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Can orbitals in a molecule be occupied via excitation like they can in atomic orbitals?
I am facing a conundrum in the way I am viewing molecule orbitals due to the existence of excited states. What I wanted to know is, since atomic orbitals can exist in excited states (i.e. you get hydrogen accessing states outside the 1s orbital and such when a high-enough energy photon is used), is this the case with molecular orbitals, where 'technically' many more states exist than shown on a molecular orbital diagram? I mean, they tend to only show the valence and core electrons, but not past that, hence the question. For example, could methane, that only whose MO diagram only shows molecular orbitals composed of 1s, 2s, and 2p orbitals, have an electronic transition that occupies a 3s orbital? And if so, how would that occur? Would it occupy one particular atom or be 'spread out' like most electrons in MO theory?
Sorry, long question, but something that is bugging me for some strange reason.

Methane does not exactly have 1s, 2s, and 2p orbitals. Instead, it has molecular orbitals composed of those. And yes, any molecule has multiple excited states that can be reached.

Okay, fixed that. Tend to make mistakes in terminology when I write in a rush.

Most of what applies for atomic orbitals in atoms can be transferred to molecular orbitals in molecules. This includes that electrons can be excited to higher lying orbitals, creating excited configurations.
Also, just as for the hydrogen atom, there are in principle infinitely many orbitals available. They are simply truncated since the higher lying ones are less important (and for numeric reasons).

Thanks for the answer. In the case of virtual orbitals, would the 'cut-off' for 'important' orbitals tend to be set really only to those that aren't outside your typical molecular orbital diagram for that reason? I mean, of course excluding the addition of things like polarization basis sets and such. That is actually why I got the question actually - was reading about post-HF methods and was wondering about what they consider as the "full CI" if there are technically infinite states... guessing it must just be the 'important ones' i.e. valance electrons, maybe some rare Rydberg ones, yes?

Usually the cutoff includes all valence orbitals (or what you can form fram all valence AOs). Things like polarization functions are about the AO basis. But also influences the MOs, since you get as many MOs as AOs are supplied. Which also means FCI depends in the AO basis set. It can only consider as many MOs as are available.

I guess I would like to ask one more addition to what we are talking about, if you don't mind - do we know the defined shapes of these molecular orbitals like we do for, say, the ones generally shown in MO diagrams? Or is it a bit more abstract since we are talking about more complex mixtures of AOs?

If am not sure if I get your question right, but occupied orbitals can be quite intuitive, e.g. core orbitals which are al,pst identical to AOs, or bonding MOs where you can relate the number of nodes to the bonding character. These orbitals which are "generally shown in MO diagrams" do come somewhere from ;) However, virtual orbitals get more abstract and occasionally change with the AO basis set. Maybe this is worth a separate, new question ...

Customarily, LCAO molecular orbital schemes are drawn using the atomic orbitals of the valence shell, ignoring both core orbitals and most higher-shell orbitals. For example:

the MO scheme of methane will contain the $\mathrm{1s}$ orbitals of the hydrogens and the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of carbon because these correspond to the valence shell
the MO scheme of $\ce{CO2}$ will use the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of both carbon and oxygen and ignore the $\mathrm{1s}$ core orbitals as well as all higher orbitals
the MO scheme of $\ce{HOCl}$ will use hydrogen’s $\mathrm{1s}$ orbital, oxygen’s $\mathrm{2s}$ and $\mathrm{2p}$ orbitals and chlorine’s $\mathrm{3s}$ and $\mathrm{3p}$ orbitals (see also the next example)
when constructing the MO scheme of hexafluoridoferrate(III), the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of fluorine are used while the $\mathrm{3d, 4s}$ and $\mathrm{4p}$ orbitals are used — these together correspond to the valence shell.

For each atomic orbital entered into the equation the solution must include exactly one molecular orbital. So having two orbitals interact will always result in a more bonding orbital (the same phases overlap) and a more antibonding orbital (opposite phases overlap). Thus, we will end up with $8, 12, 9$ and $33$ three orbitals respectively for the examples above. This already provides us with ample orbitals to excite electrons into; in the methane case, these would correspond to four $\unicode{x3c3}^*$ orbitals, three of which are degenerate due to symmetry.
Nothing is stopping us from including the core orbitals in the calculation or additional unoccupied atomic orbitals. In fact, quantum chemistry programs often include these orbitals in the calculations. The core orbitals do no harm (you will end up with a set of MOs at very low energy that are basically confined to the nuclei; linearcombining these will lead to exactly the set of core orbitals you threw into the equation). The virtual orbitals are actually typically used as polarisation functions to provide a better final result.
Including the virtual atomic orbitals will merely create more virtual molecular orbitals at the top of the energy scale. These can indeed be accessed by excitation with photons of appropriate wavelengths — in theory. In practice, they are rather far removed and photons arriving with appropriate wavelengths will also be energetic enough to do much more damage, e.g. ionise electrons completely. As an example, the electrons in molecular orbitals of $\ce{[FeF6]^3-}$ corresponding approximately to the ligands’ p orbitals already require ultraviolet photons to even be excited to the $\mathrm{t_{2g}}$ orbitals. Imagine how much energy would be needed to excite a $\mathrm{t_{2g}}$ electron to levels formed with $\mathrm{3s}(\ce{F})$ or $\mathrm{5s}(\ce{Fe})$ orbitals!


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Converting mass percentage of a solution into its density?
Is there a way of converting mass percentage of a solution to obtain its density? (With  mass percentage been given and the compounds which are mixed to form the binary solution)

No.$\mathstrut$

Related: https://chemistry.stackexchange.com/q/50269

@IvanNeretin: Huh? There are plenty of cases where an accurate solution density can be calculated from the relevant densities and mass percentage of solute. Plenty.

If we assume no significant change in volume when mixing, then the answer is yes.
Assume we have $w$ percent of a solute (suffixed as $1$), dissolved in a solvent (suffixed as $2)$. That means that $100\ \mathrm{g}$ of solution contains $w\ \mathrm{g}$ of solute and $(100-w)\ \mathrm{g}$ of solvent.
The density $d$ of the solution (for $100\ \mathrm{g}$ of solution) is calculated as:
$$d=\frac{m}{V}=\frac{100}{V_1+V_2}$$
With the respective densities $d_1$ and $d_2$, $V_1=\frac{w}{d_1}$ and $V_2=\frac{100-w}{d_2}$:
$$d=\frac{100}{\frac{w}{d_1}+\frac{100-w}{d_2}}$$
Slightly reworked:
$$d=\frac{100d_1d_2}{wd_2+(100-w)d_1}=\frac{100d_1d_2}{w(d_2-d_1)+100d_1}$$
As stated, this approximation works only if there's no significant change in volume during mixing. It works very well for example for sucrose in water solution. Should it turn out that the mixed volume $V \neq V_1+V_2$, an inaccuracy/error will arise.

There is no simple rule for that, but a linear interpolation. This interpolation is quite a good model for the components without significant intermolecular interactions, like nonpolar solvent mix.
Polar solvents and/or compounds often manifest strong interactions and solvent contractions, related to nonzero mixing enthalpy.
E.g  mixing of 1 L of ethanol and 1L of water produce ( after dissipation of mixing heat ) not 2L, but about 1.9 L.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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How can pressure build up in an equimolar gaseous reaction?
To make the question concrete, if you conduct a reaction like $\ce{H2 + Cl2 -> 2HCl}$ in a stoppered test tube, the stopper will aggressively pop off (or the glass will shatter) despite the fact that there shouldn't be any moles building up in the vessel because the reaction is equimolar. I'm sure there is an easy explanation for this but for whatever reason I can't think of it.

Ever heard about temperature?

@Ivan Almost every comment a bon mot. :D

If we consider an idealised gas behaviour and the gas state equation $$pV = nRT$$
where

$p$ is pressure
$n$ is molar amount
$R$ is universal gas constant
$T$ is absolute temperature
$V$ is gas volume.

..then it should be clear.
As the reaction is highly exothermic, temperature and therefore pressure significantly increase, as $$ p= \left( \frac {nR}{V} \right) T$$

https://en.wikipedia.org/wiki/Gas_laws#Gay-Lussac.27s_law

Yes. But the ideal gas state equation covers all 3 basic gas laws ( for isothermic, isobaric and isochoric changes ).

Yes, I just pointed out something more succinct ;)

You are right. My approach elegantly avoids my issues with  names.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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What happens when an electron in a metal is excited?
For special glasses, crystals, or gases absorb energy from an electrical current or another laser they become "excited." The excited electrons move from a lower-energy orbit to a higher-energy orbit around the atom's nucleus. When they return to their normal or "ground" state, the electrons emit photons.
I think the excited electron is always the valence electron. However In metals the valence electrons are not contained to orbitals, they exist in a sea of electrons that flow freely through the object. Does this mean that valence electrons in metals can not be excited or can not move up orbitals? or perhaps the next electron in line moves up an orbital. If so does it join the sea of electrons? (I'm just throwing out ideas here).
Maybe I'm misunderstanding what it means to be 'excited' because, as it says, energy from current can excite electrons.
Any insight is appreciated.

In a solid electrons are in Bloch wave functions. This leads to the band structure of the material. There are unoccupied states above the Fermi energy. Electrons can be excited from deeper levels into these levels. Or, it can leave the solid as, e.g., a photoelectron.

You seem to be misunderstanding what is a "sea of electrons". In fact, this is a metaphor upon a metaphor upon an abstraction. There is no sea. There is a huge bunch of orbitals. (Sure, the solid state people prefer to call them "states", but that's not really important.) The whole piece of metal is a giant molecule. It is not all that different from ordinary small molecules, except that it is very big, and many orbitals span the entire molecule (but then again, that's what they often do in normal molecules).
All these orbitals tend to have different energies. They are everywhere on the energy scale, very close to each other. You point your finger at any given energy, and you find an orbital with that energy. We can't really tell them apart. They kinda blend into a continuous spectrum. And that's what we metaphorically call the sea of electrons.
Electrons are not free, they are confined to some states. When a photon hits, any electron can get excited all right. It will move up to one of the empty states. In fact, it has an immense freedom compared to an electron in a small molecule. Instead of a few discrete states, it has an entire sea of them above. It can get excited to any of them, which in particular means that metal as a whole can absorb a photon of any energy. This, BTW, is also an explanation of metallic luster.
The excited electron is not always the valence electron (you may fire an X-ray photon at a substance and excite an electron from core orbitals deep down below), but that's irrelevant.

I think you are wrong. Only electrons near the Fermi level can get excited in a metal. Most of the states well below the Fermi level are all occupied, and due to the Pauli's exclusion principle, there cannot be more electrons with such energies. This means an electron with low energy won't be able to absorb a photon with low energy, because the resulting electron would have the same energy than an already present electron.

OK, I fixed that.

I still see the sentence "When a photon hits, any electron can get excited all right.".

Yes, and that's true (provided that photon has enough energy).

But what happens when they return to their ground state? Or do they not drop back down?

What happens when an excited electron in a hydrogen atom drops back down? Or does it not? (Hint: it does.) Same thing here.

Yes, electrons in a metal can also reach an excited state, and leave that state by emitting a photon. Metal spectra are well documented.
Doing this in a solid metal is different mainly because the density of electrons creates potential for additional collisions. Using laser ablation, we can focus energy into a very small space for a very small time. Using electricity can work for a gaseous metal (studied since early 1900's.)
The coolest way to do this is from X-ray bombardment and this technique is used to detect lead in children's toys.
With a laser, as the energy is increased, or so the length of ablation, the abundance of secondary collisions cascade through the metal and plasma is created.
Essentially, while applying electricity or radiation will excite electrons in solid metal, the effect you will notice more may be the conductivity, the heat, the incandescence, or the plasma, depending on your technique.
Penner-Hahn, James E. (2013). "Chapter 2. Technologies for Detecting Metals in Single Cells. Section 4, Intrinsic X-Ray Fluorescence". In Banci, Lucia. Metallomics and the Cell. Metal Ions in Life Sciences. 12. Springer.
Amponsah-Manager, K.; Omenetto, N.; Smith, B. W.; Gornushkin, I. B.; Winefordner, J. D. (2005). "Microchip laser ablation of metals: Investigation of the ablation process in view of its application to laser-induced breakdown spectroscopy". Journal of Analytical Atomic Spectrometry. 20 (6): 544


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Is my proof valid for the formula of elevation in boiling point?
Is my proof valid for the formula of elevation in boiling point?
proof
By Clausius clayperon equation,
$$\ln(P)=-\frac{∆H_v}{RT}+c$$
Differentiating,
$$\frac{dP}{P}=\frac{∆H_v dT}{RT^2}$$
Approximating $dP\approx ∆P$ , $dT\approx ∆T$,
$$\frac{∆P}{P}=\frac{∆H_v∆T}{RT^2}$$
Let $∆P=P_{solution}-∆P_{solvent}$ and $T^2\approx T_{boiling}^2$ also $∆T=$ Change in boiling point,
By the formula for relative lowering of vapour pressure,
$\frac{∆P}{P}=mf_{solute}$
$$mf_{solute}=\frac{∆H_v∆T_{boiling}}{RT_{boiling,solvent}^2}$$
$$∆T=\Big(\frac{RT_b^2}{∆H_v}\Big)mf_{solute}$$
But i don't know how to proceed further, can anyone help?
The actual formula differs from mine as,
$$∆T=\Big(\frac{RT_b^2}{∆H_v}\Big)molality_{solute}$$

I'm not sure that you can go further; is there a reason you expect to be able to do so?

I updated the question @a-cyclohexane-molecule


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Hyperconjugation involving other bonds
Hyperconjugation is often illustrated with C-C and C-H bonds donating electron density to the p orbital in the adjacent carbon atom. Since the concept revolves around sigma bonding orbitals donating electron density to p orbitals, would other bonds such as Si-Si or Si-H bonds be able to take part in hyperconjugation as well? If they are not able to do so, why not?

Your description of hyperconjugation is a very minimalistic one. More generally, hyperconjugation should be seen as the interaction of occupied bonding MOs with unoccupied ones — these unoccupied ones can be both antibonding MOs or empty (cationic) p atomic orbitals.
In principle, $\ce{Si-H}$ or $\ce{Si-Si}$ bonds would also be able to donate electron density in a hyperconjugative matter. In pracitce, due to silicon’s low electronegativity and greater size these orbitals simply do not overlap as well and their energy levels are such that less stabilisation may be obtainable.
Carbon-silicon bonds are of course great at providing hyperconjugation since silicon — much like hydrogen — has a lower electronegativity than carbon, meaning the occupied orbitals have a greater contribution on carbon.

According to a set of slides Ron attached in one of his previous answers, the C-Si bond is a better hyperconjugation donor than the C-C and C-H bonds because it is higher in energy, for your information

I'm sure Jan already knows that, as it is a key reason for the stability of carbocations β to silicon.

@TanYongBoon Your question does not mention the $\ce{C-Si}$ bond so I was confused about your comment at first. But it all falls into place looking at it properly ;)

Have you taken a look at the comment at my other post?

@Jan Comparing between C-C, C-Si, C-Ge, and C-Pb, which is the best hyperconjugative donor? I would think it is C-Pb because it is the weakest of the 4 bonds and thus has the highest-energy HOMO.

@TanYongBoon It should be $\ce{C-Pb}$ because the HOMO is most carbon-centred among the four.

Hi again! Another question: Do you think C-Br and C-I bonds would be good hyperconjugative donors? Would they be better than C-H and C-C bonds? I was thinking that despite they may be more electronegative, they form weak bonds with carbon and they are also not as electronegative as F and Cl.

They are not good hyperconjugators because they are polarised towards the halogen and thus away from the carbon. However, their sigma* orbitals are carbon-centred, leading to e.g. the anomeric effect.


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The cutoff keeps defaulting to 0.05 in Jmol
I am using Jmol for previewing chemical structures. I need the isoSurface value to be 0.008 for 3d printing purposes.But, the cutoff for isosurface keeps defaulting to 0.05.
How should I change it and can anyone also suggest why this keeps happening?

The syntax is
isoSurface id saSurface radius

So you should use
isoSurface id saSurface 0.008

You can use the optional id for giving it a name.
You may also want to see their reference page
Edit 1: Have you tried using DX files ?
I am not an expert in using DX to render the required results but here is the reference  :

DX, as Cube, defines a three-dimensional grid of points in space. At
  each point is a number (a "scalar value"). This set of point values is
  then used by Jmol to define an "isosurface" -- the surface separating
  points having values greater than a given cutoff from those that have
  a value less than a certain cutoff. A typical application is molecular
  orbitals. The phase parameter of the isosurface command allows bicolor
  rendering: one color for "points greater than x" and another color for
  "points less than -x". Jmol can read DX files and re-export them to
  much smaller JVXL files.

Quoted from here

Don't know if its of use but I use the following commands for making jvxl files from cube files under JSmol.  Default cutoff is 0.02.
Isosurface cutoff 0.005 filename_den.cub map  filename_esp.cub
Write isosurface filename.jvxl

OP wants it for 3D printing, can you use this ff for that purpose?

So my question I guess is does this fix the issue of it defaulting? because even if I set a cutoff inline in the terminal it still reverts to 0.02


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pH at different concentrations
In making gummies with pectin, the pH is very important and needs to be at or below 3.4.  USP indicates dissolving it in a 10% solution to test the pH, while some use a 50% solution.  I'm told the resulting pH is the same, but my rudimentary understanding of pH is that pH is the hydrogen ion concentration in solution.  Why would the pH be the same if the concentration of gummy to water is different?


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Can Methane be converted into Carbon and Two Hydrogen Atoms?
Can methane be converted into carbon and two Hydrogen Atoms?I have searched the internet for an answer to this question, and this is never discussed.
If possible can this be done easily and with how much energy?

Carbon and two hydrogen atoms is not a thing at all (a transient intermediate, at best), so what's the question?

Now you have clarified the question, the answer is yes. But it isn't a bulk process. CVD deposition of diamond (pure carbon) is done by reacting a plasma containing methane and hydrogen with a suitable surface. In effect, this converts methane into carbon (as a diamond film on the surface) and hydrogen.

There is no double bond in methane, maybe you are referring to ethene? Alkenes have double bonds, not alkanes. And moreover there is no possibly way for methane to have double bonds as the most stable form of it is CH4.
Now if you are just concerned with the breaking of C=C double bonds then the process is quite common in the transformation of organic compounds, and the energy required to do so is 614kJ/mol

There is no double bond in Methane. A double bond is written as two lines in between symbols and methane has a single carbon atom with single bonds to 4 hydrogen atoms. Methane can be split under the right conditions though. The carbon is mostly left behind during that process. If your talking about a double bond it would be written as:
C=C

Not
C-H

A double bond between carbon and hydrogen probably would not happen.

A double bond can be split though


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Why nitrogen bond is stronger than carbon bond
I hear a lot that nitrogen triple bond is the most strong in nature it have 3 empty spaces for 3 new electrons to sit in ,, but carbon have 4 empty space for electrons to sit in isn't this supposed to make stronger 4 bond ?
C-C   Energy 348
C-N.  Energy  308
reference:
science.uwaterloo.ca

So in your question you aren't necessarily very clear as to what the question is. You are talking about nitrogen triple bonds but have the energy for single bond listed.
The C-N is weaker primarily because of the dipole. The electronegativity of N is 3.04, where as in C it's 2.55. There are forces pushing this apart, and thus leading to a weaker bond strength. Furthermore, the C-C bond is a covalent bond, but the C-N bond acts more like an ionic bond than anything.


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What is this glassware between flask and condenser to collect liquid?
I've been trying to identify this piece of glassware for ages now. I've been combing the internet and can't seem find anything similar. It looks like some kind of adapter/collecter/trap that sits on the pot and the condenser goes on top.

The picture looks like an upscaled screenshot of a video hosted on YouTube. Would you mind sharing the URL? So far from the look of it I'd say it's a custom-made piece of glassware, but the context might help.

thank you.  i will find the video and post a link of it. regards, james

It is a collector indeed. For small amount of distilled target

This glassware is used to separate /remove the light phase from a biphasic (immiscible) liquid mixture.  This can be used primarily to enhance liquid-liquid extraction from the light phase to the denser phase. As a mixed distillate collects,  the denser phase will eventually siphon back through the side-arm to the flask below and the light phase remains in the collector.


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Do electrons truly flow in circuit?
I know this sounds like a stupid question, but do electrons truly flow in a circuit? It doesn't seem logical for electrons to just break away from their atoms to go move through a circuit. If they did, they would probably go through a bunch of reactions and behave oddly (right?). Do the atoms lose/gain electrons to become ions and flow through the circuit? If so, wouldn't they too big to be able to flow easily through a conductor?

There is no such thing as "their" atoms in the first place.

I'm voting to close this question as off-topic because this is not a chemistry question and more like a physics related question.

Solid state physics in particular, where electrons are in wave functions of the solid, not one atom.

The key here is the concept of metallic bonding. or, the sharing of electrons between metal atoms like those in a copper wire (a typical conductor). In this type of atomic bond, electrons no longer belong to one atom or another; they're shared equally amongst all atoms.  As such, one electron truly does go bouncing around between the atoms in the direction of the electric current.
One atom is close enough to another in metallic-bonded substances that an electron is attracted just as much to one nucleus as the next one over, so it's not so "illogical" to think that an electron might "break free" of its original nucleus.
Chemical reactions occur between atoms, not electrons, and atoms in a metal are already metallic-bonded together, so a few stray electrons wouldn't do anything to create new reactions.  However, the electrons do behave oddly!  That's the beauty of metal conductors: their electrons are free to flow wherever an electric or magnetic field dictates, meaning that they can do some pretty cool things!
The nuclei of a conductor don't become ions because every time one electron leaves, another takes its place.  Think of people at a gas station on a busy day: every time one car fills up and leaves, another pulls in almost immediately.  Thus, the metal atoms never have time to become ions, because there's plenty of electrons to keep them company.
As (nearly) always, Wikipedia has a bunch of information on the subject.

Thank you so much! This is a great answer and it answers all of my questions! I haven't heard that anywhere on line, but it makes total sense. Sorry that I may have posted this in the wrong stack exchange, but I think it does have a lot to do with chemistry and atomic bonds.


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Why is the silver nitrate chemical test more favorable than that of sodium iodide?
Two tests to determine the presence of alkyl chlorides are the silver nitrate in 1% ethanol and sodium iodide in 15% acetone tests.  These two tests both produce a precipitate when added to chloride ions (AgCl and NaCl respectively).
NaI in acetone must often be heated to 50ºC for several minutes to produce a visible precipitate.  In contrast, $ \ce{AgNO3} $ typically does not require heat and immediately produces a visible precipitate.
Why is the silver reaction so much more favorable than that of sodium?

related https://chemistry.stackexchange.com/questions/6947/why-are-nacl-and-nabr-not-soluble-whereas-nai-is-soluble-in-acetone-in-the-fin


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Where is the probability of finding an electron in 1s orbital maximum?
So the way I understand the question is that it asks us at what r (distance from nucleus) and at what angles is the probability going to be maximum. The way I see probability of finding at a point in a meaningful sense is, just like in 1D probability of finding particle at a point is taken as probability it is in dx region beside it, we find probability of finding it in a small volume near the point of interest.
Now the probability density of 1s orbital times dV in polar coordinates will end up containing a sine term (since small volume in polar cooditnates contains sine of angle with z axis). Thus the probability is dependent on not just r but also angle with z axis, which isn't expected for such a spherically symmetric orbital. I am confused here.
Of course one could say that since this is spherically symmetric, probability depends only on r and thus use only radial probability distribution and maximise it. But why is the above way which seems more general giving an apparently different answer.
Sorry if the above part isn't clear or is messy. Just to clarify things, I am not asking a "how to find" question. One way to look at my question is that if instead of maximising radial probability distribution, if I evaluate probabilities in small volume around points and maximise it, my answer should match with the one I get from maximising radial probability distribution. Consequently, I should get all points on sphere with Bohr radius but on maximising probability in small volume around a point, I get points at Bohr radius and on XY plane only (since for maxima, sine must be 1 meaning perpendicular to z axis).

You don't need $dV$. It's perfectly reasonably to ask what the probability of finding an electron is at a distance between $r$ and $r+dr$ from the nucleus. That's not any different from your 1D case.

Agree, but taking dV also should give me the same answer, but instead we get an angle dependence also. This is what I am not able to understand.

@Tyberius I did read the answers for the above question, but my question is why is there angle dependence of probability when there shouldn't be? Maybe I am comprehending things in the wrong way, but I am stuck here.

@Niket at least in the answers I saw on that post, there was no angle dependence to the probability or probability density.

@Tyberius But it does seem to have as I have written in my question. What is wrong in my interpretation? Shouldn't density at a point times small volume around that point give us probability of finding electron there? Consequently this should turn out to be same on all points on a sphere, but it doesn't.

@Niket I see what you mean. The probability density at any point doesn't depend on the angle, but the probability of an electron being in a particular region in space does depend on the angles that bound that region.

That depends on your definition of $dV$. You're in spherical coordinates so you should use the standard volume element: $r^{2}\sin\phi dr d\theta d\phi$

You are right in saying that an s orbital is spherically symmetric, hence the probability density must not depend on the angles. Also, you are right in saying that $dV$ in polar coordinates contains $\sin\theta$. So what? You don't have to multiply or divide anything by that $dV$ (nor do you need to even know what it is). Old man Schrödinger took care of that for you. $|\psi^2|$ already gives you the probability density. Just as expected, it is a function of $r$ and not of the angles.

We have probability density right there, but since we are interested in actual probability at a point (i.e. around it) and not just probability density, shouldn't we multiply density by volume to get the actual probability?

There is no such thing as actual probability at a point. If you try to define it in a natural way, it will turn out zero at any point. To have a non-zero probability, you have to integrate over some finite volume. To compare points, you have to use probability density; that's what it is designed for.

But we can use simply probability density to compare points only when all of them have same dV, which they don't have. Otherwise I get the point.

Points don't have dV. They are points. They have zero size and zero volume. dV is a feature of a particular coordinate system.

I do agree with you. But to answer where is the probability of finding  maximum, simply evaluating probability density wouldn't do the job. For 1s we may evaluate probability of finding between r and r+dr using volume of shell to get the answer. But say we have to do it for 2p. Then in this case, we cannot directly use density to compare, since even in 1s, using simply density gives points at nucleus as points of maximum probability, but there isn't any volume there (volume of shell becomes zero at nucleus), meaning there aren't any points there too. How should I do this for 2p orbital?

You have it all backwards. These are different kinds of maximum. You may ask about the point with maximum probability density, or you may ask about the spherical shell between r and r+dr having maximum probability. These questions are different and hardly related, and so are their answers. If you need a point, just use density. If it has a maximum at 0, then so be it. Who cares there's no volume there? There's no volume at any single point anywhere, and we are comparing points. Otherwise, if you are comparing spherical shells, then integrate over a spherical shell and go with that.

Let us continue this discussion in chat.


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What is the number of nodes in the molecular orbitals of cyclic H3?
A quick google search does reveal the molecular orbitals of this cyclic H3.
However, one source mentions there are are two nodes; while two of my professors said there's one node. Which is correct?
This seems to be a related question - I am not sure. There also is a question on linear H3+ - not cyclic.

There are three orbitals in the MO scheme of $\ce{H3}$. Which one are you interested in?

Particularly the two degenerate ones - the one with the lowest energy has zero nodes.

In linear $\ce{H_3}$ you have one MO with zero nodes, one with one node and one with two nodes.  When you bend the molecule around to make a cycle, the two nodes in that last orbital merge to make a second orbital with one node.  The two orbitals of cyclic $\ce{H_3}$ with one node are, of course, degenerate when the ring has threefold rotational symmetry.

And to see that we would need a detailed treatment; or is there any other way to "see" it? What I can see is: for the existence of two nodes, we would have required the two nodes to cross the opposite atoms, right?

A simple "rough sketch":  start with two parallel planes between adjacent pairs of atoms in the linear molecule (these being the nodes of the highest orbital in the linear molecule).  Bend the two bonds to make the outer two atoms come together, making the nodes follow along so they continue to pass through the molecule, and see what happens to them.  You can't avoid the merger!

Are the two orbitals with one node really degenerate in H3 or only in H3+?

@pH13 the answer is edited to "neutralize" your q.


text:
Why do we only consider relative motion of hydrogenic atoms?
In my lectures on quantum mechanics, it was said that the Hamiltonian for a hydrogenic atom could be split into relative motion and translational motion, as follows:
$$\hat{H} = - \frac{\hbar^2}{2m}\nabla^2_{cm} - \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$$
and that one only need consider the relative motion:
$$\hat{H} = - \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$$
This statement is reiterated in Atkins' Physical Chemistry (10th Edition), as well as his Molecular Quantum Mechanics (4th Edition), and various other online resources, but I am yet to find an explanation as to why.
Why can we, and why do we, neglect the translational motion of the atom?

We can do that because the variables have separated. The Hamiltonian of relative motion depends on the relative coordinates. The translational part depends on the coordinates of the center of mass. The two never mix. They don't interact; they don't even care about each other.
Separation of variables in various guises is an immensely powerful technique of solving PDE. You will encounter it time and again, including this very chapter of whatever textbook you are using, just a few lines below, when you'll start to actually solve the equation for hydrogen atom by splitting it into radial and angular parts.
So it goes.

I understand separability but the author goes on to write $- \frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r} = E\Psi$, declaring it the Schrödinger equation for a hydrogenic atom. What is the physical intuition behind seemingly totally discarding the translational motion? (i.e. how can you disregard translational motion and still set the result equal to $E$?)

This is $E$ of relative motion. Translational motion is associated with its own kinetic energy; it can be anything, so why bother.


text:
Oxidation of water by permanganate: Does this reaction happen?
Since MnO4- is a strong oxidiser, if one increased its concentration  relative to water, could it oxidise the water (Le Chatelier's Principle seems to suggest so)?
I have attempted to write the possible (redox) reaction, but I fear I've gone terribly wrong:
$$\ce{MnO4^{1-} + 2H2O -> 2H2 + 3O2 + Mn^{3+}}$$
Can someone tell me if I've constructed the equation (and balanced it) right? If not, can someone offer me some guidance in this regard?

This conversation has been moved to chat.

You have indeed gotten it terribly wrong. The key hint is that you have more than one element that would be reduced: hydrogen also goes from $\mathrm{+I}$ to $\pm0$ in your sample equation. Furthermore, in neutral or weakly basic solutions permanganate gives $\ce{MnO2}$ after oxidation, not manganese(III).
What you do want to do is to reduce permanganate to something (which I have already established would be manganese dioxide) while oxidising something in water to something. Looking at the oxidation states of hydrogen and oxygen in water, the only element that can be oxidised is oxygen, since hydrogen is already at its highest possible oxidation state $\mathrm{+I}$. Oxidising water’s oxygen leads to a few possibilities of which oxygen gas does indeed sound most likely. Therefore, we have the redox pairs:
$$\begin{align}\text{Red: }\ce{\overset{+VII}{Mn}O4-}/\ce{\overset{+IV}{Mn}O2}&&\text{Ox: }\ce{H2\overset{-II}{O}}/\ce{\overset{\pm0}{O2}}\end{align}$$
We can then determine the appropriate half-reactions by the standard procedure. Instead of walking you through it, I will present you with the final result for comparison.
$$\begin{align}\ce{MnO4- + 3e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red}\\
\ce{2H2O &-> O2 + 4e- + 4 H+}\tag{Ox}\\[1em]\hline
\ce{4 MnO4- + 4 H+ &-> 4 MnO2 + 3O2 + 2 H2O}\tag{Redox}\end{align}$$
It turns out that the final reaction equation formally doesn’t even need water. However, potassium permanganate is a shelf-stable solid under normal laboratory conditions (if light is excluded), so it is reasonable to assume that the reaction will only proceed in aqueous solution.
To determine whether the redox reaction is feasible in aqueous solution, it is easier to go with the two half-reactions we already have; thermodynamically, they must correspond to the same process with the same reaction enthalpy etc. The Nernst equation together with the inspection of the cell potential allows us to estimate whether a process will be spontaneous or not by mere inspection of the standard potentials.
$$\begin{align}E^0_\text{cell} &= E^0_\text{Red} - E^0_\text{Ox}\\
&= \pu{+1.70V} - (\pu{+1.23V})\\
&= \pu{+0.47V}\\
E^0_\text{cell} &> 0 \\&\Longrightarrow \text{spontaneous process}\end{align}$$
Thus, thermodynamics already predict the decomposition of permanganate in aqueous solution to be spontaneous under standard conditions. As mentioned, the Nernst equation will allow us to estimate whether it also is spontaneous given general laboratory conditions; I will assume $\mathrm{pH} = 7$ and $p(\ce{O2}) = \pu{0.2bar}$
$$\begin{align}E &= E^0 - \frac{0.059}{z} \lg \frac{Red}{Ox}\\[0.5em]
E &= \pu{+0.47V} - \frac{\pu{0.059V}}{12}\lg\frac{p(\ce{O2})^3}{[\ce{MnO4-}]^4[\ce{H+}]^4}\\[0.5em]
E &= \pu{+0.47V} - \frac{\pu{0.059V}\times28\times2.1\times(-4)}{12}\lg {[\ce{MnO4-}]}\\[0.3em]
E &= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\end{align}$$
We can calculate the minimum concentration from which the process will be spontaneous:
$$\begin{align}\pu{0V} &= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\\
\pu{-0.47V} &= \pu{1.16V}\times\lg [\ce{MnO4-}]\\
-0.405 &= \lg [\ce{MnO4-}]\\
10^{-0.405} &= [\ce{MnO4-}]\\
0.393 &= [\ce{MnO4-}]\end{align}$$
Thus, at any concentration higher than $\pu{0.393M}$ permanganate should start formally oxidising water or decomposing.
Whether or not this happens at exactly this concentration depends on kinetic factors, i.e. can oxygen actually be produced under these conditions, but thermodynamically it would be favourable.

What if the MnO2 and O2 were pressurised, would this force the reaction for H2.? It is less dense so I imagine it would separate like fractional distillation. Right.?

I guess like how steam bubbles form..

Energy is also being added to the MnO4 ions specifically to make the reaction occur so I believe it is possible.

@Randomrokandpetcaveman There is basically no way in which you can force these conditions towards hydrogen generation.

@ Jan: I think the final redox equation is: $$\ce{4 MnO4- + 2 H2O -> 4 MnO2 + 3O2 + 4OH-}$$ so reaction is in aqueous neutral condition. According to this article, $\ce{KMnO4}$ is not stable under highly alkaline solutions.

@MathewMahindaratne The two reaction equations are identical save for me having four protons on either side.


text:
What does the format nF10.8 mean?
I was trying to assign weights to each state in a CASSCF PES scan using nroot=2 in Gaussian. I specified 0.500000 0.500000 at the end of .com file. But it caused the program to assign the weight for the first calculation only. All the rest points of the scan had weights 0 and 0, which I don't want to have. I referred to the Gaussian manual which states

StateAverage: Used to specify a state-averaged CASSCF calculation. All states up to NRoot are averaged. This option requires the weighting of the various states to be input in format $n$F10.8 (no trailing blank line).

But I don't understand what is this nF10.8 format means and if it could help me solve the problem to assign the weight in each step of the scan. I tried writing 0.500000 0.500000 one after the another for say 50 points and I could see the required weight has been assigned. Can someone clarify the meaning of the nF10.8 format? I searched on internet but I couldn't find any helpful material.

It is a Fortran format specifier.
It means

$n$: repeat the input $n$ times, once for each weight
F: input is a floating-point number
10.8: the number is of total width 10 (including the decimal point), and has 8 digits to the right of the decimal point

This means the number cannot have more than one digit to the left of the decimal point, which makes sense since weights should all be $< 1.0$.

Thanks for the answer. So in order to write 0.500000 0.5000000 50 times, shall I write  0.500000 0.500000 <br/> 50F10.8  ?I want to know how to implement it too for the above mentioned problem.

I can't test it, because I don't have access to Gaussian, but that sounds appropriate. You say yourself in the question that it works.

@Chemist , I think that you just have to write the numbers one next to the other, first number starting at column n, the second one starting at col. n+10, etc... (using the format explained in this answer)

I mean, without any separator character and using ten characters for each number

That is probably the safest approach. To be honest, I have no idea how the scan works, and this is really a software support question. I only know what 'nF10.8' means.


text:
Is it true that organic chemistry as an independent research field is no longer developing?
One of my professors told me that organic chemistry has become a tool for biological research, rather than an active research field of its own.
Is this assertion valid? Does it describe correctly the general attitude towards organic chemistry in research institutions today?

Not sure if this actually counts as a valid (Chemistry) question here. Perhaps it's (somewhat) better off on Academia.SE?

Perhaps it is, but I wanted to hear what the chemists have to say about this.

Have you considered asking this at The Table? O:)

Bit afraid it is. ...material chemistry is another field which actually benefits of organic chemistry

I'm afraid I don't know what that is (I know it exists, but don't know what it is exactly)...

Do believe organic chemistry is alive and well. Consider, e.g., https://chemistry.meta.stackexchange.com/questions/3979/suggestions-for-synthesis-golf-targets

@hBy2Py I think it's been fairly well demonstrated that we can synthesise more or less anything given enough time and money. These games are enjoyable intellectually, but I'm not sure they're adding much value any more (sadly). But that doesn't mean organic chemistry is dead. Pure synthesis, I feel, is but there are so many other things that synthesis can be applied to. Just need to get out the bunker a little :)

How many journals are there now publishing synthetic chemistry papers? Seems like there's a new one every year. Doesn't sound like a subject in decline to me.

@Waylander A vast amount of rubbish is published. There's very little value added from another total synthesis of strychnine, or yet another way to do an allylic oxidation. I got sick of barrages of claimed yields of 95-100%, which were clearly nonsense. There's some interesting stuff, but not enough to go round. My point is there's loads of interesting applications in supramolecular chemistry, chemical biology etc begging for molecules and I think people who think of themselves as synthetic chemists should embrace that.

@awjlogan I respectfully disagree. Knowing all the walls you can run against in total synthesis from my own experience it is definitely not so that we can synthesise anything given time and money. The mere fact that total syntheses still regularly make it into Angewandte and JACS (and not just because the author is KCN) should show that there is still interesting stuff to be developed. Each one of these syntheses does add something new of value, whether it is a most elegant transformation that can serve as a template or a novel set of conditions allowing broader substrate scopes.

The closest thing to organic chemistry would now be the shape and sequencing of protiens and or DNA. That field is best left alone as it takes too much time to pay anyone to get any applicable results.
Advances in things like machine learning are promising to the unraveling of these complex structures. Humams are better at recognising the patterns hence the "fold it" game.


text:
Can I use a 1:1 mixture of isopentyl alcohol and water as mobile phase to carry out paper chromatography of a mixture of red and blue ink?

Can I use a 1:1 mixture of isopentyl alcohol and water as mobile phase to carry out paper chromatography of a mixture of red and blue ink?

We generally use ethanol water mixture as mobile phase for separating mixture of blue and red ink using paper chromatography. Can I expect better results with isopentyl alcohol and water mixture?

Who said you can mix them 1:1?

Isoamyl alcohol is only sparingly soluble in water: $28\ \mathrm{gL^{-1}}$ (Wikipedia)

The principle of chromatography is that you can always try (as per the question in the title). This can be beneficial: in my PhD thesis I had two diastereomers that were hard to separate on TLC using hexanes and ethyl acetate, but a hexanes/toluene mixture worked wonders.
However, in your case you should not expect better results. For one, as the commentators have pointed out the two liquids are not sufficiently soluble in each other. For two, your separation is already pretty good and it seems unlikely that it can be improved even more.


text:
Why do aldehydes polymerize faster in the cold?
In amateur perfumery, raw chemicals are usually stored in sealed bottles in refrigerators/freezers to reduce decomposition rates. However, aldehydes are tricky to store for long periods of time because they tend to be reactive. The two biggest decomposition routes are oxidation (which can be prevented by having a good seal on containers) and oligo/polymerization.
Folk knowledge (1, Pell Wall Perfume blog and 2, Perfumery and Flavoring Synthetics by Bedoukian page 24) claims aldehydes polymerize more quickly in the cold, and so unlike most chemicals, perfumers are warned to store aldehydes at room temperature, rather than frozen.
Why do aldehydes polymerize faster in the cold?
From thermodynamics, it makes sense polymerization occurs to greater completion in the cold, since polymerization is usually exothermic, so Le Châtelier's implies the cold will drive the reaction to the right. Likewise, since polymerization has $\Delta S <0$, cooling below the polymer ceiling temperature $T_c=\Delta H/\Delta S$ causes polymerization to be spontaneously possible.
However, I can't find any mechanistic explanation of why the rate increases in the cold, or even any other sources besides an online blog and a perfumery textbook which back up this claim. Can any polymer fans explain why cold is bad for aldehydes?

Note that "they [aldehydes] polymerizes quicker" does not mean that their rate of polymerization increases. It just means that "their rate of polymerization increases" COMPARED to another rate when the temperature diminishes.

In chemistry labs, aldehydes are typically stored as cold as possible: the more stable ones in the fridge, everything else in the freezer. This kind of contradicts your sources’ statement.


text:
Alternatives to Dalton's atomic theory
During the 19th century many scientists did not believe in Dalton's theory of atoms. What was the alternative view of matter?

related https://chemistry.stackexchange.com/questions/4989/when-did-the-atom-theory-become-indisputable

This HSM SE post might be of interest: https://hsm.stackexchange.com/questions/83/what-were-the-dominant-non-atomic-theories-of-matter-in-the-19th-century


text:
How will this reaction proceed?

Will there be a hydride shift to form a tertiary carbocation or will it proceed as secondary carbocation?
My text book shows that it would proceed as secondary carbocation
But tertiary carbocation is more stable. I am a bit confused here.
Will someone please clarify?

Depends on the reaction conditions...

Please cite the textbook.

Can you please elaborate what kind of conditions? @Zhe

Temperature, solvent, etc.

It will surely proceed as a secondary carbocation and no hydride shift will take place. Because if hydride shift takes place, bridgehead carbon will acquire positive charge i.e. will become unstable.
For detailed explanation check this out:
Why are bridge bonds with negative or positive charge on bridge head unstable?


text:
How to separate two or more gases into different containers
I know this is probably an incredibly stupid question, but I'm making a hydrogen generator for a project. The basis of it is electrolysis of baking soda and water, which apparently gives off hydrogen, oxygen, and carbon dioxide. I want to separate these three gases into separate containers. Is this possible? Also, is baking soda and water the most efficient way of doing this?

It isn't easy. Gases don't naturally separate, though in electrolysis they do appear at different electrodes (oxygen on one, hydrogen on the other) so if you don't mix them you won't have to separate them. OTOH there are specialist membranes that will selectively allow hydrogen to diffuse through them and that works on a mixture.

@matt_black I was thinking of something like that. Would it be possible to separate them using high voltage, so they would be attracted to either side?

If you're generating hydrogen and oxygen by electrolysis, the best solution is to just not let the gases mix in the first place. The two gases will be generated at different electrodes, so just trap the gases there. There is specialist glassware designed for this (e.g. this) but you can also just design something yourself.

SAFETY - You must also consider safety. A mixture of hydrogen and oxygen is explosive.

To separate hydrogen, a warmed Pd membrane is very effective... except that Pd catalytically ignites the $\ce{H2  -  O2}$ mixture. oops
A carbon nanotube membrane might work... but would be a bit pricey.
Various organic polymers have been used as hydrogen-permeable membranes, though some are poisoned by $\ce{CO2}$.
Actually, for a demonstration, $\ce{H2}$ is so much more mobile than the other gases that even a cellulose dialysis membrane might work to some extent, providing "enriched" hydrogen with some of the other gases still included.
The $\ce{CO2}$ could then be separated from oxygen by chilling. The melting point of $\ce{CO2}$ is ~217 K. Using a cooling bath of isopropanol and dry ice should get to ~195 K. It might be inefficient and redundant to use $\ce{CO2}$ to trap $\ce{CO2}$, but this would be acceptable for a demonstration, I assume.
An alternative is to compress the $\ce{CO2  -  O2}$ mix until the $\ce{CO2}$ liquefies.

I would prefer using H2SO4 and Zn or Cu so that it produces H faster as Zn and Cu are highly reactive and the process with Cu looks beautiful to and helpful in checking wheather the reaction is happening as it turns blue and u can also explain crystallization.Seperating those gases need high level equipment and can be done be coolers,high pottemtial,pressure changeable equipments

Do you talk about electrolysis or the chemical reactions of Cu and Zn with sulfuric acid? In the first case the amount of hydrogen produced is determined by the current. In the latter case copper won't give you much hydrogen.


text:
Balancing of redox reaction
I tried to balance the following redox reaction
$$\ce{S^0 + OH- -> S^2- + S2O3^2-}$$
by ion electron method, but I am unable to identify the oxidation and reduction half reactions. Please provide the correct method.

See this: https://chemistry.stackexchange.com/questions/34424/sodium-polysulfide-production-from-naoh-and-sulfur

and my answer: https://chemistry.stackexchange.com/questions/50990/reaction-between-sulfur-and-sodium-hydroxide/63794#63794

Genarally, sulfur doesn't disproportionate like that: in fact, it typically behaves very similarly to chlorine (see Balancing the redox disproportionation of chlorine by half-reactions), which upon heating with alkali disproportionates to chloride $\ce{Cl-}$ and chlorate $\ce{ClO3-}$:
$$\ce{3$\overset{0}{\ce{Cl2}}$ + 6 OH- ->[$\Delta$] 5 $\overset{\mathrm{-I}}{\ce{Cl}}${}^- + $\overset{\mathrm{+V}}{\ce{Cl}}$O3- + 3 H2O}$$
Without heating instead of chlorate $\ce{ClO3-}$ hypochlorite $\ce{ClO-}$ is formed, so this is an irrelevant process in this case. Back to sulfur: when reacting with alkali, sulphide $\ce{S^2-}$ and sulfite $\ce{SO3^2-}$ are formed as the primary products according to the scheme:
$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+IV}}{\ce{S}}$O3^2-} \tag{1}$$
Half-reactions and net equation for (1):
\begin{align}
\ce{2 S + 4 e- &-> 2 S^2-} \tag{red1}\\
\ce{S + 6OH- &-> SO3^2- + 4e- + 3H2O} \tag{ox1}\\
\hline
\ce{3S + 6 OH- &-> 2 S^2- + SO3^2- + 3 H2O} \tag{redox1}
\end{align}
Note that (red1) has been multiplied by 2 to even the number the electrons in both (red1) and (ox1). The net equation (redox1) is obtained as a sum of both half-reactions.
Now, lets assume that the reaction you've written takes place, where where thiosulfate is formed instead of sulfite. In fact, it might as well happen when boiling solutions of sulfites with sulfur for a long time:
$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+II}}{\ce{S}}${}_2O3^2-} \tag{2}$$
Half-reactions and net equation for (2):
\begin{align}
\ce{2 S + 4 e- &-> 2 S^2-} \tag{red2}\\
\ce{2S + 6OH- &-> S2O3^2- + 4e- + 3H2O} \tag{ox2}\\
\hline
\ce{4S + 6 OH- &-> 2 S^2- + S2O3^2- + 3 H2O} \tag{redox2}
\end{align}
Reaction (redox2) would be the net reaction for the process you asked about, even though I'd say it's more likely a hypothetical one. Keep in mind that for as long as the reaction takes place in aqueous solution and the redox process involves acid-base transformations (like formation of oxoanions in this case), you need to take water into account.

As far as I know (and Wikipedia confirms), thiosulphate is formed by reaction sulphite and sulphur under alkaline conditions:$$\ce{S + SO3^2- ->[OH-] S2O3^2-}$$So the second net reaction is not as unlikely as you make it sound.

@Jan Yeah, as a brutto reaction it's totally fine, I just wanted to underline the fact that it's a secondary process.


text:
Why is the copper in my pan magnetic?
My Momma bought a copper pan made for induction because it was the only one she could find that she wanted and was big enough to pan fry half a chicken.
But I noticed something odd about the copper. It seems to be magnetized. I thought that was impossible because copper is paramagnetic, not ferromagnetic. How is that possible that a paramagnetic material could be magnetized and remain magnetic even when a magnet is nowhere near it? Is it possible that they purposely put tiny amounts of iron in the copper to magnetize it and let it retain its magnetism kind of like how neodymium magnets aren't pure neodymium?

A paramagnetic material could not be magnetized and remain magnetic. So the only remaining option is "tiny amounts of iron", except that they are probably not really tiny. I believe your pan is made of iron, and copper is just a thin lining all over it.

Also, copper is diamagnetic, not paramagnetic.

Also, copper is a good conductor and will as such for a large variety of frequencies (though, not the highest ones) have an excellent susceptance by virtue of induced eddy currents. Just try dropping a magnet through a copper pipe and measure the time it takes.

Is the copper cladded on iron/steel?

For why copper is diamagnetic: https://physics.stackexchange.com/questions/59248/why-is-copper-diamagnetic


text:
Understanding the derivation of Gibbs Free Energy
In the equation: $$ΔG=ΔH−TΔS$$
I am having trouble understanding my professor's derivation of it from
$$ΔS_{\mathrm{univ}}=ΔS_{\mathrm{sys}}+ΔS_{\mathrm{surroundings}}$$
where he says $$ΔS_{\mathrm{surroundings}}= \frac{-ΔH_{\mathrm{sys}}}{T}$$
which I have learned is only true for a reversible process.
How is it possible that this heat can be reversed? And why can't we say that the change of entropy for the system is also equal to the equation. I understand that it would mean the reaction is at equilibrium, but would it not be since heat is reversible?
$$ΔS_{\mathrm{sys}}= \frac{ΔH_{\mathrm{sys}}}{T}$$

It's a simplification.

Could you give a more in-depth explanation of how this simplification makes sense conceptually?

Well, you take $\approx$ and treat it like $=$. I'm not sure what you're asking...

The surroundings are treated as a constant volume reservoir, so the first law of thermodynamics,
$$
\mathrm{d}U_\mathrm{surroundings}
  = \mathrm{d}q_\mathrm{surroundings}
  + \mathrm{d}w_\mathrm{surroundings},
$$
yields
$$
\mathrm{d}U_\mathrm{surroundings}
  = \mathrm{d}q_\mathrm{surroundings}
  \text{, or }
\Delta U_\mathrm{surroundings}
  = q_\mathrm{surroundings}.
$$
Notice that $q_\mathrm{surroundings}$ is equal to a state function, so it behaves as one, that is, it is independent whether the process is reversible or not. If the process is reversible
$$
q_\mathrm{rev,surroundings}
  = \Delta H_\mathrm{surroundings},
  \tag1\label{eq:rev}$$
if it is not reversible
$$
q_\mathrm{surroundings}
  = \Delta H_\mathrm{surroundings},
  \tag2\label{eq:irrev}$$
from \eqref{eq:rev} and \eqref{eq:irrev} follows
$$q_\mathrm{surroundings} = q_\mathrm{rev,surroundings}.$$
Therefore
$$
\Delta S_\mathrm{surroundings}
  = \frac{q_\mathrm{rev,surroundings}}{T},
$$
in particular
$$q_\mathrm{rev,surroundings} = -q_\mathrm{sys},$$
then
$$\Delta S_\mathrm{surroundings} = \frac{-q_\mathrm{sys}}{T}.$$
If the process occurs at constant pressure
$$q_\mathrm{sys} = \Delta H_\mathrm{sys},$$
therefore
$$\Delta S_\mathrm{surroundings}  = \frac{-\Delta H_\mathrm{sys}}{T}.$$

You are right. Only when the process is reversible we have:
$dS = \frac{dQ} {T}$
For an irreversible process we have instead:
$dS > \frac{dQ} {T}$
This is the Clausius inequality and when replacing $dQ$ with its expression derived from the first law ($dU = dQ - PdV $; assuming only PV work) we obtain a general criterion for spontaneous change:
$dU + PdV - TdS < 0 $
The Gibbs energy is the special case of this general criterion when $P$ and $T$ are held constant. In such case:
$dU + PdV - TdS <0 \implies d(U+PV-TS) < 0 \implies d(H-TS)<0 \implies dG<0$
Other expressions (such as the Helmholtz energy) are derived from the same general criterion.

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


text:
How did they assign absolute configuration to these cis and trans 2-methylcyclohexanols?
From [1, p. 56]:this is the 4/B rule I was talking about
This is how I practiced. In the textbook, the absolute configuration of one of the enantiomers (I am referring to the the first compound) is written as 1R,2S. But when I tried to redo it, I arrived at 1R,2R. I didn't understand where I went wrong. I think the way I am visualising the compound is not correct.

4.3.5 Nomenclature of polysubstituted cyclanes
Polysubstituted cyclanes, e.g., cyclopentanes, cyclohexanes etc. exist in a number of stereoisomers (diastereomers and enantiomers) depending on the number and nature of the substituents. The disubstituted cyclanes give two diastereomers which can be conveniently and unambiguously described as cis and trans isomers, e.g., cis- and trans-2-methylcyclohexanols (XXVII) and (XXVIII) (Figure 4.13). Each of the diastereomers in turn gives two enantiomers which can be named, following the R, S nomenclature, as R, S and S, R for the cis and S, S and R, R for the trans isomer (use 4/B rule) as shown in the Figure.

Figure 4.13 Disubstituted cyclohexanes

When I practice on this picture I always arrive at the same configuration both on C1 and C2. Am I assigning the priorities in a wrong way?
Reference

Nasipuri, D. Stereochemistry of Organic Compounds: Principles and Applications; New Age International, 1994. ISBN 978-81-224-0570-5. (Google Books)

Do you know what happens to the absolute configuration when two substituents switch places and the rest stand where they were?

You have somehow decided that the front side of the ring is actually the rear side … or something. So your drawings show exactly the wrong stereochemistry. For the hydroxy-carbon, you thus would, in theory, get the incorrect assignment (S) but you use the rule you reference to reverse that into (R). Naturally, because your initial assertion of which atoms are in which plane was incorrect, your application of the rule was also incorrect. For the methyl carbon, your inverted ring reproducibly gives you the inverted assignment (R) and you do not apply any incorrect correction.

Figure 1: the molecule in question.
In figure 1, I have reproduced the molecule in the more ‘standard’ chair configuration which gives a better approximation of the bond angles. The wedged/bold portion of the ring is in front of the paper plane, the thin solid portion behind. I have also rotated the ring by $60°$ so that the hydroxy function is on the rightmost carbon atom for easier discussion.
It should be a rather simple exercise to determine that the hydroxy-carbon (carbon 1) indeed does have an (R) configuration: the hydroxy group has highest priority, the front side of the ring second-highest, the back side third and the hydrogen pointing downwards fourth. Therefore, we must look from above the ring plane to put the lowest substituent to the back; the rotation from hydroxy to methyl side to non-methyl side is thus clockwise, therefore (R).
The same analysis performed on carbon 2, the methylated one, should give you:

carbon 1 ($\ce{CHOH}$) is highest priority; going to the back right and slightly downwards
carbon 3 ($\ce{CH2}$) is second; going to the left, slightly downwards while staying in the plane
the methyl group is third; going up while staying in the plane
the unmarked hydrogen is fourth; this goes to the front right and slightly downwards.

Thus, we should look from inside the ring as the smallest substituent is pointing towards us. However, we can also look from the smallest substituent and invert the assignment; the way we see the three groups is again clockwise, so if we were to look from inside the ring (or behind it) we would see them anticlockwise, thus (S).
Note that the wedged and bold bonds — both in the paper’s drawing and in my figure — are not there to indicate a positioning from the stereocentre but to indicate which half of the ring is in front of the paper plane and which is behind. Thus, you cannot — like you did — assume the bolded bond to be in front; it actually is in your paper plane when you are looking at the two stereocentres on either side of it.


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How can I extract cellulose in plants using acids?
Me and my classmates are doing a research project about the extraction of nano cellulose from plants to form a biocomposite, which can be a material to form an effective biodegradable material. We've done a quick research about it and most of the articles that we came across is to conduct an acid hydrolysis on the plant that we opt to use. We are kinda lost and confused as to what step or procedure we need to follow. Some articles also suggested that we can just boil the sample to extract the nano cellulose. Thank you for answering my question.

https://en.wikipedia.org/wiki/Nanocellulose#Manufacture I'm not sure that "extraction" can be done with acids only.

Thank you, @andselisk. What steps do you suggest to fully isolate and extract the cellulose?

I'm sorry I cannot answer this question, not really my field. That was just a comment based on common Wikipedia info.

You may want to take a look at Renewable biological systems for alternative sustainable energy production (FAO, 1997). Chapter 3 is on cellulosic biomass.

@RodrigodeAzevedo, thank you for the info. My team will look into that


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What is the thermal state of a reaction at equilibrium?
I was asked this question at class,

What is the thermal state of a reaction at equilibrium, the forward reaction of which, is exothermic?

I conjectured that since the forward (exothermic) reaction and the backward (endothermic) reaction occur at equal rates, the reaction at equilibrium would be athermic since there doesn't seem to be any net heat evolution. However, my teacher said this was not true.
Could someone tell me where I'm going wrong with my reasoning?

Define "thermal state"...

Well what we usually understand by thermal state at my school is the sign of enthalpy change, if it is negative we say it is exothermic, if positive we say it is endothermic, and if nil we say it is athermic.

If the sign of the enthalpy change is all you want, then even at equilibrium, the enthalpy change of the forward reaction will remain the same, and that of the backward reaction will also remain the same. Now, you can't add forward plus backward and say that it is athermic; enthalpy is defined only when we consider a certain direction of the reaction. Also, see this related question: https://chemistry.stackexchange.com/questions/43039/enthalpy-change-and-equilibrium

You're right, as far as I can see. If we look at a particular transformation $A \leftrightharpoons B$ there is a forward reaction and a backwards reaction. They are associated with enthalpy change of different signs and the same magnitude. At equilibrium the rates are the same, so the forwards and backwards reactions release and absorb the same amount of heat and it appears as if nothing is happening.
There is no separate "reaction at equilibrium" with its own enthalpy change. It's not really clear to me what "thermal state of a reaction" means, as systems have states and reactions are pathways between systems.


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What visual observations can confirm that a reactant is the limiting reagent?
I have this lab question for the lab called Copper Collection Stoichiometry, where we choose an amount of the limiting reagent (iron) for a reaction between copper (II) sulfate and iron. We are to dissolve copper (II) sulfate in water, making it blue, then add iron, and after filtrating, we should have copper left.
How do we know if iron was the limiting reagent in terms of the visual observations? We made it so that iron was the limiting reagent, but how do we know visually if it is?
I understand that in order for iron to be a limiting reagent, it should've reacted with the solution and all completely turned into copper after the experiment. However how do we know if all the iron actually turned into copper? There is no way that we can visually count every solid flake to make sure there's no iron. I also understand that if iron was the limiting reagent, then there should be still copper (II) sulfate left in the solution, as the reaction didn't completely occur. Is that the answer?

Throw in more iron and make sure that the reaction starts again. That would verify your last point.

@IvanNeretin I don't have the resources to do that, it is a school lab, and we are meant to confirm visually that iron is actually a limiting reagent

OK, then was the solution at least slightly blue in the end?

@IvanNeretin yes slightly blue, although lighter in colour than the original solution with warm water and copper (II) sulfate. Does that indicate there is still copper (II) sulfate in the filtrate? I searched up what Iron (II) sulfate looked like, however it was also slightly blue, mixed with green...

You got it right. Yes, that counts as a proof. Iron sulfate in comparable concentrations is all but colorless.

@IvanNeretin Ah I see, thank you so much, that makes alot of sense. Maybe you wish to copy that to post as an answer? I can mark it right cause reputation and stuff if you want. Thanks again

You may answer it yourself, this is allowed and even encouraged, and the idea was yours anyway.

Sounds like you do metals transmutation. ..

The colour of the filtrate after filtering out the solid is still lightly blue in colour. This indicates that Copper (II) sulfate still is present in the solution, meaning that there was not enough iron to completely react with all the copper (II) sulfate, visually confirming it to be the limiting reagent.
A solution containing only Iron (II) sulfate is colourless, so if the solution is blue in colour, then it shows that Copper (II) sulfate still is present.
Credit to Ivan Neretin for helping me.


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How to prepare a pure sample of hexane?

To prepare a pure sample of n-hexane using sodium metal as one reactant the other reactant will be:

n-propyl bromide
Ethyl bromide and n-butyl bromide
Methyl bromide and n-pentyl chloride

What is the question now? Is it which of these three is correct?

Well, one is clearly better than the others...

Why is everyone so hung up over the Wurtz reaction. It is never used in real chemistry

In 35 years lab time I have never seen it used or its use proposed.

Compared to the timescale on which the new knowledge slowly seeps from the labs to the classrooms, 35 years is not all that long.

The people who set these problems clearly have no experience in running real reactions.

Wurtz reaction is not very efficient and hard to initiate. If you use a mixture of alkyl halides you will obtain a difficult to purify mixture of products including homo-coupling from both starting materials and reduction products, so the obvious way to approach this is to use a single starting material. In this example then propyl bromide is the appropriate starting material.


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What system of nomenclature can I use to indicate the positions of specific kinds of atoms of a surface reconstruction?
I was reading a journal based on SiC reconstructions, and they seem to mention that every dangling bond in the $(1\times 1)$ unreconstructed Si face or the $(0001)$ direction bonds up with a Si adatom placed in the corresponding $T_4$ position and hence leads to a $(\sqrt{3}\times \sqrt{3})R30^o$ reconsruction.
Now, I do realize what is happening extremely clearly, but I want to know which nomenclature they use to name the position of the Si adatom on the C atoms (dark red dots) as the $T_4$ position. I would be really grateful for the help, any books or article citations in that direction would also help. I absolutely do not know whether it is a symmetry group or something else.


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How many atoms are represented in this equation?
$\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$
The answer says 26, but shouldn't it be 13? Because actually, the number of atoms involved in the reaction are 13, and once the reaction is completed, again, number of atoms is 13.
Please explain.

I believe the question is asking how many atoms are present in the written form of the equation, on both sides. In this case, you would disregard the actual reaction that is taking place, and simply count the number on the left and add it to the number on the right.

To be honest, your answer is a lot better since the atoms on either side are the same, so 26 would be double counting...

I totally agree with Zhe. One of the fundamental notions in chemistry is that chemical reactions don't create or  destroy atoms. So you have as many atoms after a chemical reaction as before. // All in all, a badly worded question.

Indeed an equation cannot represent the total number of atoms. There can be 13 n atoms. Which is not the 26 (n = 2) they intended. Very bad question.

The referenced atoms on both sides of the arrow are the exact same atoms? I like your thinking!
But, chemical equations do not refer to individual atoms.
The formula $\ce{Pb(NO3)2}$ can be read as "A molecule of this substance consists of six atoms of O, two atoms of N, ...". That is correct.
But the equation $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$, strictly speaking, does not talk about single molecules. The numbers (including the 1's that are nor written) are called "stoichiometric coefficients". The equation means: "Any number of $\ce{Pb(NO3)2}$ molecules react with the double amount of $\ce{KI}$ molecules to yield ...".
Wikipedia supports my view:

In lay terms, the stoichiometric coefficient (or stoichiometric number in the IUPAC nomenclature) of any given component is the number of molecules that participate in the reaction as written.

  ...

  In more technically precise terms, ...

... the stoichiometric coefficient is the number of molecules divided by the progress variable or extent of reaction.

...

  The (dimensionless) "units" may be taken to be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules.

"But, chemical equations do not refer to individual atoms" unless the equation represents an elementary (act of) reaction, I guess.

they didn't specify counting either the reactants or products side alone. You were to count both the reactants & products side


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Is there a catalyst for the decomposition for hydrogen peroxide that speeds it up but still makes the reaction light dependent?
I am building a Rube Goldberg machine for Science Olympiad, and one of the scoring steps is to use light to trigger a chemical reaction, which made me immediately think of the decomposition of hydrogen peroxide, but the only problem is that it doesn't occur fast enough. After doing some research, I realized that manganese oxide, the only catalyst for this reaction that I am aware of, will make the reaction occur without light. So, I am now looking for a catalyst that will speed up the reaction fast enough (to produce enough gas to inflate a balloon), but not lower the activation energy so much that it will occur without light.
Just to clarify, the competition does allow people to ask online, and there is in fact a whole community where people discuss this stuff.

There are various photocatalysts, but dunno if there's one suitable for this.

Related: https://chemistry.stackexchange.com/questions/7759/reaction-intermediates-of-mno2-catalyzed-h2o2-decomposition-reaction


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Should the concentrations in weight % of the components of a solution always add up to 100%?
There's a paper with this statement of a alkaline solution:

Nanostructured CuO films were formed by immersing the cleaned tabs into a hot $(\pu{96 \pm 3 ^\circ C})$ alkaline solution composed of $\ce{NaClO2}$, $\ce{NaOH}$, $\ce{Na3PO4 * 12 H2O}$, and DI water $(\pu{3.75 : 5 : 10 : 100 wt.\%})$.

I have looked online and read that normally the wt.% should add up to 100%. However, this statement says that DI water is at 100%. Is there something I am missing?
paper in context - https://arxiv.org/pdf/1310.2975.pdf

These numbers separated with a colon look like a ratio to me, not the absolute values.

So they are just weight ratios, and the total at 100% would just be 118.75? Thus wt% of the 1st compound would just be 3.75/118.75?

$3.75/118.75 \cdot 100%$, to be precise. That's how I see it.

To be 100% sure you can contact the authors – this is just my opinion and shouldn't be taken for granted.

Yup, I have done that as well and am waiting on their reply :)

Please add the citation to the paper so that it can be seen in context.

Could be the wrong /misuse of % m/V as well. Water is about 100 because they think 100 g gives 100 ml. There are plenty of errors like that even if might not be the current case, is better to be aware of the misuse of notation created for very diluted aqueous solutions


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Does Bent's Rule only apply to molecules where there is hybridisation?
I would just like to ask if the pre-requisite for using Bent's Rule is that the bonding in the molecule involves hybridised orbitals.
I was thinking of why the bond angle in hydrogen sulfide (~ 92 degrees) is larger than the expected 90 degrees. I thought that Bent's Rule would be able to offer a good explanation for this as the s orbital character concentrated towards the electropositive hydrogen atoms results in stronger bond pair-bond pair repulsion, increasing the bond angle. However, I realised that my entire argument may fall apart if Bent's Rule does not work for bonding involving unhybridised orbitals.

Bent's rule is not as much a rule as it is an observation. It was first applied in the context of valence bond theory and therefore used for hybrid orbitals only. At the initial stages, VBT only used s, sp, sp2, sp3, and p type orbitals. Bent proposed to allow also anything in between aiming at more flexibility for the wave function and therefore better accuracy of the description. As such it is only a tool for an already approximate view on bonding.
Most importantly, and in any and all general cases, the hybridisation state of the orbitals is determined by the molecular structure. For any case in which hybrid orbitals are used, there is also a true solution which does not use them.
Though Bent's rule has some predictive character for anything that is close to hybrid orbitals, it is basically just a shorthand for a much deeper rooted phenomenon.
However, Bent's rule is not useful in any case where there is a low s-p mixing, i.e. anything below period 2 is a stretch.

When u say "s-p mixing", do u mean s-p hybridisation? I know s-p mixing as an MO thing, but it does mean s-p hybridisation in the VB context right?

Could u also elaborate on your last statement? From what I read, u don't seem to have explained it

@TanYongBoon The s-p gap in atoms of the third period and below is in most cases too large for s and p orbitals of the same atom to mix in molecular orbitals. The hybridisation approximation breaks down at that point. That is one of the reasons for the different angles in $\ce{NH3}$ versus $\ce{PH3}$, see here.

Yes, s-p mixing is a generic term in MO theory. It refers to s and p orbitals of the same atom being a contributor to the same molecular orbital. In VB theory it used to be achieved with hybrid orbitals.


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What are bond lengths and angles for picric acid?
I am working on a project in which I am required to create a model of picric acid. I do not understand how to find the bond lengths, bond angles, shape, and atom size of this molecule. Any help in finding these measurements or in showing me how to find them would be amazing!

Homework? You should be able to find out something, if only via google. Report that, and ask what stays unclear.

I would suggest to take a crystal structure available for free from Crystallography Open Database, COD#5000146. The entry webpage already has a JSmol script loaded, so you should see the molecular structure loaded on a black area. Selecting RMB (right mouse button) -> Measurements -> Show Measurements allows you to measure linear and angular parameters with mouse clicks.
Alternatively, you can download the CIF-file (5000146.cif) and load it with any crystallographic viewer if you prefer to work offline. For beginners I'd recommend Mercury, there is a free version available for Windows, macOS and Linux:

N. B. The asymmetric unit in this structure contains two molecules of picric acids – with and without H-atoms. Obviously, you need the one with the protons assigned.


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Isobutene IUPAC naming
What is the correct name of an isomer of isobutene $\ce{C4H8}$: 1,1-dimethylethene  or 2-methylprop-1-ene? I have done an exam recently and I'm just curious about what what would be correct.

It will be 2-methylprop-1-ene , as we need to select the longest carbon chain containing the functional group i.e. alkene here.


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Does PCl3 react with ester as it does with carboxylic acid to produce an acid halide?
I did not find it anywhere in a book but I do wonder if it might react.

Yes it does. All carboxylic acid derivatives undergo nucleophilic substitution reactions. So esters, acid anyhyrides, amides all on reacting with $\ce{PCl3}$ will give acid chlorides.

Please note that there is mhchem package enabled on chemistry.SE which allows you to type chemical formulas easier and more correctly via $\ce{...}$ construction. For more info check out How can I format math/chemistry expressions here?

@andselisk I work a lot on mathematics.se so I have habits. Will surely take care from next time.

This answer is wrong.

@mithoron please provide a reference

This cannot work on amides, instead reactions leading to imidoyl chlorides, iminium chlorides, or other products, can happen. Also you should check out before answering and logic behind your answer isn't valid.

@Mithoron I am reading them from my textbook. Cengage's organic chemistry by KS Verma.

@Mithoron yes, the answerer may have been too broad, but still I would expect a reaction with esters to give the acid chloride.  Phosphorous will bond with oxygen over chlorine.


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Reaction of hydrazine and hydroxylamines with carboxylic acids
Carboxylic acids and their derivatives on reaction with hydroxylamine $\ce{NH2OH}$ and hydrazine $\ce{N2H4}$ give hydroxamic acids and acid hydrazides, respectively, through acyl nucleophilic subsitution.
Why don't they give the same reaction as aldehydes and ketones (i.e. removal of $\ce{H2O}$) giving formation of oximes as the acids also have oxo ($\ce{=O}$) group?

For simplicity, I will consider the reaction of acetic acid with ammonia, because the same principles apply. The two alternatives are:
$$
\underbrace{\ce{CH3C(O)NH2}}_{\mathrm{amide}} + \ce{H2O} \leftarrow \ce{CH3C(O)OH} + \ce{NH3}\rightarrow \underbrace{\ce{CH3C(NH)OH}}_{\mathrm{imidic \ acid}} + \ce{H2O}
$$
where the group connected via a double bond has been set in parentheses. The two products are connected by protonation/deprotonation in a tautomeric equilibrium, so the question boils down to thermodynamic stability. For most cases, we expect the C-O double bond/C-N single bond combination to be more stable than the C-N double bond/C-O single bond combination.

Is it a kind of beckmann rearrangement?

@prog_SAHIL I clarified the answer. No, it is not a Beckmann rearrangement, because no new C-N bonds are formed - they only change from double to single or vice versa. It is a simple matter of protonation and deprotonation, just like keto/enol.


text:
Percentages of radical constitutional isomers
Is it possible to use the radical selectivity factors for chlorination based on primary, secondary, and tertiary substitutions to calculate the expected percentage of each constitutional isomer formed in the chlorination reaction? If so, how?

The gist of this question is different and yet you could basically copy and paste the answer here to have it answered. If I vote for duplicate, the question will be hammered; I don’t want to do that.

@Jan Thanks! I see expected percentages but not how they were derived from selectivity factors.

Your question is a bit hard to understand, you should change the wording or add more explanation. But I believe you want to know how to calculate the percentage of a particular isomer from the relative rates of chlorination. Well the calculation is as follows:
First, derive the ratio of different isomers. To find the terms of ratio, multiply the number of hydrogens which can be substitued to form the isomer, by the rate of halogenation of that type of hydrogen.
Next, calculate percentage from ratio. This is very easy arithmetic; you can do this by creating a fraction with the value of that particular isomer on the numerator and sum of all values from the ratio on the denominator. Then multiply the fraction by 100.
The ratio of reactivity of tertiary($\ce{3^\circ}$), secondary($\ce{2^\circ}$) and primary($\ce{1^\circ}$) H-atoms for chlorination is $\ce{5 : 3.8 : 1}$  at 298K. For bromination of $\ce{3^\circ}$, $\ce{2^\circ}$, $\ce{1^\circ}$ H, the ratio of reactivity is $\ce{1600:82:1}$ at 400K
Example: Monochlorination of n-butane. Here, two isomers are formed, 1-chlorobutane, and 2-chlorobutane.

Now, substitution of six $\ce{1^\circ}$ H  (shown by blue box) results in 1-chlorobutane
and substitution of four $\ce{2^\circ}$ H (shown by red box) results in 2-chlorobutane
So, the ratio is 1-chlorobutane : 2-chlorobutane  $\ce{=6\times1 : 4\times3.8= 6:15.2}$
Calculate percentage: 1-chlorobutane is $\ce{ \frac{6}{6+15.2}\times 100\%= 28.3\%}$
and similarly, for 2-chlorobutane, percentage is $\ce{71.7\%}$

The degree notation is not generally acceptable as a shorthand for primary, secondary and tertiary; I first encountered it on this site and it seems like it has been invented in India and is only understood there.

@Jan , See https://www.masterorganicchemistry.com/2010/06/16/1°-2°-3°-4°/ and https://www.chemguide.co.uk/mechanisms/eladd/carbonium.html and also http://www.chem.ucla.edu/~harding/tutorials/cc.pdf. Clearly these websites are not Indian, and they use the degree notations. These also prove that, in fact, the symbols are quite common.Nevertheless, I will edit my post to make it better understandable.


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Interpretation of PES spectra

Given that both elements are in the same period, what information from the PES spectra indicates that oxygen would have a smaller atomic radius?
(A) The number of peaks
(B) The intensity of the peaks
(C) Binding energy of the peaks
(D) A and C
(E) B and C

Recently my chemistry teacher and I had a dispute over a test question, what would be the correct answer?
Personally, while I do agree with the fact that binding energy clearly answers the question, I also believe that the intensity of the peaks could also indicate that oxygen has a smaller radius than nitrogen, given that they are in the same period.
This is because intensity indicates the number of electrons, which indicates the number of protons, which indicates that oxygen would have a greater effective nuclear charge, and thus a smaller radius. As a result, my answer was both B and C, contrary to my teacher's answer of simply C.


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If hydrogen bonds are chemical bonds, and these bonds are broken when water changes phases, why doesn't this classify as a "chemical" change?
This is more of a terminology question. How are phase changes considered a physical property, not a chemical property when hydrogen bonds break. A chemical property can only be observed during a chemical change. Aren't hydrogen bonds a type of chemical bond, and those bonds are being broken during a phase change from ice to liquid water, for example. I understand that no intramolecular bonds are being broken. Water molecules remain intact. But we teach that a chemical bond has to break during a chemical change, but phase changes are considered to be physical.

Possible duplicate of Ripping apart plastic: is this a chemical or physical change?

The question there makes use of a different example ("ripping a plastic sheet") but the answers to it should address most of your issues too... since your question (as I interpret it) is basically "Are physical and chemical changes sharply defined?". Cheers!

The distinction between chemical and physical changes exists only in our heads.

I respectfully disagree with @paracetamol because the other question focusses on breaking covalent bonds which we should all agree are sufficiently different from hydrogen bonds.

@Jan Ah, but the physical-chemical change classification taught at school represent a false dichotomy (I don't have an issue with the type of bonding per se). But now that I've re-read the question I linked (and its answers), I realized that the answers to ron sensei's question don't make much of a case in this regard (i.e- dispelling the notion of rigid physical-chemical change classification). I stand corrected, accordingly I've withdrawn my close vote. Thanks for the pointer O:)

This would apply to metal as well.  I see the  textbook classification of chemical vs physical not overly simplified but having somehow a deeper meaning. At least a physical process never alters the minimal formula of an element/compound

You must realize that the line between physical and chemical properties is very thin. Most bonds work via the same principle, which has mostly to do with electronegativity. When looked at in reactions it has probably also to do with the according entropy and enthalpy change. Hydrogenbonds are basically bonds based on electronegative attraction between two atoms (which is a simplified few of the reality). A phase change basically when a certain energy point has been reached: since heating op a substance is equal to giving it more kinetic energy (since heat is the same as the movement of individual atoms or molecules). When you reach a certain heat there is too much kinetic energy to form hydrogenbonds the same way it as it did before it changed phase. You probably can agree that molecules or atoms moving quickly is an obvious physical property. When you look at great detail to how hydrogenbonds work, it has something in common with many other way of two objects attracting each other (for example: the formula for the gravitional force between to large heavenly bodies, is very similar to the formula for the electrical force between an electron and a nucleus).
In simple: Many of these chemical properties seem very physical when you look at them in detail. So saying that a certain property should actually be chemical because the changing of this property seems to correspond with the changing of chemical bonds, is not a realistic claim looking at how "physical"/physic based these "chemical" bonds are.

First let's clear up a misconception in the OP's Question: Water molecules remain intact. - NO!!! This is absolutely wrong. Water is sort of an odd solvent because the "molecules" are not absolutely fixed.
If pure $\ce{D2O}$  is mixed with pure $\ce{^1H2O}$, then you get $\ce{D^1HO}$ which shows that the hydrogen oxygen bond is very liable. So a hydrogen bond can easily be transformed to one of the two covalent O-H bonds by having one of the original O-H covalent bonds become a hydrogen bond. So water is a vast network of hydrogen and covalent bonds which is forever changing.
We could consider $\ce{D2O}$ and $\ce{^1H2O}$ as different compounds since they have different freezing points and different boiling points. But the point here was to somehow label the hydrogen atoms in water. In pure $\ce{^1H2O}$ the hydrogen-oxygen bond is still very liable.
Now to the overall question. The point is that most pure substances will solidify if made cold enough. Warm the substance beyond its melting point and most turn into a liquid. Heat the liquid hot enough and the liquid boils. Furthermore thermodynamics dictates that it takes a certain amount of heat to convert a certain mass of the solid at the melting point to a liquid. Also it takes a certain amount of heat to convert the liquid to a vapor at the boiling point. That is the conceptual framework upon which thermodynamics is based.
There is some acknowledgement of how various forms of bonding effect the phase transitions. For instance consider sodium chloride. The bonding energy of one $\ce{Na^+}$ ion in the solid isn't between the one $\ce{Na^+}$ ion and one other $\ce{Cl^-}$ ion but between the attraction of the $\ce{Na^+}$ ion to all the $\ce{Cl^-}$ ions and the repulsion of all the other $\ce{Na^+}$ ions. The overall factor is called the Madelung constant.

But this is an effect of water being a protic solvent. Water is also able to protonate substances with which it does not undergo hydrogen bonds such as sulphides or carbanions. Indeed, with a sufficient acidity of the corresponding alpha protons of a ketone, these can be deuterated in $\ce{D2O}$ — does that mean water hydrogen bonds to the carbon?

Suppose you heat carbon in an atmosphere of carbon dioxide.  They react; carbon monoxide is formed in the gas phase.  Cool the gas and the reaction, which is endothermic for formation of the carbon monoxide, reverses; you redeposit carbon (possibly on cooler parts of the furnace in industrial settings).
Suppose you put sugar into water and heat it up.  The sugar dissolves, and more dissolves as you heat the system.  Let it cool; the dissolution of the sugar, which is endothermic, tends to reverse and solid sugar reappears.
What is, empirically, the difference between these two processes?  "Chemical reaction" versus "physical change" is a blurred distinction.

The original post referred to phase change of a pure substance. With mixtures the situation may be different, I suspect. Otherwise why do we say that the sugar 'dissolves' - rather than 'melts' -  in water?


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Magnetic work and the first law of thermodynamics
We commonly encounter the first law of thermodynamics in the form $\mathrm{d}E = T\mathrm{d}S - p\mathrm{d}V + \mu\mathrm{d}N$, which notes the contribution of thermal, pressure-volume, and chemical work to the total energy respectively. More generally, we can write $$\mathrm{d}E = \sum_i f_i\mathrm{d}X_i,$$ where $f_i$ denotes a generalized force and $X_i$ a generalized displacement.
The pattern for all these types of work seems to be rather obvious---the generalized force is intensive, whereas the generalized displacement is extensive---but this fails when we try to take into account magnetic work, which contributes the term $-\mathbf{m}\cdot\mathrm{d}\mathbf{B}$, with $\mathbf{m}$ the magnetic moment and $\mathbf{B}$ the magnetic field (see Hill, Chandler, Blundell, etc.). Here the pattern is reversed: the magnetic moment $\mathbf{m}$ is extensive, whereas the magnetic field $\mathbf{B}$ is intensive!

How can we rationalize the unusual form of the magnetic work term?

Blundell makes an argument for this for the case of electric dipoles; I summarize it here. The relevant work term here is the electrical work $-\mathbf{p}_\text{E}\cdot\mathrm{d}\mathbf{E}$, with $\mathbf{p}_\text{E}$ the electric dipole moment and $\mathbf{E}$ the electric field.

The potential energy of the dipole in the electric field is $-\mathbf{p}_\text{E}\cdot\mathbf{E}$. If the electric field changes, the potential energy changes by $\mathrm{d}(-\mathbf{p}_\text{E}\cdot\mathbf{E}) = -\mathbf{p}_\text{E}\cdot\mathrm{d}\mathbf{E}-\mathbf{E}\cdot\mathrm{d}\mathbf{p}_\text{E}$. However, the change in the electric field will also stretch the dipole, increasing its dipole moment and doing work $q\mathbf{E}\cdot\mathrm{d}\mathbf{l} = \mathbf{E}\cdot\mathrm{d}\mathbf{p}_\text{E}$, so the net work supplied to the system is $-\mathbf{p}_\text{E}\cdot\mathrm{d}\mathbf{E}$.

I am somewhat skeptical of this argument. It seems that I could repeat the argument for other types of work---say, $PV$ work---and conclude that I should also have a $V\mathrm{d}P$ term in my energy expression.

Is this argument correct? If it is, why isn't there a $V\mathrm{d}P$ work term in the first law?
What are the conditions under which this argument holds? Do the generalized force and generalized displacement have to interact with each other in a particular manner?

It also seems to me to be of limited applicability, because the convenient cancellation of magnetic-dipole and electric-dipole terms are not replicable for other, more complicated interactions.

Is there a general, methodical approach for identifying the form of the terms in the first law for arbitrary types of work?

In a way, magnetic work is an oxymoron. Magnetic force is perpendicular to velocity, hence it doesn't do work.

@Ivan_Neretin Of course, magnetic fields do work on spins, they just do not work on moving charges.

Both are meaningful: $mdB$ is the work done on a dipole by changing the field, $Bdm$ is the work done on the field by changing the magnetization of the dipole. $B$ is closer in spirit to a potential, hence an intensive quantity, while piling more and more dipoles on top of each other is like having more matter, hence extensive.

"Magnetic work" generally refers to work done by a constant-current supply connected to a solenoid. Whether this work is done upon the thermodynamic system of interest depends on whether the system is chosen to include, in addition to the dipoles, the self-field energy of the solenoid and/or the mutual-field energy of solenoid and dipoles, defined below. For details, consider that a stationary microscopic dipole $\boldsymbol\mu$ sits in the magnetic field $\bf {B}_{sol}$ inside a solenoid driven by a constant-current supply. Integrals herein are over all space. If the field changes by ${\bf {dB}}_{sol}$ while $\boldsymbol\mu$ remains fixed, then the current supply does work only on the solenoid’s self-field energy, $\int{dV}{B_{sol}^2({\bf r})/{2\mu_0}}$. Also, the mutual-field energy of solenoid and dipole $\int{dV}{{\bf{B}}_{sol}({\bf r}) \cdot {\bf{B}}_{dip}({\bf r}) / { \mu_0}}$, and the dipole energy change by equal and opposite amounts, +$\boldsymbol\mu \cdot {\bf{dB}}_{sol} $ and -$\boldsymbol\mu \cdot {\bf{dB}}_{sol} $, respectively. In other words, the mutual-field energy does work -$\boldsymbol\mu \cdot {\bf{dB}}_{sol} $ on the dipole. If the mutual-field energy is not part of the system, then work $\boldsymbol\mu \cdot {\bf{dB}}_{sol} $ was done on the system. Otherwise, no.
In the converse case wherein the dipole changes (by $\Delta \boldsymbol{\mu}$) while the field remains fixed, the current supply does work +${\bf{B}}_{sol} \cdot \Delta \boldsymbol{\mu}$ on the mutual-field energy, and the same amount of energy leaves the dipole and flows ultimately as heat into a thermal reservoir. Namely, the dipole energy changes by -${\bf{B}}_{sol} \cdot \Delta \boldsymbol{\mu}$. If the mutual-field energy is part of the thermodynamic system, then work was done on the system. Otherwise, no. n.b. The change in dipole orientation necessarily comes about via scattering with some quantum excitation that can take energy and angular momentum from the dipole. The change in dipole orientation may be as large as the dipole moment itself, not a differential quantity.