp}}$ than $\ce{PbSO_4}$ and that they have the same stoichiometric ratio, $\ce{PbS}$ will precipitate first.
This technique is called fractional precipitation where you separate two ions, in our case $\ce{S^{2-}}$ and $\ce{SO_4^{2-}}$, by precipitating one of the ions selectively through taking advantage of their differing $\ce{K_{sp}}$ values.
This is the likely reason behind why choosing (A) itself is acceptable or choosing (A) and (C) is also acceptable.

@MLI would also like to add something, since the question feels as if it lies heavily on the qualitative side thus one may or may not observe both precipitates as in qualitative analysis, generally it is not about taking specific concentration of reagent, that is why I think the answer to this question has been given as A $OR$ A and C.

We can compare $K_{sp}$ values directly only when stoichiometries match, which they do here. Otherwise you could get different results depending on the overall concentration of the solution. IMO the equal stoichiometries should be mentioned in the answer as allowing the solubility products to be compared.


text:
Why are calcination and roasting processes carried out at a temperature just below the ore's fusion temperature?
I had read about the two important methods used for the concentration of an ore - Roasting and Calcination. I have a doubt regarding why both the methods are carried out at temperatures not above the ore's melting point.
After some research, I have been able to conclude that one of the reasons is the formation of complicated slag (i.e., difficult to remove from the ore) that tends to form if the ore is molten.
Could someone please explain what exactly happens so that a complicated slag is formed? Also, are there more reasons behind keeping the temperature below the ore's fusion temperature (any specific reasons pertaining to roasting and calcination individually)?

The purpose of roasting a/o calcination is oxidation a/o decomposition of original ore to form metal oxides.
It is desirable to keep the intermediate product solid to obtain porous product. This allows easy access for further reduction by carbon monoxide by coke or charcoal reduction process. If oxides were in form of a solidified melt, the reduction arrangement would be problematic.
Additionally, some ores or oxides have too high melting point to be practical or easily possible to melt them.


text:
Hund's rule for comparing term symbol energies in excited state
This question was asked in an exam:

The lowest energy electronic state for excited state carbon atom is,
a. $^1D_2$
b. $^3D_1$
c. $^3D_3$
d. $^3D_2$

Although we have been taught Hund's rule for determining the relative energies for various term symbols in the ground state, I haven't read anything about the rules for excited state anywhere. I couldn't not find it in the book I use i.e. Atkins, Physical Chemistry.
I am really confused about how to solve this.

summarising (a) lowest multiplicity lowest energy (b) with a given multiplicity largest L lowest energy (c) Atoms with less than 1/2 filled shell lowest J lowest energy.  Does not the excited state just means an extra electron 'upstairs' such as $s^2p^2\to s^1p^3$

But on subwhat shell are we going to apply the less than half or more than half filled subshell rule?

Hund's rules strictly assume LS coupling and application to excited states is not reliable. Of course you can try to apply them in a similar manner as for the ground state, but treat the outcome with caution


text:
What determines the consistency of hydrogen foam when reacting aluminium with sodium hydroxide solution?
I was using $\ce{NaOH}$ solution to remove hardened grease from my stove burners, and it began to generate a lot of bubbles, some of which merged into a big bubble and erupted into the surface, but most formed a very thick foam.
I left it reacting for a few minutes and it started making way too much foam, about 3 liters of it.
And then I remembered that the burners were cast aluminum, and this should be $\ce{H2}$ foam.
I took a mere spoonful and set it on fire (far from the main pot where I left it) and it produced a very loud and instantaneous bang with a fast flash instead of a slow and fiery deflagration.
This is worrying, that meant the whole foam had a very good hydrogen and air mix, if the whole thing were to explode it would cause at least a hell of a mess and would scare all my neighbors.
I guess not much $\ce{H2}$ escaped into the room, but I thoroughly ventilated the ceiling to remove any accumulated $\ce{H2}$ just in case (after dissolving all the foam in water).
Then I thought to myself, "Why didn't anyone warn me about this explosive foam? Using $\ce{NaOH}$ solution to dissolve grease is so common here".
But I couldn't find any videos on YouTube where $\ce{NaOH + Al}$ resulted in such an explosive and thick foam as mine.
Instead it just bubbles out of the solution, and setting it on fire never resulted in such a crisp bang.
And when people intentionally made foam, they didn't make it like that, instead they bubbled pure $\ce{H2}$ through a soap solution.
But my solution had no soap. Just $\ce{NaOH}$ and the aluminum pieces.
Am I missing something here?
Did the little soap produced by the reaction with the grease made all of that?


text:
Effect of impurities on melting and boiling points
I am tutoring my nephew for his science GCSEs; and his chemistry textbook states that you can tell how pure any compound is by how close its melting and boiling points are to the "official" mp and bp for that compound. It says that impurities will always lower the melting point of any compound, but - paradoxically (intuitively speaking) - will always raise the boiling point. But it just states this as a fact and offers no explanation!
I found a simple explanation for why the melting point is reduced in the case of crystalline solids:

Foreign substances in a crystalline solid disrupt the repeating pattern of forces that holds the solid together. Therefore, a smaller amount of energy is required to melt the part of the solid surrounding the impurity. This explains the melting point depression (lowering) observed from impure solids.

... although this doesn't explain why impurities in solids that are not crystalline also lower the melting point.
But I can't find any easy to understand explanation anywhere for why impurities would always increase the boiling point of any compound. I've found explanations that are specific to water (and even then, the explanations I've seen only make sense if the impurity binds more strongly to water molecules than water molecules bond to each other, which isn't always the case); and I found one explanation that says it's because of increased solution phase enthalpy, without explaining what on earth that means!
So I emailed a Professor of Physical Chemistry about this and she replied

The reason for impurities lowering the melting point yet increasing the boiling point is because the impurities stabilise the liquid phase, making it more energetically favourable. This extends the liquid range to lower temperatures (lowering the melting point) and to higher temperatures (raising the boiling point). The liquid phase is stabilised due to the entropy increase when you have the impurity (or any solute) in the liquid.  Normally solids do not dissolve the impurities, so the melting point involves transition to pure solid (with impurity separating out or staying in the liquid state).

She made it clear she didn't have time to answer any follow-up questions, and I can't make full sense of her answer, so I am hoping someone here can understand her explanation and elucidate it.
I think I understand why impurities increase the entropy of a substance in its liquid phase by more than they increase the entropy of the same substance in its solid or gasseous phases - I presume impurities don't significantly affect the entropy of a solid because the solid can't rearrange itself into different configurations without absorbing a lot of energy; and they don't affect the entropy of a gas because the gas can disperse over an almost infinite distance to all intents and purposes, so it will be almost infinitely disordered even when pure; but they do increase the entropy of a liquid because there are many more ways of arranging the molecules within a liquid if there is more than type of molecule in that liquid, and the molecules in a liquid can rearrange themselves in all those different configurations without any input of energy.
First of all, have I got that much right?
But even if I have, I don't understand why increasing the entropy of the liquid by adding an impurity to it would make it energetically more stable - and therefore both more difficult to freeze and more difficult to vapourise.
Can anyone help to explain this?
Dave

All generalities are false (as is this one). Though impurities usually lower the melting point (m.p.) by disrupting crystallization on the atomic order, consider the phase diagram of the binary alloy (amalgam) HgxAg1- x: Pure Hg melts ~-39°C, and adding even a little bit of impurity raises the melting point considerably!
And without resorting to entropic explanations, consider that a high boiling point (b.p.) solute would "hold onto" the solvent, decreasing vapor pressure. $\ce{CaCl2}$ will absorb $\ce{H2O}$ from the air, driving down the vapor pressure (and humidity) even after the $\ce{CaCl2}$ liquefies.
That said, the answer they expect on the exam is that impurities decrease the m.p. and increase b.p.

Hi DrMoishe Pippik
I can see why that would be the case if the solute binds more strongly to the solvent molecules than the solvent molecules bind to each other, but I don't understand why that would be the case if the impurity binds more weakly to the solvent molecules then the solvent molecules bind to each other?
Also, are you saying that impurities in non-crystaline solids do not lower their melting points?
Many thanks for your help
Dave

@DaveRado, do you mean why it lowers the m.p.? The impurity gets in the way of crystal formation.

Hi DrMoishe Pippik, No. I am asking why a solute would "hold onto" the solvent, decreasing vapor pressure, if the solute bonds more weakly to the solvent than the solvent molecules bond to each other.
I also asked whether you are saying that impurities in NON-crystaline solids do not lower their melting points? Thanks for your help. Dave
PS - I tried to use "@DrMoishe Pippik" above but it deleted it every time I saved, so I had to remove the "@" symbol and use "Hi" instead of "@" - any idea why?

Non-crystalline solids, e.g., glass, do not have clearly defined melting points. Consider that window glass is a mixture of substances dissolved together, so it's melting point is so broad as to be called the "softening point".

Hi DrMoishe Pippik
Thanks for the clarification regarding the melting points of non-crystalline solids.
However, my main question was about BOILING points. I can understand why a solute would "hold onto" the solvent, decreasing the vapor pressure, if the solute bonds more strongly to the solvent molecules than the solvent molecules bond to each other - but I don't understand why that would still be the case for solutes that bind more WEAKLY to the solvent molecules than the solvent molecules bind to each other. Please could you help me to understand that as well?
Many thanks
Dave

You do not need that the solvent hold on the solute. Although you put it in quotes so not sure what it meant exactly. The partial pressure is lowered even for an ideal solution. OP: consider that the behaviour does not apply if the impurity is a liquid itself.

I put it in quotes because DrMoishe Pippik put it in quotes. He said the reason a liquid containing an impurity will have a higher boiling point than the pure liquid is  that - to quote his post: "without resorting to entropic explanations, consider that a high boiling point (b.p.) solute would "hold onto" the solvent, decreasing vapor pressure." If you don't need to the solute to "hold onto" the solvent in order to increase its boiling point, then I still don't understand why impurities (almost) always increase the boiling point compared with the pure substance in liquid phase.

Also, the GCSE textbook doesn't say that non-liquid impurities always increase the boiling point, it says that all impurities increase the boiling point, compared with the pure substance. So I'm also confused why it doesn't say that the behaviour doesn't apply if the impurity is liquid itself?

@Alchimista, even if the impurity is another liquid, the vapor pressure may decrease. Consider water with glycerol impuriy: http://www.aciscience.org/docs/physical_properties_of_glycerine_and_its_solutions.pdf , p. 12. In the case of a liquid, it depends if the impurity has a lower or a higher b.p. than the solvent.

@DrMoishePippik that is clear. I think we are confusing things a bit. OP is almost equating the lowering of melting point and ebullioscioic effect. The latter are not so important, usually. Of course mixing liquid is something else and the boiling is lowered or raised depending on what component is taken as referent. I just wanted to make clear that the effect on the boiling point can be see as "dilution" of the main liquid, its chemical potential being lowered and dies not require an enthalpic term (which can be there, of course).

@DaveRado yes in fact the beginning of my comment was towards the answer not your comment. Also, if the solute bond more weekly to the (supposed) solvent you have two cases. Enthalpy dictates there is not such a solute to speak about, as it separates out, or entropy drive solubilisatiin. The delta T would be very small and dictated by "dilution" of the solvent.

I just want to put forward my explanation on this - melting point is conversion of solid to liquid . so, if the impurities are there - it definitely weakens the force of attraction between the molecules which holds them in a regular lattice usually. Hence you need to supply less energy to make it to liquid unlike with the pure substance with regular lattice arrangement.
Boiling point is like a surface phenomenon where already the liquid has mobility enough to move randomly and if impurities are also there, it definitely provides hindrance for the molecules to evaporate from the liquid surface. Imagine its just like a marathon..so crowded ! Hence more energy required to overcome the hindrance caused by the impurities.
Hence, impurity decreases the melting point and increase the boiling point I guess!
Freezing point is to bring the molecules closer. so, if impurities are there - molecules can be made into tightly packed as the impurities will also be present in the interstitial space of lattice that can help to bring it back to tightly packed solid state quickly with less energy.

Boiling is not a surface phenomena, it is a bulk thermodynamic phase transition. The rest of the answer similarly ignores thermodynamics of mixtures.

Thanks for letting me know the mistake.. But I have a doubt,  when we boil, it becomes gas ..so evapoartion is surface phenomenon right? Then why can't the impurities give hindrance to the evaporation - which leads to increase in boiling point?

When boiling water on the stove, take note of the bubbles coming up from the bottom of the pan. And do not confuse kinetic effects (such as nucleation and surface area) with thermodynamics.


text:
A solution is cloudy at lower concentration and becomes clear at higher concentration. Why?
This might be just an "everyday life science" question. I am a university professor but not in Chemistry so please educate me. Today I was adding some floor cleaner formula to water as instructed by the label. In the beginning, the solution became cloudy. However, as I added more formula, the solution became clear. I am curious why this can be and whether there is a name for such a phenomenon in Chemistry.
More details and pictures:
Just a little bit of formula added to the water; the solution was cloudy:

Then I added more formula; the solution becomes clear:

The cleaner is called "HOOVER FloorMate Grout & Tile Plus".
Out of curiosity, I checked its data sheet which states that it is a :

Blend of proprietary non-hazardous surfactants and polymers.

Here is a list of ingredients from a similar product: https://hoover.com/pages/hoover-solution-ingredients. I suspect that it is likely due to the chelating agents that remove minerals from hard water. I am not 100% sure of this though because there are a lot of chemicals.

One general explanation is that the solution added to water is more than water based. In other words it can be that an ingredient of a formulation is sparingly soluble in water and need some high enough concentration of a third compound to be dissolved. This would be the case with lots of spirits (sambuca, raki, pastis etc) in which water addition cause the flavours to separate (normal scenario). But if one will keep pouring the liquor than ethanol conc can get high again to turn the glass content crystal clear again. .

One option is you simply didn't mix properly when first adding.

At very dilute concentrations, the solvating power of water overcomes the natural tendency of surfactant molecules to agglomerate into a separate phase. The shear numerical excess of water molecules just about totally dissociates the bulk surfactant and dissolves it as separate molecules. So a mixture of surfactant and water can be a true solution - clear and homogeneous.
However, as the concentration of surfactant increases, aggregates form (micelles) which can absorb oleophilic molecules (e.g., the polymers in the formula). With sufficient surfactant, the polymer is dispersed enough and into small enough micelles to appear clear. In the second graph, the knee of the curve is at the CMC; at this point, the aqueous phase is saturated with single molecules and some are beginning to stick together, starting to be able to contain polymer.

To get to the point: Your floor cleaner needs to be concentrated enough to work well. This suggests that there is enough surfactant to exert a solvent effect on the polymer, dispersing the chains, separating the molecules, making the smallest dimension significantly less than a wavelength of light. The original floor cleaner product could appear clear to the eye. Addition of a little of the product to a lot of water reduces the micelle formation of the surfactants by favoring dissolution of single surfactant molecules: the polymer chains then have fewer surfactant molecules to surround them. Polymer molecules escape, agglomerate and become visible - but only a little, since you can reverse the process.
A more delicate detection method is to use the Tyndall effect: in a darkened room, shine a narrow beam of light through a liquid (say from a single LED flashlight). The beam may be visible, much like a flashlight beam in a smoky room, and can detect some materials at levels in the ppm range (e.g., milk in water).

I suspect your floor cleaner would be represented by the middle test tube in the series of five. Small micelles are present but the liquid seems clear, and will seem to have a small or negligible Tyndall effect. The polymers will be dissolved in the many micelles, which are too small yet to be very visible. When you dilute, you go to the left one step, the surfactant becomes more dispersed into single molecules, exerting less solvent power on the polymer. Either the micelles  grow so large as to become visible, or actually collapse, allowing the polymer molecules to agglomerate.
Adding formula back to the mixture increases the surfactant (the solvent power), redisperses the polymer molecules into tiny micelles, and the liquid becomes clear again. The key (hypothesis) is that the pure polymer dispersion would be cloudy by virtue of its large particle size (~1 micron?), but a high enough surfactant concentration makes the formula essentially clear by generating micelles much smaller than the wavelength of light.

Thanks for the detailed explanation. But as I read it, you seem to explain that the solution should be clear at lower concentration but can be cloudy at higher concentration. But what I observe is the opposite.

@max sim: you are quite correct. I left something out: your formula does not consist of just one pure surfactant in water. Let me edit to be more direct.

Your solution is probably made of aluminum sulfate $\ce{Al2(SO4)3}$, which contains aluminum ions $\ce{Al^{3+}}$, due to the reaction $$\ce{Al2(SO4)3 -> 2 Al^{3+} + 3 SO4^{2-}}$$ And the $\ce{Al^{3+}}$ ions react slowly with water according to :
$$\ce{Al^{3+} + 3 H2O <=> Al(OH)3(s) + 3 H+}$$ This equation is an equilibrium. You should see that there is only $1$ particule in the solution on the left hand side, and $3$ on the right hand side. So according to the law of mass action, diluting the solution pushes the equilibrium towards the right side, to the production of more $\ce{H+}$ ions. As a consequence, more and more $\ce{Al(OH)3}$ are produced. After a while the amount of $\ce{Al(OH)3}$ is higher than allowed by the solubility product, and the solution becomes turbid.
In the opposite direction, when dissolving more and more aluminum sulfate into water, a turbid solution is first obtained, as the reaction is pushed to the right hand side. As more and more product is dissolved, the equation is pushed in the opposite direction, so as to decrease the number of ions in solution, which also destroys the $\ce{Al(OH)3}$ precipitate, and makes the solution transparent.

The Safety Data Sheet gave a pH of 8.4.

@James Gaidis. No ! Aluminum sulfate solutions are acidic ! The p$K_a$ of the $\ce{Al^{3+}}$ ion  is $4.9$ according to The Chemistry Data Book, from J. G. Stark and H. G. Wallace, published by John Murray London, 2014,. ISBN 0-7195-3951-X

You have pointed out a conflict between the pH of the Hoover FloorMate Tile & Grout Plus and the pH of a solution of aluminum sulfate. I totally agree with you. But what is the connection between the two?


text:
Why is Aniline a poor substrate for Nitration reaction?
I have been asked to find out which one of the following is a poor substrate for Nitration reaction:

Upon encountering this question, my first reaction was the option c, as I know that the nitration mechanism involves the substitution of a positively charged Nitronium ion ($\ce{NO2+}$) and hence a $\ce{-OCOCH_3}$ group that shows $\mathrm{-M}$ effect will not be helpful for that process as it reduces the electron density in the benzene ring.
But the solution states that the answer is aniline (given in option a). I do not understand how is that possible. Even in the case when we use mixed acid as the reagent, it will form para- and meta-substituted nitro compounds. But, I do not know why and how does it behave as a 'poor substrate' in any case of nitration.
Please explain.

In the case of option c, the $\ce{-OCOCH_3}$ group will not be showing $\mathrm{-M}$ effect as the ring is directly attached to the oxygen atom. The group rather shows $\mathrm{+M}$ effect because of the two lone pairs present on the oxygen atom, increasing the electron density in the benzene ring and thus making it a good substrate for a nitration reaction.
On the other hand, as nitration is generally done with mixed acid as the reagent, the $\ce{-NH_2}$ group in aniline takes up a hydrogen atom to become $\ce{-NH_3^+}$,  which is a ring deactivating group.
Therefore, among the options given, aniline is the poorest substrate for nitration reactions.

You're making it simpler then it is. Aniline tends to oxidise, forming oligomers, unless you completely protonate it.

@Mithoron I am a high school student, so my logical reasoning is bound by my limited knowledge of the subject. It would be great if you could explain your point in detail (maybe as a separate answer or provide links to resources).

https://en.wikipedia.org/wiki/Polyaniline https://en.wikipedia.org/wiki/Aniline


text:
Why do resorcinol and iodine react to give 2-iodobenzen-1,3-diol?

I am supposed to perform the above shown reaction in the lab soon, but I don't really understand the mechanism behind it. I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which resorcinol (benzin-1,3-diol) acts as the electrophile and iodine as nucleophile. I imagine the resorcinol also electron-poor because of the -I effect caused by the electronegative Hydroxide substitutes. But isn't the +M Effect stronger? Then why should resorcinol act electrophilic?
Why is Iodine nucleophilic when there is no Lewis-acid present?
And last but not least: why is the Product 2-iodo-1,3-diol and not 4-iodobenzen-1,3-diol?

Resorcinol is very electron-rich. Think of it as 1,3-dione. Iodine is a good electrohile as the I - I bond is weak.

So you are suggesting that it is a nucleophilic aromatic substitution? But how does a weak bond alone make the iodine an electrophile?

The Iodine - Iodine bond polarises very readily giving the functional equivalent of I+ easily. The carbon between the two oxygens is an electron rich site - it is an enol both ways.

The reaction is likely an electrophilic aromatic substitiution i.e. the ring acts as the nucleophile and iodine as the electrophile.

Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer plethora of subjective questions, which should be answered and cleared to show the way to right direction.

I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which the Resorcinol (Benzin-1,3-diol) acts as the electrophile and Iodine as nucleophile.

First, resorcinol is not called benzin-1,3-diol (may be in an old book). It's correct name is 1,3-dihydroxybenzene. It is true that the reaction is an electrophilic aromatic substitution. However, the electrophile here is polarizable $\ce{I2}$ molecule because it exsists in $\ce{I^{\delta +}-I^{\delta -}}$ form under reaction conditions (where $\ce{I^{\delta +}}$, which lacks electrons is the reactive center). Aromatic nucleus with two hydroxy group is electron rich (hydroxy groups activate the ring), hence acts as the nucleophile.

But isn't the +M Effect stronger?

Yes, mesomeric effect (old resonance effect) is stronger than inductive effect. Here you have two hydroxy groups to give strong $(+M)$-effect to activate the aromatic nucleus. Hence the reaction is o,p-directive. As a result, you can expect either 2-iodo-1,3-dihyroxybenzene or 4-iodo-1,3-dihyroxybenzene or 2,4-diiodo-1,3-dihyroxybenzene or etc. as your product or mixture of them based on the reaction condition used (Ref.1 & 2).

Why is the Product 2-Iodo-1,3-diol and not 4-Iodobenzen-1,3-diol?

As I explained before, the reaction conditions govern the product ratio (see the PDF of Ref.1 I attached below for better understanding). The iodination of resorcinol is not a straight forward reaction. Slight changes of reaction conditions gives you different product. For example, the major product given in the reaction using $\ce{I2/NaHCO3}/\pu{0 ^\circ C}$ condition is 2-iodo-1,3-dihydroxybenzene as indicated in the OP's question. In a large scale reaction, the yield obtained was only 66% (Ref.3). Using the same conditions, Ref.1 reported 77% isolated yield. Keep in mind that, some of 2-iodo-1,3-dihydroxybenzene isomerized to 4-iodo-1,3-dihydroxybenzene (major) and 4,6-diiodo-1,3-dihydroxybenzene in an acidic medium at high temperature (Ref.1). Thus, temperature plays an important role in percent yield in the reaction.
On the other hand, when the electrophile ($\ce{I+}$ source is changed to $\ce{ICl}$, the main product is 4-iodo-1,3-dihydroxybenzene (4-iodoresorcinol) with 4,6-diiodo-1,3-dihydroxybenzene (4,6-diiodoresorcinol) as minor product (Ref. 1 & 2). Ref.2 also stated that use of nascent iodine $(\ce{KIO3/KI/HCl})$ has resulted in the regiospecific 2,4-diiodonation of resorcinol at room temperature (which is a rapid reaction). The $\ce{^1H}$-NMR spectrum of this reaction mixture has revealed only two products, 2,4-diiodoresorcinol as major product (56% isolated yield) and 2,4,6-triiodoresorcinol as minor product in $89:11$ ratio.
References:

Ib Thomsen, and Kurt B. G. Torssell, "Iodination of Resorcinol, 5-Methoxyresorcinol, Phloroglucinol and Resorcyclic Acid," Acta Chemica Scandinavica 1991, 45, 539-542 (DOI: 10.3891/acta.chem.scand.45-0539)(PDF).
Frederick L. Weitl, "Regiospecific 2,4-diiodination of resorcinol with nascent iodine," J. Org. Chem. 1976, 41(11), 2044–2045 (DOI: https://doi.org/10.1021/jo00873a040).
Shin-ichiro Tsujiyama and Keisuke Suzuki (submitted); David B. Guthrie, Heather M. Gibney, and Dennis P. Curran (checked), "Preparation of Benzocyclobutenone Derivatives Based on an Efficient Ggeneration of Benzynes [Bicyclo[4.2.0]octa-1,3,5-trien-7-one, 5-(phenylmethoxy)-]," Org. Synth. 2007, 84, 272-284; Coll. Vol. 11 2009, p. 488-497 (DOI: 10.15227/orgsyn.084.0272).

An electrophilic aromatic substitution reaction is one in which the aromatic ring reacts (as a nucleophile) with an electrophilic reagent. In all of the reactions that follow this naming convention, the adjectives "nucleophilic" and "electrophilic" describe the reagents, not the substrate. For example, the following reaction is a nucleophilic substitution where benzyl chloride (the electrophilic substrate) reacts with a nucleophile.

So, in an electrophilic aromatic substitution reaction, the arene (a nucleophile) is reacting with an electrophile. In your case, resorcinol is the nucleophile and iodine is the electrophile.
The nucleophilicity of resorcinol is enhanced by the basic conditions (resorcinol's first $\mathrm{p}K_\mathrm{a}$ is $9.32$).
Iodine is a reasonable electrophile with a weak, highly polarizable single bond between the two iodine atoms.


text:
Which alkene reacts faster with HCl?

Which alkene reacts faster with HCl?

The logic tells me the most stable alkene 2 which is the most substituted would react the slowest.
On the other hand, it would form the most stable tertiary carbocation which would mean it's best for this specific reaction which would make it react faster. Also, one could think about rearrangements because these could take time too. How should I answer this question?

Future is a bit too much for saying that chemistry predicts - in reasonable extent - what the products of a reaction would be.

1 and 2 will give the same cation

Since the reaction proceeds via carbocation intermediate, stability of the possible carbocation will decide the reactivity here.(Do Keep in mind, Wagner Meerwin Rearrangement)

since the cations are same, Do you think that steric factors might play a role? ig the approach of cation in first case would be less hindered ? But not sure if it applies

This is an exercise in applying the Hammond postulate (multiple times). To accurately compare reaction rates, we need to have the activation energy. We don't have this, and the actual energy of the transition state is unknowable without experimental data.
Hammond's postulate states that the structure of the transition state is more similar  to the intermediate to which it is closer in energy. For these reactions, the formation of the carbocation is endergonic, so the transition states look more like the carbocation than the alkene. We can then make some assumptions about the relative energies of the transition states from the energy change of the mechanism step if some things remain the same.
For example, alkene 3 can be protonated to form two different carbocations: A and B.

Using PM6 calculations in MOPAC, I calculated the heats of formation of alkene 3 and cations A and B.
$$\begin{array}{l|r}
& \mathrm{\Delta_r H^\circ \ (kJ/mol)} \\ \hline
3 & -42.1 \\
\mathrm{A} & 710.7 \\
\mathrm{B} & 702.6\\
\end{array}$$
Since we are starting from the same place (alkene 3), we can postulate that cation B forms more quickly than cation A because it is slightly lower in energy.
You need to do similar analysis for all your alkenes and the cations they can form. In some cases you have two alkenes that might form the same cation. In that case, the higher energy alkene would likely translate into the more rapid reaction.
You do not need to go and calculate all the energies. You should have learned some rules by now to assess the relative stability of alkenes and carbocations based on substitution pattern.