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Bleaching of chlorophyll stains using hydrogen peroxide
I have been doing some experiments concerning bleaching of chlorophyll stains from cotton clothing. Currently I have found that $\ce{H2O2}$ is fairly effective, but still leaves visible stains. Would a crude Horseradish root extract containing peroxidase get me better results?
The reason I'm asking about the crude extract and not the purified enzyme is because I don't have access to a laboratory or more complex materials and my work also aims to produce simple methods of chlorophyll degradation that could be replicated at home.

Chlorophyll is not well soluble in water.  Since your cotton clothing might have been dyed prior to the unwanted exposure to chlorophyll, you could try th following cleaning agents first at a spot less visible and if they do not remove your clothing colors, apply them on the stain:  a) vinegar, or b) pure alcohol, or c) some of luke-warm water + bile / gall soap brushed over the stain and allowed to sit there four an hour at ambient conditions prior to a normal wash.  By own experience, if you know there were stains there, you will be more likely to see them even if others do not spot them.

Horseradish peroxidase is not hydrogen peroxide, $\ce{H2O2}$, but rather an enzyme that breaks $\ce{H2O2}$ into $\ce{H2O}$ and $\ce{O2}$. In general, enzymes ending in -ase are lytic enzymes, catalyzing the breakdown of a similar-sounding substance. So, if you add ground horseradish to hydrogen peroxide, bubbles of oxygen are released. BTW, horseradish roots differ in peroxidase potency, so test beforehand with a particular batch before doing this as a demonstration... sometimes it fizzles out, or not.
Another example of lytic enzyme, amylase in the mouth, breaks apart amylose in starch, so a starchy cracker tastes sweeter as it melts in your mouth.
As @Buttonwood suggests, alcohols, ether and oils are more effective solvents than water.

$\ce{H2O2}$ is Selective
The reason why hydrogen peroxide can be used for such diverse applications is the different ways in which its power can be directed -- termed selectivity. By simply adjusting the conditions of the reaction (e.g., $\mathrm{pH}$, temperature, dose, reaction time, and/or catalyst addition), $\ce{H2O2}$ can often be made to oxidize one pollutant over another, or even to favor different oxidation products from the same pollutant.
(From http://www.h2o2.com/products-and-services/us-peroxide-technologies.aspx?pid=112)
You might try mixing 3% $\ce{H2O2}$ with vinegar, ~1/1, or with $\ce{Na2CO3}$ (washing soda), and applying these materials to first, a small test spot, and then the stain. Many times, alkali will solubilize contaminants that aren't easily oxidized - and perhaps addition of a little surface-active agent, like dishwasher soap, can assist removal of oleophilic stains. Some commercial pre-wash stain removers also contain butyl ethers as solvent and surface tension reducer.
And if the clothing is white, there is always chlorine bleach (sodium hypochlorite)!


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Why does only the cis-isomer of potassium dioxalatodiaquochromate(III) form?
A friend of mine performed a standard synthesis reaction of potassium dioxalatodiaquochromate(III) by adding potassium dichromate to ethandioic acid dihydrate and a little water. The synthesis instructions suggested that only the cis-form is created. When looking online about this, I found numerous dodgy documents corroborating this idea, such as this or this.
Why is this? I can't find an explanation as to why only one stereoisomer forms nor any mechanisms anywhere.

There is only one isomer. How could it be otherwise?

@IvanNeretin There are numerous well-known reactions across organic chemistry that lead to the production of various stereoisomers of products and both stereoisomers here are stable (in fact, there is an equilbrium between the forms set up that leans slightly towards the cis-form in aqueous solution). So why is only the cis formed in the reaction?

@IvanNeretin If it's clear, could you write an answer with a mechanism?

@IvanNeretin there seem to be a number of research papers and educational labs discussing the cis and trans isomers of this compound. I'm confused why you say it doesn't exist. This image shows the cis form. The trans would have the H2Os both axial.

It was a misunderstanding on my part; sorry about that. Yes, there are two isomers and they can be produced separately.

@IvanNeretin Can you elaborate at all? Can you explain why the given reaction only produces one isomer?

I'm not an inorganic chemist, but to my understanding, both cis- and trans-dioxalatodiaquochromate(III) ions exist (Note that some also called it as Diaquabis(Oxalato)Chromate (III)). According to what I found online is that solid state reaction gives cis-form while aqueous reaction predominantly forms trans-isomer (it is mentioned in one of your attachments). However, it's still up to debate.
Hamm in 1952 had shown that trans—cis Isomerization using UV-vis spectroscopy ($trans \rightarrow cis$; Ref.1). According to Hamm, all previous workers have prepared trans-isomer because its lack of solubility. Yet, the existence of cis- and trans-isomers has been first recognized by Werner and has presented the preparation methods of two isomers individually (Ref.2), thus it is not true that the your concept of only the cis-stereoisomer forms.
Werner had mentioned that aqueous solution of trans-isomer is stable only in the cold temperatures. That may be the reason why only cis-isomer forms in solid state preparations because the preparation needs heating at high temperature.
Using the method given by Werner, trans-dioxalatodiaquochromate(III) isomer has been prepared and its crystal structure has been studied (Ref.3 & 4).
Later studies have showed this trans-isomer converted to cis-isomer by long standing in aqueous solutions (Ref.5). Based on their results, the authors have concluded that these color changes (dark red trans-isomer to blue cis-isomer) are due to trans–cis isomerization, uncomplicated by olation phenomena and that the transformation to the cis configuration proceeds to practical completion. They also stated that:

Kinetically, the reaction (trans–cis isomerization) obeys a first-order law, and the dependence of the observed velocity constant upon temperature is found to be adequately described by the Arrhenius equation, where $A = 3 \times 10^9$ and $E = \pu{−17.9 kcal/mol}$.

Finally, in recent studies have been done on computational analysis of trans- and cis-Isomers of Potassium dioxalatodiaquochromate(III). The studies revealed that  cis-Isomer is slightly stable than trans-Isomer (Ref.6):

The trans- and cis-isomers of potassium diaquabis(oxalato)chromate (III) were studied computationally and experimentally. The structures of trans- and cis-configurations of $\ce{[Cr(H2O)2(C2O4)2]-}$ were optimized by DFT methodology with various functionals namely: B3LYP, CAM-B3LYP, TPSS, PBE, M06-L and ωB97X-D along with the more sophisticated MP2 method. The calculations show that the most stable forms for both isomers are in quartet states. The results from all DFT methods reveal that the cis-isomer is literally more stable than the trans-isomer with the lower average relative energy of $\pu{2.1 kcal/mol}$. These are consistent with the results from MP2 calculation and experimental observation. The absorption wavelengths for the excited states of trans- and cis-structures were calculated by the time-dependent density functional theory (TDDFT) method. For the experiments, the trans- and cis-isomers of potassium diaquabis(oxalato)chromate (III) were synthesized and characterized by UV-Vis spectrophotometry. Both isomers have two maximum absorption wavelengths at $415$ and $\pu{560 nm}$.

References:

Randall E. Hamm, “Complex Ions of Chromium. II. trans—cis Isomerization of Potassium Dioxalatodiaquochromate(III),” J. Am. Chem. Soc. 1953, 75(3), 609-611 (https://doi.org/10.1021/ja01099a026).
A. Werner, W. J. Bowis, A. Hoblik, H. Schwarz, H. Surber, “Über Metallverbindungen mit komplex gebundener Oxalsäure. Zweite Mitteilung: Über Dioxaloverbindungen),” Justus Liebigs Annalen der Chemie 1914, 406(3), 261-331 (https://doi.org/10.1021/ja01099a026).
J. N. van Niekerk, F. R. L. Schoening, “Structure of Trans-Potassium Dioxalato-Diaquo-Chromiate,” Nature 1950, 166, 108 (https://doi.org/10.1038/166108a0).
J. N. van Niekerk, F. R. L. Schoening, “The crystal structure of trans potassium dioxalatodiaquochromiate, $\ce{K[Cr(C2O4)2(H2O)2].3H2O}$,” Acta Cryst. 1951, 4(1), 35-41 (https://doi.org/10.1107/S0365110X51000064).
G. E. Cunningham, R. W. Burley, M. T. Friend, “Potassium Dioxalato Diaquo Chromium (III) Complexes: Trans–Cis Isomerization,” Nature 1952, 169, 1103 (https://doi.org/10.1038/1691103a0).
Songtham Ruangchaithaweesuk, Juthathip Chorkate, Thana Maihom, Potjaman Poolmee, Piti Treesukol, Tipawan Rungsawang, Phornphimon Maitarad, Bundet Boekfa, "Combined Computational and Experimental Studies of Trans- and Cis-Isomers of Potassium Diaquabis(Oxalato)Chromate (III)," Key Engineering Materials 2017, 757, 103-107 (https://doi.org/10.4028/www.scientific.net/KEM.757.103).

I was just going to say that I never meant to claim that only the cis-form occurs in reactions, simply in the reaction mentioned above. Nevertheless, an amazingly informative answer with all kinds of information previously stuck behind paywalls for me!

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Why is UV-Vis Spectroscopy not like IR Spectroscopy
UV/Vis Spectroscopy is where electromagnetic radiation in the UV to Visible range is shone at a sample to be analysed. The absorbance of each wavelength is recorded and then plotted (very similar to Infrared spectroscopy). That's why it surprised me so much.

This graph is so smooth and with only 1 peak. This is completely different to the IR graph (below):

Why is it that UV-Vis spectroscopy only produces 1 peak? Is it something to do with orbitals overlapping and hence only producing one major peak?

These spectroscopies measure different things, IR the vibrational motion of molecules in the ground state and the uv/vis primarily the electronic transitions from the ground state to each  of higher energy excited states. The latter, however,  do also show vibrational transitions, particularly when the molecule is in the gas phase and collisions with solvent do not 'smear out' the spectrum. Any decent book on spectroscopy will give you a complete description as will many phys. chem. text books.

Also, UV-Vis spectra generally have more than one peak (even the one you show here), but the peaks often happen to be broad and to superimpose.
The great difference that you note is in the changes that IR and UV radiations cause in the molecule (vibrational vs electronic excitation)

The UV spectrum has a width of a factor two, the IR of one order of magnitude.

I'm collecting ideas from the excellent comments and adding some of my own:
The typical UV/Vis spectrum has a narrow range
On the low energy side, it ends where the visible spectrum ends, at about 800 nm. On the high energy side, it ends where water and quartz start absorbing too much, at about 220 nm. Compared to an IR spectrum, it has a narrow range.
Most functional groups have no electronic transitions in that range
As far as organic compounds are concerned, the most common functional group to absorb in the UV is the benzene ring, and extended conjugated systems (e.g. beta carotene) will have electronic transitions in the visible. On the other hand, most organic functional group will show vibrational transitions in the range of a typical IR spectrum.
Peak widths
The spectrum of an electronic transition is broad because it is coupled with vibrational and rotational transitions. An example is the electronic spectrum of iodine:

Source: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/13%3A_Molecular_Spectroscopy/13-06._Electronic_Spectra_Contain_Electronic%2C_Vibrational%2C_and_Rotational_Information
In solution, the bands are not resolved, so the spectrum appears as a single broad peak.

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Adding cyanide vs nitrite for nucleophilic substitution and why does it change when we use silver salts for it
On adding $\ce{NaCN}$ to $\ce{RX}$, we get
$$\ce{NaCN + RX -> RCN + NaX}$$
And not $\ce{RNC}$. Clayden gives the explanation as follows:

Although linear cyanide (which is isoelectronic with $\ce{N2}$) has a lone pair on nitrogen and a lone pair on carbon, the nucleophilic atom is usually anionic carbon rather than neutral nitrogen as the $\mathrm{sp}$ orbital on carbon is of higher energy than that on the more electronegative nitrogen, and therefore constitutes the HOMO.

So according to Clayden, Carbon being less electronegative, has a higher HOMO than Nitrogen, thus it's lone pair attack the LUMO, to form a bond as the less the difference in energy between HOMO and LUMO, the stronger is the bond
But when we consider the same case with $\ce{NaNO2}$, so according to Clayden, nitrogen being less electronegative than oxygen, Nitrogen should have a higher energy HOMO and thus in a nucleophilic substitution reaction it is from the nitrogen side, that  the reaction should take place but ...
$$\ce{NaNO2 + RX -> RONO + NaX}$$
The reaction takes place from the oxygen side. Why?
Also when we have the silver salts of them, e.g., $\ce{AgCN}$, the reaction becomes:
 $$\ce{AgCN + RX -> RNC + AgX}$$
Similarly,
 $$\ce{AgNO2 + RX -> RNO2 + AgX}$$
Why does it change?

Mayr has a great monograph on this topic: https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201007100

I changed the title from "cyanide vs nitrile.." to "cyanide vs nitrite.."  because I felt that's what you meant. Otherwise, cyanide and nitrile are same thing in organic chemistry. If you don't like my change, please feel free to switch it back to what you wish.

Any answer that isn’t from the Mayr paper that @Zhe linked is probably wrong.

@orthocresol If I haven't already said it, thanks for directing me to that paper.

Potassium cyanide ($\ce{KCN}$) is a predominantly ionic compound ($\ce{K+}$ is Hard Acid while $\ce{CN-}$ is Soft base) and would dissociate completely to give cyanide ions in solution. Cyanide ion ($\ce{CN-}$) is considered an ambient nucleophile, which means that the negative charge can associate with either carbon or nitrogen. However, the negative charge of cyanide ion concentrated on carbon than nitrogen. Thus, reaction with haloalkanes yields cyanides (nitriles) since $\ce{C-C}$ bond is thermodynamically more stable than $\ce{C-N}$ bond (simplest explanation).
Silver cyanide ($\ce{AgCN}$) is largely covalent and it does not dissociate easily to give $\ce{CN-}$ due to soft-soft interaction ($\ce{Ag+}$ is Soft Acid and $\ce{CN-}$ is Soft base). Thus, negative charge of $\ce{CN-}$ ion is mostly available at nitrogen, and hence the attacking centre of the nucleophile will be $\ce{N}$-atom. As a result, you get nucleophiluc substitution of haloalkanes to give isocyanides (isonitrile).
Same argument can be given for nitrite formation with $\ce{KNO2}$ and nitro formation with $\ce{AgNO2}$. In this case, $\ce{KNO2}$ is a predominantly ionic compound ($\ce{K+^-O-N=O}$) where one of oxygen atoms of nitrite ion has negative charge. Therefore, the nucleophilic attack of $\ce{NO2-}$ on haloalkane would result alkyl nitrite ($\ce{R-O-N=O}$). Silver nitrite ($\ce{AgNO2}$) is, on the other hand, largely covalent and would not dissociate easily to give free $\ce{NO2-}$ ions. As a results, the lone pair on nitrogen acts as an attacking site, forming nitroalkanes (high electronegativity of oxygen compared to nitrogen prevented it from donating its lone pair, as explainnation given by Clayden).

Sir, could you explain with the MOT part ?? Clayden says that in $\ce{K+ CN- }, attack from Carbon side occurs , because it being less negative has a higher HOMO. But with $\ce{K+ NO2- }, Nitrogen being less electronegative than oxygen , should have a higher energy HOMO , thus attack from Nitrogen side should happen..... But it does not.

That's because the negative charge of nitrite ion is on one of the oxygen atoms (see the resonance structure of $\ce{NO2-}$). Negative charge with a lone pair always wins vs just a lone pair.


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Predict the number of peaks in H-1 NMR of sesamolin

Would all of the Hs in an NMR of sesamolin (5-{(1S,3aR,4R,6aR)-4-[(2H-1,3-benzodioxol-5-yl)oxy]tetrahydro-1H,3H-furo[3,4-c]furan-1-yl}-2H-1,3-benzodioxole; shown above) be non-equivalent to each other? So even for the two Hs at A, they would be non-equivalent to each other.
Follow up question (about a different molecule that is similar): for sesamin (5,5'-[(1R,3aS,4R,6aS)-tetrahydro-1H,3H-furo[3,4-c]furan-1,4-diyl]bis(2H-1,3-benzodioxole); shown below), would there be 8 peaks in an H-1 NMR?

For sesamolin, I think the pairs of protons at A and N would have the same chemical shift. In the spectrum this would appear as a singlet with an area corresponding to 4 protons due to the accidental equivalence of the A and N site. It looks like you did this in assignment of sesamin, which looks good to me.
Assigning shifts to geminal protons can be tricky. For example, the differing chemical shifts of the axial and equatorial protons in cyclohexane can only be resolved at low temperature:

Some more examples below:

Both of the images and more information on the topic are available here:
https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-02-delta.htm

That (protons at A and N) depends on wether the whole aryl group can rotate fast enough. I suspect the proton H and the proton at K would collide during rotation, and the two protons in the methyl group N are not identical. The CS difference will be small, though, so the AB spin system might collapse into a slightly broadened singlett.

Assuming the OP's original structure has been corrected to one with a 2-fold axis of rotation, the environments of A and N are the same but each methylenedioxy group has two diastereotopic H's. A'=N' and A"=N".  The "number of peaks" is not the same as asking how  many unique chemical shifts (different hydrogens) are present. Higher field will show more peaks but the same number of chemical shifts.

As one already pointed out, your sesamin molecule has extra $\ce{O}$ atom. Assuming it will be corrected, the answer for your question is yes, all hydrogens should have different chemical shifts since all of them are either chiral or diestereptropic (aliphatic hydrogens) or in different environment (aromatic protons) except for two methylene groups on the aromatic rings (A and N). A reasonable explanation is already given for Why these are different in another answer in this page. All I can do is assigning some values. I found a one research group assigning all peaks for the metabolite of sesamin found in human urine (Ref.1). I included all aliphatic proton assignment for the benefit of reader (taken in $\ce{CD3OD}$ using a $\pu{600 MHz}$ machine). I would not expect drastic chemical shifts difference for aliphatic protons in original sesamin (only minimal changes) from these values.

Reference:

Ali A. Moazzami, Rolf E. Andersson, Afaf Kamal-Eldin, "Quantitative NMR Analysis of a Sesamin Catechol Metabolite in Human Urine," The Journal of Nutrition 2007, 137(4), 940–944 (https://doi.org/10.1093/jn/137.4.940).


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Sodium metal burning off when conducting electrolysis and I don't know why
I melted down some table salt and used high voltage welder power source to try and separate the sodium and chlorine it was partially successful because I could smell and see chlorine gas but during the reaction what I think was sodium would float to the top of the molten salt and burst into flame. Is there a way to prevent this?
I was thinking maybe an airtight crucible, switching the anode and cathode or keeping a lower temperature to prevent it vaporizing. Any answer would be very helpful.

First, sodium burns very nicely in air. However, it also burns in chlorine, returning back to the $\ce{NaCl}$ with which you started.
Most welders provide alternating current, meaning the the electrode becomes alternately positive and then negative (50 times per second [50 Hz] in much of the world, and 60 Hz in North America). Therefore, both $\ce{Cl2}$ gas and $\ce{Na}$ metal are produced at each electrode and recombining.
To produce sodium metal you need a direct current supply, such as from a battery, and some means to separate the chlorine bubbling out at the positive (anode) electrode from the sodium metal at the negative (cathode) electrode. A diagram of the Down's cell should help you understand this, and you can view a video of sodium production.
As a demonstration, I've made (for a very short time) small amounts of sodium in an open crucible using a laboratory 10 volt, 6 ampere supply... One sees a small metallic bead forming under the molten salt at one wire and immediately burning when it hits the air, with sodium's characteristic yellow glare.


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Synthetic rubber coagulation
Is it possible to make crepe rubber from
combining a synthetic rubber emulsion (Styrene Butadiene Emulsion)
and a strong acid?

Could you provide a little more background, like why you think this combination might work?

Well as far as i know, crepe rubber is made by coagulating natural rubber using strong acids, i wanted to know if this applies to synthetic rubber emulsion sold by various building material companies.


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What reagents can be used to distinguish pentose and hexose sugars?

I have been asked how I can differentiate D-glucose and D-xylose using the following reagents / reactions:

Orcinol
Benedict's test
Ammoniacal silver nitrate
$\ce{FeCl3}$
$\ce{C6H5SO2Cl}$, excess $\ce{NaOH}$
$\ce{NaOH}$, $\ce{H2O}$, heat
t-Butyl chloride, $\ce{AlCl3}$
$\ce{Br2}$ in water, dark
$\ce{NH2OH.HCl}$, $\ce{KOH}$, $\ce{FeCl3}$
Resorcinol
2,4-DNPH
Iodoform test

Usually, chemical tests identify certain functional groups in molecules which react in a particular way. However, here the two compounds only differ in the number of carbons and don't have any real difference in their functional groups, so what reaction can be used to differentiate them?

react with 2,4-DNPH then a careful measurement of the melting point of the phenylhydrazone and compare against literature values

I also noticed that the compounds appear to be open-form sugars. Maybe only one of them will form a cyclic form?

I'm not going to address all of the reagents, but I want to highlight a specific approach in this partial answer.
Occasionally, knowing some history of chemistry is quite useful when thinking about how to solve problems. In this specific case, I direct you to the work of Emil Fisher (chemistry Nobel 1902), who determined the structure of all hexose sugars. He did this by building up the sugars from smaller subunits while keeping track of symmetries.
This immediately suggests a solution where one performs reduction (or oxidation) of the aldehydes. In other words, make the both ends of each molecule look the same. This is a valid strategy because the underlying symmetries of glucose and xylose are different.
The penta-ol is meso and therefore achiral, while the hexa-ol is chiral and optically active since you started with one enantiomer.
Thus, any reagent that can produce the bis-acid or bis-aldehyde without epimerizing any stereocenters along the way or any reagent that can reduce the aldehyde to an alcohol will provide a suitable method of differentiation in this specific example.

You are correct that there are few differences between glucose and xylose simply based on functional groups. The only real difference is that one has six carbons and the other five; they are members of what we normally call hexose and pentose sugars.
This is not my area of expertise, but a quick Google search suggests that these can be differentiated using what is known as Bial's test. In the presence of concentrated HCl, pentose sugars such as xylose undergo acid-catalysed dehydration to form an aromatic molecule, furfural:

Furfural can then be reacted with orcinol to form a characteristically blue or green species, which is taken as a positive result indicating the presence of a pentose:

Because hexose sugars are also capable of existing in a cyclic hemiacetal form, they are also capable of undergoing a very similar reaction. The difference is that hexose sugars do not form furfural, but rather a derivative of furfural with an extra hydroxymethyl group ($\ce{CH2OH}$):

This extra hydroxymethyl group supposedly leads to a different colour being observed (Wikipedia says: "the related hydroxymethylfurfural from hexoses may give a muddy-brown or gray solution, but this is easily distinguishable from the green color of pentoses").
According to the same Wikipedia page, the reagents you need to use are therefore:

... 0.4 g orcinol, 200 ml of concentrated hydrochloric acid and 0.5 ml of a 10% solution of ferric chloride.


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How to know if a nucleophile will attack from front-side or back-side in a sn2 reaction?
How to know if a nucleophile will attack from front-side(retention of configuration) or back-side(inversion of configuration) in sn2 reaction?
https://www.youtube.com/watch?v=L01vv1Mj64I
In this video same reaction is showing both frontside and backside.

That "frontside $S_N2$" isn't an SN2 reaction, and it is extremely unfavourable. Stupid video. Retention occurs with the $S_Ni$ mechanism, e.g. chlorination of alcohols via thionyl chloride.

I agree, this is not a useful teaching tool.

The video is an animation ,without any commetry with respect to which type of attack is energitically more feasible. Without that ,this video is not usefull

I feel like the so-called “frontside Sn2” looks more like the mechanism for electrophillic Attack to the carbon, though.

Due to the leaving group and the nature of the compound in which the nucleophile is reacted with I.e either primary or secondary haloalkane as the case may be

The video you linked to mentioned a particular research paper which you should read if interested (either just ask your institution for a copy or use you-know-what-website). The conclusion writes:
“...backside SN2-b barriers increase
along the nucleophiles F-, Cl-, Br-, and I- and decrease along the substrates CH3F, CH3Cl, CH3Br, and CH3I. Frontside SN2-f barriers show the same trends but are in all cases much higher (ca. 10-60 kcal mol-1) because of more steric repulsion as a result of the proximity between the nucleophile and leaving group.”
In this paper, they used SN2-b and SN2-f to represent backside/frontside reactions. I think that should answer your question.
There’s something that you should take note of, though: the “front-side Attack” mechanism would be the optimal one for the electrophillic Attack of methane. You might want to read that at J. Am. Chem. SOC. 1995,117, 1336-1343 (“Electrophilic Substitution of Methane Revisited”). This can be explained in molecular orbital terms (you might want to read up on the isolobal analogy if interested).
In a concerted nucleophillic substitution, the transition state is “electron-rich” and tends to adopt the most “open” geometry possible. This is anagolous to the linear geometry of the trihydrogen anion.
Meanwhile, in a concerted electrophillic substitution, the transition state is “electron-deficient” and tends to adopt a “closed” geometry, à là the trigonal planar shape of the trihydrogen cation.


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Basicity of anions

Which one of the following is strongest base?
A. $\ce{Cl-}$
B. $\ce{Br-}$
C. $\ce{I-}$
D. $\ce{CH3COO-}$

Someone please tell me the answer with proper reason.
First I thought that it's $\ce{Cl-},$ but then it clicks in my mind that it's  $\ce{CH3COO-}$ since it's a conjugate base of a weak acid.

You actually got it! The proper way to get to the (right) answer is to ask yourself which one is the conjugate base of a weak acid. If you are given more than one weak acid, you should be provided with the respective acid dissociation constants $K_\mathrm{a}$. In that case, keep in mind that the smaller (i.e. the closer to zero) the $K_\mathrm{a},$ the stronger the acid.


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Entropy in organic chemistry
Why do we not talk about entropy when deciding feasibility of an organic reaction?...We always seem to only be interested in stability of product and intermediate(essentially delta H)and draw their respective graphs of Delta H of the reaction showing the peak of intermediate..shouldn't we look at change in  Gibbs free energy for commenting on feasibility?

What do you mean when you say "feasibility?"

Spontaneity .. regarding whether the reaction proceeds through a certain pathway here

You'd have to show an example of what you're talking about, but in my experience, reaction coordinate diagrams always have free energy on the vertical axis. We do provide bond dissociate enthalpies in tables, but that's because the entropic component is not easy to generalize.

I mean we only talk about relative stabilities (and not entropy )of various intermediates while deciding which pathway the reaction will follow..why is that?

That depends on what you mean by stabilities. If you're talking about stability as measured via a free energy, that is the most accurate way to do it, since free energy is a proxy for the total entropy change of the universe, which is the specific quantity that has restrictions based on the second law of thermodynamics.

I actually mean stability in terms of intermediate stability on a high school level..like this carbocation is more stable or this resultant double bond is more stable

Unfortunately, the underlying chemistry does not change if you approach it from a high school level or not. The context is not the level but the particular effects you are trying to discuss. For the types of things you're talking about, the entropy effects are actually probably similar in the series, so bond enthalpies are generally what are looked at for simplicity.

Take a look at the second paragraph under the "Kinetic and thermodynamic considerations" section, at  https://en.wikipedia.org/wiki/Energy_profile_(chemistry)

It's a bit like the story of the person who loses their keys at night and starts searching near a lantern where it is easy to see, even though the keys were lost somewhere else. Enthalpy changes are easier to estimate than entropy changes. Luckily, often Gibbs energy of reaction is dominated by the enthalpy change.

Why is the Gibbs energy of reaction dominated by enthalpy change?..or is this like a general experimental observation?


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At boiling point, is the total pressure twice the atmospheric pressure?
We know that one definition of boiling point is that it's the temperature at which the vapor pressure is same as the atmospheric pressure.
Assuming a closed container, does that mean that at this temperature, the total pressure exerted on the liquid is twice the atmospheric pressure, i.e., the vapor pressure plus the pre-existing atmospheric pressure?
edit: Based on some comments, it appears that closed container assumption is not very helpful. The questions still remains, even without that assumption: Does the total pressure become twice the atmospheric pressure at the surface?

No, because at the expected "boiling point", 100°C, the atmospheric pressure will be two bars. In a closed vessel, water does not boil.

First of all, the "duplicate" question has discussion that claims it is not always true that boiling doesn't happen in closed containers. And secondly, the closed container assumption is tangential to the question in the title. If closed container is a bad assumption, the question still remains: Does the total pressure become twice at BP temperature, in an open system?

Closed vessel is not an assumption, it's the condition you have set in your question. Do you want it or not?

The pressure in an open vessel is, quasi by definition, ambient pressure. If it was higher, water wouldn't boil at 100°C at ambient pressure, and here everything starts to get a bit illogical imo.

Pressure is not additive
If you mix two gases of different pressure, the resulting gas does not have a pressure equal to the sum of the two pressures. Treating the gases as ideal gases, the pressure will be the average of the two pressures, weighted by the volumes. What is additive is the partial pressure of a gas mixtures, where the sum is the total pressure.
Closed container with constant volume
To simplify things, let's place a closed bottle filled with the liquid of interest into the container, adjust the pressure to 1 atm, and close the container. If we heat the container, the pressure will go up just because the gas temperature rises. If we open the bottle, some of the liquid will evaporate until the partial pressure is equal to the vapor pressure. Again, the container will have a pressure higher than atmospheric pressure. So the liquid will not boil when the temperature reaches the normal boiling point because the pressure will be above atmospheric pressure at that point, for the two reasons mentioned.
Closed container with variable volume
If the experiment is done in a closed, variable volume container that expands into the surrounding kept at 1 atm, the volume will increase as the temperature is raised (because the gas expands, and because the vapor pressure increases with increasing temperature). When the temperature reaches boiling temperature (slightly above, actually), the liquid will boil until it is all gone, and the volume will increase further during this boiling process. The partial pressure of the liquid will never reach 1 atm (because there are other gases present), so the vapor pressure will remain higher than the partial pressure throughout the boiling process.

Does the total pressure become twice the atmospheric pressure at the surface?

No. Either there is room to expand, and the gas produced by boiling will push away the other gas (and then quickly mix with it), or we keep the volume constant, and then we don't call it boiling (because the total pressure exceeds 1 atm, so the liquid will not boil. However, when you heat a liquid to the boiling point and let the system reach equilibrium, the partial pressure of that substance will be 1 atm, and if any other gases are present, the total pressure will be higher. How much higher depends on the starting conditions. If our starting condition is a completely filled closed bottle, with inert gas (e.g. nitrogen or air) already at the boiling point of the liquid and at 1 atm, when we open the bottle the total pressure will reach 2 atm.

I think you need to define what you mean by "mix," since that is not necessarily clear to a beginner reader.

@Karsten Theis ,"When the temperature reaches boiling temperature (slightly above, actually)" is from your answer.A dought though,if liquid in a closed vessel reachs boiling point ,with pressure outside at 1 atm ,would the pressure in a closed container be < 1atm or >1 atm.Are we assuming that this closed container ,its boundries are insulated from surroundings? In such a case what will be the pressure inside ?

@ChakravarthyKalyan Could you ask a new question, and make clear what you are asking? If the container is constant volume, the outside pressure has no influence on the outcome.


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How many products does this reaction yield?
How many products can we get from the monochlorination of 2,4-dimethylpentane including stereoisomerism? I have seen someone say that it yields 3 structural isomers one of which has two enantiomers making them 4 products in total, but I couldn't find the chiral carbon he was talking about.

Monochlorination of 2,4-dimethylpentane can yield 4 structural isomers.
None of those isomers have chiral carbon atoms.

It certainly doesn't yield 4 structural isomers according to my college professor and to this guy on Quora, they both say it yields 3 structural isomers yet the dispute between the two is that of one of those 3 structural isomers has an enantiomer which makes the total 4.                                        https://www.quora.com/If-Cl-is-attached-to-2-4-dimethylpentane-2-optically-active-compounds-are-possible-What-are-they

The term "products" is vague. Constitutional (structural) isomers have the same molecular formula but different atom connectivities. In your example, there are three structural isomers. Stereoisomers have the same molecular formula, the same atom connectivity but different arrangements in space. Thus, the enantiomers of 1-chloro-2,4-dimethylpentane are stereoisomers of each other. Constitutional isomers are a subset of isomers (same molecular formula) and stereoisomers are a subset of constitutional isomers.

The expected products of mono-chlorination of 2,4-dimethylpentane are depicted in following scheme:

The substrate has 12 primary hydrogens (on 1-,1'-, 5-, and 5'-methyl groups),  2 secondary hydrogens (on 3-methylene group),  and 2 tertiary hydrogens (on 2-, and 4-methine groups), as numbered in the scheme. Altogether, total of 3 different hydrogens are present, therefore, 3 different mono-chlorinated products are possible as sketched in the bottom boxes of the scheme. Although 1- and 1'-methyl groups (and 5- and 5'-methyl groups as well) are equivalent in the structure, mono-chlorination of each would give different stereoisomer (strictly speaking, enentiomers). Therefore, counting stereoisomers, you should get total of 4 different products.


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Absorption of UV Light and what color would be seen?
I am mathematically experimenting with different wavelengths of light that a fictional material would absorb. I thought of a material that absorbs a wavelength of $\pu{193 nm}$. This wavelength corresponds with the area of ultraviolet light. I am having trouble figuring out what this material would look like to a human eye. I understand that human eyes cannot see UV light. My question is, what color would this material appear as?

Ordinary window glass absorbs 193 nm light. So, if that glass does not fluoresce, i.e., emit light that has wavelength(s) in the visible spectrum (nominally 400 to 700 nm), then you see nothing different than normal.

Is it a math model of an ideal material which solely absorbs at a given wavelength, or a real-life example of a material with possible reflection, scattering, photoluminescence, …?

The wavelength is the ArF excimer laser wavelength, used in excimer laser lithography. Curious question about a deep UV wavelength.

If it is daylight you won't typically see any fluorescence. The fluorescence is happening of course, it's just swamped by the daylight that you see coming from whatever.

FYI Sunscreen absorbs UV light.

Oxygen and nitrogen molecules both absorb in the ultra violet below 200 nm where our eyes certainly cannot detect photons. If you could only see down there the sky would look black. If you look at a uv light source (besides  damaging your cornea) what you see is fluorescence, probably from the tryptophan and other aromatic molecules  in the proteins of the eye.

Kōan: What is the color of the light that is not seen?

If you have a mathematical material which only do that, then you see nothing by eyes.


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Activators for non-sulphide compounds in froth floatation
I was recently studying about Metallurgy.
Concise Inorganic Chemistry by J. D. Lee claims the process of froth floatation can also be applied to non-sulfide ores such as Malachite ($\ce{CuCO3.Cu(OH)2}$) and Anglesite ($\ce{PbSO4}$) by using $\ce{Na2S}$ as an activator.
The coatings formed are $\ce{CuS}$ and $\ce{PbS}$ respectively. However, in a previous example, they used $\ce{CuSO4}$ as the activator over a $\ce{ZnS}$ surface and a coating of $\ce{CuS}$ was formed.
Therefore, going by the same co-relation in the case of Malachite and Anglesite, shouldn't the coating be a sodium compound instead?

Partly related: Froth floatation process - why CuSO4 is activating floating character of ZnS?

Sodium compounds are soluble, they won't stay.

A previous comment noted that sodium compounds are generally soluble and would not be strongly adsorbed on malachite and anglesite.
However, the other compound formed in the reaction between Na2S and a copper or lead salt would be CuS or PbS, both of which are very insoluble. Three cases can be visualized: 1) a sodium ion approaches the Pb or Cu salt and this aquated sodium ion forces off a lattice-bound dication, or 2) a sulfide ion approaches the Pb or Cu salt and either deposits on the crystal surface, forcing a nearby OH-, CO3-- or SO4-- to leave in order to maintain electrostatic neutrality, or 3) the low (but not zero) solubility of the Cu and Pb salts allows precipitation of PbS or CuS as tiny crystals which more frequently than not become bound to the original salt by epitaxial considerations.
In the case of CuSO4 applied to ZnS, ZnS is about 20 times more soluble than CuS, so an exchange of cations will occur, with the copper going into the insoluble phase.
Case 1 seems far-fetched. Cases 2 and 3 are similar, and may occur simultaneously. The key is that the insoluble product is what sticks to the material that you want to float. And the surfactant that forms the bubbles sticks to the activator (CuS or PbS).
It's interesting that ZnS is not easily flotable. ZnS in water may acquire an aquated layer, that is, it is activated by water, which is not selected by the surfactant, so it sinks. Both the copper carbonate and lead sulfate would seem to have surfaces more compatible with water than their sulfides, too, so converting the surface is required.


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Is omega 3 a specific fatty acid?
Is eicosapentaenoic acid (EPA) omega 3 or that's just a acid that comes with a omega 3? Same question for docosahexaenoic acid (DHA).
Just want to know are both of those two acid omega 3 despite being different acids.

No, $\omega$-3 is an umbrella term, much like "acid". How can they both be called acids, despite being different molecules? Well, just like that.

There exists a range of ω-3 acids.
Fatty acids are made of a chain of carbon atoms (generally linear), with a carboxylic acid at one end. Usually the carbon atoms are linked with a single-bond, but in unsaturated acids one or more of these are double-bonds.
In order to describe the positioning of these double bonds, the carbon-atoms are numbered, starting either from the carboxylic acid end, also called the α-end,  or from the methyl-end (the other end), which is called the ω-end. Therefore, an ω-3 acid is a fatty acid with a double-bond from carbon number 3 to 4, counted from the methyl/ω-end.
Fatty acids can have several double-bonds in its chain, making it a polyunsaturated acid. An example is Eicosapentaenoic acid, which has 5 double bonds. Edit: As pointed out in comments, it is the position of the first double-bond that determines if the acid is called an ω-3 acid. Likewise: if the first double bond is from carbon number 6 to 7, the acid is an ω-6 acid.
Wikipedia has some very informative articles about both ω-3 acids and fatty acids  in general.

Actually, the ω designation refers to the position of the first double bond from the methyl end.  Thus, since the first double bond from the methyl end in EPA is at carbon 3, it is an ω3 fatty acid, not an ω3,6,9,12,15 fatty acid (the other double bonds aren't the first from the methyl end, and are thus not used in determining the ω notation).

I'll add the reason the ω notation is used is that organisms typically elongate fatty acids from the carboxylic acid end, so the ω-number of a fatty acid stays constant through this process.  Thus if you eat an 18:3ω3 fatty acid, your body can convert it to EPA (20:5ω3) or DHA (22:6ω3) by elongating it and adding double bonds.  But your body can't make DHA from, say, 18:1ω9 or 18:2ω6.

Thank you theorist, you are off course quite right when pointing out that the ω-nomeclature only refers to the first double bond. I stand corrected :-)

@FrankS So is Eicosapentaenoic acid a ω-3 instead of ω-3,6,9,12,15 acid? (If yes, could you edit?)

@Volker: That is correct. I have added an edit to the answer.


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Which of the following groups show +M and -I effect?

Which of the following group show +M and -I effect?
(A) $\ce{-\underset{\underset{\huge O}{||}}{C}-F}$
(B) $\ce{-\underset{\underset{\huge O}{||}}{C}-OR}$
(C) $\ce{-O^-}$
(D) $\ce{-OH}$

Answer is option D. D is obviously correct, but why not option C? $\ce{-O^-}$ has electrons to donate and is electronegative.

O- is an electron-rich species, which donates its charge to the ring, stringly activating it. It has a very strong +I effect!


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EPR or NMR sample that will give narrow lines at 4 K
We have made a crossed coil setup (X and Y transmit/receiving coils) with a SQUID pick up loop along Z. I have built a similar CW NMR spectrometer with inductive pickup that can detect Teflon NMR with a single scan SNR of around $1$ at around $\pu{6 mT}$ at room temperature. I expect SQUIDs to be more sensitive and the magnetization to stronger by a factor of $\pu{300 K}/\pu{4 K}.$
Now we want to see EPR or NMR lines, probably using it in a CW fashion, i.e. putting a ramp plus lock-in oscillation in X, a CW excitation frequency in Y and demodulating the SQUID signal in Z.
What sample will give us nice, narrow, distinguishable lines at $\pu{4 K}?$
We did try Teflon NMR and copper sulfate EPR and they both failed. I think the reason for the SQUID setup not working (someone who knows about samples told me), is that Teflon $T_2$ gets very short at low temperature, so the lines are very broad.
Note: we can only go up to $\pu{10 mT}$ with the X coil. Liquid $\ce{^3He}$ is off our budget.

Have you thought about endohedral fullerenes, like N@C60?

Comments are not for extended discussion; this conversation has been moved to chat.


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Use of cryolite and fluorspar in electrolytic reduction of pure alumina by Hall-Héroult process
Cryolite $\ce{(Na3AlF6)}$ and fluorspar $\ce{(CaF2)}$ are added to alumina to lower the temperature of the mixture from $\pu{2000 °C}$ to about $\pu{900 °C}$ and also to increase the conductance.
But my question is about the specific effect these compounds are producing. I mean, out of the two mentioned effects, which compound is doing what?

It is my understanding that Hall originally used cryolyte to dissolve alumina so he could perform electrolysis. The molten cryolite was both solvent and electrolyte and electrolyzing cryolite itself was not a problem. My guess is that the fluorite just helped lower the melting temperature. Fluorite (from 'to flow') has been used for centuries to help 'liquify' oxide slag, in metallurical processing, so the slag could be more readily scraped aside. Maybe someone will provide a definitive answer.

I don't know enough to post a full answer but as Ed V says $\ce{(CaF2)}$ is used "generally" in slag-metallurgical processes to control viscosity. It also provides a surplus of flourine ions, which I guess would be useful for dissolving and solvating the alumina - since the cryolite natively cannot handle much more aluminium ions, you need a surplus of fluoride.

@ChakravarthyKalyan What was the point of your edit? Units should be depicted upright and with proper spacing; representation you've suggested contradicts both rules. If you are having trouble with a mobile app not being able to render \pu{…} macros, please see the corresponding Meta post. The short version is not to use a mobile app at all as it's pretty outdated.

@andselisk ,in my mobile, these values are shown as  "undefined control sequence\pu"  twice. I can post screen shot of the same.Hence the edit.

@ChakravarthyKalyan I think I already addressed this in my comment, please read the Meta post and suggestions within carefully. If your device doesn't support formatting features available in every web browser (and mobile web browsers, too), it's not a reason to downgrade the quality of the posts.

Fluorspar is doing both, reducing the temperature of the melt as well as making the mixture conducting

According to the CRC Handbook of Chemistry and Physics, melting point of $\ce{CaF2}$ is $\pu{1418 ^\circ C}$, of $\ce{Al2O3}$ (in its corundum form) is $\pu{2053 ^\circ C}$, and of $\ce{Na3AlF6}$ is $\pu{1009 ^\circ C}$ (lowest of the all three). Thus the cryolite should melt first.
I have read mentions of the problem you asked in the book Nonaqueous Electrochemistry (you can find information about it here. For the best of my summarising ability, I can conclude that the role of the cryolite and other salts added to the reaction mixture is that they decrease the melting point of alumina. As cryolite's melting point is much lower than that of alumina as cited above, when the cryolite melts it acts as a solvent to "dissolve" the $\ce{Al2O3}$, by the process of complexation. Recall how silver chloride dissolves in aqueous ammonia or lead iodide dissolves in excess iodide. The book states that:

Cryolite is used as solvent for alumina because it has high chemical stability and a high solubility for alumina.

In detail:
When pure cryolite ($\ce{Na3AlF6}$) melts, it dissociates into fluoride ions and various lower fluoroaluminates:
$$\ce{[AlF6]^3- -> [AlF4]- + 2 F-}$$
$$\ce{[AlF6]^3- -> [AlF5]^2- + F-}$$
(etc.)
However, in the presence of alumina, the dissociation process is complicated by the fact that the products arise from $\ce{Na3AlF6}$ dissociation can react with $\ce{Al2O3}$ creating oxofluoroaluminates, and it is those complexes that directly get electrolysed (cathodically reduced) rather than the $\ce{Al2O3}$ itself. By various means of calculation and experiments (Raman spectroscopy etc.) they proposed some following reactions (or equilibria):
$$\ce{Al2O3 + 4 [AlF5]^2- -> 3 [Al2OF6]^2- +  2 F-}$$
$$\ce{Al2O3 + [AlF5]^2- + F- -> 3/2 [Al2O2F4]^2-}$$
$$\ce{2 Al2O3 + [Al2O2F4]^2- + 4 F- -> 2 [Al3O4F4]^3-}$$
...
When you add calcium fluoride (or magnesium fluoride) into the mixture, the following reactions also occur:
$$\ce{MgF2 + n F- + (2-n) [AlF5]^2- -> [MgF_{(2+n)}(AlF5)_{(2-n)}]^{(4-n)-} }$$
$$\ce{CaF2 + (3-n) [AlF5]^2- -> [CaF_{n}(AlF5)_{(3-n)}]^{(4-n)-} + (2-n)F-}$$
The electrolytes in the electrolysis process is obviously "$\ce{Na3AlF6}$", but in fact they always add an excess of $\ce{AlF3}$ alongside other salts such as $\ce{CaF2}$ ($\pu{3-5 \%}$), $\ce{LiF}$ ($\pu{0-4 \%}$) and $\ce{MgF2}$ (also $\pu{0-4 \%}$). In fact, those salts decrease the solubility of $\ce{Al2O3}$ rather than increase it; that's why they only add a little of those salts.
Then why adding those salts?
There are a couple of reasons for this. Firstly, addition of those salts increase the density of the electrolyte. Because aluminium is already produced in the liquid state (melting point $\pu{660 ^\circ C}$), if the density of the two components are too similar, it is impossible to achieve a good separation-by-weight. Kvande and Rorvik had derived an equation of the melt density as a function of temperature and salt composition:
$\pu{d = 2.64 - 0.0008T + 0.18(wt. ratio NaF/AlF_{3}) - 0.008(wt. \% Al_{2}O_{3}) + 0.005(wt. \% CaF_{2}) + 0.008(wt. \% MgF_{2}) - 0.004(wt. \% LiF)}$
As you can see from the above equation density of the mixture increases with CaF2 content.
The other reason is that those salts increase the interfacial tension at the liquid Al-electrolyte boundary. The higher interfacial tension is, the less does aluminium re-dissolve in the electrolyte (yes it can!), thus better current efficiency. For example: at the boundary this reaction occurs: $\ce{Al + 3 NaF -> AlF3 + 3 Na}$. Adding more $\ce{AlF3}$ (or other fluorides) helps surpressing this reaction, and also may increase the $\ce{Na+}$ concentration in the electrolyte - the researchers found that $\ce{Na+}$ ions are the primary charge carriers.

You've been a member of the community for some time. There is little excuse for not formatting your (often well-received) posts. I've flagged this as LQ. Also: "AlF63- -> AlF4- + 2 F-"? That's not only wrong, it's indecipherable.

I have formatted your answer. Please make sure to learn formatting by yourself. This meta post should help: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

@ToddMinehardt I know that it is evil for not formatting the post, but the truth is although I've been a member for some time I rarely spend time on this; and I am also not certainly comfortable of using math formats either - which I can improve. And in addition the thing in your comment is actually not wrong, it is just made a bit difficult in deciphering due to my lack of formating, that's all.

@ĐỨcLêHồng Understood. I've up-voted and encourage you to do a little bit of editing at least when you do visit. As I said, your contributions are uniformly good and of value to the community here.

Cryollite is used to decrease the temperature from $2000 °C$ to $900 °C$ . Fluorspar is used to increase mobility by controlling viscosity, hence in turn increasing conductivity.

Please provide a source of this information.


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Can the light reaction of photosynthesis happen spontaneous in pools of water?
I've just learnt that a part of the photosynthesis reaction is the light reaction, where sunlight is absorbed by chloroplasts, which then excites electrons to higher energy levels, allowing for water molecules to be broken down.
My question is whether this can also happen in pools of water? I mean, if all that is really needed is sunlight, then I don't see why water molecules in pools won't get excited and thus break down too.
If not, then how do chloroplasts assist in this?
And on top of that, why don't chloroplasts get damaged since this is similar to how UV rays damage our skin as it can excite electrons in cells and damage genes?

Photolysis without a chloroplast is about as efficient as driving without a car.

You are missing the other half-reaction ($\ce{CO2}$ to carbs), and the catalysts. If you add some cyanobacteria to the pool, you are all set. That is where the oxygen in the atmosphere initially came from.

Photolysis with a chloroplast is highly tuned, but not necessarily efficient (depending on how you define efficiency). However, sunlight is basically free if you can get your hands (leaves?) on it, so efficiency there is relative to other competing plants.

In the light reactions in photosynthesis the photon never interacts with the water molecule. The photon is absorbed by  one of many chlorophyll molecules which transfer the energy to a chlorophyll dimer and, after a series of electron transfer steps, generates enough energy to extract an electron from water (or a similar fuel in photosynthetic bacteria). This latter process takes place in a protein as do all the reactions in photosynthesis. The reactions have been studied in detail by ultrafast spectroscopy. The structures of the proteins are known from x-ray diffraction.

Photosynthesis is a lot more complex than just breaking down water
Yes, photosynthesis involves a reaction which partly involves absorbing photons and breaking down water. But overall it is far more complex than that as it also involves absorbing carbon dioxide and turning that into sugars and other molecules that plant can use a fuel. Breaking down water is a byproduct of that larger process as is the oxygen released by it.
The detail of the reaction involved is complex and consists of dozens of individual steps. It isn't as simple as absorbing a photon and splitting water. In fact, there are two separate and individually complex systems of proteins involved in the water-splitting part of the process: photosystem I absorbs light and generates higher energy electrons; photosystem II also absorbs light, breaks water molecules part and generates higher energy electrons. But these processes are so complex that we don't fully understand how they work (and our understanding of the structure of photosystem II was updated as recently as 2014). And that machinery is very complex involving dozens of individual proteins and components.
Given the complexity of the processes behind water-splitting in photosynthesis (which involves multiple photons and electrons) it should be clear that there is no easy way to perform the reaction.
You don't just need photons and water molecules. You need some of the most complex molecular machinery we know of and multiple photons. Sunlight + water isn't going to achieve it.


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What does pH of NH3 solution refers to?
I was introduced with an exercise which asks to find $K_\mathrm{b}$ of $\ce{NH3}$ when the $\mathrm{pH}$ of $\ce{NH3}$ solution with concentration $M_1$. Well, in order to solve it, I used the fact that $\mathrm{pOH} = 14 - \mathrm{pH}$ and the dissociation $\ce{NH3 + H2O <=> NH4+ + OH-}$ to compute $\ce{[OH-]}$ and then $K_\mathrm{b}$. However I don't fully understand what does $\mathrm{pH}$ mean, since in this dissociation there aren't any $\ce{H3O+}$ molecules. I think it relates to the conjugate dissociation of the $\ce{NH4+}$, i.e.: $\ce{NH4+ + H2O <=> NH3 + H3O+}$, so the $\mathrm{pH}$ of the $\ce{NH3}$ solution is in fact related to the concentration of $\ce{H3O+}$ which is formed from the $\ce{NH4+}$ dissociation. Is it indeed the case? I would really appreciate any explanation.

pH means the same as always. See, pure water also has pH, hence it must have some H3O+. How so? Well, just like that.

Yes, but I try to understand where the $H_3O^+$ come from since the $NH_3$ forms $NH_4^+ + OH^-$  when reacting with water and not $H_3O^+$. I'll try to emphasize my question with another example - taking $NH_4Cl$ which dissociate to $NH_4^+ + Cl^- $ , how should I understand (or maybe view would be a better term) the $pH$ and the $pOH$ of $NH_4Cl$ solution?  @IvanNeretin

For that you need to understand where the H3O+ comes from in pure H2O.

As Ivan Neretin said in the comments, $\ce{H3O+}$ is already there, via the autodissociation of water:
$$\ce{2H2O <=> H3O+ + OH-}\tag{1}$$
Now water reacts with the ammonia that you add, according to the equilibrium reaction you provided:
$$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{2}$$
This increases the hydroxide concentration further, so the first reaction is out of equilibrium, and consumes some hydronium and hydroxide by going in the net reverse direction, increasing the pH.
Alternatively, you can combine the two reactions to show that adding ammonia will lower the hydronium concentration:
$$\ce{NH3 + H3O+ <=> NH4+ + H2O}\tag{3}$$
All three reaction and the reverse reactions will occur simultaneously.

However I don't fully understand what does pH mean, since in this dissociation there aren't any H3O+ molecules.

For basic solutions, the hydronium ion is a minor species. For acidic solutions, the hydroxide ion is a minor species. Sometimes in calculations, we get away with ignoring a minor species for a while, but they are still there.


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Does a complex that has no overall charge still have color?
The iron complex $\ce{[Fe(H2O)6]^2+}$ is colored but I'm wondering, if a pair of chloride ion is inserted on its secondary valency to form [FeCl2(H2O)6], does the complex loose its color or not?

Outer sphere ions should not have a large impact on the color.

It seems the "if a chloride molecule is inserted on its primary valency" part might need clarification from your side. I suspect what you mean is a pair of chloride anions (not a molecule) replacing two water molecules within the main coordination sphere $(\ce{[Fe(H2O)4Cl2]}).$ Or are you asking about the color of $\ce{[Fe(H2O)6]Cl2}$ complex where chlorides are counterions?

The 'colour' arises from electronic transitions between the  d to d energy levels ($t_{2g}$ and $e_g$)   in the molecule. These energy levels are present whatever the ligand are, but different ligands alter their energy and so the colour observed. Sometimes this might lead to transitions  in the uv or infra red which are outside our ability to see with our eyes, but we can still measure the spectrum using spectrophotometers.

https://en.wikipedia.org/wiki/Iron_pentacarbonyl

Tetrakis(triphenylphosphano)palladium(0) or $\ce{[Pd(PPh3)4]}$ is a common catalyst in organic chemistry that features a palladium(0) core with four triphenylphosphane ligands. It is a bright yellow, sometimes described as canary yellow. See the image below (taken from Wikipedia, where a full list of authors is available); however, I feel the picture doesn’t quite do the actual compound justice.

Colour is a feature of electronic transitions. Different orbitals have different energies. If the energy difference between an occupied and an unoccupied orbital happens to correspond to the energy of a photon ($E=h\nu=\frac{hc}{\lambda}$), the photon can be absorbed and the electron excited (subject to a couple of forbidden exchanges as usually defined in the Laporte rule and the spin rule). This absorption of photons is what we perceive as colour if $\pu{400 nm}<\lambda<\pu{700nm}$.
While energy levels of orbitals are affected by the overall charge of a complex, a correct energy difference can happen independent of partial or entire charge.


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How can I obtain the wave function of the dihydrogen system, by hand using Self Consistent Field DFT calculations?
After doing some energy calculations for the $\ce{H2}$ system in Gaussian, using ωB97X-D/STO-3G level of theory, and generating a .wfn file, I wanted to understand better how the computer does the Self Consistent Field calculations.
I understand that this file produced a molecular orbital expression based on the wave functions for each $\ce{H}$ atom.
I am new to this kind of calculations and I will appreciate the help a lot.

If you have not yet thoroughly read Chapters 2 and 3 of Szabo and Ostlund's Modern Quantum Chemistry, I highly recommend it. It provides several by-hand worked examples of HF and other standard methods. It's not DFT, but it is essential starting material to understand quantum chemistry as a whole (including DFT)

The wfn file created by Gaussian only contains the final wave function in a machine readable format. The underlying mathematics are quite complex, and the algorithms the software uses is also another matter. Szabo and Ostlund's book is a very good and detailed resource. If you would like to gain a more general overview over quantum chemistry, then there are many more books available: Resources for learning Chemistry. If you have any trouble understanding these texts, please ask a new question explaining where exactly you got stuck.

Thank you a lot for oyur time and the information! I will definitely check Szabo and Ostlund's Modern Quantum Chemistry. If I have any question I will ask.


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Acid solution — lower pH when diluted
Today I came across a "thought experiment" regarding $\mathrm{pH}$ of a diluted acid solution. This idea looks somewhat logical but obviously contradicts the reality. Can you show me where is the error?
Let's suppose we have $\pu{100 g}$ of concentrated 95% (mass) sulfuric acid. That means we have a mixture of $\pu{95 g}$ of H2SO4 and $\pu{5 g}$ of $\ce{H2O}.$ Let's calculate the amount of both compounds.
$$n = \frac{m}{M}$$
$$n(\ce{H2SO4}) = \frac{\pu{95 g}}{\pu{98.08 g mol-1}} = \pu{0.9686 mol}$$
$$n(\ce{H2O}) = \frac{\pu{5 g}}{\pu{18.02 g mol-1}} = \pu{0.2778 mol}$$
Since the amount of the solvent molecules is much much lower than the amount of the $\ce{H2SO4}$ molecules, it is not possible for the acid to be completely dissolved, which leads to the relatively small amount of $\ce{H3O+}$ cations, and $\mathrm{pH}$ of the solution is relatively high $(\mathrm{pH} = -\log [\ce{H3O+}]).$
$$\ce{H2SO4 + H2O -> HSO4^- +  H3O+}$$
If we increase the amount of the $\ce{H2O}$ molecules (dilute solution), more $\ce{H3O+}$ cations can be formed, and the $\mathrm{pH}$ will decrease.
Conclusion: dilution of acid-water solutions decreases $\mathrm{pH}.$
In reality it is completely opposite. So what is the catch?

95% sulfuric acid isn't an "aqueous" solution. // At higher concentrations, certainly anything beyond 1 molar, one needs to use activities of the ions not their concentrations.

Also note, that activity coefficients will be much higher than 1, and that sulphuric acid will dissociate just to hydrogensulphate. $$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

Related: https://chemistry.stackexchange.com/questions/108842/which-rate-the-forward-or-reverse-rate-of-acid-dissociation-is-more-strongly-a

You run into all sorts of problems when the pH is outside of the range of 1 to 13. For example, in Environ. Sci. Technol. (2000) 342, p. 254-258, they say:

[...] the former National Bureau of Standards (NBS) established a set of conventions that limits measurements to 1 < pH < 13 and to ionic strength, I < 0.1. The main limitations are the activity coefficient expression, the range of defined standard pH buffers, and interferences with the reversible response of the glass H+-sensitive membrane electrode.

They go on to use standard of concentrated sulfuric acid to define the pH, using a Pitzer model to predict activity coefficients (figure from researchgate) :

In reality it is completely opposite. So what is the catch?

If you do the math, the extreme pH of -5 shown in the figure corresponds to a hydrogen ion activity of hundred-thousand. The concentration of water for pure water is "only" ~55 mol/L, so this must reflect an activity coefficient far from one.


text:
Does 2,2-dimethylpropan-1-ol give a positive Lucas test?
I learnt that the Lucas test involves the formation of a carbocation and it gives a positive test based on the stability of the carbocation formed, and hence primary alcohols do not show turbidity (condition for positive test).
Now consider 2,2-dimethylpropan-1-ol. It is a primary alcohol and I was taught that it will not give a positive Lucas test. But my doubt is, why won't a methyl shift take place making it stable and hence give a positive result?

Does this answer your question? False positive in lucas test for primary alcohol due to formation of tertiary carbocation by rearrangement?

No. It does not.
In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions
In chemical kinetics, there are two ways to deal with multiple-step mechanisms

Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)

So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.
Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test


text:
Why do the induction effects of fluorobenzene decrease electrophilic addition reactivity, conjugation effects increase selectivity?
Fluorobenzene shows signs of both the electron withdrawing and electron donating groups on benzene. Electrophilic substitution to the ring is less reactive because the fluorine is very electronegative and withdraws electron density, but it can also form a weak conjugated system, which is especially pronounced at the ortho and para positions. Why is the modification of the sigma system (induction) related to reactivity, but modification of the pi system (conjugation) affects reactivity? What prevents it from being the other way around, where fluorobenzene is more reactive than benzene, but is not ortho-para directing? What difference in reactivity vs. selectivity makes this difference?

Flourobenzene should undergo electrophillic sunstitution but not electrophillic addition.

Electrophillic addition  reactions are usually seen in alkenes and alkynes.

If I understand your problem, you should take a look at the difference between inductive and mesomeric effects: that will help you understand why fluorobenzene cannot be more reactive than benzene in electrophilic aromatic substitution. A related question: https://chemistry.stackexchange.com/questions/43245/rate-of-eas-in-chlorobenzene-and-fluorobenzene


text:
What effect does a quaternary ammonium compound have on uric acid (urine)
What effect does an Amine based quaternary ammonium compound have on urea/uric acid?
Will an amine based quaternary ammonium product react with uric acid and get rid of the smell?

The premise of this question is a little off. Urea and uric acid are both odorless; any smell is due to other compounds.

Uric acid, according to this site, is odorless:

Uric Acid is a white tasteless odorless crystalline product of protein metabolism, found in the blood and urine, as well as trace amounts found in the various organs of the body. It can build up and form stones or crystals in various disease states.

Urea, according to this site, is also odorless:

Other nitrogen-containing compounds are found in urine. A large number of different compounds can affect the smell of urine, and it can be influenced by diet (as people who’ve experienced the effects of eating asparagus can attest). Whilst the urea in urine is odourless, it can quickly break down into other nitrogen-containing compounds, such as ammonia and trimethylamine. Ammonia has a pungent smell, but a higher odour threshold than trimethylamine, which has a fishy odour – and is in fact the compound responsible for the smell of fish. These compounds are continuously created as urea breaks down, which explains why stale urine smells much worse than fresh urine.

You asked:

Will an amine based quaternary ammonium product react with uric acid and get rid of the smell?

Quaternary ammonium salts, on the other hand, might have a smell of their own. So if I understand your question, the answer is no.


text:
The influence of diffuse functions on electron density calculations
I am not a quantum chemist, but I am (starting to be) a user of electronic structure software. In the picture (screenshot) is the electron density of the same molecule. On the left I calculated the density using B3LYP/6-311++G*, on the right I used B3LYP/6-311G*.

I have visualized the densities using Avogadro. In Avogadro, I can visualize a few of the HOMO/LUMO orbitals. The screenshot is of the first LUMO for both calculations.
In the images, red indicates electron rich areas, blue indicates electron poor areas. The molecule (caffeine) was in the presence of point charges which created an electric field, explicitly polarizing the electron density.
My Question: Why does the presence of diffuse function (the plus's in 6-31++G*), make the LUMO orbital opposite in electron density? I understand allowing diffusion would change the electron density, but it looks nearly exactly opposite now (however the total density looks pretty much the same - see last image).
Note that these are screenshots of just two orbitals, the overall density plots shown below, look very similar.

I am not a quantum chemist, so please use small words...

The colours you see in the first two pictures are not electron densities, but phases of the orbitals.
The colours of a single molecule could matter if you were calculating the interactions between different molecules (in which in-phase or out-of-phase interactions could matter) but they make no difference between the molecule(s) within the same calculation.
The electron density arises from the square power of a function which is, say, positive in the red lobe and negative in the blue one.
The square power means that there is no difference between blue and red, or + and - (See Born's law). They are just with inverse phase in respect to each other, but they have no other influence on the system otherwise.
You might find in some software packages a "switch phases" button, which inverts the colours.


text:
Investigating the effect of solvents on the alkaline hydrolysis of ethyl acetate
In the following reaction, I believe a polar aprotic solvent like propanone will be more favorable since the partial negative charge on the oxygen atom in propanone will attract the sodium ion from sodium hydroxide leaving the nucleophile, $\ce{OH-}$ to freely attack the carbonyl group:
$$\ce{CH3COOC2H5 + NaOH → CH3COONa + C2H5OH}$$
On the other hand, a polar protic solvent like ethanol will be less favorable because the partial charges on the oxygen and hydrogen atoms (created due to a difference in electronegativities) in the ethanol will surround the nucleophile, $\ce{OH-}$ and would prevent the attack on the carbonyl group slowing down the reaction.
However, while performing the experiment and obtaining conductances of the solutions at different time intervals using a conductivity probe to calculate rates and $E_\mathrm{a},$ it was observed that the rate of reaction was faster in ethanol and the activation energy in ethanol was also lower proving my argument incorrect.
Can someone help?

Consider the solublity of the reaction components in the solvent, also consider whether the choice of solvent influences alternative reaction pathways


text:
Factors that influence the electrical conductivity of metals
The electrical conductivity of metals is a physical property of these materials. Much of the theory used to explain the electrical conductivity of metals is also based on physics understanding. However, I would like to ask if it is possible to incorporate some chemistry, from concepts in atomic structure or chemical bonding of metals, into explaining the variation of electrical conductivity across periods and down groups.
Based on chemistry alone, we would obviously expect predicted trends to be inaccurate. The chemistry reasoning would need to be complemented by physical theories to give a complete and accurate prediction of such a property. Recently, I have seen an over-simplistic explanation based on the number of valence electrons the metal atoms possess, equating more valence electrons to higher electrical conductivity. This was used to explain the higher electrical conductivity of $\ce {Cu}$, compared to $\ce {Ca}$, and it was clearly wrongly used.
On this note, I would like to ask, what are the factors, in terms of electronic structure of metals or their chemical bonding behaviour, that influence the electrical conductivity of metals?
Mott (1936) informs that for most of the first-row transition metals, it is the $\ce {4s}$ electrons which are delocalised and thus, contribute to the conductivity. Of course, in metallic bonding of these metals, both $\ce {3d}$ and $\ce {4s}$ orbitals overlap significantly and they should both be considered but Mott says that the $\ce {3d}$ electrons move much more slowly, due to their higher effective mass, thus contributing less to conductivity.
Mott, N. F. The electrical conductivity of transition metals. Proc. Royal Soc. Lond. 1936, 153, 699-717.

I might suggest starting through Ashcroft & Mermin's excellent Solid State Physics book - it is a big topic. But, in general, trying to directly compare chemistry (atoms and molecules) with solid state physics (crystal properties) rapidly leads to cognitive dissonance - there is just too much changing. The simple picture of atomic orbitals smoothly merging into energy bands in solids just doesn't work out most of the time.

https://chemistry.stackexchange.com/questions/57484/why-alkali-metals-have-bcc-structure may also be of interest (full disclosure - my answer). Similarly https://chemistry.stackexchange.com/questions/58847/why-is-copper-a-better-conductor-than-iron/80673#80673.

@JonCuster Thank you for your insights. Your answer in the linked post was helpful. I may explore the references/textbooks you have suggested in future.


text:
Generating colloids / nano-particles from palladium and platinum wire
I know I can generate colloidal silver from the following link
generate colloidal silver
I now have 2 palladium and 2 platinum wires and would like to make one colloidal solution of palladium and one colloidal solution of platinum both in distilled water.  I watched a video on doing this making platinum nanoparticles link but no specifics were given on how it was made, did they use AC / DC voltage, what was the amperage used, and can it be made in distilled water and if so how?
Anyone have suggestions or steps I should follow?

...Why do you want to make colloidal platinum and palladium?

@StianYttervik I want to compare their anti-bacterial effects in a colloidal solution https://www.nature.com/articles/s41467-017-02502-3

That link with the silver nanoparticles is questionable enough, and I don't want to say aloud what I think about the guy with the platinum abusing F. Mercury. I like amateur science, but it should aim to be scientific, not "look here I have no idea what I did but I'll post a video anyway". Sorry.

Sorry @RickT but you can't say you have obtained nanoparticles just for seen a black solid appear. You need to use TEM or at least (only in some cases) UV-vis technique

A reliable method to obtain metallic nanoparticles in general is to reduce a metallic ion in presence of a capping agent. For a example, a traditional method to obtain gold nanoparticles (AuNPs) is to reduce a gold salt in presence of dodecanothiol.
For example, here is a method to obtain Pd nanoparticles by thermal decomposition, still the precursor is a Pd complex and the method requires a capping agent. (TOP stands for trioctylphosphine)

Related to the videos you post, it is not common or recommendable to obtain single metallic nanoparticles by an oxidation process (oxidizing the wires for example), it goes against what a metallic nanoparticle is. Electrolytic processes can be used to obtain metallic nanoparticles, but still using a salt as precursos as you can see here. You have to end with the reduced form of the metal if you want pure metal nanoparticles like the ones you cite in Nature's article. The videos you show do exactly the opposite and I can assure you that what is formed there aren't nanoparticles. Also, you are talking about Pt and Pd and their oxidation is not a simple task.
You can't confirm that you have nanoparticles just by checking Tyndall's effect, it doesn't work like that. You need to determine the size and the shape of the nanoparticles directly, using TEM or by indirect relation between them with another property, like light scattering or surface plasmonic resonance.
Even if you can oxidize those wires you probably won't end up with metallic nanoparticles. Because to maintain them at nanometric scale you'll need a capping agent. In the best case you'll obtain the metallic oxide or hidroxide and it will be in the bulk form.
I strongly encourage you to check articles in any of the ncbi websites https://www.ncbi.nlm.nih.gov/, https://pubchem.ncbi.nlm.nih.gov/ and others which are free and peer reviewed.

I use a very similar process as colloidal silver. 30 volts, I added sodium carbonate solution (with distilled water) a pinch of sea salt, blue berry extract.... Three droppers full in a 1000 ml beaker. Two platinum wires. I'm letting it go about 5-6 hours then adding my reducing agent. ( Manuka honey in distilled water) one dropper full. Sorry measurements aren't super precise but maybe someone else can be specific for times, amounts and measurements so you can calculate the ppms. I'm kind of winging it. Good luck! Same for palladium. Also for iridium, indium and copper. Different extracts but that's your choice. I like watermelon in copper, and turmeric extract in silver. Blueberry is good in the others... platinum will take the longest, 8-12 hours. I watch for the color change and clean wires every hour or so.


text:
Procedure to assign the irreducible representation of a derivative operator
How can we assign an irreducible representation to a derivative operator like
$\displaystyle\frac{\mathrm d}{\mathrm dx}$ or the whole gradient operator $\nabla$ in a given point group?
Since the Hamilton operator can be assigned to the totally symmetric representation, it seems like there is a way to to do so for derivatives or operators in general, but I have no clue how to do so. I am well aware how to determine the representation for normal functions, but I have never seen it done for operators.

You can get better insights by posting this question in MathOverFlow. You may find mathematical researchers there. https://mathoverflow.net/

Did you post your question on mathoverflow? Google shows this answer as one of the first results. It would be helpful to others if you could link to the question.

I have not posted it on mathoverflow.net. I don't know enough mathematics to ask properly. I only know the physical chemistry approach and i am pretty sure that in pure mathematics the technical language is different. I would prefer it to be answered here since it is a topic that every chemistry student encounters even if it isn't treated in detail or brushed over very quickly.

The transformation of a derivative operator under a linear transformation $\hat R $ is given by
$$
\frac{\partial}{\partial x_i} \rightarrow \sum_j R_{ji} \frac{\partial}{\partial x_j} = \sum R^T_{ij}\frac{\partial}{\partial  x_j}
$$
where $R_{nm}$ are the matrix elements of the matrix representation of the transformation operator in regard to cartesian coordiantes. The matrix is defined as the coordinate transformation matrix that takes the initial coordiantes to the transformed coordinates.
$$
\hat R : x \rightarrow  Rx
$$
The character of an representation with the derivative operators as basis is therefore the same as the character of the corresponding coordinate representation, since the trace of a matrix and a matrix transpose are identical. The derivative operators form thus a basis for the same irreducible representation as their coordinates and belong to the same irreducible representation.


text:
Does double ended fatty acid exist?
A fatty acid consists of a carbon chain, with a carboxylic acid on one end, and just a hydrogen on the other.
Does a substance with two carboxylic acid ends and no hydrogen end exist?

I'm not sure this is compatible with the site topic. (It's from scientific curiosity,  not for homework. And relief of scientific curiosity may not be a valid "actual problem".)

Plenty of them, e.g. https://en.wikipedia.org/wiki/Adipic_acid, https://en.wikipedia.org/wiki/Sebacic_acid

Yes and they are called bolaamphiphiles or bolaphiles


text:
DFT vs. MP2 for stacked dimer
Recently, I looked at these two papers analyzing the excited-state properties of  modified DNA bases (2-aminopurine and 8-vinyl-A) and how they are influenced by stacking with natural nucleobases:

Jean, J. M.; Hall, K. B. 2-Aminopurine fluorescence quenching and lifetimes: Role of base stacking. Proceedings of the National Academy of Sciences 2001, 98 (1), 37–41 DOI: 10.1073/pnas.98.1.37.
Kenfack, C. A.; Burger, A.; Mély, Y. Excited-State Properties and Transitions of Fluorescent 8-Vinyl Adenosine in DNA. J. Phys. Chem. B 2006, 110 (51), 26327–26336 DOI: 10.1021/jp064388h.

Briefly, the authors considered dimers comprising the respective modified base and each of the natural bases. The geometries of the individual monomers were optimized using MP2/6-31G(d,p) while maintaining their relative position within in the dimer. Afterwards, the authors examined the excited state properties with TDDFT (B3LYP/6-311+G(d)) to identify dark states that could indicate fluorescence quenching of the modified bases.
I'm not an expert in theoretical chemistry and so I was wondering how the choice of methods here was motivated. In particular, why was MP2 used for geometry optimization and not DFT? Does it have to do with a better description of dispersion interactions by MP2?
For their TDDFT calculations the authors used the 6-311+G(d) basis set. I want to try out some similar calculations using ORCA. According to the manual and online resources from the developers, Ahlrich's basis sets seem to be preferred over Pople style basis sets. Could I expect similar (or better/worse) accuracy with def2-TZVP? It's also triple-zeta and contains polarization functions on heavy atoms, but lacks the diffusion functions that were present in the studies mentioned above. Would these be important anyway in such a case were only valence transitions are investigated?

Interesting question, but the goal of this site collect general problems/answers that are useful for many people/ It would improve your question if you would briefly summarize the methods and models the two paper use. Most people will not download the paper and read just to understand what you are asking and "Some modified DNA bases" is not enough to understand which interactions might be important from the authros point of view. Also, it would be better if you were focusing on a single question (eg DFT vs MP2) instead of many smaller ones.

Thank you, Greg. I've added a short paragraph describing the methods used in the papers.

Both papers you refer to used methods which were basically state of the art at their time (Gaussian 98/03 compared to Gaussian 16 today). But that is almost twenty (or fifteen) years ago. Nowadays we thankfully have more developed methods available, and one should check whether the results found back then are still consistent. An (admittedly also a bit outdated) overview can be found here: DFT Functional Selection Criteria.
Long story short: You cannot reliably perform calculations on complexes where non-covalent interactions are dominant. You need to to use a method that treats dispersion at least at an empirically derived level. A popular example is Grimmes DFT-D3 program, which is also available in many ESS codes. (Starting point is The Journal of Chemical Physics 2010, 132 (15), 154104.) In the semi-empirical method xtb a newer version is already implemented.
For that reason I guess (since I only read the methods sections) they used MP2 for molecular structure optimisations, as it recovers dispersion interactions to some degree.
Whether or not DFT is suitable for your investigations, I cannot say. A thorough research of the recent literature will certainly offer far better insight.
The Pople basis sets, while an excellent example to understand the structure of them, are generally a terrible choice for almost everything practical. There are multiple basis sets available, which produce more consistent results, and are better optimised to perform in modern ESS codes. If implementations are missing from you ESS program of choice, you have a high chance to find it in the Basis Set Exchange Library.
For ORCA there are most (if not all) Ahlrichs' def2-variations are available, see the ORCA input library.
I would expect augmented basis sets (ma-def2-XVP or def2-XVPD) are necessary to investigate such complexes.

Thank your for the detailed answer, Martin!
When you say the results should be checked for consistency, do you mean that the geometries should be reoptimized at different levels of theory (e.g. SCS-MP2 or a double hybrid like B2PLYP with D3(BJ) correction, etc.) before evaluating the excited state behavior with different methods?
As far as I understood the methods description in the papers, the author's first optimized the monomers individually and then placed them in the dimers in a fixed orientation. Could an improvement of the geometry really be expected in this case?

@PracticalChemist In those cases it is doubtful to expect a 'better' description of the molecular structure. I don't really understand, why they optimise the monomer structures individually just to place them into a fixed position (from x-ray iirc); I'd assume similar results could have been achieved with single point on x-ray. However, since there are methods available that cover non-covalent interactions, I'd assume the overall description, also in terms of TDDFT or excited states would be better. However, I'm not really an expert on this; more recent literature is more helpful, I guess.

I came accross this reference that is complementary to @Martin-マーチン response and you may find helpful:  [Lars Goerigk and Nisha Mehta, A Trip to the Density Functional Theory Zoo: Warnings and Recommendations for the User, Australian Journal of Chemistry, 2019.] (https://doi.org/10.1071/CH19023