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Resonance structure and stability
Does the existence of resonance structures always translate as stability?
I've been taught that resonance structures generally "spread out" charge in order to compensate for deficiency or excess of electrons. So one could assume that as the number of those structures increases, so does the stability of the molecule, since nature favours it.
However, considering an ester and a ketone and comparing them as to the acidity of their α-hydrogens, things seem to me a bit complicated.
We can draw the structures shown below if I'm not mistaken:

So, couldn't someone suggest that it should be easier for the ester to lose its hydrogen since there is one more structre, compared to ketone? I know that ketones are more acidic and I do understand that the last ester enolate structure is not stable at all compared to the others. But doesn't the existence of it, however brief, affect favourably the stability of the ester enolate?

Resonance is a concept that is necessary to describe some observed phenomena due to the incompleteness of the model (the Lewis structure). From these diagrams alone it is impossible to judge even on a qualitative level how pronounced resonance is; it does not tell you how important a single structure of the set is to the description. Related: https://chemistry.stackexchange.com/q/51632/4945

I see. So regarding the question of acidity?

Your last structure illustrates exactly why it's weaker not stronger - it's impeding delocalisation of negative charge from carbon.  BTW "existence of it, however brief" is wrong - it on't exist, but is contributor to resonance hybrid.

The number of Resonance structures can be a way of describing stability of some molecule relative to the other ones, as it is found to be helpful in understanding many basic ideas about natural phenomena. But higher number of resonance structures can't be the ultimate inference for describing every phenomenon we observe.

For example, when we are concerned about acidity of the esters and the ketones, resonating structures do contribute to the stability of the conjugate base. But we ultimately are bothered about how much the negative charge generated in the conjugate base after losing the proton  is delocalised. More resonating structures in case of esters don't simply mean that the negative charge is more delocalised. It is the cross-conjugation which plays the role of prohibiting the electronic delocalisation in esters and also that is absent in ketones resulting in their increasing acidity.

So, In many cases, it may appear that higher number of resonance structure will help in favour of the phenomenon, but it's always not the case. We may actually have to take in account of the real effect of Resonance to judge the ultimate result.

authors: Αντώνιος Κελεσίδης


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What is the benefit of using Am-241 in smoke detectors even though it is a radioactive element?
I see the labels denoting Am-241 in my smoke detectors. I am having a hard time understand why Am-241 is used in smoke detectors even though it is a radioactive element. It seems to me like a lot of trouble to use a radioisotope in consumer products and notably extra effort given that Am-241 is a synthetic isotope. Why would Am-241 be used in the manufacture of smoke detectors?
The only mitigating factor I can see is that it has a relatively long half life.

You would have to explain how to make a good smoke detector without a radioactive element.

"even though it is a radioactive element" - can you elaborate on why you think this is a negative? By going through your exact concerns you may be able to answer your own question.

My body has carbon in it, some of which is naturally C-14, a radioactive isotope of carbon. Still, it'd probably be best for me to hang onto that carbon. :) Point is, don't get too concerned just because something is radioactive. The amount might be very limited, or in the case of a detector, they manage it properly to limit risk. A question like this can be one piece of figuring out when to worry about it, and when not to. It's about pros and cons, too, how many people (if any) are harmed by a bit of radioactive material in a detector vs. how many are harmed when there is no smoke detector?

The usage in ionizing smoke detectors requires a radioactive isotope to work.  In addition to a sufficient half-life to make a smoke detector with a suitable service life, Am-241 emits mostly alpha radiation (which is used for the ionizing mechanism) with relatively little of the more hazardous and useless gamma radiation.  See the Applications section in  https://en.m.wikipedia.org/wiki/Americium.

It is possible to ionize air directly using an electrical potential difference. But this requires high voltages and wouldn't be supported by a battery.

The radioactive substance is required for the smoke detector to work
Americium 241 based smoke detectors work because of the radioactivity of the 241Am isotope. The basic mechanism inside the detector relies on measuring the amount of ionisation in two compartments in the detector. One compartment is a closed sample of air; the other is open to household air. Detectors in both compartments measure the number of ions created by the alpha radiation from the 241Am by measuring the current between two electrodes. If the household air contains smoke particles, the current is altered in the open compartment because the smoke particles (rather than oxygen or nitrogen molecules in air) tend to pick up the ionisation and this significantly alters the current. This makes a very sensitive detector for smoke particles.
So the whole point of the 241Am is to provide a source of the ionising radiation. And this isn't a problem as you don't need much of the isotope to do the job and it is easy to manufacture in a safe form (were the 241Am is safely contained in a gold foil and can't escape). And the half life is longer than the lifetime of the detector so isn't really a constraint.
There are other types of smoke detector. Some use scattering of infra-red light by particles in air. But they are more sensitive to things like water vapour or mist which give more false alarms.

When we talk of radioactive substances, the main three types of radiation are alpha radiation, beta radiation and gamma radiation.
Alpha radiation is the radiation of a particle which can only travel a very short distance, except in a vacuum.  But if an alpha emitting substance is inside your body it will do a lot of damage.
Beta radiation can travel further in air.  It is a much smaller particle, and does less damage inside you than alpha, but more than gamma.
Gamma radiation can travel through almost anything (except thick lead), but the particles are very small.  We are all being hit by some gamma radiation all the time.  In high doses it is deadly. When your body is hit by gamma, a small proportion of the gamma is stopped: it is that portion that is stopped that causes the damage, the majority that passes straight through does no harm.  And our bodies can cope with/repair small damage.
Diagnostic medicines are given containing gamma emitting substances for the purposes of performing scans.  The type of substance given would be one with a short half-life... long enough to have the scan.  Because it is a fairly small amount of gamma radiation the benefit to the patient outweighs the risk of damage.
You could have a ton of alpha emitting substance sitting in your living room and it probably would do you no harm until you started to crush it up and inject it inside you.  But if it was a high-dose gamma emitting substance you wouldn't be reading this.
Smoke detectors use alpha radiation.  When buying a smoke detector for installation, remember that as long as you don't try to crush up the smoke detector and try to eat it or inject the tiny amount of emitting substance, a life insurance policy probably isn't necessary.  Finally, it isn't a good idea to break the detector, crush up the emitting chemical, get it somehow into a liquid, and inject into the wife (or husband), because it would be tricky to explain it away as an accident, either to her or, in the preferred scenario, to the legal authorities.

Photons are blocked by water vapour, but the alpha particle radiation from A-241 continues through only stopped by smoke.
It's therefore much more sensitive to a single particle of smoke. Water vapour on the other hand, is only able to stop the photons of a photo-electric smoke detector, which uses no Americium-241 at all.
Americium is also likely the most toxic substances on the planet around with a teaspoon able to kill hundreds of thousands if taken intravenous. It's mixed with Gold and beaten into an aluminium container. You could go around burnt buildings to collect them perhaps, with a radiation detector.
The batteries last longer.

This doesn't answer the question.

@matt_black It's actually the only answer which attempts to answer the (implicit) question: What's the unique property of Americium that we need it in smoke detectors (even though it's radioactive)? If I understand correctly, the alpha particles aren't stopped by water vapor which would (wrongly, I assume) trigger a light based detector. Admittedly the answer could be better worded.

Yes, but the answer doesn't explain why or how smoke detectors work. eg "It's extremely sensitive..." What is extremely sensitive? Plus other random but irrelevant facts that don't point usefully to an answer.

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What is the reason behind electron transition in the flame test
I am a beginner in this topic but I am aware that when flame test (for alkali and alkaline earth metals) is done there is a s-s transition of electrons from the ground to their excited state.
But I don't understand why the electron needs to transit back to their ground state if I consider that the temperature of the oxidising or non luminous flame (that is the energy supplied to the electron) remains constant.
So my question is
Why does the electron release the absorbed energy and transit back to the ground state even though a constant external energy is supplied to it?
In other words, what is the need for the electron to absorb the energy , then reach the excited state and release the energy to transit back to the ground state. Instead why does it not remain in the excited state after absorbing the required amount.

I'm voting to close this question as off-topic because it belongs on Physics.SE

@A.K. I was not sure like I had said I just began reading this topic. But if it fits better in Physics.SE...can the question be migrated?

Questions can indeed be migrated, but I disagree that this question belongs on Physics.SE.

Related but probably not a dupe: https://chemistry.stackexchange.com/q/31476

The electron releases its energy because it can. It happens because it has a non-zero probability to do so. If it would never happen it would have zero probability.
Whether the electron just got excited prior to emitting some light does not matter. Particles do not have a memory. Instead such processes simply happen based on whether the requirements are met (e.q. absorbing energy is only possible if energy is available) and on probabilistic rules (quantum mechanics).
Consider some particles in thermal equilibrium: Even the though the temperature does not change on a macroscopic level, the particles still constantly exchange heat with each other. This is called dynamic equilibrium.
The difference to the flame test is, that the particles exchange energy not only by heat, but by light as well.
Exchange of heat is basically a collision of two atoms, so this is a 2-particle process: its rate dependents on the likelihood of those 2 particles "hitting" each other.
Emission of light is a 1-particle process: it can just happen spontaneously. However, absorbing light is a 2-particle process: a photon needs to intercept an atom in order for the atom to absorb its energy.
Heat is the source of energy, and it allows for redistribution between all particles. But for the light, the emission happens much more often than absorption, because 1-particle processes are much more likely to happen than 2-particle processes. Therefore we see the overall emission of light until the heat source is gone.

This may also go into the direction of Fermi's golden rule, where they realized that using Bohr's atomic model and simple physics could not predict why some emission lines tend to be brighter than others, or in other words why some transitions seem to happen more often than other's. This is also where the transition probability  or transition rate is introduced.

Thermal energy is not constant on an atomic level. Heat energy is due to the motion of particles, and is transferred through collisions, or through radiation, which is also quantized. Each collision has a chance of raising electrons to an excited state, where it may linger for a while in a metastable state, or fall back to a lower state quickly. In either case, energy is released on falling back (such as the yellow sodium line).
This might be more easily understood in the case of a glow discharge. Electron collisions ionize atoms, which then revert to a lower state.

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In chemistry what do they mean by disorder when in the context of entropy?
Entropy is often the amount of disorder is present in a system, but when they refer it to the degree of disorder in a system what do they mean. It just seems the term is quite vague and not very descriptive. Does it refer to the number of choices a molecule possesses in interacting with other molecules in a system or does it mean something else?

Rather than using 'amount of disorder', which as you say is vague and is often abbreviated simply to 'disorder', it is more revealing to think of entropy as a measure of the number of different ways 'particles' can be disposed among the available energy levels.

What you have said about entropy is generally correct. For a given energy it is related to the number of accessible states the system can be in. If a system has 100 energetically accessible configurations then we would say it has more entropy than a system with only 10. You can call this disorder because you have no knowledge of which of the 100 (or 10) states the system is actually in.
That said, entropy has a very well define mathematical meaning. I'll start with the punchline and then breakdown what it means. Entropy ($S$) is equal to the natural log of the multiplicity function ($W$, or $g$ sometimes) times boltzmann's constant ($k_b$):
\begin{equation}
S = k_b \ln(W)
\end{equation}
The multiplicity function $W$ is the number of states in the system of a given energy. So what does this mean? Let's imagine we have a magnet with three unpaired electrons which can either be pointed up or down. (e.g. $\uparrow \uparrow \uparrow $,  $\uparrow \downarrow \uparrow$, etc. ) Now let's say that there's a magnetic field applied so that up and down have different energies. Each $\uparrow$ adds $\epsilon$ to the energy while each $\downarrow$ subtracts $\epsilon$ from  the energy. This system can now exist in states ranging from +3$\epsilon$ to -3$\epsilon$.
Let's look at the E = +3$\epsilon$ and -3$\epsilon$ states first. In both cases, W($\pm 3\epsilon$) =1 becasue there is only one configuration for each energy, $\uparrow \uparrow \uparrow $ and $\downarrow \downarrow \downarrow$  respectively. Thus, using the formula, S = $k_b \ln(1) = 0$. Thinking of entropy as disorder, this makes sense. If the energy is $\pm3\epsilon$ then we know exactly which state the system is in and there is no chance whatsoever.
Now let's look at the case of E = $\pm \epsilon$. Here, we have multiple options for each state. In the E = +$\epsilon$ there are three possible states: $\uparrow \uparrow \downarrow$, $\uparrow \downarrow \uparrow$, and $\downarrow \uparrow \uparrow$. Similarly, for the E = -$\epsilon$ case there are also three states: $\uparrow \downarrow \downarrow$, $\downarrow \uparrow \downarrow$, and $\downarrow \downarrow \uparrow$. Thus, $W(\pm \epsilon) = 3$, and the entropy is $S = k_b \ln(3)$
$k_b$ is called The Boltzmann Constant and is just a scaling constant. It doesn't contain any physics and exists only because we happen to use different units for temperature and energy. In SI it's equal to ~$1.38×10^{−23}$ J/K, but you can equivalently measure temperature in energy units and then $k_b =1$ (with no units).
The physical insight is that entropy entirely a consequence system being allowed in multiple states.


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Is there a case when an atom of some element (e.g. carbon) does not have its expected number of protons (e.g. 6)?
Page 12 of the Study Guide for Campbell Biology, 11th Edition has the following question:

A carbon atom has 6 neutrons. How many protons are present in the nucleus of a carbon atom?
A. 12
  B. 8
  C. 6
  D. Can't tell; more information is required.

The answer given on page 431 is D. I think the answer should be C since carbon has 6 protons.
What type of additional information might be required?
I understand that elements are defined by their number of protons, but maybe there is some strange edge case like a moment during nuclear fusion? Is there a case when a carbon atom does not have 6 protons?

The answer may technically be (d), because without any external knowledge of the Periodic Table, there is no way to infer the number of protons from the number of neutrons...

@orthocresol That's an interesting idea, thanks.

@orthocresol it still seems to be a poor question because they do include the reference to carbon. If they just said an atom, that would be better.

All isotopes of carbon have 6 protons. That of course is what makes the atoms carbon. The number of neutrons can, and does, vary. Pure and simple the question has a horribly wrong answer.

A carbon atom has 6 neutrons. How many protons are present in the nucleus of a carbon atom?

Obviously this wording would make any astute student say the answer is 6 since carbon by definition has 6 protons. You are correct, elements are defined by their number of protons. Key word being defined, changing the number of protons changes the element.
I believe the key here that your book is getting at is you only know that you have an atom from a carbon sample that has 6 neutrons but nothing else. I think the question is trying to evaluate if you are aware that proton numbers are independent of neutron numbers thus knowing there are 6 neutrons does not help you determine protons, especially given one of the options is:

D. Can't tell; more information is required.

If this is what they wanted then it was a very poor choice of words to say a carbon atom in the question then ask for number of protons. They should have said:

Given an atom (or nucleus) with 6 neutrons, how many protons are present in the nucleus of the atom?

Maybe an even more well-formulated question would be "atom A has 6 protons, and atom B has 6 neutrons, which elements do A and B correspond to?", which has the bonus of reinforcing the notion that only proton count matters. I presume all chemistry tests/study guides come with a periodic table.


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Extracting sulphur from hot springs with medieval technology?
For the sake of a dungeons and dragons game. Could sulphur be deposited on/extracted from hot springs water through evaporation or other proess?

Sulfur does naturally occur around hot springs and volcanos. It would be very reasonable to assume that a hot spring could have natural deposits of sulfur near it that could be collected without any special effort.

Time to dig into the ancient alchemical texts or modern "witchy" stuff. The ancient knowledge of substances and plants would be lost were in not for herbal healers and Wiccans.
Sublimation of Sulfur to get the "flowers of sulfur"
I would love to try this, but my income levels limit my "craft" to making tea!
https://distillatio.wordpress.com/2013/02/13/sublimation-of-sulphur-fantastic-photos/


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Benzophenone Ketyl color origin
I was recently preparing the benzophenone ketyl radical by reacting it with sodium in toluene. Although this compound is really well known and it seems somewhat obvious and accepted in literature that the radical causes a color I am still unable to find anything on the actual transition. There are some luminescence experiments but at least to my research there doesn't seem to be a good explanation on the actual color.
My suggestion would be that perhaps occupying the former LUMO of the benzophenone to generate a SOMO causes the orbital to shift in energy. I always confuse when something is stabilized or destabilized but since it seems to be in interaction with the two benzene rings I'd suggest that by occupying the orbital you can also lower it in energy a bit and thus promote the excitation of a former HOMO electron into the singly occupied SOMO with visible light.
According to what I found there are various mechanisms depending on the environment around the radical orbital how it can stabilize itself and thus lower the gap between the two orbitals (former HOMO and new SOMO).

TL;DR: The primary transition responsible for color is apparently the ${\pi{-}\pi}^*$ transition of the conjugated ketyl radical anion. This absorbs usually around the green to orange portions of the visible spectrum, leaving behind the purple-blue, and sometimes green color.

Original answer (background info):
According to Wikipedia/Benzophenone, the ketyl radical is formed by first, transition of the ketone from a singlet (double bond) to a triplet (diradical) state, and then abstraction of an electron from a donor to form the ketyl radical anion.
As I understand, only the radical anion is colorful. Both ref 1 and Wikipedia/Ketyl indicate that color disappears if the radical anion is allowed to oxidize (back to the ketone, which should be in equilibrium with the diradical). So my guess is, when an extra electron is added to the triplet state, this enables new electronic transitions that produce color.
These two sources also indicate both solvent and substituent effects: In the Wikipedia article, the solvent is toluene, and color is deep blue for benzophenone. In the paper, the solvent is isopropyl alcohol, and color is greenish-blue for benzophenone, and varies from blue to green depending on substituents. This also suggests that conjugation plays an effect, since the substituents can directly affect only the phenyls. But this, in turn, affects the energy levels of the double bond.

Edit:
A Google search for "benzophenone ketyl spectrum" (including the image search) gives several results of varying relevance, such as refs 2-4. In particular, from ref 2 (dealing with uranium ketyl radicals):

The electronic absorption spectrum of 2 in toluene features an intense absorption band ($\lambda_\text{max} = \pu{562 nm}$, $\epsilon = \pu{5680 M^{-1} cm^{-1}}$) in the visible region (Figure 8, top). This absorption gives rise to the ketyl complex’s distinctive and intense purple color and is assigned to the ${\pi{-}\pi}^*$ transition of a highly conjugated ketyl system. The transition energy of $\pu{562 nm}$ is blue-shifted from that of the sodium benzophenone ketyl radical, which exhibits the ${\pi{-}\pi}^*$ transition at $\pu{618 nm}$.

Ref 3 talks about $D_1 \leftarrow D_0$ and subsequent $D_n \leftarrow D_1$ transitions, and provides an energy level diagram. I assume $D_n$ refers to the doublet states that form upon abstraction of a hydrogen by (or addition of an electron to) the triplet diradical.
Finally, ref 4 adds:

Bzp is known to have a high ISC quantum yield following UV excitation, but until recently the mechanism of population of the lowest triplet excited state ($T_1$) after photoexciting to its $S_1$ (${nπ}^*$) state remained uncertain. ... These investigations indicate that the dominant pathway involves fast ISC from $S_1$ to $T_2$ (${\pi\pi}^*$) as an intermediate state, with the subsequent rate of population of the lowest vibrational levels of $T_1$ controlled by the time scale for dissipation of excess vibrational energy to the solvent bath after internal conversion (IC) from $T_2$.

(Bzp = benzophenone, ISC = intersystem crossing). It then proceeds to discuss these transitions in more detail.

J. Am. Chem. Soc. 1933, 55 (1), 391–395; doi:10.1021/ja01328a057
J. Am. Chem. Soc. 2008, 130 (20), 6567–6576; doi:10.1021/ja801007q
J. Phys. Chem. A 2004, 108 (40), 8147–8150; doi:10.1021/jp047058a
J. Am. Chem. Soc. 2019, 141 (38), 15222–15229; doi:10.1021/jacs.9b07047


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Soda mixed with lemon juice
I mixed 1 tbsp of soda and 1 squeeze of lemon juice in a glass. There was strong reaction with a lot of foam. What caused such a reaction?

Acid bas reaction

If you did this experiment without knowing this reaction will follow, and discovered this independently, then congrats, you're good little experimenter. It does not harm to learn at least some basic chemistry beforehands. Mixing some other household items can be more dangerous.

Soda is sodium carbonate. Lemon juice contains citric acid. What you witnessed was a classic reaction of carbonate with acid to generate carbon dioxide, water and the salt of the acid. It can be a vigorous reaction with a lot of gas evolution - which caused the foaming.

Actually baking soda is sodium bicarbonate.

The OP did not specify baking soda


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Radioactive elements
Radioactive elements emit energy as far as I understand, but what I don't get is how do you know if an element emits more than any other?
That is, how does one measure how much energy an element emits. And is there any general page with some statistics about emission amounts for all the elements?

Do you want to know about the danger from a certain amount and form of specific radioactive nuclids, or the absolute amount of energy that a mole (or kg) of it would emit while it decays completely (or during one hour/ it's halftime/ ... ) ?

More than a century of nuclear physics has yielded a lot of information...

We can measure both the number of particles emitted and their energy
Radioactive elements emit several types of radiation. The major types are alpha particles (2 protons and two neutrons or, in other words, a helium nucleus), beta particles (which are electrons) and gamma radiation (basically high energy X-ray photons). These have different abilities to do harm regardless of their energy (so even paper will block most alpha particles but you might need a large lump of lead to stop gamma radiation).
The first thing that is easy to measure is the amount of radiation. Sensitive detectors can measure every individual radioactive emission passing through them. So we can measure the count of emission events. More radioactive elements have a higher emission count. All other things being equal, a lower half-life equates to a higher emission count. The SI unit for this is the becquerel which is one emission per second.
Measuring the danger from radiation is much harder than just measuring how much of it there is as different types of radiation do different amounts of damage to people or equipment. Alpha particles are much nastier than the same number of beta particles, for example. So the units for this are more complicated and there are more of them. The simplest is the gray (which is 1J of energy absorbed per kg of stuff). But the damage done depends on more than just the energy so there is another unit for biological exposure that accounts for that called the sievert.
The actual energy per emission can be measured by using a variety of instruments. Specific radioactive decays have characteristic emission energies which are often measured in electron volts (eV) which are an energy unit suitable for describing the energy of very small things (an eV is about 1.6×10-19<\sup> joules). Gamma photons have an energy of about 40 keV, beta particles ~500 keV and alpha particles ~5,000 keV.
How dangerous a particular isotope is depends on a combination of things: the energy of the particles it emits; how easy it is to stop that particle before it hits you; and how much damage that particle will do to you if you absorb it.

Do you know of any list or paper that contains each elements amount of radiation?

@Tordin wikipedia has a long list here. I'm sure there are references to more detailed information.


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Are the bonds in the monomers at the end of a polymer filled or unfilled?
In polymers the bonds between monomers are changed, usually from double to single1, to form a long-chain polymer with the now single bond joining another monomer. But in the monomers at the very ends of a polymer chain, there is an unfilled bond, a carbon atom with only three bonds filled. During the polymerization process are those unfilled bonds left unfilled, or is there some process that fills those bonds with, say, another hydrogen atom?

I say this because I don't know if there are any polymers made where the bond inside the original monomer is a triple bond. In all the commonly used plastics (those that have recycling symbols) the original monomers have double bonds between the carbon atoms.

You can polymerise ethine, but the problem is the same. You need to find out how the termination step of your polymerisation works. There are numerous possibilities for every of the different polymerisation types.

And initiation step for another end.

@Karl, I didn't realize there was a termination step involved in polymerization. Of course, that makes sense.

Apparently, this question got a couple of downvotes. I'd be curious as to why. Was the question worded in some way that made it ambiguous? If you thought it was stupid, why? If you thought it was ignorant, well...yeah. That's why I asked it.

The term "(un)filled bond"  is a bit unorthodox. ;-) The term you are looking for is valence. https://en.wikipedia.org/wiki/Valence_(chemistry)

@Karl, the term "valence" to me means the number of electrons an element wants to share or donate with another element. So, should I have said something to the effect that there are valence electrons left to be filled instead of my original verbiage? I thought of using "valence" in my question's title but couldn't figure out how to word it without it becoming too verbose. I'm sorry if that offended you.

Valence is not the same as valence electron(s).  The former is the order of connectivity in a molecule, the latter refers to the number of electrons a single atom has to create valencies.

You are asking about polymers made by radical polymerization.

But in the monomers at the very ends of a polymer chain, there is an unfilled bond, a carbon atom with only three bonds filled.

During polymerization this is true, the radical will continue to react with monomer and propagate the polymerization, but the radical does not sit in the bulk. There are a number of ways which an additional atom amy bond with the carbon radical in a process called termination. Typically this is done by combination or inhibition.
Combination is the process where two radicals come together to form a bond leaving no free radicals to continue the reaction. This may occur when two radical chains combine or when a selected radical is added to cease the reaction. This radical could be the other half of the initiator which began the radical polymerization.
Combination of two polymerizing chains.

Combination of polymerizing chain with radical
For inhibition, a radical combines with a molecule that can form a stable radical such as oxygen or butylated hydroxytoluene (BHT).
 Radical chain combining with oxygen to form stable, non-ractive radical.
In these cases the radical will eventually find a stray radical or atom to bond with (like water), the unstable carbon radical will not exist permanently.

I say this because I don't know if there are any polymers made where the bond inside the original monomer is a triple bond. In all the commonly used plastics (those that have recycling symbols) the original monomers have double bonds between the carbon atoms.

Polyacetylene uses a triply bonded monomer in its production.

The issue is not limited to radical polymerisation. All chain reactions have an active centre that needs to be terminated in the end.

@Karl yes, but because the OP said "* there is an unfilled bond, a carbon atom with only three bonds filled*", I'm lead to believe they are talking about radical polymerization.

All polymerisation reactions must be started (by creating an active centre), then the polymer chains grows (by attacking, in most cases, a C=C bond, which then becomes part of new active centre that get's shifted down the chain), and then the chain must terminate, either by recombination with another radical (in case of a radical polymerisation), or by transfer of the active centre (radical, anion, cation, metal complex) to another molecule.
The latter can lead to a new active (polymerising) chain (a chain transfer reaction) or to the formation of an inactive species (actual termination of the polymersiation reaction).


text:
Sn + HNO3: When will the metastannic acid H2SnO3 appear?
I am dissolving tin in nitric acid.
What reactions will take place depending on the conentration of $\rm{HNO}_3$ and temperature?
Particularly, I am interested when the metastannic acid $\rm{H_2SnO_3}$ is likely to be produced, according to
$\rm{Sn + 4 HNO_3 → H_2O + H_2SnO_3 + 4 NO_2}$
P. S.
References to books or scientific articles are welcome.

It is a common belief that tin dissolves in dilute nitric acid to produce tin(II) nitrate. The reactions are addressed in a previous answer of mine. However, employing a more hot and concentrated nitric acid yields an unstable compound called hydrated stannic oxide also known as β-stannic acid. The compound also has a α-form which is later converted to β form. The relevant information can be found here:

[...]tin dissolves slowly, in very dilute nitric acid to produce
  stannous nitrate (but) when hot and more concentrated acid is
  employed, is very unstable and quickly decomposes, yielding the form
  of hydrated stannic oxide known as β-stannic acid. Probably
  α-stannic acid is first produced [...] which then passes into the
  β-form. The product is, therefore, liable to contain both α- and
  β-forms, and the pure β-acid is obtained by dissolving this product
  in sodium hydroxide solution, and then adding excess of concentrated
  soda, which precipitates sodium p-stannate whilst the α-salt remains
  in solution. Pure β-stannic acid, or β-metastannic acid, is then
  obtained by decomposing the sodium salt with acid. It is also formed
  by the hydrolysis of its sodium salt at 60° C., and by boiling a
  dilute solution of stannic chloride, with or without the addition of
  nitric acid.[...]
[...]Gelatinous, precipitated β-stannic acid has the empirical
  composition $\ce{SnO2.4H2O}$, when air-dried $\ce{SnO2.2H2O}$, and
  when dried in a vacuum, $\ce{SnO2.H2O}$; [...]

Thank you! What if I use heated dilute (say, 10%) nitric acid? Will stannous nitrate decompose in this case?

@bcp In the previous answer of mine, it is said that the reaction products will be a mixture of stannous nitrate and stannic nitrate but the proportion will vary according to temperature and concentration of acid. So, i suppose, hot solution of nitric acid will probably produce more amount of stannic nitrate.


text:
Does anything get precipitated in the half cells during the action of a salt bridge?
In the Zn/cu cell, once the reaction starts, excess ions begin to accumulate in both the half cells. To maintain the electrical neutrality, a salt bridge is used. how does the salt bridge actually work?
For example, if we use NaCl salt bridge, does ZnCl2 get precipitated in the anodic half cell?

It could, but ZnCl2 is very soluble in water. 432.0 g/ 100 g (25 °C) .

Salt bridge works by letting the excess ions move someplace else, so in the end there is no excess.

The function of the salt bridge is to supply appropriate ions to the solutions to maintain electrical balance.
At the cathode, as reduction occurs, positive ions are taken from solution and deposited on the Cu cathode. To replace this, positive ions from the salt bridge will preferentially move into that side of the cell.
At the anode, neutral elemental Zn is oxidizing into Zn+2 ions and negative ions from the salt bridge will move to this half of the cell to maintain charge neutrality.
Very nice diagram and description from U Texas website here


text:
Why in 2-hydroxybutanenitrile is there no syntax saying where in the compound the nitrile is?
I was asked to name the compound in question, and it seems as though the nitrile could be at either end from the name alone.
What removes the ambiguity? Could the nitrile not be either 1 or 4, since there’s an even number of carbons in the chain?

The nitrile can only be at the end of the chain, and you start the numbering from that end because it takes priority

According to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book),

P-14.4 Numbering
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);

Because the nitrile is considered the principal characteristic group, the triply bonded carbon is given the lowest possible locant number '1'.

P-14.3.4 Omission of locants [...] Locants are omitted in preferred IUPAC names in the following cases:
P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

The terminal locant '1' for the suffix nitrile is therefore omitted in the preferred IUPAC name (PIN).
As long as you are following the IUPAC guidelines for the generation of a PIN, there is no ambiguity, because the nitrile must be given the locant '1'. It is just not written explicitly in the PIN.


text:
The intrinsic electron-withdrawing nature of alkyl groups
What you are about to read may be a very mind-boggling paragraph but please do not regard it as nonsense. Please think through it thoroughly.
In Chapter 6 of Organic Chemistry (4th ed.) by Maitland Jones Jr. and Steven A. Fleming, the following is written (p. 237):

In 1970, Professor John Brauman (b. 1937) and his co-workers at Stanford university showed that in the gas phase the opposite acidity order obtained. The intrinsic acidity of the four alcohols of Table 6.6 is exactly opposite to that found in solution. The acidity order measured in solution reflects a powerful effect of the solvent, not the natural acidities of the alcohols themselves. Organic ions are almost all unstable species, and the formation of the alkoxide anions depends critically on how easy it is to stabilize them through interaction with solvent molecules, a process called solvation. tert-Butyl alcohol is a weaker acid in solution than methyl alcohol because the large tert-butyl alkoxide ion is difficult to solvate. The more alkyl groups, the more difficult it is for the stabilizing solvent molecules to approach (Fig. 6.22). Of course, in the gas phase where solvation is impossible, the natural acidity order is observed.

Let me provide some context. The author is discussing the acidity of alcohols in the gas phase and in the aqueous phase. In Table 6.6 as shown above, the pKa values are provided to show the decreasing acidity of the alcohols as the alkyl portion increases in size. The author explains this by saying that it is not really due to the electron-donating effect of the alkyl groups, concentrating the negative charge on the oxygen atom. Rather, he argues that it is due to the decreased extent of solvation due to the steric hindrance provided by the bulkiness of the alkyl groups. This is a reasonable rationalisation.
However, the authors then go on to discuss the acidity of the alcohols in the gas phase and mention that the acidity trend is reversed in this situation. The alkyl groups now increase the acidity of the alcohol, rather than decrease it. The authors then state that it is due to the electron-withdrawing nature of the alkyl substituents. They even provide a qualitative molecular orbital picture of this "electron-withdrawing" interaction, as shown in the following paragraph (p. 238):

Alkyl groups have both filled and empty molecular orbitals (see the problems at the end of Chapter 1 for several examples). A pair of electrons adjacent to an alkyl group can be stabilized through overlap with the LUMO of the alkyl group. (Similarly, an alkyl group stabilizes an adjacent empty orbital through overlap with the alkyl HOMO.)

I would like to ask if anyone has ever heard of such an unconventional rationalisation and whether there is a sound theoretical basis for the postulations made above by the authors. If the postulations are indeed inconsistent with current chemical understanding, please provide a clear response detailing the areas of conflict.

What's supposed to be "unconventional"? That one may have, say  -CH3 < -H < X in one situation and -H < -CH3 < -X in another? Or some hyperconjugation?

@Mithoron The unconventional part is that alkyl groups which are normally said to be electron-donating are electron-withdrawing here.

Ah, seems you didn't got it, all this +/- I has nothing to do with actual electron density transfer. H is arbitrarily put as "0" and substituents making stronger acid as "withdrawing" but actually C-H bonds are polarised to carbon and C-C can be symmetrical. All this is hardly "real" science, just a little something for highschoolers  to have any relations for acidity etc.

Check out hyperconjugation, or the anomeric effect

Related: https://chemistry.stackexchange.com/questions/42535/is-methanol-more-acidic-than-water

I wonder if the pKa for water (15.7) is a typo. Usually we think it to be 14.0 (at 25C, anyway). And I wonder how much the gas phase acidity changes over this series.
Acidity in the gas phase means ROH must dissociate to RO- and H+ but doesn't put the proton onto anything else. Since the proton isn't solvated, it must have been knocked loose, say in a mass spectrometer, where the fragmentation potential can be measured. That just seems quite different from our usual idea of "acidity". But OK, it's an observation of a characteristic of a series of related compounds.
The tendency to designate specific places for electrons (LUMO, HOMO) is understandable, but sometimes this description is uncertain and unnecessary. In a gas phase RO- ion, the negative charge is not 100% around the nucleus of the oxygen atom. Surely it will become unsymmetrical and drift where it might find a little positivity, like on the carbons and the hydrogens. Alkyl groups are polarizable and stabilize both anions and cations (https://onlinelibrary.wiley.com/doi/abs/10.1002/ejoc.200700004).
It's simpler to explain the acidity changes in water and in the gas phase by referring to the size-dependent polarizability of the alkyl groups, but using molecular orbitals may have seemed better for publication.

I never knew that the effect is stabilising for both the cation and the anion. I guess it is not right to say that alkyl groups are electron-releasing all the time. Thanks for the insights. I think I'll just award you the points.

Note that the table compares very small acidities, and the difference for different alkyl substituents is small (although the table for the gas phase data is missing), so the contribution of the alkyl groups to the total electron withdrawing effect (stabilizing the loss of a proton) is small. One doesn't think of carbon as electron withdrawing relative to oxygen, but it does have a higher electron affinity than H. The point, in any case, is to explain relative acidities rather than absolute acidities.
Consider then the general gas phase dissociation reaction
$ X-O-H \ce{->} X-O^- $  + $ H^+$
or alternatively the electron affinity of $X-O$ radicals:
$ X-O^. + e^- \ce{->} X-O^- $
A simple but intuitive physical model that suggests why $[tBuO^-]/[tBuO^.]$ should be larger than $[MetO^-]/[MetO^.]$ in the gas phase is the particle in a box model. If you assume the charge density is uniform in the alkyl groups (reasonable for hydrocarbons), then the extra electron has more room to move about (a bigger "box") in the tBu ion. Since the energy levels are inversely proportional to the size of the box, then the energy of the electron will be lower (more stable) in the larger tBu ion. The greater delocalization in the $tBuO^-$ ion lowers the energy of the electron relative to the case of the methoxy ion, so that the ionization energy is lower and acidity highest for tBuOH in vaccuum.
Arguing in term of LUMOs might be more accurate but a simple delocalization model is sufficient in this case.
It would also be interesting to compare acidities in an organic solvent. See e.g. this article.

That is an interesting perspective. Could you find a source to support yourself?

What would facilitate this delocalisation of charge, considering that the carbon and hydrogen atoms are less electronegative?

Molecules are composed of electrons and positively charged nuclei. While electrons repel the lone electron on the oxygen, the nuclei attract it.  The electrons are happy to buzz about over a larger volume, this lowers their energy, particularly if there are some attractive charges such as protons. This is a variation on the theme that atoms can bond to form molecules despite individual atoms being neutral.

You might consider comparing electron affinities of other compounds, See e.g.  https://pubs.acs.org/doi/abs/10.1021/j100021a060?journalCode=jpchax . Or consider even atomic EAs: https://en.wikipedia.org/wiki/Electron_affinity. Despite the larger size, larger atoms with identical nuclear charge often share comparable EAs, which can be explained as being due to the size of the larger atoms (delocalization).

Perhaps use of the term "delocalization" is a bit loose, since this typically refers to conjugated systems. But the idea is that (i) we are comparing similar compounds, and (ii) electrons are not localized, so when we speak of more electronegative elements attracting electrons keep in mind that the electrons are still buzzing about over as large a volume as possible where there are attractive interactions.

Two of the links above lead to an article discussing electron affinities of completely unrelated compounds.

You are right, I included the link just to provide an additional line of thought on where to look further, not to answer directly your question.

There is absolutely nothing unconventional about this rationalisation, although it appears to not be widely known.
Carbocations are stabilized by hyperconjugation. This is an orbital overlap that can be quantified. The question is whether hyperconjugation forms part of the polarizability (redistribution of electrons). Polarizability seems (to me) to be quite a nebulous term.
Carbanions are also stabilized by orbital overlap, which could also potentially be covered under the same term.
Electronegativity is a poor argument to use to explain the extent and direction of stabilization.


text:
According to MO theory, why is the dilithium cation more stable than the dilithium anion?
According to molecular orbital theory, $\ce{Li2}$ fills up two electrons in the $\sigma_\mathrm{g2s}$ molecular orbital;
$\ce{Li2+}$ fills up one electron in the $\sigma_\mathrm{g2s}$ MO, and
$\ce{Li2-}$ fills up two electrons in the $\sigma_\mathrm{g2s}$ MO and one electron in $\sigma^*_\mathrm{u2s}$, resulting in the same bond order for the ions.
I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_\mathrm{g2s}$, therefore there is no pairing energy.
I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?
Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same. Even though $\ce{Li2-}$ is unstable by filling an antibonding orbital, it will be offset by filling two electrons into the bonding orbital and it has the same bond order as $\ce{Li2+}$ does.
What am I missing now?

First of all some visual aid: an MO Diagram for $\ce{Li2}$.

The concept of bond order allows a very rough estimate of stability. Here we get $\mathrm{BO}(\ce{Li2})=1$, $\mathrm{BO}(\ce{Li2+})=0.5$, and $\mathrm{BO}(\ce{Li2-})=0.5$. So in first approximation, $\ce{Li2-}$ and $\ce{Li2+}$ are equally stable, while $\ce{Li2}$ more stable. But we can consider some more details:

I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_{g2s}$, therefore there is no pairing energy.

Adding a second electron to some orbital, would at first double the energy of that bond. But it also introduces some electron-electron repulsion between the two. This however is usually smaller than the gained bonding energy. So the contribution of the $\sigma_\mathrm{g2s}$ contribution in $\ce{Li2}$ is a little less than double its contribution in $\ce{Li2+}$. Please take this last statement with care, because things are more complex than that.

I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?

Basically yes. The thing is, that the energy split of bonding and anti-bonding orbitals is not symmetric with respect to the originating atomic orbitals. The anti-bonding orbital is a little bit higher, than what the bonding orbital is lowered:
$$
E(\sigma^*_\mathrm{u2s}) - E(\mathrm{2s}) >
E(\mathrm{2s}) - E(\sigma^*_\mathrm{g2s})
$$
This is because if you add more electrons, you add more electron-electron repulsion. In case of $\ce{Li2-}$ this means the electron in the anti-bonding orbital is more destabilizing (with respect to $\ce{Li2}$), than what the additional electron in the bonding orbital is stabilizing (when going from $\ce{Li2+}$ to $\ce{Li2}$).

Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same.

Not just almost. The electronic wave function must follow the symmetry of the molecular structure, therefore the weights of the orbitals have to be exactly the same. But this argument is not really helpful for this question.


text:
Why cyclohexane-1,4-dicarbaldehyde does not give aldol condensation?
I was recently came to know as a fact that cyclohexane-1,4-dicarbaldehyde does not give the aldol condensation reaction in spite of possessing α-hydrogens. I was told that this is due to steric reasons, but I am trying to come up with a more specific reason, which does not require prior knowledge of this fact. Can anyone guide me towards a better explanation?
Edit 1: I am able to understand that there is absolutely no scope for an intramolecular aldol condensation. But what I am asking is whether even the aldol REACTION would take place, as basically the exact wording of the book is " cyclohexane-1,4-dicarbaldehyde gives zero aldol reaction products". So I am confused that what part is aldol reaction and/or condensation

I presume you mean intermolecularly, not intramolecularly. Any CONDENSATION reaction, by definition, must lose a small molecule, in this instance water. With only one alpha-hydrogen, the best you can do is an aldol REACTION (aldol = ALdehyde alcohOL), which is reversible. You need at least two alpha-hydrogens to effect elimination of water.

@user55119 But once it has undergone aldol reaction, it will have formed a beta-hydroxy aldehyde, from where it can surely undergo $\ce{E1CB}$ reaction to form the alpha-beta unsaturated ketone(I think).

It seems that you are considering an intramolecular reaction. Forget it! Any aldol product will reverse to the dialdehyde owing to strain issues. n-Butyraldehyde can undergo an aldol CONDENSATION because it has two alpha-hydrogens vincinal to the aldehyde group. Removal of the first hydrogen leads to the aldol product. Removal of the second hydrogen effects elimination of hydroxide (net loss of water).

@user55119 So by that reasoning, isopropyl aldehyde will undergo intermolecular aldol reaction, but not condensation. Have I followed you properly?

It's isobutyraldehyde. Depends on the conditions as to where the equilibrium lies.  To obtain the aldol of acetone, acetone is boiled in a Soxhlet extractor (look it up, it's like a percolator coffee pot) with Ca(OH)2 in the cup. This way, 4-hydroxy-4-methyl-pentan-2-one is formed and returned to the base-free pot faster than elimination can occur. The dimer has a high enough bp that it doesn't come in contact with the base again.

@user55119 I'm quite confused now. I was told that isopropyl aldehyde will not undergo any kind of aldol reaction or condensation as the alpha hydrogen is negligibly acidic. What part of the mechanism is aldol reaction ,and which is aldol condensation? Also,  how does it have anything to do with the number of alpha hydrogens present?

Your claim is worded very poorly, and taken in the broadest sense, is not true. However, it's more complex than an outright yes or no.
If we're interested in fundamental reactivity, it suffices to look at cyclohexanecarbaldehyde, as there is no reason that the dialdehyde should be any different (apart from the likelihood that it gives very messy reactions in practice). The main issue is that "aldol reaction" is a very unspecific term, which is why you (and everybody else) is having trouble explaining it.

Firstly, it can refer either to condensations where you eliminate a molecule of water after the initial addition of an enolate, or to additions where the product is a β-hydroxyketone.
Formally, any condensation involving cyclohexanecarbaldehyde as a nucleophilic component is not possible, because the α-carbon is tertiary. This includes the self-condensation, where it acts as both the nucleophile and electrophile. After forming a new bond, the α-carbon will become a quaternary centre, which cannot take part in elimination of water.

However, if CyCHO is the electrophilic component (i.e. a different enolate reacts with CyCHO), condensations are still possible. The crossed aldol condensation with acetone is well-documented in the literature:

If we go back to simple aldol additions, which still count as aldol reactions, there are many examples of CyCHO acting as the electrophilic component in an aldol addition. Here is a nice example (Tetrahedron: Asymmetry 2000, 11 (15), 3211–3220):

On the other hand, aldol additions with cyclohexanecarbaldehyde acting as the nucleophilic component are fairly rare. This reflects a general trend that aldol reactions generating quaternary centres (i.e. aldol reactions involving α,α-disubstituted enolates, such as the enolate of CyCHO) are traditionally very difficult to accomplish. This is mainly because of steric hindrance – the reacting centre of the enolate (the α-carbon) is bulky.
The above is a good rule of thumb, but there still are examples of CyCHO acting as the nucleophilic component in aldol reactions. Many of them involve secondary amine catalysis (example below taken from Org. Lett. 2003, 5 (23), 4369–4372), although there is one example of the lithium enolate of CyCHO attacking a carbonyl group.

There are also plenty of examples where the silyl enol ether of CyCHO reacts with π-electrophiles. Silyl enol ethers are weaker nucleophiles than lithium enolates, so generally Lewis acid catalysis is required to form more powerful electrophiles. Still, these are considered to be aldol-type reactions (specifically Mukaiyama aldols).


text:
How do I have to interpret the input to the OCC flag in Molpro?
I'm trying to understand the occupied orbital specification in Molpro. They're specified by the OCC flag, while it's stated in the documentation, that
OCC, m1, m2, ..., mn;
where the mᵢ are the number of occupied orbitals (including core/frozen and closed).
Let's say I want to compute energies in $\ce{N2}$, which has $D_\mathrm{2h}$ symmetry point group, i.e. the following character table:
$$D_\mathrm{2h}\\
\begin{array}{ccc} \hline
\text{No.} & \text{Name} & \text{Function}\\
1& \mathrm{A_g}    & s  \\
2& \mathrm{B_{3u}} & x  \\
3& \mathrm{B_{2u}} & y  \\
4& \mathrm{B_{1g}} & xy \\
5& \mathrm{B_{1u}} & z \\
6& \mathrm{B_{2g}} & xz \\
7& \mathrm{B_{3g}} & yz \\
8& \mathrm{A_u}    & xyz \\
\hline
\end{array}
$$
Then, what exactly does the OCC specification in the following code mean?
{hf-scf;
 occ,3,1,1,0,2,0,0,0;
 wf,14,1,0;
}

In my understanding, it specifies:

3 orbitals of $\mathrm{A_g}$ symmetry, i.e. $s$ orbitals
1 orbital of $\mathrm{B_{3u}}$ symmetry, i.e. $p_x$ orbital
1 orbital of $\mathrm{B_{2u}}$ symmetry, i.e. $p_y$ orbital
2 orbitals of $\mathrm{B_{1u}}$ symmetry, i.e. $z$ orbital

What I find strange with this interpretation is the choice of orbitals - why would I choose 3 $s$ orbitals, while only one $p_x$ and 2 $p_z$?

They specify molecular orbitals, not atomic orbitals; you have to first build the linear combinations. 1s+1s has Ag symmetry, as does 2s+2s, as does 2p+2p (X/y/z depending how the coordinate system is chosen)...

@Martin-マーチン Thank you! Now that makes more sense... Would you mind to write down the answer with explanation of the whole N2 case?

You are correct in your interpretation of the OCC flag (that command corresponds to $3$ electrons in $\mathrm{a_g}$ orbitals, $1$ in $\mathrm{b_{3u}}$, $1$ in $\mathrm{b_{2u}}$, and $2$ in $\mathrm{b_{1u}}$). (Because these are orbitals, not overall wavefuction symmetries we use lowercase letters)
Your confusion seems to be arising from visualizing the orbitals which you are including. Remember that these representations refer to the symmetry of the molecular orbital, not atomic orbitals. $\mathrm{a_g}$ refers to the set of orbitals which have no phase change under all of the symmetry operations, so that corresponds to the $\sigma(\ce{1s})$, $\sigma(\ce{2s})$, and $\sigma(\ce{p_z})$, orbitals. Similarly, $\mathrm{b_{1u}}$ refers to the filled $\sigma^*(\ce{1s})$, $\sigma^*(\ce{2s})$. Finally the $\mathrm{b_{3u}}$, and $\mathrm{b_{2u}}$ orbitals refer to the filled $\pi(\ce{p_x})$ and $\pi(\ce{p_y})$ orbitals.
As we know, the occupation of $\ce{N2}$ is:
$$
  \ce{\sigma(1s)^2 \sigma(2s)^2 \sigma^*(1s)^2 \sigma^*(2s)^2
  \pi(p_x)^2 \pi(p_y)^2 \sigma(p_z)^2},
$$
and using the $D_\mathrm{2h}$ point group we can rewrite this as:
$$\mathrm{1a_g^2\ 2a_g^2\ 1b_{1u}^2 \ 2b_{1u}^2 \ 1b_{3u}^2\ 2b_{3u}^2\ 3a_g^2}.$$
Hopefully, this rewriting of the occupation pattern in the chosen symmetry makes the choice of orbitals clear.
To really understand which orbitals fall under which irreducible representation you need to visualize them (or draw them in the case of a dimer) and then pull out the character table for the point group.
You can find a $D_\mathrm{2h}$ character table on the homepage of Dr. Achim Gelessus (Jacobs University Bremen).

Please do not combine MathJax with normal text, it'll break on many devices (especially with the responsive design.

Group theory tables are available as the linked meta post. Warning: Uses MathJax heavily, very slow site.

Just a typo - $\pi(p_y)$ orbital has a $B_{2u}$ not a $B_{3u}$ symmetry.


text:
If a sample to be carbon-dated is wrapped in a piece of paper, will you still be able to effectively measure the radioactivity of the sample?
If a sample to be carbon-dated is wrapped in a piece of paper, will you still be able to effectively measure the radioactivity of the sample?

Need more info. Can you take it out of the paper? Does the paper have the same age as the sample?

Hardly, with or without paper wrap.
Most of the radiation will be absorbed by the sample itself, you'll have a hard time modelling the absorption to derive the actual remaining activity of the carbon. Artifacts are usually not a piece of completely homogeneous material. Also you'll have a hard time finding anyone with a suitable Geiger counter, if you want to do more than guesswork.)
Modern radiocarbon methods are destructive, you take a number of tiny (<1 mg), representative samples out of your artifact, and directly do mass spectrometry on them, without waiting for radioactive decay.


text:
What exactly is meant by 'multi-configurational' and 'multireference'?
Some topics here have touched on this before (see 1, 2, 3), but I haven't found a clear definition yet.
I would like to know what exact property of the wave function these terms refer to. It would also be helpful to have a clear definition of 'reference' and 'configuration'. I'll try to explain below where my problems are in clearly understanding/defining these terms:
Starting with Hartree-Fock, it is obvious that the wave function in HF is both single-reference and 'single-configurational': there is only one Slater determinant.
Going to configuration interaction methods, the wave function now becomes a linear combination of several Slater determinants. The additional Slater determinants are excitations of the ground state determinant: virtual orbitals from HF are taken and replace previously occupied orbitals in the determinant. However, these orbitals still have the same coefficients as in HF - only the coefficients in front of the Slater determinants are optimized for the linear combination. If I'm correct, 'configuration' here refers to one particular Slater determinant - a method is, therefore 'multi-configurational' if the wave function that is used has two or more (different) Slater determinants, correct? That also means that none of the CI methods (be it CIS, CISD,..., or Full CI) are multireference methods?
Continuing with CASSCF, this method is basically Full CI limited to the chosen active space of orbitals. It is therefore multi-configurational. At the same time, it is also often referred to as being 'multireference'. The only difference to CI, however, seems to be the optimization of the coefficients in the Slater determinants themselves, hence, this must be the defining criterion for 'multireference'? What does 'reference' here refer to?
Now there is also multireference-CI. From the above definition, I would expect this to be a form of CI where I also optimize the orbitals, but that does not seem to be the case. The Wikipedia article on MRCI starts with:

In quantum chemistry, the multireference configuration interaction (MRCI) method consists of a configuration interaction expansion of the eigenstates of the electronic molecular Hamiltonian in a set of Slater determinants which correspond to excitations of the ground state electronic configuration but also of some excited states. The Slater determinants from which the excitations are performed are called reference determinants.

This is confusing to me: Is 'excitation of the ground state electronic configuration' vs. 'excited states' referring to the optimization of the Slater determinants themselves? Or does 'excited state' refer to configurations with different total spin? That would be a different definition of 'reference', but then CASSCF would only be a multireference method if it uses the corresponding SA-CSFs, regardless whether the Slater determinants are optimized or not?

Excitated S.D. are built by replacing orbitals of some reference det. If a method uses many of them is multiconfigurational. Often those SD are share most orbitals with the ref. det., and are derived from last one, so only one ref. det. is involved (monoreference). If the excited orbs. are derived from more than one reference, then the method is multireference.

Your problem seems to be with the terminology used in CI methods, so let me go through the different terms you mentioned:

A configuration is a certain occupation of (molecular) orbitals. Mathematically configurations can be represented in 2 ways. The first one is the Slater Determinant (SD), an anti-symmetrized product of spin-orbitals. Slater Determinants, however, are not eigenfunctions of the spin operator $\hat S^2$, but the electronic wave function needs to fulfill this requirement. Therefore one constructs spin-adapted Configuration State Functions (CSF) as certain linear-combinations of SDs. Using CSFs instead of SDs usually makes a calculation more stable. "Configuration" is a more general term, which does not explicitly say whether an SD or CSF is considered.
Multi-configurational just means the method considers more than one configuration.
Reference means we have a designated configuration from which the excitations are generated. Single-reference thus means we only have one such configurations (usually the HF configuration), e.g. in CISD or CCSD. Multi-reference means we have more than one configurations to generate excitations from.

So multi-configurational just means we have many configurations, while single-/multireference says something about how those configurations are selected/generated.
How does single-/multireference apply to the different methods?

In HF we do not generate any excitations, therefore this concept does not really apply here. But I think one could still call it single-reference and single-configurational.
CISD and CCSD are the typical examples of single-reference methods. Nothing special here.
In FCI and CASCI we don't restrict the excitations to certain degrees (Single, Double, etc.), instead, we just take all of them. The concept of having one (or more) reference configurations is a possible perspective, but it is not really necessary. FCI can be viewed as having one reference configuration and taking all excitation degrees (up to the number of electrons available), which would make it single-reference. But from another perspective, we can argue that every MRCI wave function is just a truncation of the corresponding FCI wave function. So if FCI covers everything and more than MRCI does, would it not be multi-reference as well? Again, I would just say the concept does not apply here.
MCSCF (CASSCF) is a combination of CI (CASCI) and SCF, where SCF means we optimize the orbitals as well. This is not done in CI(SD...), CC(SD...), MRCI, etc. Other than that, the above arguments apply in the same way to the CI space as before.

Multi-Reference CI and CASSCF
As argued above, personally I would not consider CASSCF to be multi-reference, as the concept does not really apply. But I can see why people would consider it as such.
One usually does CASSCF because single-reference methods fall short of describing the wave function correctly, in a qualitative way. Such systems are then called strongly correlated and CASSCF can treat that strong correlation (also called static correlation) which is missing in single-reference calculations. In turn, however, CASSCF is missing that kind of correlation single-reference method can treat well, called weak or dynamic correlation.
Multi-reference is now the approach to combine both, strong and weak correlation, by doing first a CASSCF, and then using those configurations as a reference space, to generate all the excitations from.
Orbital optimization, however, is usually not feasible for dynamic correlation, because dynamic correlation usually means to have a lot of configurations and therefore huge CI space. With static correlation there are much fewer configurations, so doing orbital optimization additional to CI optimization is feasible and yields qualitative improvements of the orbitals. Therefore this is done in CASSCF, but not in MRCI.
"Excited states" in MRCI
What is meant here by "excited states" are actually "excited configurations" (with respect to the HF configuration). If you, for example, choose a CAS as your reference space, then all configurations, except for the HF configuration itself, are excited configurations.
Please note, that "excited state" is not a good choice of words here since it is actually something different: a state is a result of a CI calculation and is physically observable. On the other hand, a configuration is a basis function you put in a CI calculation and therefore is more of an abstract mathematical tool. However, you can approximate a state as a configuration to some extent, this is for example done in HF.

Very clear answer! The CASPT2 method then, which does second order perturbation theory on the CASSCF wave function (which is the reference configuration) is a multi-reference method. Is this right?

@TippeTop: Yes. Although it is "reference configurations", plural. Maybe just a typo ;)

Oh yes, that's right! :)

Does a state include many configurations with same state symmetry? And reference means reference configurations rather than reference states?

Yes to the first question, although I would not use the term "include", better say "all configurations of same symmetry form the basis for a state of that symmetry".
About the second question, I think that would depend on the context.

I would like to add to the above description.  The reference states involve the spin function.  For two electrons there are four spin functions, aa, (ab + ba)/sqrt 2, and bb for the triple state and (ab - ba)/sqrt 2 for the single state. (a is spin up and b is spin down.)  For three electrons the spin functions are aaa, (aab + aba + baa)/sqrt 3, (abb + bab + bba)/sqrt 3, bbb for the quartet state. There are two spin functions for the three electron S sub z = 1/2, S = 1/2 state: (-2aab + baa + aba)/sqrt 6, and (aba - baa)/sqrt 2.  Including one of these spin functions in the wave function is a single reference wave function. Including both of these spin functions is a multireference wave function. For four electrons S = 0, S sub z = 0 state there are two spin functions: (ab - ba)(ab - ba)/2, and (2aabb +2bbaa - abab - baba - abba - baab)/sqrt 12.  The first of these is the spin function used in the single determinant Hartree - Fock wave function.  Both of these spin function are used for multireference situations.  Ruedenberg discusses the spin function in some of his MCSCF papers, such as Ruedenberg and Sundberg, in Quantum Science, J.-L. Calais et al., Springer, 1976, pp 505 -515.

This answer, or comment, could profit from some MathJax and some paragraphing.


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How can I make an alloy wetter?
I'm trying to make a Bismuth-Indium based low-melt alloy wetter (specifically, better at wetting to solid metals and glass). I've tried adding gallium, which obviously works, but due to its metal-attacking nature it can't be used here.
I know that antimony and tellurium are often used as surfactants, but I've only read of them acting as surfactants in organic solvents, and can't find anything on whether alloying them with a liquid metal would improve its wetting ability. As I don't have any Sb/Te at present, I'd rather ask if this has any chance of helping before trying them.
If not, is there anything else I could incorporate into the alloy - possibly a non-metal/metalloid - which would improve wetting?

I don't believe Sb and Te can act as surfactants in organic solvents. What you have read must have related to metal alloys.

Be careful with tellurium. I have heard in ( highly recommended ) the Episodic Table of Elements  podcast, that Ingesting tellurium, or even holding it in your hand for a quick second, will cause your breath and body odor to reek with a somewhat garlicky smell that is guaranteed to repel vampires — and anyone else who crosses your path, too. The body metabolizes this element into the extremely pungent dimethyl telluride. ....   it will generate breath so foul that birds will fall from the sky, the city gates will be closed, and new widows will weep.

The surface of glass has many Si-O-H and Si-O-Si bonds; deeper inside would be Na and Ca and Al. An active metal might engage those surface atoms. Na might be too reactive (forming a dross of NaOH on the surface of your alloy, or reacting with water at the glass surface). Aluminum or zinc or perhaps magnesium might be active enough to contribute some sort of adhesion to glass.
Tin oxide is bound to glass surfaces for electrical conductivity. Perhaps it could be an intermediate layer, bonding glass and your alloy. And silver is deposited on glass to make mirrors. Your alloy might well stick to a silver deposit. So, while this doesn't actually make your alloy wetter, it could solve your problem of adhering to glass.

Titanium and chrome are great sticking layers onto glass. Of course, that just changes the wetting problem to a different interface...

Silver can wet glass directly but is usually applied via a chemical reaction depositing silver on the glass surface. Modern silvering processes improve this by pre-treating the glass with chemically deposited tin.

If I remember correctly, thermometers that use galinstan alloy as the fill liquid, in place of mercury, use a tin oxide coating on the inside capillary surface, to prevent wetting.

@EdV Tin oxide may be an alternative to gallium oxide the corresponding wikipedia entry mentions to prevent wetting.  Interesting entry, both for the crosslink as (potential) application of this material for liquid mirrors as well as anode in brilliant microfocus X-ray tubes.

@Buttonwood Ah, thanks for the correction! I had not seen that linked wikipedia entry or the applications beyond thermometry.

@EdV I don't exclude tin oxide being of use for galinstan, since typical floating glass (naturally) is anyway doped by tin on the side it was floating on the molten tin (on occasion this side may be identified by its fluorescence in UV); and for ITO, where doping is intentional. Re wikipedia, the index of its chemicals is public, and these are easier retrieved on the Wikipedia Chemical Structure Explorer.


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Are the stereoisomers of nucleosides actually anomers?
An anomer is an epimer at an (hemi)-acetal carbon like $\ce{R2C(OR')2}$.
Why is the 1'-position of a nucleoside referred to as the anomeric center despite the lack of acetal functional groups?

They're just like sugars, just with nitrogen instead of O. One could call them hemiaminals maybe.

Well sure, but I don't think that justifies calling the isomers 'anomers'.

The C1 position of carbohydrates and carbohydrate derivatives is generally referred to as the anomeric center and compounds differing only in their configuration at this position are called anomers. In nucleosides, the 1'-position may not be part of a hemiacetal or acetal functional group, but it is at the same oxidation state. The IUPAC Gold Book defines anomers as:

Diastereoisomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.

The fact that the definition includes "glycosides" and "related cyclic forms of sugars, or related molecules" points towards why it is acceptable to describe the 1'-position as the anomeric center. Furthermore, this position in nucleosides is at the same oxidation state as a hemiacetal or acetal. Therefore, there are structural and chemical reactivity similarities between these functional groups and the 1'-position of nucleosides.
The correct name for this R1C(OR2)(NR3R4)R5 functional group is a hemiaminal ether. Hemiaminals have a free hydroxyl group: R1C(OH)(NR2R3)R4.
Also of note is that the anomeric effect refers to the preference for polar groups to be axial on pyranose derivatives, where the polar group may be not only oxygen substituents (eg. hemiacetals, acetals), but also other derivatives such as halides (eg. glycosyl halides). In all these cases, C1 would be referred to as the anomeric center.

To add to the correct anomer definition cited by Mr. Organic Chemistry Explained, also note, that from the carbohydrate chemistry nomenclature standpoint, these are N-glycosides, more correctly N-glycosyl compounds (or glycosylamines), and the α/β (not a/b, like in your image) anomeric configuration notation is used in the names, see in Nomenclature of Carbohydrates (Recommendations 1996):

2-Carb-33. Glycosides and glycosyl compounds
(…)
2-Carb-33.6. N-Glycosyl compounds (glycosylamines)
N-Glycosyl derivatives are conveniently named as glycosylamines. In the case of complex heterocyclic amines, such as nucleosides, the same approach is used.
Examples:
(…)

1-β-ᴅ-ribofuranosyluracil
  (trivial name uridine)


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Solving two electron integral numerically
Is there any software package in python or C to evaluate two electron integral repulsion integral in quantum chemistry of atoms. I am trying to solve the helium atom by constructing many body basis.Using those basis i am attempting to evaluate matrix elements.Also please suggest some references to understand more about two electron integral.

There are https://github.com/evaleev/libint and https://github.com/sunqm/libcint for calculating integrals over GTOs, if that is what you mean. For standard literature take a look at the books by Szabo and Ostlund as well as Helgaker, Jorgensen and Olsen, both about electronic structure theory.

The canonical text on quantum chemistry is Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund. Every electronic structure theorist I have ever met keeps a copy of that book within arms reach of his/her desk.
As for software, I would recommend PySCF. It is an entirely free python library for quantum chemistry with the most critical routines written in C and is an excellent platform to develop on. It's also robust and fast enough to be used for large calculations should you want to scale up. (https://github.com/sunqm/pyscf)
PySCF uses libcint to perform the integrals, and contains a low-level interface to the integral evaluator. An example of how to use these methods to make a df-HF calculation is shown in the docs here: https://sunqm.github.io/pyscf/tutorial.html#molecular-integrals. A complete list of available integral routines from the libcint package is shown here (it's very comprehensive): https://sunqm.github.io/pyscf/gto.html#module-pyscf.gto.moleintor

https://github.com/sunqm/libcint is the actual library used by (and I assume written for) PySCF for these integrals.


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Feasibility of application of Boyle's law in the case of varying number of moles
While reading up on ideal gases and the Boyle's law, I encountered a question in my textbook wherein a balloon was being filled with helium (assumed to be ideal) till it was just about to burst. The pressure and volume coordinates at some intermittent time of the process, and the limiting pressure at the bursting point were provided, and the volume at that point was asked. The book solved it by applying Boyle's law at the two $\ce{P,V}$ coordinates. Also, it was given that this entire activity is being carried out at room temperature.
So, if the amount of gas in the system is constantly changing, then can we apply Boyle's law in this case? If not, then can somebody provide a solution?

It may help to quote the actual question from this textbook.

A balloon is filled with Hydrogen at room temperature.It will burst if pressure exceeds 0.2 bar.Of at 1 bar pressure, the gas occupies 2.27L volume, up to what volume can balloon be expanded.

As per the textbook , They simply applied P1V1=P2V2 as per Boyles law

Possible duplicate of Volume change inside a balloon upon decreasing the the outer pressure

You are correct that using Boyle's Law wouldn't work for the problem as described. I assume the author of the question had some thought of what they wanted the question to be about, but didn't make sure the real situation made since. There is an early question that addresses this same problem, so I marked as a duplicate. @PrakharPandey


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Aluminum and aluminum oxide
I have a new grain mill that I am hesitant to use because the rollers are anodized aluminum.  As I understand it the anodized aluminum is actually aluminum oxide and supposedly non-toxic.  I realize that the aluminum/alzheimer connection is in doubt (but I have seen it fluctuate as so many "certainties" do).  I do not wish to take a chance playing with aluminum in my food, so can you answer the following questions:
Given that eventually the aluminum oxide would wear off, exposing me to the aluminum.  Would the aluminum form a new protective oxide and if so, how fast would the aluminum re-oxidize and reform a protective barrier? (instantaneously, days, weeks?)

Pretty much instantaneously. On the other hand, very little food has a hardness exceeding alumina, so the wear will be minimal.

I am not an expert in food safety, so understand this is for information only.
Aluminum, when exposed to the atmosphere, will immediately oxidize.  Aluminum oxide is not very soluble an will likely pass straight out of the body when ingested.  Aluminum oxide is in certain toothpastes as an abrasive.
If aluminum is exposed to fluoride, chloride, or sulfate there could be an issue if it converts to a soluble form.  As long as grinders are cleaned according to the manufacturer's recommendations this should not be an issue.
The amount of aluminum oxide which wears off a grinder is related to the abrasiveness of the material being ground.  Stainless steel wears about six times less than aluminum.  A titanium grinder is even better.


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Slater Determinant in Python package
Is there any library functions which I could use in python to construct slater determinant? Can someone suggest me something to fix this issue ? Is it really possible to determinant with functions as elements of the determinant ?

Maybe https://www.sympy.org/ can be of help.


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How am I supposed to answer this?

First the question gives no units or that these are covalent radii or what, so I assume they are in nanometers and assume an atomic radius. I check my periodic table app and it says a 92pm atomic radius, 75pm covalent and 155pm van der waal for the N atom, which happens to be the first option I crossed out. Now what? BTW, this question is from a previous national aptitude test.

If you remove the outermost electrons to give the ion do you think the atomic radius goes up or down?

In problems with self-contained data you should look for consistency, not "truth."

@Waylander of course the radius goes down, even I know that. But the question asks about adding electrons not removing them.

Effectively, the question is giving the radii in some sort of reduced units, where it's expressing the size of each species relative to the size of something else.
Thus one can interpret the choices in this question as providing relative sizes.  So, for instance, choice "A" says the ion is 2.60/.77 = 3.38 times the size of the neutral atom, and so on.
Given your knowledge of the actual sizes of the ion and the neutral atom, you should now be able to choose the best answer.

But I don't know the size of the $\ce{N-3}$ ion.


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Significant Figures Calculations and Precise Instruments
I understand the rules for significant figures and how to read instruments, except for one thing.
I was told that our digital scale is particularly precise to the 1,000ths decimal place.
So in multiplication/division, the answer should have the same number of significant figures as the term with the least number of significant figures.
So if I weigh something on a balance accurate to the 1000th's place that happens to read 0.300g. Then I should record that weight as 0.300g.
The confusion for me is that when we determine significant figures from abstract numbers we first have to determine the precision(significant figures) of each term. Well, 0.300g is just one sig-fig. So if I were to then divide it by a volume to find the density and the volume is 0.111 Liters the result would be 0.03g/ml.
What was the point in writing it down initially to the precision of the instrument (0.300g instead of 0.3g) if in terms of calculations we treat it as a low resolution(imprecise/low-sig-fig) number? OR am I doing it wrong, and the significant figures are determined via the precision of the instrument.
My understanding of the purpose convention of significant figures it to let people know the quality of the measurements. Well if you have a high precision instrument, and happen to weigh something that has only one-sig fig isn't this misleading?

I think you are doing it wrong. Given 0.300g, I would read that as having 3 sig figs while 0.3g would be 1 sig fig. It sometimes helps to write these in scientific notation to see the difference (i.e. $3.00\times10^{-1}$ vs $3\times10^{-1}$)

If you are saying 0.300, then you are expressing certainly about that second zero. But then it's significant, isn't it?

I am doing it wrong. I was mixing the rules for decimal vs no decimal. Woops. I appreciate everyone's answers. Since my question was malformed I'm not going to credit an answer but I will upvote everyone's participation here.

If your balance weighs to a thousandth of a gram, and you happen to get a weight of 0.300 grams then you have 3 significant figures. The first zero before the decimal point isn't significant. But the three, and the two trailing zeros are.
This is a bit different than math class. You actually measured the two trailing zeros. However you can't increase the precision of the measurement by tacking on say 3 more zeros and turning the "measurement"  into 0.300000.
In the olden days when using a slide rule, or log tables, carrying three significant figures in calculations was as good as could be done. Now with calculators carrying a few extra digits in intermediate calculations ins't a problem. So I'd recommend that you carry two extra figures in intermediate calculations to try to reduce rounding errors, and then round the final result to the appropriate number of figures.
So:
$$\dfrac{0.300\text{ grams}}{0.111\text{ liters}\times 1000\text {ml/liter}} \approx  0.0027027\text{ g/ml}\ce{->[rounding]}2.70\times10^{-3}\text{ g/ml}$$


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Building a Salophen-Complex with Iron(III) and Cyanate

I prepared the compound in (1). Now I want to prepare the complex with a cyanate ligand (2).
Is it possible to prepare Fe(OCN)3 or should I form the chloro-complex first and then change the Ligand?
I searched for available salts, however there is no iron salt. There are only potassium cyanate, sodium cyanate and other salts. Is there a particular reason why iron salts are not available?


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Why are solids and liquids ignored in the equilibrium constant?
If a reaction is taking place and is going towards equilibrium, wouldn’t the reactant be consumed to form products and vice versa? If this reactant/product is a solid or liquid, wouldn’t the number of moles when the reaction reaches equilibrium decrease/increase from its initial number of moles? Wouldn’t this change in moles from start of reaction to point of equilibrium be important to the equation, since it’s a ratio of products to reactants at equilibrium?

Well there isn't quite enough information to answer the question. A specific problem would allow a precise answer. However, in general, products in another phase don't appear in the equilibrium expression. So in a solution, gaseous or solid products wouldn't be used in the equilibrium expression.

A solid or liquid that should take part in a solution or gas phase reaction has to dissolve or evaporate before it can react in any meaningful way.
If you add more solid solute to a saturated solution, its concentration does not change, and so doesn't the equillibrium constant of a reaction of the dissolved solute.
(The solute might also react directly out of the solid. Theoretically. But this different reaction needs much more energy than the one from solution, and it will only add a minuscule shift to the total equillibrium, even if your solid has a huge surface. Surely neglectable, most likely not even measurable.)