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Does ferrocenium chloride ([Fe(C5H5)2]Cl) exist as a stable compound?
I am interested in using ferrocenium chloride ($\ce{[Fe(C5H5)2]Cl}$) for a synthesis application, but I couldn't find any sources referring to its existence, even though some websites and a figure in the ferrocene article on Wikipedia (Figure 6) indicate that it can be produced by reaction with FeCl3.
So my question is, is ferrocenium chloride an existing, stable compound at all? And if it is, does anyone know where more information about it can be found?

In practice, tetrafluoroborate is used. Likely a weakly coordinating ligand is needed if you want to have ferrocenium as a separate ion.

Figure 6 does not in fact specify ferrocenium chloride as thevproduct with FeCl3. It shows the cation, which could form with a complex counterion.

Ferrocenium appears to be best known with relatively low-nucleophic ions, as opposed to monatomic halides. As Mithoron comments, the most commonly available ferrocenium salt is not any chloride but the tetrafluoroborate.
Toma and Šebesta[1] report:

Soon after the discovery of ferrocene, its easy oxidation to ferrocenium cation by ferric chloride or benzoquinone was described [2]. Thus, the first isolated ferrocenium salt was ferrocenium tetrachloroferrate. The ferrocenium ion can be prepared with various counterions, such as tetrafluoroborate, hexafluorophosphate, hexafluoroantimonate and tetraphenylborate.

Thus ferrocene oxidized by ferric chloride gives $\ce{Fc^+FeCl4^-}$,not the simple chloride, as the isolated salt, and other counterions are also polyatomic having low nucleophilicty. The best known pure halide of ferrocenium is the iodide, which is actually the triiodide [3] as the iodide ion formed by ferrocene reducing elemental iodine complexes with additional iodine.
References

Š. Toma and R. Šebesta (2015). "Applicatiins of Ferrocrnium Salts in Organic Synthesis". Synthesis, Vol. 47, 1683-1695.

G. Wilkinson, M. Rosenblum, M. C. Whiting, R. B. Woodward (1952). _J. Am. Chrm. Soc., Vol. 74, 2125.

Tsur Bernstein and F. Herbstein (1968). "The crystal structure of ferricinium triiodide, (C5H5)2FeI3". Acta Crystallographica Section B Structural Crystallography and Crystal Chemistry. 24. 1640-1645. 10.1107/S0567740868004784.

Ferrocenium chloride certainly exist and it has been observed during σ-π rearrangement of σ-complex intermediate. Iron(III) chloride reacts with sodium cyclopentadienide ($\ce{NaCp}$) in the presence of THF at -80 °C to form di/tri-σ-cyclopentadienyliron complex which are unstable and rearrange to ferrocenium  chloride and ferrocene at -50 °C.

(Sic; the source misprinted charge on the chloride ion.)

It further notes:

We can anticipate that a ligand-associate (above -60 °C) Fe(III) complex represents a d5 distorted octahedral, d2sp3, configuration that is labile and of moderate stability. Thus a σ-π rearrangement giving an extremely stable ferrocene. Fe(II), d6 undistorted octahedral d2sp3 configuration is easily visualized. Similarly, a stable Fe(III) d5 ferricenium chloride also may be anticipated.

Reference:

Introduction to Metal π-Complex Chemistry, M. Tsutsui, Springer Science & Business Media, 2013

Also note that the reaction between ferrocene with iron(III) chloride does not give ferrocenium chloride, rather it is a µ-oxo compound $\ce{[Fe^{III}(η-C5H5)2]2[(Fe^{III}Cl3)2O])}$ which is named ferrocenium µ-Oxo-bis[trichloroferrate(III)]

Too bad the authors rendered chloride ion with two negative charges in the feoorcenium chloride on the upper right. I make a note of this in the answer.

I find this reaction scheme suspicious. "d2sp3 configuration" doesn't do much good too.

It maybe, of course, that the chloride ion goes on to complex with additional ferric chloride so that all ferrocenium chloride, like all glory, is fleeting.


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Stability of H3MnO4, H2MnO4, HMnO4
I cannot seek out the reason why while $\ce{H3PO4}$ (phosphorus oxidation +5) is relatively stable and obtainable but $\ce{H3MnO4}$ (the same oxidation state +5) is rarely mentioned and perhaps unobtainable. $\ce{HMnO4}$ could be found on Wikipedia, so I suppose that the stability would decrease by the following order:
$$\ce{HMnO4 > H2MnO4 > H3MnO4}$$
Here is my perspective: I would link the two compounds $\ce{H3MnO4}$ and $\ce{H3PO4}$ via $\ce{H3AsO4}$. I think the key factor that could explain this is the polarity of $\ce{X-O}$ bond. $\ce{Mn(V)}$ has lower electronegativity than $\ce{As(V)}$ so the $\ce{Mn(V)-O}$ bond is more polar and attracts water to give better hydrolysis. The electronegativity of $\ce{Mn(V) < Mn(VI) < Mn(VII)}$, it would probably be the reason why $\ce{HMnO4}$ is the most stable among acids.
Is there anything wrong with my ideas?

It's a matter of redox, not some "hydrolysis".

Manganese and Phosphorus have nothing in common. Why would you link the two acids $\ce{H3PO4}$ and $\ce{H3MnO4}$ ? What is the meaning of "link" ? And why link via $\ce{H3AsO4}$  ?

I was making observation on the stability of oxoacids whose center atoms are similar in oxidation state, which is +5 in the case of H3PO4 and H3MnO4

One difference between manganese(V) and phosphorus(V) is the former can undergo disproportionation with higher manganese oxidation states available.
Not only $\ce{Mn^VO4^{3-}}$ but also $\ce{Mn^{VI}O4^{2-}}$ are stable only as solid salts with an appropriate counterion such as $\ce{K^+}$, or in highly concentrated alkali solutions. Otherwise they tend to disproportionate forming $\ce{Mn^{VII}O4^-}$ and $\ce{Mn^{IV}O2}$. Thus we never can reach a solution acidic enough for $\ce{H2MnO4}$ or $\ce{H3MnO4}$ to be identified. Phosphorus(V) cannot disproportionate in this manner with no higher oxidation state of phosphorus available by typical chemical processes, so $\ce{H3PO4}$ is more plausible.
The $\ce{MnO4^{3-}}$ ion has been substituted for phosphate in synthetic apatites (Wikipedia, citing [1] and [2]). (Manganese in natural apatites would more likely appear in the +2 oxidation state as a trace substituent for calcium.)
References

K. Dardenne, D. Vivien, and D. Huguenin (1999): "Color of Mn(V)-substituted apatites A10((B, Mn)O4)6F2, A = Ba, Sr, Ca; B= P, V". Journal of Solid State Chemistry, volume 146, issue 2, pages 464-472. doi:10.1006/jssc.1999.8394

Grisafe, D.A. and Hummel, F.A. (1970): "Pentavalent ion substitutions in the apatite structure, part A: Crystal chemistry". Journal of Solid State Chemistry, volume 2, issue 2, pages 160-166 doi:10.1016/0022-4596(70)90064-2

Can I ask the reason that Mn(V) and Mn(VI) oxoanion is only stable at highly alkali solutions?
I was thinking that Mn(VII) oxoanion is more stable due to the stabilization of negative charge, which could be represented by resonance structures. While Mn(V) and Mn(VI) is less stabilized by the decrease of Mn=O bonds. Is it right to assume so?

Wikipedia's article on the Mn(V) species suggesrs that a protonated species is an intermediate species in the disproportionation. Also tge disproportiinations consume protons and thus are less thermidynamically favored in stringly basic solutions.

Manganese doesn't like to be in +5 and +6 O.S. and the acid form of hypomanganate and manganate anion cannot exist due to rapid disproportionation. So, they exist as salts of alkali metals. Manganese in +7 O.S. is quite stable and hence permanganic acid can still exist in aqueous solution although it can deproptonate or decompose in solution to manganese dioxide, oxygen, and water, with initially formed manganese dioxide catalyzing further decomposition.
Note that manganic acid can exist in a metastable form during the formation of manganate salts but decompose to permanganic acid as it has been noted in an old book(some of the terms are outdated):

Manganic acid in a state of purity has not as yet been obtained; it is known only in combination with bases. In order to obtain it, a strong base, for example, potash or the nitrates of such bases are exposed to a read heat in conjunction with peroxide of manganese (assume MnO2). Whether the operation be performed with or without the contact of air, a manganate is produced; with the contact of air, the peroxide of manganese absorbs oxygen and transformed into manganic acid; without the contact of air, according to Mitscherlich, the peroxide of manganese is decomposed into manganic acid and deuteroxide of manganese (assume Mn2O3)
[...]
The manganates in solid state are so intensely green that they often appear black. Their aqueous solution are intensely green also; manganic acid easily decompose to hypermanganic acid (assume HMnO4). All acids, even weakest, impart immediately an intense red color as the green solution of manganates convert into hypermanganic acid, which is accompanied by the formation of black precipitate of peroxide of manganese, with hypermanganic acid further decomposition (if further heat is applied)

Ref.: CHEMICAL ANALYSIS by H.Rose, W. Tegg Publ., 1848

"Peroxide of manganese". I guess in 1848 they could not distinguish between peroxides as in MgO2/CaO2 and dioxides as in TiO2/MnO2.

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Energy Decomposition Analysis for SN2 involving ambident nucleophiles and an alkali metal counterion
I am wanting to perform an EDA on transition state structure of the nucleophilic attack of an ambident nucleophile (2 atoms that could be used in nucleophilic attack on the same molecule) on a benzyl halide. I currently have access to Gaussian16 and ORCA, as well as any other open source software that is available. I am trying to understand what factors should be of greatest importance to these transition states, but I am unsure of what EDA to use to probe this question.
I am aware of SAPT, EDA-NOCV, and ALMO-COVP for energy decomposition analyses, but was hoping to hear of any more methods I could incorporate (that I can get access to) and/or thoughts on what dominating factors in this process (i.e., which EDA method is more applicable). I'd also like to fill any gaps in knowledge, if anyone has the time to lend me a hand.
Thanks, and all the best.

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Cesium Chloride Density Gradient Centrifugation and Isotopes
In Cesium Chloride Density Gradient Centrifugation, as used by Meselson and Stahl, Cesium ions can be seen as strongly affected by centrifugation, so much so that they can overcome their ionic bonds with water.
Amidst an equilibrium of ionic bonds and centrifugation, cesium forms a gradient, with higher concentrations near the bottom of the tube, and vice versa.
If so, will there be an enrichment of trace Cesium-135 near the bottom of the centrifugation tube compared to the top? Cesium has the dominant isotope of Cesium-133.
I am exploring this question regarding my inquiry into isotope enrichment. I am asking this question because I can see the gradient is mass-dependent (Cl ions do not form such a gradient as they are much lighter), but have seen no results in the literature.

Regarding why I am investigating this over other dominant methods: yes, this method still requires high RPM and intensive energy usage. Yet, it has the clear advantage of only needing a couple of high-speed centrifuges. Maybe even just one.

IMHO, It would have much lower enrichment degree than centrifugation of gases. Cl- must form the same gradient as the average gradient of both Cs isotopes, otherwise coulombic force would overrule centrifugation.

@Poutnik Hello, thank you for your answer, and I apologize for the late reply due to exam season. I understand what you are saying: that the coulomb force would prevent the efficient creation of Cs gradients in the first place, and I think this fact will be useful in furthering this research. Once again, thank you for your contribution.

The ratio of masses of 135Cs to 137Cs is smaller, by a factor of 5, than the ratio of masses of 14N to  15N.
While what you suggest is theoretically possible, the thermal motion of water, which must be not much below 273 K, despite freezing point depression, would make separation of isotopes in liquid water unfeasible. It might actually be more energy efficient to use a mass spectrometer!
Sorry I couldn't put a better spin on that experiment.

Thank you for your answer. There is a crucial paper regarding this topic I want to discuss in private. Is there any way to do this on Stack Exchange?

@YoungJunLee, there is a chat section, https://chat.stackexchange.com/?tab=site&host=chemistry.stackexchange.com. However, I do not have expertise to be of help.


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chemical to prevent recrystallization of zinc and iron sulphate in solution
500 gm zinc sulphate and 280 gm iron sulphate  ( total weight 780gm ) were taken in flask and water is added to make final volume 1 liter. the solution is heated to dissolve the zinc and iron sulphate in water completely. after heating any water loss is compensated by adding water and making final volume to 1 liter. the problem is after cooling down the solution zinc and iron sulphate is getting recrystallize to some extent and crystals are settling down at bottom and side of flask. is there any chemical that can be added to prevent the recrystallization of salt. i want to keep the iron and zinc sulphate fully dissolved in water even after solution cools down.
any suggestions or help is appreciated.

The unit gram has the symbol g, not gm, similarly the kilogram has the symbol kg, not kgm.

It would not change the answer, but you have not specified the iron oxidation state and generally the hydrate form of both salts.

What a user asks about may confuse and hide what the user wants to do and what the user needs to know.  A broader context, background, purpose could help in that.

It seems there is no way but using more water, as you try to violate their solubility at room temperature.
Note that both salts mutually suppress their solubility, that is usually  published as  $\pu{g/100 mL}$ of water.
The first effect decreasing solubility is the common sulphate anion, leading to lower equilibrium activities of cations, what means lower effective solubility. It is similar as decreasing of solubility of slightly soluble $\ce{PbCl2}$ by adding $\ce{NaCl}$ or $\ce{Pb(NO3)}$ to the solution.
The second effect is a general salt effect, as there is ion competition for being hydrated and dissolved.
There is possibility you could slow down the crystallization by addition of thickeners like ethylenglycol, glycerine, polyvinylalkohol or sucrose, but the addition would further decrease the solubility.
Another option worthy to try are surfactants.
In any case, additives may interfere with the purpose of the solution.


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XRD experiment: how to interpret data and solve crystal structure from 2theta vs intensity values
I have just performed an XRD (X ray diffraction) experiment (powder), but the XRD machine I have used is not able to describe crystal data, I have $2\theta$ and intensity in arbitrary units in a UXD file. My knowledge in XRD is limited, only conceptual, so I don't know exactly to solve a structure by manual calculations, I find these procedure to be very difficult. My question is if there is any software I can use to solve the XRD data into a crystal CIF file, so I can interrogate the database COD (Crystallography Online Database) to obtain my molecular model. The compound I have synthetized is not a novel/unknown product, so it is possible to find it in databases (it is Graphene Oxide).

You need just the unit cell parameters, not a CIF file.

And how do I determine unit cell parameters from XRD file? Or I need to know them from another source?

You can find an example calculation here https://chemistry-maths-book.com/chapter-9/Fourier-2D-Diffraction.html  as well as an explanation of x-ray diffraction in crystals

If your data is digitised you can use Rietveld refinement.


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Thin layer Chromotography Separation
I am confused about the basic understanding when it comes to TLC. I was under the assumption that less polar compounds move farther on the plate due to it having fewer IMFs. But I also heard that a more polar eluent will make the compound not move far. I'm not sure how to overlap these two findings which would have a better separation.
For example, I was given a question where we would match the eluent to the solvent plate. Some choices were 100% hexane, 85% ethyl acetate in hexane, 50% ethyl acetate in hexane, and 10% ethyl acetate in hexane. I thought the plate for 100% hexane would have spots near the solvent front because it is nonpolar, but those spots were closer to the start line. Visa versa for 85% ethyl acetate in hexane as the spots were closer to the solvent front even though it is very polar.
What is the overlap in IMFs of compound found in the spots vs the eluent?

The whole question and discussion is meaningless without knowing the stationary phase coated on the TLC. What is IMF?

Aside of distribution of analyte between the phases, there is the factor of affinity of the mobile phase, or it's components, to the stationary phase.
If the stationary phase is polar and if the mobile phase contains a polar enough component, it has tendency to wash everything out.
It can be frequently seen the system has the second solvent front and the stationary phase looks differently when it passes.
It is clear that ethyl acetate is washing out your analytes.


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Can products of Claisen Condensation undergo the reaction again?
I have a reaction of ethyl acetoacetate with 1-bromopentane in ethanol and sodium ethoxide. I understand that the main product is formed by SN2 mechanism between ethyl acetoacetate and 1-bromopentane, but can the acetoacetate also undergo Claisen Condensation in this case? It has two alpha-hydrogens, so it should be technically possible. Or is it due to stereochemistry that the Claisen Condensation can´t appear with ethyl acetoacetate?

Going to depend on the reaction conditions - how much pentyl bromide and ethoxide was added. 2 eq of ethoxide and 1 eq of bromide gives you the very stable anion of the pentyl acetoacetate, a 2nd eq of pentyl bromide gives you the neutral di-pentyl acetoacetate. More ethoxide then creates the possibility of Claisen.

It's not stereochemistry, but steric hindrance that hampers disubstitution. It could still happen to some extent, though, if you're not careful.


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What is the reason of producing *spark* in the volcanic reaction (decomposition of ammonium dichromate)?
In a scientific demonstration of volcanic eruption(decomposition of ammonium dichromate), the green chromium(III) oxide produces an orange spark to mimic the eruption of lava from a volcano.
I would like to know how the spark is produced. Is it due to the release of photon when the excited state returns to the ground state?

The mechanism of spark shining is the same as the mechanism of the orange glow in the core of the "volcano" - thermal radiation of solid bodies.
Sparks are just dust particles of mostly Chromium(III) oxide, going upwards due intense thermal convection and by local mechanical effects of self-burning of crystals.

To prove that point, use a diffraction grating or hand-held spectroscope -- if the spectrum is continuous, it's thermal, and if there are lines, then I'd be excited to see the results of that experiment -- post it here as an answer.

Sure. Send me the device.

That comment was for the OP ;-) BTW, hand spectroscopes are ~US$2.5 at Amazon and AliExpress.

@DrMoishePippik It was hard to imply without the explicit user reference. :-) // The point is, sparks of any kind are not gas based to have a line spectrum.

P.S. Obviously except electric arcs and similar, that are rather a plasma

And thermoluminescence in spark particles, which also has peaks: https://www.edinst.com/photoluminescence-thermoluminescence-of-phosphors/ BTW, not everything radiates as a black body, just to put that in the limelight.

Let us continue this discussion in chat.


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Determining High Spin or Low Spin Configuration: Influence of Weak and Strong Field Ligands
Can a coordination complex be classified as high spin if it possesses the same number of unpaired electrons in both weak and strong field ligand environments? I am curious about the factors that determine the spin state of a complex and how it may or may not change in response to different ligand strengths. Any insights into the interplay between ligand field effects and the spin state of coordination complexes would be greatly appreciated.

Is the statement "With strong field ligand, Cr(III) octahedral complex will be high spin"(it is $d^3$ so there will be no effect of WFL or SFL on number of unpaired electrons) correct or incorrect

High spin and low spin are the relative terms in splitting of electrons to the $t_{2g}$ and $e_g$ orbitals. According to the molecular orbital theory (MOT), the ligands approach the metal symmetrically along the vertices of an octahedron with the metal at center. Considering the axes of approach of ligand to be $\pm x$, $\pm y$ and $\pm z$, the ligands interact with the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (along with other valance orbitals), splitting them into $e_g$ bonding and $e_g^*$ antibonding orbitals. In this case, $e_g$ is the symmetry of the orbitals. Rest of the $d$-orbitals of the metal, $d_{xy}$, $d_{yz}$ and $d_{zx}$ does not interact with the ligand and stay as non-bonding orbitals of $t_{2g}$ symmetry.
Thus, according to MOT, the interactions of metals with the ligands make the d-orbitals split. The electrons fill up the orbitals one by one, and the number is such that the number of electrons in the $d$-orbitals of the metal exactly match up with that to be filled up in the $t_{2g}$ and $e_g^*$ orbitals, assuming the outermost $s$-orbital of the metal is filled and each ligand provides a contribution of two electrons.
When a "strong field ligand" couples strongly with the metal, the $e_g^*$ orbital has higher energy than the exchange energy (pairing energy) of electrons when compared to $t_{2g}$. In this case, the electrons, specifically the fourth and fifth electron to be filled, find it energetically more favourable to pair up in $t_{2g}$, thus reducing the spin quantum number and hence making a low spin complex.
On the other hand, a "weak field ligand" couples weakly with the metal orbitals, and thus the exchange energy is higher compared to the split between $t_{2g}$ and $e_g$. Thus, the fourth and fifth electrons prefer occupying the $e_g^*$ orbitsl in this case, thus making it a high spin complex.
A table with the electronic configuration for ideal strong and weak field ligands are given below:

No. of $d$-electrons
$t_{2g}$ SF
$e_g^*$ SF
Unpaired $e^-$ in SF
$t_{2g}$ WF
$e_g^*$ WF
Unpaired $e^-$ in WF

1
1
0
1
1
0
1

2
2
0
2
2
0
2

3
3
0
3
3
0
3

4
4
0
2
3
1
4

5
5
0
1
3
2
5

6
6
0
0
4
2
4

7
6
1
1
5
2
3

8
6
2
2
6
2
2

9
6
3
1
6
3
1

10
6
4
0
6
4
0

For a full diagram of the MO for octahedral metal-ligand interactions, please check this link.


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How can H2 emit light with only 2 molecular orbitals?
I know that light is emitted when an electron is excited to a higher energy state and then decays back to its ground state, emitting a photon whose energy corresponds to the difference in energy of the two energy states.
I also learned that if a H2 molecule absorbs a photon with an energy equivalent to the energy difference between the bonding and anti-bonding orbital, then the electron will be promoted to the anti-bonding orbital, causing the bond between the H atoms to break (since there will be no stabilizing interaction).
My question is how can a H2 molecule emit light if the only possible higher energy state that the electron can go to is the anti-bonding oribtal, and that will break the bond (which doesnt release photons)? What am I missing here?

The light could be emitted when the dissociated atoms recombine, or perhaps tge electron returns to the ground state before the molecule can actually dissociate.

When an electron is raised to a higher level, that absorbs energy. When it falls back to a lower energy state, it releases energy.

Experiments show that there are several electronically excited states of increasing energy, each of these also has vibrational & rotational  levels from which emission can occur.

As 1s is not the only orbital in H atoms. transition to the antibonding sigma orbital is not the only option.

Gerhard Herzberg (Nobel Prize in Chemistry, 1971) literally wrote the book on the spectra of diatomic molecules: https://en.wikipedia.org/wiki/Gerhard_Herzberg. So molecular spectroscopy is complex and the little first year chemistry MO diagram is not particularly useful in this regard.


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Oxidation states of Thiosulphate

I was given this question and don't understand why the two sulphur atoms have different oxidation states - and how to actually figure it out. Could someone please explain it to me? I've attached the reasoning that was provided but it isn't very clear to me. I understand that oxygen is usually -2 (with exceptions) and that the molecule itself should have a net charge of -2 but I thought that then meant that each sulphur atom would have an oxidation state of +2.  Any help would be greatly appreciated!!

If you count oxidation state formally, than it is done by moving all electrons of a shared electron pair to the more negative atom or dividing it between atoms if they have the same oxidation state. This, however, gives rise to question for which resonance structure one calculates oxidation state, which might give different values for thiosulfate. On another hand, thiosulfate is isoelectronic to sulfate, so it is natural to use same oxidation numbers for same positions as in sulfate. This approach is often used for organization of compounds into groups.

If the last paragraph in your quotation is meant to be the book's answer, I would stain that text with my intensely colored, anthocyanin-rich cranberry jam*. I would find either $-1$ and $+5$ or $0$ and $+4$ more defensible.
-1 and +5
In this model we use the raw electronegativities of the atoms to render the sulfur-sulfur bond nonpolar, so the outer sulfur atom would have the same $-1$ oxidation state as it would in the disulfide ion $\ce{S2^{2-}}$. The central sulfur atom, losing two electrons to each oxygen atom to satisfy $-2$ for the latter, ends up at $-1+6=+5$.
0 and +4
Here we take a slightly more advanced approach, arguing that the sulfur-sulfur bond is inductively polarized by the oxygen atoms withdrawing electronic charge from the central sulfur. In response the sulfur-sulfur bond has its electrons polarized towards the central atom as if that atom were more electronegative. Thereby the outer sulfur is rendered with an oxidation state of $0$ and the inner one with $-2+6=+4$. This assignment has the advantage of dovetailing with a well-known chemical property of thiosulfate: a generally redox-inert acid such as hydrochloric acid causes it decompose to elemental sulfur (oxidation state $0$) and sulfur dioxide (sulfur oxidation state $+4$).

*Cranberries are actually white on the inside when fresh. The bulk color turns red only when the anthocyanin dye, which is not colored in its reduced form, is exposed to air as the cranberries are crushed or popped.

Lol yes I would love to stain the textbook too. Thank you for the response, that makes a lot more sense to me!!

@KokoFaen If the answer helped, please accept it by clicking on the checkmark.

There's one in every crowd. Would be good to know why the downvote, even if that is as much of a luxury as a free lamborghini.

@OscarLanzi maybe they don't like cranberry jam ;)

@NilayGhosh that 's because they're thinking of the canned stuff. Make your own and then you'll like it.

I'm getting rwo downvotes,so must be missing something. Do tell.


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Does Cu+ have a greater ionic radius than Sr2+?
Although Strontium is in group 2, reducing the number of electrons as it becomes ionized makes it group 18, period 4 in terms of electrons. Therefore, ionized Strontium (Sr2+) is in the same period as Cu+. Cu+ has four shells and Sr2+ has four as well. The deciding factor for which ion has the greater ionic radius is the number of protons. The fewer protons given that the comparing ions are in the same period means that it has a greater radius. Sr2+ has more protons whilst being in the same period as Cu+, therefore the increase in nuclear charge provides a greater force of attraction between the nucleus and the electrons, decreasing the ionic radius. Does this mean that Cu+ has a greater ionic radius than Sr2+? Why is it said in my workbook that Sr2+ has a larger radius than Cu+?

The wikipedia table of ionic radii says r(Cu+)=91 nm, r(Sr62+)=132 nm. Not the decreasing radius trens along the periods link and that ions for the first two groups are relatively big.

I think I got it. I did not consider the amount of shells the ions concluded for some reason, which is the obvious thing to do. Cu+ has an electron configuration of 1s2,2s2,2p6,3s2,3p6,3d10 and Sr2+ has an electron configuration of 1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6. Putting it in a way that looks better in order to visualize the shells, it would look like this respectively:
(1s2),(2s2,2p6),(3s2,3p6,3d10) and (1s2),(2s2,2p6),(3s2,3p6),(4s2,3d10,4p6).
As you can see, Cu+ only has 3, whilst Sr2+ has 4. As Sr2+ has more inner shells, Sr2+ is larger.


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Is it possible to make acids and bases in amphiprotic solvents other than water
In such solvents, could there be chemicals which could donate/accept protons, and if so could their solutions display acidic/basic properties?

Organic chemistry is filled with examples of chemicals that accept or donate protons in solvents other than water

You may find useful reviewing of search results for
site:libretexts.org OR site:stackexchange.com OR site:wikipedia.org OR site:hyperphysics.phy-astr.gsu.edu acids bases amphiprotic solvents

Certainly! There are various definitions for acid and base, such as Brønsted–Lowry, that describe acid/base reactions as the transfer of protons ($\ce{H+}$), without requiring hydroxyl ions ($\ce{OH-}$), and Lewis, which goes beyond $\ce{H+}$ and deals with electron exchange.
An example of a Brønsted–Lowry acid/base system is ammonia as the solvent. The ammonium ion ($\ce{NH4+}$) corresponds to the hydronium ion, and the amide ion ($\ce{NH2-}$) to the hydroxide ion.
An example of a Lewis acid is $\ce{BF3}$, an electron acceptor rather than a proton donor.


text:
Cyanoacrylate glue and citric acid
I got some cyanoacrylate glue on my fingers yesterday which dried up quickly and would not fall. My daughter (n.b. she's 7th grade, no chemistry education but wants to be a chemist and reads a lot by herself) suggested I try citric acid. Her reasoning was that it's a harmless acid and "might dissolve the triple bond", or maybe hydrogen bonds will form between the "many hydroxides, which will make the glue more soluble".
So, after literally sticking my finger in a lemon for 15 min. half of the white substance was gone and the rest just gradually fell off in the next 1-2 hours.
My question is, what really happened? What was the reaction?
My meager organic chemistry knowledge from highschool is very inadequate for answering this.

Not sure exactly why the acid helped. But it is usually advised to immerse the glued body parts to water as warm/hot as one can withstand for minutes. The help is often said to be due sweating of glued skin.


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Will H+ ions undergo reduction at cathodes in a chemical cell?

Metal P is the cathode, and it should attract anions.  In this cell, anions present are Cl- and OH-, but they are reducing agents, reduction should occur at cathodes.  So will H+ ions undergo reduction at the Cathode, even though it is a cation?

Why should cathodes attract anions?

I thought positive electrodes attracted negative ions

And cathodes are positive because..... ? // BTW you have not specified if it is a galvanic cell or an electrolytic cell.

Its a chemical cell, I guess its the same as a galvanic cell

A chemical cell is too broad term. Both mentioned cells are chemical cells. // You may find useful reviewing of search results for
site:libretexts.org OR site:stackexchange.com OR site:wikipedia.org OR site:hyperphysics.phy-astr.gsu.edu electrolytic galvanic cell cathode anode potential

It it was a galvanic cell and if the copper electrode served as anode, the metal P would have to be a metal with higher standard reduction potential like silver and the electrolyte would have to be some soluble salt of that metal, like $\ce{AgNO3}$.
Then, there would be
dissolution of the copper anode:
$\ce{Cu^(s) -> Cu^2+(aq) + 2 e-}$
metallic silver deposition on the cathode:
$\ce{Ag+(aq) + e - -> Ag(s)}$

If is was an electrolytic cell and if the copper electrode served as anode, then
there would be dissolution of the copper anode again:
$\ce{Cu^(s) -> Cu^2+(aq) + 2 e-}$
but evolution of the gaseous hydrogen on the cathode:
$\ce{2 H+(aq) + 2 e- -> H2(g)}$

Note that any static potential difference of electrodes causes very tiny migration of ions in the opposite directions until the potential gradient in the solution is cancelled by the equal counter-gradient due charge dislocations.
Note that both anions and cations are present at both electrodes. the potential difference makes just very tiny difference in that.
If the is ongoing net oxidation/reduction on electrodes, it disrupts the above equilibrium, and there is ongoing ion migration, caused by electrostatic gradient and and concentration gradient.

Poutnik's answer is perfectly correct. But the three following informations should be added.
$1)$ It must be mentioned that, in order to get a galvanic cell with an $\ce{Cu/Cu^{2+}}$ anode and a $\ce{Ag/Ag+}$ cathode, the two electrode compartments must be separated by a membrane or a bridge. If these compartments are not separated, the following reaction will immediately occur on the copper plate $$\ce{Cu + 2 Ag+ -> Cu^{2+} + 2 Ag}$$ and the cell will not work at all.
$2$) Also the anodic electrolyte must not contain $\ce{MgCl2}$, because the silver ions $\ce{Ag+}$ present around the anode will react with the $\ce{Cl-}$ ions to produce a precipitate of $\ce{AgCl}$ according to the following equation : $$\ce{Ag+ + Cl- -> AgCl}$$
$3$) Anyway the cell will not work a long time, because after a while, some $\ce{Cl-}$ from the cathodic compartment will cross the membrane and react with the silver ions. So the $\ce{Ag+}$ ions will disappear for two reasons : a) normal behavior of the anode, b) precipitation of $\ce{AgCl}$.
So the cell description must be entirely modified to be able to deliver some current.


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Products formed upon pyrolysis of alkanes
What sort of products will be formed when we subject a higher alkane (say decane) to pyrolysis by heating at 773-973 K in the presence of $\ce{H2/Ni}$? I know that lower aliphatic alkanes and alkenes will be products that will be formed but will other products such as aliphatic alkynes and saturated and unsaturated cyclic compounds be formed (say cyclopropane and cyclopropene)?

Pyrolysis of alcanes is often carried out with heavy hydrocarbons from kerosene or higher fraction of petroleum and mineral oils. It produces a mixture of lighter alkenes and alkanes. Alkynes are not produced significantly. But with $\ce{H2/Ni}$, aliphatic alkenes are reduced to alkanes. The proportion of aromatic compounds is about $10$% in petroleum. No cyclopropane ou cyclobutane are obtained by pyrolysis of petroleum.

@Maurice Oh yeah I completely ignored the fact that we were using H2/Ni so any alkynes or alkenes would be reduced to alkanes, I just thought it was some sort of reagent that is necessary to use. BTW are any non aromatic or anti-aromatic cyclic compounds formed upon pyrolysis of petroleum or is it just aromatic compounds?

Does it not basically vary from compound to compound? I've studied in my 11th year about two different large-sized alkenes and we were expected to remember those two as where the chain broke was quite random and had no logic

I would like to send an image of my book too but the comments section won't help

I think you're talking about hydrocracking.

Pyrolysis of alkanes (cracking) involves breaking of $\ce{C-C}$ and $\ce{C-H}$ bonds to form free radicals. These radicals join to form smaller alkanes. So, there is higher chance of forming straight alkanes rather than unsaturated alkynes. But still there are chances of forming alkynes (where the alkenes undergo dehydrogenation) which are considered undesirable impurities in the cracking process.

In the last stage of cracking, the alkanes can undergo cycloaddition and aromatization to form cycloalkanes/cycloalkenes and benzene.
Ref.: Gholami, Zahra & Gholami, Fatemeh & Tišler, Zdeněk & Vakili, Mohammadtaghi. (2021). A Review on the Production of Light Olefins Using Steam Cracking of Hydrocarbons. Energies. 14. 8190. DOI: 10.3390/en14238190
My other answer: Is reforming and isomerization the same thing?

Did you see my comment?^ It appears to be hydrocracking - much different results.


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Is the amine nitrogen of methylglycinediacetic acid (MGDA) protonated?
MGDA is a tertiary amine with 3 acetic acid groups. I would like to know whether the amine Nitrogen of MGDA is protonated or not for a given pH value. I could not find any literature data which specifies the molecular form of the amine see here.
Conjugated acids of tertiary amines have a pKa of around 10.8. Even if the pKa value reduce a bit in MGDA (due to electron withdrawing carboxylates), I feel the pKa should still be high enough for the amine to be protonated. However, none of the reported structures of MGDA reports this amine to be protonated. can this be due to steric effects ?

The same was observed for GLDA, a very similar amine + carboxylate structure See here

related https://chemistry.stackexchange.com/questions/86137/why-do-the-amino-groups-in-edta-deprotonate-first-in-contrast-to-amino-acids

It's not a structure of acid but salt. Neutral acid would be mostly a zwitterion - sterics would only slightly hamper protonation.


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Tollens' Test - Clean Up
In school, we had an experiment to find out the biomolecules contained by the different food-stuffs. After doing Tollens' test which turned out to be positive, our teacher instructed us to clean the test tube by using crystals of $\ce{K2Cr2O7}$ and concentrated $\ce{H2SO4}$. Then the solution in the test-tube turned green-yellow with an odour similar to that of Arlin oil (Ayurveda product, I really don't know what adjective I should describe it as).
When I asked my teacher about it, she told that it helps in cleaning the test tube from the silver mirror but did not know the chemistry behind it. Can anyone please explain to me how it works, with chemical equations supporting the same?

This mixture, often called "chromic acid," is a common way to clean glassware. It is a strongly oxidizing acid which reacts with most organic material, so it is able to remove most types of organic sludge. You are likely smelling some mixture of the oxidation products.
It is very important to clean up Tollen's reagent, as it produces an explosive upon standing. More info here, including an accident report. However, chromates and dichromates are carcinogens and need to be disposed of as hazardous chromium waste. Making the spent (or unused) reagent safe only requires a dilute acid, so in most school labs, the spent Tollens reagent is disposed of with dilute acid and lots of water, producing fewer hazardous byproducts. This does not remove the silver mirror, which could be removed with nitric acid... or you can just dispose of the test tube. Chromic acid is overkill and most people would not risk exposure to chromium(VI) to save a test tube.
This paper [1] is very old, but it describes the reaction between silver, sulfuric acid, and potassium dichromate. I haven't looked beyond the first page (I don't want to pay for a paper from 1837!), but it tells you want to expect. Perhaps the ability to react with the silver mirror is why your teacher uses chromic acid despite the hassle and risks.
[1] R. Wazington, Esq., "On the action of chromic acid upon silver, and its combinations with the oxide of that metal," The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 1837 (first page).

Further from the paper that you've shared, '...and is changed to a dark crimson ; this salt is the bichromate of silver... During this operation the colour of the supernatant liquid becomes much deeper, until the orange red tint of the original solution has become of a dark mahogany hue, and it then passes gradually into a deep green." So did I notice an extremely fast reaction? As I could see only green and not all these details, given the lack of slo-mo cameras in the 19th century...

Plus it would also be nice if you could include the equations, given that glucose solution was the sample for the Tollen's test.

@HarikrishnanM regarding your first comment - that is a possibility along with other dichromate salts it is generating. All of them are orange.

This answer will focus on the equations involved in the reactions of the questions. For the explanation please refer anon's answer.
Equations:

Tollen's test of formaldehyde (works same for glucose, only difference being that instead of formic acid, gluconic acid is formed)

$\ce{HCHO + [Ag(NH3)2+]OH- -> HCOO- + Ag \downarrow + NH4+}$
Here Ag forms the silver mirror.

Removal of Silver mirror with acidified $\ce{K2Cr2O7}$

$\ce{Ag + H2CrO4 -> Ag2CrO4 + Cr^3+}$
$\ce{Ag2CrO4}$ is a crimson red insoluble compound. $\ce{H2}$ gas is evolved.

Oxidation of HCOOH formed to $\ce{H2CO3}$ followed by further decomposition into water and $\ce{CO2}$ gas along with reduction of Cr(VI) to Cr(III)

Formation of Ammonium Chromate - yellow

From all this, the chromates formed are precipitated by the formation of chromic acid in the reaction $\ce{K2Cr2O7 + H2SO4 -> H2CrO4 + other products}$.
The smell may be explained by the ammonium ion formation in Tollen's test and the colour changes may be explained by the different precipitates that are formed throughout.
SOURCES: Wikipedia, ChemicalAid, ACS Publications, NCERT.
Answer updated in accordance to inputs from comments.

Chromic acid is a tremendous oxidizing agent and whatever organics the Tollen's test generates will react with it along with the residing silver and in that process, the chromium which is initially in +6 O.S. will be reduced to +3 O.S. which defines the gradual change in color to green. This is a necessary steps as untouched test tubes after performing Tollen's test can be dangerous..

@NilayGhosh post has been verified and edited accordingly. Any other suggestions are welcome.

Ag dissolution in K2Cr2O7/H2SO4 would lead to Cr(III), not H2. Otherwise, Ag would get dissolved in HCl too, what is not the case.

@Poutnik inputs have been incorporated.

Well, not really. The reaction is not enumerated (You have just replaced H2 by Cr^3+) and the comment remains wrong.

Well @Poutnik can you please edit it?

I can. But it is better for you if you do so yourself.

I'll check as I am a bit confused with all the info input

@Poutnik, please help me out. I am certainly clueless on your comment.
Edit 1 rectified Nilay Ghosh's comment, edit 2 rectified yours

@Poutnik please edit this post out, I am exhausted from thinking what mistakes I made. Please help


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Why do bonds interact with other bonds and exhibit overlapping effect?
In carbocations, alkenes and carbon free radicals during hyperconjugation, overlapping of molecular orbitals take place. So how and why does this happen?


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In JJ thomson's cathode ray experiment why is the effect of gravity on the electron not considered?

Explaining the setup:
The experiment is described in the picture.
Instead of the magnets in the picture imagine two circular coils on both the sides with current running through it, this creates a magnetic field perpendicular to the loop which can be predicted using Biot-Savart law. Run the current in the loop in such a way so that the magnetic field is going into your screen, away from you. Without the magnets the beam is deflected up towards the positive plate proving the charge on the beam is -ve as +ve and -ve attract.
$q(v \times B) = \textrm{Magnetic force}$
so the magnetic force is acting downwards.
The q/m ratio can be determined as well by combining both the magnets and the charged plates and changing the electric and magnetic fields until there seems to be no deflection on the beam so net force acting on beam must be close to 0 and we solve taking all the forces, more details here.
I had a problem with how Thomson finds the q/m ratio which is why the gravitational force on the electron is not considered? Only the magnetic force and electric force is considered. I thought maybe gravitational force is negligible? I included gravitational force and got an expression which relates q/m to a constant value
Magnetic force + Gravitational force = Electric force
or
$$qvB +m \times \pu{9.8 m/s^2} = qE$$
$$v = E/B - (\pu{9.8 m/s^2} \times m)/(qB)$$
When we only have the magnetic force acting on the beam, it will always act perpendicular to  the velocity due to the magnetic field going into the screen, so it will constantly change the direction of velocity  but not its magnitude and the beam will move in a circle:
Centripetal force = Magnetic force
or
$$mv^2/r = qvB$$
$$mv/r = qB$$
$$\frac{m(E/B - (\pu{9.8 m/s^2} \times m)/(qB))}{r} = qB $$
$$\frac{mE}{Br} - \frac{\pu{9.8 m/s^2} \times m^2}{Bqr} -qB = 0$$
This can be treated as a quadratic equation where the variable is m. If we have an equation $ax^2 + bx + c = 0$ its roots are $\frac{-b +- \sqrt{b^2 - 4ac}}{2a}$, so that
$$m = \frac{ \left( -E/Br +- \sqrt{ \frac{E^2}{B^2 r^2} - \frac{4 \cdot \pu{9.8 m/s^2}}{r}}\right)Bqr  }{2 \cdot \pu{9.8 m/s^2}}$$
$$\frac{m }{q}= \frac{ \left( -E/Br +- \sqrt{ \frac{E^2}{B^2 r^2} - \frac{4 \cdot \pu{9.8 m/s^2}}{r}}\right)Br  }{2 \cdot \pu{9.8 m/s^2}}$$
But I don't know the magnetic and electric fields. I need the values of electric field of the charged plates and magnetic field of the coils at which there is no net deflection and the radius of the circle the beam goes through when only the magnetic field acts on it, provide the source as well. Maybe there is some other way to understand why gravitational force is not included if so could you explain that way.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

The question should be clear, Why is effects of gravity not included in JJ thomson's finding of q/m ratio of electron where q is its charge and m its mass. What details are missing?

Pick some electron energy, and find the velocity and then the time spent traversing the experiment. From that time you can directly calculate the distance the electron 'falls' during the experiment. Say 1keV electron and 1 meter. I get a vertical displacement from gravity of ~10 femtometers, but may have made a mistake.

@JonCuster Yes there is an answer on physicsstackexchange which i linked which has the same answer. That is the best answer I feel as we can also say by just looking at the beam that the deflection is negligible in absence of electric and magnetic fields.

Various versions of Thomson's original article (Philosophical Magazine, 44, 293-316 (1897)) can be accessed freely online, for instance at Google books, or here (link liable to rot).
But you can work around not knowing the details of the experiment. Interestingly, you don't even need to know the magnitude of the magnetic field applied (therefore current).
Starting from your two equations we can  compute $q/m$ (solving a quadratic equation in q instead of m):
$$\frac{q}{m} = \frac{E}{2B^2r}+\left[\frac{E^2}{4B^4r^2}-\frac{g}{B^2r} \right]^{1/2} \\ = \frac{E}{2B^2r}\left(1+\left[1-\frac{4gB^2r}{E^2} \right]^{1/2}\right) \tag{1}$$
We can use the fact that
if we assume g is negligibly small then we know the solution is
$$\frac{q}{m} = \left(\frac{E}{B^2 r}\right)_{g=0}$$
Based on this, if we define
$$\left(\frac{q}{m}\right)_{\textrm{approx}} = \frac{E}{B^2 r}$$
we can use a Taylor expansion assuming the gravitational effect to be very small relative to the electromagnetic ones to recast Eq (1) as
$$\frac{q}{m} \approx \left(\frac{q}{m}\right)_{\textrm{approx}}\left[1-\frac{g}{E}\left(\frac{m}{q} \right)_{\textrm{approx}} \right] \\  \approx \left(\frac{q}{m}\right)_{\textrm{approx}} \left[1-\frac{mg}{qE} \right]$$
This result suggests using known constants to evaluate
$$E >> \frac{mg}{q} = \pu{5.57e-11 V/m}$$
as a condition for the value of the applied electric field above which g can be neglected. Using reasonable values for the distance between the electrodes we see that the applied potential required to satisfy the condition is bound to be quite small, much smaller than the range for instance reported in examples online, which are in the order of hundreds of volts.

If you insist on wanting to know the current applied, work backward using the balance condition to solve for the applied magnetic field, knowing the range of electric fields that may have been used.

Can you explain how you got $\frac{q}{m} \approx \frac12\left(\frac{q}{m}\right)_{\textrm{approx}}\left[2+\frac{mg}{qE} \right]$

I made a mistake in the centripetal force = magnetic force equation, I didn't include effects of gravity there Im trying to include it. If I can then there should be a change in the equations

@Saif Perform a Taylor expansion on $(1+x)^{1/2}$. // Yes, I thought about the additional term in centripetal force balance, but that leads to messy equations I believe. Consider my answer a first-order approximation. Also, consider Ian Bush's comment. This is a different way to justify ignoring the g term.

I'm not familiar with Taylor expansion so I tried learning it.
$$\frac{q}{m} = \frac{E}{2B^2r} +(\frac{E^2}{4B^4r^2}(1 - \frac{4B^2rg}{E^2}) )^{1/2} $$
$$ \frac{E}{2B^2r} +  \frac{E^2}{2B^2r}(1 + x)^1/2 $$
$$ \frac{E}{2B^2r}+ \frac{E^2}{2B^2r}(1 + \frac{1}{2}x)$$ Here you are using the approximation around x = 0?
so then I substituted for x I was typing this as you edited it, and got what you got in the new edit

It should be $\frac{E}{2B^2 r} (1 + x)^{1/2}$

Yes, a few errors crept in but it's correct now I believe. It seems the approach used by Thomson to determine e/m differs from that described by the equations in your post. His approach used magnetic deflection and measurements of heat generated and total charge of the cathode rays. Either way, Thomson's article provides you with estimates of the magnitude of electric fields used to deflect the cathode rays during the determination of the conductivity in the evacuated tubes.

Really how with heat, I'll look at the article. Anyway Ian's answer is good as it can be confirmed gravity's effects is negligible by just looking at the beam when there aren't any electric or magnetic fields and seeing the beam is not deflected at all + this answer  is good as well. Though I still want to know how and what would happen if I could somehow include gravity's effects in the centripetal force but ill try it on my own.

I linked here from my answer to the HSM SE cross-posted version


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Matching a Slater-type wavefunction with 6-31G basis set for first and second row atoms Gaussian basis set
I am trying to match the slater wavefunction with 6-31G basis set for first and second row elements. I have matche hydrogen, helium and lithium atom 1s orbital with Slater wavefunction using STO-6G but unable to match with 6-31G basis set.

What you mean exactly by matching? You must review how Slater Type Orbitals (STOs) and Gaussian Type Orbitals (GTOs) are build. You must  consider what type of orbitals are used to describe the radial part of the orbital, and  distinguish between primitive and contracted orbitals. Basically, with STOs you get only an orbital of each type of the "available" orbitals (minimal basis set).

continue...  In the case of GTOs you get and extended basis set. For 6-31G b. s. (split valence b.s.), you get one orbital for each of the internal electrons but you get two orbitals of each type ($\mathrm{s}$, $\mathrm{p_x}$, $\mathrm{p_y}$, $\mathrm{p_z}$,...) for the valence electron orbitals.


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Can a multimeter be used to measure PH?
I would like to feed my garden a solution of water and Trisodium Phosphate (TSP), but I am uncertain of the correct amount of TSP to water ratio. Too acidic or too basic of a solution can harm my plants. Since TSP is very basic, it should be more conductive, or inversely, has less resistance than pure water which has a high resistance.
I figure I might be able to use a multimeter to measure the pH, but how I would calibrate and interpret the readings. A resistance of zero would mean it's very basic, but it could also mean it's very acidic. Since regular tap water is already fairly conductive, the readings would require some scaling using a resistor in series with the probes.
Alternatively, would supplying a small voltage (3V) in series with the probes while measuring  of the solution to determine its polarity and resistance and thereby its pH? A negative reading, implies it's acidic whereas a positive reading implies its basic.

Trisodium phosphate may be very bad idea unless very diluted. It is too basic for ordinary life, similarly as sodium carbonate.

The answer to the question in the title is "no." A multimeter cannot measure pH. Dissolving any ionic compound, whether acidic, basic, or neutral, will lower the resistance. There is no way to tell the difference. If you are asking whether a multimeter can determine the concentration of TSP in your water, then yeah, but why not just weigh it when you make the solution? Besides, the pH of your fertilizer is not especially useful for determining what happens when it is applied to soil. Soils generally have some buffering capacity.

A multimeter measures resistance, i.e., the blocking of electron (or ion) flow. A pH meter measures the voltage difference between glass-encased electrodes.
As you might realize, silica-based glasses are not very conductive, so a meter with very high input impedance must be attached to those probes.
Your proposed circuit might measure something, and perhaps what the circuit measures might have some relationship to ion concentration and ion mobility, but it would not measure pH, per se.

Confused. Why would the meter need to have a high impedance? Wouldn't it need a lower impedance if the electrodes are not conductive?

See Ohm's law for resistance in series.

@user148298 An ideal voltmeter has infinite impedance. The next best is having the impedance several orders higher than the impedance of the measured circuit.

You cannot use a multimeter to measure the solution pH. What you could do is to measure it's conductance, measuring AC current going through the solution from AC low voltage ($\le \ \pu{12 V}$) source in defined geometry.
Carefully add small amounts of the trisodium salt until reaching maximal acceptable pH, measured by pH kits or paper strips. Note it's conductance (or AC current)
Then you can indirectly control pH of future prepared solutions by measuring it's conductance or AC current. This way you can avoid expenses for regularly spent pH strips or kits. OTOH, pH kits may last long.
But there remains factual objection - why trisodium phosphate? It is too alkalic and brings sodium to soil, what may not be desired.

I heard TSP was soft banned as an ingredient in household cleaners precisely because it's a powerful fertilizer, leading to the overgrowth of algae which leads to the death of fish in streams and lakes due to asphyxiation.

It is valid for all phosphates, But they were mostly used in solid formulas for wash machine. The question remains why trisodium phosphate? Other phosphates are more suitable as fertilizer. High pH would lead to phosphate precipitation by soil cations.

Why AC and not DC current?

AC is used to measure conductance to avoid deposition of chemicals on electrodes, what affects measurement. Solutions do not have simple resistance/conductance values like solids.

I heard TSP was soft banned as an ingredient in household cleaners precisely because it's a powerful fertilizer, leading to the overgrowth of algae which leads to the death of fish in streams and lakes due to asphyxiation.

Why AC and not DC current?


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Small (personal) scale sea salt production?
I was looking to try to make a small setup for producing small amounts of sea salt. Ideally a series of evaporation chambers, with a heavy brine as the finished product, that could then be baked. May need one (or more) heating elements? What sort of equipment would be needed to achieve this, and how would I go about setting it up? I live in a super tiny apartment, and I don't have a whole lot of space, so the smaller the better. Trying this out as a little project, to potentially upscale in the future. Thanks so much!

What a user asks about may confuse and hide what the user wants to do and what the user needs to know.  A broader context, background, purpose could help in that.

Consider the monetary cost, and energy cost, and possibly harm to the environment, to use a heat source other than solar to evaporate water. Use mathematics: How much energy does it take to raise water from room temperature to boiling., and from boiling to vapor? How much does sea salt cost? How much does electricity cost?

Commercially, it's done by solar evaporation. Take sea water, put it in the sun, wait. In your apartment, even if you don't have direct sunlight, surely there is a warm spot somewhere. The only equipment you need is a tray to hold the water.
You should filter the water first to remove solids like sand, algae, jellyfish, etc. that no one wants in their salt.
The large operations have multiple ponds at different stages of concentration -- this reduces the number of ponds you have to mine, and allows some solids to settle out. But you can do just fine with a single "pond".
In general, commercial operations don't evaporate the water all the way to dryness, which does mean you end up throwing out some of the brine. Sodium chloride is the major component of seawater, and therefore crystallizes out preferentially compared to the unwanted stuff like magnesium, sulfate, heavy metals, marine biotoxins, fluorochemicals, etc. If you evaporate all the water, all that stuff ends up in your salt too. Maybe get your salt tested for these contaminants if you're going to eat a bunch of it or feed it to the public.


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Crystal lattice structure from single crystal x-ray diffraction pattern
Is there a good reference describing the inverse problem of determining the crystal lattice, unit cell etc., from a single crystal x-ray diffraction pattern and chemical composition? I am looking for one that goes into details on how the fourier transforms are used, how the atomic numbers are used etc. Any sort of publication/book/lecture notes/video will be helpful. Thanks!

Here is a start: https://youtu.be/9n1u8ymc8aw?feature=shared

And here a more comprehensive source: https://global.oup.com/academic/product/fundamentals-of-crystallography-9780199573653?cc=us&lang=en&

G. Rhodes 'Crystallography made crystal clear- a guide for users of macromolecular models' (publ AP 2006), D. McKie & C. McKie, 'Essentials of Crystallography'  (publ.  Blackwell 1992 ), C. Hammond, 'The basics of Crystallography and Diffraction' (OUP 2003) ,  R. James ' The optical principles of the Diffraction of X-rays' (publ Bell & Sons 1965) Classic text, v thorough.

Thank you, @porphyrin and @Karsten!


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Reactivity of Alkyl Halides with E2 with same number of substituents in the end
I have a question regarding the reactivity of alkyl halide for E2. I know that the general rule is tertiary alkyl halide reacts the fastest with E2, and primary alkyl halide reacts the slowest.
I was told that the reasoning behind this trend is due to stability of product based on the number of substituents the alkene have. However, I wanted to know if this trend still holds if both a primary alkyl halide and a tertiary alkyl halide results in the same number of substituents.
For example, consider the two pairs below:

Here, the first one is primary alkyl halide, while the second one is tertiary alkyl halide. Therefore, I could decide automatically that the latter will have a higher reactivity toward E2. However, if I draw the E2 product for both alkyl halide, I get the same alkene.

The book this question is from states that the tertiary alkyl halide will have a higher E2 reactivity, but provides no further justification.
Am I missing something important here? I cannot seem to understand why the 2nd compound has a higher E2 reactivity? I am guessing that maybe it could be due to the fact that it is more difficult for the OH to get in proximity of the beta hydrogen of the first compound. However, I wanted to see if there is a more clear justification for why the tertiary alkyl halide hear has a higher E2 reactivity.
Could it be that a primary alkyl halide is more susceptible to an SN2 reaction? However, does that mean it loses reactivity toward an E2?
It would be great if I could get an explanation for this! Thank you :)

What you are missing is that the tBu cation is more stable and more likely to form

Wouldn't that be more relevant for E1 reaction? E2 is concerted and therefore I thought the formation of cation wasn't a contributing factor.

Compounds don't get to chose the mechanism. I think it would be closer to E1 for t-butyl no matter how you tried to manipulate it.

Given that the base is a strong base, shouldn't it be E2?


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Why does glucose react with hydroxyl amine and HCN but doesn't give Schiff's test or react with NaHSO3?
I can't understand this because if there isn't a free CHO group then it also shouldn't react with NH2OH and HCN.
On the same note can anyone also explain why pentaacetate of glucose not react with NH2OH? Why isn't there a free CHO in this case?

Look at the structure. There is a hemiacetal which does a lot of the same reactions as an aldehyde group. This is why glucose reacts with Hydroxylamine and HCN as cyanide and hydroxylamine are both very good nucloephile. Sulfite is, by comparison, a much poorer nucleopphile.

Oh ok I get it now. I believe in the pentaacetate form there is no free OH group to make it an aldehyde and therefore it doesn't react with hydroxylamine. Can you correct me if I'm wrong?

You are correct see - https://chemistry.stackexchange.com/questions/22269/does-glucose-react-with-bradys-reagent

Alright thank you


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Static Equilibrium in an open and closed system
I've had a bit of confusion on whether static equilibrium can exist in an open system.
My textbook states (on modelling static equilibrium through water in a flask):
"Static equilibrium is shown by flask X as it is open, and the liquid can evaporate and leave the flask. This process of the liquid turning into a gas and leaving the flask shows a static equilibrium because eventually the flask will be empty."
My chemistry teacher states:
"Static equilibrium is defined as when the forward and reverse rates of a process are zero, and concentrations of species stay constant. Since in an open system there will constantly be species entering and leaving the system, the concentrations of species will not stay constant and hence will not reach equilibrium."
Can someone please clarify?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

It may be useful to review search results for site:chemistry.stackexchange.com OR site:chem.libretexts.org static dynamic equilibrium steady state open closed system

The flask would be in equilibrium only after most of the water evaporates -- there would always be an exchange between air (assuming it has some humidity above zero) and the water adsorbed to the glass surface.
For that matter, does the text state the inital temperature of the flask vs. the dew point of the air? If the flask is colder than the dew point, water would condense, not evaporate, until the flask warms.
Your chemistry teacher seems to have expressed equilibrium far better than the text, and you might mention to her or him that the text is unclear -- you might not be the only student with that question.


text:
When forming an epoxide with m-CPBA, what happens when we wash our mixture with Na2S2O4?
The m-CPBA is reduced by the $\ce{S2O4^{2-}}$, but why does it leave the mixture then?
Why does the reduced m-CPBA leave with the oxidised  $\ce{S2O4^{2-}}$?

The pKa of MCPBA is~7.6, the pKa of 3-Cl-benzoic acid is ~3.8 so is much easier to remove by an aq. base wash


text:
How can dimethyl maleate be converted to dimethyl fumarate using piperidine?

To move from the Z isomer to the E isomer, the  π bond will need to break to allow free rotation. Once the bond is broken, I think that forming the trans (E) alkene will be more stable as it has less steric hindrance.
I think you could add the piperidine to the C=C bond to give some sort of amine but I do not know how that group could be removed to give the alkene again.

Welcome to ChemSE! Your question will likely be considered as homework and be voted down. You need to make some effort to answer the question yourself. Take a look here about homework policy.

@Viraj How stable do you think the addition product is?


text:
Does H2, Pd/C reduce a conjugated ketone?

Would H2, Pd/C be a sufficient reagent for this reaction to proceed?
The question is from a test which asked what would be the most suitable reagent for the above reaction

LiAlH4 plus a lanthanide salt is the way to go. I don't think any of these conditions are good options  see https://pubs.rsc.org/en/content/articlelanding/1986/p1/p19860001929

I faced the same question today in the test! Think it should be option b.

@HarikrishnanM Why would you guys use a reagent good for C=C when you got C=O reduction?

@HarikrishnanM  I don't think simple hydrogenation with Pd/C will give you the allylic alcohol. See this answer  https://chemistry.stackexchange.com/questions/19628/enones-and-hydrogenation

I checked the answer key given by a famous coaching institute. It is given as 'b' option as $\ce{LiAlH4}$ is a very strong reducing agent and reduces everything it finds. Thus $\ce{H2 / Pd + C}$ is the way to go

Also allylic alcohols are more stable because of the resonance effect and thus it is justified.

Zn-Hg / HCl is the reagent for Clemmensen reduction and will remove the =O group altogether hence it's not suitable.
Na / liq. NH₃ makes it undergo Birch reduction and reduces the aromatic ring

By the way, this is the National Standard Examination in Chemistry, the 1st part of the qualifier to int'l olympiads in India.

As with B option, H2/Pd will also reduce the double which we don't want..
C, LiAlH4 does not reduce double bond( exception incase of cinnamic acid as double bond is in conjugation with phenyl ring as well as carbonyl carbon) and will reduce ketone to alcohol. Hence Option 3 is most accurate.

Answer is option (c) according to the answer key uploaded by the IAPT agency.

$\ce{Na / liq. NH₃}$ is the reagent of Birch reduction and will reduce the aromatic ring. So, it's not the right way.
$\ce{Zn-Hg / HCl}$ will result in Clemmensen reduction and will completely remove the carbonyl oxygen. We don't want that.
$\ce{LiAlH4}$ looks a good enough reagent to go about this reaction as it correctly reduces the ketone group and also gives our desired product.
$\ce{H2 + Pd}$ will lead to hydrogenation, as pointed out rightly by some members.

Thus it is (c).

Big difference between those people who write exam questions and those who actually run reactions every day. This answer is incorrect. Hydrogenation under these conditions reduces the double bond first.

I don't think IAPT judges purely based on the answer key. Surely going to challenge this question if isn't already 'deleted' or given as a bonus

Let us wait then...

@Harikrishnan M: I see that you are a 12th grader; I admire your enthusiasm. You have much to learn. From my personal experience, hydrogenation with Pd/C will give the saturated alcohol. $\ce{LiAlH4}$ is the best solution as long as it doesn't react like cinnamaldehyde with $\ce{LiAlH4}$ and give the saturated alcohol. ....continued....

....continued....Na/$\ce{NH3}$ will afford the saturated ketone upon protonation of the enolate of the ketone. There will be no Birch reduction because there is no alcohol present. Too bad the constructor of the question didn't offer $\ce{NaBH4}$ or $\ce{LiA(O-t-But)3H}$ as. options.

@user55119 I also was looking for $\ce{NaBH4}$... on not finding it I was pretty confused

And the fact that it was emphasised to me lot of times to sparingly use $\ce{LiAlH4}$ as it was very strong and reduces everything

The answer key has been released, and I have updated my answer accordingly.


text:
Why is quinuclidine more basic than triethylamine?
I am trying to compare the basic strength of two similar amines

Usually this is done by comparing the relative stability of the ion formed on the protonation of nitrogen's lone pair.
This can be done by comparing electronic factors, steric factors, solvation of the ion, etc.
In this example, the stabilisation of the ion due to the +I effect of the groups is the same, as the groups are almost identical.
How will the solvation of the ions formed differ as they have slightly different structures?
Are there other factors that one can consider to help theoretically conclude the relative basicity here?

"steric factors" (more precisely locked conformation)

Thing is they are so similar, that it's better to call it a draw and be done with it. Similar amines have little differences in pKa and they are beyond any rules of thumb.

@Mithoron The data indicates that quinuclidine is more basic than tri ethyl amine, the thread you have linked compares 1-azabicyclo[2.2.1]heptane  and tri ethyl amine, where tri ethyl amine is more basic. The explanation provided by the accepted answer would suggest the bicyclo compound being less basic, which is not the case here


text:
How to compute solvent reorganization energy in Marcus' non-equilibrium approach?
I try to solve the following equations for some system solvated in the water. The goal is to obtain value in kcal/mol.  Unfortunately, I cannot reach the desired output. May someone help me?

The  variables are:
n = 1 (single electron exchanged)
rA = 3.68Å
rD = 2.90Å
R = 6.58Å
ε∞ = 5.2 F/m
ε0 = 78.408 F/m

My try:
\begin{gather}
\lambda \ =\ ( ne)^{2}\left(\frac{1}{2r_{A}} +\frac{1}{2r_{B}} -\frac{1}{R}\right)\left(\frac{1}{\epsilon _{\infty }} -\frac{1}{\epsilon _{0}}\right) \notag\\
 \notag\\
\lambda \ -\ reorganization\ energy\ [ ?\ kcal/mol] \notag\\
 \notag\\
n\ -\ number\ of\ exchanged\ electrons\ [ =1] \notag\\
e\ -\ electron\ charge\ \left[ =\ 1.602\times 10^{-19} C\right] \notag\\
r_{A} -\ radius\ of\ the\ acceptor\ cavity\ [ =3.68Å] \notag\\
r_{D} -\ radius\ of\ the\ donor\ cavity\ [ =2.50Å ]\\
R\ -\ intermolecular\ distance\ [ =6.18Å ] \notag\\
\epsilon _{\infty } \ -\ optical\ dielectric\ constant\ [ =5.2] \notag\\
\epsilon _{0} \ -\ static\ dielectric\ constant\ [ =78.408] \notag\\
 \notag\\
\mathbf{Conversion:} \notag\\
 \notag\\
r_{A} \ =\ 3.68Å \ =0.368nm \notag\\
r_{D} \ =\ 2.50Å =0.250nm \notag\\
R=6.18Å =0.618nm \notag\\
\epsilon _{\infty } \ =\ 5.2\times 8.854\times 10^{-12}\frac{F}{m} =4.604\times 10^{-11}\frac{F}{m} \notag\\
\epsilon _{0} \ =\ 78.408\times 8.854\times 10^{-12}\frac{F}{m} =6.942\times 10^{-10}\frac{F}{m} \notag\\
 \notag\\
 \notag\\
\mathbf{Calculation} : \notag\\
 \notag\\
\lambda \ =\ 2.566\times 10^{-38} \ [ C] \ \left(\frac{1}{0.736\ [ nm]} +\frac{1}{0.500\ [ nm]} -\frac{1}{0.618\ [ nm]}\right)\left(\frac{1}{4.604\times 10^{-11}\frac{F}{m}} -\frac{1}{6.942\times 10^{-10}\frac{F}{m}}\right) \notag\\
\lambda \ =\ 2.566\times 10^{-38} \ [ C] \ \left( 1.359\left[\frac{1}{nm}\right] +2\left[\frac{1}{[ nm]}\right] -1.618\left[\frac{1}{nm}\right]\right)\left(\frac{1}{4.604\times 10^{-11}\frac{F}{m}} -\frac{1}{6.942\times 10^{-10}\frac{F}{m}}\right) \notag\\
 \notag\\
\lambda \ =\ \left( 2.566\times 10^{-38}\right) \ [ C] \ \times 1.741\left[\frac{1}{nm}\right] \times \left( 2.028\times 10^{10}\right)\left[\frac{m}{F}\right] \notag\\
\lambda \ =\ 9.06\ \times 10^{-28}\left[\frac{C\times m}{nm\times F}\right] \  \notag\\
\lambda \ =\ 9.06\ \times 10^{-28}\left[ 10^{9}\frac{C\times nm}{nm\times F}\right] \  \notag\\
\lambda \ =\ 9.06\ \times 10^{-19}\left[\frac{C}{F}\right] \  \notag\\
 \notag
\end{gather}

Change your units to SI,  nm not angstrom then you need to multiply $e$ by $1/(4\pi \epsilon_0))$ where $\epsilon_0$ is the permittivity of free space $8.8\times 10^{-12}$ F/m. The answer will be in J then you can convert.  $\lambda$ is typically $1 $eV $\sim 8000$ cm$^{-1}$. The dielectric constants are relative numbers, water about $78$.

@porphyrin Thanks for the answer. I will change units to SI, but why shall I multiply electron charge? Also,  is the reorganization energy in kcal/mol, following Marcus theory.

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of essential parts.
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@Poutnik Done as required

@porphyrin May you elaborate? I have created online on mathcha, would appreciate help. Been struggling with that for few days already...
https://www.mathcha.io/editor/L4KX2TX8UkvcJXtQ5e8x6IQ9jyZPTL4kdnJu98Q942

In SI units then conversion.
see Harrison et al Chemical Physics 116 (1987) 429-448 for equations
nm  = 1e-9
q   = 1.60218e-19
eps0= 8.854e-12
rA  = 3.68/10*nm
rD  = 2.90/10*nm
R   = 6.58/10*nm
einf= 5.2
e0  = 78.408
lam= (q**2/(4*np.pi*eps0))*(1/(2*rA)+1/(2*rD)-1/R)*(1/einf-1/e0)
print('{:s}{:g}{:s}{:g}{:s}{:g}{:s}'.format('lambda ', lam, 'J or
',lam*6.022e20,'kJ/mol or ', lam*6.022e20/4.18, ' kcal/mol') )

lambda 6.47513e-20J or 38.9932kJ/mol or 9.32852 kcal/mol

Thank you ! I will try to decipher this now :)

And basically - if there are more electrons exchanged I just and only need to multiply q^2 by the number of electrons exchanged. Nowhere else?

I'm not sure about more that one electron. If two this could mean from the same orbital then the reorganisation energy one could argue would be the same as potentials are the same, from different orbitals then its probably a 2 -step process with different $\lambda$ anyway.  BTW the code above is in python.


text:
When is chromatography regarded as a viable industrial purification strategy?
I understand the basics that chromatography is a fairly slow and expensive process compared to direct synthesis. When a synthesis route is not feasible, the economics become favorable. Not being in the chemical industry, I do not know the specifics of these economics or how common industrial scale chromatography is.
Three applications I know of are sucrose separation from beet molasses, insulin from genetically modified E.coli, and B-Carotene from algae. There are surely many others especially in the pharmaceutical industry, but I have not been able to find a list of specific chemicals.

Within pharma the scale-up chemists put a lot of time in to minimising chromatographic separations for drug substances.

Most organic chemicals are purified and separated by column chromatography / TLC.

Questions that request lists are frowned upon. You could reformat the question so that it addresses an underlying concept, for instance, "when is chromatography regarded as a viable industrial purification strategy?" And adding some details regarding your own findings.

@BuckThorn, Thanks for the feedback!


text:
Effect on Metal-Phosphine Bond if Metal centre gets oxidised
Looking at the Octahedral complex, the chromium is a $d^6$ complex and, if the Chromium bond gets oxidised, it will have lesser electrons to donate to phosphine's $\sigma^{*}$ orbital.
This effect is similar to the CO bond. If the metal centre is oxidised, there is lesser electrons donated to the $\pi^{*}$ orbitals.
And when there is lesser electrons donated to the ligand's antibonding orbital ($\sigma^{*}, \pi^{*}$) orbital, it means that their Bond Order will be higher which stabilises the molecule (shorter bond length). However, this is only true for CO-Metal bond, length, but not true for the Phospine-Metal bond length. The P-M bond length would actually be weaker after oxidization of the metal centre. WHy is this the case?


text:
Why is the standard enthalpy of formation temperature-dependent in JANAF tables?
I'm unsure why $\Delta_\mathrm{f}H^\circ$ within the NIST-JANAF Thermochemical Tables [1] is temperature-dependent. Wouldn't the $\Delta_\mathrm{f}H^\circ$ represent the enthalpy of formation at the reference temperature (298.15 K) by default? If not, what does it represent?
The following screenshot of the online page for water is provided as an example:

EDIT:
I believe my point was missed, so I will provide another example.
The following is Equation 13.17b within Fundamentals of Engineering Thermodynamics
$$ Q - W = \sum_P n \left( \bar{h}_f^\circ + \Delta \bar{h} - \bar{R} T_P \right) - \sum_R n \left( \bar{h}_f^\circ + \Delta \bar{h} - \bar{R} T_R \right) $$
If I wanted to evaluate the $\bar{h}_f^\circ + \Delta \bar{h}$ term for gaseous water in a hypothetical reaction at a temperature of 800 K, how would I do so?
Would it be (-241.826 + 18.002) kJ/mol or (-246.443) kJ/mol?
Reference

Thomas C. Allison. NIST-JANAF Thermochemical Tables - SRD 13, 2013. DOI: 10.18434/T42S31.
Michael J. Moran “Reacting Mixtures and Combustion.” Fundamentals of Engineering Thermodynamics, Wiley, S.l., 2020, pp. 519–519.

In context of standard enthalpies of formation, there is no standard temperature. More at https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation

What is the difference between ΔfH and H-H° in NIST–JANAF tables?

@andselisk, see my edit. That other post did not answer my question, but I wasn't clear with what I was asking.


text:
Separation of benzoic acid and naphthalene from its mixture
The best method for the separation of naphthalene and benzoic acid from their mixture ?
a)chromatography b)crystallization
c)distillation.  d)sublimation
many sources says answer is crystallization and others say sublimation....I am confused rn...which is the correct one...

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

There is no best method and there are multiple best methods.  As being the best one depends on the applied criteria and their weight. The criteria may be cost, time, manual effort, complexity of the process, equipment demand,  efficiency of separation, risk of contamination or decomposition and some more. The weights depend on priorities. // Crystalization and sublimation are the  candidates.

The answer is e), none of the above. Ether extraction of neutral naphthalene from the sodium salt of benzoic acid in aq. NaOH. Acidify, extract benzoic acid.

Chromatography is good for separating very small amounts of chemicals but is not, in general, good for separating large (macroscopic) amounts of substances.
Both naphthalene and benzoic acid are solids at room temperature so distillation doesn't seem very viable.
Recrystallization might work, but you'd need to have only one compound crystallize. Separating based on crystal structure isn't viable. However you might be able to keep benzoic acid in a basic solution while crystallizing naphthalene.
Naphthalene is used in moth balls and so has an appreciable vapor pressure. Thus sublimation would be the best choice from those given.
However if I had to do this, I'd try some sort of two phase separation in a sepatory funnel. I'd extract the benzoic acid into a basic solution, then acidify the basic solution to get benzoic acid again. Evaporate the solvent from the naphthalene to recover it.
Benzoic acid and naphthalene are such different chemicals that it is hard to conceive how you'd end up with a mixture of the two. The point of using a two phase liquid extraction would be to increase the purity of the benzoic acid. I'd worry that benzoic acid might be coating some naphthalene particles preventing those naphthalene particles from subliming.
In either case, a simple melting point test of both after separation would give a good indication of their individual purity.


text:
SMARTS: recursive search
i have a question regarding the usage of recursive SMARTS. I'm using python and rdkit for my coding. My goal is to get the substructure of carboxylic acids which are bonded by an ethyl-group to an aromatic compound. So far so good. With the following script I'm only getting substructure hits for methyl-linkers, although I'm defining an ethyl linker (CCa) . I mean i get my desired outcome by just adding another "C" but i still wanna understand the problem.
smiles_list = ["OC(CCCC1=CC=CC=C1)=O", "OC(CCC1=CC=CC=C1)=O", "OC(CC1=CC=CC=C1)=O", "OC(C1=CC=CC=C1)=O"]
mols = [Chem.MolFromSmiles(smiles) for smiles in smiles_list]
smarts_carboxy= "[C$(CCa)](=O)[OX2H1]"
patt = Chem.MolFromSmarts(smarts_carboxy)
highlight_scaffold = [mol.GetSubstructMatch(patt) for mol in mols]

Draw.MolsToGridImage(mols, highlightAtomLists = highlight_scaffold, molsPerRow=5, subImgSize=(300,300))

Thanks in advance for your help!

[C$(CCa)] means that you are looking for a carbon and the found carbon is bonded to a carbon that is bonded to an aromatic atom.
The recursive part starts always with the atom you are looking for and is followed by a condition, so there is only one carbon in the linker.
Adding another C sets the conditon to two carbons.

Awesome, thanks for your answer! From time to time SMARTS can be a bit confusing....


text:
Do we group reversible reaction with all reactants and product solid as homogenous or heterogeneous equilibrium?
My personal thought is that they should be in heterogenous equilibrium as we define it as all reactants and products being of different phase (phase-their must not be a visible boundary) and solids clearly have a visible separation.

It can be either way, depending on reaction.

Good luck to identify one such reaction.

Tin pest?

Well there are reactions with both solids such as reaction between carbon(graphite) and carbon(diamond) and the reaction lies under heterogenous equilibrium as graphite and diamond are different phases of carbon. I want to know, should this be the case for all solid reactants and products reaction?


text:
What is the difference between standard Gibbs energy and Gibbs energy of formation?
Good day guys,
I am reading a book on electrochemical engineering, I came across some definitions of Gibbs energy and I am a bit confused as the book does not go into a lot of detail.
The book first introduces Gibbs energy of a reaction
$$\Delta G_{Rx} = \Delta G_f^{products} - \Delta G_f^{reactants}$$
and it mentions that $\Delta G_f$ is the gibbs energy of formation for a compound. And it refers me to a list in the appendix on Gibbs energy of formation values at ambient/standard conditions as $\Delta G$ and it does not mention pressures.
Later on in the book, the standard gibbs energy for a cell reaction is introduced as:
$$ \Delta G^o_{Rx} = \sum_i s_i \Delta G_{f,i}^o $$
where $s_i$ are the stoichiometric coefficients  and $\Delta G_{f,i}^o $ is the Gibbs energy of formation at standard/ambient conditions.
What exactly is the difference between $\Delta G^o$, $\Delta G_f $ and $\Delta G$ ? Does $\Delta G^o$ imply standard conditions and $\Delta G_f $ at some condition ?
Is $$\Delta G_{Rx} = \Delta H - T\Delta S = \sum_i s_i \Delta G_{f,i}^o $$ ?

Here are the two definitions again:
$$\Delta G_{Rx} = \Delta G_f^{products} - \Delta G_f^{reactants}\tag{1}$$
$$ \Delta G^o_{Rx} = \sum_i s_i \Delta G_{f,i}^o \tag{2}$$

What exactly is the difference between $\Delta G^o$, $\Delta G_f $ and $\Delta G$ ? Does $\Delta G^o$ imply standard conditions and $\Delta G_f $ at some condition ?

The $^\circ$ added to a thermodynamic values does imply standard conditions. So you could rewrite equation 1 for standard conditions:
$$\Delta G^\circ_{Rx} = \Delta G{^\circ_f}^{products} - \Delta G{^\circ_f}^{reactants}\tag{1*}$$
Now, (1*) and (2) say exactly the same thing.

What is the difference between standard Gibbs energy and Gibbs energy of formation?

I will rephrase the question as "what is the difference between standard Gibbs energy of reaction and standard Gibbs energy of formation?". Typically, you would look up the standard Gibbs energy of formation, as those are the ones available. (Importantly, the standard state also sets partial pressures and concentrations, as the Gibbs energy is dependent on those). The symbols you show are for the Gibbs energy of reaction, a molar quantity (units Joule per mole). A change in Gibbs energy in general could also have dimensions of energy (units Joule), so it is important to differentiate.
So what is the difference?
The standard Gibbs energy of formation refers to a small set of standard Gibbs energies of reactions: Those with a single product and a coefficient of one, with reactants all elements in their allotrope present and most stable at standard conditions and the given temperature. This tells you that the standard Gibbs energy of formation for elements (the most stable allotrope in its physical state at the chosen temperature) is zero by definition.
Here is one such reaction:
$$\ce{1/8 S8(s) + 3/2 O2(g) -> SO3(g)}\tag{A}$$
For reaction (A), the standard Gibbs energy of reaction would be equal to the standard Gibbs energy of formation for sulfur trioxide.
I can write the reaction in a nicer form:
$$\ce{S8(s) + 12 O2(g) -> 8 SO3(g)}\tag{B}$$
For reaction (B), the standard Gibbs energy of reaction would be 8 times the standard Gibbs energy of formation of $\ce{SO3(g)}$, as equation (2) or (1*) will tell you.
We could have a different reaction that makes sulfur trioxide:
$$\ce{SO2(g) + 1/2 O2(g) -> SO3(g)}\tag{C}$$
To calculate the standard Gibbs energy of reaction (C), you would need to know the standard Gibbs energy of formation of $\ce{SO2(g)}$ as well.

Just to make sure I understood correctly, the notation $\Delta G_f$ of products/reactants refers to the Gibbs energy of formation under non-standard conditions ? If standard conditions are used we denote this with the superscript (º) ?

Yes, or at least conditions that may include non-standard conditions. The definition is still valid if you happen to find yourself at standard conditions, of course, but also if not. I think in the textbook they were just lazy and did not include the superscript $^\circ$.


text:
Is there a supercritical state for ice?
I know there is such a thing as a supercritical fluid, where the vapor and liquid phases are indistinguishable, but is there such a thing as a supercritical state for ice? Another way of asking this is: what happens to ice when it is subjected to pressures above the critical pressure, 217.75 atm, in the case of water?

Contrary to most substances, increasing the pressure of solid water turns it into liquid. Then, even if you surpass the critical pressure, it would not be a supercritical fluid if you do not surpass the critical temperature.

See the water phase diagram on Physics SE

You ask three things at once, thinking they are one thing, which they are not.

Does the solid/liquid line end with a critical point, like the liquid/gas line? No, it just keeps going on and on.
Does anything happen to ice at water's critical pressure? Nothing at all; liquid water does not have an abrupt change of any properties at that point, neither does ice. The critical pressure is only important in relation to the liquid/gas line. For the solid/liquid line, it is about as non-special as a random point on a highway in Arizona, with tumbleweeds and Morricone music, with hundred miles behind you and hundred miles ahead.
Does anything eventually happen as we increase pressure more than that? Why, yes, a lot of wonderful things happen (see the diagram below). We'll get other crystal form of ice instead of the one we were used to, and then another, and then yet another, and then more. Some of them melt at fairly high temperatures. Come to think of it, having multiple crystalline forms is not a rare thing at all. If anything, water is special in that it is better studied than most other compounds, and that's why so many forms are known.

(source: Wikimedia Commons)
So it goes.
(P.S. Yes, I've read Cat's Cradle too.)

That is absolutely fascinating! I will continue doing research on the myriad variations of ice.

Water does not have a known critical point apart from the familiar one involving the usual liquid and vapor phases at about 374°C, 22 MPa.
Cerium is a different story. The phase diagram pictured below shows two fcc phases, labeled $\alpha$ and $\gamma$, that are distinct at ambient pressure but merge into a single fcc phase at roughly 250°C and 2 GPa.

Source
The discussion of physical characteristics given by Wikipedia implies that at low pressure the lower-temperature $\alpha$ has a higher metallic-bonding valence (from more involvement of the $4f$ orbitals in the bonding) making it denser and more conductive than $\gamma$, but this metallurgical difference decreases upon compression and thus the phases merge:

Cerium has a variable electronic structure. The energy of the 4f electron is nearly the same as that of the outer 5d and 6s electrons that are delocalized in the metallic state, and only a small amount of energy is required to change the relative occupancy of these electronic levels. This gives rise to dual valence states. For example, a volume change of about 10% occurs when cerium is subjected to high pressures or low temperatures. It appears that the valence changes from about 3 to 4 when it is cooled or compressed.1

Cited Reference

Johansson, Börje; Luo, Wei; Li, Sa; Ahuja, Rajeev (17 September 2014). "Cerium; Crystal Structure and Position in The Periodic Table". Scientific Reports. 4: 6398. Bibcode:2014NatSR...4E6398J. doi:10.1038/srep06398. PMC 4165975. PMID 25227991.

@Mithoron what did you do? It looks like my illustration was changed for an identical one.

tsk, tsk you didn't check the code. I only added description.


text:
How can we prevent retrogradation of starch?
Are there any additives which can prevent retrogradation of starch? I have prepared 13% of starch solution along with 2% of NaCl. The results were bit better. The synthesized gel was flowable with some bubbles inside it, perhaps 2% of salt was more than enough. Coming back to my question, are there any better substances which can be used for this purpose?

I just saw a mention that stuff like Na+ accelerates it.


text:
Carrying out electrolysis of very dilute NaCl for a long time
I find a similar question in stack exchange (What will happen if the electrolysis of Brine takes place for a long time?), yet it does NOT seem to respond to my question.
Suppose I am electrolysing a very dilute sodium chloride solution. At the anode, the hydroxide ions will get oxidised into oxygen gas, while at the cathode, the hydrogen ions will get reduced into hydrogen gas.
Now, if I leave the set-up ongoing for a long time, is it true that all the water will be consumed and I will obtain a beaker of molten sodium chloride (given that heat is constantly provided)?
My lecturer pointed out that this is impossible to happen. When a lot of water is consumed, then I will obtain concentrated sodium chloride solution, namely brine. So, at the anode, the chloride ions will get oxidised to give chlorine gas, while at the cathode, the hydrogen ions are still reduced to give hydrogen gas.
Can anyone provide me with the whole picture of the case provided by me? Please be noted that I am talking about very diluted NaCl solution instead of brine.

Well, your scenario becomes this earlier scenario with time, so it kinda does answer.

The redox potentials cited by Harikrishnan are all valid for ionic concentrations equal to $1$ M. As soon as these concentrations are different from $1$ M, these potentials are modified according to Nernst's law. This is the case here, because in $\ce{NaCl}$ solution, the concentration, for example, of $\ce{H+}$ ion is not $\ce{1}$ M. It is around $\ce{10^{-7}}$ M. The same sort of deviation may occur for all ions in this electrolysis, including the oxidation of chloride ion to chlorine, as the present solution is "very dilute". As a consequence, all potentials should be recalculated to be adapted to a "rather dilute" and nearly neutral solution of $\ce{NaCl}$. Unfortunately this cannot be done here, as the concentration of $\ce{NaCl}$ is not known.
The potentials will change, but the only useful result is that $\ce{H2}$ is still produced at the cathode according to : $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ Depending on the concentrations, both $\ce{Cl2}$ and $\ce{O2}$ are produced at the anode, according to the following equations : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ $$\ce{2 Cl- -> Cl2 + 2 e-}$$ Note that the ions $\ce{H+}$ from the anode react with the $\ce{OH-}$ from the cathode, and produce new water molecules. If the discharge of $\ce{Cl-}$ is momentarily forgotten, electrolysis by $4$ electrons consumes $6$ $\ce{H2O}$ and produce $4$ $\ce{H+}$ plus $4$ $\ce{OH-}$ that recombine to form $4$ $\ce{H2O}$. As a result, water is consumed. As the amount of $\ce{Na+}$ ion is not destroyed, the solution becomes more concentrated in $\ce{NaCl}$. It also contains a little bit of $\ce{NaOH}$, as there are two possible reactions at the anode, and only the one starting with $\ce{H2O}$ produces $\ce{H+}$ ions. So the solution becomes slowly basic. After a rather long time, all water is destroyed, and solid $\ce{NaCl}$ plus $\ce{NaOH}$ remained in the electrolysis cell, as Freeby Freeby mentioned it. The trouble is that, at the end, you cannot prevent $\ce{Cl2}$ from reacting with $\ce{NaOH}$ according to : $$\ce{Cl2 + 2 NaOH -> NaClO + NaCl}$$ so that the solution is a bleach, and the evaporation of bleach gets a disproportionation according to : $$\ce{3 NaClO -> NaClO3 + 2 NaCl}$$
As a final result the powder obtained on the long range, out of an electrolyzed solution of $\ce{NaCl}$, is made of $\ce{NaCl}$ plus a few percent of $\ce{NaOH}$ and $\ce{NaClO3}$

I understand the complexities of the situation now Maurice! Thank you. Further, it's nice that our answers both complement each other.

Thank you Harikrishnan. You should also know that, in the last three lines of your initial answer, you have interchanged the words ANODE and CATHODE. The anode must produce electrons. The cathode mut consume electrons.

Yes Maurice I will edit it.

Looking at your answer I get an exciting application. If this process was made in brine it would result in the formation of bleach in higher yield is it not?

@Harikrishnan. The electrolysis of NaCl solutions is the standard way to produce bleach in industry.

During the electrolysis of aqueous sodium chloride solution, the products are $\ce{NaOH}$, $\ce{Cl2}$ and $\ce{H2}$ . In this case besides $\ce{Na+}$ and $\ce{Cl–}$ ions we also have $\ce{H+}$ and $\ce{OH–}$ ions along with the solvent molecules, $\ce{H2O}$.
At the cathode there is competition between the following reduction reactions:
$\ce{Na+ (aq) + e– → Na (s) : E = – 2.71 V}$
$\ce{H+ (aq) + e– → ½ H2(g): E = 0.00 V}$
The reaction with higher value of $E^⊖$ is preferred and therefore, the reaction at the cathode during electrolysis is the 2nd one.
But H+ (aq) is produced by the dissociation of H2O, i.e.,
$\ce{H2O (l) → H+ (aq) + OH– (aq)}$
Therefore, the net reaction at the cathode may be written as the sum of above 2 reactions:
$\ce{H2O (l) + e– → ½H2(g) + OH–}$
At the anode the following oxidation reactions are possible:
$\ce{Cl– (aq) → ½ Cl2 (g) + e– : E = 1.36 V}$
$\ce{2H2O (l ) → O2(g) + 4H+ (aq) + 4e– : E = 1.23 V}$
The reaction at anode with lower value of $E^⊖$ is preferred and therefore, water should get oxidised in preference to $\ce{Cl–}$ (aq). However, on account of overpotential of oxygen, the 1st reaction is preferred. Thus, the net reactions may be summarised as:
$\ce{NaCl (aq) ->[H2O] Na+ (aq) + Cl– (aq)}$
CATHODE: $\ce{H2O (l) + e– → ½H2(g) + OH–}$
ANODE: $\ce{Cl– (aq) → ½ Cl2 (g) + e–}$
NET REACTION: $\ce{NaCl(aq) + H2O(l) -> Na+(aq) + OH-(aq) + 1/2H2(g) + 1/2Cl2(g)}$
Source: NCERT [Refer page 85 / 23rd page in the PDF.]


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When C2 axis of symmetry is present then is the compound meso?
If suppose a compound has C2 axis of symmetry then is it confirmed that the compound will be meso or it can have an enantiomeric pair? And which of following conditions are sufficient enough to declare a compound meso:- Plane of symmetry, Centre of symmetry, axis of symmetry and alternate axis of symmetry?


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Ca(OH)2 not dissolving even as much as it should
according to solubility charts Ca(OH)2 should dissolve at 20-25 oC at about 1.6g per litre.  I have tried 0.8, 0.4 and 0.2g in 500ml and even the 0.2 does not completely dissolve, even left overnight. (in fact I would say most of it is undissolved on the bottom of the flask).  rm temp is around 19-20oC, using lab grade ca(OH)2 and distilled water Any thoughts/suggestions?

Could part of it be already CaCO3 ? How does the not dissolved part behave in an acid? Another alternative is Ca(OH)2 can be covered by CaCO3 that kinetically slows down the dissolution.

Well, you should heat it and mix instead of wasting time - some things won't just do by itself.

Check for calcium carbonate with a little vinegar reation. The water may also contain dissolved calcium salts or CO2 whhich converts the hydroxide to the carbonate.