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The melting and boiling point decreases down the group up to group 14 (not including transition metals), but the trend reverses from group 15. Why?
In groups 1, 2, 13 and 14, the melting and boiling point decreases down the group with a few exceptions. In group 15 the melting/boiling point increases up to Arsenic and then started decreasing. In groups 16, 17 and 18 it increases down the group, but with few exceptions.
Groups 1 and 2 are held by metallic bonding and the strength of metallic bonding decreases down the group, so this explains the first two trends. Carbon doesn't actually melt or boil, it sublimes but the point still stands that it's sublimation point is much higher than Silicon's boiling point.
Oxygen exists as $\ce{O2}$ and sulphur as $\ce{S8}$, so this explains the large difference in boiling points, but atomic orbitals become more diffuse down the group, so the bond strength and thus, the melting points should decrease but that doesn't happen.
Almost all other periodic trends like ionisation enthalpy or density are fairly consistent across the groups. Why is this different? I couldn't find more information anywhere. What's happening here?

Describe in details how you "couldn't find more information anywhere.", as it is often just a figure of speech for "I have not thought nor searched thoroughly enough, but saying that would look bad."  // Who says transition of declining slopes to inclining ones cannot go through a peak shape? Who says chemistry of elements cannot affect it?

@Poutnik, I did search for previous questions on this site and I got vague answers like van Der Waals forces, nothing related to thisunfortunately

Why just this site? You know, effort not shown can be considered as effort not done. You  did not think/search deeply, if chemical bonds and element chemical character did not come to your mind. For group 15, it comes from triple bond of N2 through covalent single bonds of P, As to metallic bonds of Sb, Bi. Why should one even consider monotonous trends ?

@Poutnik, edited to add more information like you had suggested.

Go through these questions: https://chemistry.stackexchange.com/questions/33815/trend-in-the-melting-point-down-the-group ... https://chemistry.stackexchange.com/questions/114425/reasons-for-low-melting-point-of-gallium .. https://chemistry.stackexchange.com/questions/35716/why-there-is-an-exception-in-melting-and-boiling-point-in-p-block ... https://chemistry.stackexchange.com/questions/41340/why-do-the-melting-points-of-group-15-elements-increase-upto-arsenic-but-then-de .. https://chemistry.stackexchange.com/questions/114857/question-on-melting-points-of-group-17-18-and-1

The problem is, we want to see a trend in everything. There are various factors that govern melting point and boiling point which is the reason of perceived anomalies. "The equation which has more number of variables, is harder to solve."
For melting point, few of them are: Crystal system, size of atom, atom-atom distance, distance between two layers of atoms, metallic character, metallic bond strength, inert pair effect, relativistic effect, etc.
Note that, down a group, metal-metal bond strength decreases due to increase in size but metallic character increases because now, $\ce{e-}$ abstraction is easy from that metal.
Group 1 and 2: All have metal-metal bond but the strength of bond decreases. Thus, m.p. decreases.
Group 13: Except $\ce{B}$ all are metals. The m.p. decreases from $\ce{B}$ to $\ce{Al}$ then decreases for $\ce{Ga}$ and then increases till $\ce{Tl}$.
The low m.p. of $\ce{Ga}$ is due to existance of $\ce{Ga2}$ dimers in the orthorhombic crystal. The higher m.p. of $\ce{In}$ and $\ce{Tl}$ may be due to body centered tetragonal and hexagonal close packing structures, respectively.
This explains the boiling point trend in group 13.
Group 14: Except $\ce{C}$, $\ce{Si}$ and $\ce{Ge}$, all ($\ce{Sn}$ and $\ce{Pb}$) are metal. The m.p. decreases till $\ce{Sn}$ and then increases for $\ce{Pb}$.
$\ce{Ge}$ has face centered diamond cubic lattice, $\beta~\ce{Sn}$ has body centered tetragonal lattice and $\ce{Pb}$ has face centered cubic lattice. The higher m.p. of $\ce{Pb}$ than that of $\ce{Sn}$ may be due to this factor and a lower van der Waals radius.

In group 15 the melting/boiling point increases up to Arsenic and then started decreasing.

Group 15: The m.p. increases upto $\ce{Sb}$ then $\ce{Bi}$ has a low melting point. $\ce{As}$ sublimes at $\pu{887 K}$ and $\pu{1 atm}$.
$\ce{As}$, $\ce{Sb}$ and $\ce{Bi}$ all have layered structure crystalising in rhombohedral lattice. The atomic radius of $\ce{Sb}$ and $\ce{Bi}$ are very similar. The low m.p. may be due to higher atom-atom and layer-layer distance than those in $\ce{Sb}$ as explained here.
Group 16: The m.p. increases upto $\ce{Te}$ then $\ce{Po}$ has a low melting point.
$\ce{S}$ is orthorhombic, grey $\ce{Se}$ is hexagonal, $\ce{Te}$ is rhombohedral and $\alpha~\ce{Po}$ is simple cubic.
Another reason for low m.p. and b.p. of $\ce{Po}$ is explained here.
Group 17 and 18: None of them are metals. The m.p. and b.p. increases uniformly with the increase of van der Waals forces.
I know that this is not a complete answer but you can do your own findings based on this.

Why does metallic bonding weaken down the group?


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About Real Gas and Ideal Gas - Select the correct statement(s):

Select the correct statement(s):

At Boyle's temperature a real gas behaves like an ideal gas at low pressure.
Above critical conditions, a real gas behaves like an ideal gas.
At high pressure van der Waals's constant 'b' dominates over 'a'.
For Hydrogen gas 'b' dominates over 'a' at all temperatures.

Answer in textbook is given: 1 & 3
Can someone please explain me why 4 is not included in answer. My teacher told me that Due to very small size of hydrogen and helium atoms there value of "a" is negligible. So shouldn't "b" always dominate "a" for hydrogen and helium atoms ("a" and "b" are van der Waals' constants).

Hint: If A4 is true, H2 compressibility factor is always > 1. Analyze if it is true.

source: Stack Exchange

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

sort: votes


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How does a molecule (for example, ClF₅) have a steric number of more than 4 if SP³ hybridisation is the maximum?

"The sum of the number of atoms bonded to a central atom and the number of lone pairs formed by its nonbonding valence electrons is known as the central atom's steric number."
Wikipedia's entry about "Steric number"

I just started learning hybridisation and stuff for fun, and I just found out this "oddity".

This earlier answer relating to $\ce{IF5}$ may be helpful.

The steric number of $\ce{ClF_5}$ is 6, and this is possible because $\ce{sp^3}$ hybridization is not the maximum, or the 'limit' to hybridization. In the case of $\ce{ClF_5}$ the hybridization is $\ce{sp^3d^2}$, the hybridization involving  the $\ce{3s}$,  $\ce{3p}$ and  $\ce{3d}$ orbitals.

Image credits: https://techiescientist.com/clf5-lewis-structure/

3d orbitals are not involved in bonding in ClF5. The best representation is partial covalent bonds (ie bond order <1) combined with ionic interactions.

Supporting @Andrew see https://chemistry.stackexchange.com/questions/4760/why-do-compounds-like-sf6-and-sf4-exist-but-sh6-and-sh4-dont/49844#49844

https://chemistry.stackexchange.com/q/76726/4945

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


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Why does fcc lithium melt with increased pressure?
According to the phase diagram of lithium given here, the fcc phase formed at moderately high pressure melts at decreasing temperature upon adding more pressure, an effect better known with the ordinary solid phases of water, silicon, and germanium among a few other materials.
In water, the directionality of the hydrogen bonding between molecules leads to a low-density solid structure, which, in effect, collapses under pressure as the directional bonding breaks down in the liquid. The covalent bonds in silicon and germanium, with their tetrahedrally coordinated diamond-type structure, behave likewise as these elements switch to a metallic structure on melting.
But a low density diamond-type structure seems implausible in fcc lithium since the fcc phase has to be formed under pressure and thus should be denser than the eight-coordinate bcc structure. So what structural feature in fcc lithium enables a water-like, or silicon-like, collapse to a liquid under more pressure?

Some research I have done on this problem suggests that it is the atoms themselves that break apart and collapse. When we account for the distribution of electrons the fcc structure in lithium is not really close-packed and this is collapsible.
Go outside the lines
Magnesium silicide ($\ce{Mg2Si}$), commonly used as a strengthening agent in aluminum alloys, is black. Sodium sulfide, fully isoelectronic and isostructural with magnesium silicide, is white (yellowish when polysulfide impurities are present). This color difference arises from the unusual electronic properties of magnesium silicide, which make that compound a narrow-gap semiconductor (thus absorbing visible light as electrons cross this gap) while sodium sulfide is an insulator.
Panke and Wöfel [1] studied the electron distribution in magnesium silicide, finding (among other things) that in the plane of the magnesium atoms there is a small local maximum in the electron density distribution at the interstitial sites within squares of atoms (see picture below). Magnesium silicide thus has a partial electride character. This electron distribution imparts electron mobility and thus electrical conduction in the cubic magnesium-atom lattice.

Irresistible force, movable electrons
Frost et al. [2] compute that fcc lithium develops a similar character as it is compressed. The valence electrons are partially transferred to octahedral interstitial sites, leaving behind the small and dense lithium cores. This electron transfer into interstitial sites increases with increasing pressure, although not to the point of forming a complete electride salt in the fcc phase (some other, higher-pressure phases are full electrides). The illustration below from this reference shows this electron transfer and localization in lithium. The variance in electron density between the atoms is similar in magnitude to that in ambient-pressure magnesium silicide, except here the maximum applies over all three dimensions instead of just two.

Thus in fcc lithium, the interstitial electron density acts as spacers between the lithium cores which become positively charged and mutually repelling. The lithium cores shrink away from the close-packing that would ordinarily be expected in an fcc metal, and thus the structure may be potentially collapsed if enough pressure is applied to force the cores to re-intermingle with the valence electrons.
References
1.
Panke D., Wölfel E. (1972) "Distribution of Valence Electrons in Magnesium Silicide." In: Sirota N.N. (eds) Chemical Bonds in Solids. Springer, Boston, MA. https://doi.org/10.1007/978-1-4684-1686-2_1.
2.
Mungo Frost, Abraham L. Levitan, Peihao Sun, and Siegfried Glenzer. "Equation of state and electron localisation in fcc lithium".
Journal of Applied Physics 123, 065901 (2018); https://doi.org/10.1063/1.5020296.


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Are spontaneous/induced dipoles the same as van der Waals forces?
I am puzzled by the names of dipoles.
What is the difference among permanent dipoles, spontaneous dipoles, van der Waals forces, and intermolecular forces?
Does "The intermolecular force" imply van der Waals and dipole forces?

Induced dipoles are created by external electrostatic field. Permanent ones do not need such a field. // The van der Waals forces  are usually described as a combination of the London dispersion forces between "instantaneously induced dipoles", Debye forces between permanent dipoles and induced dipoles, and the Keesom force between permanent molecular dipoles whose rotational orientations are dynamically averaged over time. https://en.wikipedia.org/wiki/Van_der_Waals_force

So, as dipoles are not forces, they cannot be the same as the vdW forces..


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Vibrations of metal complexes (polyatomic molecules)
I am reading about vibrational spectroscopy of polyatomic molecules and specifically about metal complexes. What I don't understand is what happens on the vibrational modes of the ligand (e.g. water) when coordinates to a metal ion.
When the complex is formed then it doesn't make sense to speak about the vibrational modes of water because it is not anymore a triatomic molecule. But we use IR spectroscopy for identification of functional groups. So if the identity of the ligand is lost upon coordination how can we study the functional groups that have been coordinated?
In an IR absorption spectra of a complex the wavenumbers where absorption happens includes that of the ligand's (e.g. symmetric stretch of water). So that means that the water (or any other ligand) still exhibits its own vibrational modes. Shouldn't the new wavenumbers be bigger as they involve more atoms and therefore more energy is needed for the vibrational modes?
In summary, I want to understand why it is still possible to identify ligands upon coordination and why the energy of the vibrational modes is not proportional to the number of atoms. For example should we expect methane's vibrational modes to be higher in energy than the vibrational modes of pentane? An intuition based answer will be nice.

If you recall, the infrared spectrum of a compound has several regions. One of the region is the functional group region, where you see the vibrations of a functional group of a molecule.
Then you have a fingerprint region which corresponds to the vibration of the whole molecule. As expected, this region is unique to the molecule. So, if your complex has water, you should see broad OH vibrations, and then in the fingerprint region you have more complex vibration particular to the molecule involving the bound ligands.
You should expect change in frequency of a bound ligand vs. a free ligand. Since the spectral bands are so wide in solids/liquids that the shift is not greatly visible.

After coordination the water can't behave like a triatomic molecule. It is bonded to the metal ion or whatever is coordinated. Why it still exhibits its vibrational modes?

Yes, -OH can still vibrate. We are not talking about HOH.


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Vapor - Liquid Equilibrium (VLE)
Most fundamental equation for VLE is $$ \mu_i^L = \mu_i^V $$
It states that for every component chemical potential must be equal in both liquid and vapor phase at equilibrium. However, in my thermo textbook, this equation is derived for isolated systems while usually when dealing with VLE, systems are closed since heat is usually exchanged with surroundings like in distillation for example. Does this equation hold also for closed systems and if so how can it be derived?

The question rather is, if the system is in equilibrium or not. Boiling liquid definitely is not. More closely the system approaches equilibrium, more closely the relation of respective ch.p. approaches equality.

That isn't the question. Question is what is written in the description.

In fact, it is. Is the closed system in equilibrium or not? If not the equation can be true just approximately, depending on nature and extend of the deviation.

And if it is in equilibrium, it is functionally equivalent to isolated system, as energy exchange is zero.

We are dealing with closed system at equilibrium. Why should energy exchanged be zero at VLE in closed system? We can have water vapor in equilibrium with liquid and add heat (latent heat)to the system which will make liquid evaporate, but this won't change the fact that chemical potential off the water in two phases is the same.

If you add to system in eq. heat dQ, liquid will have  ch.p. increased wrt vapor by dmu, until dn of liquid evaporates and both phases have the same ch.p. again. The more macroscopic this heat addition or rate is, the bigger difference of ch.p. of both phases will be. Ch.p. difference is the power leading to phase changes.

Hmm, interesting. Adding heat can change internal energy and create some work by the first law. I am not sure how would it affect chemical potential of phases or how to justify that liquid's chemical potential will increase as you say?

What do you think the difference between an "isolated" and a "closed" system is?


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What are all the steps on the organic chemistry diagram called The Boat?
When I learnt A-level chemistry, I was told to use a diagram the teacher called The Boat.
It was a diagram illustrating all the chemical reactions that you can do to simple organic compounds. For instance, the mast of the boat, showed reduction/oxidation of alcohol ⇌ aldehyde ⇌ ketone, if I remember correctly?
I'm looking for a copy of this exact diagram (so please do not recommend others), to reference, if I ever need to discuss these reactions.
My assumption is that this was not an illustration that only existed at my secondary school, and that my description is enough to identify it by. If not, please do let me know in a comment.

Only heard of it in reference to a conformational variation vis a vis chair. To be clear: you are referring to a diagram in which all reactions for one compound are displayed in a boat-like structure? This may or may not be something your instructor was taught and is not universally known.

This is definitely nothing to do with the chair molecule, which is all I could find on a quick pre-question Google. It was for a large number (15-20) different molecules, and for instance showed how to go from ethanol/alcohol to  a carboxylic acid in one direction, etc.

Remember where was it in? What textbook?

@mithoron it was always printed out on paper/hand drawn as part of the teaching so I don't know.

I found something called an oxidation ladder which might be along the right lines (nothing with boat components though) - does this help move in the right direction?

@toddMinhardt that's the right sort of thing. Looking at related pins it was a reaction map for certain. I may have to try and get back in touch with them

I hadn't heard of a "boat" diagram, but the alcohol-aldehyde-ketone "mast" rang a bell, so I went hunting...
Is this the "boat" diagram you're talking about (or similar to it)?
https://o.quizlet.com/MGXYzJ7s01UHNZq3TDAVEw_b.png
(retrieved from https://quizlet.com/513370563/organic-chemistry-reaction-pathways-diagram/)
I can see how the alcohol/aldehyde/ketone interactions would make something akin to the mast and boom of a sailboat, though the rest takes a bit more squinting to work into place. :)
edit: The following link also came up while searching, though the reactions you mention aren't in the mast of the boat in this diagram; if your "boat" was drawn from the side rather than from the top, however, this might have been something akin to what you'd been shown.
https://www.tes.com/teaching-resource/the-organic-boat-7513303

It's much more likely I misremembered the mast being made of those, because that is the spitting image of what I remember from school.


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Best practices to obtain small quantity of sealed solution without oxidising entire solution?
I'm ordering liquid based chemicals which only come in large bulk quantities. These solutions have a shelf life of approximately 2 years unopened, but 2 - 3 weeks once opened and oxidation kicks in. To put things into perspective, they come in 5 litre bottles to dilute to 50 litres but I only require 100ml per use. So going through 5 litres in 2 - 3 weeks is not realistic whatsoever.
So this got me thinking if there is a way I could obtain small quantities of a solution without at all exposing the entire solution to the atmosphere? The first thing that comes to mind is to obtain the solution within a vacuum chamber, then finding a way to vacuum seal the bottle back up.
Would that be a good practice or are there better ways to work around this problem?

In a lab environment this would be done by sealing the 5L bottle with a septum under a dry nitrogen atmosphere. To extract solution you would use a syringe and needle through the septum with a second needle supplying nitrogen into the 5L bottle. Harder to do this at home!

A simple solution would be to use a glove bag (these are readily available from chemical suppliers) flushed with an inert gas like nitrogen to transfer the solutions from the large bottle into a number of small 100mL sealable bottles.

You might also join a photo hobbyist group, perhaps through a nearby school, to share the cost of chemicals. Learning together is more fun, too. Sorry, this is not a technical answer.

Could an electricity based induction sealer be used within the inert atmosphere glove box to seal a foil liner over the smaller bottles?

@srb633 This is possible (because all what the device does, is providing heat locally, like a pair of curling irons).  For a stationary installation, you e.g., pass the power cord across one of the feedthroughs like for a balance (an example with a laminator is shown on fig. 12 here).


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Why is ester more reactive to nucleophile than carboxylic acids?
I was given just this statement at face value, but I think it refers to nucleophilic reactions where the carbonyl is the electrophile.
From what I understand, the $\ce{-OR}$ in ester is more electron donating inductively than the $\ce{-OH}$ in $\ce{-COOH}$ (due to the alkyl groups in $\ce{-OR}$). So this should make the carbonyl carbon less positive, making it a weaker electrophile, making it less reactive no?
The resources I read say that it's because the $\ce{-OR}$ is a better leaving group than $\ce{-OH}$. But alkoxide ($\ce{RO-}$) is a stronger base than hydroxide ($\ce{OH-}$), so why would it leave first?

Remember that a carboxylic acid is an ACID. It donates a proton, forming an anion and will protonate anionic and basic nucleophiles rendering them non-nucleophilic. If there is an excess of the nucleophile, this means that it is attempting to attack a species that is negatively charged; this is not a favourable interaction. Should it succeed the leaving group is not -OH it is O-


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On what basis are the prefixes α and β given to black phosphorus?
My textbook says the following about black phosphorus:

α black phosphorus is formed when red phosphorus is heated in a sealed tube at $\pu{803 K}$. β black phosphorous is prepared by heating white phosphorus at $\pu{473 K}$ under high pressure.

National Academy of Sciences USA confirms that Bridgman successfully obtained black phosphorus for the first time in 1914 by conversion from white phosphorus at a pressure of $\pu{1.2 GPa}$ and an elevated temperature of $\pu{473 K}$. This probably means that the nomenclature is not on basis of the date of synthesis. Then on what basis are the prefixes given?
Edit: After searching a bit more, I could only find this terminology in Indian higher secondary text books (not even university ones), and I have a nagging suspicion at the back of my mind that it's an outdated term. Anyways I'll leave the question up since I would still like to know the history behind it, and why it stopped being called so.
Reference:

Renaissance of black phosphorus, Xi Ling, Han Wang, Shengxi Huang, Fengnian Xia, Mildred S. Dresselhaus, Proceedings of the National Academy of Sciences Apr 2015, 112 (15) 4523-4530; DOI: 10.1073/pnas.1416581112

In the past, assignment of alpha, beta, gamma, etc. to polymorphs was historical or/and related to (thermodynamic) stability.  See e.g., this related post.

For future reference:  1) Since mhchem is not universally understood, it is better to refrain from its use in titles of questions and answers.  It is fine for bodies of questions/answers/comments of chemistry.se.  2) For a contemporary eye, it is odd to still read temperatures provided in  $\pu{^\circ{}K}$ as if there were degree kelvin.  It is better to either use $\pu{K}$ (like $\pu{77 K}$) for kelvin, or $\pu{^\circ{}C}$ (like $\pu{100 ^\circ{}C}$) for celsius, or $\pu{32 ^\circ{}F}$ for fahrenheit (note the spaces) instead.

See this question: https://chemistry.stackexchange.com/questions/151307/stability-of-%ce%b1-%ce%b2-%ce%b3-allotropes-at-different-temperatures/151355 (this is also applicable to your sulfur post).

The earliest discovered molecular form of the elemental phosphorus is white phosphorus $(\ce{P_\mathrm{W}})$, which was isolated in 1669 by a German alchemist, Hennig Brand (Ref.1). Later in 1914, Percy Bridgman has introduced the phase transition of white phosphorus to black phosphorus $(\ce{P_\mathrm{B}})$ under high pressure (Ref.2). The irreversible transformation was observed when white phosphorus was left at $\pu{200 ^\circ C}$ under $\pu{12000 kg cm-2}$ $(\approx \pu{12000 bar})$ pressure for about $\pu{15 min}$. Ref.2 remarks the observation as follows:

On taking the apparatus apart the next morning it was found that the phosphorus had been transformed indeed, but not to the familiar red form, but to another modification that is apparently quite new. The new form is grayish black like graphite, breaks with a similar greasy looking fracture, and will mark on paper. It is stable in the air although slightly deliquescent, is ignited with difficulty by a match, and cannot be exploded by a blow from a hammer, as red phosphorus can. It is a conductor of electricity and a rectifier. The most definite characteristic is its high density, 2.69 against 2.34 for the red variety, and I.9 for the yellow. The experiment was repeated twice, successfully each time. Once it was necessary to go to nearly I3,ooo kgm. to start the transformation. The time necessary to force the reaction to start may also vary considerably. An attempt to similarly transform the commercial powdered red phosphorus into the new variety by subjecting it to 200o and I3,ooo kgm. for over half an hour was without result.

According to this excerpt, it is pretty sure that Bridgman has produced the first batch of $\beta$-$\ce{P_\mathrm{B}}$, which conduct electricity:

This excerpt is also proved that the way of how to make black phosphorus reported in Lumen Learning:

Black phosphorus is obtained by heating white phosphorus under high pressures (about 12,000 standard atmospheres, or 1.2 gigapascals). In appearance, properties, and structure, black phosphorus resembles graphite — it is black and flaky, a conductor of electricity, and has puckered sheets of linked atoms.

Accordingly, it is clear that  $\beta$-$\ce{P_\mathrm{B}}$ is prepared by heating white phosphorus at $\pu{473 K}$ under high pressure (not black phosphorus at $\pu{473 K}$ under high pressure as stated in your textbook). I think it is a genuine mistake.
Moreover, according to literature, now more than five allotropes of phosphorous have been known:

To learn more about black phosphorus, I think Ref.3 would be helpful.

References:

Mehdi Ghambarian, Zahra Azizi, and Mohammed Ghashghaee, “Functionalization and Doping of Black Phosphorus,” In Black Phosphorus: Synthesis, Properties and Applications; Inamuddin, Rajender Boddula, and Abdullah M. Asiri, Editors; Springer Nature Switzerland AG: Cham, Switzerland, 2020, pp. 1-30 (ISBN: 978-3-030-29555-4).
Percy W. Bridgman, “Change of Phase under Pressure. I. The Phase Diagram of Eleven Substances with Especial Reference to The Melting Curve,” The Physical Review 1914, III(3), 153-203 (DOI: https://doi.org/10.1103/PhysRev.3.153).
Rui Gusmão, Zdenek Sofer, and Martin Pumera, “Black Phosphorus Rediscovered: From Bulk Material to Monolayers,” Angew Chem., Int. Ed. Eng. 2017, 56(28), 8052-8072 (DOI: https://doi.org/10.1002/anie.201610512).

This is great information, but doesn't actually address the specific question, which is the source of the "alpha" and "beta" designations.


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Why does sulphur exist in two forms in solid state, monoclinic and rhombic? Why in one favoured over another at a certain temperature and pressure?
Both Rhombic and Monoclinic sulphur consist of $S_8$ molecules packed together. Yet the former is more stable below $114^{\circ}C$ at 1 atm and spontaneously converts to monoclinic sulphur above that temperature.  Does this have something to do with the crystal structure? Gibbs equation tells us that structures with higher entropy are preferred at a higher temperature.
Even at the same temperature we can see that rhombic form is more common at higher pressures. Is pressure decreasing the entropy and favouring one form over another?

Does that mean a monoclinic structure has higher entropy and lower enthalpy, if that's true why is entropy higher? Or is my hypothesis wrong and it's based on something completely different.

"Does this have something to do with the crystal structure?" Yes. ;-)

@Karl, and could you give me a hint please : )(or even the answer if you want to)

Well, have you had a look a the two crystal stuctures of sulphur? (I haven't, that's your homework. ;) )

@Karl I suppose the monoclinic structure will have lesser symmetry, but if this answer is to be believed it's not that straightforward. https://physics.stackexchange.com/questions/43492/does-high-entropy-means-low-symmetry  Now I'm in high school and the only book for physical chem I know about is Atkins, maybe you can suggest some other one : )

@Karl, even if my first hypothesis about the entropy being higher is somewhat correct, I feel that I don't exactly know the reason, and I'm clutching at straws lol

Have a look at these questions: https://chemistry.stackexchange.com/questions/141580/which-allotropes-of-sulfur-exist-naturally ... https://chemistry.stackexchange.com/questions/80823/does-sulfur-have-various-allotropes-at-room-temperature ... https://chemistry.stackexchange.com/questions/72807/are-allotropes-of-sulfur-two-different-phases ... https://chemistry.stackexchange.com/questions/8438/reaction-between-zinc-and-sulfur ... https://chemistry.stackexchange.com/questions/27833/why-does-sulfur-form-s8-instead-of-s3/27877

Alternatively, you can look at the phase diagram of sulfur.


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How to identify 'water' mistakes in protein crystal structures: Are B-factors and occupancy good indicators?
When solving the structure of a protein from diffraction data (x-ray crystallography), it can sometimes be difficult know what should be modelled into small spheres of electron density that are not part of the protein structure, e.g. most likely solvent. Such 'density spheres' are in most cases water molecules, but could also contain a transition metal (e.g. Mg, Fe, Zn etc) or an ion such as SO4, Na, K or Cl depending on the crystallisation conditions. Although it is often obvious for a crystallographer when a large SO4 ion should be modelled instead of a small water molecule, it can be more difficult to evaluate finished structures from the PDB database when conducting a literature review.
Question: How can one quickly identify suspicious water molecules that might have been modelled wrong - or what one should be looking for if one suspects a particular water molecules to be something else. From a PDB file, could one simply look at B-factor and occupancy of the water molecules and identify 'suspicious' molecules that are most likely something larger?
In example:

Would it be correct to be suspicious of a water molecule with a very low B-factor, as this can indicated high coordination and thus most most likely should be a metal or ion of some kind instead of water?
If one suspect that an ion should be modelled in, how does one know if it should be Na, Cl or some other atom if there are many different ion's in the crystallisation condition? (The coordination of metals are not always perfect in proteins, and thus this might always be a speculation or more protein-class/fold dependant).
Could one say something about the water molecules based on occupancy with or within looking at the B-factor? Or are there some other good indicators one should be looking at?

If you use a published .pdb chances of such mistake should be slim, I guess. BTW the question may be better of on Biology of Bioinformatics.SE

(Small molecule X-ray perspective:) Isn't there a validator like checkcif for .pdb which warns if e.g., $\ce{Cl-}$ too close (less than vdW radius) to other atoms? Like this one? Such a check may be more valuable with the difference Fourier map/structure factors (basically observed vs. modeled electron density) if this information retained. Recently, Jmol improved in this perspective (entry, an early demo).

I equally expect shape and volume render the discern of a $\ce{SO^{2-}_4}$ from a spherical ion (despite the often lower resolution in protein crystallography vs. small molecule crystallography) possible, even if disordered if there are suitable programs available for proteins as they are for small molecules (e.g., DSR).

I suppose it's hard to guess from a pdb file only. You'd need to check the data (map).

Thanks for the input! I just came across a couple of structure that I found to be poorly done when looking at the actual density maps - and this had made me a bit paranoid about structure quality when my structural analysis is important for an article. Its time consuming to look through the density maps of each individual structure when reviewing many different ones - and since B-factors and occupancy is available info in the pdb file, the process is more efficient if they could be used as an indicator of when one should also check the actual maps.  But maybe the work just has to be done.

A similar question asked on another SE site: https://mattermodeling.stackexchange.com/questions/11598/how-easy-is-it-to-distinguish-water-from-ions-in-xray-crystallography

How can one quickly identify suspicious water molecules that might have been modelled wrong - or what one should be looking for if one suspects a particular water molecules to be something else.

You can check how many ligands the water has, and what the distances are. Most water molecules can safely be ignored, except for ones in an active site or so. When the protein is in solution, it is covered by water (and other particles making up the solvent) on the entire surface. In a crystal, you are missing some of these water because of crystal contacts to neighboring proteins, and because many water molecules are not sufficiently fixed in place ("ordered") to give significant electron density peaks.

Such 'density spheres' are in most cases water molecules, but could also contain a transition metal (e.g. Mg, Fe, Zn etc) or an ion such as SO4, Na, K or Cl depending on the crystallisation conditions.

As alluded to in the comments, most of your examples have much higher electron count than water, so they are unlikely to be confused with water. Sodium, magnesium, flouride and ammonium ions are closest in electron count, and you would have to use the nature and location of the neighboring atoms to make a call. See e.g. this paper.

[OP in comments] Thanks for the input! I just came across a couple of structure that I found to be poorly done when looking at the actual density maps - and this had made me a bit paranoid about structure quality when my structural analysis is important for an article. Its time consuming to look through the density maps of each individual structure when reviewing many different ones - and since B-factors and occupancy is available info in the pdb file, the process is more efficient if they could be used as an indicator of when one should also check the actual maps. But maybe the work just has to be done.

You should carefully look at those regions in the crystal structure (and density) that are important for your own work. The original paper might not have had your angle of interest, so there might be areas in the structure that were modelled with moderate attention (as opposed to the detailed scrutiny an active site would get when it comes to identifying bound ions that look like water). You should also look at which substances were used in crystallization. If there was a high concentration of - say - chloride, it is likely that you have some chloride ions bind somewhere on the protein surface.

Would it be correct to be suspicious of a water molecule with a very low B-factor, as this can indicated high coordination and thus most most likely should be a metal or ion of some kind instead of water?

Yes. It also could indicate an atom with more than 8 electrons (e.g. potassium).

If one suspect that an ion should be modelled in, how does one know if it should be Na, Cl or some other atom if there are many different ion's in the crystallisation condition? (The coordination of metals are not always perfect in proteins, and thus this might always be a speculation or more protein-class/fold dependant).

You look at the coordination sphere and the electrostatic potential. For cations you could have ligands like glutamate or aspartate, while you don't expect those for anions.

Could one say something about the water molecules based on occupancy with or within looking at the B-factor? Or are there some other good indicators one should be looking at?

Occupancies of water molecules below 1.0 are a red flag. You need very well-ordered waters in a high resolution structure to have enough data to support modelling of partially occupied sites. As for B-factors, you can order you water molecules by B-factor and number of hydrogen bonding partners. The lower the B-factor, the more consistent the B-factor of water and neighbors, and the higher the coordination number of the water, the more likely you would find a water there in reality (i.e. in solution in the cell).

[from the comments] You could check structures from PDB-REDO. They automatically remove waters that are not in the electron density map, and I think in some cases the removed waters are replaced with missing side chains.

If you go through the trouble and compare the deposited structure with the PDB-REDO database, I would not trust the PDB-REDO more than the deposited structure but rather treat all parts of the model that shows differences with caution.


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Determining R-S configuration with lowest priority groups in plane of paper
I was studying R-S nomenclatures and determining configurations of molecules when I came across a certain problem, which had the 4th priority group on the plan of paper (not on a dash or wedge)
According to this: https://www.masterorganicchemistry.com/2016/10/20/introduction-to-assigning-r-and-s-the-cahn-ingold-prelog-rules/#five
All I need to do to determine the R-S config in these cases is to switch the group in the 4th place with the one in the back, thus "flipping its configuration", then determine the R-S for that molecule and the actual result is the opposite. I know why this works and it has worked for me in the previous problems that I have attempted. However, in this question:

For the left compound (looking at the carbon on the bottom), clearly, $\ce{Br}$ has priority $1$, $\ce{Cl}$ is $2$, $\ce{F}$ is $3$ and the other carbon attached to it is clearly $4$ in priority which is in the plane of paper. Swapping the groups in priority $4$ with that in the back i.e., $\ce{Cl}$, and going from $1,2,3$, I get S as the absolute configuration of the flipped molecule, so the actual configuration should be R. But the answer does say that S is indeed the configuration of both the carbons in the 1st compound and R is the configuration for both carbons on the right molecule.
Why does the swapping of the groups in back and lowest priority not work here?

You seem to know how to prioritize groups. Follow the instructions here and believe in yourself.  https://chemistry.stackexchange.com/questions/151831/r-or-s-configuration/151842#151842

"the 4th group on wedge, dash, or on the plane of paper" This is not the right question to ask. You don't need to, and should not, memorise a series of rules for each individual case. Instead, you rotate the molecule until it matches the description given in the rules (lowest priority group facing away). To do this, you either build a model which you can physically rotate, or learn to draw out the rotation on paper, or learn to visualise it in your head.

Techie5879: If you have followed my suggestion in the Comment above, you probably have answered this question yourself. Now that you no longer "flip" structures, you should have the R/S assignments for A and B (vide infra). Because each carbon in A and B bears F, Cl and Br and both A and B have R and S configurations, they are identical meso compounds. I was not familiar with the term "homomer", so I looked it up.

Homomers : Homomers are the identical representations of the same compound. It means that when the same compound is represented in different ways, then all the representations are known as homomers of each other. These representations are superimposable on each other. (Italics added.)

If you pickup A (mentally) and superimpose the S-carbons onto B, pairing halogens (F with F, etc.), then the R-carbons will superimpose halogens but they will not match halogen for halogen. In this context they are not homomers. That is to say, they are different conformations of the same compound. Because A and B are identical compounds, they cannot be enantiomers or diastereomers.
I would appreciate any constructive feedback as to whether or not my interpretation of "homomer" is correct or whether or not the term serves a meaningful purpose. Exclude any applications to polymers or biochemistry.
ADDENDUM: Per the OP's request, the Fischer projections of A, aka B, are provided here.

Yes your interpretation of homomers is correct. Also, this is the same answer I am getting, but the answer in the book says they are enantiomers and apparently drawing the Fischer projection gives opposite configurations at the carbons? Could you also update your answer with what the Fischer projection of the two molecules would be?

The person who wrote the question was not necessarily the person who provided the answers. They may have simply placed a mirror plane between A and B and said, "Oh look! they are images. Therefore, they are enantiomers." Nonsense, they are identical meso compounds. I could draw Fischer projections but they are more suitable for carbon chains greater than two.

Change of mind. The Fischer projections of the hexahalide are linked in the ADDENDUM.

Okay, this matches with my conclusion then. I'll wait a bit for another answer, if I see any change of conclusion as to whether there are enantiomers or homomers, and if not I'll accept this answer. Hope thats fair.

Fair enough. BTW, I should have said above, "Oh look! They are mirror images".

One way to assess the configuration without having yo swap or ritate functional groups us to use the concept of even and odd permutations.
Say you have ranked the substituents on a stereocenter and a group other than Group 4 is oriented to the bottom. Proceed as follows:

List the rankings of the three groups directed to you in clockwise order.

Attach the bottom group at the end of the list.

Count the pairwise interchages that you would need to get your permutation to the "standard" 1234. An even count, called an even permutation, corresponds to R and an odd permutarion corresponds to S.

Let us look at the left carbon in molecule A. Bromine ranks 1, chlorine 2, fluorine 3 and the other half of the molecule, attached through carbon, ranks 4. So 2 is directed to the bottom and the other groups in clockwise order are 134. Thus you render your permutation as 1342.
To get from 1342 to 1234 you then need to interchage the 2 with the 4 and then with the 3; if you try to interchage 2 directly with 3 first you have 3 and 4 in the wrong order. Thus you need two interchanges, an even number, and so the left carbon is configured R.
Can you work out the other centers with this method?


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Reaction between sodium hydroxide and aluminium oxide
I am a 10th grade student.
Is it possible to know the reason behind the formation of sodium aluminate when sodium hydroxide and aluminum oxide reacts. Is there an intuitive reason for this other than mere observation.
I would like to know an easy explanation as I am a 10th grade student

Does this answer your question? Why is Aluminium oxide amphoteric?

The strange property of $\ce{Al2O3}$ is due to the position of aluminium in the periodic table. As it is not obvious, I have to start a rather long explanation. Sorry !
Let's start with the definition of acids used in the beginning of the $20$th century. At this time an acid was a substance containing a Hydrogen atom and able to produce $\ce{H^+}$ (or $\ce{H3O^+}$) in water. Their formula were beginning with $\ce{H}$, and hydroxide had formula ending in $\ce{H}$. The main property of the acids was that they could react with Hydroxides like NaOH, to produce salts, according to the general equation :
Acid + Hydroxide -> Salt + Water.
With this definition, an acid is made of Hydrogen, maybe some Oxygen and another element which is electronegative. All non-metals are in this group. But there are a couple of metallic elements which are nearly at the border of the limit metals - non-metals in the periodic table. These elements deserve to be discussed.
Aluminium is in this border group. It makes an hydroxide $\ce{Al(OH)3}$ which usually works as a hydroxide and sometimes as an acid. Usually it works as a hydroxide in reaction with strong acids, like $$\ce{Al(OH)3 + 3HCl ->  AlCl3 + 3 H2O}$$ $$\ce{2 Al(OH)3 + 3 H2SO4 -> Al2(SO4)3 + 6 H2O}$$
But $\ce{Al(OH)3}$ is a rather weak hydroxide. It does produce extremely few $\ce{OH-}$ ions in water. And such weak hydroxides can sometimes behave as weak acids, if they are in contact with a strong acid. Matter of fact, $\ce{Al(OH)3}$ reacts with NaOH, which is a strong hydroxide, as if it was a weak acid $\ce{H3AlO3}$. The equation can be : $$\ce{H3AlO3 + 3 NaOH -> Na3AlO3 + 3 H2O }$$ or, depending on the concentrations : $$\ce{H3AlO3 + NaOH -> NaAlO2 + 2 H2O}$$ Today we know that the aluminate ion is either $\ce{[Al(OH)4]^{-}}$ or $\ce{[Al(OH)6]^{3-}}$. But it does not matter for the present development.
Now you should understand that the strange reaction of aluminium hydroxide is due to the position of aluminium at the border of the separation metals - non-metals in the periodic table.
Your question was not related to aluminium hydroxide, but to aluminium oxide. All that has been said about aluminium hydroxide can also be said about oxide, because in water, the following equilibrium can occur : $$\ce{2 Al(OH)3 <=> Al2O3 + 3 H2O}$$ so that $\ce{Al2O3}$ may react with acids and hydroxides exactly like $\ce{Al(OH)3}$

Strange? Arguably, most metals are amphoteric to some extent.

Elements from columns $1$ and $2$ are not amphoteric

That doesn't contradict what I said. Then again, Be is amphoteric.

Be is at the border, as its neighbor the boron B is already a non-metal.

@maurice boron is a metalloid, no?

Elements are metals or non-metals, For me, Boron is not a metal. As a consequence, it is a non-metal. In my dictionary, metalloid is an old name for non-metal


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Doubts regarding Mole concept and Avogadros number
So I am really confused about simple topics like mole and Avogadro
Here is my oversimplified view
So mass of 1 atom of elements in grams was not in whole numbers and it was difficult using em for calculations for ex (avg) atomic mass of carbon is 2.6566962 × 10^-23 g g, weight of hydrogen atom is 1.6735575 × 10^-24 g, mass of carbon is 1.9944235 × 10^-23 g. (these were measured using mass spectrometer
So one way was to use a different unit instead of gram. Elements could be measured relative to each other. So mass of one atom on an element could be assigned 1 and others could have mass relative to that
So scientists decided to assign Carbon which is an element with 6p 6n and 6e a value of 12u
since electron has negligible weight as compared to proton and neutron which are almost equal in weight, it can be said
1u~ mass of 1 proton + 1 neutron = mass of 2 neutron or proton
Note:i know that mass of an element cannot be said as sum of mass of its subatomic particles because when subatomic particles come together to form the element some mass is released in form of energy but we will ignore that for simplicity
1 amu = 1.66054e-24 g
q)i have a question y is carbon used to define amu then why not say that 1 amu is equal to 1/16 the mass of oxygen or 1/20 the mass of neon, is it because carbon is found quite commonly
q)is the current value of amu the most ideal, would it be better if the value of amu was instead double of what it is? Basically what would be the problem if any if hydrogen was given the value 2u and others were relative to it.
Now coming to Avogadro's number/constant/mole
So like there's dozen which is defined as group of 12, usually used to measure fruits/bananas
why 12? because Twelve is convenient because it has the most divisors of any number under its double, a property only true of 1, 2, 6, 12, 60, 360, and 2520.
So Scientists needed a term to describe a group of atoms/molecules/subatomic particles
since it almost impossible to measure mass of 1 atom of different elements u could instead measure a large constant number of atoms of elements
This was called mole now u had to decide the value of mole it couldnt be 10 100 or 100000 as they were too small
So they decided that the number of atoms of carbon(it goes without saying 12C) in 12g of Carbon will be used as a standard and they named the number obtained Avogadro's Number
So in simpler terms mole is
Lets says u have 1 oxygen atom (which has 8 electrons 8 protons and 8 neutrons) mass of 16u = 16 * 1.66054e-24 g. If u take 6.022 * 1023 of sch oxygen atoms then then the mass will be equal to 16g
how does this makes sense/ what is the the correct explanation
what is the need of mole if the ratio of masses of atoms of two elements remains same irrespective of the number of atoms
what does gas laws have to do with avogadros constant
what is the difference between avogadros number and constant and why doesnt one of em have a unit per mole
Some of the questions might be dumb but i will be really grateful if someone could help me out

I think you are asking many questions under a the same post. I would recommend adding them as different questions. And you might also find some of your questions' answers if you browse this site thoroughly.

Please edit for grammar, punctuation, and style per the guidelines for this site.

I will try to answer to your first question, related to the choice of a reference atom.
The choice of the reference atom for measuring atomic masses has been a long and tedious discussion that had lasted during the whole $20$th century. There has been plenty of congresses devoted to this choice. In the beginning the reference was Hydrogen, which was given the value $1.00000$. After some time, there was doubt about the proportion of $\ce{^2H}$ or deuterium in hydrogen, which was changing from spot to spot on the planet, so that the H atom had not the same mass everywhere.. So after a couple of congresses, the scientists decided to replace hydrogen by $1/16$ of the mass of oxygen as a reference for atomic masses. Later on, there was also difficulty about the proportion of the isotope $\ce{^{18}O}$. So, after many congresses, scientists decided to choose the isotope $\ce{^{12}C}$ so that the unit of atomic mass is $1/12$ times the mass of this $\ce{^{12}C}$ isotope. Why to choose this one (and not another isotope) was a hard discussion, where solving technical difficulties had an important part. I cannot enter this discussion, which would take hours to develop, because there was ass many possibilities as the number of atoms available. So please take it for granted. It is not important to know the series of reasons that had been used to convince the specialists of mass spectrometry about the choice of $\ce{^{12}C}$


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Why does steam store large amounts of energy?
We all know that steam can be used to perform mechanical work and steam has a high capacity for energy storage. But why does steam have such high energy capacity?
I tried searching for answers on the net but couldn't find any. So I came up with my own theory:

Steam is gaseous form of water and water is made up of H2O molecules.
Since the molecule consists of hydrogen, there will be a presence of hydrogen bonds between every two hyrogens of every two neighbouring molecules. Steam is formed when these bonds are broken and each H2O molecule moves further apart from their neighbouring molecules. But hydrogen bonds are strong. Thus in order to break these bonds, large amounts of energy must be supplied to liquid water inorder to convert it to steam. And this large amounts energy is thus stored in steam. Thus making steam a large energy capacitor.

Is my theory right?

Yes. Your theory makes sense.Why not ?

What do you mean by “energy capacity”? Most liquids can be evaporated in order to perform mechanical work via the volume expansion of the phase change

@Andrew energy storage capacity compared to other liquids that donot possess hydrogen bonds?

Consider 2 mechanical analogies: 1/ A ball laying on the top of a heap above the flat. 2/ A ball laying on the flat next to a hole. Which scenario reminds you the relation of energies of water and vapor ? Does vapor store big amount of energy like 1/, or is it rather a big amount of energy is released, forming a liquid, like in 2/ ? // And yes, the high value of water heat of evaporation is caused by energy needed to break hydrogen bonds.

So do you mean the enthalpy of vaporization? or maybe the specific heat capacity? Neither of those are directly related to the mechanical work that can be done by expansion, although they are relevant to the process.


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How are Silanol groups functionalized unto mesoporous silica structures?
As described in this research paper:
Immobilization of Jacobsen type catalysts on modified silica, Jairo Cubillos , Edwing Grajales, Santiago Vásquez, Consuelo Montes de Correa, Rev. Fac. Ing. Univ. Antioquia N.° 57 pp. 38-48. Enero, 2011 (link)
The Jacobsen type catalysts are attached unto the silica through a reaction between Silanol that is already part of the silica and 3-APTES. However, there is no mention on how these Silanol molecules were already present in the silica. Any idea?

As mentioned in the answer, silanol groups are are normal terminal groups on silica, but you mention "modified" silica... So are you asking how was it modified? It's quite possible it was processed, for example to get good surface density of the groups.

The bulk of silica is indeed silicon dioxide, but at surface there is a discontinuity. The dangling O, exist as OHs, as silanols. Therefore silica surface itself is full of silanols, which are used for functionalizing silica. Native silica looks like the following with various types of silanols. APTES is a very common chemistry in the silica modification chemistry.
Note silicon dioxide, like quartz may not be easily to functionalized because the surface area is very low and therefore the number of silanols is also low.

Ref:High-Performance Precoated Stationary Phases, Colin F. Poole, in Instrumental Thin-Layer Chromatography, 2015


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Using cooling to temporarily lock in deformation in plastics?
I need to shrink PEEK or solid polyimide tube, $\pu{1mm}$ wall thickness and $\pu{15mm}$ OD over another fragile component. Is it possible to stretch the tube over a mandrel, lock in the deformation using solid $\ce{CO2}$, slip over the component, then shrink by allowing temperature to rise to room temperature? Could also use acetone plus dry ice or liquid nitrogen if necessary.

Seems worth trying... see https://www.nature.com/articles/srep44936


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Activation of Grignard reagent
For activating Grignard reagents, iodine or 1,2-dibromoethane are usually added, and these lead to the formation of $\ce{MgI2}$ or $\ce{MgBr2}$ respectively . Why can these $\ce{MgI2}$ and $\ce{MgBr2}$ activate Grignard reagents? If possible, please show me the literature.

These ingredients are not "activating the gringard reagent" they are activating the magnesium. Some compounds we want to make Grignards from are not very reactive and it helps to activate the magnesium surface so those reactions happen faster. Small, reactive molecules are good at this. Though sometimes even this isn't enough.

Have you checked Wikipedia?

Typically the reaction to form Grignard reagents involves the use of magnesium ribbon. All magnesium is coated with a passivating layer of magnesium oxide, which inhibits reactions with the organic halide. Many methods have been developed to weaken this passivating layer, thereby exposing highly reactive magnesium to the organic halide. Mechanical methods include crushing of the Mg pieces in situ, rapid stirring, and sonication. Iodine, methyl iodide, and 1,2-dibromoethane are common activating agents. The use of 1,2-dibromoethane is advantageous as its action can be monitored by the observation of bubbles of ethylene. Furthermore, the side-products are innocuous:
$$\ce{Mg + BrC2H4Br → C2H4 ^ + MgBr2}$$
The amount of Mg consumed by these activating agents is usually insignificant. A small amount of mercuric chloride will amalgamate the surface of the metal, enhancing its reactivity. Addition of preformed Grignard reagent is often used as the initiator.
Specially activated magnesium, such as Rieke magnesium, circumvents this problem. The oxide layer can also be broken up using ultrasound, using a stirring rod to scratch the oxidized layer off, or by adding a few drops of iodine or 1,2-Diiodoethane. Another option is to use sublimed magnesium or magnesium anthracene.

With the formation of ethylene, the oxide layer is somewhat etched exposing the underlying metal which can react with the bromoderivative. See this previous chem.SE question.
For literature, see:
Reactivity of Magnesium Surfaces during the Formation of Grignard Reagents
Cynthia E. Teerlinck and Walter J. Bowyer, The Journal of Organic Chemistry 1996 61 (3), 1059-1064, DOI: 10.1021/jo951626o


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Acetamide and Nitrous acid Reaction Mechanism
My book says that when acetamide is treated with nitrous acid, ethanoic acid, nitrogen and water are formed but it provides no mechanism. I googled but could not find any mechanism. My guess was that the reaction might start through the protonation of the nitrogen, but after that I got lost. How does this reaction proceed?

It's a generic reactions of amino groups with nitroso cations

Amides can be hydrolysed in acidic and basic medium. Nitrous acid facilitates acidic hydrolysis of amides. The reaction can be written as:
$$\ce{CH3CONH2 + HNO2 -> CH3COOH + N2 + H2O}$$
The main reactive group of the reaction is nitrosonium ion generated from the in-situ reaction of sodium nitrite and hydrochloric acid at 0 °C.

This nitrosonium ion then attacks the lone pair at nitrogen atoms and forming an intermediate, N-nitrosoamine which undergo proton shift by a sequence analogous to interconversion of a ketone to an enol to form a diazoic acid.

This diazoic acid decompose in acidic medium to diazonium ions:

Diazonium salts can be regarded as combinations of carbocations  $\ce{R+}$  with  $\ce{N2}$  and, because of the considerable stability of nitrogen in the form of $\ce{N2}$, it is expected that diazonium salts to decompose readily with evolution of nitrogen and formation of carbocations.

Since our starting reactant was acetamide, the carbocation would be acetyl cation,  $\ce{CH3CO+}$ which in fact will react with the water molecules to form  acetic acid. Some of the proton generated will drive the reaction forward while some of them will react with hydroxide ion to be the source of water molecules.
$$\ce{CH3CO+ + H2O -> CH3COOH + H+}$$
Secondary amides give  N -nitroso compounds with nitrous acid, whereas tertiary amides do not react.

For reference, I used following chemlibretext articles:

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/23%3A_Organonitrogen_Compounds_I-_Amines/23.10%3A_Amines_with_Nitrous_Acid
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/24%3A_Organonitrogen_Compounds_II-_Amides_Nitriles__Nitro_Compounds/24.04%3A_Hydrolysis_of_Amides

In many reactions (like the basic hydrolysis of amides you linked above), the Oxygen atom in the carbonyl group takes away an electron in one step, and then gives it back in the next step. Why doesn't the nitrogen atom take away the electron directly?

@Tahsin I have discussed acidic hydrolysis not basic hydrolysis. Which step are you referring to? I can't understand.

I'm talking about the alkaline hydrolysis of amide in your second reference link. It's just below the acidic hydrolysis. I understand that you were talking about acidic hydrolysis, I understood that part.


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Identify E/Z Configuration

[(2E)-4-ethylhept-2-en-3-yl]cyclopropane why it is named E
Where according to cahn ingold prelog rules At left side Methyl is prior
and on the right side the butane should be prior than cyclopropane

One side of the double bond is decorated by the methyl group and the hydrogen atom.  The other side is decorated by cyclopropyl and the the branched alkyl chain.  For either side you apply the CIP rules, and then, by the book, the double bond either is (E)-, or (Z)-configurated.  If you are unsure how to trade with double bonds/cyclic substituents, see e.g. the answers here on chemistry.se) and then post your solution as self-answer.

Welcome to ChemSE, M. Mostafa. While @Buttonwood has given you ample direction to solve the configuration of alkene 1, visual inspection reveals a "butane" chain at C1 (arbitrary numbering) versus a cyclopropane ring. Accordingly, the temptation is to assign a (Z)-configuration to the alkene because butane (C4)>cyclopropane (C3) and CH3>H. But you and others would fall into a trap. According to the CIP Rules the cyclopropane ring is deconstructed into an open chain species as illustrated in the digraphs. Starting from C6 (red dot, non-duplicate atom) in alkene 1, two paths [6$\rightarrow$7$\rightarrow$10$\rightarrow$(6) and  6$\rightarrow$8$\rightarrow$9$\rightarrow$(6)] are traced around the cyclopropane ring terminating in duplicate atoms [red dots, (6)]. Duplicate atoms have the same atomic number as non-duplicate atoms but the former are attached to three atoms of atomic number zero.
Sphere 3, which contains C5 and C12 in one substituent and C9 and C10 in the other, allows for a one-to-one comparison. C5 and C9 are a tie because {C,H,H}={(C),H,H}. But C10>C12  {(C),H,H}>{H,H,H}. Therefore, the cyclopropane ring outranks the "butane", or hexyl chain. Structure 1 is an (E)-alkene.


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How much solvent do you add for a 1:20 dilution, and why is it called 1 to 20?

A 1:20 dilution implies that you take 1 part of stock solution and add 19 parts of water to get a total volume of diluted solution equal to 20 times that of the stock solution

So is 1:20 the way it is because the ratio is solute:solvent/diluent, right? So the solvent/diluent is just n-1 of the solvent?

I would make a final solution with conc 1/20 of the initial one. See Poutnik point (2).

Dilution ratio 1 : N has 2 different, conflicting meanings:

(1) The ratio of the 2 mixed volumes ( typical for 1 : 1 to 1 : 3, for everyday/household chemistry )
(2) The ratio of the final and initial concentrations (typical for 1 : 5 and more, for using calibrated volumes and exact concentrations ).

Consult the context of your question, but the quote obviously considers the latter case. 1 : 20 means the solute concentration is 20 times lower than before dilution ( the second row below).
1 : 20(1) means 1 : 21(2) ( approximately(*) )
1 : 20(2) means 1 : 19(1) ( approximately )
(*) - There is one trap: It is better to say you take 1 volume of the stock solution and while mixing, you fill up by solvent the final volume N. As volumes are not generally additive. By other words, by mixing $\pu{10 mL}$ of solution and $\pu{190 mL}$ of solvent, you may not get $\pu{200 mL}$.

I read more articles and I (think) finally get it. The ratio is simply just the solute ratio to the solvent ratio.

Be sure about that ......


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Intermolecular forces between polyatomic ions and water
I was wondering what kind of intermoleuclar attraction exists between nitrate ion $\ce{NO3-}$ and $\ce{H2O}$?
Is it ion-dipole attraction or is it hydrogen bonding? Since there is nitrogen and oxygen, are both molecules highly electronegtive? Does each oxygen on the $\ce{NO3-}$ attract one water each?

Well, it's pretty obvious there's hydrogen bonding. Average number number of h-bonds per oxygen isn't constant.

It's a good question. I would start by drawing resonance forms. Then again bonding with the solvent probably stabilizes individual forms (causes resonance to break down)?

All mentioned forms take part. Ion-dipole interaction on longer distance, bond dipole-dipole interactions at shorter distance and finally hydrogen bonding.

Thank you for the responses. I had a deeper look into the hydration of nitrate ions and some associated papers. The oxygen with its negative charge is the hydrogen bond acceptor.
Following on Poutnik's answer, how does ion-dipole interaction and hydrogen bonding exist at the same time?  From my understanding ion-dipole is the electrostatic interaction between and ion and a charged dipole; In this case,  nitrate ion and the partially positive hydrogen in water.


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What does the prefix tetrahydro mean?
I'm wondering what is the meaning of tetrahydro when used as a prefix.
For example, THC (tetrahydrocannabinol) and also THF folic acid (tetrahydro folate)  or  THF (tetrahydrofuran).
I couldn't recognize any detail in their structure that implies a "TETRA HYDRO" character and could not determine in which position in the molecule those "hydro"'s should appear.
Does "hydro" here means water? Or hydrogen? Or maybe a complex with four molecules of water?
Sorry for the lack of knowledge,  I couldn't find any information on the web about this type of nomenclature....

Hydrogen. It means the hydrogenated version of an aromatic with 4 hydrogens added. Generally used where the hydrogenated version does not have a separate name though there is some crossover e.g. tetrahydronaphthalene/tetralin. To be completely systematic the name should specify which bonds are saturated. e.g. tetralin/1,2,3,4-tetrahydronaphthalene

It is partly because many compounds were not originally named in an entirely systematic way. A very simple example is the popular solvent tetrahydrofuran. Named as a tetra-hydrogenated derivative of furan (a five membered ring, C4H4O). The simple names make sense if you know the history and survive because the full systematic names are often very unwieldy.

Specifically in this case: the 'proper' name for tetrahydrofuran is 'oxolane', which is not really unwieldy per se, but definitely obscure. Other systematic names can be pretty clunky.

The chemical prefix "hydro" is a confusing one, because the Greek root "hydro" in English words usually refers to water, as in "hydrophilic" (hydro=water; philic=loving) for example. Another example relevant to this question is "hydrogen" (hydro=water and gen=making).
In chemical names, however, "hydro" is a shortened form of "hydrogen" that retains the meaning "hydrogen" rather than its root meaning of "water". Thus, it means that a hydrogen atom has been added to the molecule whose name follows the prefix. Greek root multipliers like "tetra" are used if more than one hydrogen atom has been added, so "tetrahydroX" means that four atoms of hydrogen have been added to X.
As a simple example, tetrahydrofuran (top) and its root compound furan (bottom) are shown below (images from wikicommons). The white balls represent hydrogen atoms, black are carbon and red is oxygen.

According to the Gold Book and IUPAC 1988 Recommendations (Ref.1):

Hydron: The general name for the cation $\ce{H+}$; the species $\ce{H-}$ is the hydride anion and $\ce{H}$ is the hydro group. These are general names to be used without regard to the nuclear mass of the hydrogen entity, either for hydrogen in its natural abundance or where it is not desired to distinguish between the isotopes.

Thus, the following names comprise general names, to be used without regard to the nuclear mass of the hydrogen entity, either for hydrogen in its natural abundance or where it is not desired to distinguish between the isotopes:

The atom, $\ce{H}$: hydrogen
The cation, $\ce{H+}$: hydron
The anion, $\ce{H-}$: hydride
The group, $\ce{-H}$: hydro

The specific names pertaining to specific isotope of the forth category (hydro) are protio for $\ce{^1H}$, deuterio for $\ce{^2H}$, and tritio for $\ce{^3H}$ (Ref.1).
In IUPAC nomenclature, R-3.1.2 states that:

If the name of the parent hydride implies the presence of the maximum number of noncumulative double bonds (see R-2.4.1), other states of hydrogenation can usually be indicated by use of the prefix "hydro-" together with an appropriate numerical prefix signifying the addition of hydrogen atoms. This operation is regarded as the reduction of double bonds; thus, hydrogen atoms can only be added in pairs (by use of "dihydro-", "tetrahydro-", etc.). "Indicated hydrogen" (see R-1.3) is included if required by the parent hydride.

R-1.3 Indicated Hydrogen: Under certain circumstances it is necessary to indicate in the name of a ring, or ring system, containing the maximum number of noncumulative double bonds, one or more positions where no multiple bond is attached. This is done by specifying the presence of an "extra" hydrogen atom at such positions by citation of the appropriate numerical locant followed by an italicized capital $H$.

Some examples for these rules are given below:

References:

J. F. Bunnett and R. A. Y. Jones, "Names for hydrogen atoms, ions, and groups, and for reactions involving them (Recommendations 1988)," Pure & Appl. Chem. 1988, 60(7), 1115-1116 (DOI: https://doi.org/10.1351/pac198860071115).


text:
Do different types of MOs have different occupancies?
There was a question in my textbook regarding the number of electrons in a molecular orbital.

The maximum number of electrons in a sigma, pi, and delta molecular orbital are respectively?

Now, I believe that any molecular orbital can be occupied by a maximum of two electrons, and so the answer might be 2 electrons for sigma, 2 electrons for pi, and also 2 for delta orbital.
Is my answer correct, or are there other concepts that need to be applied? Is it true that any molecular orbital can occupy a maximum of only 2 electrons? Or is it possible to have more than 2 electrons for delta molecular orbital?

Without knowing the question verbatim and its source (citation) it is basically impossible to answer this. In molecular orbital theory, essentially a restricted Hartree Fock approach, the answer is two for all. I am going to close this question so that an appropriate [edit] may be made.

I you don't know what was his point then how we are supposed to? Delta orbitals aren't special.

I got the point, but edited it just because it said the question needs clarity. I edited it so that others can find it easier to understand the question verbatim, which was a bit confusing earlier. Thanks .

If this friend has some scientific point to make, then it should be given. But an assertion that is not backed up by any facts is not helpful in explaining the question.

Firstly, note that the labels $\sigma$, $\pi$, and $\delta$ aren't universally applicable to MOs; it depends on the molecular geometry. These labels are mostly useful for linear molecules. Non-linear molecules often have MOs that are labelled differently. Methane is a decent example. Other examples include water and ammonia.

Restricting ourselves to linear molecules, where the labels do apply: some of the confusion likely arises because of the "simplest" case of a homonuclear diatomic molecule ($\ce{X2}$), where $\sigma$ MOs are always "alone" and $\pi$ MOs always come in pairs.

So it's tempting to think that $\pi$ MOs can fit four electrons. This is not correct, though. The fact is that there are two $\pi$ MOs that happen to have the same energy, and each can fit two electrons.
Every MO can contain up to two electrons, period.
(Well, at least at this level of knowledge. If you are doing electronic structure theory, you may find that this statement is oversimplified; but you can work that out for yourself.)
Pictures are taken from Housecroft & Sharpe's Inorganic Chemistry, 4th ed.


text:
Why is the sum of the square of the orbital coefficients 1
I'm reading up on molecular orbital theory and LCAO in Ian Fleming's "Molecular Orbitals and Organic Chemical Reactions" and I don't understand the logic here:
"When there are electrons in the
orbital, the squares of the c-values are a measure of the electron population in the neighbourhood of the atom
in question. Thus in each orbital the sum of the squares of all the c-values must equal one, since only one
electron in each spin state can be in the orbital."
I get the first part but i don't see how c values are related to spin and wouldn't total spin in an orbital equal 0? Any help would be appreciated but, since I haven't studied quantum mechanics yet, a simple as possible explanation would be best.

The short answer is that $\sum |c|^2$ is basically equal to "the number of electrons in the orbital". You might think that this should be 2, not 1. But note that each molecular orbital actually comprises two different "spin orbitals". Each "spin orbital" contains one electron of a particular spin - so there is a spin-up orbital which can hold one spin-up electron, and a spin-down orbital which can hold a spin-down electron.
The value of $\sum |c|^2$ is calibrated with respect to these "spin orbitals", so the value has to be 1.
Obviously, this raises far more questions: What is a spin orbital exactly? What is a molecular orbital exactly? Why does one MO have two spin orbitals? (Well, you might have some intuition as to that.) Why does $\sum |c|^2$ represent electron density? Why is the number taken with respect to the spin orbitals and not the molecular orbitals? I could provide answers for these, but unfortunately, no further detail is possible without going into quantum mechanics. If you want to truly understand this, there is no way around it. Of course, it doesn't mean you have to learn it right now. The only point is that it will only make more sense if and when you do it.
When you do, you will find that these topics are covered in standard QM texts.

I was always confused when the definition of an MO was a “one-electron wave function” given that an MO can hold 2 electrons. This makes much more sense now.


text:
How to compare the stability of 3,4-Dimethylcyclopentene and Cyclopentane, 1-methyl-2-methylene
Two alkenes are given as follows:
,

How do I compare their stability?
Denoting the first compound as (A) and the second as (B).
Both the alkenes have same number of $\alpha$ hydrogens, which is $3$. Hence in terms of stability by hyperconjugation they should be similar. However the structure (A) contains $2$ carbons in $sp^2$ hybridisation "inside" the ring, whereas the compound (B) contains only one such ring carbon. Now since the typical bond angle of $sp^2$ is given as $120^\circ$, it is was intuitive for me to think that structure (A) would suffer a higher angle strain since the angles of a pentagon are $108^\circ$, however it seems incorrect since the solution of my paper did not consider this. Please help me to figure out what is that I am doing wrong and what the correct order should be, or whether they have equal stabilities.

Heat of Combustion?

The question stated to tell which alkenes are more stable than (A) and (B) was one of the options. They mostly likely did mean to ask the "Stability" of the double bonds, so heat of hydrogenation I think.

The problem does not state the stereochemistry in A.  Hydrogenation of B will likely give a mixture. The stereochemistry of A should be defined. Heat of combustion is the best bet.

@user55119 How should I be comparing the heat of combustion? The number of carbons is the same so that factor is ruled out.

Heats of formation are computed from heats of combustion. Unfortunately, these compounds are not listed at the NIST site. https://webbook.nist.gov/chemistry/. Sometimes heats of formation can be estimated using desmotic reactions. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/homodesmotic/homodesmotic.html


text:
Electrophilic Substitution of Substituted Benzene
I know that phenol doesn't undergo the Friedel-Crafts Alkylation and Acylation reactions, because the reagent $\ce{AlCl3}$ forms a complex with phenol, as the oxygen atom donates it's lone pair to the Lewis acid.
Now, if I use $\ce{OR}$ group substituted instead of $\ce{OH}$, will the same comple, formation occur, or will the reaction take place?

Still the same problem remains unresolved you removed O's acidic H but increased O's donating capacity. So the answer is no.

If R=CH3, anisole reacts with succinic anhydride in the presence of AlCl3 in a FC reaction. There are numerous methods for demethylating anisole and its derivatives.

Friedel-Crafts reactions of anisole is known and the relevant acylation is performed in most academic institutions during undergraduate sophomore organic chemistry laboratories. For instance, Aluminum Chloride Catalyzed Friedel-Crafts reaction of anisole with Epoxide has been published (Ref.1). The authors have performed the reaction in various solvents involving nitromethane, which is a stronger base and would have strong interaction with the Lewis acid catalysts comparable to that with the substrate:

The reaction of anisole with propylene oxide by aluminum chloride has been studied in various kinds of solvents. While the normal Friedel-Crafts reaction yielded 2-(methoxyphenyl)-1-propanols, 1,1-bis(methoxyphenyl)-propanes were also formed in this reaction; these were not obtained by the subsequent reaction of mono(methoxyphenyl) products with another anisole. The yields of 1,1-bis(methoxyphenyl)propanes increased by increasing the basicity of solvents. The mechanism of the formation of 1,1-bis(methoxyphenyl)propanes can be explained by the isomerization of epoxide to aldehyde and the subsequent condensation with anisole. Various kinds of epoxides also gave 1,1-bis(methoxyphenyl)alkanes in nitromethane.

Further, in earlier studies, the Friedel–Crafts alkylation of anisole with 2-phenylethyl chloride and 2-phenylethanol has resulted in a product, p-methoxydibenzyl. The result has showed isotope position rearrangement corresponding to an essentially 50% rearrangement of the $\ce{C^{14}}$-labeled atoms from the $\ce{C}$-1 to the $\ce{C}$-2 positions in the original chloride and alcohol when the substrates are 2-phenylethyl-1-$\ce{C^{14}}$ chloride and 2-phenylethanol-1-$\ce{C^{14}}$.
References:

Inoue Masashi, Sugita Toshio, and Ichikawa Katsuhiko, "Aluminum Chloride Catalyzed Friedel-Crafts Reaction of Anisole with Epoxide Accompanying Isomerization," Bulletin of the Chemical Society of Japan 1978, 51(1), 174-178 (DOI: https://doi.org/10.1246/bcsj.51.174).
C. C. Lee, A. G. Forman, and A. Rosenthal, "Rearrangement Studies with $\ce{C^{14}}$: III. The Friedel–Crafts alkylation of anisole with 2-phenylethyl-1-$\ce{C^{14}}$ chloride and 2-phenylethanol-1-$\ce{C^{14}}$," Canadian Journal of Chemistry 1957, 35(3), 220-225 (DOI: https://doi.org/10.1139/v57-033).


text:
What gives dilute/conc nitric acid/ conc sulfuric acid their oxidizing property?
I know that dilute sulfuric acid and very dilute nitric acid don't have any oxidizing tendency while dilute/conc $\ce{HNO3}$ and conc $\ce{H2SO4}$ do.
Why do these more concentrated acids have this property?

Nilay, The question is interesting- none of the so-called answers in the links are true answers. Yes, there is a lot of hand waving there. Electrode potentials do not apply to conc. acid, so they better not be invoked here. I don't know the answer either.

Oh, and this here post is supposed to be for some "canonical answer"? And why it shouldn't go to the earlier question? Well, then at least clean up this "oxidizing property" which is hardly a good wording.

https://chemistry.stackexchange.com/questions/600/why-is-nitric-acid-such-a-strong-oxidizing-agent?rq=1 https://chemistry.stackexchange.com/questions/51140/weird-wikipedia-section-on-oxidizing-behavior-of-nitric-and-sulfuric-acids https://chemistry.stackexchange.com/questions/38822/why-iron-reacts-differently-with-concentrated-and-dilute-sulfuric-acid

https://chemistry.stackexchange.com/questions/97090/which-is-a-better-oxidising-agent-concentrated-nitric-acid-or-dilute-nitric-aci https://chemistry.stackexchange.com/questions/17067/how-does-copper-reduce-dilute-nitric-acid-to-nitric-oxide-and-concentrated-nitri https://chemistry.stackexchange.com/questions/51921/nitric-acid-acting-as-oxidising-agent

Related: Why is nitric acid such a strong oxidizing agent?

Thanks for all the feedback ;) Now I have found out the answer on the Internet.
For the diluted ( or very diluted, for nitric acid) form of these acids, the anions are quite stable as there are sufficient water molecules that form bonds with them. Moreover, there's the concentration effect-- hydrogen ions are more abundant as they come fro two sources-- water and acid. As a result, hydrogen ions act as an oxidizing agent instead.
For the conc form of these acids, relatively less water molecules are present to form bonds with anions to stabilize them. Moreover, hydrogen ions are not as abundant as the anions. Also, nitrate ions and sulfate ions are stronger oxidizing agents in comparison to hydrogen ions. Thus the conc form of acids exhibits oxidizng properties.
As for why nitric acid have oxiding properties even at diluted form, it is due to the fact that nitrate ions are more unstable than sulfuric ions. Nitrogen has to maintain a positive (+) charge to retain the structure of nitrate. Nitrogen is a fairly electronegative atom and it prefers to gain electrons than to lose one. This makes NO3- unstable. Therefore NO3- is reactive and readily acts as an oxidizing agent. While sulfate ions, while reaching their maximum oxidation state, have a neutral charge.
It may not be the most accurate and perfect answer tho. Feel free to correct me if I have made a mistake. Many thanks!

Actually, the sulfur atom is not neutral in the sulfate anion since there actually are four bonds in sulfate(hence the formal charge on sulfur being 6-4=2).

https://pubs.rsc.org/en/content/articlehtml/2015/sc/c5sc02076j  according to this open-source article quantitatively defining hypervalency, what makes nitrate more unstable than sulfate is that the former is hypervalent while the latter is not; the former since nitrogen and oxygen are similar in electronegativity, the latter since sulfur is much less electronegative than oxygen.

Oxyanions have a central atom surrounded by a number of more electronegative O atoms and the central atom is significantly positively charged. They are subject to reduction. There are three main factors first the electronegativity of the central atom the more electronegative the stronger the oxidant, second the more O's the stronger the oxidant and third the stronger the X-O bond strength the weaker the oxidant. Simple examples nitrate is stronger than carbonate, nitrite, and sulfate for each of the reasons. Sulfate and perchlorate are at least kinetically less reactive because their bonds to O have significant double bond character from charge feedback into 3d orbitals.  The second factor in their reactions is that these bonds to O must be broken; this requires that an excess of H3O+ be present to form water.  At low acid concentrations the anions are not oxidizers at high acid concentrations they are progressively stronger and the kinetics also improve.
The statement has been made that electrode potentials do not apply. That is incorrect! Standard potential at all activities = 1 need to be adjusted by actual activities and concentrations according to the Nernst equation. Ignoring activities and just using concentrations will give some idea of what is going on.


text:
Anomaly in Faraday's constant measurements
I was reading someone's biography, where the writer stated his early interest in Faraday's constant  measurements (this is set six decades ago). My understanding was that Faraday constant is a true constant but it turns out it is not or it was not considered a true constant for a long time. Different experimental measurements give different results, and the reasons were not fully understood yet! As per the biography, the difference was due to fundamental reasons. Now we are talking about very very small differences but in the world of constants this is not normal. In order to further see the status, as per the author NIST used to state two values for the Faraday's constant Link to the essay on ResearchGate

The Faraday constant as a fundamental physical value has its peculiar
features, which make it standing out of the other physical
constants. According to the official documents of NIST 1, this
constant has two values: $$ \begin{array}{l} \mathrm{F}=96485.33289
 \pm 0.00059 \mathrm{C} / \mathrm{mole} \text { and } \\
 \mathrm{F}^{*}=96485.3251 \pm 0.0012 \mathrm{C} / \mathrm{mole} .
 \end{array} $$ The second value refers to the "ordinary electric
current". The values are determined according to different
experimental techniques. The difference amounts $0.000008 \%$, which
is a tiny difference, however, such a discrepancy is not
characteristic for other physical constants.

The author's link to NIST is no longer valid, and today the Faraday's constant value is stated to be exact on the NIST website. This change must be fairly recent, but no textbook of physical chemistry, new or old ever stated this problem of getting different results by different methods. Has anyone seen this point in mainstream books? I never saw that and it is hard to trace this issue. One cannot even do a exact search of "ordinary electric current" associated with Faraday's constant.

Regarding mentioned different measurement techniques, there can be still some not yet addressed mutual bias/systematic error. Similar case is for Gravitational constant, where measurements have much better precission than reproducibility.

Interesting. Since the elementary electric charge is now exact by definition, and likewise the Avogadro number, this appears to force the Faraday constant to a single value. Maybe the former two values of the Faraday constant was due to human beings: dueling committees of experts. ;-)

@EdV, I will probably ask the author, he is alive. It seem this was something fundamental. He mentions that this something fundamental related to the solvation of electron. People may have artificially fixed it just like Carbon-12 to avoid these headaches.


text:
Can Mg/ether reduce carbonyl groups?
One of the ways of preparing Grignard reagent is:
$\ce{R-X ->[Mg/ether] R-MgX}$
However let's assume that a carbonyl group is present along with a halide in the substrate. Will $\ce{Mg/ether}$ react with the carbonyl group in a way similar to that by $\ce{Zn}$ in Clemmensen reduction? Or would $\ce{Mg/ether}$ react with the carbonyl group in any way at all?

How do you mean by similar way to ZN in clemmensen? I don't quite understand, could you explain a bit more?

@Buraian metals like Na,Mg,Zn can release electrons which ends up reducing or reacting with carbonyl compounds in many cases. However the reaction and end product varies

It is known that $\ce{Mg}$ will reduce ketones. The articles (Ref.1 and 2) in the References section confirm this fact.
If you want to form the Grignard reagent of a molecule that also happens to contain a carbonyl group, then the usual practice is to protect it first. A cyclic ketal/acetal, which are stable under basic conditions, is often the first choice for this.

References:

Ji Young Kim, Hak Do Kim, Min Jung Seo, Hyoung Rae Kim, Zaesung No, Deok-Chan Ha, and Ge Hyeong Lee, "Reduction of ketones to corresponding alcohols with magnesium metal in absolute alcohols," Tetrahedron Lett. 2006, 47(1), 9–12 (DOI: https://doi.org/10.1016/j.tetlet.2005.10.140).
Magnesium - Organic Chemistry Portal

I have added few words to make some sense. If you don't like what I did, please feel free to roll back. Good finding though.

I think that the formation of the Grignard will be inhibited. The formation of a Grignard is an autocatayltic reaction. It is well known that one of the best ways to get a Grignard synthesis going is to add some of an existing solution of a Grignard reagent.
I reason that as soon as any of the Grignard forms then it will be consumed by the reaction with the ketone. As a result the reaction of the alkyl (or aryl) halide with the magnesium will be so slow that for all intents and purposes we can say "nothing happens"
It will be like trying to make a Grignard in ether that has a trace of water, I once tried to make methyl magnesium iodide and failed. I then discovered that the ether was not as anhydrous as I thought it was. I am sure that if something is present that destroys the Grignard as soon as it forms then the rate of the reaction of the magnesium with the alkyl or aryl halide is so slow that as a synthetic reaction we can regard the reaction as being "useless".
When I tried to make the Grignard in the slightly damp ether, I tried adding iodine, I tried ultrasound and even these things were not able to get the synthesis of the Grignard going. I have always found that if I exclude water strictly from the reaction (bake all the glass equipment overnight before use and sodium dried ether) that the synthesis of a Grignard never fails. What I do is to stirr the magnesium under nitrogen for a long time. I then add a little ether and some of the alkyl halide and I leave it with no stirring for a while. This normally gets it started.

I don't think "nothing happens". I think starting materials will be consumed but nothing useful comes out of the reaction mixture.

Mg in ether , benzene  and possibly other solvents couples ketones to give pinacols [1,2 diols]. I would imagine were a halide present in the molecule the results would be very exciting putting two exothermic reactions together.  A similar reaction is the acyloin condensation that couples esters using Na metal.  I suspect the acyloin reaction can happen readily electrochemically. The Reformatsky reaction of an alpha bromoester with zinc and a ketone reacts as an insitu Grignard attacking the ketone but not the ester.  The idea of insitu Grignard reactions is intriguing.

Having run the pinacol reaction of acetone as an undergraduate, I know that the reaction of magnesium with acetone in benzene requires the use of a mercury salt to help things along. The mercury(II) chloride we used would have reacted with the magnesium to form magnesium amalgum which will be more reactive to the acetone than the magensium turnings would have been. I know of a greener way of doing a similar reaction which uses aquerous ammonium chloride to activate the magnesium. I think without the mercury or another activating agent that the reaction will not go.


text:
Why is it sometimes kJ only, and in other times kJ/mol? What's the difference?
Here is an example in my book.  Computing the enthalpy of formation $\Delta{}H_\mathrm{f}$ via multiple enthalpies of reaction $\Delta{}H_\mathrm{r}$ I notice some of the equations run by $\pu{kJ/mol}$ and others by $\pu{kJ}$ only.  I'm puzzled about the conceptual difference between these values.  How do they relate to each other?

Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own explicit  effort.  How do I ask a good question.  Asking should be rather the last than the first idea coming to your mind. // Using photos/screenshots of text ) instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused nor referred in answers.

If you understand difference between $\pu{g}$ and $\pu{g/mol}$, you understand difference between $\pu{kJ}$ and $\pu{kJ/mol}$. // Another thing is, if the units are correctly used. We cannot explain why they were eventually used incorrectly. Always compare what you know with what you read, do not take the latter automatically as right.

The book(s) are being sloppy and you are right to be confused. For a given reaction the heat is computed for the reaction as written, that is, for the molar amounts specified in the reaction equation. Sometimes reactions are written on a per mole basis, with the extent of reaction expressed as a molar amount. Heats of formation are typically defined as molar quantities, defined on a per-mole basis, for instance  kJ/mol. The short of it is that people (myself included) are sometimes sloppy, and designation as a molar quantity and use of appropriate units is sometimes neglected.

In the bottom example, for instance, the heat of formation of carbon disulfide should be declared in $\pu{kJ/mol}$ units. And yes, in the case of reaction (5) in one case it is given in per-mole units but not in another. That's sloppy.

The keyword is dimension analysis.
For instance, in example 1-7, the aim is to determine the standard enthalpy of formation $\Delta{}H_\mathrm{f}$ for $\ce{SO2}$ (note the subscript is f) determined with equations 1, 2, and 3.  After rearrangement $3 \rightarrow 4$ you add the equations to obtain equation #5 which however describes the enthalpy of reaction (note subscript is r) to yield $\pu{2 mol}$  of $\ce{SO2}$ instead of the conventionally tabulated enthalpy of formation (again subscript f) for $\pu{1 mol}$ of $\ce{SO2}$.
Thus, $$\Delta{}H_\mathrm{r} = \pu{2 mol} \times \Delta{}H_\mathrm{f} = \pu{2 mol} \times(\pu{-296 kJ/mol}) = \pu{-592 kJ}.$$ One may argue the publisher was sloppy when writing 2 instead of 2 mol in this line.
You may retain that $\Delta{}H_\mathrm{r}$ as a special case (mathematically speaking, a subset) of the more general $\Delta{}H_\mathrm{r}$.  That is to say, for a reaction yielding $\pu{4.2 mol}$ of $\ce{SO2}$, the reaction enthalpy $\Delta{}H_\mathrm{r}$ is:
$$\Delta{}H_\mathrm{r} = \pu{4.2 mol} \times \Delta{}H_\mathrm{f} = \pu{4.2 mol} \times(\pu{-296 kJ/mol}) = \pu{-1243.2 kJ.}$$


text:
Intuition behind EMF independence of temperature for unit molarity solutions
The Nernst equation is given as:
$$E= E^o - \frac{RT}{nF} \ln Q$$
Now, it turns out that if all the liquid components in the cell are unit molar i.e: of molarity one, then $Q=1$ and hence $\ln Q=0$ making the EMF temperature independent.
This puzzled me, is there an intuition on why a mixture of unit molarity solution has no dependence on temperature for their EMF?

This question could be equivalently asked in terms of Gibb's Free energy, why does Gibb's Free energy become temperature independent if each constituent has molarity of  one:
$$ \Delta G = \Delta G^o + RT \ln Q$$

Comments are not for extended discussion; this conversation has been moved to chat.


text:
Why do we have to use octahedral voids for formula making in solid state chemistry?
Allow me to take an example to explain myself.
Q) What is the formula of a compound in which the element Y forms a ccp lattice and an atom of X occupy ${1\over3}$rd of tetrahedral voids?
Now to find the formula we first say that the space occupied by octahedral voids are $2$*space occupied by tetrahedral voids = $2\over3$.
No. of atoms of element X = $2\over3$*no. of octahedral voids
$\implies$ No. of atoms of element X = ${2\over3} \times 4$ = $8\over3$
Ratio of X and Y = ${8\over3}:4$ = $2:3$
$\therefore$ formula is $X_2Y_3$.
Coming to my question - Why can't we use space occupied by tetrahedral voids?

We don't have to use the octahedral voids. This is nonsense.

This is how my teacher and my textbook shows it. I'm new to all this so I don't know a whole lot.

We have twice as many tetrahedral voids as lattice-forming atoms, hence 2*Y. We only use $1\over3$ of them, hence the number of $X={2\over3}Y$, which gives the formula $\ce{X2Y3}$. That's how I show it.

Aren't you using the concept I used indirectly? Coz number of lattice forming atoms = number of octahedral voids (as far as my knowledge goes).

That's right, which also explains why we've got the same result. It is just that we don't have to mention the octahedral voids anywhere.

Ohh okay. Thank you for helping. I appreciate it!

@HarshDarji if you got your answer, feel free to self answer it. Self-answer are appreciated and it also help future readers.


text:
How to prepare an emulsion (compounds in wt%)
I am trying to replicate an experiment to generate microparticles but I found myself confused about how to prepare the ingredients as some need to be weighed and dissolved before and I think the author is referring to weight/weight percentage. I would like to know if the g of mass synthesis and polymeric content are informative or not.
I would really appreciate any insight you have.

The distinction made in the table is confusing as for I do not see which info it does provide. Moreover why the sum doesn't amount to 100 is a mystery to me. I won't be able to follow the table :( if not interpreting and taking everithing in grams and using the same proportions (but it won't be m/m %)

@Alchimista. I get your point. However, I do not understand why the author labeled this table as recipe...