text:
Reaction between Ammonium chloride, Concentrated sulphuric acid and copper turnings
During a salt test, dry test for acid radical , mixed ammonium chloride with concentrated sulphuric acid and heated the mixture and added copper turnings to the mixture, the mixture on heating turned yellow.
Sulphur Dioxide was also evolved as detected by its odour.
Bubbling across the copper turnings were observed.
Wanted to know the chemistry behind the reaction, composition of the yellow compound.

Be aware that laconic, not elaborated questions without hints of solving attempt are usually frown upon and they get often closed.  OTOH, well elaborated questions have higher probability to  receive acceptable answers.  How do I ask a good question?.

Mixing ammonia, metals, and acids has been known to produce explosive things like nitrides, fulminates, and ammonium nitrate.  Great care should be taken. Potential side reactions should be thoroughly researched.

A possible process expressed in ionic and summary formula forms:
\begin{align}
\ce{4 Cl- + 2 H+ + H2SO4 + Cu &-> CuCl4^2- + SO2 + 2 H2O}\\
\ce{4 NH4Cl + 3 H2SO4 + Cu &-> (NH4)2[CuCl4] + SO2 + 2 NH4HSO4 + 2 H2O }
\end{align}
As oxidative behavior of hot concentrated $\ce{H2SO4}$ is well known:
$$\ce{Cu + 5 H2SO4(l) -> Cu^2+(solv) + SO2(g) + 2 H3O+(solv) + 4 HSO4-(solv)}$$

Check out oxyhemocyanin nitrogen-copper-oxygen chemistry!

Poutnik's explanation is consistent with experimental observations and yellow color of chloro copper complex. This is a known complex of copper. The alternative proposed hypothesis in another answer suggesting the formation of $\ce{NOCl}$ or $\ce{NO}$ is not correct because it contradicts a well known analytical reaction employed in Kjeldahl method. Here nitrogen containing compounds (say meat, food) are digested with concentrated sulfuric acid at very high temperature, near the boiling point, and the end product is ammonium sulfate! So it is known that sulfuric acid will not oxidize ammonium ion to nitrogen oxide or other complexes even under extreme conditions.

Kjeldahl method "yields lower protein results when CuSO4 used as a catalyst".  Why?

An interesting reference here.  It even has pictures.

I do not trust Wikipedia for scientific purposes and their claim is also incorrect. Wikipedia is a excellent starting place to search not the end. The second reference has nothing to do with sulfuric acid. A senior good old scientist told me to read at least 12 papers before forming an opinion. One can start from a very reliable reference, Determination of Kjeldahl Nitrogen in Fertilizers by AOAC Official Method: Effect of Copper Sulfate as a Catalyst. They say there is no difference!

I ran Kjeldahls years ago.  I appreciate your point of view.  I will continue to research this.  Good old evolution disposes of ammonia by forming urea.

Some bacteria can oxidize ammonia with a "red copper protein" enzyme. (page 2).

It might be nitrosyl chloride.  At elevated temperatures ammonia can be oxidized to  NO, which can oxidize to NO2, and further to nitric acid.
In acidic solution, chlorine can combine with NO to form NOCl, a yellow gas.
Since essentially a mixture of aqua Regia results, ammonia and copper may both be subsequently oxidized.
It is certainly hoped this was run in a well ventilated hood.

There is also another possible reaction as mentioned in Poutniks answer between copper and H2SO4 --> Copper sulphate and SO2 gas.

Ammonia cannot be oxidized to NO in solution even by sulfuric. This speculation is incorrect.  The elevated temperature is around 1000 degrees.

@Achem I remind you that equilibration chemistry only requires that product can be produced and more will be if it is removed as NOCl.  Note that Poutnik explanation is not favored either, but can occur in sufficient quantities to be significant. (Unless you have a better explanation).  Further more formation of NO can be catalyzed at lower temperatures.  NOCl is known to occur in aqua regia.

Further related reading


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Propagating production of chromic acid consuming only Cr$_2$O$_3$ & H$_2$O$_2$
I only have access to a small amount of pure chromic acid & I need to produce more. On the other hand, what I have is ample $\ce{Cr2O3}$.
Would adding $\ce{Cr2O3}$ powder to chromic acid & $\ce{H2O2}$ & heating it up convert all the $\ce{Cr2O3}$ to chromic acid?
1.) $\ce{Cr2O3 + 9H2O + 6H+ -> 2[Cr(H2O)6]^{3+}}$
2.) $\ce{[Cr(H2O)6]^{3+} + 3H2O2 -> CrO3 + 9H2O}$
3.) $\ce{CrO3 + H2O <=> H2CrO4}$
Then the resulting $\ce{H2CrO4}$ will propagate reaction back to (1), consuming only $\ce{Cr2O3}$ & $\ce{H2O2}$.
Looks to like it could work. Do you guys think this needs special apparatus for applying high pressure or high heat?

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Does hand dishwashing liquid degrease as well as add "grease"?
When preparing a copper etching plate for etching, it must be degreased completely before putting on the etching ground. Otherwise the acid may "bite" through the ground ("foul biting") in spots not desired. Whiting has been used for degreasing and alcohol has also been used successfully. I was wondering whether using hand dishwashing liquid as a preliminary measure (before the alcohol or whiting) to remove oil-based ink and other substances both degreases -- as I think it may do when washing dishes -- but also adds "grease."

What do you mean by “grease” being added?

What is "whiting"  ?

@Buttonwood Hmm, that is understandable, but the used context is confusing Whiting has been used for degreasing, like pointing to something else.

@Poutnik "Whiting is used by the craftsman as a cleaning agent to help remove the residues from the work after the soldering operation. Whiting is one of the ingredients in polishing compounds, putties, paints, etc. This soft non-abrasive powder is usually applied with a stiff scrub brush. Its use is recommended particularly on copper foil shades to remove the flux and gum, and to give a high shine to the glass and metal." by https://www.whittemoredurgin.com/catalog.aspx?chemicals-consumables-notions-stained-glass/whiting. CaCO3 (chalk?) by https://www.merriam-webster.com/dictionary/whiting

There is a large set of chemicals which either can act as a moisturizer and/or emollient by chance, or which are intentionally added to the detergent for this function.  You find them in cosmetic shampoos, as well as dishwashing liquids advertised as "especially friendly to the skin of your hands".  It can be difficult to get rid of them, one reason why labs active in analytical chemistry (e.g., characterization of surfaces)* are careful when using them / their dishwashers finally rinse the glass ware with deionized water.
To extend Poutnik's answer, the typical suspects are (for example) palmitates, stearyl alcohol and stearyl ethers, squalene, decyl esters, polyethylene glycols (PEG).  There equally are some telling generic names for them e.g., lamesoft (BASF), or tegosoft (evonik).  At small scale, you can remove them with clean distilled water, followed by a rinse with e.g., isopropanol/rubber alcohol.

* Surface tension, wettability of surfaces, shape of drops sitting on surfaces relate to surface energies, a topic in physical chemistry.  One approach to characterize them is literally to place a drop on a surface and to check its shape:

(edit of an image by Chiu et al.)
Especially with the advent of cell phone cameras, this is an affordable experiment for the undergrad labs.  As reported by Chiu et al., students take a couple of photos, fed them into the freely available ImageJ program supplemented by the plugin about contact angles (a link to youtube for a brief tutorial), and determine the contact angle (enclosed by the blue straight lines added to the illustration above).  Recording multiple drops per sample improves your evidence by statistics, especially if comparing different samples/PCBs.
For process/quality control, one can hence imagine to record this characteristic angle between a surface of a reference sample deemed clean enough, and distilled water.  On any subsequent PCB, one would drop again a drop of distilled water to record anew a couple of photos (and by this, contact angles) about different spots as a check before the board enters the next stage of soldering, coating, etc.
Chiu, Y.-C.; Jenks, M. A.; Richards-Babb, M.; Ratcliff, B. B.; Juvik, J. A.;  Ku, K.-M. Demonstrating the Effect of Surfactant on Water Retention of Waxy Leaf Surfaces. J. Chem. Educ. 2017, 94, 230–234; doi 10.1021/acs.jchemed.6b00546.

Total "degreasing" of skin is undesired and would be harmful in long term.
For that reason, skin washing formulas contain auxiliary components, partly lipophilic, intended to adhere on skin and regulate the level of skin "degreasing". And they partly "distract" the surfactant not to do skin fat emulgation so thoroughly.
The overall effect interferes with the goal to have clean surface.
It is better to use formulas advised to use with gloves (like liquid formulas for manual dishwashing).
Or even better - using pure surfactants intended for labs (like sodium laurylsulfate, or some biodegradable alternatives).
Or, organic solvents as mentioned, but then there can be material-solvent compatibility issue.

I believe whiting is calcium carbonate. Thanks all for the thorough answers and comments.

@Sketcher Thanks. Interesting. I have originally thought it was some procedure.  :-)

Isn't the subject the copper etching plate, not (human) hands? 'hand dishwashing liquid' as a compound noun? Or just dishwashing liquid - "...is primarily used for hand washing of glasses, plates, cutlery, and cooking utensils in a sink or bowl.".

@PeterMortensen You are right :-) I must have read it wrong. But it would not change the answer much, just putting less stress on the additives.

I initially thought of "grease" in layman's tactile terms as a wet, somewhat viscous and slippery, possibly oily liquid. Yes, the copper plate is the subject, although human hands do manipulate it, sometimes with plastic gloves. In any case, I believe now that a household soap product will not effectively degrease the copper. Thanks again for the comments, which helped to clarify, particularly in light of the skin-related additives in the soap.

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Is there any literature regarding the neutral hydrolysis of nylon 66?
I am working on a project for my thesis regarding the neutral hydrolysis of nylon 66 at high temperatures and pressures (265C,40 bar) however the only literature I could find relates to alkaline and acid hydrolysis. Is there any data that links to my project?

This looks relevent  https://hal.science/hal-01082759/document

Recently asked: https://chemistry.stackexchange.com/questions/171318/how-to-separate-nylon-66-salt-hexamethylenediammonium-adipate-into-adipic-acid

also check: https://projekter.aau.dk/projekter/files/414254361/TEPE4_1003_Thesis.pdf


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Aromaticity of [14]annulene
Our teacher had taught us that cis-trans-[10]annulene is non-aromatic. The reason he gave is due to repulsions between hydrogens, the structure is no longer planar

But later he told us that [14]annulene is aromatic. But isn't there a repulsion between hydrogens in [14]annulene too.

In fact, there are 4 hydrogens repelling each other, so why is it still planar and aromatic?

It is not quite planar and not quite aromatic.

Of possible interest regarding [10]-annulene. Go here.

If you have a molecular model kit, use it to see the difference.
In [10]annulene, the inward-directed hydrogen atoms are separated by six carbon atoms along the ring and thus, in your conformation, are forced onto the same side of the hexagon where steric hindrance becomes severe. Other confirmations prevent that, but overstress the sigma bonds within the ring.
In [14]annulene, the inward-directed hydrogens are separated by only four or five intervening carbons, so they are pushed towards each other along diagonals rather than sides of the hexagons. The difference in length of the diagonals versus the sides makes the steric hindrance less severe.
Even so, [14]annulene has only limited aromatic character. [18]annulene, with a still more sterically favorable arrangement of inward-directed hydrogens, is more strongly aromatic.


text:
Are there natural enzymes that catalyze different reactions under different conditions?
Are there natural enzymes that catalyze different reactions under different conditions? For example an enzyme undergoes a conformational change under certain pH and starts catalysing a different reaction.

see "moonlighting proteins", for example: https://royalsocietypublishing.org/doi/10.1098/rstb.2016.0523


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Are there official-like guidelines regarding chemical discoveries (e.g., drugs) by rational thought, rather than by accident?
Some even very significant discoveries have been accidental, but if science is supposed to be rational, then accidents should be oddities for this if they cannot be quantified.
Are there generally accepted (or official-like) guidelines about what the role of "accidents" are or can be in professional chemistry?
What (for example, tests, procedures, standards, ...) must one fulfill additionally in order to have "accidents" progress into use?

Drug design always looks more rational in restrospect

Be careful about format of your questions. This is Q/A site, focused on particular answers of reasonable size and scope. Too general or opinion based questions, leading to chat are not suited for this site.

Take drugs. Many are discovered accidentally. But how they are discovered is irrelevant to their being used. To get used they have to pass many rigorous tests of efficacy and safety. How they were discovered is irrelevant to this process.

I think you use the word "science" for two distinct concepts: the scientific method, on the one hand, which includes Descartes' rational doubt; and knowledge and technical advancements, on the other hand. This double meaning of "science" confuses your argument that "science is supposed to be rational". The method is rational, the doubts are rational; but it doesn't make sense to say that knowledge is rational. Knowledge of a drug is something we either have or don't have, and whether we acquired this knowledge following an accident or following a hunch is a bit irrelevant to rationality.

Sticking with the specific example of drug discovery, in the United States, the FDA has two general requirements for approval of a new drug: 1) it must be relatively safe for the consumer and 2) it must be effective at treating something, ie outcomes are more favorable than for patients given a placebo. There is no evaluation of how the drug was identified as potentially useful as a treatment, nor is it even necessary to know how it works. (A well known example of this is lithium treatment for bipolar disorder. It's been used for decades, and there's still a lot of debate about how it works.)
For example, if a company is testing a drug as a potential treatment for high blood pressure, and they notice that all recipients of the drug experienced extra hair growth, they might change course and begin evaluating it as a cure for baldness. If another company developed a competing drug based on decades of study of hair follicle physiology, the two drugs would still face the same evaluation process.
So for the case of drug discovery, there is no additional requirement for the acceptance of "accidental" discoveries.

Please use 1. and 2. to indent your points. Otherwise, I don't know where the 2nd one ends

Hmm, is not the Q rather about discoveries in chemistry in general, using drugs just as an example? The answer is rather about pre-clinical and clinical proper and controlled practice in medicine, leading to the eventual approval. A drug approval and drug discovery are  very different processes. Sure,  clinical testing can discover yet another potential usage of the drug (like Viagra, originally tested as a cardiology drug), but that is not a typical discovery. Another thing is a compound being discovered as a potential drug to be tested/approved is rather discovery of medicine than chemistry.

Nature of matter is too complex to be fully predictable, so what is not predictable at current chemistry level is  discovered accidentally, not being foreseen until encountered.
The first paragraph is incorrect, being rational and making accidental discoveries does not exclude each other. Ability to predict unknown phenomena is rather rare (like prediction of antimatter).
Ability to predict behavior of some molecule within the known is easier.
But all is very case dependent, there is hardly any general approach.

In several last decades, applying advances of quantum chemistry, there have been developed a screening method of evaluation of various compounds before them being synthetized yet, eliminating the ones not being promising. One of the leading scientists on this field was the late prof. Rudolf Zahradník Researchgate.net - scientific-contributions 1959-2013, a former head of the Czech Academy of Science.
In recent years, there are being developed AI powered screening computational methods, based on learn structural patterns of biological objects, organic molecules and their relations and interactions.

Yes so maybe there's acceptance of uncertainty in discovery, but there's still much other things to do before it can join the body of science as a valid part of it? So what would these be, given an accidental discovery?

It looks your topic targets rather the whole science then chemistry itself.
You should explicitly mention in the question the summary of results of your prior research about scientific approach, before asking. As the site is not substitution for own searching for the answer.

Well, or the criteria for scientific method in chemistry. And this should be much easier answered by a practiced professional than a literature research by amateurs.

The criteria for scientific method are the same across the science. Only objects and methods differ. // The amateur research provides results from many professionals. The point is, the site rather hesitates to provide answers that can be easily found elsewhere, not to make redundancy on internet, sometimes on the site itself. That is the reason why the OP is supposed to do own offline/online search for the answer before asking.

But they are not necessarily followed in practice, for example: https://en.wikipedia.org/wiki/Replication_crisis

Sure, but it is not specific for chemistry. It is issue of economic funding of science. Nobody gets grants for replication of results of others.

Let us continue this discussion in chat.


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Why is the theoretical value of the electrode potential different to the experimental value for Cu and Al for a galvanic cell by using Nernst equation
I've construct a galvanic cell with the Al strip putting in the Aluminum nitrate solution and Cu strip putting in the CuSO4 solution. KNO3 is used as the salt bridge in this experiment. The question is about how does changing the concentration of the copper sulfate solution affect the electrode potential in this galvanic cell. I use 5 different variables of the concentration of the copper sulfate, which are 1.5M, 1M, 0.5M 0.25M, and 0M which is distilled water.
I choose using 1M of concentration of CuSO4 as the example to calculate the theoretical value of the electrode potential by using Nernst equation. The following picture is my working process and my data table. I'm just wondering about do I do something wrong in my work. If I didn't make mistakes in my working process, why is the experimental value so differ from the theoretical value? When the copper sulfate is being replaced by water, why does it still have a voltage? When I replaced both copper sulfate and aluminium nitrate to water, why does it have a voltage?

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 at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

For some reasons, many amateur chemists are not aware that aluminium is one of the worst choices to experimentally verify theoretical electrode potential values.

Don't use an aluminium electrode. In the presence of air,  aluminium is always covered by a thin, continuous, colorless, insulating and waterproof layer of alumina (Aluminium oxide $\ce{Al2O3}$). If some accident happens that rubs and removes part of this this layer, the metallic surface will be immediately reoxidized by air. So the metallic electrode is never in contact with the aqueous solution.
If the aqueous solution is acidic, this alumina layer is partially dissolved, and aluminium metal is in contact with water. It will immediately react with water according to  :$$\ce{2 Al + 3 H2O -> 2 Al(OH)3 + 3 H2}$$ so that the potential taken by this electrode follows Nernst's law for the $\ce{H2/H^+}$ redox half-cell. The two electrodes are then : $$\ce{H2 -> 2 H+ + 2 e^-}$$ $$\ce{Cu^{2+} + 2 e- -> Cu}$$ And Aluminium does not play a role in this cell. The potential of the Hydrogen electrode depends only on the pH of the solution in contact with the aluminum plate.
If the pH of the solution is $5$, the $\ce{H2/H^+}$ electrode has a potential $$\pu{E} = - (0.059/2)\times5 = - 0.15 \mathrm{V}$$Coupled with the $1$ molar $\ce{CuSO4}$ electrode, it produces cell voltage of $0.15 + 0.34 = 0.49 $V. This is nearly exactly what you found in your measurements with $1$ M $\ce{CuSO4}$. Your pH was not exactly $5$.

Another variable is the partial pressure of hydrogen, affecting H2/H+ potential, as hydrogen could be mixed with other gases, like nitrogen or oxygen, assuming the water is at least partially saturated by air.

Of course the partial pressure of hydrogen may affect the hydrogen potential. But it is hard to take this effect into account in the proposed set-up.

Surely it is. I have mentioned it as just another reason why not to wonder why the potential differs from E(Al/Al^3+).


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Any way to predict what quantities of heptane and chloroform will match the density of water?
I am trying to make a solution of heptane and chloroform that matches the density of water.
Is this something which must be done experimentally as it is generally unpredictable? Does a software exist to computationally simulate this?
My end goal is to have a vertical interface rather than a horizontal one and then to make a pickering emulsion.
Thanks

As a "first shot" prediction, assume density as linear interpolation of solvent densities, according to their mass fraction. If it does not fit, determine the real density (e.g. weighting of measured volume) and use it as the 3rd point for quadratic interpolation to get the water density.

https://materials.springer.com/thermophysical/docs/ve1_c47c91 has data

Thanks to Jon Custer, now you have a full set of densities against mole fractions of chloroform and n-heptane mixtures in this site. Just for my curiosity, I have plot the relevant graph:

The data is given for temperature at $\pu{25 ^\circ C}$ and the set is in good agreement $(R^2 = 0.9994)$. Since $x$ in the equation is the density, and assuming density of water at $\pu{25 ^\circ C}$ is $\pu{1.0 g mL-1}$, you may get that density by mixing $(-1.1887 + 3.727 - 1.9736) = 0.5647$ mole fraction.
I also found a reliable full data set of densities against mole fractions of chloroform and n-hexane mixtures as well (Ref.1). Just in case if you want to use hexane instead of heptane, I included that as well:

Reference:

Mehdi Hasan, Ujjan B. Kadam, Apoorva P. Hiray, and Arun B. Sawant, "Densities, Viscosities, and Ultrasonic Velocity Studies of Binary Mixtures of Chloroform with Pentan-1-ol, Hexan-1-ol, and Heptan-1-ol at (303.15 and 313.15) K," J. Chem. Eng. Data 2006, 51(2), 671–675 (DOI: https://doi.org/10.1021/je0504459).


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If Ethyne has the lowest C-H bond length (highest C-H bond energy) compared to Ethane and Ethene, why is it the most acidic?
I am not an expert in chemistry, but these are the few facts I know from Standard XII chemistry books (these facts could be wrong or incomplete or modified by now).
Facts I know -

Ethane has 'sp$^3$' hybridisation, Ethene has 'sp$^2$', and Ethyne 'sp'.

The % of 's' orbital present follows in each of the hybridisation
follows the order sp > sp$^2$ > sp$^3$.

Since 's' orbital has greater penetration (to the nucleus) depth, the H atom in Ethyne comes the closest to the Carbon atom nucleus than in Ethene and Ethane, hence C-H bond length of Ethyne is the lowest, followed by Ethene and then Ethane.

On the other hand, since 's' orbital has greater penetration depth, 'sp' hybridised orbitals has the most Electronegativity, followed by 'sp$^2$' and then 'sp$^3$'. Because of this, Ethyne is most acidic, since the carbanion (HC $\equiv$ C $^-$)  is more stable than (H$_2$C = CH$^-$) and (H$_3$C $-$ CH$_2$$^-$). Ethyne can release $H^+$ more easily than Ethene and Ethane, hence Ethyne is most acidic, followed by Ethene and then Ethane.

Following a similar reasoning of the nature of 's' orbital, one finds that the C-H bond energy of Eythne is the highest, followed by Ethene and then Ethane.

Question -
How can Ethyne be the most acidic and yet have the shortest C-H bond length. Isn't there a general rule that "the shorter the bond length, the stronger the bond strength, and so the weaker the acid"? Shouldn't shorter X-H bond lengths (X is some atom) mean that X and H has greater bond strength and so it does not dissociate much into $X^-$ and $H^+$ ions, thereby not producing much $H^+$ ions?
Is there something wrong with the Facts that I stated or is the statement of my Question not true in general?

Note that names of elements or compounds are not generally written with a capital letter.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

The difference in bond lengths is not dramatic: ethane: 109.1 pm, ethene: 108.6 pm, ethyne: 106.3 pm. https://cccbdb.nist.gov/exp2x.asp?casno=74840&charge=0, https://cccbdb.nist.gov/exp2x.asp?casno=74840&charge=0, https://cccbdb.nist.gov/exp2x.asp

You can find a good and simple explanation here.


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Number of neutrons in a chemical question?
Consider the equation $\ce{H2 + H2+ -> H + H3+}$
What is the atomic structure of the $H_3^+$ ion? I understand that the H$_3^+$ ion contains 3 protons and 2 electrons. However, how many neutrons does it contain? Couldn't $H$ be any isotope of Hydrogen, so we don't necessarily know how many neutrons?
In general, will we never know the number of neutrons within a chemical equation?

Unless otherwise stated, assume the most common isotope of hydrogen, thus no neutrons, to be predominant. Of course, there could be an occasional deuterium atom mixed in, but the no-neutron, all-hydrogen-1 structure dominates.
Note that the neutron count has an impact on mass and thus on vibrational frequencies, which in turn may impact the ability of this species to dissipate excess heat and enable the formation of stars. Our existence might well depend on the spectrum being what it is specifically with (the predominant form of) three hydrogen-1 (protium) atoms.

I object. Unless otherwise stated, I would assume H (as well as any other atomic symbol) to mean the mixture of isotopes with natural abundances.

@IvanNeretin I try different wording.

Now I agree.$;$

The #  of neutrons isnt very relevant to chemical properties because the pulling and pushing is done between protons and electrons(EM force).
Since this question is about stoichiometry most likely you need to assume H to be the mixture of stable isotopes with natural abundances.

With hydrogen there are important effects of doubling the mass on the nucleus from protium to deuterium on molecular vibrations. As noted in my answer it may have impacted the heat dissipation required to form stars.


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Why ionic radius increases with negative charge?
Here's my understanding:

An ion with a negative charge has gained electrons. Hence the negative charge.
With a greater negative charge, there should be more attraction towards the positive charge.
As a result the electrons should be pulled towards the positively charged electrons.
Ionic radius should decrease.
Further an ion gains an electron to achieve a full shell. So the number of shells remains constant.

I understand that this is wrong. However, what is wrong with this intuition? Why does the ionic radius increase when the charge becomes more negative and decrease when the charge becomes more positive?

You are ignoring the electron repulsion, that's what is wrong.

The addition of electrons does not mean attraction increases on each electron..
if you know about slater's rule, you will have a better understanding. When you add electrons an atom forms ions, which in fact become greater in size than an elemental atom because of electronic repulsion between the electrons. Hence, Ionic Radius Increase with the increase in negative charge

This would be correct if you could increase the charge of an electron from -1 to -2. That is not how ions acquire a negative charge, though. (In contrast, nuclei come with different charges, which is why the inner electrons in sodium are closer to the nucleus than those in lithium, for example.)

With a greater negative charge, there should be more attraction towards the positive charge

The pulling and pushing doesn't happen collectively to all electrons at the same time, it happens to each electron individually. So adding an electron to an atom doesn't increase the pushing and pulling to other electrons. On the other hand the nucleus is attracted more to the electrons, but the symmetries of the orbitals of an atom prevent the nucleus from moving right-left, up-down or inside-outside.
And of course the electron-electron repulsion pushes electrons away from each other so the overall radius is increased.

The number of protons and neutrons remain the same when you add an electron to an ion. So the Effective nuclear charge doesn't change, only the number of electrons changes. So, the effective nuclear charge which can hold 'n' number of electrons isn't sufficient hold 'n+1' number of  electrons. So the nucleus does not attract the electrons. But due to the inter-electronic repulsions, the electrons get away from each other, so they get farther from each other.


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Order of reaction with and without the presence of catalyst
When ethene reacts with hydrogen to give ethane in the presence of Ni, the order of the reaction is 2, if I am not wrong, please correct me if I am. But if we remove the catalyst, and simply allow ethene with hydrogen to react (assuming the reaction does indeed take place without the catalyst, albeit slowly probably), will the order of the reaction change? Because it might undergo via a different mechanism. If it does change, what is the order of this reaction without the catalyst?

" if I am not wrong" You are and it's almost a given anytime someone says it here ;-> There's no order of reaction for heterogeneously catalysed reactions.

And if you hear someone say some order was found, it's just an approximation, at best.

The problem with reactions orders is that while people should use activities it is common to use concentrations, as a result things can become a bit strange.

The reaction of hydrogen with ethylene (ethene) will be very very slow. The HOMO of hydrogen and the LUMO of ethylene are such that there will be no net orbitial overlap. As a result the gas phase reaction of H2 with ethylene will be impossible.

There might be a possibility that we could react hydrogen and ethylene in a free radical reaction but I will ignore that possibility.
The order of a reaction normally relates to a homogenous reaction, it normally relates to the rate limiting step (rate determinign step, RDS). The order of the reaction normally is determined by the number of molecules of a given species which must react in this step. It is worth noting that equilbria might be hidden from the casual viewer, cosndier the order of the reaction of NO (nitric oxide) and oxygen in the gas phase.
The problem of reactions on metal surfaces is that the reaction is not of free ethylene and hydrogen, but instead it is of metal bound molecules. The most simple model for dealing with the adsorption of gases onto surfaces is the Langmuir isotherm model.
I have dug up some real life data for alkene hydrogenation and the order of the reaction is not simple (Journal of Molecular Catalysis A: Chemical
Volume 320, Issues 1–2, 1 April 2010, Pages 34-39). On sulfided NiMo the order of reaction for the alkene (olefin) is about 0.25 while the order for hydrogen is about 1.2.
If you read Kinetics of ethylene hydrogenation over a supported nickel catalyst, Ho-Peng Koh and R. Hughes, Journal of Catalysis, Volume 33, Issue 1, April 1974, Pages 7-16 you will see that the reaction is first order with respect to hydrogen but about 0.25 order with respect to ethylene. The order with respect to ethylene changes as a function of temperture.
I think that you are thinking about a very difficult problem, I think it is one which a lot of PhD students would struggle with. So if you are having trouble then please do not panic.
If you want to understand the problem then there are several ways to deal with it. You need to consider the adsorption of two substances onto the metal surface. What has to happen is that the hydrogen molecules must first adsorb as H2 molecules, they then dissociate into hydrogen atoms which are on the nickel surface.
Next the ethylene which is adsorbed onto the nickel has to react with a hydrogen atom to form a ethyl group. The ethyl group then has to react with another hydrogen atom to form ethane. The ethane will be far less able to bind to the surface so it will fly away.
You have have to consider what is the slow step on the nickel surface, also the binding of one of the substrates can reduce the number of active sites which are able to bind to the other one. The langmuir model is pretty good and it has three core rules.

A site is either empty or occupied by a thing
All sites are identical
What happens at one site has no effect on another site.

If you want to get an insight into alkene hydrogenation, one thing you could consider is thinking about the chemistry of Wilkinson's cataylst, this is likely to be one of the best small molecule models for the hydrogenation of alkenes on a metal surface.
The cycle for Wilkinson's catalyst is this four step cycle.

Martin, if you read the literature references on heterogenious hydrogenation cats, you will find that people have measured the hydrogenation rate as a function of the partial pressure of hydrogen and the concentration of the alkene. The values for the two cats I ahve looked up are not round numbers. In the 1973 paper rate = k ([H2]^q)([Alkene]^r). If you look at the rates of reactions in homogenious cataylsis (
INORGANIC & NUCLEAR CHEMISTRY LETTERS, 1972, vol 8, page 1) you will see that the kinetics of wilkinsons's cat are complex. A bit like enzyme kinetics in my view

The mechanism I posted for RhCl(PPh3)n is the normally accepted cycle for the cataylst, I think that it is unreasonable to insist on including the transition states. It is a given that all reactions have a transition state. Should we include a step where diffusion occurs to bring the molecules together, if you consider things like polymerisation of styrene to high conversion then it is clear that the rate of diffusion of things is important.

I am sure that the activation of H2 by the Rh in the cycle is a single step reaction which will occur via a transition state. If you consider the kinetics of homogenious cats they do vary a lot. Compare and contrast the Monstanto (Rh) and Cateva (Ir) processes for methyl carbonylation (acetic acid production). For the Rh chemistry the RDS will be oxidative addition of the MeI while for Ir it is an insertion reaction if my memory holds true.

Thanks a lot!
It has been quite useful


text:
Why do non-metals not have delocalised electrons, whilst metals do have delocalised electrons?
Why do non-metals not have delocalised electrons, whilst metals do have delocalised electrons? I understand that delocalised electrons is defined as “electrons that are not bound in place to a single atom or a single bond between two atoms”, and I think that delocalised electrons are necessary for the bonding of two elements (please correct me if I am wrong). If I am correct about my latter statement, shouldn’t both metals and non-metals have delocalised electrons (because they form ionic bonds)?
Clarification about my understanding of delocalised electrons would be much appreciated.
Edit: I would appreciate it if someone could comment on how I could improve my post please since it is being downvoted. That would be grateful since I am new to Chemistry Stackexhange, thank you.

See that "or a single bond" clause? This is about electrons delocalised between two atoms, and that's what typically happens in non-metals.

Are you aware that you are making an incorrect generalization? Or perhaps you should be clear whether by "metals" you are referring to elements or materials with particular properties?

Well, instead of simply asking what delocalisation means, you asked... the thing you did and apparently wasted 50 rep.

@BuckThorn No, I am not aware I am making an incorrect generalisation. That is why I was hoping someone could post an answer to correct my incorrect statements and help me clear my understanding.

The Wikipedia pages on delocalization and metal bonding are a start, but maybe somebody here will contribute an answer.

You should probably also be aware that the explanation of particular observables (such as conductivity/delocalization) are not always unique. The mechanisms that explain electron delocalization in metals, aromatic compounds and conducting organic polymers may differ.

@IvanNeretin - the band structure of, say Al2O3, is well described using standard solid state physics Bloch formalism. The electron states are non-localized. There just aren’t very many electrons at all in the conduction band.

Delocalization of electrons is used as an explanation for different properties. The OP asks why the electrons in non-metals are less delocalized than in metals, although some delocalization is required to bond the atoms of metals and non-metals alike. To answer the question, we need to measure a property that is tightly related to delocalization of electrons.
Thermal conductivity measures how well heat (thermal energy in motion) can pass through a material under a temperature differential. In metals, both electric current (flow of charge) and heat transfer (flow of thermal energy) are primarily carried by electrons. (Ref 1) We could expect, then, that materials with extensive delocalization of electrons would have high values of thermal and electric conductivity, and conversely, that materials with small values of thermal and electrical conductivity would have minimal delocalization of electrons, and these materials would be categorized as non-metals.
The data have been categorized in different ways for the whole periodic table (Ref 2). Several graphs illustrate the thermal

and electrical conductivities for the whole periodic table.

Both graphs use the same axes, and the blue vertical line marks manganese, at. wt. = 25 . The non-metals are bunched at the lower right corner of both graphs, but surprisingly, there are many metals that are not much better conductors than the non-metals. Use of a log scale for the vertical axis gives an even sharper separation and makes the class of metals look more uniform.

Nevertheless, there are differences in thermal and electrical conductivities that cannot be ascribed simply to a sea of a certain number of electrons. Other factors must include bond energy levels, including empty levels. The existence of other classifications, like metalloids and semiconductors (and doping) might suggest other mechanisms that might affect electron mobility and delocalization. A look at the plot by atomic number (below) could point to other factors.

The lanthanides stand out as a group, all of which have relatively low, but not zero, conductivity.
Some properties of the elements that might be related to electron mobility, like electronegativity, show no unique regions. While there are sharp distinctions between some non-metals and some metals, it seems that there are always a few borderline cases.

Different ways of graphing the data can exaggerate (and simplify) the differences, whereas linear plots might just blur distinctions. But the conclusion must be that delocalization of electrons is a variable that ranges from near zero to near complete and depends on factors in addition to a classification into metal and non-metal.
Edit: A new theory of metal bonding suggests that instead of visualizing cations of metal floating in an anionic sea of electrons, the spaces between metal cations may be considered voids which contain orbitals which localize some or all of the electrons in the sea. These orbitals are compressible and/or rearrangeable under pressure (see image and black box of explanation):

The images are from Chemistry World (Ref 3), which refers to the article at Ref 4, which is behind a paywall (no images in the abstract). In any case, this new model, which is based on extensive computations, gives a new view to the cation-in-a-sea-of-electrons picture, and could help us understand the mechanism of electron mobility in pure metals, similar to how mobility is understood in doped semiconductors.
Ref 1. https://materion.com/-/media/files/alloy/newsletters/technical-tidbits/issueno-104-electrical--thermal-conductivity.pdf
Ref 2. https://periodictable.com/Elements/029/data.html
Ref 3. https://www.chemistryworld.com/news/new-theory-provides-answers-to-why-metals-have-the-structures-that-they-do/4017056.article?utm_source=cw_monthly&utm_medium=email&utm_campaign=cw_newsletters
Ref 4. Y Sun et al, Proc. Natl. Acad. Sci. USA, 2022, DOI: 10.1073/pnas.2218405120


text:
How does O2- have the largest radius

Why is the correct option O^2-, aren't all of them equal since they all have the same number of energy shells and same number of electrons in them

Let's start with a simpler question. $\rm H$ (neutral) and $\ce{He+}$: what's bigger? They have the same number of electrons (one).

Well, 2- ions rapidly become 1- ions, so I consider it a trick question.

@JonCuster Unless there is meant a ionic radius in ionic compound context.

Sure, context matters there.

F-: There is only one electron added, there is inter-electronic repulsion due to which the radius increases
Mg2+: Two electrons are removed, but the number of protons remain the same, so nucleus pulls the electrons towards itself, and the radius decreases slightly
Na+: An electron is removed, so there is more effective nuclear charge, and pulls electrons towards nucleus, so again radius decreases.
O2-: Two electrons are added there is more inter-electronic repulsion, so the radius increases (and it is more than F-, because here two electrons are added but in F-
only one electron was added)
In these questions where we compare the radius of isoelectronic species, i.e. species with same number of electrons, we compare the ratio of Nuclear charge (Z) to the number of electrons(e) ratio.
If Z/e ratio increases the radius decreases,and if ratio decreases radius increases.


text:
Activation energy of the reaction between magnesium carbonate and hydrochloric acid
I've been working on a project that uses some abnormal methods to calculate the activation energy using only one trial of an exothermic reaction.
However now that I've completed the lab, I want to find a literature value or at least another lab which calculated the activation energy before. Balanced equation below:
$$\ce{MgCO3(s) + 2HCl(aq) -> MgCl2(aq) + CO2(g) + H2O(l)}$$
I've checked through all of the responses to Is there a database on chemical reactions, similar to NIST, but far more complete? However, all the replies focused on organic chemistry or purely gaseous reactions.
Where can I find a database or lab which measured the activation energy of similar/the same reaction?

Then better use something in your project that has actual chance of working, as far as Ea is concerned. Like a reaction between things in the same phase.

Is there any "activation energy" (above a ground state) at all?  This is an acid/base reaction.  There does not seem to be any rate limiting step at all, rate would be based on concentration of reactants.  However 2 H2 + O2 -> 2 H2O is reversible at high enough temperatures (and does have an Ea).  This may be more what you are after.  One interesting possibility: would HCl react with MgCO3 at near absolute zero temperatures?

Robert DiGiovanni - I was under the impression that everything technically had an activation energy because at absolute zero (theoretically) there could be no reaction because nothing was moving. Therefore there has to be activation energy, even for an exothermic reaction, just to get started.

Mithoron - if I assume that the MgCO3 had to dissolve in the water before reacting, then is there a way to find the Ea of MgCO3(aq) and HCl(aq)?


text:
What would the maximum energy density of a power source based on carbon-14
Various ways of generating electricity from small radioactive power sources have been long known.
The Voyager space probes launched in the 1970s use thermoelectric generation from plutonium sources to generate ~100W of power to keep the probes running when they go very far from the sun. Some small devices have used other sources and other conversion mechanisms to prove power for very low power applications like pacemakers where long lasting power that doesn't need to be replaced is an advantage.
But recent claims suggest a breakthrough where devices powered by carbon-14 could replace the batteries in mobile phones (Nano diamond Batteries web page highlights stories claiming their produces could "revolutionise mobile phone batteries").
These claims seem like uncritical hype but should be easy to check.
Ignoring engineering issues (like conversion efficiency) how does the energy density of a power source based on C-14 compare to a mobile phone battery?
(I'm guessing a simple calculation based on the radioactive properties of C-14 could give a decent answer and that any battery experts can tell us what a Li-ion battery achieves so we can compare them. Also, nuclear power sources are not "batteries" but they are compact power sources so please don't be distracted by the terminology).

The isotopic power of C-14 is 0.001305 W/g. That should be the theoretical upper limit even for an ideal system.

Some energy would be carried away by electron antineutrinos. For given thermal energy, the efficiency of energy conversion by Seeback effect thermocouples  in space probes (usually with $\ce{^{238}Pu}$ with $\pu{0.57 W/g}$)  is reportedly like 10%. There are prototypes using the Stirling engine with up to 40% efficiency, but with doubts about the long term reliability due moving parts.

Furthermore, the specific activity of C-14 is 1.6485E+11 Bq/g and the exemption limit is 1E+7 Bq. This gives you another limit if you do not want a safety license to be required for such a battery.

@Poutnik neutrinos are excluded from the value I gave above; so it's the energy available to the decay heat problem.

@Loong That's exactly the sort of data needed: A kg of C-14 would produce about a watt, which is less than the peak consumption rate of a typical mobile phone even if conversion were perfect. Do you have the calculation behind the W/g number or is it just a well-known fact?

@matt_black When calculating myself: The half life of C-14 is 5700 a, which gives you a decay constant of 3.853E-12/s or 2.321E+12 Bq/mol or 1.657E+11 Bq/g. The maximum beta energy is 0.15648 MeV, but the mean beta energy is only 0.04945 MeV = 7.923E-15 J. So the isotopic power is 1.657E+11 Bq/g * 7.923E-15 J = 1.313E-03 W/g. The value given by my old Nuclides 2000 is 1.3050E-03 W/g.

BTW, the title speaks about energy density, but the body and comments rather about power density. // It is better to call related quantities as specific energy/gravimetric energy density (J/kg), specific power/gravimetric power density (W/kg),  energy density/volumetric energy density (J/m3), power density/volumetric power density W/m3).

@Poutnik I'm happy to be corrected on the appropriate terminology to use for the comparison with conventional batteries. I suspect the relevant one is gravimetric power density in this case. Correct me if I'm wrong.

Just a summary: as @Loong and others state, the theoretical maximum power would be ~1 mW/g. If the phone requires ~1 W average for transmit (if I remember correctly,cell phones can use ~4W for talking to more distant towers), 1 kg of C14 would be needed in a 100% efficient cell. Not only would it be heavy, keeping the 156 kV beta particles contained would take some heavy shielding. On the other hand, you would not need a night-light.

@DrMoishePippik The maximum range of the 156.48 keV beta radiation from C-14 in plastic is less than 1 mm. No heavy shielding is required.

@Loong, the beta particles themselves are short range, but the Bremsstrahlung radiation produced by their stopping, i.e., hard x-rays, is quite a bit more penetrating!  https://physicsopenlab.org/2017/08/02/bremsstrahlung-radiation/ . Consider that color CRT's, using acceleration of just 40 kV or so, required heavy leaded-glass screens -- upt to 2 kg of Pb! https://computer.howstuffworks.com/question678.htm

@DrMoishePippik That's correct. That's why you use low-z material like plastic for the shielding of pure beta radiation so that the contribution of hard x-rays is significantly reduced. The fraction of the beta energy released as bremsstrahlung is then only maybe 0.02 %. The dose rate from a 1E+13 Bq source in 1 m distance without additional shielding of x-rays would be about 0.04 mSv/h. No problem in a controlled area, but certainly not suitable for public use.

The maximum theoretical power density of isotopically pure $\ce{^14C}$ can be estimated from the nuclear decay data of $\ce{^14C}$ as follows.
The half-life of $\ce{^14C}$ is $T_{1/2}=5.70\cdot 10^3\ \mathrm a$ 1. Thus, the corresponding decay constant is
$$\tau=\frac{\ln2}{T_{1/2}}=3.9\cdot 10^{-12}\ \mathrm s^{-1}$$
and the molar activity is
$$A_\mathrm M=\tau N_\mathrm A=2.3\cdot 10^{12}\ \mathrm{Bq\ mol^{-1}}$$
Since the molar mass of $\ce{^14C}$ is $M=14.0\ \mathrm{g\ mol^{-1}}$, the specific activity of $\ce{^14C}$ is
$$a=\frac{A_\mathrm M}M=1.7\cdot 10^{11}\ \mathrm{Bq\ g^{-1}}$$
The energy of the beta decay of $\ce{^14C}$ is $E=156.48\ \mathrm{keV}$ 1. However, the beta spectrum is continuous because the total energy of the decay process is divided between the electron, the antineutrino, and the recoiling nuclide. The mean beta energy only amounts to $E_\mathrm\beta=49.45\ \mathrm{keV}$ 1.

Since
$$E_\mathrm\beta=49.45\ \mathrm{keV}=7.923\cdot 10^{-15}\ \mathrm J$$
the power density (so-called isotopic power) of $\ce{^14C}$ is
$$e_\mathrm\beta=aE_\mathrm\beta=1.3\cdot 10^{-3}\ \mathrm{W\ g^{-1}}$$
How does this compare to a typical mobile phone battery?
An iPhone 14 pro has a Li-ion battery with capacity of about 3,200 mAh. At a voltage of 3.6 V, this gives a capacity of about 11.5 Wh.
It weighs just over 200g. While the power consumption varies a lot depending on what the phone is doing these have been estimated to peak at over 4W for some other smartphones. We can estimate the average power consumption by noting that a typical iPhone needs to be recharged every one or two days. 11.5 Wh over a 48h time period suggests an average power consumption of about 240mW.
To provide the average power (ignoring conversion efficiency and any other engineering issues) this would require about 185g of C-14. But to provide the peak power (say 4W) would require more than 3kg of C-14 which is 15 times heavier than the total weight of an existing iPhone 14 pro.
This would be truly a revolution in mobile phone batteries, but not the good sort.

References
1 ICRP Publication 107. Nuclear Decay Data for Dosimetric Calculations. Ann. ICRP 2008, 38 (3).

Overall efficiency after accounting for the isotope separation is pretty lousy…

By way of comparison: 100 g C-14 is about 1.7E+13 Bq which approximately corresponds to the annual C-14 production of a CANDU reactor. The global atmospheric inventory of C-14 from natural sources is about 1.4E+15 Bq or 8.4 kg.


text:
Why does the FTIR-ATR spectrum for hexamethylenediammonium adipate not have a peak for the N-H bond?
I have synthesized pure hexamethylenediammonium adipate by reacting equimolar quantities of adipic acid and hexamethylenediamine in ethanol. The diammonium salt, being insoluble in ethanol, was crystallized out and the mixture filtered. Cold ethanol was used to wash the residue to remove any unreacted acid/amine. The salt was then dried in a vacuum oven to remove any ethanol.
This is the structure of the salt:

The FTIR-ATR spectrum of the salt after oven-drying does not have a peak for the N-H bonds (at around 3500-3300 cm^-1). Is this due to the positive charge on the nitrogen?

Your anticipation of absorption bands of $\ce{-NH2}$ close $3300\ldots \pu{3500 cm^-1}$ is about free and neutral amine. Your intended product is about an (organic) ammonium, $\ce{C-NH^+_3}$.  Both the additional hydrogen as well as the additional positive charge, formally on nitrogen alters the bond strength $\ce{N-H}$.  By consequence, the position of the absorption band will change.  (Assuming the vibrations still were the same as a movement, maybe the additional charge equally altered the change of the dipole moment during the vibration; if this were the case, the intensity of the absorption band could change, too).
The first literature reference I found is by Vouyiouka et al.  Note, their setup used differs from yours; instead of a neat sample (your experiment), it is in nujol; instead of ATR (your experiment), it is by transmission (KBr crystal).  It still is indicative to read their report which states

the characteristic peaks of the PA-6,6 salt at $\pu{2210 cm^-1}$ corresponding to the strength vibration of the ionized amine groups ($\ce{-NH^+_3}$), at $\pu{1521 cm^-1}$ corresponding to the deformation vibration of $\ce{-NH^+_3}$, and at $1648$ and $\pu{1451 cm^-1}$ corresponding to the vibration of the $\ce{COO-}$ group.

The absorption bands seem present, the one reported at $\approx \pu{2210 cm^-1}$ partly covered by a broader signal around $\approx \pu{(2230 \pm 20) cm^-1}$.  On occasion, the intensity of an absorption band for the same material is weaker for the solid (here: crystalline) sample, than for sample in the liquid state.  Do you have complementary literature references for recorded IR spectra (since your setup uses the ATR technique, preferably literature references equally using the ATR technique)?  Can you ascertain the diamine used was in good shape, as well as the sample of adipic acid?  Since ATR-IR spectroscopy does use only a small sample, and does not require long time of preparation/experimentation, recording two ATR-IR spectra of the starting materials for comparison with literature as well as the product obtained would be an additional option.
Vouyiouka, S. N.; Koumantarakis, G. E.; Papaspyrides C. D. Preparation and Solid-State Polyamidation of Hexamethylenediammonium Adipate: The Effect of Sodium 5-Sulfoisophthalic Acid. J. Appl. Pol. Sci. 2007, 104, 1609–1619 doi 10.1002/app.25762.

A few hints about improving the record / the presentation of the record with ATR-IR:

Contrasting to pressing a recording an IR spectrum in transmission mode, in the ATR mode, infrared radiation shortly has contact with your sample.  The penetration depth of the radiation (before returning into the spectrometer) depends on the refractive indices of the sample and the window material (Ge more often seen for routine instruments, coated/non-coated diamond more often for research instruments), and the varied frequency of the incident infrared radiation.  Both influence the intensity of the absorption bands eventually plot in the spectrum; ascertain that this transmission correction is applied prior to the background correction.

The preferred representation of the IR spectra depends on the audience.  Chemists are used to decreasing wavenumbers from left to right (as you did) on the abscissa, yet to plot the percentage of (remaining) transmittance as ordinate.  This yields «hanging absorption bands».  By appearance of your plot, you possibly engage ThermoFisher Nicolet's software Omnic; there, the adjusting swap only requires a few additional mouse clicks.

In your representation, the abscissa continuously decreases linearly in wavenumbers.  This unnecessarily extends a region less interesting for (organic) chemistry on expense of the more interesting region for (organic) chemistry of the chemical fingerprint, i.e., $\approx 1200\ldots \pu{700 cm^-1}$.  To ease visual interpretation it is better to split the abscissa into two, or three regions.  (In case of Thermo's Omnic, either a) at $\pu{2000 cm^-1}$, or b) at $2200$ and $\pu{1000 cm^-1}$; in each case marked by a vertical dashed line.)  Again, this will not add, nor remove data to your spectrum recorded.  However, it will ease to recognize individual absorption bands.  Annotations on peaks will be easier to read, too.

As usual, retain a copy of the raw, unprocessed data, as well of the data after application of transmission correction, background correction, peak pic, etc.  To prevent a vendor lock-in by Thermo's binary data .sp format, consider .jcamp-dx and .csv in plain text.  This allows you (and others past you) to replicate your work, as well as the re-use of the data in subsequent experiments.

Upon further review, a small peak seems to lie close to the predicted 2210 cm^(-1) frequency compared with the sloping background.

@OscarLanzi You are right. Hence I run a digitization with the freely available WebPlotDigitizer by Ankit Rohatgi (intermediate screen photo, each dot is a point traced). Because the image is a bit tilted, and because the spectrum (without peak pick) was recorded under different conditions than the reference data, I added some safety margin around the absorption band.


text:
The first electron affinity of an atom
I've read that the first electron affinity is always negative, meaning that energy is always released upon the addition of this first electron. However, I don't think this will apply to noble gases. Will adding an electron to a noble gas release energy?
Could someone please clarify electron affinity?
Edit: Even though there are no stable negative ions of noble gases. I assume it is still possible to add an electron. Yet it would require energy. Is this true, meaning the affinity is positive?

See https://en.wikipedia.org/wiki/Electron_affinity_%28data_page%29  // Note that for historical reasons, Ea is positive when energy is released, the opposite compared to enthalpy change.

There are no stable negative ions of noble gases. Helium is metastable for long enough to use in accelerators, but that is it.


text:
What is the correct unit of a RI detector used for HPLC analysis?
I analyze sugar compounds in soft drinks with HPLC using refractive index detection (RID) on Agilent 1260 Infinity Binary LC (System User Guide (PDF)). I am not sure which unit to put next to the $y$-axis of the chromatogram.
Is it RIU, μRIU, or something entirely different?

The detector  is measuring change in refractive index of the solution.

Yes, this is for the UV detection. Did you check the model and make of the detector? Is it https://www.agilent.com/en/product/liquid-chromatography/hplc-components-accessories/hplc-detectors/1260-infinity-ii-refractive-index-detector

Good question. Now that you have mentioned the instrument name (Agilent 1260 LC), it is easy to locate Agilent's "Refractive Index Detector Agilent InfinityLab LC Series User Manual." It is always good to mention the entire set-up. Some detector display microvoltages on the y-axis on UV detectors.
Intuitively, you can see refractive index can not be negative (you have a negative dead time dip) and refractive index values cannot range in thousands. The signal is definitely not absolute refractive index (n). It is basically $\Delta n$, and in the jargon of detector, it is "Refractive Index Units". Note that refractive index has no units.
The reason for asking for the model was to check what the manual says about the output and what is the scaling factor. It is nano-refractive index units, and the detector is calibrted with respect to a certain concentration of a glucose solution vs. pure water.
On page 19, the manual states,

When sample elutes from the column into the sample cell the refractive
index of the cell contents changes. The change in refractive index
deflects the light beam as it passes through the flow cell resulting
in an unequal amount of light falling on each diode. The change in
current from the diodes that this causes is amplified and used to
produce the calibrated detector signal. This signal expressed, as nano
Refractive Index Units (nRIU), corresponds to the difference between
the refractive index of sample in the sample cell and the mobile phase
in the reference cell.

Secondary note: You have overloaded peak shapes, they should not front. Dilute the sample.

Thank you so much for taking a look at my issue! So, the unit of this specific RI detector is nano Refractive Index Units (nRIU), did I understand that correctly? (I'm not a native English speaker and this is just my high school project)

Yes, it is plotting the difference of refractive index of mobile phase and your analyte multiplied by 10$^9$.


text:
Unit of equilibrium constant in formula for Gibbs free energy change at arbitrary temperature
We know the relation for Gibbs free energy change of a reaction at any stage and any temperature
$$\Delta G=\Delta G^{\circ}+RT\ln Q$$
where Q is reaction quotient. Subsequently, we get the rate constant as
$$\Delta G^{\circ}=-RT\ln k_{eq}$$
My question is how do we deal with the units of $k_{eq}$, for example, if we have the reaction
$$N_{2}+3H_{2}\rightleftharpoons2NH_{3}$$
Rate constant at equilibrium can be written in terms of partial pressure as
$$k_{p}=\frac{P_{NH_{3}}^{2}}{P_{N_{2}}\cdot P_{H_{2}}^{3}}$$
I could calculate partial pressure in Pascal, torr, bar and so on which will change $\Delta G$. Please give clarification on the convention as to how we write $k_{p}$ when partial pressures in different units change $k_{p}$ according to equation above

Have you tried searching on SE for 'equilibrium constant unit'? There are several questions which address this.

The partial pressures are in bars.

Why is it in bars? Is it convention or does it depend on the unit of the universal gas constant that we chose? @Chet Miller

The equilibrium constant has to be dimensionless because the argument to the log  has to be dimensionless. This means that each pressure has be made dimensionless by divided by 1 atm. or 1 bar, or 760 torr or whatever units you are using.

It is in bars because $\Delta  G^0$ is defined as the change in G from pure reactants at 1 bar to pure products at 1 bar.


text:
Water vapor in CO2 - a chemical reaction?
I am doing some work on measuring water content of a stream of CO2 at ambient temperatures (25°C).
My question is this: Am I dealing with purely a mixture of gases, or is there some chemical reactions involved? If so, to what degree? I assume there is a reversible reaction to H2CO3
If this is the case, it might alter the measurements.
Analysis techniques include optical and microwave

That would be "$\ce{H2CO3}$" and it's not detectable even in water saturated with $\ce{CO2}$.

@Mithoron Not even by NMR?

NMR is too darn insensitive. You would need IR and/or mass spectrometry combined. The relavant paper is "Ludwig, R. and Kornath, A., 2000. In spite of the chemist's belief: Carbonic acid is surprisingly stable. Angewandte Chemie International Edition, 39(8), pp.1421-1423."

@Mithoron, Who said that H2CO3 is not detectable?

Gaseous carbonic acid ($\ce{H2CO3}$) is reportedly more stable than one could think, but  water vapor autocatalyzes decomposition of eventual traces of gaseous carbonic acid to carbon dioxide and more water vapor. So chances to meet carbonic acid, aside of very small equilibrium concentration in liquid water, is being trapped in deeply frozen matrix, or in interplanetary/interstellar space.

We will not get a significant reaction in a low-density gas phase; the gas mixture is essentially $\ce{H2O + CO2}$. Carbonic acid can be made, but not by reaction of carbon dioxide with water at ambient pressure.
Carbon dioxide does react slightly with liquid water to form a small amount of $\ce{H2CO3}$. But the ability to form this acid at all from carbon dioxide and water depends on the hydrogen bonding between its hydroxyl groups and surrounding water molecules. That requires the water to be densely packed in the medium, which in ordinary conditions means not in the gas phase.


text:
How can I make a polyurethane formulae more adhesive?
For a project I'm working on, how can I make polyurethane more adhesive and able to stick to surfaces more effectively, not necessarily super-glue like but just stickier properties. What other chemicals may I need to add, or would I need to also alter the main 2 ingredients? (isocyanate and polyol)
Also interested in making the formulae more medically stable, safe for skin and deeper wounds.
I would appreciate any input or answers!

Look at a chemical data safety sheet for the isocyanate of interest, e.g., https://www.amcsupplies.com.au/manuals/ISF%20136%20Isocyanate%20-%20MSDS.pdf, and decide if it's suitable for use as a wound adhesive. Certain words such as "carcinogen" and "toxic" might be good clues.

polyurethane has been used as a wound treatment method in recent experiments so it is safe for wounds, I was mainly looking for ways to make it more adhesive


text:
Calculate the pH at the equilibrium point in an acetic acid sodium hydroxide titration
You perform a titration of $\ce{CH3COOH}$ with $\ce{NaOH}$. Let the original concentration of acetic acid be $\pu{0.2 M}$. The volumes of $\ce{CH3COOH}$ and $\ce{NaOH}$ are the same. Then the equivalence point will be reached when equivolume of $\pu{0.2 M}$ of $\ce{NaOH}$ has been added to the solution. At the equivalence point, the $\mathrm{pH}$ will be above 7. How does one calculate the exact $\mathrm{pH}$ value of the solution, which will be at the equilibrium point? The reactions we have are:
$$\ce{CH3COOH + OH- <=> CH3COO- + H2O} \tag1$$
($\ce{Na^+}$ is a bystander or spectator ion, and therefore I ignored here).
The equilibrium constant for the given reaction is $K_1 = 1/K_b(\ce{CH3COO^-}) \gg 0$. We also have the following reaction:
For the reaction, $$\ce{CH3COO- + H3O+ <=> CH3COOH + H2O} \tag2$$, the equilibrium constant is $K_2 = 1/K_a(\ce{CH3COOH}) \gg 0$.
It is argued on other posts here that the first reaction "essentially" goes to completion and because $\ce{CH3COO-}$ later reacts with $\ce{H3O+}$ in the reaction $(2)$, the $\mathrm{pH}$ will be above 7. However, this motivation seems flawed to me. When the second reaction occurs, that will decrease the concentration of $\ce{CH3COO-}$, which will trigger reaction $(1)$ to move to the right. This will decrease the concentration of $\ce{OH^-$ ions. I know that the rate at which the concentration of $OH-}$ decreases in reaction $(1)$ and the rate at which the concentration of $\ce{H3O+}$ decreases in reaction $(2)$ are not equal due to the different equilibrium constants for these two reactions. However, I don't understand how to calculate this $\mathrm{pH}$ using the above logic.
Could anyone show how to calculate the $\mathrm{pH}$ that this titration will have at the equivalence point using logic above? Thanks!

You have not specified what volumes of each reagent are being mixed together.

Sorry, I have updated the question now.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

Why have you omitted the reaction of acatate ion with water, being a weak base?

@Poutnik that is incorporated into the first reaction, right?

But it does not matter if you flip a reaction around, since the equilibrium constant will also change (and become 1/K_eq). What am I missing?

Use the Henderson Hasselbach equation to get a better understanding.  I know of no reason to try to calculate pH after essentially all of the acetic acid is neutralized (half of it is neutralized at pH 4.72). After that, one would want pH to increase ASAP (it will) until the indicator changes.  Even split drop technique with dilute NaOH will not land on a lucky 7.0 very often.

If the expected pH is alkaline, the hydroxide concentration will be higher than the hydronium concentration. Thus, working with K$_1$ is easier (you can neglect the hydronium concentration and focus on the major species acetate and the minor species acetic acid and hydroxide). As the last step in the calculation, you can estimate the hydronium concentration from the hydroxide concentration.
The logic given by the OP does not follow a simple  path to the solution, so I would not use it. Instead, you can first let reaction 1 go to completion, and then realize you overshot and go reverse a bit. That way, it looks like a straightforward weak base calculation.
Here are the concentrations of species after you add the two solutions, pretending that acetic acid does not partially dissociates, and that it does not react with sodium hydroxide:
$$c_\mathrm{acetic\ acid} = \pu{0.1 mol L-1}$$
$$c_\mathrm{acetate} = \pu{0.0 mol L-1}$$
$$c_\mathrm{hydroxide} = \pu{0.1 mol L-1}$$
The latter ignores water auto-dissociation.
Now we pretend the reaction goes to completion:
$$c_\mathrm{acetic\ acid} = \pu{0.0 mol L-1}$$
$$c_\mathrm{acetate} = \pu{0.1 mol L-1}$$
$$c_\mathrm{hydroxide} = \pu{0.0 mol L-1}$$
Again, we are ignoring the auto-dissociation of water.
This is not at equilibrium because acetate is a weak base. You can calculate the hydroxide and acetic acid concentrations (still ignoring the auto-dissociation of water) using the usual method you learned for weak bases. Then, you can calculate the concentration of hydronium ions, neglecting that when hydronium ions form, hydroxide ions also form (you can do that because the hydroxide concentration is much higher than the hydronium concentration, so the error is very small).
Once you have a set of approximate equilibrium concentrations, you can plug them into the equilibrium constant expression to verify that your approximations made sense (i.e. the equilibrium constants calculated are close to those known or given).

Do you mean that you only consider reaction Nr. 1, and based on the equilibrium constant for that reaction, you calculate the concentration of [OH^-] at equilibrium? Then you estimate $[H_3O^+]$ as $[H_3O^+]=K_w/[OH^-]$.

@UserE Yes, exactly.

Why is $c_{acetic\ acid}=0.1$ in your first equation? Should not it be 0.2 M?

When you mix it with the other solution (any solution, really), it gets diluted.

At the equivalence point, there is excess neither of acetic acid neither of sodium hydroxide. The solution is equivalent to the solution of sodium acetate of the concentration $\pu{0.1 M}$, as both acetate and sodium ions are in the doubled volume now.
For all evaluation below, I assume the temperature $T_\mathrm{C} = \pu{25 ^\circ C}$. Everywhere is also silently assumed the ion activity coefficients are equal to 1.
We know that an alkaline salt of a weak acid with the acidity constant $K_\mathrm{a}$ acts as a weak base with the basicity constant $K_\mathrm{b} = K_\mathrm{w}/K_\mathrm{a}$ (and $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$):
$$K_b = \frac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}= \frac{[\ce{HA}]K_\mathrm{w}}{[\ce{H+}][\ce{A-}]} = \frac{K_\mathrm{w}}{K_\mathrm{a}} \tag{1}$$
For diluted solutions of weak acids, there is the approximating equation:
$$\mathrm{pH} \approx 0.5(\mathrm{p}K_\mathrm{a} - \log{c}) \label{2} \tag{2},$$
assuming conditions:
$$c \gg [\ce{H+}] \gg [\ce{OH-}]\tag{2a}.$$
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}  \overset{[\ce{H+}] \gg [\ce{OH-}]}\approx \frac{[\ce{H+}][\ce{H+}]}{[\ce{HA}]} \overset{c \gg [\ce{H+}]}\approx \frac{[\ce{H+}][\ce{H+}]}{c} \\
\implies [\ce{H+}]=\sqrt{K_\mathrm{a} \cdot c} \implies \log{([\ce{H+}])}=0.5(\log{(K_\mathrm{a})} + \log{(c)}) \implies \eqref{2} \tag{2b}$$
Similarly, for weak bases, there is the approximation:
$$\mathrm{pOH} \approx 0.5(\mathrm{p}K_\mathrm{b} - \log{c})\tag{3},$$
assuming conditions:
$$c \gg [\ce{OH-}] \gg [\ce{H+}]  \label{3a} \tag{3a}.$$
From it there is derived the approximated equation for $\mathrm{pH}$ of solution of a salt of a weak acid and a strong base at given molar concentration:
$$\mathrm{pH} = 14 - \mathrm{pOH} \approx 14 - 0.5(14 - \mathrm{p}K_\mathrm{a} - \log{c}) = 7 + 0.5(\mathrm{p}K_\mathrm{a} + \log{c})  \tag{4},$$
assuming the conditions \eqref{3a} are met.
For the particular case in the question:
$$\mathrm{pH} \approx  7 + 0.5(4.75 + \log{0.1}) \approx 8.88 \tag{5}.$$
Checking $\eqref{3a}$:
$$\pu{0.1 M}  \gg \pu{10^{-5.12} M} \gg \pu{10^{-8.88} M} \tag{6}$$

Hmm, I have intentionally used t for Celsius temperature in contrary of T for absolute temperature. I have also intentionally used  A/B instead of frac for easier reading and text flow. Similarly, I have not intentionally used MJ on standalone inline pH and pOH, but that can be subject of various opinions. And, no access seems better than niether access to me.

You can always roll back edits, and keep it like that or edit more yourself. I think on this site, the philosophy is a bit different than on Wikipedia ("no editor owns an article – any contributions can and may be mercilessly edited and redistributed"). Here, if you have a substantially different answer, you just post your own. However, we edit questions extensively when warranted, and we do sometimes edit answers as well.

@Karsten I know and I did partial reediting of this.  Personally, I hesitate to change intentions of the posts of others. I fix obvious errors or apply formatting that is out of the poster knowledge scope, but otherwise rather suggest changes.

@Poutnik what is equation 2 for theorem? I haven't seen it before. I tried googling it but found little. What is it called?

@UserE It is not a theorem and I am not sure if it has a particular name. It is equation for calculation of pH of diluted weak acid, simplified under specific conditions (2a), allowing simplification (2b). It is regularly taught in lectures about pH of water solutions of weak acids, bases and salts.

@Poutnik Ok. I understand. In equation 5, though, why is c=0.1 M?

Ohh, my bad. Forgot that … Thanks!


text:
How do I know if a reaction is favorable? (High School student)
I am doing a homework question where I have to react NaCl + Something to get NaOH, and I can't find a reliable source online as to whether a reaction is possible.
For example, I thought of NH4OH + NaCl -> NaOH + NH4Cl, but I don't know if a reaction would occur, and I decided it is appropriate to ask this question since I had the same question on different equations that I managed to do.

Well, $\ce{NH4OH}$ is basically just another name for ammonia. https://chemistry.stackexchange.com/q/10230/4945 on the question at hand,I don't see an easy way to convert sodium chloride to sodium hydroxide. Could you please include to to exercise verbatim (with citation) here.

@Martin-マーチン isnt $NH_{3}$ ammonia?

Hey! I think I found a solution: Mg(OH)2 + 2 NaCl -> 2 NaOH + MgCl2↓ and therefore it is possible, since we have a precipitate.

No, I don't think you have. Magnesium chloride is pretty well soluble in water. And yes, @Volpina, ammonia is $\ce{NH3}$, as the linked question explains

$\ce{Mg(OH)2}$ is the precipitate as in Milk of Magnesia. BTW, a Phillips screwdriver is Milk of Magnesia and vodka.

Your reasoning is valid -- i.e., a precipitate would force the reaction in the direction that produces $\ce{NaOH}$... but $\ce{MgCl2}$ is highly soluble, about 50 g/100 ml, according to this useful Wikipedia solubility table.
You can use that table to find less soluble chlorides, though... Search on chloride and look for an insoluble one.
Persist... most chlorides are soluble. Too bad electricity was not allowed... or is it?
BTW, please edit your question to include your comment. If you don't show your thought processes, it might appear you are looking for someone to do your homework for you ;-) Also, notice my slight edit: a great many reactions are possible, but you need to find one that is favorable, i.e., favors going in the direction you want (technically, equilibrium favors the product).

I am doing a homework question where I have to react NaCl + Something
to get NaOH, and I can't find a reliable source online as to whether a
reaction is possible.

I am glad that you could not find any, because there is no such "something" that will convert $\ce{NaCl}$ into $\ce{NaOH}$ directly, except electricity. See chloro-alkali process.
However, long before electricity were widely available, a process called Solvay process could convert NaCl to sodium carbonate. Reacting water soluble sodium carbonate with lime aq. $\ce{Ca(OH)2}$ gave us $\ce{NaOH}$ solution.

A reaction is favorable, if and when the change in free enthalpy G is negative. So try to find the free enthalpies of formation of products and reagents. Do calculate the free enthalpy change, taking into account the concentrations of the different stuffs. The reaction  is possible if $\Delta $G is negative. If it is positive, the reaction goes in opposite direction. This reasoning does indicate whether the reaction is possible or not. It does not give any information about the rate of the reaction, which can be very quick or rather slow.

Just to little complement AChem - before electrolytic production of $\ce{NaOH}$, one of industrial methods was the combination of

Solvay process of soda production:

\begin{align}
\ce{NaCl(aq) + NH3(aq) + CO2(aq) + H2O(l) &-> NH4Cl(aq) + NaHCO3(s)}\\
\ce{2 NH4Cl(aq) + Ca(OH)2(s,aq) &-> CaCl2(aq) + 2 NH3(g) + 2 H2O(l)}\\
\ce{2 NaHCO3(s)&->[heat]Na2CO3(s) + H2O(g) + CO2(g)}\\
\end{align}

Caustification of soda:

$$\ce{Na2CO3(aq) + Ca(OH)2(s,aq) -> 2 NaOH(aq) + CaCO3(g)}$$

There is an elegant lab method of the direct conversion, but too expensive, suitable for hardly more but demonstration:
$$\ce{2 NaCl(aq) + Ag2O(s) + H2O(l) -> 2 AgCl(s) + 2 NaOH(aq)}$$

Going from $\ce{NaCl}$ to $\ce{NaOH}$ you exchange the chloride anion for hydroxide. To do this, you might think of using something that releases hydroxide ions and takes the chloride ions out of solution by forming an insoluble compound. The only soluble hydroxide whose chloride is insoluble is the extremely toxic thalium (I) hydroxide. The reaction would be as follows:
$\ce{NaCl(aq) + TlOH(aq) -> NaOH(aq) + TlCl(s)}$
Though the intended solution is most likely electrolysis, as implied by the other answers.


text:
How much group representation theory is used in chemistry research?
I am a mathematician working in group theory and representation theory. When I try to explain these subjects to non-mathematicians, I often fail to provide a convincing reason to study these. So I am trying to learn how these theories are used outside pure mathematics. So how much group theory and representation theory do chemists really use in their research? I heard that there is something called "point group" that describes the symmetry of a molecule, but do you use any "advanced" group theory/representation theory results (something you don't expect to see in an undergraduate textbook) to study this point group thing or any other group that you use in chemistry? For example:
-Has anyone ever tried to use "modular representation" (representation over fields with positive characteristic) or "projective representation" (a group homomorphism into projective general linear group instead of the usual general linear group) in a chemistry research? If not, do you think chemistry will need such thing in the future?
-Do you ever need to think about representations with high dimension, or only work with at most 3-dimensional representations?
-How complicated are the groups you see in chemistry? Are they mostly finite subgroups of O(3) or other similar groups? Do you ever get an infinite group or a finite but complicated group, like being nonabelian and having long composition series, or having very large order?
-Can you recommend some high quality chemistry research papers involving a lot of group/representation theory, or some reputable journals where I can find such papers?

I am not a mathematician so I'm not sure if I understand all of the above. These are quite a few questions up there. Is probably too broad to be reasonably answered in our format. I think you should start with some introductory texts on quantum mechanics, physical and theoretical chemistry, and spectroscopy to get an overview. At the very least there's plenty of maths in there. Then follow the publications referenced within. This list will get you started: https://chemistry.stackexchange.com/q/37303/4945

Another hint I'd like to leave is to use the site search. There are about 100 questions with the tag [tag:group-theory]. You'll probably then find https://chemistry.stackexchange.com/q/6231/4945, which I think would also be a good place to start, too.

Wigner and x-ray crystallographers certainly use group theory. As do many solid state physics folk because of the tie to crystal structure.

Some of the publications' titles indicate the relevance of group theory e.g. Raman Spectroscopy, Group Theory, and Computational Chemistry: A Physical Chemistry Laboratory Experiment on para-Difluorobenzene 2022JChemEduc2129, or that you can anticipate it is relevant to the article, one way, or an other, e.g. From Outside In: Stretching Students’ Conceptual Understanding of Molecular Symmetry with 2D and 3D Manipulatives, 2023JChemEduc1063. Both found with "Group theory" in ACS' search box.

You need to look for info on  Point Groups (for molecules) and Space Groups for crystals. Some more mathematical texts are D. Bishop 'Group Theory and Chemistry' (publ Dover), M. Tinkham 'Group theory and quantum mechanics' (publ Dover) and W. Borchardt-Orr 'Crystallography' (publ Springer)

@porphyrin There is a small typo as the keys are so close: not Borchardt-Orr, but Borchardt-Ott (only because I once read the book).

@Buttonwood, thanks , as you guessed its my stubby fingers

In addition to the 119 questions with the "group-theory" tag here (which Martin pointed out), there's another 66 of them here and another 21 of them here.

I'm doing a PhD in organic electrochemistry. Let me first give you some literature that is usually used for studying chemistry here in Germany, probably in order of increasing complexity:

Ingo Peter Lorentz: Molekülsymmetrie und Spektroskopie (German) (ISBN 978-3110364927)
F. Albert Cotton: Chemical Applications of Group Theory (ISBN 978-0471510949)
Dirk Steinborn: Symmetrie und Struktur in der Chemie (German) (ISBN 978-3527284184)
Yves Jean: Molecular Orbitals of Transition Metal Complexes (ISBN 978-0198530930)
Thomas Albright: Orbital Interactions in Chemistry, 2. ed. (ASIN B010WF0GKG)

For a scientist, symmetry is the most important concept that exists. Symmetry can be found in all parts of science: chemistry, physics, astronomy, biology and much more. It's one of the few concepts that play an important role in both science and art as well. No other concept plays such an important role in chemistry.
Group theory is something that is often poorly understood (and poorly teached). As a pratical chemist who stands in the lab, I think about molecules and their symmetry all day. Since chemists prefer a visual type of thinking over strict group theory, we don't have group theory in our mind all the time. In general, we have the feeling that symmetry is our friend because it can really simplify a lot of complicated problems, both of practical or theoretical nature. Most important, no matter what spectroscopic method you use, the symmetry of a molecule has direct implications on the spectrum that you obtain from it.
Examples
Here are 10 examples where group theory is the indispensable foundation:

Let's start with a little quantum mechanics: Atomic orbitals and molecular orbitals are classified by group theory. Even if you are not a theoretical chemist, you should have learned that orbitals can only overlap if their symmetry and spacial orientation can match (and their energy is similar). Looking at the most simple molecule, the dihydrogen molecule $\rm H_2$, we know that each hydrogen atom has a single occupied 1s orbital. The overlap of both gives a bonding $\sigma$ orbital and an antibonding $\sigma^*$ orbital. The bonding $\sigma$ orbital is occupied by the two electrons, resulting in a lower energy of the system. This is the reason Helium prefers to stay as $\rm He$, not as $\rm He_2$: In that case, we would have 4 electrons and the antibonding orbitals would be occupied as well, resulting in a higher energy of the system. For more complex molecules, we need to make a symmetry adapted linear combination of orbitals. You can read more about that in "Molecular Quantum Mechanics" by Peter Atkins. An impressive example of how symmetry can simplify the discussion of orbitals is the molecular orbital diagram of ferrocene.

The most important spectroscopic method in all of chemistry is nuclear magnetic resonance spectroscopy (NMR spectroscopy). NMR spectroscopy has advanced chemistry more than any other method before or since. When organic chemists look at an NMR spectrum, we use the symmetry of a molecule to match the signals of the spectrum to the molecule. Doing so is standard in any review process when submitting a paper. If we look at isopropanol for example, we know it has $\rm C_{2v}$ symmetry, meaning that the 6 protons of the two $\rm CH_3$ groups will give us one singlet signal of height 6. The proton of the hydroxyl group gives us one singlet of height 1. Chemists don't think often about the point group, they think visually about the symmetry elements of the molecule.

Symmetry plays a major role in the synthesis of molecules. Just to give one example, a symmetrical ketone will give you the same enolate while an unsymmetrical ketone can give you two different ones. If two different ones can be obtained, it is important which one you obtain in order to synthesize the correct molecule. E.J. Corey and Robert B. Woodward pioneered this.

We also use Mulliken symbols: In coordination chemistry, we often deal with electronic spectra (UV-Vis). We learn for example that $\rm Mn^{2+}$ is almost colorless because Crystal Field Theory tells us it has the electronic configuration $\rm t_{2g}^3 eg^{2}$. This electronic state is classified by group theory as a $\rm {}^{6}A_{1g}$ state. In electronic transitions, the transition dipole moment must not be equa to zero. This means electronic transitions must be accompanied by a change in orbital symmetry. Because both $\rm t_{2g}$ and $\rm e_{g}$ are gerade (g), the transition is weak (LaPorte selectrion rule). Also, the electronic transition would require a change of spin of one electron, meaning spin multiplicity would change.

A molecule (inorganic or organic) can only be chiral if is does not contain an improper axis of rotation. In a chemists way of thinking, this means a carbon atom has 4 different substituents. The entire discussion of isomers is based on symmetry: If two molecules behave like mirror images of each others, they are called enantiomers.

In Infrared Spectroscopy, a molecular vibration can only be seen if the molecule has a permanent dipole moment. In Raman spectroscopy, a change in polarizability is required in order to see a molecular vibration in the Raman spectrum. Discussing which molecular vibration can be seen in which spectrum is done using group theory, which is very difficult to do for the average chemist. Symmetry also plays an important role in Mössbauer spectroscopy.

Solid state chemistry heavily relies on X-ray diffraction and other diffraction methods such as neutron diffraction. There are 230 space groups which can classify the crystal structure of a crystalline solid compound. Based on symmetry discussions, the reflexes in X-ray diffraction patterns can be interpreted. I know little about that, but I know it requires a lot of skill and experience to obtain good and correct results. (Note that diffraction methods are no spectroscopic methods, because spectroscopy is the energy-resolved analysis of the interaction of electromagnetic radiation with matter).

A very important thing in coordination chemistry is the Jahn-Teller-effect: Any molecular system which is not linear and has a degenerated electronic state will result in a breakdown of symmetry which changes the molecular structure and reduces the symmetry of the molecule. For example transition metal complex $\rm [Cr(H_2O)_6]^{3+}$ is not octahedral ($\rm O_h$), because its electronic ground state is degenerate according to Crystal Field Theory. The z-axis is elongated in the complex, meaning the symmetry reduces from $\rm O_h$ to $\rm D_{4h}$, because each orbital with a z-component is lowered in energy. The overall every of the complex is then lower. A similar case is $\rm [Ti(H_2O)_6]^{3+}$, but here the complex is compressed on the z-axis to resolve the degeneracy. This also lowers the energy by going from $\rm O_h$ symmetry to $\rm D_{4h}$ as well.

The Woodward–Hoffmann rules and the isolobal analogy are concepts which classify the symmetry of molecular orbitals to explain chemical reactions or structure of molecules. Hoffmann's Nobel lecture called "Building Bridges between Organic Chemistry and Inorganic Chemistry" is a must read for chemists.

The symmetry of orbitals and the symmetry of crystal structures can be combined to understand and explain properties of solids such as electronic or optical properties (electronic band structure). This plays a huge role for the understanding of semiconductors and other advanced materials. Also, the Goodenough-Kanamori-Anderson rules (GKA rules) are used to explain magnetism in solids based on symmetry and orientation of spin orbitals.

Summary

Chemists think about the structure of molecules and solids a lot. The structure of a molecule, its symmetry and the implications of this on the spectroscopic data you can obtain from the molecule is linked inseparable.
Chemists sometimes apply orbital theory if required (some more some less, depending on the field you are working in). The analysis of the symmetry of the molecule or the symmetry of the molecular orbitals is the foundation for this.
Symmetry is indespensable for chemistry, because entire fields of chemistry were founded based on symmetry considerations. A nice example is how Alfred Werner came up with coordination chemistry simply by making symmetry considerations. When Werner wrote down his ideas based on symmetry, he did not yet conduct one single experiment in the lab to prove any of this (!)
Symmetry plays a major role in the synthesis of molecules or solids.
Group theory is still poorly taught in my opinion. While there are plenty of good books about quantum mechanics or relativity, group theory remains a gap. I myself could construct the molecular orbital diagram of ferrocene based on symmetry analysis, but I could not tell you how to use the projection operator to obtain the molecular orbitals of the ammonia molecule for example.

+10 for a solid answer! I gave a 300 point bounty here only because I'd already spent the bounty and it was the best answer out of the ones that came, but I was hoping the question could get a better answer, do you think you can write one?

I understand that symmetry plays a very important role, but I still don't see how deep the mathematics involved is. But hopefully searching with these keywords will give some idea. Thank you!

@user1271772 I'm sorry but I don't think I can improve the answer given there. I am not a theoretical chemist.

@Absol I think a strict treatment of group theory to understand of solve a problem in chemistry remains the job of a theoretical chemist.

I am a mathematician working in group theory and representation
theory. When I try to explain these subjects to non-mathematicians, I
often fail to provide a convincing reason to study these.

A brief discourse on history is usually sufficient to emphasize the importance of group theory in a talk or a lecture. So, you may have to start from the time when group theory helped to solve difficult chemical problems, especially in vibrational spectroscopy. The audience won't be convinced if you begin with a technically dry introduction. To be honest, most of those pioneers were hardcore physicists such as Mullikan, Placzek (never got credit for his work) Wigner and many others. You can also mention the severe opposition to group theory in early physics. Innovative ideas are always unwelcome by a certain group. Group theory users were called Gruppenpest (Group theory pests). I do not think you can find "pure" chemists (chemists who did not have mathematical training but were trained in chemical reactions) who woke up one day and started using it. Most of the work has been borrowed from traditional physics.
Here is the list of papers/books or works that you can start to convince the audience of the importance of group theory to chemical applications.

Group Theory: And its Application to the Quantum Mechanics of Atomic Spectra by Eugene P. Wigner (Book by a Nobel Prize Winner).

The “Gruppen Pest” Yesterday, Today, and Tomorrow, by Brian G. Wybourne (article)

Zur Deutung der Spektren mehratomiger Molekule by L.Tisza, Again like Plakczek he got scant credit for his pioneering work.

Search the work of these authors. A couple of years ago I was compiling the history of group theory symbols used in the chemical literature out of curiosity. No claims to understanding Wigner's work.
From here, at least, you can search for their names and follow their citation chains to see the latest users of group theory.
P.S. Do not worry about a topic at an undergraduate level. Today's undergraduate curriculum is so superficially broad that students have now heard of everything, and they think they know everything, but often the reverse is true. Even some PhDs have this mentality. I had one undergraduate ask me if I knew string theory (of course not). So there is no need to start with a very esoteric or arcane application.

For convenience, but without further checking: 1.  https://vdoc.pub/documents/group-theory-and-its-application-to-the-quantum-mechanics-of-atomic-spectra-4llpk214bnd0 this answers to be the publisher: https://www.sciencedirect.com/bookseries/pure-and-applied-physics/vol/5/suppl/C 2. Brian G. Wybourne, The “gruppen pest” yesterday, today, and tomorrow https://doi.org/10.1002/qua.560070706  3. Tisza, L. Zur Deutung der Spektren mehratomiger Moleküle. Z. Physik 82, 48–72 (1933). https://doi.org/10.1007/BF01342114 (you should be able to comprehend German for this)

Thank you, I'll try Wigner's book.


text:
Acidity of Benzenepropanenitrile
Which is the more acidic hydrogen in the parent chain of this compound?
I am confused while considering the various electronic effects.

Which site do you think is most acidic? Have looked for carbanion pKa tables?

Review the guide How to ask. Not following the guidance may lead to objections, question downvoting or even question closure.

Sure I'll follow it @Poutnik .I'm new here,thanks for letting me know.

I would have called it 3-phenylpropanenitrile, but let's get to the real chemistry here.
Clearly the candidates for most acidic hydrogen atom are those on the saturated carbon atoms between the phenyl ring and nitrile (CN) group. If you choose to remove a proton from the carbon next to the phenyl ring, the negative charge can be dispersed into the phenyl ring via resonance. If instead you choose to remove a proton from the carbon next to the nitrile group, the negative charge can be delocalized into the nitrile group.
You were probably taught by rote whether the nitrile group or the phenyl ring is better at taking up the negative charge, but there is a better way to find out. This table gives the $\pu{pK_a}$ values of some simple compounds corresponding to different environments for removing a proton -- the lower the number, the stronger the acid and the better the conjugate base is for taking up the negative charge.
We find that acetonitrile (proton next to a nitrile group) gives 25 whereas toluene (proton nect to a phenyl ring) gives 41. Though neither of these is appreciably acidic in water, acetonitrile could be much more so in some other solvents.
So the hydrogens next to the nitrile group of your given compound, 3-phenylpropanenitrile, would be the most acidic.