text:
Determining which compound is more 'ionic'
I was going through my chemistry textbook (IB Pearson), and it explicitly stated that the higher the absolute difference between the electronegativity of elements in a binary compound, the more 'ionic' that compound is, which does make intuitive sense given that a higher electronegativity difference leads to the compound being more 'polar.'
However, the book came up with the following example:

Which of the following pairs will form the most ionic compound?
1) Be and F (the electronegativity difference is $|4-1.6| = 2.4)$
  2) Si and O (the electronegativity difference is $|3.4-1.9| = 1.5)$
  3) N and Cl (the electronegativity difference is $|3.2-3| = 0.2)$
  4) K and S (the electronegativity difference is $|2.6-0.8| = 1.8)$
$\ce{K2S}$ will be the most ionic compound as the difference in electronegativity is the greatest.

I don't get how this works, the difference in electronegativity between beryllium and fluorine is visible greater, so why does the book say that the binary compound formed by K and S is more ionic?

https://chemistry.stackexchange.com/questions/57172/bef2-covalent-or-ionic

https://chemistry.stackexchange.com/questions/17064/is-kf-the-most-ionic-compound?rq=1

Tl, dr -- upon further review we can't really tell whether potassium sulfide or beryllium fluoride is more ionic without some quantitative details.
There is more involved in the balance between ionic and covalent bonding than electronegativity differential.  You have to consider molecular orbital structure, too.  One of the answers to this question describes how molecular orbital structure can drive highly ionic bonding when electronegativity differences are modest (as in salts with aromatic ions) or even the "wrong way" (as in alkalides).
With single atoms, as pointed out in the other question, you cannot expect 100% ionic bonds even when molecular orbitals favor it.  Yet there can still be an impact which we can see by comparing organolithium compounds with organic fluorides.  The former acts as a ready source of anionic carbon, while the latter are notoriously unreactive towards the displacement of fluoride ions, as if the carbon-lithium bond had much more ionic character.  Yet the electronegativity differentials in the carbon-lithium and carbon fluorine bonds are about the same (oppositely directed with respect to carbon, of course).  The difference comes from lithium having a diffuse valence orbital that overlaps poorly with carbon, whereas fluorine is more compact and overlaps well, thus promoting electron sharing between both atoms in the bonding orbital.
Similarly, potassium has a diffuse valence orbital, even more so than lithium, which overlaps poorly with most nonmetals.  The ionicity of potassium sulfide, already considerable based on electronegativity, is thus enhanced even more.  Beryllium and fluorine, with good covalent overlap, would have a more covalent interaction than what might be expected from their electronegativity difference.
Since potassium sulfide has poor overlap and beryllium fluoride has a somewhat higher electronegativity difference, the two effects oppose each other when the compounds are compared.  We cannot tell from these qualitative arguments which one is more ionic.

"One of the answers" is a funky way of referring to an earlier answer of you. ;) [Btw. did you perhaps mean to link to it directly? I really needed to think twice before I got where you wanted to go with that paragraph.] // I generally agree with your conclusions; have a look at "Is beryllium difluoride covalent or ionic?" which suggests that $\ce{BeF2}$ is indeed very, very ionic.

Yup, you got me.  I had no better way to "hide" what I was referring to.  :-).  Woujd have been interesting to see what the referenced method would say about potassium sulfide, I suspect this would be found above 90% ionic too.

Why hide it? I quite like your answer, and it is very fitting indeed. Most of us re-use our answers if they are related, there is nothing wrong with that; and I think it's better than repeating a lot of the same arguments.

It seems the examples were thrown together without considering that someone would try to understand them.
The rule does make sense, but there are several factors involved here, and they are not mentioned. The problem is not even that you are comparing apples and oranges; it's more like fruit salad vs vegetable stew!
Let's simplify: Keep one of the elements the same and change the other. N and Cl have a difference of 0.2 and the compound is covalent (with lots of charge separation). If we look at KCl, the difference is 3.2 - 0.8 = 2.4, so the difference is larger and suggests a larger ionicity. That difference suggests why KCl completely ionizes in water.
That extra factor involved is theoretically comparing "degrees of ionicity" by calculations vs seeing the actual ions in water (well, observing the changes that we ascribe to ionization, like freezing point depression) from KCl. Comparing the calculated differences is a mental mastication, but if you just change one variable at a time, you can make some useful explanations or testable predictions.


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Can you achieve plasma charge?
This question might be too common, but I'm very curious to understand and find out if this is possible.
We see a lot of sci-fi movies and TV series where they have plasma guns, in reality is this something achievable ?
I'm very familiar with the concept of "anything is possible", but I have no idea about chemistry. All I know is what I read on wikipedia about Plasma or "Phase of Matter".
Can someone give me some clear explanations on what's possible and what not ?

Can you achieve plasma charge ?
How dangerous is it ?
Has someone already achieved this but perhaps keeps it secret ?

Please don't criticise me, it's just a question :)

What is plasma charge?

@IvanNeretin Did you really just ask me this ?

By "plasma charge", do you mean to transfer electric charge with a plasma??

@DrMoishePippik Yes.. like a "plasma gun", https://en.wikipedia.org/wiki/Plasma_weapon

If you felt the need to write "Please don't criticise me, it's just a question" then you should make so much effort that you wouldn't  feel it instead of writing this comment. BTW you're making your question as unclear as possible - "achieve" plasma "charge"? Instead of "make a plasma rifle"? Not to say this has little to do with chemistry and a lot with physics and even more with engineering.

I did not "feel the need" of writing that, I simply added it because people around stack websites understand have a tendency of down voting and criticising instead of actually helping or trying to answer the question. And my question is pretty clear, no I do not and did not and will not ask "how to make a plasma rifle". That is unethical, impractical and not to mention illegal. I asked about exactly what the gentlemen below answered me. So stop wasting your time and my time, if you don't plan on answering the question, thanks.

No, your question is not clear. Science requires being specific, and goes to great lengths including technical 'jargon' to be precise. 'Plasma charge' is not a technical term that is recognized in physics or chemistry.

A plasma discharge certainly can transfer charge.
Plasma guns are commonly used to reduce static charge, e.g. when pulling plastic wrap from a roll or for cleaning dust from vinyl discs. Small ones are available from about US$100 and larger plasma guns for more.
Or you could make your own plasma gun, using a piezoelectric generator from a US$1 lighter. Rather than let a spark jump across the circuit, connect a terminal of the generator to a needle. Squeezing the crystal creates a plasma or ion wind, and in a dark room, you'd see a faint purple glow from the needle tip.
If by "plasma charge", you mean some fictional weapon, it would be possible but of limited use. In air, a plasma does not propagate very far, unless there is a powerful ionizing force, e.g. lightning or high-powered radio transmitter, ionizing the path. If you wanted a weapon, then the high-power radio would be more effective than plasma. There is an exception: an electron beam at a few megavolts in low-pressure air self-focuses because of ionization along its path and can propagate a few meters. Of course, in "space", i.e. at very low pressure, the plasma would disperse in a puff of gas and ions.
Next time you visit a doctor or dentist, you might see a plasma gun in action!

Amazing answer and information! Thank you. And as you can imagine I was relating to the fictional weapon, however I had no idea that "plasmas do not propagate very far". Thanks for all the links, I`m checking them all now :)

source: Stack Exchange

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

sort: votes


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transpiration of trees and the cooling of the air
The change from the liquid state of water to the water vapor requires a large latent heat to loosen the hydrogen bonds. The process of transpiration in the plant involves exactly this change of state. Here, the water evaporates from the lower leaf surface and the leaf gives up some of its heat. In this way, the leaf is prevented from overheating that would result from exposure to sunlight. Sweating in animals involves this same latent heat aspect of water.
If trees use transpiration to decrease their temperature, shouldn't the air around them  be warmer? The leaves are cooled as water evaporates, and the air should be warmer.
I always experience the opposite as I approach a forest. I'm not sure if the effect could be attributed to the reduced irradiation due to foliage.

I am not a botanist but I don't think trees use transpiration primarily to decrease their temperature. In general they don't regulate their temperature in an active way (they may use antifreeze compounds though), in that sense they are poikilotherms.

Related: https://chemistry.stackexchange.com/questions/112043/is-heat-just-a-change-in-temperature/112057#112057

You seem to misunderstand the thermodynamics of evaporation: "Evaporation is an endothermic process, in that heat is absorbed during evaporation." It appears you are confusing evaporative cooling with a heat pump. In a heat pump, the compression half of the cycle changes the vapor back to liquid, which does make things warmer.
Consider boiling water at sea level: Water liquid at 100°C becomes water vapor at 100°C, requiring ~540 cal/g energy to change state from liquid to vapor at the same temperature. At lower temperatures, as water evaporates, thermal energy is withdrawn from the liquid, cooling it below ambient temperature. "Swamp coolers" work on that principle.

The air at ground level in a forest will be shady - therefore considerably cooler than in full sun. That is probably the major cause/effect of the noticeable temperature change on entering a forest.
The leaves will be warmed by sunlight, cooled by evaporation. At the first bit of evaporation, the air could be warmed a little as the leaves are cooled a little. But it's not at all clear that evaporation of water could become the driving process and cool the leaves and warm the surrounding air to a noticeable degree.
You could wet a dishtowel and swing it around to see how much it warms the air. The towel will become noticeably cooler as water evaporates, but the air is so much larger a reservoir of energy that a change in its temperature will be dispersed over  a large volume and will be insensible. Perhaps the effect could be measured in a confined volume with no stirring and very sensitive thermometers.

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


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Decomposition of total Interaction energy with SAPT or EDA-NOCV
I want to perform energy decomposition analysis for some Hydrogen bonded complexes. I came to know that ORCA and Psi4 allows EDA-NOCV an SAPT(0,2,3) respectively. But I do not know which one would be better, and how to perform them.

Although an answer to this question would be opinion-based, I recommend an article that does both techniques and compares them. Hint: numbers are different but conclusions are rather consistent.

Although this is a personal opinion, SAPT0, especially F-SAPT(0) seems like a powerful tool for performing EDA and decomposing the energies for different functional groups.

@user3786990 could you share link?

@PrasantaBandyopadhyay https://pubs.acs.org/doi/abs/10.1021/acs.jctc.7b01053 is one example.

SAPT(O) is an amazing tool for partitioning the interaction energies, it yields meaningful values and data. ETS-NOCV is also very good for partitioning the interaction energy, but it does it in a much different way, it yields values that correspond more to the orbital interactions, and more so the sigma and pi orbital mixing (although orca will not partition more than just the overall orbital mixing). ETS-NOCV is really good for giving qualitative orbital pictures for electron donation and back-donation. For hydrogen bonding, i'd suggest SAPT


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Calculating pH of a saturated weak acid/base solution
How can you calculate the $\mathrm{pH}$ if you know for example the $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{b}$ value of a weak acid/base and how much acid/base can dilute in water. Let's take an example:
Quinoline has $\mathrm{p}K_\mathrm{a} = 4.5$ and $\pu{0.6 g}/\pu{100 ml}$ can dilute in water. I can easily calculate that the concentration of the saturated solution is $\pu{0.0465 M}$. So if the reaction is
$$
\ce{C9H7N + H2O -> C9H7NH+ + OH-},
$$
why $[\ce{OH-}]\neq \pu{0.0465 M}$? I know that I can get the correct answer by making an ICE-table and use that given $\mathrm{p}K_\mathrm{a}$ value but why is it necessary? It seems weird as after making the ICE-table it seems that all quinoline isn't diluted…

Quinoline with pKb=14-pKa= 9.5 is a very  weak base. Why there should be almost 0.05 M OH- like if it had been strong base ? pH =~ 14 - 0.5(pKb - log c)=9.25 + 0.5.log c.

Yeah I know that it's obviously wrong but if Quinoline dilutes to water, shouldn't there be OH-? Or if not, what are there if Quinoline is diluted?

There is OH-. It is always there, if there is water. But there are 2 chemical equilibriums. Dissociation of water and protonization of quinoline. Together with mass and charge balance, there is the set of equations for the set of variables.


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Size of atomic shells
Is the distance of some $n^{th}$ shell ($n$ $\neq$ outermost shell) from the nucleus of an atom different for different elements?
If so, then how much is the difference or how could we calculate it?

Yes, the size* of the $n$-th shell will differ between elements. The reason is that the nuclear charge will shrink each shell as it goes up. Additional electrons cannot fully compensate this effect because they do not perfectly shield the core.
The difference can be measured (see below) or calculated using quantum chemistry with flexible basis sets. Results from the latter can be validated against X-ray spectroscopy. However, even the Bohr model will yield different radii for the first shell when increasing the nuclear charge.
*The definition of the size is not trivial. One can use the van-der-Waals radius, ionic radius or determine the isosphere that contains e.g. 90% of the electron probability density as long as one is consistent about it.


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Qualitative determination of the presence of cations
In "Introduction to Analytical Chemistry" I have been asked to propose various tests and chemical methods to determine whether a sample contains a series of metallic cations.
These metallic cations are as follows: $\ce{Cu^{2+}}$, $\ce{Pt^{2+}}$, $\ce{Pt^{4+}}$, $\ce{Hg^{+}}$, $\ce{Fe^{3+}}$.
Obviously, I'm not asking you to do the exercise for me. But, if it is possible, I would like to be indicated some book or manual, if it is possible that it is in Internet, to be able to solve it correctly. Also, I hope they can give some indication of how they would solve it.

The red Jander-Blasius is your friend.

Please consult Vogel, A Textbook of Qualitative Inorganic Analysis. It is one of the classics in the field. Also think of atomic emission spectroscopy. Search these keywords and you would find an answer.

You can recognize Fe(III) by adding some thiocyanate ion to the solution, which must be at least a little acidic. If it contains Fe(III) ions, it will turn dark red. This test is specific to iron.
Copper(II) ion can be recognized by adding some potassium iodide to its solution, which must be at lest a little bit acidic. A brown precipitate should appear which is made of a mixture of copper(I) iodide and iodine I2. If you add an equal volume of an alcane like cyclohexane and stir, the iodine I2 will pass into the organic phase which yields a purple solution.
I don't know the properties of Pt ions.
Mercury(I)+ ion may be recognized by adding some HCl. It will produce a white precipitate of Hg2Cl2. The only ion producing the same precipitate is the silver ion Ag+.


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Why can't you have compressed oxygen and nitrogen in the same tank?
I just watched a movie about the Apollo missions. They wanted an oxygen/nitrogen atmosphere on the spacecraft. But they said it would be too heavy and expensive to use an oxygen tank and a nitrogen tank and also equipment to monitor and control the mix ratio, so they went with pure oxygen. Why couldn't they put oxygen and nitrogen together in one tank at the right ratio so then they would only need one valve to maintain pressure?
Thanks Karl. It's so obvious now that you mention it! Guess I'm too new here and it won't let me reply to your comment.

Because oxygen is consumed, and if you replace consumed oxygen with a mixture of oxygen and nitrogen, maintaining the atmospheric pressure, you end up with too little oxygen after a (rather short) while. A human consumes about 1 cubic meter of oxygen per day, and the volume of the these spacecraft was in the range of 12 cubic meters if I remember correctly.

Sorry, six cubic metres. https://en.wikipedia.org/wiki/Apollo_Lunar_Module

Do you mean compressed air? That exists.

I'm voting to close this question as off-topic because it's based on false premise.

Ever fill your tires at a gas station? What are the twp main component gases in tge tank through which the air passes on the way to your tires?


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What happened to the Klopman - Salem equation?
I am relatively new and interested in the field of theoretical chemistry.
I just stumbled across the Klopman - Salem equation (https://en.wikipedia.org/wiki/Klopman-Salem_equation).
In daily life, I only use DFT for bonding energies and I also do not find too many papers about the Klopman - Salem equation, why?
What are the advantages and disadvantages of this theory? Are there any theories, that improved this concept?

Although its importance from a "conceptual" point of view (it justifies both the frontier orbitals theory and the hard and soft acids and bases theory) this equation is very rarely used in theoretical calculations because it comes out from multiple approximations (i.e. rigid molecules, Hammond's postulate, no solvent contribution, no entropic contribution to the transition state) and its simplicity in terms of computational weight doesn't overcome its scarce accuracy.


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Is there a free, super simple app or website for IUPAC names from structures?
I am looking for a free, very simple tool for IUPAC names of organic molecules.
I teach high school chemistry, and it would be fantastic to have an app or web tool that students could use to draw molecules and verify the systematic names of branched alkanes and simple alcohols, aldehydes, ketones, possibly acids; simple ethers, esters, cycloalkanes would be nice but aren't strictly necessary.
I do have MarvinSketch, but I find it too complex for absolute beginners, both to install and to use.
Chemdraw JS does not work well on mobile.
Suggestions?

Meta dupe: Software to name compounds; Main dupes: Identify a molecule by structure; Resources for learning Chemistry.

@andselisk is there an app that does the opposite?

@MarkC Yes. For example, ChemDraw JS and MarvinSketch can both convert name to a structure.

@MarkC   In addition to the apps andselisk mentioned, you can also enter the name here:  https://opsin.ch.cam.ac.uk

A bit more intuitive to interact with may be Chemdoodle Webcomponents which is a JavaScript library set-up to integrate well within webpages and mobile devices like phone or tablet, too.  It comprises 2D/3D sketcher, a NMR predictor, and is freely available (open source GPL license).  Derived from Chemdoodle, there is a structure -> name utility speking multiple dialects (e.g., IUPAC / CA, Hantzsch nomenclature), too.
Example:

The interaction with Chemdoodle is nicely explained by Pritt Balagopal here on ChemSE.

Beautiful, but very low "maximum number of free transactions."

Possibly, but not necessarily you did not find yet one setting you up a webserver with a webpage along the tutorial (https://web.chemdoodle.com/installation/download/ and https://web.chemdoodle.com/tutorial/introduction/).  If your school already has a site license (e.g., Illinois, http://www.chem.illinois.edu/clcwebsite/ChemDoodle.html) than you may add running the program from a cloud (https://web.chemdoodle.com/docs/ichemlabs-cloud-services/) without additional charge, too.

That looks like it requires more IT knowledge than I have...

Disclaimer: I made this app.
You can try out this app:

iOS: https://apps.apple.com/en-us/app/nomenklaturhelfer-neptun/id1570934959
Android: https://play.google.com/store/apps/details?id=de.pschwind.planetary.nomenclature

Pro side:

Free molecule naming for alkanes & alcohols in free draw mode (bottom right)
If you want, also some "levels" and tutorials on how to name alkanes

Contra side:

Nomenclature of alkenes, alkynes and some other functional groups is not free
No naming of cycloalkanes and other more complex structures

ChemAxon Ltd. offers a web tool at https://chemicalize.com/#/drawing .
Usage of "Chemicalize" requires a ChemAxon account, which is free but requires a name and e-mail to create.
Chemicalize is very easy to learn and use.
IUPAC names, as well as other basic properties are generated for molecules including up to 12 heavy atoms.

The Chemicalize application requires registration with an institution and has a great disadvantage for free usage: the drawn molecules must contain 12 or less non-hydrogen atoms. We must to pay for molecules that contain more than 12 non-hydrogen atoms.

Very intuitive app, easy and free app for PC is ChemSketch, you can download it, I use Windows version and it is fantastic there are short videos (on YouTube) on how to use it. I will explain few things but it is easier to watch a video.
To get a IUPAC name for a certain molecule you just draw the molecule, select that molecule and click Tools and then click Generate Name.
You can easily draw and use whole periodic table within the vertical tool bar. You can draw double and triple bond just by double/triple click on the single bond. When it comes to selecting, 3D models and many other options also very easy, it can even tell you certain things about isomers...
I am really amazed by the app!! The only thing that is negative about the app is that I am in university and I work with molecules larger than 50 atoms. You get a warning that you can not generate name for molecules that have more than 50 atoms, it says how to solve it but I think you might have to upgrade then and pay for the app.


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no positive $+$ or $-$ or negative label for $\pi$ orbitals
I want to ask a question about the symmetry for $\pi$ orbitals.
We were learning today about the deduction of an MO diagram for a typical $\ce{M-M}$ system with $\ce{D_{\infty h}}$ and I was confused on the labelling of the $\pi$ orbitals.
When labelling the $\ce{dz^2}$ orbitals, the following checklist was produced:

orbitals along the bond $\rightarrow$ $\sigma$
symmetry change through $i$ $\rightarrow$ g
no phase change for $\sigma_{v}$ $\rightarrow$ +
overall symmetry $\sigma_{g}^+$

but when we were labelling the $\ce{d_{yz}}$ and $\ce{d_{xz}}$ orbitals, the following information was given, without further explanation:

no $+$ or $-$ label for $\pi$ orbitals

I worked out that the $+$ or $-$ relates to how the phase changes after reflection in the $\sigma _{v}$ but still couldn't work out why I can't apply the signs to the $\pi$ MOs.
Why is there no $+$ or $-$ notation for these $\pi$ MO's?

Pick the yz plane as $\sigma_v$. What happens to the phase of $d_{yz}$ when $\sigma_v$ is applied? What happens to $d_{xz}$?

@Tyberius for both I think there is no change in phase. If I am correct, can't I just write $+$ anyway?

You are right for the $d_{yz}$, since the mirror plane is in the same plane as the orbital. But for $d_{xz}$ pair, the mirror plane bisects each orbital, flipping the lobes around the z-axis. So each of the π orbitals have different phase behavior, so in total it can't be + or -. The same is true for any of the nonsigma orbitals, as can be seen in the $D_{\infty h}$ character table.


text:
Ponderomotive Effect on Ionic Liquids
Would an alternating electric field cause net motion of ions in a liquid in the case of the electrodes being placed outside the liquid?
Generally speaking, what happens if you apply an external electric field to an ionic liquid, taking into account that there are no electrodes places inside the liquid itself, but are separate from the liquid container in the same manner you have a parallel plate capacitor.

Yes, due to electrostatic coupling. The conductors and the liquid would form a capacitor, causing back-and forth motion of the ions (and in polar molecules). This is an effective way to heat something without contact, as in dielectric heating.
This capacitance enables liquid crystal displays and E-Ink to display images using little or no current when not actively switching.


text:
Is there an uncertainty associated with the value 0 K for absolute zero?
When I say absolute zero, I’m not talking about the hypothetical temperature 0 K; I’m talking about the temperature at which a thermodynamic system has the lowest energy.
Everywhere I look, sites maintain that this should be –273.15 ºC, but I can’t find any uncertainty quoted with this prediction. Could someone shed some light on this?
To add a comment that I left on theorist’s answer:
I know that the Celsius scale is defined to be an easy conversion from the kelvin scale. What I mean to ask is the following: The Boltzmann constant $k_{\rm B}$ has a fixed value; the reference temperature $273.15\;\rm K$ is also a fixed value used in the relation $t/\mathrm{°C}=T/\mathrm{K}-273.15$. Therefore, how can we know that a thermodynamic system has the lowest energy at exactly $0\;\rm K$? If the triple point of water $T_{\rm TPW}=273.16\;\rm K$ has a standard relative uncertainty of $3.7\times10^{-7}$, then why shouldn’t $T_{\rm A0}$, especially since it has never been achieved? Something has to give and not be by definition, so to speak.

As another answer explains, absolute zero is defined in the Kelvin temperature scale as precisely $\pu{0 K}$. But this is not the entire story, as all measurements have an associated uncertainty (all thermometers are imprecise), see e.g. the wikipedia article describing the International Temperature Scale of 1990 and the Provisional Low Temperature Scale of 2000. For instance the second wikipedia article links to a International Committee for Weights and Measures' document describing the uncertainties associated with the calibration of the temperature scale based on measurements of the melting pressure of $\ce{^3He}$.
Recently the SI value of the Kelvin was changed. This is explained in the wikipedia:

The definition of the kelvin underwent a fundamental change. Rather than using the triple point of water to fix the temperature scale, the new definition uses the energy equivalent as given by Boltzmann's equation.
Previous definition: The kelvin, unit of thermodynamic temperature, is 1/273.16 of the thermodynamic temperature of the triple point of water.
2019 definition: The kelvin, symbol K, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $1.380649×10^{−23}$ when expressed in the unit $\pu{J⋅K−1}$, which is equal to $\pu{kg⋅m2⋅s−2⋅K−1}$, where the kilogram, metre and second are defined in terms of h, c and $\Delta ν_{Cs}$.
The kelvin may be expressed directly in terms of the defining constants as:
$$\pu{1 K} =  \frac{1.380 649 \times 10^{23}}{(6.626 070 15 \times 10^{-34})(9 192 631 770)} \frac{h \Delta \nu _{Cs}}{k}$$

So what does this mean for the definition of $\pu{0 K}$ ? That doesn't change, that is an absolute, thermodynamic definition. However, evidently the triple point of water is now no longer exactly $\pu{273.16 K}$. You can think of this as shifting the uncertainty from the size of a step to the number of steps that take you from absolute zero to the triple point. But both before and after the redefinition, $\pu{0 K}$ has meant $\pu{0\pm  0 K}$, that hasn't changed.

$^\ast$you can't actually reach absolute zero, so you can only extrapolate to 0 K and provide an uncertainty for this extrapolation, i.e. you could estimate how far your measured temperature is from absolute zero, and provide uncertainties associated with your estimate.

Our temperature scale is fixed around the reference temperature $T_0=273.15;\rm K$ and a change of one kelvin of temperature results in exactly a change of $1.380,648\times10^{-23};\rm J$ of thermal energy according to the International Bureau of Weights and Measures.

It's all well explained in International Committee for Weights and Measures documentation such as:  https://www.bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9-EN.pdf

@gen-zreadytoperish That's not true, as Buck Thorn quoted. Kelvin is a unit of absolute temperature - zero means zero, just like zero meters means no distance and zero seconds means no time. There's no uncertainty, it's not a result of measurement, it's zero by definition (regardless if it's possible to reach or measure e.g. zero distance). The "reference temperature" you're talking about is the uncertain part: "At the time of adopting the present definition TTPW was equal to 273.16 K with a relative standard uncertainty of 3.7E-7 based on measurements of k made prior to the redefinition".

@gen-zreadytoperish So, the zero value is fixed. The Boltzmann costant is fixed. The thing that changes with improved measurements and what-not is the triple point of water. If we suddenly realized we were all stupid and the triple point of water is actually 283.16 K (oops!), that's the only thing that changes - not the Kelvin scale, not the zero-point, not the Boltzmann constant.

This might also be helpful: https://en.wikipedia.org/wiki/Boltzmann_constant; also, look up the ideal gas thermometer

There is no uncertainty in this value, because the Celsius scale is defined in terms of absolute zero, rather than visa-versa.   I.e, the Celsius scale is defined by specifying that -273.15 C (exactly) corresponds to absolute zero, and +0.1 C (exactly) corresponds to the triple point of Vienna Standard Mean Ocean Water (VSMOW).  This locks in both the size (i.e., the spacing) of Celsius degrees (each Celsius degree is 1/273.16 of the way between absolute zero and the t.p. of VSMOW), and the offset between the Celsius and Kelvin temperature scales.
N.B.: This definition has been changed as of 2019 (see other answer).

Well, the water triple point is at +0.01 Deg C, resp. at 273.16 K.

I know that the Celsius scale is defined to be an easy conversion from the kelvin scale. What I mean to ask is the following: The Boltzmann constant $k_{\rm B}$ has a fixed value; the reference temperature $273.15;\rm K$ is also a fixed value used in the relation $t/\mathrm{°C}=T/\mathrm{K}-273.15$. Therefore, how can we know that a thermodynamic system has the lowest energy at exactly $0;\rm K$? If the triple point of water $T_{\rm TPW}=273.16;\rm K$ has a standard relative uncertainty of $3.7\times10^{-7}$, then why shouldn’t $T_{\rm A0}$, especially since it has never been achieved?

It is the other way. The Kelvin scale has been defined to be an easy conversion from the Celsius scale. The same for the Rankin scale from the Fahrenheit scale.

@Poutnik Historically, it is the other way. But today, Kelvin is the fundamental scale. The scale is defined in terms of the Boltzmann constant, which is a fundamental constant in the SI system of units. But even then, from the very beginning, Kelvin was specifically designed to be an absolute temperature scale, while both Celsius and Fahrenheit are relative. Zero Kelvin was always zero by definition, "infinite cold".

@gen-zreadytoperish It's not the result of any measurement. The system at lowest thermodynamic energy has zero Kelvin by definition. That's the whole point of having an absolute temperature scale. The thing that changes is how that translates to Celsius or relative to the triple point of water or whatever. This was originally determined theoretically based on Kelvin's calculations; but regardless of what measurements happen, zero will always be zero, and one degree Kelvin will always be change of thermal energy of 1.380 649E−23 J, exactly, without uncertainties, by definition.

@Luaan I do not object. In fact, I have written something similar in my answer that I decided not to post or keep.

If you want to define your own temperature scale you need at least to fixed (no uncertainty) points. Celcius assigned 0 degrees to a mixture of ice and water at atmospheric pressure and 100 degrees to the boiling point of water. If I understand your question correctly you are asking how can one put a "thermometer" at absolute zero if we can never reach that temperature? The answer is we can't. As mentioned by Buck Thorn the definition of the Kelvin changed recently (on May 20th of 2019 to be precise) and is now based on the Boltzmann constant. Before this definition, the Kelvin temperature scale was based on several reference temperatures which covered the range from 0.9 mK to 1357.77 K. By comparing with these reference temperatures, the absolute temperature could be determined within a certain measurements uncertainty. The scale was defined such that it provides a 0 temperature that is as close as possible to the absolute zero. The triple point of water was defined in such a way that an increase of 1 Kelvin corresponds to an increase of 1 Celcius. This consistency is an important part of redefining base units.
Since the new SI base units came into effect earlier this year the Boltzmann constant was fixed. To this end the committee charged with defining the new SI base units asked the Committee on Data for Science and Technology (CODATA) to collect very precise measurements of the Boltzmann constant that rely on several measurement techniques. The value that was obtained was very precise, even more precise than the SI committee needed. Why? Because the committee wanted to make sure that in the new system the triple point of water was still 273.16 K. So they rounded off the experimental Boltzmann factor to a relative uncertainty of about 1e-6. As a consequence, the new scale is compatible with the old scale, but extends over a much wider range as it is now linked to a measurement of the thermal energy $E=k_\text{B}T$. Of course when you now measure the thermal energy of the triple point of water you get a measurement uncertainty. Note that in principle we could get rid of temperature and use a measurement of the thermal energy instead. In that sense the Boltzmann constant is similar to the Avogadro constant: it's a conversion factor that has mostly historical (and practical) reasons and is not a "true fundamental constant of nature".


text:
Why does heating an atom make it emit certain frequencies?
We're going over quantum basics in chemistry right now and I'm very confused.
Electrons can only accept in discreet quanta to move up an energy level, right? And they reflect other forms of light that don't supply energy in their specific quanta, right? And flame is just infrared electromagnetic radiation, right?
Then when heated by flame, why do the electrons move up to a next energy level? Shouldn't they need a certain wavelength to move up - one that isn't provided by the flame, given the emission for the atom is in the visible light range?
Thanks for any help.

Please note discrete vs discreet (https://www.dictionary.com/e/discreet-and-discrete/)

Flame is not infrared radiation. It is a luminous chemical reaction.

Heat energy is distributed and some of the energy might match the amount required to excite an electron
There are two parts to the answer.
Firstly, there is more to emission than electrons leaping to a new energy level. Molecules can be excited by making them move faster (kinetic energy), vibrate at a different frequency (vibrational energy), rotate faster (rotational energy) and so on. Vibrational energy is quantised (so are others but sometimes the levels are so close together that, in practice, we see a continuum) but the wavelengths of light that match changes are in the infra-red region of light.
One of the effects of putting heat energy into a system is to distribute that energy across all the possible excitations. Each will show a Boltzmann like distribution with some molecules having a lot and other having a little.
Secondly, if there is enough heat a small number of molecules might just, randomly, have enough to match an electronic transition. Perhaps two very fast-moving molecules bang into each other and that matches a specific electronic transition. This means that, even if the average energy from the heat source is too low to cause an excitation, there may be a small number of individual molecules that have enough energy.
The reason why heat sometimes causes even electronic emissions is because the heat energy is distributed statistically across many molecules and excitation modes. So in a large statistical sample some molecules will have the right energy to match the transition. Don't forget those statistics when thinking about what is happening at the molecular level.

Matt, I am afraid this reasoning that "so in a large statistical sample some molecules will have the right energy to match the transition." is incorrect. I think nobody $knows$ the exact mechanism of this transfer of energy to excite the atoms. The student is asking about atomic emission. It is rare to heat a molecule (in the gas phase) and cause it to show a spectrum. The question is how are we heating the atomized matter in the gas phase. If it is a flame, not only thermal excitation and collisions, but chemical reactions may also cause electronic excitation (e.g. $\ce{NH3-O2}$ flame).

Also, the temperature would determine how many lines we would see in a spectrum. In a plasma (inductively coupled plasma) certainly there is no chemical reaction but there is a crazy collision process of free atoms with ions and electrons. What is mechanism of this collisional heating? Again, perhaps nobody knows.

Absorbing photons is just one of ways atoms and their electrons, or molecules, can gain energy. Other ways are e.g mechanical collisions and chemical reactions.
Note that atoms do not reflect light they do not absorb, but let it pass through like if they were not there. If they had reflected it,  air would have been an opaque fog. A very small number of photons are dispersed by the Rayleigh scattering, what gives the sky it's blue colour.
A flame  contains mixture of

atoms
atomic ions
free electrons
molecules
radicals(molecules with unpaired electrons}
molecular ions ( and ion radicals )

Atoms and atomic ions do have spectral lines in their emission or absorption spectra, related to particular energies, frequencies and wavelengths ( somewhat broaden, mostly  by the Doppler effect ).
But molecules, radicals and molecular ions have rather broad, diffuse bands.
Photons in UV, VIS and near IR have energies typical for electron excitations/transitions.
Far IR and MW have energies needed to change molecular bond vibrations and bond/molecule rotations, as these are quantized as well.
High energetic collision and mainly chemics reactions frequently lead to a  molecule with excited electron. This energy gets dissipated either as mechanical energy passed to other molecules, or, it can be emitted.
Special cases are free electrons and solid matter in flames, that are not limited to particular frequencies at all.
Electrons can emit photons of any frequency (limited by electron energy ).
Solids ( smoke, large scale atomic structures ) have continuos emission spectrum.

Thanks for the answer! Could you explain a little more why near IR and UV photons cause electron transitions? Do these photons have the right quanta for the atoms they hit?;

Yes, they do. But particular wavelengths for particular species. Or wavelength bands.


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How to increase the conversion of a reaction between an ammonium salt and ethylenediamine?

I've been trying this reaction in my lab and there is the formation of the substituted ethylene diamine but along with that, I am obtaining some amount of RCH2N(CH3)2. The reaction is done under solvent-free condition using only excess of ethylenediamine. After the reaction, it is washed with water and the compound is extracted with DCM. Can anyone suggest me a method to ensure complete conversion of the ammonium salt? Increasing reaction time or temperature isn't showing any effect.

You are asking your nucleophile to attack the most hindered site. My suggestion would be alter the leaving species by using something more hindered than trimethylamine. Can you use, say, pyrrolidine to form the intial amine, quaternise with ethyl iodide then react with ethylene diamine?

@Waylander Thank you for responding. Basically what you are saying is that I've to use more sterically hindered amine. I can try that but the problem is the R group here is ferrocene which is already quite bulky. So, the formation of RCH2N(CH3)2 might be inevitable. It's a good suggestion though. I can try it.

Can you do this a slightly different way? If you take the ferrocene dimethylamine (your undesired side product), oxidise to the N-oxide and then acylate you could do a Polonovski rearrangement to give the aldehyde. Reductive amination of this with mono-BOC ethylenediamine would give a much cleaner reaction.


text:
Is there a way to extract the diamagnetic and paramagentic components from NMR output from Gaussian16?
I'd like to be able to split the NMR value in Gaussian16 into its diamagnetic and paramagnetic terms. I'm coming over from ADF, where this can be done easily. Is there a straightforward command to give Gaussian16?
Thanks!

Please include an example input file, so that contributors have a chance to work on what you are working on; and maybe explain with this example what you expect to get.


text:
Reaction mechanism with benzylchloride (KMnO4 and Soda lime)

So I know that $\ce{KMnO4}$ converts any side chain irrespective of chain length to -$\ce{COOH}$ group and Soda lime converts that into the respective alkane. But in benzyl chloride, the alkyl group side chain also has a chlorine atom attached. How would the mechanism proceed?
It is given that Benzene is the product.

You have water present, benzyl chloride is a reactive species. Consider the possibility that the benzyl chloride hydrolyses to benzyl alcohol early in the reaction

@Waylander Ok, but then how would KMnO4 react with a side chain containing alcohol?

Alcohol to acid is a std permangate oxidation

@Waylander Ah i haven't gotten to that part i guess

This is not a magical reaction. Yet, it is a two step reaction as depicted in diagram below:

As @Waylander's comment elsewhere, benzyl chloride might have converted to benzyl alcohol before get oxidized to benzoic acid in acidic $\ce{KMnO4}$ solution (the first step). This oxidation is well known (see here). The second step is heating (may be $\gt \pu{200 ^\circ C}$) of solid benzoic acid in the presence of solid soda lime to cause the decarboxylation. This reaction is well described by Dr. Jim Clark (here) and also here.

Acid permanganate oxidation of alcohols is mentioned here  https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.06%3A_Oxidation_of_Alcohols

@Waylander: I included your site in my answer. Thank you for the insight! :-)


text:
How to know which reduction and oxidation reactions occurred?
We just had a lab experiment on galvanic cells and we used 0.5 M and 1 M iron(III) nitrate solution at the anode (iron electrode) and 1 M and 0.5 M copper(II) nitrate solution at the cathode (copper electrode) for runs 1 and 2, respectively.
How can I determine which redox reaction occurred? There are many possible reactions $(\ce{Cu^2+ + e- -> Cu+}$ or $\ce{Cu^2+ + 2e- -> Cu}).$ The experimental cell potential we got was 0.48 V and 0.44 V. Temperature that we recorded was 28 °C and 29 °C for runs 1 and 2.

Here, Fe(III) will be reduced to Fe(II), and Copper(0) will be oxidized to Cu(II).
The standard potential for Cu2+/Cu is +0.34 V, and for Fe3+/Fe2+ it is +0.77 V.
Nernst's law gives the potentiels for a 0.5 M solution. It is +0.34 - 0.009 V = 0.33 V pour the Copper electrode, and 0.77 - 0.018 = 0.75 V for the Fe3+/Fe2+ electrode.
With this choice, you get nearly your results.
It you had chosen another system of half-reactions, you will get results which are far away from your measurements.

Question, why did the Fe undergo reduction despite it being in the anode? I thought that in galvanic cells, anodes undergo reduction?


text:
Nitration of tri-substituted benzene with Acetic acid and HNO3

So I looked it up on the internet and it says that this is to be a nitration reaction. But how is the $\ce{NO2+}$ formed? I simply cant seem to be able to do the first step of the mechanism, that is the generation of the electrophile.
I thought that as $\ce{HNO3}$ would be the stronger acid, it would lose $\ce{H+}$ that would get added to the acetic acid and it would lose water to form a the electrophile that gets added to the ring. But apparently this is not the case. Why? Could someone show the mechanism?

NO2^+ may be formed by the equation : 2 HNO3 + AcOH --> NO2^+ + AcO- + H3O+ + NO3^-

Your reaction goes way back to 1930s when Ingold and coworkers have done pioneering work on  nitration of aromatic compounds using nitric acid in organic solvents in place of sulfuric acid (vide infra). Let's see how this has been developed:
Historical perspectives: The earliest recorded experiments on the kinetics of aromatic nitration has been performed in 1884, when the nitration of benzene and some of its simpler derivatives with various amounts of nitric acid have been examined at various temperatures (Ref.1). No solvent has been used (just nitric acid), and thus, the reaction mixtures were heterogeneous. The first nitration of aromatic substances with nitric acid in sulfuric acid was obtained in 1904, when the velocity constants in the nitration was studied (Ref.1). In their punlication in 1938, Ingold and coworkers have reported the kinetics of nitration using nitric acid in organic solvents (other than sulfuric acid), specifically nitromethane as the solvent (Ref.1). In this studies with nitromethane as the solvent and nitric acid in constant excess, they demonstrated zeroth order kinetics for the nitration of benzene and activated aromatic nucleous such as toluene. They have observed that, when less reactive compounds such as chlorobenzene were used, there were some dependence of the rate on the compound. Hence, they have continued these experiments with considerably less reactive compounds such as 1,4-dichlorobenzene and ethyl benzoate to show that less reactive compounds have established first-order kinetics on nitration.
Originally, Benford and Ingold thought that their results were peculiar to nitromethane, yet they have observed ail the same phenomena with acetic acid as solvent (your case!). Based on these data, Ingold and coworkers proposed the formation of nitronium ion ($\ce{NO2+}$), which may be the rate-determining if it is sufficiently rapidly removed by the aromatic compound used as the substrate. They proposed the actual mechanism of the formation of nitronium ion as follows (Ref.2):
$$\text{In sulfuric acid:} \: \: \ce{HNO3 + H2SO4 <=> H2NO3+ + HSO4-}\\
\text{and} \: \: \ce{H2NO3+ + H2SO4 <=> NO2+ + H3O+ + HSO4-}$$
$$\text{In organic solvents:} \: \: \ce{2 HNO3 <=> H2NO3+ + NO3-}\\
\text{and} \: \: \ce{H2NO3+ <=> NO2+ + H2O}$$
This mechanism is well established now. Based on these fact, I hope I answered your question. In their further work, Ingold and coworkers have proposed the mechanism for nitration of aromatics based on the nitronium ion formation (Ref.3). In that publication, they proposed existence of nitronium ion through rapid equilibrium with concentrated nitric acid, and that mechanism is believed to be the case to date.
References:

Gilbert A. Benford, Christopher K. Ingold, “177. Influence of directing groups on nuclear reactivity in oriented aromatic substitutions. Part V. A dilatometric investigation of the kinetics of aromatic nitration especially in nitromethane solution,” Journal of the Chemical Society 1938, 929-955 (DOI: 10.1039/JR9380000929).
E. D. Hughes, C. K. Ingold, R. I. Reed, “Kinetics of Aromatic Nitration: the Nitronium Ion,” Nature 1946, 158(4013), 448-449 (https://doi.org/10.1038/158448c0).
R. J. Gillespie, E. D. Hughes, C. K. Ingold, D. J. Millen, R. I. Reed, “Kinetics and mechanism of aromatic nitration,” Nature 1949, 163(4146), 599-600 (https://doi.org/10.1038/163599b0).


text:
Relation between the oxidation state of central atom and its coordination number in a complex compound?
We know for a complex say $\ce{[Co(NH3)3Cl3]}$ we have $\ce{Co^{3+}}$  and its coordination number here is 6. And for this if we know either oxidation state or coordination number then other is fixed. So is there a way to figure out relation between them and for other compounds like $\ce{[Ni(CO)4]}$, $\ce{K2[PtCl4]}$, and $\ce{[Pt(NH3)2Cl3]}$.

Your premise " if we know either oxidation state or coordination number then other is fixed" is completely not true.

The hexaaquacobalt(II) complex $\ce{[Co(H2O)6]^2+}$ is octahedral.
The tetrachloridocobaltate(II) complex $\ce{[CoCl4]^2-}$ is tetrahedral.
knowing the oxidation state does not allow prediction of the coordination number or vice-versa.

How [Co(H2O)6]2+ is possible, considering the outer electronic configuration of cobalt (4s2)(3d7)(4p0) if it is in its +2 oxidation state i.e. (4s0)(3d7)(4p0) it's not possible for it to accommodate 6 ligands and thus to have coordination number of 6

@PrashantSingh Of course it is possible, it is known, isolated and characterised. $\mathrm d^7$ is a perfectly fine electronic configuration of a transition metal. Indeed, you can find all eleven possible configurations from $\mathrm d^0$ to $\mathrm d^{10}$.

There is no relation between the oxidation state of the central atom and its coordination number. The complex compound you just mentioned is a good example. It was [Co(NH3)6]Cl3 and it contains Cobalt at the oxidation state +3. In water the corresponding ion will be formed : [Co(NH3)6]^3+. But a similar ion may be formed with Cobalt(II) and the same number of 6 NH3 molecules as ligands. It will yield the ion [Co(NH3)6]^2+ and the salt [Co(NH3)6]Cl2.

But considering the outer electronic configuration of cobalt (4s2)(3d7)(4p0) if it is in its +2 oxidation state i.e. (4s0)(3d7)(4p0) it's not possible for it to accommodate 6 ligands and thus to have coordination number of 6.


text:
Analytical Organic Chemistry - Do you agree with my answer to this C-13 NMR question?
Question 2b:

Because of the 184.7 signal, I think it's either an ester or carboxylic acid. There isn't a DEPT signal based on 184.7 because the carbon is not bonded to any hydrogens. Since there are some carbons not being shown, I figure there is probably some symmetry going on. From what I know, the DEPT (+) signs indicate the presence of either a CH or CH3.  Since it's kind of hard to make a symmetric ester, I settled on the molecule: methylpropanoic acid. The 2 methyl groups at the end of the chain are symmetric, being represented by 1 peak.
Do you agree with this reasoning?

First, this is a homework question. You should show your effort to solve this problem , thus, we can geet involve helping you. You have shown effort in the problem 2a (I believe since your values are not given), but part of that is completely wrong. For example, look at the reasoning given for no peaks shown in DEPT spectrum. Please show your honest effort and show reasonable reasoning. So far, I see at least no knowledge of DEPT in you!

I edited the post.

Good analysis. But it's just not methylpropanoic acid (yout naming is incorrect). It is 2-methylpropanoic acid.


text:
Will ground glass joints relieve pressure non-catastrophically?
I need to heat a fluid for seven days at a time. The fluid goes in round bottom flask that is suspended in a water bath. The test fluid will decompose and generate a small amount of gas over time.
Gas generation is variable and dependent on what material I am testing in this material compatibility experiment.
I have experience in plumbing and fluid systems but not chemistry. In a plumbing system I would add a relief valve to relieve the pressure generated over time, without allowing potentially dirty air to contaminate the experiment.
I have a ground glass joint. I found a source that says the ground glass plug will lift to relieve pressure, even if I put a plastic clip on it, and the plastic clip would ensure the plug would not go flying.
I still have concern since this is an environment I am not familiar with, and I am not keen on exploding glassware in the facility. Past experimenters have placed small beakers upside down on top of the flask, and have also used aluminum foil as a cover (I will not do this so the vapor contact with foil does not affect the experiment results). I tried it with a beaker but a moderate amount of fluid condensed on the surface, which was not ideal.
Will I be fine securing a ground glass joint with a plastic clip, should I use an upside down beaker, or is there a more obvious solution?

What about sticking a balloon on an outlet of your setup, allowing the gas to expand into it?

Made clear if you want to protection the material inside or if you are interested in collecting gss. In the first case simply let the system to atmospheric P by connecting a simple trap with cotton or calcium chloride. Beware calcium chloride can clog on long run. In line with Tgr_Vinz answer. Otherwise DrMoishe Pippik

What gas is generated? If it's nothing corrosive etc. then the balloon suggestion is a good one. Place a rubber septum into the ground glass joint to seal it, and pierce it with a wide-bore needle attached to a balloon filled with an inert gas (N2 or Ar).

I am testing hydrogen peroxide for compatibility with various materials, so there will be gaseous oxygen, water vapor, and hydrogen peroxide vapor. If I use balloons, the hydrogen peroxide vapor will react with it and show a higher rate of decomposition than reality. Going to think on the reflux condenser a bit but that seems like a good solution. Not sure if I am allowed to link here, but "Long Term Storability of Hydrogen Peroxide" pages 3-4 (there is a legal pre-published copy online) describe the test, although I am trying to improve it.

A balloon can be connected to the flask with a long, thin polyethylene tube, say ~1 mm diameter (you can draw it out with gentle flame heating). Pressure can be relieved by slow gas diffusion rather than a free-flowing connection between the flask and the balloon.

If I didn't miss any point, you want to make sure that:
-the system is not in direct contact with the environment, in order to keep it clean
-pressure doesn't build up dangerously
-the non-gaseous product is not lost
What about a reflux condenser? Even unstoppered, it makes loss of material very unlikely. If you are concerned about contamination, you can add to it an inert atmosphere or a tube packed with cotton, according to your needs

I think this is a good solution for one of the two steps of the test process.
For the other step, I have some 50 mL volumetric flasks on a hot plate with temperature feedback. They have a 12/21 joint. I have to heat the 90% hydrogen peroxide to 100C for 24 hours, so water bath was not a good option, various oils were fire hazards, etc. Open to suggestions on other setups, but what is the proper chemistry tool to act as a mini-condenser in a joint that size?

Ground-glass joints are notorious for seizing at the wrong time, particularly if reactants are alkaline.
If pressure might change, I'd add some relief mechanism, such as a water-trap or rupture disk. Without knowing your setup, something as simple as a balloon might work, if not attacked by reactants.

I routinely use a positive pressure of argon with a balloon in an otherwise sealed reaction setup where all joints are (greased) ground glass joints, potentially secured with clips. It is rather rare for them to give in to the balloon pressure but at least once the weakest link popped open (a single stopper that was not secured properly). Depending on your setup, the weakest link may be one that gives way non-catastrophically or it may require enough force to turn catastrophic—it is not something I would want to bet on if I have a precious experiment running. Especially since heating a fluid to me almost automatically implies reflux condensor.
If exchange of air with the outside atmosphere is not a problem, your options include:

any glass tube with a joint and a $90^\circ$ angle that you can put on top to prevent contaminants falling down from above
a classical drying tube, filled e.g. with calcium chloride to prevent significant water drawing
a plastic stopper with a large disc-like top that is one size smaller than the ground joint

If gas exchange is not desired (i.e. you have some kind of inert gas in your vessel) then I would opt for one of many possible relief valves

the type commonly used for alcoholic fermentation, filled with a suitable liquid
special pressure-relief glassware that fits on joints exists
a balloon can help relieve the pressure but it has limits

Neither list is complete and with a little MacGyver skill I am sure you can build an appropriate device with whatever you have in your laboratory.

If I understand the set up and purpose, just use a rubber stopper with a needle (any) through it.
I do expect that nothing you put in hydrogen peroxide shall react violently, as for you are doing compatibility tests.


text:
Is it possible to create an arbitrary, stable, unique chemical fingerprint?
Context: I want to be able to hash real-world objects
Is it possible to create a substance containing a combination of elements (or isotopes?) in a random and difficult to reproduce (but simple to verify) combination?
For example if the "fingerprint" was 69% X and 31% Y this information can be easily hashed but it should be very difficult to reproduce with exactly that combination. This substance would then be combined with the object in an unobtrusive way and it should be simple to verify that that object has that exact "fingerprint".
In my mind using DNA/RNA in some way seems like a good solution but I believe it's not very stable. I feel it would be useful only if it lasts at least 20-50 years or so.

There are a variety of approaches for high-reliability supply chain management that work under vaguely analogous processes. But I don't know the details.

Related:https://en.m.wikipedia.org/wiki/Molecular_memory

I believe that oil is being tagged with DNA. A small amount of DNA is added to the oil and the source of the oil can be identified if it's spilled.

See also https://en.wikipedia.org/wiki/Taggant#Explosive_taggants

I don't think DNA is added to oil, it won't be soluble. It may naturally have some. Oil has a complex chemical signature which is being studied by gas chromatography, liquid chromatography and very advanced mass spectrometry techniques.

I think one of the points with hashing is simplicity. You could compare the original documents line by line but hashing is just much faster (although perhaps I am thinking more about checksums and the like). A lot of substances can be tested for origin by obtaining a composition profile. This is obviously not a unique identifier, in the way that checksum is. A composition profile is necessary, not sufficient, to show provenance. But simply spiking with a combination of stable molecules as you suggest might be good enough to ensure a unique fingerprint.


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Relative Energies between LGOs and 2s orbital of oxygen in H2O MO Diagram

In this MO diagram, we find that the LGOs of the two premixed H1s orbitals are significantly higher in energy than oxygen's 2s orbital. Although oxygen's $Z_{eff}$ is considerably higher, is it really enough to make it so that the 2s orbital of oxygen is lower than the in-phase combination of two 1s orbitals?

One of those oxygen p orbitals should not remain at the same level. There is one which points to the middle between the two hydrogen atoms and which thus obviously forms a bonding interaction, much stronger than that of the 2s orbital.

You probably already understand that the oxygen 2s orbital is lower in energy than the hydrogen 1s to begin with. For the unconvinced reader, consider the fact that the ionisation energies of oxygen and hydrogen (1314 and 1312 kJ/mol respectively) are quite similar, which implies that O2p and H1s have (somewhat) similar energies.
The question is then about how much lower in energy the H1s + H1s bonding combination is, relative to the H1s orbital itself. However, the hydrogen atoms are quite far apart: using the H–O–H bond angle (104.45°) and H–O bond length (95.84 pm) quoted on Wikipedia, the H–H distance can be calculated as being 151.5 pm, over twice as long as in dihydrogen itself (74 pm). This means a very poor overlap between the 1s orbitals, so you should not have an in-phase combination that is significantly lower in energy than the original hydrogen 1s orbital.
For the same reason, the "antibonding" LGO should also have roughly the same energy as the original 1s orbitals. The only reason why we are looking at bonding & antibonding LGOs is because of symmetry; it doesn't have anything to do whether they are actually bonding or antibonding to an appreciable extent.


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Bond length comparison in substituted phosphorus pentahalide
What would be the comparison between $\ce{P-F}$ and $\ce{P-Cl}$ bond length in  phosphorus tetrafluoride chloride $\ce{PF4Cl}$ and phosphorus trifluoride dichloride $\ce{PF3Cl2}$ in the equitorial hybrid orbital arrangements?

While determining the bond length between the two compounds I am using Bent's rule, since in the case of $\ce{PF4Cl}$ there are two $\ce{F}$ atoms in equatorial region and thus they decrease their %s character in turn resulting in a larger bond length and since there are two $\ce{F}$ atoms here the %s would be relatively less as compared to the $\ce{P-F}$ bond in $\ce{PF3Cl2}$, so this decrease in %s character in flourine atom should result in increase of %s character in $\ce{P-Cl}$ bond and reducing the bond length and $\ce{P-F}$ bond length would be greater in $\ce{PF4Cl}$.

8. Bond lengths (pm) in substituted phosphorous pentahalides are given below:
$$
\begin{array}{lcc}
  & \ce{PF4Cl} & \ce{PF3Cl2} \\
\ce{P-F_{eq}} & x & y \\
\ce{P-Cl_{eq}} & a & b
\end{array}
$$
$$\text{(I)}~x > y \qquad \text{(II)}~y > x \qquad \text{(III)}~a > b \qquad \text{(IV)}~b > a$$
Choose correct code:
(a) Only I
  (b) II and IV
  (c) I and III
  (d) Only IV

So if I'm assuming things right  my answer about the $\ce{P-F}$ bond length is right, but here in the option for $\ce{P-Cl}$ bond length is that it's greater in $\ce{PF4Cl}$. Why? Can someone please explain?

Could you please cite the source of the problem you've posted a photo of (e.g. author(s), title, edition, publisher, year, page number(s) and ID(s), if it's a textbook)?

This question is from a book called V Joshi by cengage publications. It only mentions the correct option but no explanation for the question so I'm totally clueless what's happening here

It can be analysed based on Bent's rules. Basically P-F bond will have more p character and hence p-Cl more s character.

According to (1) and (2), the values in your table would be as follows (all values in Å):
$$\begin{array}{c c c}
& \ce{PF4Cl} & \ce{PF3Cl2} \\
\ce{P-F_{eq}} & 1.541 & 1.546 \\
\ce{P-Cl_{eq}} & 1.999 & 2.004 \\
\end{array}$$
we have $x < y$ and $a < b$, which means that (b) is the correct answer

According to Bent's rule:

atomic s character concentrates in orbitals directed towards electropositive substituents

Bonds with larger s-character are shorter in nature. Keeping these two points in mind and analyzing the structures, in $\ce{PF4Cl}$, almost all the s-character is directed towards $\ce{Cl}$, which means that $\ce{P-Cl}$ bond length in $\ce{PF4Cl}$ is less than that in $\ce{PF3Cl2}$. Similarily, considering that the $\mathrm{sp^3d}$ hybridization can be considered as $\mathrm{sp^2+pd}$, in the equatorial plane, the s-character shared between the two fluorine atoms in $\ce{PF4Cl}$ would be more than the s-character given to the single fluorine atom in $\ce{PF3Cl2}$, hence the shorter bond length.
Another perspective is that the above set of arguments are largely hand-waving, as Bent's rule is at best empirical and the bond length differences are $\pu{\approx 5 pm}$, which is very small. An easier question would be to compare bond lengths between $\ce{PCl5}$, $\ce{PF3Cl2}$ and $\ce{PF5}$, but for this question, I don't have a 'solid' explanation per se, other than the lengths from the references and the elaboration of Bent's rule above.
References:

Leiding, Jeff, et al. “Bonding in PF2Cl, PF3Cl, and PF4Cl: Insight into Isomerism and Apicophilicity from Ab Initio Calculations and the Recoupled Pair Bonding Model.” Theoretical Chemistry Accounts, vol. 133, no. 2, Feb. 2014, p. 1428. doi:10.1007/s00214-013-1428-7.
French, Richard J., et al. “Dichlorotrifluorophosphorane (PCl2F3): Molecular Structure by Gas-Phase Electron Diffraction and Quadratic Force Field.” Inorganic Chemistry, vol. 24, no. 18, Aug. 1985, pp. 2774–77. doi:10.1021/ic00212a014


text:
Determining the hydrogen peroxide content in acetic acid
For a project I have to determine the peroxide content in some samples, for which we plan to use glacial acetic acid. Unfortunately, I don't know the peroxide content of my acetic acid and because we need a high level of precision (0.1 ppm at least), this is important to know. So, I wanted to ask if you know how we could determine the peroxide content in our acetic acid.

When you ask for a precision of 0.1 ppm, I assume you actually mean the detection limit?
A very common method for quantitative analysis of peroxide content is iodometric titration: the peroxides react with potassium iodide forming free iodine which is subsequently titrated with thiosulfate. It is basically at straight forward method, which only requires none or little preparation of the samples. But it is usually performed manually, and it could be a problem to obtain a DL of 0.1 ppm.
More sensitive methods (HPLC/fluorescence/spectrophotometric) methods requires on the other hand standard-curves made from samples with known content.
With the risk of being self-promotive, I can refer you to a recent review I have written about quantitative analysis of peroxoacetic acid [1].
Maybe you can find some inspiration and references there?
References

Sode, F. Analytical Methods for Peroxo Acids – a Review. Anal. Methods 2019, 11 (27), 3372–3380. https://doi.org/10/ggbpt8. (ResearchGate)

why are you linking to Researchgate where one cannot actually read the article, rather than using the DOI to link to RSC?

Articles in RSC is behind a paywall and the first year after publication articles are under restriction to how they may be made publicly accessible by the author. But via Researchgate (or similar fora) I am permitted to send copies to anyone who asks.

Thank you @andselisk for the advice. I have edited my answer.

@FrankS No prob, I took a liberty to edit it a little for better readability and kept previous RG URL for those who use it.


text:
Explanation for the reaction of gold and iodine(iii) solution
The Experiment
A few days ago I saw a video of dissolving gold in the condition of $\ce{ICl3/HCl(aq)}$. In the video, $\ce{ICl3(s)}$ is added to a test tube, followed by adding conc. hydrochloric acid. The solid is dissolved in the solution and the solution turned yellow. No green color was shown.
A gold wire is then placed in the solution for a short period of time. Putting the wire on the starch indicator produces a color of blue.
As a comparison, a piece of titanium was placed in the solution and then indicated by starch. No blue was produced.
The Speculation
After adding $\ce{HCl(aq)}$ to $\ce{ICl3(g)}$, I can think of 3 types of reactions may occur.
1) Lewis acid-base reaction.$$\ce{ICl3(g) +Cl-(aq)<->ICl4-(aq)}$$ As I'm not sure whether $\ce{ICl3}$ is a strong Lewis acid and whether $\ce{HICl4}$ is a strong acid (I suggest it is), I don't know it may present as a molecule in the solution.
ii) Disproportionation. As no purple color was seen, which leads to no iodine was produced.
iii) Hydration. $$\ce{ICl4-(aq) +5H2O(l)<->HIO2(aq) +4Cl-(aq) +3H3O+(aq)}$$
This reaction is unlikely to occur as $\ce{HIO2(aq)}$ is unstable and will disproportionate but no disproportionation production could be observed.
The OP of the video claimed that the equation of the reaction of gold and the solution is $$\ce{2HICl4 +2Au->2HAuCl4 +I2}$$

Questions
  a) Does the solution of iodine(iii) really contains much $\ce{ICl4-}$, and it is the cause of oxidation of the gold wire?
  b) Does the equation claimed above really occur?
  c) Is there any paper that investigates the electrode potential of $\ce{ICl4-/I2}$? Hence we can support the correctness of the equation thermodynamically.

That seems to be OK, at least approximately.


text:
Azeotropic Distillation - Am I interpreting this VLE diagram correctly?
I've been looking at ways to re-concentrate a formic acid-water binary mixture with the VLE diagram below.

Am I correct in saying that to concentrate a solution of say , 40% formic acid by weight I could (excluding all recycle streams for simplicity) implement a distillation column at 1 bar to produce a distillate at the azeotrope (78%) and then pass this distillate into a second column operating at 0.1 bar to distill further to 95%?
Aditionally, would operating the 1bar column for a distillate of say only 70% (past the 0.1 bar azeotrope) then performing the remaining concentration in the low pressure column result in any significant energy savings due to less steam required at the lower pressure?
Lastly, if I was to START with a solution of 85% formic acid, already beyond both azeotropes, I could just perform simple distillation in a single column at either pressure point to obtain the 95% product?


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Why do bubbles form when a reactive metal is placed in a less reactive metal's salt solution?
What is the gas that if formed when a reactive metal is placed in a salt solution of a less reactive metal? Theoretically, it should just be a simple displacement.
I know that this question has been asked on a different page, but there Hydrogen formation was the accepted answer, which I don't think is the case (see observations).
Observations:
When a strip of magnesium ribbon was placed in a concentrated copper sulfate solution, the magnesium ribbon piece started bubbling.
Magnesium didn't bubble as violently when placed in plain water of the same temperature.
The gas was bubbled through a soap solution and the bubble was passed near a candle flame. It just popped and nothing happened.
Pure hydrogen gas was obtained by placing a Mg ribbon in acid, and bubbles were obtained. These exploded with a bang when passed over a flame (learnt the hard way not to do this!).
Regular distilled water was used everywhere.
Thanks a lot in advance for any help.

When you dip magnesium ribbon into a copper sulfate solution, theoretically (and realistically) you do get a simple replacement reaction. Then reality sets in: you have a metallic anode (magnesium) with little cathodes (copper) all over it. At that point, the magnesium just overreacts. Well, it reacts faster.
The size of the bubbles will cause some difference in the popping ability of the bubbles. Bubbles could pop differently if some were pure H2 and others were mixed with air.

The bubbles are still hydrogen.  Transition metal salts are generally much weaker acids than whatever you put your magnesium ribbon into for your hydrogen control.  So you get less gas for the candle flame to react with when you use the transition metal salt, which I am sure you saw.  That the decreased amount of gas still popped, which is basically a low-intensity bang, should be deemed a positive test for hydrogen.

The bubble just popped due to the heat of the flame. The gas in it did not burn in any sort of way. Additionally, there was as much, if not more gas evolving from the Mg/CuSO4 mixture than the Mg/Acid mixture (I used acetic acid). It isn't possible for the gas to be a mixture of H2 and something else, is it?

I don't see any chance of this being anything other than hydrogen. Popping individual bubbles can be deceptive. Try to collect a whole beaker (with proper safety precautions, that is).

Let me try collecting more gas. I'll post a video if I can. Is there any possibility of liberating any dissolved gases in the solution? I read this as an answer to another question...


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What does a gasoline fuel contain
I would like to ask you a quetsion about hydrocarbons and gasoline. If you mix a octane with ethanol, what kind of bonds would you gate as in how  will the atoms react. Can you in someway explain it as polar and nonpolar bond. I would so appriciate any answer.

If you mix ethanol and  octane, you'll get a mixture; there is no chemical reaction.

And just to be clear about the common misunderstanding: gasoline fuel doesn't contain much octane.

I'm not sure why this is getting so many downvotes. Seems like a reasonable question.

The most common constituents of gasoline are branched alkanes (25-40%), mostly having 5 or 6 carbons, and aromatics (20-50%) (mostly alkylbenzenes like 1,2-, 1,3-, and 1,4-xylene). Under standard conditions, there is no chemical reaction between these stable hydrocarbons and ethanol. It should probably be a more-or-less homogeneous mixture of these hydrocarbons and ethanol, with possibly some transient hydrogen bonding between the alcohol groups of ethanol.
However, there are some physical effects of the ethanol. The presence of ethanol in the gasoline allows more water (from moisture in the air) to dissolve in the gasoline. This is because the ethanol can form hydrogen bonds with water and is miscible with water. If too much water is in the tank, phase separation will occur leading to droplets of ethanol–water, which can cause corrosion of the gas tank and contribute to fuel line freezing, among other issues.


text:
Working principle of Fisher-Price Light Sketcher
I was just searching around in my store room and I found this Fisher-Price Light Sketcher. I am just amazed how the light from the pen draws on the board without even contact! I searched a lot but couldn't find a working principle of this thing. Probably the light emitted by the pen is black light. Can you tell me how it works? Thanks! The product: https://service.mattel.com/us/productDetail.aspx?prodno=L3244&siteid=27
The text mentions that Light from the child-safe pen reacts with special ink on the product drawing surface. How does this work? And is this type of light safe for human skin?

I could come back to this if no other user will answer. It is uv-excited fluorescence in the visible, perhaps followed by phosphorescence. Alternatively emission from a structural change, true photochromism. Sometimes the term photochromism is used even for the first standard mechanism, especially in solid matrixes and application like this. About safety, I think we can trust Mattel and governments enough.

Can you explain further please?

Are you familiar with fluorescence? The general principle is uv excites the molecules that after non radiative relaxations emit in the visible. Is not truly phitochromism, which should be reserved to colour changes related to the absorbing molecule adjusting its structure upon photoexcitation, thereby changing absorption spectrum. This is also possible but it should lead to a more permanent trace.

Can you write an answer?


text:
Why does Nitrogen Oxide NO exist?
The nitrogen oxide $\ce{NO}$ exists and has an uneven number of valence electrons. This is "impossible". Why does it exist? Why is it not dimerized?
Nitrogen has 5 valence electrons and oxygen 6. Whatever your skill, you cannot get a noble gas configuration for both atoms in a molecule like $\ce{NO}$ .
Of course, $\ce{NO}$ dimerizes at very low temperatures. But why only at low temperatures? $\ce{H}$ atoms and alkyl radicals dimerize easily at room temperatures, and even above. Why is it so hard to dimerize $\ce{NO}$? Why is the bond $\ce{N-N}$ in $\ce{N2O2}$ so weak?
Edit : The following assumption has been published : "This question already has answers here:
Why doesn't nitrogen monoxide dimerize? (3 answers)"
I regret to say that none of the three proposed references do answer the question : Why is the bond $\ce{N-N}$ so weak in $\ce{N2O2}$

https://chemistry.stackexchange.com/questions/32265/which-relatively-simple-molecules-violate-the-octet-rule https://chemistry.stackexchange.com/questions/116841/what-is-dot-sign-in-no

Oxygen basically has 2 unpaired electrons. Why doesn't it polymerize?

I changed the duplicate target to one that was more appropriate.


text:
Is there a hypervalent molecule which has no contribution of d-orbitals at all?
I thought that d-orbitals were used to "extend the octet rule" in these molecules until I looked it up on Google and saw that they actually contributed only a little to the molecule. A little, but not zero.

Is there a hypervalent period-3-element molecule that has mathematically been proved not to involve d-orbitals at all?
Do they have a role, any role, in the chemical properties of the final molecules?

Answer to question 1: yes, pretty much all of them.

@ regulars: I wonder if this can serve as a canonical d-orbital (non)-participation dupe target =D

Partial answer to 2: They have an effect on the polarisation of the electron density. ||  This could well be a canonical question; @Jan you're a regular, I'm counting on you ;)

@Jan really? Because one Chemistry.SE post I saw (Google the words "d-orbital SO3", quote marks and all) said that d-orbitals contributed to 1% of the mathematical hybrid, 1% but still not 0%, in sulfur trioxide...

The explanation for this is that these are most likely polarisation functions, which do not correspond to actual orbitals (even in the MO framework not). If you would remove those functions when you attempt to solve for the wave function, you will still get a very, very good approximation. And since orbitals are a mere mathematical concept, you can include or exclude them at will; we don't know what is right.

@Martin-マーチン I fear others are more qualified to write a good answer … ^^'

The conclusion is left to the reader. I believe that there is no useful value in involving d-orbitals in the description of "hypervalent" molecules while remaining on the level of the octet rule.

A typical example for when the octet rule seems to fail is sulfur hexafluoride.
The naive approach
Let's do a quantum chemistry calculation. Using a PBE-D3/def2-TZVP geometry for $\ce{SF6}$, we will take a look at the orbital coefficients of the basis functions centered on $\ce{S}$ from a HF/def2-SVP single point calculation:
                11        12        13        14        15        16
             -1.81184  -1.68415  -1.68415  -1.68415  -1.63226  -1.63226
              2.00000   2.00000   2.00000   2.00000   2.00000   2.00000
             --------  --------  --------  --------  --------  --------
0S   1s      0.039494 -0.000000 -0.000000 -0.000000  0.000000 -0.000000
0S   2s      0.243289 -0.000000 -0.000000 -0.000000  0.000000 -0.000000
0S   3s     -0.378889  0.000000  0.000000  0.000000 -0.000000  0.000000
0S   4s      0.060237 -0.000000 -0.000000 -0.000000  0.000000 -0.000000
0S   1pz    -0.000000 -0.083520 -0.002820 -0.008492 -0.000000  0.000000
0S   1px     0.000000 -0.008939  0.022821  0.080342 -0.000000  0.000000
0S   1py    -0.000000  0.000390 -0.080789  0.022991  0.000000  0.000000
0S   2pz    -0.000000 -0.137305 -0.004635 -0.013960 -0.000000  0.000000
0S   2px     0.000000 -0.014696  0.037518  0.132082 -0.000000  0.000000
0S   2py    -0.000000  0.000641 -0.132816  0.037798  0.000000 -0.000000
0S   3pz     0.000000  0.015799  0.000533  0.001606  0.000000 -0.000000
0S   3px    -0.000000  0.001691 -0.004317 -0.015198  0.000000 -0.000000
0S   3py    -0.000000 -0.000074  0.015282 -0.004349 -0.000000  0.000000
0S   1dz2   -0.000000 -0.000000  0.000000  0.000000  0.091291 -0.003993
0S   1dxz   -0.000000 -0.000000 -0.000000 -0.000000  0.000000  0.000000
0S   1dyz    0.000000  0.000000 -0.000000  0.000000  0.000000  0.000000
0S   1dx2y2  0.000000  0.000000 -0.000000 -0.000000  0.003993  0.091291
0S   1dxy    0.000000 -0.000000  0.000000  0.000000  0.000000  0.000000

Alright, on MO 15 there is a d-orbital coefficient of 0.09. That's a d-orbital contribution. End of story.
Well, no. Let's do a few more calculations. Same as above, but let's dispense with the d-functions on $\ce{S}$ (HF/def2-SVP-d).
                11        12        13        14        15        16
             -1.88534  -1.73860  -1.73860  -1.73860  -1.67047  -1.67047
              2.00000   2.00000   2.00000   2.00000   2.00000   2.00000
             --------  --------  --------  --------  --------  --------
0S   1s      0.043781  0.000000  0.000000 -0.000000 -0.000000  0.000000
0S   2s      0.271803  0.000000  0.000000 -0.000000 -0.000000  0.000000
0S   3s     -0.432659 -0.000000 -0.000000  0.000000  0.000000 -0.000000
0S   4s      0.055247  0.000000 -0.000000 -0.000000  0.000000 -0.000000
0S   1pz    -0.000000  0.091235  0.002771 -0.009073  0.000000 -0.000000
0S   1px     0.000000  0.009479 -0.023157  0.088248  0.000000 -0.000000
0S   1py    -0.000000 -0.000375  0.088712  0.023319 -0.000000 -0.000000
0S   2pz    -0.000000  0.153047  0.004648 -0.015220  0.000000 -0.000000
0S   2px     0.000000  0.015901 -0.038846  0.148037  0.000000 -0.000000
0S   2py    -0.000000 -0.000629  0.148816  0.039118 -0.000000  0.000000
0S   3pz     0.000000 -0.004655 -0.000141  0.000463 -0.000000  0.000000
0S   3px    -0.000000 -0.000484  0.001181 -0.004503 -0.000000  0.000000
0S   3py    -0.000000  0.000019 -0.004526 -0.001190  0.000000 -0.000000

So the orbital order and symmetry are identical and the orbital energies are very similar. Thus: Are the d-orbitals of vital importance to this system? Let's do the same thing after adding a f and a g-function to $\ce{S}$ (HF/def2-SVP+fg). Turns out there are maximum f and g orbital coefficients in the bonding orbitals of $0.024$ and $0.017$, respectively. Does that mean we now have to consider spdfg-hybrid orbitals?
Let's do the same kind of calculations for $\ce{CH4}$. For def2-SVP (one d-function on $\ce{C}$), we find a maximum d-orbital coefficient of $0.022$. d-orbitals are now important for $\ce{C}$? (For def2-SVP+fg, the maximum f and g-orbital coefficient are actually below $0.01$.) It turns out just looking at numerical results is not going to help us getting very far.
Stepping back
We need to remember that hybridization and nice MO diagrams are cleaned up, idealized versions of quantum chemistry calculations. They are mainly educational tools for undergrad students and zeroth-order approximations to be kept readily available in a chemist's head. I cannot think of a purpose for d-orbitals on $\ce{S}$ in that regard. If you can, continue to use them. I will cite the (unfortunately German-only) book of Riedel$^{[1]}$, which states in the 5th edition (p. 114, translated by me):

In conclusion: The significance of d-orbitals in nonmetal [main group] chemistry is not settled. The pros and cons of d-orbital participation are being theoretically debated.

In the 6th edition, however (p. 111):

For a long time, it was supposed that in nonmetals of the 3rd row and below, d-orbitals participate in hybridization and that these hybrid orbitals (such as d$^2$sp$^3$ or dsp$^3$) form bonds.

This reflects the shift in the interpretative thinking on results similar to the ones above. Unfortunately, no sources, primary or otherwise, are being cited in Riedel's book.
A few words on hybridization
We also need to remember that hybridization does not occur except in our minds. It is not part of the quantum chemical calculations themselves or of observable phenomena. Rather, it is a posteriori approach to making quantum chemical results more accessible to other chemists. As such, it is useful and there is a place for it. It introduces a language that helps in the description of a lot of (organic) chemistry. But it cannot succeed as a standalone theory and application beyond the 2nd row of the periodic table fails rather easily. (Consider the bonding angles of $\ce{H2O}$ (102°) and $\ce{H2S}$ (92°). The former fits nicely with sp$^3$ hybridization, the latter looks more like a result of p-s $\sigma$-bonding.)

$^{[1]}$ Erwin Riedel: Anorganische Chemie. De Gruyter, New York & Berlin. 5th edition 2002, 6th edition 2004. With contributions by Christoph Janiak.


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melting and boiling ...really equilibrium?
Why are melting and boiling considered equilibrium processes even though the amount (concentration)  of both phases keep changing i.e from solid to liquid and so on?

There can only be an equilibrium if the process is adiabatic.

If the amount of phases is changing, this is not an equilibrium.

By equilibrium I suppose you mean reversible.

I think what you are asking is this:  Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant).  By contrast, equilibria for phase transitions don't require specific ratios of products to reactants.   [For instance, at the phase transition between ice and water, ice and water can be in equilibrium with any relative amounts of ice and water.] Why is this?
Consider a gas-phase chemical reaction, say
$$\ce{A(g) + B(g) <=> C(g)}.$$
At equilibrium, the chemical potentials of all species are the same.  Associated with this is an equilibrium constant,
$$K_p(T) = \frac{\displaystyle\frac{p_\ce{C}}{p^\circ}}{\displaystyle\frac{p_\ce{A}}{p^\circ}\cdot\frac{p_\ce{B}}{p^\circ}},$$
which specifies a strict relationship between the partial pressures of the reactants and products.
Alternately, consider a solid-liquid phase transition,
$$\ce{A(s) -> A(l)}.$$
At the melting temperature, the chemical potentials of the two phases are the same.  However, by contrast with the example gas-phase reaction, this is true independent of the relative amounts of $\ce{A(s)}$ and $\ce{A(l)}.$
The question then arises: why the difference between the two?  The answer is that the particular mathematical form of the equilibrium constant—and the attendant constraint it imposes on the relative amounts of reactants and products—arises from the entropy of mixing, and there is no entropy of mixing in a phase transition!  At equilibrium, in the example gas-phase reaction, the entropy of mixing just balances the relative free energies of the reactants and products.  That's why we have an equilibrium.  If it weren't for the entropy of mixing, reactions would always go completely to the side whose pure components had the lower free energy (which is exactly what happens in a phase transition).
By contrast, there is no entropy of mixing term in a phase change, because the difference in phases keeps the two components separated.  And without an entropy of mixing term, the relative amounts of the two phases become irrelevant to their relative chemical potential, and thus irrelevant to the equilibrium state (by which I mean any relative amount of the two phases is allowed at the phase transition temperature).
This also explains why, when we are (say) somewhat above the melting temperature, we don't have a small amount of solid in equilibrium with a large amount of liquid (as would be the case if this behaved like a chemical reaction).  The reason for this is there's no entropy of mixing to favor this mixed state over the liquid alone, i.e., over the pure substance with the lower chemical potential (which, above the melting temperature, is the liquid state).  [Or more accurately, while there is an entropy of mixing, because the two phases aren't mixed at a molecular level—phase is a bulk property (i.e., a collective property of many molecules), not a molecular one—the entropy of mixing term (which is essentially an exchange entropy) is negligible in this case.]
N.B.: Just for clarity, let me emphasize that, as a substance is melting or boiling, it is not at equilibrium.  It's not at equilibrium during a phase transition until there is no macroscopic change in the system, i.e., until it settles into a fixed ratio of one phase to another (and that it can be at equilibrium at any ratio).
*Incidentally, this is also why, when we have a mixed-phase chemical reaction where one phase can't mix with the others, that phase isn't included in the equilibrium expression.  For instance, for
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)},$$ we find that
$$K_p(T) = \frac{p_\ce{CO2}}{p^\circ}$$

How does this address boiling?

@BuckThorn  It's the same thing.  Imagine the liquid and gaseous forms of water together in an adiabatic piston at T= 100C and p = 1 atm.  Under these conditions, the chemical potentials of both phases are the same, and we have phase coexistence.  Now we temporarily make it non-adiabatic and allow some heat to flow in, converting some of the liquid water to gas (but keeping the temp at 100 C).  We are still at equilibrium between the liquid and the gas, even though the relative amounts have changed.  I.e., equilibrium, for phase coexistence, requires no specific ratio of liquid to gas.

Why does it have to be adibatic at all? At the BP T and p, no net heat transfer should occur even if it is not adiabatic.

@BuckThorn It may not need to be adiabatic.  The point is that, to keep a certain ratio of liquid to gas, there needs to be no heat flow.  So I specified adiabatic to give a clear example.  But, as you say, if the conditions are isothermal, classically there should likewise be no heat flow.  Except I'm wondering about this:  If we consider fluctuations, it seems possible that, under isothermal conditions, the system could drift from its initial conditions, attaining a new equilibrium state at that same T and p (i.e., here there would be heat flow, as a result of this "thermal random walk").

Two different phases of a substance in contact with each other in a closed system at some uniform temperature and pressure (thermal and mechanical equilibrium) will be in equilibrium if the chemical potential of the substance is the same in both phases. It turns out that at its boiling point, a liquid has the same chemical potential as its vapor at that pressure and temperature. Similarly, the liquid and solid phases share the same chemical potential at the melting T and p. Provided surface effects are ignored and the quantities are macroscopic, any proportion of the two phases will be an equilibrium state at that T and p. Another way of saying the same thing is that, for a total amount of substance, at phase equilibrium the free energy of the substance will be independent of the amount of each phase. Since there is no change in free energy when performing a transition from one mole ratio to another, the process is reversible, which also implies equilibrium.
As an illustration, consider that at a given temperature, a condensed phase (liquid or solid) will be in equilibrium with vapor at a particular pressure (this is in accordance with Gibbs' phase rule). You can change the relative volumes (or fractions) of the two phases and they will remain in equilibrium at that p and T. In addition, because any of those different fractional compositions represents an equilibrium point, you can perform a reversible transformation by progressively changing the relative amount of each phase at constant T and p, all the while remaining "in equilibrium". You would not say, however, that the two systems of different composition are in equilibrium with each other at that point, even though you might envision some way of coupling them somehow in which case you could say that.

Clarification: As rightly pointed out in other answers, a process is not an equilibrium state. Therefore, "boiling" and "melting" processes are not equilibrium states, they are processes. The game is given away by the use of the gerund. However, melting point and boiling point are terms used to refer to states (not processes) at which liquid and solid or gas and liquid are in equilibrium with each other. The best analogy I can think of is moving an object along a perfectly flat surface: sitting still is an equilibrium state, while moving the object over the surface is a process that potentially  follows a series of equilibrium states (if performed reversibly it proceeds along a series of equilibrium states).

[OP] Why are melting and boiling considered equilibrium processes [...]

They should not be considered equilibrium processes. If melting is defined as the process where there is a net change from solid to liquid phase, this is not an equilibrium. If boiling is defined as the process where liquid turns into vapor (rolling boil with bubbles forming below the surface), this is not an equilibrium either.
Wikipedia, in the article on melting point retrieved 10/25/2019, makes a statement that is perhaps too general:

[Wikipedia] At the melting point the solid and liquid phase exist in equilibrium.

If you slowly transfer heat to the system, its temperature will remain at the melting point but solid will turn into liquid, which shows that the system is not at equilibrium.
On the other hand, it is correct to say "when the solid and liquid phase exist in equilibrium, we call the temperature of the system the melting point of that substance". Similarly, when the liquid and vapor phase (both pure) exist in equilibrium, the temperature is called the boiling point for the prevailing pressure.
Not adiabatic, no thermal equilibrium
When you actually melt a substance, you are transferring heat into the system and the system is not a thermal equilibrium. Typically, the overall temperature is higher than the melting point. Also, it is not uniform because the melting process is endothermic, influencing the local temperature when the substance gradually melts. Similarly, we typically boil liquids by heating them from the bottom, with a temperature gradient inside the sample (even if you use a microwave, the temperature does not rise uniformly). Complicating matters, there are supercooled or superheated liquids that have a kinetic barrier to starting the process of freezing or boiling, respectively, so freezing or boiling might not even occur as you reach the freezing and boiling point.

[OP] Why are melting and boiling considered equilibrium processes even though the amount (concentration) of both phases keep changing i.e from solid to liquid and so on?

Let's look at the second part of the statement. If both phases are pure, the concentrations (or activities) do not change at a given temperature, so that part of the statement is not accurate (the amounts do change while solid is melting or liquid is freezing/crystallizing). So at the melting temperature, you could have more liquid present or more solid, and in both cases the system would be at equilibrium if not heat is transferred in or out. Both systems would have the same reaction quotient $Q = 1$. Another way of defining the melting point is to say it is the temperature where the equilibrium constant for the phase transition is equal to one, $K_\mathrm{eq} = 1$.
In an adiabatic system, if the temperature is a bit higher than the melting point, some solid will melt until you reach the melting temperature. At that point, the system will be at equilibrium, and no net changes occur. If the temperature is a bit lower than the melting point, some liquid will freeze until you reach the melting temperature. At that point, the system will be at equilibrium, and no net changes occur.
Take-home message
You can have a liquid/solid system at equilibrium at the melting temperature, but in this case no net melting occurs. You can have a liquid/vapor system at equilibrium at the boiling temperature, but in this case no net evaporation or boiling occurs.


text:
Nucleophilic Substitution Reaction of Alcohol

What is the final product?
According to me, it's option (3). But given answer in the booklet is (2).
Given primary alcohol undergoes SN2 reaction to produce isopentyl bromide and the latter again undergoes SN2 to form isopentyl cyanide.
Can someone please help?

Carbocations will rearrange to place the positive charge on more substituted carbon atoms.

@ringo Carbocation formation doesn't occur in SN2 mechanism.

The alcohol/HBr reaction is not SN2.

It absolutely does for primary or methyl alcohols.