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Why can radioactive contamination be spread by people?
So I am reading a book called "Voices from Chernobyl" where witnesses, nuclear plant workers, firefighters and other persons involved in the 1986 accident give testimony of their experiences.
The very first chapter goes along the lines of the wife of a firefighter that was dispatched for duty on the very day of the incident. He suffered acute radiation and was hospitalized. His wife remain with him throughout the last days before he died. One thing it was mentioned in the book very clearly is how nurses constantly tried to warn his wife not to touch him, hug him or even share objects with him.
In this article, it is mentioned that a person who was exposed to radiation should not be a danger for others once his clothes have been disposed.
So, my question is, I am getting confused between the two readings. Is a person who has been exposed to a big dose of radiation a danger for other people? Why?

Things like sweat can carry radioactive elements to the surface of the skin. Plus the victim is emitting radiation, and even a little distance helps.

@TAR86: thank you for that link. I shamelessly included it into my answer :-)

With all due respect to Svetlana Alexievich for her writing skills and moral standing, she's no expert in nuclear physics, so not all details are necessarily accurate.

The question reminds a bit to HBO's mini series «Chernobyl» (https://www.en.wikipedia.org/wiki/Chernobyl_(miniseries)), and in particular of the snippet around 2:30 min here: https://www.youtube.com/watch?v=EV6G1GpTxjE.  A counter balance to the series: https://www.youtube.com/watch?v=m1GEPsSVpZY.

TLDR: (a) "radioactive contamination" doesn't necessarily spread any more easily than any other kind of contamination, but some kinds can be dangerous in extremely small amounts if it gets inside your body. Also, (b) You don't get contaminated by exposure to radiation. You only get contaminated by touching the radioactive stuff which, at accident sites, can include radioactive dust, radioactive smoke, and radioactive liquids and gasses.

https://chemistry.stackexchange.com/questions/82706/radioactivity-of-materials

@IvanNeretin and even if she were, I understand that the book is a (literarily edited) collection of interviews/monologues/witness reports. This limits severely what can be done against misunderstandings that the witnesses have and express in the interviews. However, a reader of the book needs to be aware of this.

I asked the same kind of question about the series on Physics and got a reply there: https://physics.stackexchange.com/questions/485397/irradiated-persons-as-sources-of-radiation

You may be interested by https://en.wikipedia.org/wiki/Goiânia_accident.

IMHO the edited title substantially changes the question.

Summary: there is not necessarily a contradiction between the two.

Radiation is not contagious, and
a person who has been exposed to ionizing radiation is not dangerous to other people once they are not contaminated with radioactive material any more, but
while they are still contaminated with radioactive material, they may pose a danger (highest danger is to themselves, though).
In any case, other people are dangerous to someone who suffers radiation illness (infection risk).

Moreover,

Thinking radiation contagious is (still!) a fairly widespread mistake, and
I'd think misunderstanding the purpose of the instructions (she being dangerous for her husband rather than the other way round) quite likely in a highly stressful situations such as the husband dying. I.e., I wouldn't expect her to spend time then on double-checking whether she correctly understood the reason behind the instructions and ask, "Sorry, sister, is this because he's dangerous to me or because I'm dangerous to him?".

Thus, it should not be surprising to meet any such misconceptions in such a book. Nor would I be surprised if readers misunderstand the book this way even in case the firefighter's wife is and was perfectly aware that her husband did not pose a radiation danger to her. I'm writing the remainder of this answer mostly assuming that the mistake is in the book rather than on OP's side.
There are basically only two possibilities clear a misunderstanding that happened back then in a book of witness reports: the witness saying "nowadays I know I was dangerous to him, not he to me. But back then I didn't know that."
or the editor putting a footnote explaining the misunderstanding and that it was widespread at the time.

Ionizing radiation and contamination with radioactive substances
Radiation is not contagious in the sense that the firefighter's exposure to ionizing radiation did make him radiate himself. That being said,

Induced radioactivity exists. But that needs very particular conditions to happen to any practically relevant extent. In the sense of the question, you won't be able to produce any measureable effect in a living organism, not even a living organism that is about to die of the radiation.
The firefighter of course radiates like any other living organism. Humans radiate with about 4 kBq due to containing about 15 mg of $\ce{^{40}K}$.

But: chemical contamination, including radioactive substances, can be transferred from one body to another, and they can be incorporated and accumulated where they cause much damage.
(I also wouldn't call that contagious, since the total amount of radioactive material does not increase, but if you consider, say, crystal violet or methylene blue contagious "since if you touch someone, the ones touched by them will be violet/blue", then you can also call the radioactive substances contagious)
However, any radioactive substance contamination that is transferred to a practically relevant amount by touching can also be washed off - and that is the first thing to do for decontamination besides taking off potentially contaminated clothes. Any injury that means that the area cannot be thoroughly washed can also not be touched (for reason of the injury).
Thirdly, during the decay of such radioactive material, other substances form, which may be far more difficult to get rid of (see radon example below). Here, one may say that one can "catch" a contamination that one doesn't easily get rid off again.  However, any such conamination would not be transferred from the firefighter to his wife.

X-rays/γ-rays: these are ionizing electromagnetic rays, i.e. high energy photons. They cause damage when being absorbed, either by directly damaging some biomolecule or by forming OH⋅ radicals/ROS which in turn cause further damage.
The radicals in themselves are nothing very special - they occur all the time as side products of our energy metabolism and we have powerful mechanisms to cope with them. Part of radiation illness is that these mechanisms are overwhelmed.
So, after exposure to X- or γ-rays, we have radicals inside the body, but no "foreign" nuclei, and no body surface contamination. I.e. nothing that became radioactive because of the exposure to high energy photons.

A somewhat different example related to Chernobyl and Fukushima would be incorporation of radioactive  $\ce{^{131}I}$ in the thyroid gland. In particular, if someone with iodine deficiency incorporates iodine, pretty much all of it will end up in the thyroid gland. If that available iodine is $\ce{^{131}I}$, their thyroid gland will subsequently be exposed to large radiation doses. This incorporated radioactivity does include γ radiation of which a part leaves their body.
$\ce{^{131}I}$ is administered in radiotherapy in doses where the patients are e.g. kept in hospital for approximately 2 days (at least here in Germany) so as to not contaminate the wastewater with the radioactive $\ce{^{131}I}$ they excrete in their urine. Such patients are also advised to avoid close contact for e.g. a week after treatment in order to not cause accidental exposure to others, in particular children and pregnant women.. Such radiotherapy treatements use dosages in the 100 - 400 Gy range to the thyroid. And the guidelines are of course with a safety margin.  A quick search for $\ce{^{131}I}$ radiation doeses to the thyroid in  Ukrainian children after Chernobyl got me to Brenner et al: I‐131 Dose Response for Incident Thyroid Cancers in Ukraine Related to the Chornobyl Accident. The largst dose category is > 3.0 Gy, and a diagram has a point a bit below 5 Gy, so 1 - 2 orders of magnitude below the radiotherapy doses.
My conclusion from this is that even in case the firefighter got a $\ce{^{131}I}$ dose to kill off his thyroid, the wife giving the dying husband several close goodbye hugs 10 - 14 days after the exposure would be unlikely to pose a significant threat to her health due to radiation from his thyroid (and under the particular circumstances, the $\ce{^{131}I}$ she ingested after the accident would be a far more important concern for her health).
Again, I would not describe this as "contagious" - but your word use may vary.

In this guest post to cancer letter, R. P. Gale discusses some of his experiences as an MD at the famous hospital 6 (the Soviet Russian radiation clinic) treating the radiation illness patients from Chernobyl with at particular view to the HBO series.

Another error was to portray the victims as being dangerously radioactive. Most radiation contamination was superficial and relatively easily managed by routine procedures. This is entirely different than the Goiania accident, where the victims ate 137-cesium and we had to isolate them from most medical personnel.
Lastly, there is the dangerous representation that, because one of the victims was radioactive, his pregnant wife endangered her unborn child by entering his hospital room. First, as discussed, none of the victims were radioactive—their exposures were almost exclusively external, not internal. More importantly, risk to a fetus from an exposure like this is infinitesimally small.

Valid Reasons for not allowing the wife close to the husband that have nothing to do with radiation being "contagious".

Radiation illness: the bone marrow is rather radiation sensitive, and leukopenia (too low leukocyte counts, a type of immune suppression) are a typical part of radiation illness.
A radiation illness patient is thus at a very high risk from infections.

Radiation illness often comes with burns (the skin is most exposed, and for α and β radiation, almost all damage happens in the skin). Already "normal" severe burns are doubly difficult in terms of infections: the skin damage means that the normal protective barrier against microorganisms is broken down in those areas, and in addition there is a severe immune suppression (after initial inflammatory response). Infections cause half of the deaths after severe burns

Both are very valid safety reasons, just for the firefighter rather than for his wife's safety. Saying that the wife can go close to the firefighter would amount to saying "He's anyways going to die within the next days - it doesn't matter whether he catches an additional sepsis."

"Contagious" radiation as wrong but possibly valid concern after the Chernobyl accident
So from a scientific point of view radiation is not "contagious". Nevertheless, there is a still widespread, though mistaken fear of this. Personally, here and now I count this pretty much in the tin foil hat corner. But OTOH, in the situation in the Ukraine(ian SSR) immediately after the accident I think it a more understandable concern since the possibilities to check whether this concern is valid or not were severely limited. Not only for the general population, but even for the medical staff.  In such a situation, it is a valid decision to err on the side of caution.

How much did the hospital staff know about radiation medicine? Had they any experience with radiation injuries of this kind? There was a major accident, no internet, and a very restrictive information politic already without disaster.
In the video interview (thanks to @TAR86), Alla Shapiro describes that the "hush up" included that the medical staff was deliberately hindered/prevented from accessing medical information about radiation. She also explains that their medical training did not include radiation..
Note however, that she was at a clinic in Kiev. P.R. Gale describes much better expertise at hospital 6 in Moscow (where most of the radiation illnesses were treated).

The post you link says that thorough washing is sufficient for the Fukushima evacuees, and the experts linked above agree for the Chernobyl fire fighters.

However, Shapiro also says that there was a whole lot of this fear in the general population that radiation/radioactivity could be "contagious"/someone who was exposed to radioactive radiation would radiate themselves and thus pose a danger*.
In a situation where probably almost everyone realized that the government tried to hush up a big problem (correct assessment), and in addition any relevant information being locked in the "poison cupboard" of the libraries and not accessible (no way to obtain factual information): would you trust "information" that these patients are not "contagious" wrt. radiation, i.e. information is very much what the government wishes or would you tend to err on the safe side?

In addition: How much did the hospital staff know about what actually had happened and what the firefighter had actually been exposed to?
With those activities to hush up the Chernobyl incident, the hospital staff may have been unsure about what else that firefighter had been exposed to besides high doses of radiation.
Remaining contamination with radioactive material (including incorporated material) can be measured comparatively easily. However, I have no idea whether such instruments were available at the hospitals, say, in Kiev, to measure remaining contamination of their patients: a) the available instruements may have been needed more urgently at the site of the power plant and b) also political coniderations/hushing up may have been standing against that. I'd expect the Moskow hospital to have all kinds of instruments (but that may be my predjudice)

* I'm pretty sure that a non-negligible fraction of the population here in Germany would express that fear if you ask them. Including medical staff, and even after Chernobyl and Fukushima.
If you want an example: have a look at this post (in German) about radiation treatment for food on a web site by the offical consumer protection organizations

 Werden Lebensmittel mit ionisierenden Strahlen behandelt, wird die Strahlenmenge genau dosiert. Die Energiemenge ist so gering, dass die Lebensmittel nicht radioaktiv werden und sich nur leicht erwärmen.

My translation and my emphasis:

When food is treated with ionizing radiation, the amount of radiation is accurately dosed. The energs is so small that the food does not become radioactive and only heats up slightly.

While it is of course true that the food does not become radioactive, that sentence IMHO does insinuate that this could [easily] be the case with higher doses. And one comment (out of a total of 14) clearly indictates that the writer thinks radiated potatoes will radiate themselves.

While some of the comments about whether the firefighers not being radioactive might be true, it is clear that the authorities at the time though they were (hence the precautions). Unless we have definitive measurements we can't be sure. Moreover, they were exposed to extraordinary amounts of material from an exploded nuclear core  without any proper protection so general observations from  other incidents might well not apply here.

In addition to HBO's series «Chernobyl», because one accent of your answer is about the medication «at least here in Germany», you might be interested to know the Swiss precautions for those in vicinity (50 km) of the (now former) nuclear power plants on Swiss terain. Every 10 yr each inhabitant gets his / her blister of KI by mail: http://www.iodtabletten.ch/de/home (German), http://www.iodtabletten.ch/fr/home (French).  Looking westwards, vigilance is kept (old Fessenheim, or Superphenix) and the many still running plants (https://en.wikipedia.org/wiki/Nuclear_power_in_France).

Another problem here is that the instructions to the Firefighter's wife were very clearly not for his protection alone. The authorities clearly thought the firefighters were a risk to others (they might have been wrong, but they didn't bury them in lead-lined coffins for no reason). Theoretical arguments are not relevant to what they did at the time and are probably not relevant anyway given the extreme nature of the exposure they were subjected to.

It's worth noting that it's not just misguided fear and lack of information - after all, while the opinion of the survivors can be mistaken, it's all based on the fact that the child of Ludmila Ignatenko did die four days after being born after being diagnosed with cirrhosis of the liver and congenital heart disease. The only question is whether it is related to some radiation damage during the hospital visit, or perhaps solely to the exposure Ludmila had before leaving the area of Chernobyl.

Radiation isn't "contagious" unless you are heavily contaminated and can cause damage to others
The firefighters in Chernobyl attended the fire before anyone fully appreciated just how bad the accident was. The core was exposed and lumps of the extremely radioactive graphite moderator were spread around the area where they worked (the series had one scene where a fireman picked up a graphite lump for a short period and suffered extreme radiation burns just a short time later). The firefighters were not wearing protective clothing suitable for protecting them from radiation or from the radioactive particles scattered around the site. Their clothes were heavily contaminated–so much so that they are still a hazard today (they are still in the abandoned hospital's basement). But, since the clothes were not designed to protect the firefighters from bodily contamination, their removal does not mean the firefighters' bodies were radiation free.
This, I think, is the key to understanding how cautious the hospitals were at letting outsiders near the firefighters. They were radioactive because of their extensive exposure to dust and radiation from the reactor core. As far as I can remember, the series didn't show any radiation measurements on the firefighters but it strongly implied they were still "hot" and that staying near them was hazardous for others.
So it isn't so much that radiation is "contagious", it is just that, in this circumstance, the firefighters were themselves quite radioactive. Less radioactive than their clothes, but still radioactive enough to be dangerous.

Do you have a source to how radioactive the firefighters (i.e. the contaminations that stuck to them after washing) were? I was looking for some, but only found various not-so-trustworthy looking sites they had to be buried in special coffins - ranging from leaden coffin to being wrapped twice in zinc foil [!?] in wooden coffins, and to whether a big concrete block was needed [but not sufficient] to protect the graveyard visitors. In contast, a post (see quotation in my answer) by one of the MDs helping to treat the patients claims that there wasn't much of a problem in that respect.

I don't have a reliable source. But the thinking at the time that the firefighters needed to be buried in sealed, lead-covered coffins is accurate. The Chernobyl series shows this and it did happen. So, fairly clearly,  the authorities at the time thought they were dangerously radioactive. Hence the precautions.

@matt_black it does not and did not to be opinion based - it's trivial to put a dosimeter next to a patient or a corpse and verify if it's radioactive or not. It may be tricky to find evidence of whether and how it was done, and how radioactive (if any) the corpses were, but I'd presume that the local officials would not go out of the way and order a concrete-covered burial unless the corpses were actually still radioactive.

I forgot to mention that this woman was pregnant and she lost her newly born after just a few hours of giving birth. Both the book and the series seem to imply it was because she had been with her husband going against the recomendations of dtaying away from her husband. Itis exolained that the baby was born wirh a damaged liver due to the exposure and that saved his mother's life. How can that be explained?

@Peteris "Trivial" to put a dosimeter next to a patient or a corpse... If you have a dosimeter, as well as the knowledge to use it and interpret its readings with confidence... not so trivial, considering the time (pre-internet).

Radiation and radioactivity are often mixed up. A matter like uranium can be radioactive. It means that it contains atoms that can emit radiations made of particules that travel away with great speed in the surroundings. These radiations may snatch electrons in the surroundings, which cause serious troubles in the target. But at the end of their path, they disappear as matter or are transformed into ordinary electrons, or harmless helium atoms. So there is no radioactive material in the target.
A radioactive matter is like an army. Its soldiers are dangerous, because they can fire and shoot bullets that may kill a victim. But after the shot, the bullet is not dangerous any more. After the shot, the soldier is still dangerous, as he may shoot again, but the bullet and victim are not dangerous.
Radiation is not "contagious". Radioactivity may be considered as "contagious" : radioactive substances may be transmitted from place to place, like an army that can move.
A person who had touched a radioactive substance like uranium, may have absorbed some uranium atoms through the skin. This person becomes radioactive and then emits radiations. Apparently the Chernobyl firefighter has been irradiated when fighting the accident. He was touched by the radiation and became a victim of his duty during his job. But, bad luck, he had also the opportunity of touching some uranium from the exploding power station. And then, this uranium will stay on or in his body, and will emit radiation later on in the future. This person becomes radioactive, not because of the radiation he has received during his job, but because of the uranium that entered his body when he touched a piece made of uranium in the rubble of the power station. That is why nobody should touch him, because if this uranium is glued on the skin, it may be transmitted to somebody else.

If substantial amounts of the contaminant can be transfered by touch (e.g. [sweaty] skin to other person's skin), they can also be washed off. Off the patient in the first place, and the wife should also wash e.g. her hands before and after holding his hand.


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if I have concentrated acetic acid (99%) what are the out put gases during electrolysis?
If I use concentrated acetic acid (glacial acetic acid($\ce{CH3COOH}$)) as an electrolyte. What ions are produced at the anode and cathode?
Can I get ethanol with it ($\ce{C2H6O}$)? If not, what would be the products?

I´m sure you can tell what one product is. As for the other electrode, what is the primary product, when you have abstracted one electron? The second intermediate should then be equally obvious, if you know a bit about organic chemisty. And the third and final is then trivial. ;)

And btw., this is a well known named reaktion. Google?

Upon electrolysis, organic acids will lose the carboxyl group and dimerize.
This is known as the Kolbe electrolysis:
https://en.wikipedia.org/wiki/Kolbe_electrolysis
In the case of electrolysis of acetic acid, ethane and $\ce{CO2}$ will be formed.

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Boiling point elevation with multiple solutes
I've encountered a problem where glucose and $\ce{NaCl}$ are mixed with water and $\Delta T$ is to be calculated.
I've calculated  $\Delta T$ for both glucose ($i = 1$) and $\ce{NaCl}$ ($i = 2$) separately and added them.
Is this procedure correct or should I have treated them as one solute and used $i = 3$ as the Van't Hoff factor?

You cannot use i=3 unless both solutes have the same concentration, your initial method is correct assuming you have done the calculations properly.

why not ? doesn't ΔT only depend on the molality of solutes regardless what they are or their concentrations ?

It does matter on their concentrations, you wouldn't need to care about the what the particles are, but you need to consider the number of the particles.. molality is a metric of concentration btw ;-). You would get the same answer if you were to find the $\Delta T$ of each individual ion with i=1 and add all the $\Delta T$

I'm not sure if I understand...  Suppose you add 5mol of glucose and 10mol of NaCl in 1kg of water. Then their combined molality is  b=15 mol/kg and their combined i = 3...AH i see now, you cannot assume i =3 because NaCl contributes twice as much, only if the concentrations are the same then you can combine. Is this right ?:)

What is the Van't Hoff factor? It basically decides the number of particles formed for every one particle we dissolve. For a 100% dissociating binary electrolyte like $\ce{NaCl}$, the value is 2. Split that into $\ce{Na+}$ and $\ce{Cl-}$, for each of which the value of i=1 since it can't dissociate further. For a given amount of $\ce{NaCl}$, the amount of $\ce{Na+}$ and $\ce{Cl-}$ would be equal to the initial amount of $\ce{NaCl}$ present.. Adding the $\Delta T$ for the two components, we get the same value as if we had taken just 100% dissociating $\ce{NaCl}$.

Exactly. The edit just popped up now.. Too late I guess :)

What is i = 1, i = 2?

@The99sLearner the vant hoff factors.

Take b = b(glucose) + 2b(NaCl) then times with boiling point constant, is that correct? Is that = 'take i = 3'?

@The99sLearner That is not "take i=3" which is what OP later on understood. you can "take i=3" only when b(glucose) = b($\ce{NaCl}$)..

When there is more than $1$ solute in the solution, the total elevation in boiling point ($\Delta T_\mathrm{b,total}$) or the depression in freezing point ($\Delta T_\mathrm{f,total}$) is defined as the sum of each individual value for each type of solute.
What is Van't Hoff factor? According to Chem Libretexts, the Van't Hoff factor is defined as:

$$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \tag{1} \label{1}$$

Now, why does this work? Let's take an example to see how it works.
We take a simple case as what you have mentioned in the question:

Find the elevation in boiling point of a solution of $\pu{1 m}$ glucose and $\pu{2 m}$ $\ce{NaCl}$ [$100$% dissociated] in water.
$\mathrm k_\mathrm {b,water}= \pu{0.52 ^\circ C kg mol-1}$

We can do this in two ways. One, where we take $i = 2$ for $\ce{NaCl}$. Second, we state that $\ce{NaCl}$ dissociates into $\ce{Na+}$ and $\ce{Cl-}$ and take the three types of solutes to be present instead of two [going from basic principles].
Method $1$
Here we would have two different solute particles. Finding the value of $\Delta T_\mathrm{b}$ for both glucose and $\ce{NaCl}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,NaCl} \\
&= \mathrm{k_b}[1\cdot1 \,+2\cdot2] \\
&= 5\cdot\mathrm{k_b} \\
\end{align}
Method $2$
Here we would have three different solute particles. Here, $\ce{Na+}$and $\ce{Cl-}$ would have a value of $i=1$ and their individual molality would be identical to that of $\ce{NaCl}$ since the salt was $100$% dissociated in solution
Finding the value of $\Delta T_\mathrm{b}$ for glucose, $\ce{Na+}$ and $\ce{Cl-}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,Cl^-} +\Delta T_\mathrm{b,Na^+} \\
&= \mathrm{k_b}[1\cdot1 \,+1\cdot2 + 1\cdot2] \\
&= 5\cdot\mathrm{k_b} \\
\end{align}
Therefore, both methods yield the same answer. The second method is actually the derivation for value of $i$ for a dissociating salt.
Now, you can apply $i_\text{total}=i_1+i_2$ only when the concentrations of the given solutes are the same. The proof is similar to how the second method proves the first above. In such a scenario, you can assume them to be one solute since colligative properties don't care about how many types of solute there are but rather care only about the amount of solute particles present in the solution.

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Is the green product from Iodine reaction with Aluminium coated plastic, hydrated Aluminium Iodine
I found a package containing some Iodine I bought a while ago.  It was contained in a plastic jar placed inside an Aluminium coated plastic pocket.  The pocket had strange green worm like trails over its surface.  In the most advanced place, it actually seems to have removed the Aluminium and left the plastic transparent with a green / yellow tinge.  I wonder what reaction was occurring here?
Aluminium and Iodine should produce Aluminium Iodide which is white.  Why it appears green from the outside of the pocket is not clear.  Perhaps it is the hydrate form which is yellow so with the plastic colour somehow appears green?
Why this pattern advanced in strange worm like patterns is also not clear to me.
As an aside, I assume this was the Iodine I purchased, it is present as a solid spheres which appears through the brown plastic bottle as shiny black/brown.  When it was shipped I assumed it would be safe to leave as it is.  It has never been exposed to high temperatures or sunlight.  Apparently I need to research a better storage system than that provided to me as the wax seal they appear to have tried to apply round the neck of the jar has not worked to contained the Iodine.

Please leave a comment if you're going to downvote... otherwise I have no idea how to improve the question... or what to avoid in the future apart from: Avoid StackExchange, which would be sad!

welcome to Chemistry Stack-Exchange !

I think the worm like patterns are to attributed to the packaging case textures as seen on the image

Is the appearance of the iodine within the glass container ok?  With a small-scale reaction / analysis you already know and may reference to, does this iodine still (re)act as anticipated?  If so, you could continue with the earlier anticipated task and deploy the iodine as intended.

@Buttonwood I'll confirm that and get back to you but from a visual inspection of the contents inside the bottle it looks fine.

@AJP actually that was in reference to the first comment :P

Hahaha XD @pipiedipie  :)  I've caught up now!  :)

No. Aluminium salts, even basic ones, are transparent (unless anion is colored, and iodine isn't colored)
Of common metals, blue-green hues are typical for copper. The exact compound, however, is very debatable. it's definitely not a simple copper iodide, as it is very pale yellow-brown, so it is something resulted from subsequent reaction with air.
Another source of greenish hues is iron, but those are typically unstable long-term in air unless the compound is in a big chunk. Glasses often have greenish hue due to presence of iron and are stable, because iron there cannot easily be oxidized. On the other hand, greenish solutions of Iron (II) are readily oxidize by air with formation of brownish hues.
Finally, green hue are characteristic to chromium (III). But I don't see how such an alloy could be used here.
My bet is on copper.
Note, the coloring you observed might be cause even by very small amount of coloring ions.

Just speculating, but Aluminum is commonly employed in products as an alloy. For example, per a source, to quote:

Aluminium alloys (or aluminum alloys; see spelling differences) are alloys in which aluminium (Al) is the predominant metal. The typical alloying elements are copper, magnesium, manganese, silicon, tin and zinc.

As such, here my guess is a small presence of copper ions to account for the green hue (other Cu related green compounds include, for example, Verdigris). Also relatedly, my experience with ammonia complexes of Cu, for example, indicates an ability of the metal complex to display strong coloration at even very low concentrations.
As to exactly what inorganic/organic iodine related compound/complex, may possibly be present speculatively with copper, I will leave to others.


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Determining electronic configuration
One writes the electronic configuration in terms of shells, spdf subshells, orbitals, and spin of electrons. There are exceptions to principles we know. How does one confirm that the way of writing is correct? How did scientists understand the exceptions in case of nobium and platinum in electronic configuration? Is it possible to take an atom and find which shell its electrons are in? How did this spdf electronic configuration (along with exceptions) be first understood.

Excellent question...but I have been searching for this answer for almost a decade without a hope. I have asked spectroscopists, physicists... No luck. Leave Pt and Niobium...talk about our simple Na. My question was and is...given an experimental emission spectrum, tell me how people developed electron configurations. Finally, a Canadian physics professor said, that these orbitals and electron configurations are bogus! It is good until the hydrogen only.

In my personal opinion, all this information is hidden in old physics in German and very early physics papers which nobody bothers to read now but they are happy to "teach" configurations. Even Hund's Linienspektrum und periodisches System der Elemente has never been translated.

I now feel it can also be determined from atomic spectra.Each line in emmision spectrum conveys information about location of electrons.

compenthusiast, you cannot talk about the location, but you can talk about the energy levels.

I meany energy levels by location.Just like saying "it is located in 2s orbital" or something.

Probably, I believe the answers lie in Hund's book which is still in German :-)

I think its time to google translate.

Is there a source of a free pdf?

compenthusiast....Well, I really wish it were that easy ...I have written two detailed papers on German machine translation and problems/solutions associated with it. It will not be that straightforward especially when there are plenty of equations embedded in the text.

I have discussed this problem once in the past with an Italian colleague, teaching in Palermo University, and who was in charge of training the Italian team at the International Chemistry Olympiad. I remember he told me that this order can be established by plotting the square root of the successive ionization potentials of any atom versus its atomic number. Of course there is one point for H, two for He, three for Li, etc. If you report these √I values versus the atomic number Z, you will find that all highest values (relative to last $1$s electron) are steadily increasing, of course. But these points remain alined across the page. Likewise, all next to highest values (also relative to $1$s) are alined, and surprisingly the slopes of both lines are equal. The first line starts at H, and the second at He. But they are parallel, after my Italian colleague.
Now if you consider the next but two highest √I values, the curve starts at Li, and the corresponding points ($2$s) are also alined through the diagram, with a slope which is nearly the same as the first two curves. Likewise the points corresponding to the next but three values (starting at Be) are alined and the line is parallel to the preceding line.
Something new happens with the next points. If you consider the fifth √I values by decreasing numerical order, these points start with Boron. And they are alined across the diagram. They correspond to $2$p electrons. New surprise : the points are still alined. But the slope of this line is much smaller than before. The slope seems to be typical of p electrons.
If you join the sixth √I values, stqrting with Carbon, the curve is still a line, with the same slope typical for p electrons. Same thing for the $7$th, $8$th, $9$th and $10$th electron. All lines are parallel.
First difficulty occurs with the $11$th √I values, starting with Na. The curve joining these points is much steeper than for the preceding electrons. It has the same slope, typical of the $1$s and $2$s electrons. So, this line cuts preceeding lines. As a consequence, when studying the points through Mg, Al, Si, etc., it is not easy to decide which √I value belongs to what.
I have not done it myself, but my colleague told me that this job may be carried out through all the periodic table, and that it is possible to recognize a bundle of steep lines for all s electrons, another bundle of not so steep lines for p electrons, another bundle of still less inclined lines for d electrons, etc. and the same with f electrons.
Apparently, this job explains why there are irregularities in the electronic configurations through the periodic table.
If some reader will try to check this statement, I would be pleased to hear of his job.

Maurice, I think the ionization energies can show a periodic rend, but it may not tell us anything about the valency configuration. All this experimental knowledge about electron configuration is lost. It is a lost art.

@ Farooq. You are right. The experimental knowledge about electron configuration is lost. As you said yesterday, somebody should translate Hund's publication.

Thanks for answering .However it still seems this question hasn't got a full proof answer.

I just found a link that might be related:https://pubs.acs.org/doi/abs/10.1021/ed073p617

In my opinion ,probably chemical properties(in what ratio elements react) of atoms might have paved way to the determination of valance of atoms.However this does not clarify inner configuration.

@compenthusiast, Nice article but this is NOT the historical way electron configuration were determined.


text:
Is structural formula of 3-butanol possible?
Recently in my question paper.There was question asking me to draw the structural formula of 3-butanol.
Is the formula of 3-butanol possible?

No. There is only 1-or 2-butanol because the naming conventions dictates that you give the funcctional group the lowest possible number, so it is for example 1-butanol not 4-butanol.

I could tell you the structure, but IUPAC would have to kill me. ;-)

3-butanol has an umambiguous structure, but by convention, we choose not to allow that name to reduce the possible names for a compound.

did it actually say "3-butanol" or was it "tert-butanol"? tert-Butanol (or t-butanol) is short for "tertiary butanol" which literally translates to "third butanol", so it could be mistranslated as "3-butanol" if your text is a translation. In any case, I would assume they mean t-butanol.

If tert-butanol, then of course it's 2-methylpropan-2-ol.  One carbon atom and then three methyl groups and a hydroxyl group joined to it.

Formally, the structural formulas of n-butanol, n=1,2,3,4 are possible, similarly as n-ethanol, n=1,2.
But due the symmetry, there is redundance, as 4-butanol is identical to 1-butanol, 3-butanol to 2-butanol and 1-ethanol to 2-ethanol. Therefore IUPAC rules use just 1-butanol, 2-butanol and ethanol.
Using eventually 3-butanol for tertiary butanol ( or t-butanol ) aka 2-methyl-2-propanol is incorrect.


text:
What is a Herzberg?
Whatman lists the flow rates for their filters in a unit called a 'Herzberg':

What is this?
I've been searching around and I did find information about Herzberg flow rate testers, for example this old paper on filtration flow rate measurement and this summary of filtration terminology, but I can't seem to find two sources that agree with eachother.
The clearest looking description I found was that second link above, which states:

Whatman quantifies liquid flow rate for its range of filters by using a Herzberg flow rate tester. Prefiltered deaerated water is applied to the test filter (effective area $\pu{10 cm2}$) at a constant hydrostatic head ($\pu{10 cm}$). The rate of the flow is measured in seconds per $\pu{100 mL}$. Flow rate can also be measured by the modified ASTM method which uses a quadrant folded filter held in a wire loop. It is not considered to be as reliable or consistent as the Herzberg test.

If that is a reliable description then it looks like "375 herzberg" means that it took $\pu{375 seconds}$ for $\pu{100 mL}$ of water with a $\pu{10 cm}$ constant head to flow through a $\pu{10 cm2}$ filter, which means that more Herzberg = less flow.
Although that's a little confusing too because the $\pu{2.0 \mu m}$ filter flow rate above is shown as 375 Herzberg, but these larger $\pu{2.5 \mu m}$ filters are listed as 1870. Now the $\pu{2.5 \mu m}$ filters are $\pu{200 \mu m}$ thick compared to the $\pu{2.0 \mu m}$ filters at $\pu{160 \mu m}$, and also I have no real understanding of how any of this works, but my initial reaction is "how could the larger pore size have a $5 \times$ slower flow rate?".
Is that correct? If not, what is this unit?
(I realize that second source is clear, but since I can't find consistent descriptions I mostly just want to make sure that's what it means.)

Herzberg filtration speed: The time taken to filter $\pu{100 mL}$ water at $\pu{20 ^\circ C}$ through a filter area of $\pu{10 cm2}$ at a constant pressure of $\pu{5 cm}$ water column (Thomas Scientific).
The values are given in seconds. So, the example OP has given has Herzberg filtration speed of $\pu{375 s}$. That means the relevant filter paper takes $\pu{375 s}$ to filter $\pu{100 mL}$ of water under above Herzberg filtration conditions. Thus your understanding is partially correct (except for $\pu{10 cm}$ water column):

If that is a reliable description then it looks like "375 herzberg" means that it took $\pu{375 s}$ for $\pu{100 mL}$ of water with a $\pu{10 cm}$ constant head to flow through a $\pu{10 cm2}$ filter, which means that more Herzberg = less flow.

Keep in mind that the temperature is important (Ref.1). This measurements are done at only $\pu{20 ^\circ C}$. This reference described Herzberg's research as:

Herzberg used a constant-head apparatus, based on the principle of the Mariotte flask, which forced water through a horizontal disk of filter paper $\pu{10 cm2}$ in area and discharged it into a measuring flask. With a head of $\pu{5 cm}$ of water at $\pu{20 ^\circ C}$, the instrument was used to measure either the time of filtration of $\pu{100 ml}$ of water or the volume of water filtered in $\pu{1 minute}$. He reported a range of filtering rates of $23$ to $\pu{760 ml/min}$ per $\pu{100 cm2}$ of area for 30 filter papers (Ref.2).

Thus, it is safe to assume that the definition given in Thomas Scientific is based on these conditions.
References:

Herman Bogaty, Frederick T. Carson, "Measurements of rate of flow of water through filter paper," Journal of Research of the National Bureau of Standards 1944, 33, 353-362 (PDF).
Wilhelm Herzberg, In Papierprüfung (Paper testing); Fifth Edition, Springer Verlag: Berlin Heidelberg, Germany, 1921 (ISBN: 978-3-662-23207-1).

can you cite a source, since your definition contradicts the one provided by OP (5 cm pressure vs 10 cm)?

Thanks; your source is actually one of the two links I cited in my question ("this old paper on filtration flow rate measurement"). The thing is, it describes Herzberg's work, but otoh it doesn't appear to define a named unit based on that work (although I could not find an English version of Papierprüfung, but I haven't looked hard yet). Plus the 5cm vs 10cm head thing. :)  Hmm...

@ Andrew: I have edited to give additional infomation. Although Thomas Scientific is a vendor, they have followed exactly what Herzberg has done to find various flow rates of various filter papers. I think that would clear OP's doubts about definition.

Seems legit, or at least as convincing as it'll ever be.


text:
Affordable, non-contaminating flexible tubing for common acids
Which tubing material is affordable, flexible, widely available, and suited for transporting very pure acids ("puriss." products) in their most concentrated form without contaminating them?
PTFE seems compatible with most if not all of them, but I don't think it's very flexible, and I'm not sure whether PTFE hoses would be affordable (nor where I could buy some).

Usually if you require very high purity chemicals (like semiconductor grade) there is a good enough reason to spend enough money to get the proper tubing.

@Jon_Custer: What is the proper tubing?


text:
Photoluminescence Spectra of Solids
What information can we gain with the photoluminescence spectra of solids?  The bandgap can be determined for example, but are there any other useful properties we can learn?  I care about this in the context of semiconductors but other answers are also appreciated.

Bandgap, defect states and dynamics, excitons, carrier dynamics, ...

Can you provide a source for some of this?  I am finding it hard to get a good source, even knowing roughly what I expect to find out.

Pankove's classic Optical Processes in Semiconductors is a good starting point.


text:
Are there reactions where the stirring direction would make a difference?
I'm not a chemist. I don't really do anything with chemistry, so I know little of the subject.
I know a bit about chiral molecules. For example, I know that homochirality can affect light polarity.
I'm just curious: Are there chemical reactions where the direction of the stirring (clockwise vs. anti-clockwise) would make a difference? A simple explanation would be appreciated.

While influence on chirality is next to none, mixing direction may be important in some processes, for example making ice cream with liquid N2 :D

Here is a situation where the direction of stirring may be significant! You are adding a compound to a vat that is also slowly draining (which induces a cyclic rotation). If you stir in the direction opposite to the flow, more turbulent and likely better mixing. Note: the particular 'right' direction varies depending on your particular hemisphere (Northern or Southern). Interestingly, an answer by Buttonwood below does mention the hemispheric factor, albeit, contextually not usually significant.

I am afraid introducing Coriolis forces brings more troubles than benefit. Apparently vortexes have rotation independent of the hemisphere.

Coriolis forces due to Earth's rotation are large only when the vortex is as large as a typical weather system.  Buffalo Wild Wings commercials aside, you can't set up a weather system in a small space, so the Coriolis effect has little impact in your reactor.

Aren't molecules so small that no matter what direction you stirred, the molecule would only see the tangent (i.e. a straight line) for all intents and purposes?

I feel that if the reaction vessel is itself chiral (its geometric shape I mean), it could be possible that stirring in different directions could change a reaction's outcome, regardless of the compound being chiral or not. For example, consider a biphasic liquid system in some chiral container where stirring in one direction causes approximately laminar flow, whereas stirring in another direction causes turbulent flow. Maybe something like a rotational equivalent of a Tesla valve.

I’m voting to close this question because no.

Stirring directions only make a difference if you're Harry Potter.

The question is far from magic only if one considers a lot of details. Although the answer is NO the " in principle" issue might come from in depth thinking. It is nice that this question has received upvotes and not a superficial treatment.

TBH I first thought of this as I was stirring my hot coco. Curiosity comes from everywhere.

@NicolauSakerNeto A propellor mixer (see Buttonwood's answer below) is chiral, and you don´t need a biphasic mixture to see an effect, it´s already going to make some difference for e.g. heat transport. It depends solely on a variance in buoyancy of the components.

No, for three reasons:
Firstly, you need a relatively large molecule and/or a high viscosity and/or a strong shear/extensional flow, before the molecule even notices that it´s in a nonuniform environment. The molecule, if it has an elongated form, will then align perpendicular to the normal vector of that flow field. That only occurs for sufficiently concentrated solutions of polymers, or liquid crystals.
Secondly, you need three vectors to be able to differentiate a left and right hand. You only have the normal vector of the flow field, and the preferred direction of your molecule. Or in other words, your reaction will be indifferent to you watching it from the top of the beaker or through the bottom. ;)
Thirdly, to get a third vector into your system, you need the normal of the flow field to change significantly, in a specific direction (using, say, a propellor instead of a stirring bar), on the lenght scale of your molecules. That is impossible with a macroscopic stirrer.

could gravity be a factor? I guess not, as molecules are too light

@NooneAtAll https://chemistry.stackexchange.com/questions/80816/does-gravity-affect-the-rate-of-a-chemical-reaction

@Noone AtAll it depends on the meaning. Not really on a single act, the rest might be layering and concentration gradients. But again, there would be no practical meaning. But once I saw a calculation on the effect of gravity on electrons flows in a photovoltaic panel... :)

@NooneAtAll Gravity per se certainly not. Buoyancy differences of course, if your upward and downward flow paths are different. As was already noted above, chemical engineers do worry about such things. ;)

In short: if the direction of stirring were a both reproducible and highly significant parameter for the synthesis of chiral molecules, the manufacturers would offer a back- and forward direction of stirring by default all across their stirrers.  But no, I'm not aware that there is such an effect.
Note, however, beside the stirring plate in the (small scale) lab driving a magnet in the beaker

(reference)
larger scale mixing uses overhead mixers.  These look similar to a drilling machine and a replaceable, axle-mounted propeller:

(reference)
These occasionally have the option to alter the direction of stirring, but only because the mounting of the stirring blades may be chiral, screw-like (examples).
Side note: At best of my knowledge, there isn't a significant and reproducible influence of the geographic position (latitude), either.  Thus, the synthesis of and with chiral molecules is independent if performed in Boston (northern hemisphere), Perth (southern hemisphere), or Quito (almost equatorial).

There actually will likely be modifications to synthesis protocols in Quito, but not because of the Coriolis effect. Quito is at 2850m elevation, leading to a nearly 10C reduction in the boiling point of water.

These two intriguing papers conclude chiral environments can be induced by directional stirring, where the direction of the chirality depends on the direction of stirring.  And other work has shown that chiral environments steer the direction of chiral synthesis.*
So: If [direction of stirring] -> [direction of chirality of environment] -> [direction of chirality of synthesis], then the direction of stirring could affect chiral synthesis.
https://chemistry-europe.onlinelibrary.wiley.com/doi/full/10.1002/cphc.201200003
"Recently, several reports have suggested that the vortex flow of a solution of an achiral molecule gives rise to a CD signal, which is dependent on the stirring direction. This article introduces types of molecular architecture and material designs that show stir‐induced chirality....Supramolecular assemblies can be deformed to a chiral structure in a vortex flow which shows a CD signal even if the assemblies are achiral. ...Recently, we have shown that the stirring‐induced chirality caused by the host can be transferred to an achiral guest alignment. If such a transfer of chiral information could be realized on a molecular level from the vortex flow through a supramolecular aggregate to individual molecules, the consequences will be useful to a broad field in science and industry."[1]
[1] Okano, Kunihiko, and Takashi Yamashita. "Formation of chiral environments by a mechanical induced vortex flow." ChemPhysChem 13.9 (2012): 2263-2271.
https://chemistry-europe.onlinelibrary.wiley.com/doi/full/10.1002/chem.201100713
"This article introduces types of molecular architecture and material designs that show stir‐induced chirality.....Here we report that a solution of an achiral cationic gelator can vicariously transfer the macroscopic chirality of the vortex stirring towards an achiral molecule (in fact, a molecule showing racemic conformations)." [2]
[2] Okano, Kunihiko, et al. "Emergence of chiral environments by effect of flows: The case of an ionic oligomer and Congo red dye." Chemistry–A European Journal 17.34 (2011): 9288-9292.
*See for instance the following, which shows an example of how a chiral environment, created by a chiral solvent, "steers chiral synthesis":  https://cen.acs.org/synthesis/catalysis/Solvent-steers-chiral-synthesis/97/web/2019/07

On a larger scale, the direction of stirring can be used to optimize homogenization of complex mixtures. For instance, paint-like products can be mixed with a propellor-type blade. When the direction of rotation orients the blade pitch so as to force liquid down and around, a vortex is created which may dip so low as to allow air to be incorporated into the mix. Depending on the viscosity of the product, the air incorporated might rise to the surface rapidly when the stirrer is turned off (which is OK), or it might be trapped in the product, making it unsuitable for customer use. But automatically using the blade direction which does not produce that kind of vortex requires more attention to the homogenization process, and often does not do as well.
There are high-shear blades which have notches angled to produce best results when spun in one direction, even tho the blades are flat (except for the notches and bent-up sections) and look quite symmetrical.

Chemical engineers are so much more aware about mixing than chemists: http://encyclopedia.che.engin.umich.edu/Pages/MaterialsHandling/Mixers/Mixers.html#ABGO3XKWBU8K42UW13D5

@Buttonwood Can you update the link? It doesn't work.

https://encyclopedia.che.engin.umich.edu/mixers/


text:
Do we have some "rule of thumb" to estimate the miscibility of metals?
For example, for ordinary fluids, we have polar and apolar ones. The "rule of thumb" is that polar fluids mix with each other, and also apolar fluids mix with each other, but polar fluids don't mix with apolars.
Of course, it is not a very exact rule. For example, bensene mixes in water - but it does it badly. While the also polar ethanol mixes water very well.
What is the case with liquid metals? Does some analogous - possibly more complex - category structure exist?

Most liquid metals mix with each other. Cases of non-miscibility are rare.

@IvanNeretin Thanks! But, for example, molten uranium does not mix with molten iron. Iron does not mix with gold. And so on.

@IvanNeretin - rare, perhaps, but common enough once you've seen enough phase diagrams. Lead in particular is prone to miscibility gaps. Then you get some weird things like Ag-Cu has a eutectic, Cu-Ni is mainly completely soluble (with a solid miscibility gap at low temperatures), but Ag-Ni has a huge liquid miscibility gap that only disappears around 3000K. And all are fcc metals. So, no, there is no 'rule of thumb' that works.

@NilayGhosh Amalgams are a specific form of mixtures, i.e. when one of the parts is mercury. It is a sub-set of the question, i.e. if we would only see the miscibility of only the water with other liquids. The "rule of thumb" I described, would work: polar liquids mostly mix with water, apolar ones don't.

https://chemistry.stackexchange.com/questions/35014/miscibility-of-molten-metals  https://chemistry.stackexchange.com/questions/42248/why-are-pd-and-pt-completely-soluble-in-cu-but-ag-only-partly-in-cu-when-formin

Bottom line: there are no simple rules of thumb.
To demonstrate this, I will show a round robin of phase diagrams for fcc metals ($\ce{Ag}$, $\ce{Cu}$, $\ce{Au}$, $\ce{Ni}$) with no intermetallic compounds - just fcc solid and liquid.

$\ce{Ag}$-$\ce{Au}$, full mutual solubility, liquid and solid enthalpies of mixing pretty close:

$\ce{Ag}$-$\ce{Cu}$, with limited mutual solid solubility and a eutectic:

$\ce{Ag}$-$\ce{Ni}$, no solid solubility, a huge liquid miscibility gap:

$\ce{Au}$-$\ce{Cu}$, complete mutual solid solubility plus a eutectic:

$\ce{Au}$-$\ce{Ni}$, solid miscibility gap at lower temperatures, but complete solid miscibility above that, a slightly irregular liquidus:

$\ce{Cu}$-$\ce{Ni}$, solid miscibility gap at lower temperatures, pretty ideal liquidus:

Bottom line: even for nominally 'simple' combinations of near-ideal fcc metals coming up with a workable rule of thumb just doesn't work.

After visualising a simplified version of your answer, I found this line: $\ce{Ag-Au-Cu-Ni}$. Metals on this line mix well with their neighbors. Maybe some real number parameter, like an $x$, could be ordered to them, and then we could say, they mix well if their $x$-parameters are close?

...we could even append iron to the end of the list. Nickel mixes well with iron, but not with the others.

@peterh-ReinstateMonica - Indeed, Fe-Cu has limited solubility and a bit of an odd liquidus (nearly a small miscibility gap), Fe-Ag has a huge liquid solubility gap, and Fe-Au has a small eutectic and almost a range of complete solubility in the fcc phase across the composition range. Fe-Ni has complete fcc solubility, and even an ordered fcc low temperature phase.

There are about 75 experimentally observable metals in the periodic table. That means 2775 miscibility disagrams. Is it somewhere available? I think, some AI could find the most relevant and most simple ruleset quite easily.

...it would not need even AI, simple gradient walking would be enough.

Well, the ASM (https://matdata.asminternational.org/apd/index.aspx) is the main keeper of all things for phase diagrams. The Calphad calculation method has lots of adherents, and a journal (and that is where my diagrams are generated from, out of ~1000 I've implemented).


text:
Inductive effect of acetyl group
Question:

Since the acetyl group provides $-M$, methyl $+I$ and chlorine $-I$, it is obvious that the correct order is as given in option $(a)$
However while solving such questions I would compare the inductive effect as well as resonance effect. Thats where I wondered what could be the inductive effect acetyl group has to offer

We have a policy which states that you should show your thoughts, effort and attempts to understand underlying principles and solve the question. It'll make us certain that we aren't doing your homework for you. ( As homework is considered literal homework, self-study questions, puzzles, worked examples etc.) Please provide your full reasoning or thoughts on this, otherwise, the question may get closed. See Homework

Yes I'm aware of that. This is a straightforward question. I couldn't decide whether it shows +I or -I.

I didn't get you. Which carbon is electron deficient and why? The other is amino group

So, it would be better if you could post an image of the question and what your thoughts were. is it  $\ce{-COCH3}$? and $\ce{-NH2}$. the groups attached. If so then the carbon attached is electron deficient and so has a -I effect

You should draw the resonance structures. In the para- position, the inductive effect is dwarfed by the resonance contribution.

Even if the inductive effect is small, what could it be?


text:
Grasping the concept of Electronic Spin, Effective Spin and Fictitious Spin
Trying to learn alone some aspects of quantum mechanics is, sometimes, a struggle. Reading the excellent paper by Piwowarska [1] I was hoping to, finally, understand what is the origin of the so-called “fictitious spin $1/2$” in a $\mathrm d^7$ cobalt ion.

Fig. 2 Visualization of distinction between the electronic, efective, and fctitious spins and respective spin Hamiltonians in terms of the spin levels of $\ce{Co^2+}(\mathrm{3d^7})$ ions

If my cobalt is high spin $(3/2$ is the electronic spin $S),$ then the effective spin $(\widetilde{S})$ is also $3/2,$ but where is the fictitious spin $({S}')$ $1/2$ arising from if the molecule is, in fact, high spin?
Reference

Piwowarska, D.; Gnutek, P.; Rudowicz, C. Origin of the Ground Kramers Doublets for $\ce{Co^2+}(\mathrm{3d^7})$ Ions with the Effective Spin 3/2 Versus the Fictitious ‘Spin’ ½. Appl Magn Reson 2019, 50 (6), 797–808. DOI: 10.1007/s00723-018-1080-4. (Open Access)

Here's what I believe they're trying to say.
First, note that the discussion is mostly limited to high spin complexes (3 unpaired electrons out of the 7 d electrons), so the figure you reproduced is only representative of high spin states.
Second, recall that when D > 0 (ie left side of figure), the five d orbital energy levels split so that we have three t2g orbitals that are lower in energy than two e2 orbitals because of crystal field splitting. When D < 0 (right side of figure), the reverse is true, two e2 orbitals are lower in energy than three t2g orbitals.
In the absence of a crystal field, there is no difference in energy across the five orbitals, so for the case of 7 d electrons, we have a 4A state, which is the pre-split state in both cases in the figure.
In typical simple crystal field models, D>0 is associated with octahedral (i.e. OH) coordination and D<0 with tetrahedral (TH) or cubic coordination, but the authors give examples of TH with both D>0 and D<0, so the sign of D should not be assumed based on geometry here (nor should geometry be assumed based on sign).
Looking at the figure, then, we can start with the right side (because it's easier to understand) and see the transition from 4A to a state of D<0. That is the application of the crystal field. Since D<0, we know that the two e2 orbitals are lower in energy than the three t2g orbitals. Since we have high-spin d7, our electronic state is e24t2g3, with all three t2g electrons unpaired.
What we see in the figure is that there are two pairs of energy levels, with the states with spin $\pm \frac32$ at lower energy than those with spin $\pm \frac12$. The spin $\pm \frac32$ and $\pm \frac12$ values refer to the sum of the spins of the three unpaired electrons. The $\pm \frac32$ states are those in which all three t2g electrons have the same spin (either all $+\frac12$ or all $-\frac12$), while spin $\pm \frac12$ means that one of the three has a different spin than the other two. We know from Hund's rule that unpaired electrons in degenerate orbitals have the lowest energy when they all have the same spin direction, due to the exchange energy term. That is the source of the splitting in energy between the two possible total spin states.
We now get to the "fictitious spin" part. This is actually relatively simple. It has nothing to do with actual spin and simply indicates the number of experimentally relevant spin states, but is expressed in $\frac 12$ increments to have the familiar form of a spin value. Thus, fictitious spin of $\frac12$ just means two possible states, as one would observe for an electron with possible spins $\pm \frac12$. A fictitious spin of 1 would mean three possible states, as we would have for the case of two unpaired electrons.
In the figure, the fictitious spin is $\frac12$ because under the experimental conditions, only the two lower energy levels (ie actual spin of $\pm \frac32$) are observed. The two states with spin $\pm \frac12$ are higher enough in energy that they do not occur.
The left side can be interpreted similarly, except that here, when D>0, the three t2g orbitals are lower in energy than the two e2 orbitals, and the electron configuration is t2g5e22, with one of the t2g and both of the e2 electrons unpaired. Now, since the unpaired electrons are not in three degenerate orbitals, the lower energy states have the unpaired t2g opposite spin to the unpaired e2 electrons (which by Hund's rule again are the same spin).
The fictitious spin is interpreted in the same way on this side of the figure, simply meaning that only two states are accessible under the conditions of interest.


text:
Controlled burn with colored flames (Gender Reveal)
Is there a single solvent that can dissolve chemicals/elements that when burned, will produce the color of its respective chemical/element?
If so, what chemical/element can be dissolved in this solution and burned to produce vibrant colors that can correspond to any of the following colors:

Blue
Green
Pink
Violet

A hue formed by Blue and Green OR Pink and Violet will also work.

A single solvent? Probably not. Different elements have dramatically differing solubilities in most solvents, and so you would have to tune the solvent to the element at least to some extent.

However, for the specific task of identifying a solvent that would (a) burn and (b) dissolve chemicals suitable for giving either a blue/green color or a pink/violet color, then yes, I think it should be feasible.
Aqueous ethanol solutions at a concentration above about $20\%$ w/w are readily flammable at room temperature, and the aqueous base should provide a suitable base solvent for many salts. Vodka might be one good readily-available source of an ethanol-water solvent for this purpose.
To achieve the pink/purple color, potassium chloride should be a good option, as it's freely soluble in water and the potassium gives a purple color in the ethanolic flame.
Getting the blue color might be a bit trickier, as the most common element giving a blue flame is copper, and most copper salts aren't especially soluble in water at neutral pH, or in ethanol. What might work is to acidify the ethanol solution with a high-concentration vinegar stock (I've seen products available at up to $45\%$ acetic acid), and then add the copper salt. Copper sulfate pentahydrate is probably a good option there, as it's commercially available as, e.g., an algaecide. You'd have to experiment with the recipe to find something that burns readily with the desired color, but that doesn't smell too strongly of vinegar.

Did you mean to refer to potassium chlorate or will potassium chloride work for this process?

@josephlevenson As best I understand it, potassium chloride should suffice. The color is coming from the potassium ion, not any combustive/oxidative properties of the counterion.

Great information, Thank you! I will let you know how it turns out. If you know of any other reactions that can be used to reveal gender, i would be really interested to hear your ideas. I have a very large property where fire or fumes/smoke are no threat. your input is greatly appreciated.

I would prefer lithium salts for a red flame. Potassium produces a purple color, but it is not intense. Lithium salts produce much more intense red flames. And copper chloride is more volatile than copper sulfate. So I would recommend copper chloride to get a blue flame.

@Maurice I have experimented with the lithium and it does produce a very nice and aggressive flame. I have placed an order for the copper chloride. Should this be dissolved in the same manner as explained in the original answer? I will be doing this later in the day while the sun is still slightly shining and would require the same burning effect that the lithium gives off.

@ Joseph. If I were you I would start in advance by producing a regular vertical flame (nearly colorless) with a methane or propane or butane burner. A big stable flame, as long as possible. And then, just when you want the rest of the world know the signal, I would spray an aqueous solution of lithium chloride or of copper chloride horizontally through the flame (at the base of the flame). And repeat the spraying operation as often as desired. Each time it will produce a puff of colored light, which is as big as the length of the flame. It does not last more than half a second. But if you are two or three people equipped with such a spray near the flame, each one may send a puff one after the other. The emitted light will nearly look like a continuous red or blue-green flame. This is not dangerous, as only aqueous solutions are used.

That is what I wanted to suggest.

@maurice Unfortunately I feel as though i am running short on time and would not be able to build a burner large enough to display for a group of people back in a field. Do you think i can put a few inches of alcohol in a wide and short steel drum and use this for the same process? Would this flame be hot enough for use with aqueous solutions? can i try to diffuse oxygen into the alcohol for a hotter burn or maybe add some kind of agitation?

@ Joseph. What does mean "a group of people back in a field" ? Three, five, ten, twenty people ? Back ? Back ? In a field ? Who are these people ? Customers ? Members of the family ? Colleagues of the army ? Second, it is not necessary to build a big burner. What for ? How far is the "group of people" ? A regular burner gives a flame 10 to 20 cm high. Is it not high enough ? I must recognize that I am not a specialist of chemical shows. I have never tried to produce huge colored flames. My demonstrations are made for a group of maybe twenty students (one class) in a classroom.

@ Joseph.  You may try to use a wide steel drum as you say. But I would not play with the fire in adding oxygen, air with or without agitation. Be prepared to extinguish the fire by covering it with a blanket, in case suddenly all the alcohol gets vaporized and burns all of a sudden, producing a huge flame.

@Maurice This is a gender reveal for my first born. There will be about 30 family members. The reveal will take place in a large open dirt field. The group will be as far back as required for safety depending on how large i can make the flame and keep it stable considering  wind conditions are like at the time of the reveal. I have also considered mixing up a batch of candy rocket fuel mixing in what ever salt to give off the appropriate color. This could be placed in a 5 gallon metal drum and anchored to the round with the flames faced towards the sky to present the colors to the crowd.

@ Joseph. OK. So, as the birth is not for tomorrow afternoon, why don't you try both colors with whatever drum you want ?

@Maurice I have just spend a lot of time and money buying different chemicals and i cant get acceptable results. This is why i was reaching out for help. I understand and respect the safety aspect of these reactions but i need a very aggressive reaction that will set off with a display of colors

@ Joseph. OK. Aggressive reactions may also be aggressive to you. So try to use a drum filled with alcohol or with something else. I have no experience about candy rocket fuel. it should work if you add some lithium chloride or copper chloride.


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Having trouble naming hydrocarbons with the IUPAC nomenclature rules
First of all, thanks a lot for taking the time. I'm homeschooling and don't currently have any proper teachers, so I thought this would be a good place to ask. Its an easy question I think, but it has me very confused.
In my chemistry book, they showed this hydrocarbon
Translated from Spanish, it would be something like 4-ethyl -1,2 dimethylcyclohexane. Now, my question is, according to the rules, when there is more than one different type of ramification, you number the carbons giving the radical with alphabetical preference the lowest value, in this case, ethyl is first in the alphabet than methyl, so shouldn't it be 1-ethyl- 3,4 dimethylcyclohexane? They would all still be between the first 4 carbons, so the rule of lowest value wouldn't really change. Unless, the di, indicating the 2 methyl radicals counts and gives it alphabetical preference, but to my understanding, those prefixes aren't considered when giving alphabetical preference. Thanks for your time and have a great day!

The first name has the correct alphabetical order of groups: ethyl before dimethyl. The "d" of dimethyl is not involved in alphabetizing. The numbering is correct because the one-to-one comparisons of locants is {1,2,4 < 1,3,4}. The correct name is 4-ethyl-1,2-dimethylcyclohexane. For a brief set of IUPAC rules applicable to this question see, https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm


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How much oxygen is used up by the browning of an apple?
This is inspired by a recent popular question on Cooking.
Let's assume that I take some apples, cut them up, and place them in a hermetically sealed container. Then the container goes into the fridge. The apple pieces will start to get brown, because some of the apple molecules are reacting with oxygen (enzymatic browning). Obviously, since the container is now sealed, this reduces the total amount of oxygen in the container.
Now I had an interesting discussion in comments with another user - their intuition told them that this browning reaction will use up the oxygen so quickly that the apple will brown less than when held in a place with a normal atmosphere. My intuition went to the other direction - that so little oxygen will be used up, that the apple will brown as much as when exposed to unlimited quantities of air. I tried to look up some numbers, but didn't trust my rusty high-schoo-chemistry knowledge enough even for back-of-the-envelope calculations. So here my question: how much of the oxygen will be used up? Especially, will browning stop earlier than with unlimited air?
An interesting side question would be: if the browning does use up significant amounts of oxygen, which will stop the browning first, the system running out of oxygen, or the system running out of the necessary enzymes? But that's an optional question, since getting data on the enzymes may be too complicated.
I am aware that there are too many degrees of freedom for an exact answer. I am perfectly happy with an answer which describes the magnitude of the effect in a way that is easy to grasp, or gives a range for the expected oxygen concentration (even if it's a wide range). To make it more answerable, here are some assumptions:

let's define a container with 100 g of apple and 100 ml air
since I expect that only molecules on the apple's surface will react, I would be interested in a discussion of the two possible extremes - either an apple cut in half (let's assume that's 40 cm^2) or an apple hacked into many small pieces for maximum surface, but not pureed (I don't know how to calculate the surface, does chemistry offer a formula for the minimum particle size before air stops being able to flow through the particles?)
the container is held at 4 Celsius (I hope we can ignore the time in which it cools down from kitchen temperature - but it is filled at kitchen temperature, I don't know if the effect on the total amount of oxygen is significant)
we can keep the apple for a maximum of 5 days before we declare it inedible
we are not using any methods to slow down browning. Just cut the apple, close the container, let it sit undisturbed.
I found a source measuring the amount of phenolic compounds in apples: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4665438/#:~:text=In%20general%2C%20the%20polyphenolic%20contents,per%20apple)%20%5B22%5D. As far as I can tell, the relevant measurement is the 75.7 to 93.0 mg of total phenolic compounds per 100 g of apple flesh, so let's round to 80 (I'm somewhat puzzled by the abbreviation GAE, maybe you know what it is?). I am aware that it lists many different compounds, but I hope that the reactions they do use up a relatively similar number of oxygen atoms, up to a small numeric constant (so maybe compound A reacts with 2 atoms of oxygen and compound B with 3 atoms), so that should be enough for estimating a range.

I think your intuition is OK - even so little air wouldn't be much problem.

@Mithoron I hope you don't mind that I would like something more than an intuition, even though I admit that a chemist's intuition is much more valuable than mine. My nagging "but what if..." voice will be satisfied much better by somebody doing a quick-and-dirty calculation sketching out the magnitude, and besides, maybe there are effects here which I am overlooking completely.

Sure, sure. BTW it's one of things that experiment would tell best and pretty easily ;)

Good question that might serve as the basis for some experiments, and could even be done at home in safety by students! They might use the amount of air as a variable, going from ~none (e.g. heat-shrink plastic wrap) to 100 ml. Antioxidants, such as ascorbic acid, could also be a variable. And when they're done, they can even eat the somewhat-browned apples ;-)


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Why isn't the mass of the things dissolved in water not added to the mass of the water in an enthalpy calc?
Take a change in enthalpy question. When $\pu{5 g}$ of powder is dissolved in $\ce{100 ml}$ of water, the added substance raises the temperature of the water by $\ce{5 ^\circ C}$ and you are told to find the enthalpy change of the system. Since $Q=mc\Delta T, c=4.19, \Delta T=5$, but why does the mass equal $\pu{100 g}$ rather than the combined mass of water and the dissolve powder which is $\ce{105 g}$?

Hi and welcome to Chemistry SE! Going forward, please use $\LaTeX$ to format your posts.

Let's get specific here. The "system" consists of $\pu{100 g}$ of water ($c_w = \pu{4.18 J/gK}$) and $\pu{5 g}$ of powder ($c_p$). If the initial temperature of the powder is $T_p$, final temperature is $T_p'$ and that of the water is $T_w$, then the enthalpy change of the system would be:
$$\Delta H = H_f - H_i \\
= m_wc_w(T_w+5-T_w) + m_pc_p(T_p'-T_p)$$
Due to the system being in equilibrium, the final temperature of the powder $T_p' = T_w + 5$:
$$\Delta H = m_wc_w\times 5 + m_pc_p\times(T_w+5-T_p)$$
As you can see, the enthalpy change obviously depends on the value of $T_w$, $T_p$ and $m_p$. Let's make some realistic assumptions: the "powder" is $\ce{NaCl}$ ($c_p = \pu{0.85 J/gK}$) and $T_p = T_w + 10$ (reasonable to assume, if the water is at $\pu{15 ^\circ C}$ and salt at $\pu{25 ^\circ C}$). Substituting values, we see that
$$\Delta H = 2090 - 21.25 = 2068.75$$
As a percentage, the change in enthalpy of the powder is
$$\frac{\Delta H_{\text{powder}}}{\Delta H} \times 100 = \frac{21.25}{2068.75} \times 100 \approx 1\%$$
therefore, You don't count the enthalpy change of the powder because it's negligible. If you had to account for change in enthalpy of powder as well, you would have been given the specific heat constant and temperature of the powder in the problem.

Note that if we make unrealistic assumptions, the enthalpy of powder would become significant. If you drop $\pu{5 g}$ sodium chloride at a temperature of $\pu{1000 K}$ into $\pu{100 g}$ water at $\pu{298 K}$, then almost all of your water would evaporate and you would have a very different calculation on your hands (one that involves pressure as well).

Who says you can linearly superimpose the enthalpies of the pure substances to get the enthalpy of a non-ideal solution?  That is not  valid.  It is only valid for an ideal solution.


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Backbonding in phosphorous pentoxide
I read this today in a book that $\ce{P=O}$ in $\ce{P4O10}$ consists of a coordinate bond and pπ-dπ backbonding, but why does this happen?
Can't phosphorus share its lone pair with one of the lone pairs available in oxygen to form the double bond?

Get a better book.

Phosphorus can share its lone pair with oxygen to form a bond $\ce{P-O}$. The same thing happens  in a lot of compounds of phosphorus at oxidation number $+5$. This bond may be considered as double or as simple, according to the book you are reading.

There are two different phosphorus oxygen bonds in this molecule. Please clarify which one you are talking about. Please also cite the source for that claim.

Coordinate bonds are covalent bonds where both bond electrons stem from only one of the two bond partners. They are formed when a Lewis base donates two electrons into accepting orbitals of the Lewis acid. That's not what you're looking at in this case, where both $\ce{P}$ and $\ce{O}$ contribute one electron to the common $\sigma$ bond. The interesting part is about everything that is to be described beyond the bond order of $1$. For example, everything that contributes to PO bonding beyond that single $\sigma$ bond.
I will further regard to phosphane oxides $\ce{P(=O)R3}$, where the question was originally asking about $\ce{P=O}$ bonds in $\ce{P4O10}$. The problem might be reduced to PO bonding in phosphoric acid, as $\ce{P4O10}$ is nothing more than its anhydride. The general observations regarding the $\ce{PO}$ bonding situation are unaffected. Kutzelnigg noticed in $1977$ already:

[The PO] bond is somewhere between single and triple and [...] only by chance it may happen to be a double bond.

What we know about the '$\ce{P=O}$ bond':

Its length is remarkably shorter than the P-O bond ($\pu{1.43 Å}$ compared to $\pu{1.60 Å}$ for $\ce{P4O10}$ according to Wikipedia).

It is highly polar.

Its bond strength of about $\pu{540 kJ/mol}$ is higher than for a corresponding single bond.

A critical review regarding the bonding situation in phosphane oxides was given by Gilheany. If it comes to Lewis valence formula, one might find three reasonable structures:

None of those is suitable to represent all theoretical and experimental findings regarding the characteristics of the $\ce{PO}$ bond!
Some scientists prefer to use the zwitterionic representation A, where the predominant $\sigma$ character of the $\ce{PO}$ bond is highlighted, while its dipolar character is also taken into account. If one chooses to write a Lewis structure as shown in B, one implies (besides one  $\sigma$ bond) the contribution of a $\pi$ bond between oxygen p orbitals and phosphorus d orbitals. However, the $3$d orbitals of phosphorus are no suitable $\pi$ accepting orbitals for $2$p orbitals of oxygen, as they are quite high in energy. Besides compatible symmetry, atomic orbitals (AOs) of the atoms in question require to be somewhat similar in energy to allow for a combination to form molecular orbitals (MOs).
If one allows $\pi$ back bonding from oxygen into $\sigma ^*$($\ce{P-R}$) (anti-bonding orbitals between P and its substituents R), one might come up with Lewis valence formula C. However, it should be noted that the $\pi$ bond order for $\ce{Me3PO}$ is only $0.7$ - and not $2$, what is implied by structure C!


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What is vacuum-assisted ultrasonication?
This quote from a methods section of a paper:

[samples were] ultrasonicated for approximately 30 min to ensure
adequate dispersion of the components throughout the mixture. The
solution was then degassed under vacuum-assisted ultrasonication for 1
h to ensure removal of residual oxygen from the mixture...

I know what ultrasonication is. I know what degassing is - bubbling N2 through the solution.
When the author is suggesting vacuum-assisted ultrasonication am I right in my understanding that what they are suggesting is to use a pump to reduce the pressure of the solution during ultrasonication? I imagine this aids gas release by further reducing the partial pressure of the solution.

Is there special equipment for this? I'm not sure the membrane I use
to seal the cuvette will be sufficient hold a strong vacuum.
Should I lower the pressure enough for the solution to boil?
What rate of nitrogen is reasonable? Is this like to defeat the purpose of the vacuum?
What difference will this make compared to ultrasonication and n2 flow through the solution at ambient pressure  - is the vacuum likely to be worth the effort?

The are caveats with vacuum sonication. The key question is what volumes are we talking about, what is your solvent and what is the apparatus? We routinely do this for HPLC but our eluent volumes are about 1 L.

What is the solvent? This is important, as different solvents behave differently - think about DCM vs toluene in a rotary evaporator. I don't know any practical example of vacuum degassing, but as no N2 is mentioned, I think they just applied a vaccuum to a container and put it in ultrasounds. If that is the case, then any RBF, or Schlenk tube, or similar would do it, I guess. But a setup like that works with a high-bp solvent; for things like DCM things get harder

The solvent is actually a viscous polymer - lauryl methacrylate (LMA) (80 wt%) and
ethylene glycol dimethacrylate (EGDM). Would expect around 50-100ml.

Vacuum degassing is also commonly used for viscous materials, such as when mixing PDMS with catalyst.  Without it, the crosslinked polymer product (e.g a cast film) is often full of bubble cavities.


text:
Additive for water to increase the viscosity to a value between 5 and 20 cP
Is there an additive that I can add to water that will increase the viscosity to a value between $5$ and $\pu{20 cP}$? Ideally, I would like the  mixture to be newtonian and since I will be working with gallons of mixture, it would be nice if I didn't have to add too much of the stuff.

5 to 20 cP is a lot! I don't think simple addition of organic solvents can achieve that. What is the purpose? Polyols like glycerin are water miscible and they might be Netwonian.

Do you have some own ideas about this? If you work with gallons of liquids, you must have some experience. (Btw. 1 cP = 1 mPas = pure water)

Without providing the purpose and acceptable additivce properties, the question is very unclear.

You wasn't clear about your purpose to increase viscosity of water (water has viscosity of $\pu{1 mPa\cdot s}$ or $\pu{1 cP}$). However, whatever your reason, xanthan gum would be the materiel you are looking for. Kyoto Electronics has published a note on measurement of viscosity of xanthan gum solution. According to their results (average of 5 tests), 0.1% xanthan gum solution has shown viscosity of $\pu{19.1 cP}$. Five sample averages of 0.2% and 0.3% solutions have given viscosity readings of $\pu{57.54 cP}$ and $\pu{134.6 cP}$.
Accordingly, you can use range of percentages (0.01-0.1%)of xanthan gum to get water viscosity of $\pu{5-20 cP}$. The amount of material to add to water is minimal as you wanted. In addition, Casson plot (shear rate versus shear stress) of 0.1% xanthan gum solution is a straight line. Hence this solution is believed to be a Newtonian solution.

The key problem is that the OP wants Newtonian aqueous solutions i.e, viscosity should remain constant as a function of shear rate. Xanthan gum solutions may be non-Newtonian.

@M. Farooq: Each solution (0.1%, 0.2%. and 0.3%) has a constant shear rate ($52.81, 22.87$, and $\pu{10.69 s-1}$). Do they Newtonian?

Dear Mathew, We need to a see a curve of viscosity vs. shear rate in order to decide whether the fluid is Newtonian or non-Newtonian. If it is a straight flat line, then the solution is Newtonian. Don't worry because the OP has still not bothered to clarify the purpose or the shear rate range.

Thank you to everyone who responded. I've read about xantham gum, but haven't tried it.  CMC is the additive I was thinking of, but I believe that CMC solutions are non-newtonian.


text:
What's the driving force of this reaction?
The question is to find the major product in this reaction-

And this is the answer given-

The first step seems to be a kind of intramolecular SN2 reaction. But how can it occur on this carbon which has a lot of steric hindrance. What is the driving force of this reaction? I have never seen such a reaction that occurs solely on giving heat except, of course, dehydration reactions.
Edit- I am asking about the driving force of the first step.

"What is the driving force of this reaction?" Are you asking for the driving force of the first step, or the entire reaction? "under what topic will it be?" Well, that's not for me to decide, that's for your teacher or your textbook to decide. "I have never seen such a reaction that occurs solely on giving heat" There are lots more out there which I'm sure you'll see in due course. Also, you can't "give heat" to a reaction. You just heat it, or it "occurs upon heating".

Also, I've searched, and I can't find any reports of this reaction in the literature. As far as I can tell, nobody has even made the starting material before. Where is this question from? Is it a reaction that you or somebody else has done?

@Orthocresol It would most likely be a question in a book. [I assume that they were asking OP to use anchimeric effect here]. Also, the curly arrow given here seems to wrong. It should have attacked the carbon connecting the Br and not the adjacent C-C bond

@Safdar Indeed,IIRC the source for this is Himanshu Pandey.

I don't have access to literature sources as of now,and so I can't comment on whether this reaction has actually been done. But what I can say on a purely theoretical basis is that being an intramolecular reaction increases the chances of collisions as compared to an intermolecular case of trying an SN2 with a tertiary halide. Also, not a lot of nuclear rearrangement is taking place in the first step,so that may contribute to the speed of the reaction as per the principle of least motion. But yes, would the methylpyrrole break aromaticity to even do this this reaction is questionable as well

@YusufHasan You recall correctly.

@orthocresol I edited the question and removed the "under what topic" part.

Alright. Indoles are quite nucleophilic and like to do this sort of intramolecular attack on electrophiles, so the SN2 step doesn't strike me as being all that bad, also taking into account what @YusufHasan said (which is right). I don't know what would happen after that, though. I don't know why the reaction should go in the forward direction, or why the product should be favoured over the reactant (after all, there is a reverse pathway involving the same steps), or whether you could even have an alkyl migration in the intermediate cyclopropane (this is also a known pattern with indoles).

Now, the problem with reactions that have never been done before, is that you end up digging for an explanation for something that may not even be real. So I really don't want to come up with some explanation that sounds authoritative but has no meaning.

Looks like a (very hacky) SN-NGP (Nucleophilic substitution with Neighbouring Group Participation) question.