text:
Why can't ligands with "locked structures" undergo chelation?
The following quote is from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha), from the chapter "Coordination Compounds", page 177:

...it must be noted that $\ce{NH2NH2}$ and $\ce{N(CH2CH2)3N}$ cannot act as chelating ligands due to the formation of a three membered ring and locked structure respectively.

But, as per the accepted answer to the question Are all complexes with a polydentate ligand examples of chelation?, when a ligand forms more than two bonds to the central atom, the result is a chelated complex. This is also inline with the IUPAC Gold Book's definition of Chelation:

The formation or presence of bonds (or other attractive interactions) between two or more separate binding sites within the same ligand and a single central atom.

As per my understanding, it doesn't seem to exclude any ligands as mentioned in my textbook.
I can understand that since $\ce{NH2NH2}$ forms a three membered ring, there would be ring strain in the structure and hence the process of chelation is not energetically feasible compared to four membered (or higher) rings. I think this would be the same for all such ligands with donors adjacent to each other, as I have not encountered any such species so far.
However, I'm unable to understand why $\ce{N(CH2CH2)3N}$ (1,4-Diazabicyclo[2.2.2]octane) cannot undergo chelation.

From its molecular structure, I can imagine that the three ethylene groups act as springs to accommodate any ring strain while bonding with the central atom and so there must be no hinderance to undergo chelation. The following molecular structure is what I think might be if $\ce{N(CH2CH2)3N}$ undergoes chelation with cobalt:

I felt given the size of the cage it would be difficult to fit the metal atom inside it and hence thought that it might be off-centred as shown in the image above.
So, it would be helpful if you could clarify whether my reasoning for the inability of $\ce{NH2NH2}$ to chelate is correct or not and why is it not possible for $\ce{N(CH2CH2)3N}$ to undergo chelation? Further, it would be helpful is you could tell why is there a contradiction between the book's statement and IUPAC definition? Is the statement from the book incorrect (because IUPAC can't be incorrect)?

@Mithoron: I considered the metal atom to lie off-centred and not at the centre of the cage. If you can't imagine, I can try to draw a diagram. If you think that must be at the centre it would be helpful if you can explain why that's so?

And how you imagine turning lone pairs inside out? IIRC this is called piramidalisation  and I doubt such structure was ever found for cases similar to that. BTW I guessed you'd think it somewhat off-center and it's still inside and doesn't make sense.

@Mithoron: I've added the molecular structure of what I thought might be if DABCO undergoes chelation. Hopefully it clears the fact what (and why) I meant by "off-centred". After reading the answers, I can see somehow this is not possible.

The simple answer for your query: the lone pair is practically immobile, and can't orient itself to chelate.
To understand this, we first have to understand the structure of DABCO. The rings force a conformational rigidity on both the nitrogens, and therefore the fluxional process of nitrogen inversion is inhibited in these structures. Now, as the accepted answer here states, in this process, the entire system swings from an $\ce{sp^3}$
hybridized state to a planar $\ce{sp^2}$ transitional state back to $\ce{sp^3}$. In this transitional process, the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases.
However, due to the restrictions imposed by the structure, one can imagine that both the nitrogens of DABCO are perenially stuck in this planar transition state of the inversion process. Hence, the lone pair in each of them resides in an almost pure p orbital, while the bonding scheme for the central atom would be roughly $\ce{sp^2}$ .(Note: This just a suggestion for visualizing the geometry of the molecule. Please don't take it as an actual mechanism)
Hence, it would like this from a side view:

Blue:Nitrogen|Pink: p orbital containing lone pair (source)
So as you can see, there is not much possibility of having the metal atom lie off-center, as then the lone pairs will have to re-align either above or below the plane of the screen. But the lone pairs are stuck on the plane of the screen as shown.
Trying to fit the metal atom inside the cage will of course, be sterically hindered( although some other reagents like crown ethers are used for trapping cations in a such a fashion)
So although they can exhibit denticity of one like the other answer stated, chelation might be a far-fetched game for DABCO
As Martin pointed out in the comments,perhaps a little more introspection would be required to look at the orbital picture of this molecule. I will be citing a few findings from a research paper which has analysed the molecule based on frontier MO theory and thereby debunks a few of the classical valence bond hybridization predictions:
Electron momentum spectroscopy of the frontier electrons of DABCO does not support an sp3 hybrid lone-pair description

Abstract
The highest occupied molecular orbital (HOMO) and next-highest occupied molecular orbital (NHOMO) valence orbital electron density distributions of 1,4-diazabicyclo[2.2.2]octane (DABCO) have been investigated by electron momentum spectroscopy, a technique that probes the orbital-like nature of valence (frontier) electron transfer out of a molecule.
In contrast, Pauling's widely used and taught valence bond (hybridization) model, which is equivalent to a localized molecular orbital description, does not correspond at all to the experimental measurements. It follows that, for considerations of electron transfer, the "lone pairs" of DABCO are not localized or hybridized, but rather exist as nondegenerate orbitals that are delocalized differently over the molecular framework
Introduction and background
A first-order through-space treatment (2, 3) that the antisymmetric(AS) antibonding combination of 2p functions on the nitrogen atoms(in a simple LCAO-MO description) will comprise the HOMO, while the symmetric (S) bonding combination will comprise the NHOMO. However,when the through-bond interaction is turned on,the frontier orbital energy ordering is predicted to be reversed i.e. the HOMO is bonding(S) , while the NHOMO is antibonding(AS)
Results and Discussion
The EMS binding energy spectrum confirms the earlier PES observations(4, 5) that there are two outer valence IPs(7.52 and 9.6 eV, respectively), and not one corresponding to the presence of degenerate nitrogen lone pairs in DABCO.

NHOMO=Next Highest Molecular Orbital
What all this evidently shows is that both the lone pairs are in fact,not degenerate(as classical hybridization would have predicted both of them to be in $\ce{sp^3}$ hybrid orbitals.)
In fact,both of them lie in inherently different orbitals with different ionization energies, that is,the HOMO and NHOMO, which are constituted by in-phase interaction of 2s orbitals and out-of-phase interaction of 2p orbitals respectively as a first order LCAO approximation
Here is an MO diagram for DABCO:

secondary interactions with the in-phase combinations of the CÐC s and s orbitals which reverse the order of the HOMO and HOMO-1. The orbitals symmetries are specified in D3h(source)

Thanks for the answer! I think we can also say that the p orbitals of nitrogen atoms are along the $\ce{N-N}$ axis of DABCO ring using symmetry arguments. This is evident when we consider the three dimensional molecular structure provided in the PubChem link in your post. So, if at all DABCO must undergo chelation it would be possible only when the metal atom is at the centre of the cage. I hope the rotation about the $\ce{N-C}$ bonds are restricted so that it can't align as shown in the question.

Yep,you summed it up pretty much correctly. The $\ce{N-C}$ bond rotation is restricted, so the alignment you have proposed is not possible as such. I doubt there are enough transition metal atoms that have the ability to fit into that cage and also coordinate with two donor atoms at the same time,but yeah, it's theoretically possible to chelate within the cage. But as Anirudha's answer cited from various sources,that DABCO's compounds are generally linear and monodentate. So it is

The lone pairs are most likely in sp³ like type orbitals due to pyramidalisation of nitrogen at the bridge heads.

@Martin - マーチン♦ I can't find a source for it as of now,but I distinctly remember a question in our college exam which mentioned the lone pair of DABCO in a p type orbital,and the hybridization to be $\ce{sp^2}$. Also,I have learnt here on Chem.SE that geometry drives hybridization,not the other way around. The $\ce{sp^2/ p}$ orbital conjecture seems much more capable to be of explaining the bonding situation than the $\ce{sp^3}$ hybridization,as I feel that the latter can't account for a perpendicular lone pair. Am I wrong in postulating this?

@Martin - マーチン♦ Ah,I found something! Electron momentum spectroscopy of the frontier electrons of DABCO does not support an sp3 hybrid lone-pair description

Well, I said it is due to pyramidalisation. Nitrogen is not planar coordinated, so it cannot have sp² orbitals (in a first order approximation). Everything beyond that is also beyond a discussion in the comments. It seems it is more complicated and hybridisation not a good model to explain it...

@Martin - マーチン♦ Perhaps you're right. I will edit the answer reflecting some findings from this research paper

My copy of Concise Inorganic Chemistry, 5th edition (J.D.Lee, Wiley publishers)1 has nothing contrary to say about hydrazine or 1,4-diazabicyclo[2.2. 2]octane (henceforth referred to as DABCO). It is most likely that the author of your adaptation has used his liberty to add these facts in your copy.
This then raises the question: is the author correct? Hydrazine derivatives such as Benzhydrazides and Salicylhydrazides 2 have been proven to show chelation. Pure hydrazine, however, is unlikely to show chelation due to excessive ring strain. Also, no literature has been found which shows hydrazine undergoing chelation. DABCO, on the other hand, forms complexes with Mercury halides3 with a denticity of 1. This ligand also does not show chelation, and fitting a large metal atom even off-center is highly unlikely due to large steric hindrance and the geometry of the lone pair, which is perpendicular to the structure and not favourable for chelation. Most complexes formed by DABCO tend to be linear in nature4 and no rings are formed.
In short, the author is correct in stating that hydrazine and DABCO do not form chelates.
EDIT: You may also want to read more on a class of compounds called Cryptands, specifically Cryptand-222. This compound displays the 'cagelike' ligation you were looking for in DABCO.
References:

Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers &
Blackwell Science, 1996.
Issa, R. M., et al. Coordination compounds of hydrazine derivatives
with transition metals. I. Metal Chelates with Benzhydrazide and
Salicylhydrazide. Zeitschrift fur anorganische und allgemeine
Chemie, vol. 354, no. 1–2, Sept. 1967, pp. 90–97.
doi:10.1002/zaac.19673540118.
Shan, Zeng-Mei, et al. Synthesis, Crystal Structures, and
Characterization of Three Mercury(II) Halides Inorganic–Organic
Hybrid Compounds with 1,4-Diazabicyclo[2.2.2]Octane Ligand.
Inorganica Chimica Acta, vol. 366, no. 1, Jan. 2011, pp. 141–46.
doi:10.1016/j.ica.2010.10.023.
Thorp-Greenwood, Flora L., et al. Three-Dimensional Silver-Dabco
Coordination Polymers with Zeolitic or Three-Connected Topology.
Crystal Growth & Design, vol. 14, no. 11, Nov. 2014, pp. 5361–65.
doi:10.1021/cg501231v.

Thanks for the answer! Regarding "It is most likely that the author of your adaptation has used his liberty to add these facts in your copy.": Only the text in first blockquote is from the textbook, others are my own interpretations :)

@GuruVishnu I understood that. I couldn't find the text in the first block quote in my copy of J D Lee, which is why I said that.


text:
Prove Law of Chemical Equivalence
A somewhat infamous law (to me at least), that states:
$$a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$$
for any and every balanced chemical equation -
$$a.A+b.B = c.C+d.D$$
where $i$ denotes the stoichiometric coefficient(/no. of moles) of the chemical species $I$ and $I_{eq}$ denotes the number of equivalents in 1 mole(/formula mass...?) of the chemical species $I$.
Am I wrong about any of the facts I've presented above? Please mention any errors you spot.
Assuming that I did fine, my simple question is,

What's the mathematical proof for this law?

I tried surfing online, but all I came across (with my not-so-deep searches) were verbal explanations of the law.
Is there no mathematical proof to this law whatsoever? I'll be satisfied to just find a proof, even if I don't understand the complicated math going into it (for I'm an eleventh grader).
Edit to provide an example:
$$\ce{2NaOH + H2SO4  -> Na2SO4 + 2H2O}$$
For A, $a \cdot A_{eq} = 2 \ (moles) × 1 \ (equiv.\ per\ mole) = 2$
For B, $b \cdot B_{eq} = 1 \ (moles) × 2 \ (equiv.\ per\ mole) = 2$
And this follows for other species, hence showing $a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$

You are missing something. Where is i and where is I?

@M.Farooq I believe he means the $i$=lower case symbols and $I$=upper case symbols.

It's not so much that this relationship has a mathematical proof, but more that it proceeds from the law of conservation of mass/matter.

@BaiduryaMathaddict, I am still not following your choice of notation. You are using A, B, C, D for molecular species as well as number of equivalents. That is why I would still say that you further explain the notation withany real example.

A real example can be useful to understand your notations. Anyway this law is simply the direct consequence of the conservation of number of atoms of the same element on both sides of the chemical equation.

An example
Here is a balanced equation:
$$\ce{2 H2 + O2 -> 2 H2O}$$
Both sides of the equation show four atoms of hydrogen and two atoms of oxygen, so it is balanced. The coefficients are $\nu_{\ce{H2}} = 2$, $\nu_{\ce{O2}} = 1$ and $\nu_{\ce{H2O}} = 2$.
If 20 moles of $\ce{H2}$ and 10 moles of $\ce{O2}$ are used up in a reaction forming water (you can figure out 20 moles are formed), I can write:
$$\frac{n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{n_{\ce{O2}}}{\nu_{\ce{O2}}},$$ where $n$ is the amount that is used up in the reaction.
I can not write:
$$n_{\ce{H2}} \cdot \nu_{\ce{H2}} = n_{\ce{H2}} \cdot \nu_{\ce{H2}},$$
as you will find when substituting the values from the example.
I should also not write:
$$2 \cdot \ce{H2} = 1 \cdot \ce{O2} $$
Hydrogen atoms are not oxygen atoms, no matter what we multiply it (nuclear reactions aside)
What about $I$ and $i$?
If you call the stoichiometric coefficients $i$ instead of $\nu$, and you want the equation below to be true:
$$i_{\ce{H2}} \cdot I_{\ce{H2}} =  i_{\ce{O2}} \cdot I_{\ce{O2}},$$
you have to make sure that $I$ is proportional to $\frac{1}{i}$, i.e. define them by
$$I_{\ce{H2}} = \frac{ \mathrm{const} }{ i_{\ce{H2}} } $$
$$I_{\ce{O2}} = \frac{ \mathrm{const} }{ i_{\ce{O2}} }$$
If you look at definitions of "number of equivalents" (a concept that is outdated), you will find that this is the case. Once that is in place, it is a matter of showing that for any non-zero $a$ or $b$,
$$\mathrm{const} \cdot \frac{a}{a} = \mathrm{const} \cdot \frac{b}{b} $$

Source: https://www.askiitians.com/iit-jee-chemistry/physical-chemistry/stoichiometry-and-redox-reactions/law-of-equivalents.aspx

@BaiduryaMathaddict Thanks for the edits (balance vs balanced). The edits to the mhchem code were unnecessary, so I rejected the edit and fixed the mistake (less work for me). If what follows an underscore is another backslash command, you don't need the braces around it, i.e. $Z_\delta$ is fine and $Z_{\delta}$ is harder to read. You do need it in $Z_delta$ vs $Z_{delta}$.

@BaiduryaMathaddict Sorry, I don't see the problem. Maybe my rendering engine is different (phone vs computer sometimes makes a difference), or some other subtlety. For me, your edits and the original looked identical.

@BaiduryaMathaddict I tried to edit it accordingly. It is hard to test without seeing the problem, but I think I got them all (and found another formatting problem). I learned something today!

There are a couple of problems with the main gist of the question. First of all we do not need to invoke the concept of moles here. Equivalents originated in 1700, survived until the 1960s and vanished slowly. Since the govt. did not bother to update chemistry syllabi, this concept is still lingering in some areas. It is creating havoc because students have to deal with moles and then equivalents and then they try to find relations between them. Gram equivalents do not need the crutches of moles, because they pre-date moles by centuries.
This is why I was continuously asking you to improve the notation because there is no need of moles in the notation of your query.
$$\ce{aA + bB -> cC +dD}$$
$$\ce{2NaOH + H2SO4  -> Na2SO4 + 2H2O}$$
You can now parse this reaction as follows:
1 equivalent weight of NaOH will react with 1 equivalent weight of $\ce{H2SO4}$ to produce 1 equivalent weight of $\ce{Na2SO4}$ and 1 equivalent weight of $\ce{H2O}$
Now, let us see what equivalent weights we are talking about  (I assume you know how to calculate equivalent weights from the chemical equation)
equivalent weight of NaOH = 40
equivalent weight of sulfuric acid = 49
equivalent weight of sodium sulfate = 71
equivalent weight of water = 18.
Does it make sense if I say that 40 g of NaOH will react with 49 g of sulfuric acid to produce 71 g of sodium sulfate and 18 g of water?
Now, equivalents and equivalent weight are two different things. Equivalents is similar to calculating "moles" and equivalent weight is analogous to "molecular weight".
So, equivalents = weight in gram/ equivalent weight
So let us say, someone tells you that we have 0.1 equivalents of NaOH, how many equivalents of sulfuric acid are needed?
The answer is 0.1.
In terms of grams, you would need 0.1 x 40 = 4 g NaOH and, 0.1 x 49 = 4.9 g sulfuric acid.
Using your notation and formulas, I will arrive at wrong numbers!
[Post Edit].
Now that you have clarified the notation using my example, and their units it makes perfect sense. Earlier I told you that equivalents do not need the mole concept. They are a parallel system.
Now the formula $$a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$$ connects the mole with the concept of equivalent weight (not equivalent).
It originates from the observational fact that molecular weight or the formula weight of any substance is an integer multiple of equivalent weight, i.e.,
$$\mathrm{Formula\,weight} = \mathrm{n\,x\,equivalent\,weight}$$.
How many moles are there in 1 formula weight? 1, right? The value of $\mathrm{n}$ is derived independently based on the chemical properties of the substance.
$$\ce{2NaOH + H2SO4  -> Na2SO4 + 2H2O}$$
You can now read this reaction as follows:
Fact no. 1: 1 equivalent weight of NaOH will react with 1 equivalent weight of $\ce{H2SO4}$ to produce 1 equivalent weight of $\ce{Na2SO4}$ and 1 equivalent weight of $\ce{H2O}$
Fact no. 2: 2 mole of NaOH will react with 1 mole of $\ce{H2SO4}$ to produce 1 mole of $\ce{Na2SO4}$ and 2 mole of $\ce{H2O}$
How would we connect the two facts:
We can say, in 1 formula weight of NaOH there is only 1 equivalent weight of NaOH,
In 1 formula weight of $\ce{H2SO4}$  there are 2 equivalent weights of sulfuric acid and so on. So your formula works and produces 2 in each case this reaction.
Will this formula work or fail in disproportionation or comproportionation reaction, I don't know. Most likely, it fails there.

"My question has more to do with why 1 equivalent of a species reacts with exactly 1 of another" Now this is a different question. It has nothing to do with the original post. Please post this as a separate question (with you background understanding and search). Also mention that you don't wish to invoke moles and it is a hitorical query rather. Your current question will be closed soon. Only one more person has to vote for its closure as being unclear out of (4/5 close votes). You see if I use your (corrected) formula, I get wrong numbers. So your formula is not correct.

If you can edit the question with this particular example using the numbers in this answer, it would be helpful to understand your formula. I am not a moderator, but I can see the reason for closure votes.

Please also edit the moles issue. You don't need to bring it in moles. That is adding to confusion.

@BaiduryaMathaddict, If you are satisfied with this part of the question, accept it. Otherwise post a new question, with an example. I was telling that if you post an example you will get better answers. The moment you did, you query became understandable. I understand your question as to why 1 equivalent of NaOH will react with 1 equivalent of H2SO4. This is a separate story.

source: Stack Exchange

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

sort: votes


text:
Semiconductor Li-ion batteries
I have recently come across the concept of semiconductor Li-ion batters, for instance carbonised polyacenic semiconductor (PAS) Li-doped materials.
I am not sure how common it is for a battery to be formed of a semiconductor, and are their any other examples?
Do all Li-ion polymer batteries involve a semiconducting component?
How do these differ from metallic batteries?
Is there an advantage?

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


text:
An automated approach to visualize several xyz structures
I have several .xyz files (each composed of several molecules) and I'd like to find an automated way to visualize and save them. I'd appreciate any guidance.

What do you mean by "visualize", e.g. what is the desired output format? Also, what kind of information needs to be visualized?

Have you seen How to get 3D molecular structures for 270 compounds?; Visualizing atoms in XYZ file with colors representing partial charges; Is there any molecule visualizing software for Python?

Thanks for the reply. By visualization I mean I need something which reads my set of xyz files and saves their "images" automatically. I mean I don't want to save the figures one by one. About the posts you have referred me to, I feel they are not fully relevant cause I have the 3D coordinates (se don't need to generate them at this stage) and don't want to save the figures (extracted e.g from ASE) one by one. Ideally, I need some sort of TCL script for vmd where it can read xyz files sequentially and saves the images automatically.

Wouldn't the orientation be a problem? If you can't see the individual files and rotate them, then you might get a picture that's in an odd orientation.

You can use VMD to do this.
Note that the following is only tested on Linux with VMD version 1.9.2, i don't know how it works out on Windows or Apple OS.
The following is from a script that i used to render a trajectory. The first thing to do is to load your xyz file as trajectory. I.e. in a format so that vmd allows you to "slide" through the structures in its normal view. Then execute the script in the tlc console. The script iterates over the steps in incremenets of 2 and assumes that you have 2000 structures. The iteration variable is i. The actual command that changes the frames is the animate goto $i command. You would need to adjust that to fit your requirement. I used the tachyion renderer and saved the output figures in tga format. You will most likely have to adjust the path to the renderer binary which is the path /software/vmd-1.9.2/lib/vmd/tachyon_LINUXAMD64 in my case. VMD saves first a .dat file that contains the scene and then applies the renderer to this dat file. This is why i am setting two path variables, first fin for the .dat file and then fout for the actual tga file.
for {set i 0} {$i < 2001} {incr i 2} {
animate goto $i
set fin [format "pic_%04d.dat" $i]
set fout [format "pic_%04d.tga" $i]
render Tachyon $fin "/software/vmd-1.9.2/lib/vmd/tachyon_LINUXAMD64" -aasamples 12 $fin -format TARGA -o $fout -res 1920 1080
}

The nice part about this is that you can implement other commands during your loop, like a zoom or a rotation. All you have to do, is find out what the tcl command in vmd is.

I recommend using software called ChemCraft. The software isn't free; however, they offer a 150 day trial period. I typically use this program to open formatted checkpoint files from Gaussian calculations to view orbitals. It has plenty of other functions, including the ability to open multiple .xyz files and save multiple images.
The default colors are a bit yucky, but I recommend using one of the following from the Display menu: CPK, GaussView, or Spartan. Those will color atoms with the typical scheme: Nitrogen (blue), Oxygen (red), etc.
I hope this helps!

There are also programs such as jmol, avogadro, and pymol (which can be obtained for free if you compile it yourself). There just isn't really any good reason to pay for a software if you all you want to do is use it for visualization.

If you want to use Python, you could

Use ase to load the xyz file structure by structure, see here.
Then use existing molecular visualization tools, such as the kind built into ase, see here. Or use other such as pyxtal. These can be saved as images directly using Python scripts.
Write a short script to iterate over all your structures.

Alternatively, Ovito can also do this, but requires a somewhat steeper learning curve.


text:
What is the naming convention for Sn and Pb in inter-metallic compounds?
How would one go about naming inter-metallic (specifically, transition post-transition) compounds containing tin and lead?
It is usually easy figuring out the naming of such compounds as they follow a simple naming convention, for example ruthenium germanide, niobium aluminiumide and so on.
But what about tin? Would you get ruthenium stannide? Stannumide? Or something completely different? Same goes for lead, but I feel like I've heard the name aurous plumbide referring to $\ce{AuPb2}$ before.

@Mithoron Though Sb isn't unreasonable to include? 'Stibnide'?

@ChristianS.O. -- AFAIK neither Sn nor Pb is a transition metal, so I removed that tag.

Worked in metallurgy ( R&D Dept) my whole career and this problem never came up. Generally called intermetalllics and maybe a likely composition.

@hBy2Py Sb would be antimonide, example Nickel Antimonide

After doing some digging I've found evidence to support that tin based inter-metallic compounds should be named 'stannide' such as in the paper by Fickenscher et al. [1].
Reference

Fickenscher, T.; Rodewald, U. C.; Niehaus, O.; Gerke, B.; Haverkamp, S.; Eckert, H.; Pöttgen, R. The Stannides $\ce{RE3Au6Sn5}$ $(\ce{RE} = \ce{La}, \ce{Ce}, \ce{Pr}, \ce{Nd}, \ce{Sm})$ – Synthesis, Structure, Magnetic Properties and $\ce{^{119}Sn}$ Mössbauer Spectroscopy. Zeitschrift für Naturforschung B 2015, 70 (6), 425–434. DOI: 10.1515/znb-2015-0050.


text:
High-temperature DFT
Why is it so difficult to perform DFT calculations that consider temperature?
I have seen that time-dependent DFT is needed to model systems at high temperature.
Why is this the case? What about finite temperature functionals? Why is it also acceptable to use zero-temperature functional with TD-DFT?
I have also read that thermal fluctuations can result in the occupation of electronic states (excited states) above the ground state.
I have tried to rationalise the fact that the electron density changes with temperature as:
If the ions in the until cell move, their degree of orbital overlap and electron density will change. The degree of orbital overlap influences the band-structure, which then impacts all of the electronic and physical properties.
Is this accurate?

One thing I will request you to note here is the definition of temperature. Temperature, in broad sense, is defined as the average kinetic energy (for equilibrium conditions, I hope you are speaking of that) of a system. The contribution mostly arises from translational kinetic energy, and the fluctuation in temperature is proportional to $\frac{1}{N-1}$. (I don't remember if it has some power or something.) In other words, a particle has infinite temperature fluctuation. So if you want to do finite temperature DFT calculations, you will need an ensemble of particles, and it will only add the cost.
In case you have infinite computational power (or at least enough to support no of molecules in the order of Avogadro number), you can, in principle, go for DFT at a finite temperature.

If you are worried about occupation of the higher states, you can calculate the higher states using Kohn-Sham Hamiltionian (these are just eigenfunctions of the Hamiltionian), and then calculate the occupation at a temperature using the formula for the concerned ensemble. Since the states of the KS hamiltionian are its eigenstates, these should not change if you have a static distribution of electrons. The eigenstates mean that these are decoupled by the same KS Hamiltonian, and in principle, you are not allowed to take an electon from one state to other without external purturbation that couples the states.

However, at a given temperature, everything is dynamic. This invokes that the states have finite lifetime, which from Uncertainty principle, imposes broadening in energy, i.e. it is not an eigenstate anymore. This imply there can be coupling between different states, and hence electrons can jump between states without external perturbation. You can externally add broadening to the states using broadning matrix, that mimics the temperature to some extent.

For TDDFT, you actually introduce an alternate electric field, and then this purturbation is used to calculate coupling between the concerned states. Thus, TDDFT calculates the excitation using light. It has nothing to do with the temperature.

Thank you for your great explanation. Can you please explain how in this paper (given below) has calculated the overall energy barrier at two different temperatures (0 K and 298 K). They explained "The overall energy barrier at 0 K for the LH and ER mechanisms is 0.63 and 0.77 eV, respectively. At 298.15 K, the overall energy barrier for the LH mechanism decreases to 0.58 eV, whereas that for the ER mechanism increases to 0.88 eV, which implies that CO oxidation on Ni-Gr prefers to proceed via the LH mechanism kinetically." DOI: 10.1039/C6NJ00924G

Temperature is difficult to capture in any electronic structure theory (not just DFT) because almost all of them invoke the Born-Oppenheimer approximation. The BO approximation assumes the electrons instantaneously equilibrate to the current nuclear configuration (geometry). The difficulty arises then because most theories assume the nuclei are stationary as they solve for the electronic structure (either the wavefunction, or in the case of DFT, the density). As you cannot clearly define temperature without nuclear motion, these methods cannot capture temperature. As Mitradip notes, you would need to use ensemble or finite-temperature DFT to recover a description of temperature. AFAIK, you cannot use just TD-DFT to capture thermal effects. You need to recover a thermal distribution (which the ensemble achieves).
As for the thermal excitation of electronic states:
You would only get electronic excitation if the thermal energy was enough to surpass the energy of an excited state. Unless you are at very high temperatures or have systems that have low-lying electronic excited states, it's not so much of a concern as is the excitation of vibrational (and of course rotational) states.
However, if you do have significant thermally induced electronic excitation, then inclusion of thermal effects would certainly affect the predicted observables of your material under study.

Is it possible to calculate the total energy of a system at a finite temperature using vasp? Does VASP implement finite temperature calculations?


text:
Grignard alkoxide intermediate as starting material for ether or ester synthesis
I'm having a Grignard reaction between a ketone or aldehyde with RMgX in mind.
The average workup uses water to hydrolize the intermediate alkoxide. But could this intermediate be used without isolation to prepare an ether or ester? For example by simple addition of RX, tosyl chloride, acetyl chloride ...
I don't know if the compoud is reactive enough, however. But it is an alkoxide and you probably wouldn't even need additional base for e.g. HCl scavenging. Maybe someone here knows more about the reactivity or maybe there are some literature examples?

Alkoxide, unless hydrolysed, acts as a very good base and is even used for the synthsis of ethers. However the alkoxide obtained in this case will have a yield that is quite low since the O- anion forms a salt as OMgX, hindering its ability to attack the alkyl group R-X. Thus, it is generally avoided.

You could always do a transmetallation by treating with excess LiCl and then work up with your alkyl halide.

Transmetallation is very good idea. It can be done without much effort. One can simply even add NaCl and then proceed ahead


text:
Is C4O5 possible?
I have searched all over the internet and I have found no specific answer. I just want to know if C4O5 is possible and what would it be called.

One could write many isomers with such formula. None would be realistic, most  probably.

+1 This seems like a reasonable question and has an interesting answer; I don't understand the aggressive down voting and insta-closing, or how that benefits the site.

Welcome to Stack Exchange @Liam, don't let this bother you. In the future you might add a little more detail to your question and when you say "I have searched... and have not found..." go ahead and include the results of what you did find and explain why it's not sufficient. Usually questions that show some effort are better received than those that don't show it. You may have searched hard but unless you include some evidence, some people will still assume (fairly or unfairly) that you haven't.

@uhoh Not my downvotes, but I guess the concern would be that without any other detail/context, this question could just as easily be "is $\ce{C_xO_y}$ possible?" or really even "is $\ce{X}$ possible" where $\ce{X}$ could be any compound that someone could possibly dream up.

@Tyberius I'm not sure how that's going to be helpful to me, if you have some helpful advice for the OP it might be better to comment to them directly

@uhoh The OP should be pinged automatically as well. I was making a counter point to it being a "reasonable question" in its current form.

@Tyberius do you mean the actual question about the molecule, or do you mean the way it is currently composed as a Stack Exchange question post? I think that the former is quite reasonable and the latter can be addressed with helpful comments addressed to the OP which include ways to improve it. My position on these won't change unless the question is edited.

Tetracarbon pentoxide, $\ce{C4O5}$, does not appear to be known, but Wikipedia reports what has become a surprisingly large variety of oxocarbons.  They include simple carbon monoxide and carbon dioxide, various extended linear structures, single and multiple rings, oligomers and polymers, even fullerene oxides.  Some are surprisingly stable:  mellitic anhydride, $\ce{C_{12}O9}$, can be melted at 161°C, and the powerful pi acceptor $\ce{C_{10}O8}$ is stable in dry air up to 140°C.

C4O4 and C4O6 is known but not C4O5.

C5O5 too, but only in trace amounts.

$\ce{C4O5}$ can be found in a Chinese patent art(Ref. 1). It is called Oxolane-2,3,4,5-tetrone both in the patent and PubChem. In the patent,  $\ce{C4O5}$ is shown as compound 2 and appears to be a compound which undergoes further reaction with itself eventually giving a desired product.
A google translation  of the relative portion of the patent, from Chinese to English follows:

Route 1, the synthetic route is shown below, 2,3-dioxosuccinic acid
(1) is heated in acetic anhydride solvent and passed through the ring
reaction to obtain furan 2,3,4,5-tetraketone (2), compound (2) ) React
with alkylamine in THF and thionyl chloride to obtain
pyrrolidine-2,3,4,5-tetraketone (3) containing solubilizing alkyl
chain[...]

Clearly, compound 2 was synthesized and it further reacted with itself. $\ce{C4O5}$, although obscure, does exist.

References:

CN108976392 - "Conjugated polymer photoelectric material containing
quinoxaline imide condensed ring and application thereof" (link)


text:
Shouldn't the pressure of gas above a liquid slightly decrease when some of its molecules dissolves in the liquid?
Lets say a liquid and gas above are in equilibrium.
(1) Now if I suddenly increase the pressure of the gas it will take some time for some molecules of the gas to enter the liquid and get dissolved. When its finally done and equilibrium is reached shouldn't the pressure of the gas be a little lesser?
(2) Conversely if I violently shake a carbonated water bottle shouldn't the pressure increase a tiny little bit?

In many sources I have read that the pressure doesn't change at all when shaking. It does increase a little bit because of the energy we are giving while shaking.

What I want to know is a detailed explanation of the above two main questions.

Are you asking about obvious facts you mention, or some sources you vaguely mention, that are either wrong, or you misunderstood their meaning - probably about equilibrium vapor pressure, which is constant?

Is the new equilibrium established instantly when i increase the pressure of the gas above the liquid which was already in equilibrium @Mithoron

My question first.

Do you know that there are fluctuations if you see in micro scales? Your equilibrium is undergoing fluctuations. I am asking that if I suddenly change the pressure above the liquid from say 2units to 4 units in my measuring scale, it must take some time for some molecules to get into the liquid and dissolve. That will lower the concentration of gas above the liquid be it very small. And so shoudnt the final pressure be a bit lesser than 4 units @Mithoron

Are you taking about liquid vapor pressure or pressure by air above the surface. Your vessel should be closed (constant volume) to make this any sense.

@Mithoron If I had to guess, I think Shyam is asking (at least in #1) about the dynamics that would occur before the system has fully settled to equilibrium. I.e., it looks to me that they're interested in the case of a rapid increase in pressure, followed by relaxation to a new equilibrium; not one that follows a (minimally ir-)reversible path between states.

1) If you induce a sudden chance in pressure you will indeed take the system out of equilibrium. Lets consider a closed recipient containing some liquid and some gas above. If you suddenly inject more gas into the recipient, the gas pressure is initially going to raise fast and then going to decrease slowly while part of the gas is dissolving, until the system reaches equilibrium again. However, this new equilibrium will have a higher gas pressure AND greater amount of gas dissolved in the liquid when compared to the initial equilibrium. Of the amount of gas you injected part will be dissolved and part will stay above the liquid, contributing to the higher gas pressure. It makes no sense to have pressure under the initial value, this would only be possible if the system hadn't been at equilibrium at the start.
2) Shaking doesn't increase the pressure, unless you shake the bottle long enough to cause a considerable increase in the temperature inside, but that would be quite an impressive amount of shaking. If you could take the system out of equilibrium by releasing some of the dissolved gas from the liquid phase, then that would increase the pressure, but shaking doesn't accomplish that, and I believe it wouldn't be possible to shift a system out of equilibrium like that without transference of matter or heat and without a change in volume or temperature.
There is a very interesting video by Veritasium on youtube about this bottle issue, I recommend you to watch it:
https://www.youtube.com/watch?v=K-Fc08X56R0&t=3s

Why should decrease when it does decrease ?
The gas with liquid  vapour, previously in equilibrium with liquid, with suddenly increased pressure:

Vapour gets over-saturated and condenses until the saturated vapour pressure is reached again.
Gas starts dissolving and gas partial pressure decreases, until the new equilibrium between its partial  pressure and it's molar fraction in liquid is achieved ( Henry's law ).
The kinetics of both processes is about exponential, asymptotically reaching new values of pressures.

Suddenly opened still and shaked carbonised drink bottles:

Shaking the bottle increases its temperature and equilibrium gas pressure by conversion  of  mechanical energy to thermal one via friction. But the increase is very  light and can be neglected for very most cases. Definitely for the difference in visual effects after opening.
Shaking itself does not increase equilibrium nor current pressure. But it does one essential thing. It creates inner liquid-gas contact surface where $\ce{CO2}$ can evolve when pressure suddenly drops.
If pressure in a standing still bottle ( without bubbles ) suddenly drops, liquid gets suddenly over-saturated, but waits for a trigger to burst. Within the volume, gas had no seed where bubbles  can grow. See the related Bubble chamber  The only gas formation is at liquid surface, that does not lead to  visible effects.
If previously shaken, there are already miriads of bubbles that rapidly start growing, leading to violent gas formation.
So it is all rather about the different kinetics.


text:
Why rutile structure have primitive unit cell instead of body centered?
In the literature (Wikipedia), I read that $\ce{MgH2}$ have a structure of rutile. Then, I looked at its space group which is $P4_2/mnm$. From this notation, I understood that it possesses a (tetragonal) primitive unit cell. However, when I looked at the structure of $\ce{MgH2, Mg^2+}$ ions are seemed to be ordered in body centered. Indeed, at each corner there are $\ce{8 Mg^2+}$ ions and in the middle, there is another one. Then, why is this called $P4_2/mnm$ given that it resembles more to body centered? Is there any smaller unit cell hidden here?
Thank you in advance.

it looks like a body centred cubic but it's not, because the sides of the cell are of different sizes. It's a rectangular cuboid not a cube

Its always called bcc (body centered cube), but this is not a cube but tetragonal so be that.

Note that the hydrogens on the top and bottom faces are not in the same positions as the hydrogens inside the cell.

You said you read that $\ce{MgH2}$ has a structure of rutile. Actually it is $\alpha$-$\ce{MgH2}$, which forms at room temperature and atmospheric pressure and have rutile type structure. The crystal structure of $\alpha$-$\ce{MgD2}$ was studied as early as 1963 (Ref.1). The abstract states that:

A magnesium deuteride preparation of composition $\ce{Mg(D_{0.9}H_{0.1})2}$ is found to be tetragonal with $a = 4.5025, c = \pu{3.0123 \mathring A }$. The atomic positions are: $\ce{2 Mg}$ in $(000)(\frac12 \frac12 \frac12)$, $\ce{4 D + H}$ in $\pm (xx0)(\frac12 +x, \frac12 –x, \frac12)$ with $x = 0.306 \pm 0.003$. The bond lengths are $\ce{Mg-6D} = \pu{1.95 \pm 0.02 \mathring A }$. The structure is of rutile type.

The authors state that the choice of $\ce{MgD2}$ in place of $\ce{MgH2}$ because hydrogen atoms had no measurable effects on diffraction intensities (note that they were doing neutron diffraction studies). These studies were a followup to confirm their previous studied on $\ce{MgH2}$ (Ellinger, et al. in 1955; Ref.2). In that studies on $\ce{MgH2}$ with X-ray diffraction, it has been shown that $\ce{MgH2}$ crystallizes in the tetragonal system with $a_\circ = \pu{4.5168 \mathring A}$ and $c_\circ = \pu{3.0205 \mathring A}$. The measured density of the crystals was $\pu{1.45 \pm 0.03 g/cm-3}$, which has shown that there are two molecules in the unit cell (Ref.2). The calculated X-ray density is $\pu{1.419 g/cm-3}$.
The last statement answers your question. The rutile type structure provide two $\ce{MgH2}$ molecules in the unit cell. There are one $\ce{Mg}$ atom in the center (shares with one unit cell) and eight other $\ce{Mg}$ atoms at corners, each of which shares it with eight unit cells. Thus, number of $\ce{Mg}$ atoms in a unit cell is: $1 \times 1 + 8 \times \frac {1}{8}=2$. There are six $\ce{H}$ atoms in the unit cell. If you look closely, you'd realize two of them in inside the unit cell (shares with one unit cell) and rest are on two opposite surfaces, two in each surface (two $\ce{H}$ atoms in each surface share with two unit cells). Thus, number of $\ce{H}$ atoms in a unit cell is: $2 \times 1 + 4 \times \frac {1}{2}=4$. Hence, number of $\ce{MgH2}$ molecules in a unit cell is $\ce{Mg2H4 # 2 MgH2}$.

The X-ray powder pattern shows that two $\ce{Mg}$ atoms in the unit cell are in a body-centered configuration, i.e. $\ce{2 Mg}$ in $(000)(\frac12 \frac12 \frac12)$. A consideration of all tetragonal space groups shows that there is one and only one possible choice of positions for the four $\ce{H}$ atoms leading to a plausible structure. This is,
$$\text{space group }  P4/mnm \ \left(D^{14}_{4h} \right)\\
\ce{2Mg} \text { in } (000)(\frac12 \frac12 \frac12)\\
\ce{4H} \text { in } \pm (XX0)\left(X + \frac12, \frac12-X, \frac12\right) \ \text{with } X=0.306$$
In this structure (rutile type), each magnesium is coordinated to six hydrogens at a distance of $\pu{1.95 \mathring A}$ and each hydrogen is coordinated to three magnesiums. One $\ce{H-H}$ distance is $\pu{2.49 \mathring A}$ and the others are $\pu{2.76 \mathring A}$. The distance of $\pu{2.76 \mathring A}$ compares favorably with the diameter of $\ce{H-}$ ion ($\pu{2.72 \mathring A}$) as found in $\ce{LiH}$. One short $\ce{H-H}$ distance is characteristic of one anion-anion distance of the rutile type structure.

By these statements, it can be concluded that the structure of $\alpha$-$\ce{MgH2}$ is indeed rutile type (not body-centered cube).
In another point of view, one would also suggest that the structure of $\alpha$-$\ce{MgH2}$ is face-centered cube, based on equal distances of $\ce{6 Mg-H}$ (see $\color{blue}{\text{blue box}}$  in following diagram; opposite face $\ce{H}$ atoms are highlighted in blue, red, and gray for visualization). However, again, calculations lead to $\ce{MgH3}$ in the unit cell, which would rule out the possibility.

The above diagram shows the crystal structures of other modifications of $\ce{MgH2}$ along with $\alpha$-$\ce{MgH2}$ (Ref.3 & 4). These other modifications are formed in high pressure situations. For example, when low-pressure phase $\alpha$-$\ce{MgH2}$ was heated at $\pu{1070 K}$ and in $\pu{2 GPa}$ pressure, the high-pressure phase $\gamma$-$\ce{MgH2}$ was formed in the orthorhombic (space group: $Pbcn$) structure (Ref.3):

The high-pressure phase $\gamma$-$\ce{MgH2}$ was formed by heating the low-pressure phase $\alpha$-$\ce{MgH2}$ in a multianvil press at $\pu{2 GPa}$ pressure to $\pu{1070 K}$ for $\pu{120 min}$ and successive rapid quenching. Investigation by X-ray and neutron powder diffraction on the deuteride at ambient conditions revealed that it crystallises with the orthorhombic $\alpha$-$\ce{PbO2}$ type structure (space group $Pbcn, Z=4, a=4.5213(3), b=5.4382(3), c= \pu{4.9337(3) \mathring A }$ (hydride); $a=4.5056(3), b=5.4212(3), c= \pu{4.9183(3) \mathring A}$ (deuteride) at $T=\pu{295 K}$). The deuterium atoms surround magnesium in a distorted octahedral configuration with bond distances $\ce{Mg-D} = 1.915(3), 1.943(3)$ and $\pu{2.004(3)  \mathring A }$. The rutile structure of $\alpha$-$\ce{MgH2}$ was re-evaluated.

The bonding nature of $\ce{MgH2}$ is analyzed in recent studies with the help of charge-density, charge-transfer, electron-localization-function, and Mulliken-population analyses, which clearly have shown that all polymorphs of $\ce{MgH2}$  are to be classified as ionic materials with $\ce{Mg}$ and $\ce{H}$ in nearly 2+ and 1−  states, respectively (Ref.4).
References:

W. H. Zachariasen, C. E. Holley, Jr., J. F. Stamper, Jr., “Neutron diffraction study of magnesium deuteride ,” Acta Chrystallographica 1963, 16(5), 352-353 (https://doi.org/10.1107/S0365110X63000967).
F. H. Ellinger, C. E. Holley, Jr., B. B. McInteer, D. Pavone, R. M. Potter, E. Staritzky, W. H. Zachariasen, “The Preparation and Some Properties of Magnesium Hydride,” J. Am. Chem. Soc. 1955, 77(9), 2647–2648 (https://doi.org/10.1021/ja01614a094).
M. Bortz, B. Bertheville, G. Böttger, K. Yvon, “Structure of the high pressure phase $\gamma$-$\ce{MgH2}$ by neutron powder diffraction,” Journal of Alloys and Compounds 1999, 287(1-2), L4-L6 (https://doi.org/10.1016/S0925-8388(99)00028-6).
P. Vajeeston, P. Ravindran, B. C. Hauback, H. Fjellvåg, A. Kjekshus, S. Furuseth, M. Hanfland, “Structural stability and pressure-induced phase transitions in $\ce{MgH2}$,” Phys. Rev. 2006, B73, 224102 (https://doi.org/10.1103/PhysRevB.73.224102).

Thank you very much for your kind and detailed answer! I just have one final ambiguity about the rutile structure. I understood that there are 2 MgH2 molecules in the unit cell and one short H-H bond in MgH2, which are all characteristic of rutile structure. However, then, by looking at the structure, can we not say any rutile structure is body centered instead of primitive? Really, from the structure, I am not convinced how rutile structure is then primitvie. Again, thanks a lot for your help.

AS authors have noted, there is one and only one way to explain the arrangement of ions in crystal structure, which is rutile type. As I show you, six $\ce{H}$ arrange around one $\ce{Mg}$ such that it could have gotten for unit cell. But, if it is the unit cell, each $\ce{H}$ shares with two unit cells and one $\ce{Mg}$ shares with one cell. Thus, each unit cell contains $\ce{MgH3}$, which is impossible. Do you see my logic?

Yes, I do totally agree with you and see your logic. I see that the only possibility to define the arrangement is rutile type and that MgH3 (i.e. face centered unit cell) is not possible. However, what I am not clear now is why rutile is primitive (simple) unit cell instead of body centered, given that in the unit, we definitely do find another Mg atom (making 2 Mg per unit cell) whereas by definition, primitive unit cell should only have one Mg. I am not sure if my question is clear enough. And also, thank you very much.

Continuing from my last comment, I had a guess that the Mg at the centre of the unit cell actually has a different direction of polyhedra (formed with 6 hydrogen) compared to other 8 Mg ions at the corner, thus it is not contained in the same unit cell. Is my supposition correct?

Not correct. Those 8 magnesium have same environment in next unit cell.

Why rutile structure have primitive unit cell instead of body centered?

For it to be centered, the centering operation has to apply to all atoms, not just the magnesium atom. This is not the case, as the picture demonstrates for one of the hydrogen atoms. The orange arrows show the putative translation operation, which applies for $\ce{Mg}$ but not for $\ce{H}$:

Thank you very much, now this is more clear!


text:
Problem interpreting CV diagram in a paper
I am an undergraduate materials science student. I am currently preparing to conduct some research over the summer on alkaline electrolyte membrane fuel cells and have been reading some papers to get background information (this is a very new area to me). The surface chemistry section of the paper involves several cyclic voltammetry plots, and I am having trouble understanding a part of one of them.
To preface the diagram, the author states:

For the Pd-black-coated electrode, a pair of symmetrical peaks is observed
  between −1.0 and −0.82 V and is assigned to the formation/
  desorption of proton−palladium species...

I interpreted these peaks to be the ones enclosed in the black box below:
Next, the author makes the following statement:

The [proton−palladium species] formation/desorption peaks are smaller for the
  Pd−Ni or Pd−Ni−P electrodes compared with the Pd-coated
  electrode.

To my uneducated eyes, this appears to be incorrect, as at least the Pd-Ni-P peaks appear to be larger than the Pd-blk peaks. How should I go about correctly interpreting this diagram? The paper can be found here, and the diagram and statements in question are in section 3.3: Surface Electrochemistry of Catalyst-Coated Electrodes. Any guidance is greatly appreciated, and I am sorry if my question is repetitive or the answer is obvious. If additional information is needed, please let me know.


text:
Electrophilic attack of X+ on double bond
Is it the double bond attacking the $\ce{X+}$ ion or the other way around?
Also, does it form a cyclic transition state if it isn't bromine or chlorine?
Eg. cyclohexene $+ \,\ce{Cl+ ->}$ cyclic transition state
But would the same thing happen if I put any other electrophile?

These kind of reactions are called Electrophilic addition reactions. These are usually given by alkenes and alkynes. In your example, the electrophile is $\ce{Cl+}$. The alkene will act as a source of π electrons and will donate it to the electrophile which is in need of electrons. So, alkene is called as nucleophile in this reaction.

Is it the double bond attacking the X+ ion or the other way around?

It depends on how you look at it. We can say that alkene polarizes the $\ce{Cl-Cl}$ bond by pushing electrons into the anti-bonding molecular orbital. This weakens the bond and hence electrophile is getting generated. In the next step,the electrophile attacks the double bond [or double bond attacks the electrophile] and hence electrophile gets added across across the double bond
Why does the 3-membered cyclic transition state form?
The cyclic transition state is just a temporary relief for the electrophile that's just been added. The π molecular orbital donates electrons to the electrophile.
While the p-orbitals in the π-bond overlap with the electrophile, each p-orbital forms a σ-bond.

But would the same thing happen if I put any other electrophile?

Generally,electrophilic addition reactions take place through carbocation intermediate.
But this reaction can take place through cyclic 3-membered intermediate,if the electrophile has a lone pair.
Apart from $\ce{Cl+,Br+}$ even $\ce{Hg^2+}$ (when the reagent is $\ce{Hg(OAc)2/H3O+}$) and $\ce{NO+}$ (when the reagent is $\ce{NOCl}$) can form the 3-membered cyclic intermediate.

This is my first answer.Can anyone verify whether what I have written is correct? Please suggest some tips on answering and tell what I can improve on,so that I can write better answers in the future.Thanks :)

Include references if you have them

thanks at lot actually this helped :)

@Waylander,Thanks for the suggestion.Unfortunately,I don't have any such references for this.If you have them, you can add them so that the answer is complete.Thanks

@ganuwoahh Thanks for the feedback.It's nice to hear that it helped you :)


text:
Can PDMS silicone be produced with dichloromethane?
I was taking a simulated admission test for a Technology Institute (Brazilian ITA) and one of the questions asked for the production of PDMS silicone via these described steps:

Reaction of $\ce{SiO2}$ with carbon, freeing $\ce{CO2}.$
Non-volatile product of (1) reacting with dichloromethane in a $\ce{H2}$-rich atmosphere.
Product of (2) reacting with water giving PDMS and $\ce{HCl}.$

The proposed solution was:
$$
\require{enclose}
\begin{align}
\ce{SiO2 + C &-> Si + CO2}\tag{1}\\
\ce{Si + 2 CH2Cl2 + 2 H2 &-> Si(CH3)2Cl2 + 2HCl}\tag{2}\\
\ce{n Si(CH3)2Cl2 + n H2O &-> \enclose{horizontalstrike}{\;(}OSi(CH3)2\enclose{horizontalstrike}{)_n} + 2n HCl}\tag{3}
\end{align}
$$
However, this route does not follow the Rochow process (direct process) with chloromethane on step number 2. This created confusion among the students who took the test because it was hard to know if this route even existed and that $\ce{HCl}$ had to be dispensed on step number 2 to balance the equation.(Weird to come up with this by yourself.)
The question that remained among the confusion is:
Can PDMS be produced via this dichloromethane route?
Haven't found any information directly related to this on the internet.


text:
Energy lost due to stability
We used to say that when an atom is stable it is in its minimum energy state.Every shell has its fixed energy. So from where does the energy come which is lost by an atom to achieve stability.
Do electrons lose energy? If yes then will there be any effect of the lose of energy on the speed of electron revolving around the nucleus.
Thank you
Sorry for the English.

There is no rotation to begin with.

Sorry but, I didn't get what you mean to say

Electrons do not "revolve around the orbit". Don't try to apply Kepler problem at the atomic scale, it doesn't work like that.

@andselisk Sommerfeld was chemistry's Kepler ;)

@andselisk I wrote it by mistake sorry for that. Well I have edited it now.

See, the atoms have the energy they need to have, no more, no less. You conclude that they must have had greater energy (why?), which they later lost, and wonder where did that energy come from. Why, really?

I think you didn't understand my question properly. I am saying that when an atom is stabilised it is in minimum energy state. So I am talking about the difference in the energy states in its destabilised form and stabilised form.

You see an atom in the minimum energy state. Why do you assume it ever was in any other state?

@Ivan Neretin I think I wrote something in this question which makes it confusing. Well I got my answer from. Thanks you for giving me your time. I even think your question about why do I assume can be cleared by reading his answer. He is talking about the same thing as I am.

Let call it rather the atom ground state and the atom excited state.
Same excited states can be kinetically quite stable, like atomized hydrogen in interstellar space with mutual parallel (excited state) and antiparallel ( ground state) proton and electron spin. This spin flipping is the source of the famous 21 cm hydrogen spectral line, used as the scale for the Pioneer cosmic sons plaquette.
Excited atom states have higher energy of electrons than ground states. This energy consists of kinetic and potential energy.
The energy exchange with the rest of universe during excited/ground state switching is via absorption/emission of photons, or via collisions/interactions with other molecules, atoms or subatomic particles.
If we consider classical central force like gravity or electrostatics, we know that lower energy means higher mean speed, as potential energy decrease is twice as much as kinetic energy increase.
In the quantum realm, we cannot speak about particular orbit, momentary values of kinetic or potential energy, nor the momentary position nor velocity nor speed.
We can speak about electron energy, orbital and spin angular momentum
and related distribution of the probability of electron occurrence.

In lithium atom if we anyhow manage to remove the two inner electrons from the s orbital but not interrupting the p orbital electron can we say it is the excited state of hydrogen.

Lithium atom with 2 1s and 1 2p electron already is an excited lithium atom.. If we ejected the 2 inner 1s electrons e.g by energetic photons, it would be excited $\ce{Li^2+}$ ion. Ion with lithium kernel is not an atom of hydrogen, but Hydrogen-like excited 1 electron system with different orbitals due higher kernel charge.


text:
Calculating the biological standard potential
I have an electrochemical cell with following overall reaction:
$$\ce{5 Sn^4+ + 2 Mn^2+ + 8 H2O <=> 5 Sn^2+ + 2 MnO4- + 16 H+}$$
I have defined the half reactions as:
$$
\begin{align}
\ce{Sn^2+ &-> Sn^4+ + 2 e-} \\
\ce{MnO4- + 8 H+ + 5 e- & -> Mn^2+ + 4 H2O}
\end{align}
$$
I have trouble calculating the biological standard electrochemical potential. This is what I've tried:
The standard potential can be found from tables:
$$E^\circ = \pu{1.51 V} - \pu{0.15 V} = \pu{1.36 V}$$
Then I use Nernst equation in which all species except $\ce{H+}$ are in their standard states, so their activities are all equal to $1:$
$$
\begin{align}
E &= E^\circ - \frac{RT}{vF}\ln(10)\cdot \mathrm{pH} \\
  &= \pu{1.36 V} - \frac{\pu{8.314 J mol-1 K-1} × \pu{298.15 K}}{10 × \pu{96485 C mol-1}}\ln(10) × 7 \\
  &= \pu{1.31 V},
\end{align}
$$
which is wrong. Is there something wrong with the calculations or the understanding?

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

Unfortunately, the OP's Nernst equation is incorrect for several reasons. The balanced net redox equation is
$$\ce{5 Sn^{4+} (aq) + 2 Mn^{2+} (aq) + 8 H2O <=> 5 Sn^{2+} (aq) + 2 MnO4^- (aq) + 16 H+ (aq) \tag 1}$$
and n = 10 electrons for the reaction as written. The reaction quotient, Q, is
$$\ce{Q = \frac{[Sn^{2+}]^5 [MnO4^-]^2 [H+]^16}{[Sn^{4+}]^5 [Mn^{2+}]^2}              \tag 2}$$
and the Nernst equation is
$$\displaystyle{E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln(Q) = E_{cell}^\circ - \frac{RTln(10)}{nF} \log(Q)         \tag 3}    $$
The OP's question states that "all species except $\ce{H+}$ are in their standard states, so their activities are all equal to 1". Hence, in this very contrived example, Q reduces dramatically:
$$\displaystyle{Q = [H^+]^{16}             \tag 4}$$
The OP notes that $\displaystyle{E_{cell}^\circ }$ = 1.36 V (from tables). With T = 298.15 K, $R = \pu{8.31446261815324 J mol-1 K-1}$ and $F = \pu{96485.3321233100184 C mol-1}$, then
$$\ce{\displaystyle{ \frac{RTln(10)}{nF} \approx \frac{0.05916 \ V}{n} }        \tag 5}$$
So the Nernst equation is simply
$$\displaystyle{E_{cell} \approx 1.36 \ V - \frac{0.05916 \ V}{n} \log(Q) = 1.36 \ V - \frac{0.05916 \ V}{10} \log([H^+]^{16})      \tag 6}    $$
Hence, using $\ce{pH = -log[H^+] = +log(1/[H^+])}$, the Nernst equation can be expressed as
$$\displaystyle{E_{cell} \approx 1.36 \ V - 16 \times \frac{0.05916 \ V}{10} \log[H^+] =  1.36 \ V + 16 \times \frac{0.05916 \ V}{10} pH      \tag 7}    $$
If pH = 7, then $\ce{E_{cell} \approx 2.02 \ V}.$
It is better to use expression (3) in order to avoid getting too tricky. This is where the OP went wrong: forgetting the "16" exponent on the hydrogen ion concentration and also missing the fact that the negative sign before the pre-log correction term must be subsumed into the pH definition. In a realistic example, it will not happen that everything is in standard state except the hydrogen ion.
What about dividing the net ionic equation, i.e., equation (1), by 16? Then n = 10/16 = 5/8 and all the exponents in the Q expression also get divided by 16. This is no problem if everything is in standard state except the hydrogen ion. Otherwise, it is just facilitating making a silly mistake.

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!


text:
Why does BCF (boron carbofluoride) not exist?
$\ce{BCF}$ (boron carbofluoride, or carbon borofluoride, not to be confused with another so-called BCF molecule that has nothing to do with it) should exist as a vapor molecule with the bonds $\ce{B≡C-F}$, it's simple; but it doesn't seem to exist, or I can't find any information (I just found a heavy thesis where $\ce{H2··FCB}$ —a theoretical molecule with Van Der Waals forces— is computationally studied, but it turns to be quite unstable even with the help of $\ce{H2}$).
Instead, other molecules like $\ce{C2O}$, $\ce{FBO}$, $\ce{NCF}$, vapor $\ce{BN}$, etc, do exist.
Why not $\ce{BCF}$ or $\ce{BCCl}$?

Even C=B-F would be much more viable.

@Mithoron  You are right about C=B-F, now I've read a few papers which state that computationally it is 100 - 200 kJ/mol more favorable than BCF.  Anyway even C=B-F seems to be quite unstable.  So in your opinion, as a thought experiment, if we could engineer to obtain BCF, we would just get C=B-F?  It's a sound idea.

FCB molecule wouldn't survive perhaps even single collision with another FCB, and probably even H2 - instant addition on first occasion. Your FCB would be absurdly strong Lewis acid.


text:
Exothermic dissolution
When I was learning about salts that show exothermic dissolution, I noticed that the same salts used in Daniel Cell( due to similarity of size of anion and cation) were undergoing exothermic dissolution. Is there any relation between the similarity in size of anion/cation and the type of dissolution it undergoes?

There is no easy general pattern.
The dissolution can be formally divided into 2 steps:

breaking the ion lattice into free ions in vacuum
hydration these "naked" ions by water.

Both lattice forming and hydration enthalpies are negative and their absolute value increases with increasing ion charge and decreasing ion radius.
The relation of their values and the overall enthalpy sign  depends on the particular element chemistry, that modifies hydration and lattice enthalpy values.
There are some trends, but hard to generalize.
You can compare sodium and potassium salts of big anions with delocalized charge, like $\ce{NO3-, ClO3-, ClO4-}$. Dissolution of respective potassium salts is significantly more endothermic, compared to sodium ones. Sodium ion is smaller and is more intensely hydrated with lower enthalpy of hydration.
This difference is bigger that difference of their lattice enthalpies. It is probably due anion size,  packing of bigger potassium cation makes less difference than for chlorides.

So there is no way to predict type of dissolution other than looking at their enthalpies?

There is always way of learning empirical facts and patterns.. Chemistry students often search for deriving predictions to avoid it, but chemistry is in many aspects a stubborn old empirical man. It could be predicted by intensive application of quantum chemistry, sometimes relativistic one, but learning empirics is much easier.

For simple ionic salts, the enthalpy of dissolution is given by
$$\Delta H_{\text{disso}} = \Delta H_{\text{lattice}} + \Delta H_{\text{solvation}}$$
where $\Delta H_{\text{lattice}}$ is the lattice dissociation enthalpy (positive) and $\Delta H_{\text{solvation}}$ is the solvation enthalpy (negative).
For exothermic or endothermic dissolution, we have to compare the lattice and solvation enthalpies. Generally, salts with a small cation and anion (think Lithium Fluoride) have very low solvation enthalpies but also have very high lattice enthalpies (given by the Born-Lande equation), hence their enthalpies of dissolution are relatively higher than salts with larger cations/anions, such as Caesium Fluoride.
This graph1 illustrates the comparative hydration enthalpies of fluorides of $\ce{Na+, K+, Rb+, Cs+}$ (graph a) and also the hydration enthalpies of sodium salts of $\ce{F-, Cl-, Br-, I-}$ (graph b). Notice the decreasing trend: dissolution of $\ce{CsF}$ is more exothermic than $\ce{NaF}$ and similarily for $\ce{NaI}$ and $\ce{NaF}$.

Hydration enthalpies for gaseous ions are calculated as described by Latimer et al.2 in his paper.
Further reading:

Wikipedia article on Enthalpy Change of Solution
Chemistry LibreTexts: Enthalpy Change of Solution

References:

Yıldıran, Hüseyin, et al. “A Theoretical Study on Calculation of Absolute Hydration Enthalpies for Some Univalent Ions.” Ionics, vol. 14, no. 6, Nov. 2008, pp. 541–43. doi:10.1007/s11581-008-0214-3.
Latimer, Wendell M., et al. “The Free Energy of Hydration of Gaseous Ions, and the Absolute Potential of the Normal Calomel Electrode.” The Journal of Chemical Physics, vol. 7, no. 2, Feb. 1939, pp. 108–11. doi:10.1063/1.1750387.


text:
What is the "‡" symbol meaning in a reaction mechanism?
I was studying hydroboration from Clayden's Organic Chemistry [1, p. 1281] and the transition state had “‡” symbol in the upper right corner of the activated complex:

We know that this is not the whole story because of the stereochemistry. Hydroboration is a syn addition across the alkene. As the addition of the empty p orbital to the less substituted end of the alkene gets under way, a hydrogen atom from the boron adds, with its pair of electrons, to the carbon
  atom, which is becoming positively charged. The two steps shown above are concerted, but formation of the C–B bond goes ahead of formation of the C–H bond so that boron and carbon are partially charged in the four-centred transition state.

What does this symbol mean? I have seen it before as a superscript at the Gibbs energy symbol.
References

Organic Chemistry; Clayden, J., Ed.; Oxford University Press: Oxford; New York, 2001. ISBN 978-0-19-850347-7.

The symbol is called "double dagger" (sometimes also "double cross") and is used to denote transition state (a maximum in an energy diagram; also often denoted with "*" or "TS") or a related physical property.
Note, however, that a transition state and an intermediate are two different terms.
The symbol has the peculiar origins: as written by H. Eyring, it's entirely the merit of the secretary of the department, Miss Lucy D’Arcy [1, p. 9]:

The symbol, ‡, first appeared in the second paper dealing with the activated complex.
  In that manuscript a star was used to designate the activated state but
  Miss Lucy D'Arcy, the departmental secretary, lacking a star used a plus and minus sign.
  Believing the typesetters would interpret this as a star it was left and by
  their option the cross, ‡, has become the almost universally used sign for the activated state.

Also, note that according to IUPAC recommendations [2, p. 1080]:

In accordance with previous IUPAC recommendations (IUPAC QUANTITIES (1988)) the symbol ‡ to indicate transition states ("double dagger") is used as a prefix to the appropriate quantities, e.g. $Δ^‡G$ rather than the more often used $ΔG^‡.$

In the future, I suggest to use services like Shapecatcher or Detexify to find the name of the symbol.
Finding its usage in chemistry afterwards is normally a trivial task.
References

Eyring, H. Models in Research. International Journal of Quantum Chemistry 1969, 3 (S3A), 5–15. DOI: 10.1002/qua.560030705.
Muller, P. Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). Pure and Applied Chemistry 1994, 66 (5), 1077–1184. DOI: 10.1351/pac199466051077. (Free Access)

The double dagger symbol is just to signify that something is or is related to the transition state of a reaction. It appears above Gibbs energy when it is referring to the Gibb's energy of activation (or the difference in Gibb's energy between the reactants and the transition state). It can also appear above the enthalpy symbol, H, when referring to the enthalpy of activation (or the difference in enthalpy between the reactants and the transition state).


text:
Does the property of additivity for the amounts of different substances hold up for ideal gas law?

A mixture contains $\pu{8.00 g}$ each of $\ce{O2},$ $\ce{CO2},$ and $\ce{SO2}$ at STP.  Calculate the volume of this mixture.  Which of the gases would exert the greatest pressure and why?

Here is my work:
$$
\begin{align}
n(\ce{O2}) &= \frac{\pu{8.00 g}}{\pu{32.00 g mol-1}} = \pu{0.250 mol}\\
n(\ce{CO2}) &= \frac{\pu{8.00 g}}{\pu{44.01 g mol-1}} = \pu{0.182 mol}\\
n(\ce{SO2}) &= \frac{\pu{8.00 g}}{\pu{64.07 g mol-1}} = \pu{0.125 mol}
\end{align}
$$
$$n_\mathrm{tot} = \pu{0.250 mol} + \pu{0.182 mol} + \pu{0.125 mol} = \pu{0.557 mol}$$
$$pV = nRT \quad\implies\quad V = \frac{nRT}{p}$$
$$V = \frac{\pu{0.557 mol}\times\pu{0.0821 atm L mol^-1 K^-1}\times\pu{273 K}}{\pu{1.00 atm}} = \pu{12.5 L}$$
Considering $p = nRT/V,$ the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure. This will be $\ce{O2}$ because it has the lowest molar mass/volume.
Is this how handling the amounts of substance work (can different substances be added like this)?

You are using (intuitively) Dalton's law of partial pressure when you add the moles. Ideal gas law doesn't know the molecular formula. All it considers is the number of particles.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

You are assuming a pressure of 1.00 atm. Please note that, for about 40 years, STP corresponds to a pressure of 100 kPa = 1.00 bar. Also note that the use of the unit standard atmosphere (atm) is deprecated.

The question would be better phrased as which gas contributes most to the total pressure? The contribution is referred to as the partial pressure. The sum of the partial pressures is the total pressure.
The formula you would use to compute the partial pressure $p_i$ of gas component $i$ is either of the following:
$$p_i=\chi_i p = \frac{n_i}{n}p = \frac{n_i}{V}RT$$
where $n_i$ is the amount of component $i$ in the gas and $\chi_i$ its mole fraction. So the argument "the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure" is not quite right. It suffices to say that the gas that contributes the greatest amount of particles will contribute (exert) the greatest pressure. If you mix identical masses of each gas, that will be the gas with lowest molecular mass.


text:
Reaction of acetylene with hypochlorous acid
In reaction of acetylene with hypochlorus acid chloro enol is intermediate, which further reacts with hypochlorous acid to form dichloro  ketone. My question is that why chloro enol does not tautomerise to form ketone. And if the chloro enlo further reacts to form dichloro ketone, then this must be the minor product. And mono chloro ketone should be major product . But thisdoes not happen . Why ?

Do you mean the reaction of ethyne or the reactions of alkynes in general?

The chemistry of HOCl with various compounds is touched upon in this Science Direct piece https://www.sciencedirect.com/topics/chemistry/hypochlorous-acid .

The enol probably does tautomerize but it doesn't last. Some of the enol remains in enol form at equilibrium, and that reacts further with the hypochlorous acid.  This causes some of the keto form to revert to the enol, which then gets consumed again by the hypochlorous acid, and so on.  The selective reaction of the acid only with the enol ultimately pulls all of the keto form back to the enol and then the final, fully reacted product you see.


text:
IUPAC nomenclature of protonated aminoacids

What is the IUPAC systematic name for the above compound.

So I know that carboxylic acids have highest priority. Now the length of carbon chain is 7. So I think the name should be 7-ammonioheptanoic acid. Am I correct?

Since this is a cation, I would suppose there should be "-ium" somewhere. For the neutral molecule, 7‐aminoheptanoic acid is correct IMO.

@andselisk Ig, OP used ammonio- which is used as prefix for $\ce{NH4+}$ more commonly. Name seems correct to me, don't know if it's legitimate name according to IUPAC rules..

@Zenix My bad, you are correct, IUPAC Blue Book from 2013 indeed allows to denote $\ce{-\overset{+}{N}H3}$ with ammonio-.

ChemDraw dubiously says the name is 6-carboxyhexan-1-aminium ...

Interestingly, Ammonium salts come above acids in the seniority table for functional groups (Reference: Wikipedia). Also, the ion in question has an alkyl group attached to it, which makes the suffix aminium rather than the commonly used ammonium. (Reference: Gold Book, Wikipedia)  This means that the correct name of the compound is 6-carboxyhexan-1-aminium. This was also checked with ChemDraw online, as mentioned in the comments.

"suffix" or "prefix" (or am I turned around somehow)?

Suffix=after root word, Prefix=before root word. So in this name, the prefix is carboxy and suffix is aminium


text:
Reactions of diborane, silane, methane, aluminium, boron, sodium, phosphorus trichloride, white phosphorus and iron with water at RT

How many of the following will evolve $\ce{H2}$ gas when dissolved in water under room conditions?
$\ce{B2H6},$ $\ce{SiH4},$ $\ce{CH4},$ $\ce{Al},$ $\ce{B},$ $\ce{Na},$ $\ce{PCl3},$ $\ce{P(white)},$ $\ce{Fe}.$

Initially, it seemed easy to me and I got 7 (all except $\ce{PCl3},$ $\ce{CH4}).$ Correct me if I am wrong here.
However, the answer given was 3, which given were most likely $\ce{B2H6},$ $\ce{SiH4}$ and $\ce{Na}.$
What did I do wrong here? Does the temperature condition hold that much difference here? How do you predict products like these?

Hi Storm CR, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

I have visited them,How do you expect me to make this post better.It would be very much appreciated if you could tell me what I did wrong there^

$$\ce{B2H6 + 6H2O -> 2H3BO3 + 6H2 \\
SiH4 + 4H2O -> Si(OH)4 + 4H2 \\
2Na + 2H2O -> 2NaOH + H2 }$$
The rest do not react with water under ordinary conditions. $\ce{Al, Fe, B}$ react only with steam or when they are red hot. White $\ce{P}$ undergoes easy hydrolysis when coordinated with ruthenium, as illustrated by Vaira et al. [1]. $\ce{CH4}$ does not undergo hydrolysis with water and $\ce{PCl3}$ does not give hydrogen on hydrolysis.
Reference

Vaira, M. D.; Frediani, P.; Seniori Costantini, S.; Peruzzini, M.; Stoppioni, P. Easy Hydrolysis of White Phosphorus Coordinated to Ruthenium. Dalton Transactions 2005, 0 (13), 2234–2236. DOI: 10.1039/B504795A.


text:
Resonance hybrid of nitrosobenzene

In nitrosobenzene, there are two ways that the $\ce{-N=O}$ group can show mesomeric effect.

It can show positive mesomeric effect by becoming $\ce{=N+=O}$
It can show negative mesomeric effect by becoming $\ce{=N-O-}$

So will the electron density on the benzene ring increase or decrease?

I think it should decrease because in the first case we have a $\delta^+$ charge on nitrogen and a $\delta^-$ charge on carbon, and in the other case we have a $\delta^+$ charge on carbon and a $\delta ^-$ charge on oxygen.Is this reasoning correct? Or will there even be such an decrease?

hint: try drawing the resonance structures for the full benzene ring in each case.

$NO$ can show both ±M effect, however -M effect is stronger. It is a deactivating group, yet ortho para directing towards EAS. Have a look at https://chemistry.stackexchange.com/questions/57044/why-is-the-nitroso-group-a-deactivating-group-for-electrophilic-aromatic-substit

Your reasoning is not quite correct. Since electron density on $\ce{N=O}$ bond is more around $\ce{O}$ than $\ce{N}$ (electronegativity of $\ce{O}$ is greater than that of $\ce{N}$), $\ce{N}$ should have a slight $\delta+$ charge, which demands electrons from adjacent $\pi$-system. As a result, the electron density on the phenyl nucleus of nitrozobenzene is less compared to that of benzene. This phenomena is shown as a ring deactivating factor. However, this fact may be hard to prove practically because of the instability of nitrozobenzene.
Yet, in computational studies, one set of descriptors of electron structure of aromatic compounds may come from the pEDA/sEDA approach (Ref.1). The pEDA and sEDA descriptors are defined as populations of the $\pi$- and $\sigma$-orbital electrons, respectively, in a given planar molecule or its planar part (Ref.1 & Ref.2):

The abstract of Ref.2 states that:

An application of quantum chemical modeling allowed us to investigate a substituent effect on a $\sigma$ and $\pi$ electron structure of a ring and the nitro group in a series of meta- and para-$\ce{X}$-substituted nitrobenzene derivatives ($\ce{X =}$ $\ce{NMe2}$, $\ce{NHMe}$, $\ce{NH2}$, $\ce{OH}$, $\ce{OMe}$, $\ce{Me}$, $\ce{H}$, $\ce{F}$, $\ce{Cl}$, $\ce{CF3}$, $\ce{CN}$, $\ce{CHO}$, $\ce{COMe}$, $\ce{CONH2}$, $\ce{COOH}$, $\ce{NO2}$, and $\ce{NO}$). The obtained pEDA and sEDA parameters (the $\pi$- and $\sigma$-electron structure characteristics of a given planar fragment of the system obtained by the summation of $\pi$- and $\sigma$-orbital occupancies, respectively) of the $\ce{NO2}$ group and the benzene ring allowed us to reveal the impact of the substituents on their mutual relations as well as to analyze them from the viewpoint of substituent characteristics. The decisive factor for dependence of pEDA on sEDA of the ring is electronegativity of the atom linking the substituent with the ring; in subgroups an increase of sEDA is associated with a decrease of pEDA. The obtained mutual relation between pEDA($\ce{NO2}$) and pEDA(ring) characteristics documents strong resonance interactions for electron-donating substituents in the para position. The observed substituent effect on the $\sigma$-electron structure of the nitro group, sEDA($\ce{NO2}$), is significantly greater (∼1.6 times) for meta derivatives than for the para ones.

Although, there is no general correlation between these two contributions, pEDA(ring) and sEDA(ring), to the description of the electronic structure of the transmitting ring, it's worth comparing effect of $\ce{NO}$ substituent with the known effect of $\ce{NO2}$. It is clearly the population of the $\pi$-electrons are lower in nitrobenzene when $\ce{NO}$ group substituted in para- or meta-position of it compared to that of second $\ce{NO2}$ group substituted in either position (see $\color{green}{\text{green circle}}$ of the plot). Both $\ce{NO}$ and $\ce{NO2}$ groupd have shown $\pi$-electon withdrawing effects on the ring compared to nitrobenzene itself ($\ce{X = H}$; $\color{blue}{\text{blue circle}}$). Therefore, it is safe to say that the electron density on the phenyl ring decreases in nitrozobenzene compared to that of benzene. This conclusion is supported by the positive Hammett substituent constant ($\sigma$-value) listed for $\ce{NO2}$ substituent in benzoic acid: $\sigma_{p-\ce{NO}} = 0.91$ and $\sigma_{m-\ce{NO}} = 0.91$. By comparison, $\sigma_{p-\ce{NO2}} = 0.78$ and $\sigma_{m-\ce{NO2}} = 0.71$ (The authors referred Ref.3 for these $\sigma$-values).

References:

Halina Szatylowicz, Anna Jezuita, Tadeusz M. Krygowski, “On the relations between aromaticity and substituent effect,”  Structural Chemistry 2019, 30, 1529–1548 (https://doi.org/10.1007/s11224-019-01360-7).
Halina Szatylowicz , Anna Jezuita, Krzysztof Ejsmont, Tadeusz M Krygowski, “Substituent Effect on the σ- And π-Electron Structure of the Nitro Group and the Ring in Meta- And Para-Substituted Nitrobenzenes,”  J. Phys. Chem. A 2017, 121(27), 5196-5203 (https://doi.org/10.1021/acs.jpca.7b03418).
Corwin  Hansch, A. Leo, R. W. Taft, “A survey of Hammett substituent constants and resonance and field parameters,”  Chem. Rev. 1991, 91(2), 165–195 (https://doi.org/10.1021/cr00002a004).

My teacher just told me that whenever a substituent shows both +M and -M effect, we should just take the -M effect. Is that true?

Not necessarily.


text:
Naming of [Pt(py)4][PtCl4]

Why is $\ce{[Pt(py)4][PtCl4]}$ called tetrakis(pyridine)platinum(II) tetrachloridoplatinate(II) and not tetrapyridineplatinum(II) tetrachloridoplatinate(II)?

I thought you use tetrakis for ligands that contain numerical prefixes. I don't understand how pyridine contain numerical prefixes.
I saw this in this site.
The same applies for copper(II) glycinate in IUPAC naming.
Is this because of the property of the ligand?
I know this seems to be a silly question, but I can't understand why this is.

Tetrapyridineplatinum would be one Pt atom with one ligand and only one called tetrapyridine.

Related: How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?

Py and Cl both ligands are monodentate . Therefore, the coordination number of Pt in both the cases are 4 and Pt in general shows +2 oxidation state for 4 coordination number. Again py is neutral ligand whereas Cl is anionic (-1). So, $\ce{[Pt(py)4]^2+}$ is positively charged and $\ce{[PtCl4]^2-}$ is negatively charged. In IUPAC nomenclature, for a coordination complex, the name of cation is written first and of anion at end (cation or anion may be simple or any coordination entity). Therefore the second name, that you told can't be accepted.
Now , the prefixes bis, tris, tetrakis are used-

When the name of the ligand is lengthy or complicated (polysyllabic or multiplicative prefixes are already present in the name of the ligand)
In cases of organic ligand.
In all other cases where using bi, tri, tetra, ... creates ambiguity because in that case a single name may indicate more than one compound .

In case of py first it is an organic ligand and despite that if you use bi/tri/tetra that may indicate a different ligand (for example bipyridine and pyridine are two different ligands) . So, the prefix tetrakis will be used here.

bpy is a ligand, not bipy. Also could you site some references?

Everybody knows the first condition I told  but the second and third condition are nothing but derived from the first . No organic exists without without and numerical term within it and for any nomenclature ambiguity may arise using bi/tri... whenever the ligand is with numerical term or its bi/tri form exists.


text:
Are HCP and Primitive Hexagonal Unit Cell the same thing?
In HCP, the [2D] Hexagonal packed layers are arranged in ABAB form. The same happens in Primitive Hexagonal Unit Cell. So does that mean they are the same? If not what differentiates one from the other?

Hi and welcome to SE. It's unclear what you are trying to ask. Please edit your post to explain clearly what your question is.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

Where is the difference between a crocodile and a dolphin?

@IvanNeretin I made changes to the question. I hope now it's okay.

@Zenix NO! it's not a homework question, edited the question just now. Could you answer the question now??

@AniruddhaDeb I edited the question, is it okay now??

the same happens in primitive hexagonal unit cell

Herein lies the gap in your concept: The same does not happen in primitive hexagonal unit cell. Here's an image to clear things up

The left one is a primitive hexagonal unit cell whereas the right one is a HCP lattice. To use your terminology, a simple primitive cell has packing of the type AAAA.. whereas a HCP cell has packing of the type ABAB..
I highly recommend reading your textbook or referring to some more material on solid state chemistry, such as this SlideShare presentation on basic crystallography.

So from 2d hexagonal packed layers we can make 1) HCP and 2)CCP but not Primitive Hexagonal Unit Cell??

Yes you can. How to do so is an exercise left for the reader.

the second part of the question, can Primitive Hexagonal Unit Cell be made in the same way by arranging the 2D hexagonal layers in AAA type??

The answer to this last question of yours is already here on this page, a few lines above.

There is no such thing as an HCP lattice in terms of classifying symmetry. HCP is a primitive hexagonal structure with two atoms in the primitive cell (one per asymmetric unit). See https://chemistry.stackexchange.com/a/134572


text:
Contradictory Properties of Mercuric Cyanide
It is said that mercuric cyanide [$\ce{Hg(CN)_2}$]  is water soluble and also is a non-electrolyte.
How is this possible?

It does not dissociate into ions? It is not a salt?

Would be nice to add a reference rather than state "it is said..." Just because someone says so doesn't make it true. Then again Maurice's answer seems a pretty convincing explanation of why it might be true.

The simple answer is that it is not a ionic compound.

@BuckThorn I didn't add any references as almost all of the information I found was word-of-mouth and the rest were too vague or indirect to be cited.This paper simply speculates based on experimental data.

Sugar is also water-soluble and a non-electrolyte.  Most acids that are water soluble are only weak acids and thus close to being non-electrolytes.  What's the fuss over this one case out of many?

Mercury cyanide behaves like Mercury chloride $\ce{HgCl_2}$. Both are soluble in water, and both do not dissociate in water. These compounds are not salts. The bonds $\ce{Hg-Cl}$ or $\ce{Hg-CN}$ are more covalent than ionic. When the atom Hg is included in compounds it often behaves as a non-metal.
This is due to a relativistic effect. In all elements with atomic number higher than $70$, the outer electrons move at a speed that is not far from c, the velocity of light in vacuum. Of course everybody knows that electrons do not move like planets in the atom. But the relativistic corrections are significant "as if" the electrons were moving. And one of the effect due to relativity is that the length decreases at velocities going near to c. Pekka Pyykko has shown that this entails a contraction of the relativistic electron orbitals. And this contraction is proportional to the sum of the two quantum number n and l.
Mercury has the configuration {$\ce{Xe}$} $\ce{4f^{14} 5d^7 6s^2}$. The three external orbital have a sum n+l equal to $8$ for $4$f, $8$ for $5$d, and $7$ for $6$s. So the electron $4$f and $5$d (with n+l = $8$) are so much contracted that they nearly disappear in the internal core of electrons. It must be nearly the same for the $6$s electrons. And the mercury atom looks a bit like Xenon when seen from outside. That is why it is "nearly" a noble gas : its boiling point is rather low with respect to its neighboring atoms.
All the chemical properties of Mercury are due to its two $6$s electrons which are rather difficult to extract, as they are partly included in the inner core. These electrons "prefer" making covalent bonds rather than ionic bonds, when included in molecules with $\ce{CN}$ group or $\ce{Cl}$ atoms.
Pekka Pyykko, Relativity and the Periodic System of Elements, Accounts of Chemical Research, Vol. 12, No. 8, (1979), p. 276 - 281.

@Poutnik. All I can say is that your interpretation is not Pekka Pyykko's opinion.

"Dirac's opinion on the role relativistic quantum mechanics would play for chemical systems is wrong for two reasons. First, electrons in s and p atomic orbitals travel at a significant fraction of the speed of light. Second, relativistic effects cause indirect consequences that are especially evident for d and f atomic orbitals."
Pyykkö, Pekka (1988). "Relativistic effects in structural chemistry". Chemical Reviews. 88 (3): 563–594. doi:10.1021/cr00085a006.

@Poutnik So you are saying that the importance of Quantum/relativistic effects is overstated in Chemistry?

@Alan Whitteaker Not at all. How did you came to such conclusion ?

@Poutnik I thought so because of the way your comment was structured.

@Alan Whitteaker Well, non relativistic quantum computation of mercury melting point provides the result +80 Deg C. Relativistic one has the error within 1 Deg C. Make your conclusion.

"In all elements with atomic number higher than 70, the outer electrons move at a speed that is not far from c, the velocity of light in vacuum.".  No.  It is the innermost 1s electrons that acquire significantly relativistic mechanics.  The effect of that condition then cascades into the quantum states of all the other electrons even though the latter are not themselves strongly relativistic.

@Oscar Lanzi. Pekka Pyykkö says exactly the contrary in his publications. He says that the outer electrons are "buried" in the inner core of the heavy atoms. And this explain why Mercury is "nearly" a noble gas (boils at low temperature, produces dimeric $\ce{Hg_2^{2+}}$ ions, like noble gases), why Thallium has often +1 as oxidation number (like alcaline metals, may replace K in alums) , why Lead often behaves like an alkaline earth metal ( Oxidation +2, insoluble PbSO4), etc.

Yes the outer (s) electrons are buried, but those are not the relativistic ones.  Outer electrons, even if they are buried, are not buried deeply enough to draw as much energy from the electrostatic field as their own mass energy, which is what would make them relativistic.  The really buried inner s electrons are the relativistic ones.

@Oscar Lanzi. If the outer electrons are buried in the core of the atom, that is a proof that they are relativistic. If they were not relativistic, they would have remained in the outer layers.

Cannot come to agreement and I don't want to be dinged for an extended discussion.  So I'm out.


text:
Illegal valence meaning
I'm using the rdkit package to process some molecules.
I noticed that when creating a molecule from a SMILES string the library does some syntactic and some semantic checks. This question regards the semantic part (sanitization). (Just in case someone wants to take a look, the checks are described in detail here).
In the sanitization process the explicit and implicit valence of all atoms is computed and it fails when atoms have illegal valence. I don't really understand what "illegal valence" mean in this context. I tried using some compounds with charged atoms to see if they have "illegal valence" but they didn't. Examples: Cc1cc(SCC(=O)c2ccc(Br)cc2)cc(C)[o+]1, O=[V]([O-])([O-])[O-] where the Oxygen in both cases is charged (once positively and once negatively).

What does "illegal valence" mean? Is it just using more electrons than the ones actually available in the atom? e.g. 2 bonds with a hydrogen.

Furthermore, if the previous is true, I suppose that even combinations that are potentially unstable (in practice) with a lot of charged atoms would still be considered "semantically valid"?

Thanks!

The positive oxygen charge is tripping your program. I noticed a similar thing happening in ChemDraw online.

What do you mean by "tripping"?

Are you able to share the molecules which produced the error? It's difficult to say whether this is a bug in the RDKit or just an incorrect SMILES string. 'Illegal valence' is likely to be a reference to an incorrect number of valence electrons.

No molecules produced any error. I'm trying to understand when this error is triggered :)


text:
Hyperconjugation - concept clarification
As I currently understand it, Hyperconjugation occurs in carbocations and alkenes via the interaction of the electrons in the MO of a C-H bond at an adjacent carbon to bring stability to the system.The interaction requires for in phase overlap and occurs when the C-H bond is in the same plane as the vacant p orbital. (or the pi bond in case of alkenes) It is thus not seen if the beta hydrogens are in the plane perpendicular to them.
If this is true, then how does hyperconjugation work, if at all in cyclic systems where free bond rotation is impossible and thus the orbitals will never align perfectly? What conceptual bridge am i missing here?

Cyclic systems aren't planar. The system orients itself in such a way that the vacant p orbital is in plane in order to gain the ability to engage in hyperconjugation which stabilizes it very well.


text:
Ortho effect in cyclohexane

Compare the acidities and basicities of the compounds in the first and second
  row respectively.

In aromatic compounds the methyl group would cause an "ortho" effect. Here we don't have an aromatic compound. Would there still be something like an ortho effect?

I think you can answer your own question if you look at the reasons behind the ortho effect in aromatic compounds.

We could assume ring flipping.

I think I got it, but I might be wrong so be prepared.
There will be difference because of the way the carbons in the ring are hybridized. An sp2 hybridization state would make conformations impossible. An sp3 carbon allows both axial and equatorial positions for each group. So there would be no way to ensure that the bulky groups always interact.

The ortho effect is an effect that is closely linked with planarity. Only aromatic/anti-aromatic rings are planar and only the groups attached to aromatic rings can stabilise themselves through the resonance effect. However, in non-aromatic rings like cyclohexane, there is no planarity. The carbons are also sp3 hybridised. So the interaction between the bulky groups will not matter anyways.


text:
Bond rotation energy comparison of aromatic compounds

  Which of the above compounds has a lower bond rotation energy?

Here is my progress: We basically need to find which compound has more single bond character. Now both compounds will break the double bond, and give the electron pair so as to make both cycles aromatic. But what do I do after that?

related https://chemistry.stackexchange.com/questions/95249/do-there-exist-aromatic-compounds-similar-to-azulene

Huh, there was this exact case recently, but it might be deleted now. I wager that double bond to cyclopropenyl has more unfavorable angles (60 when 120 deg. is usual).

Tropylium cation(the top ring on the right image after bond rotation) is EXTREMELY stable. Probably the most stable carbocation, among the "usual" examples. It's even more stable than cyclopropyl methyl carbocation.

@user600016 Is there a reason to why it is so stable?

Highly conjugated  aromatic system. PS: take a look at. https://chemistry.stackexchange.com/questions/48264/stability-comparison-between-the-tropylium-and-tricyclopropylcarbinyl-carbocatio

Cleavage of the double bond will lead to both compounds having two rings that are aromatic, however the amount of resonance energy(hence stability) will not be the same.
Both systems will have a Cyclopentadienyl anion, but one will have a Cyclopropenium cation
while the other has a Tropylium cation . As the Tropylium cation has more resonance structures, it will have a higher resonance energy and hence greater stability than the Cyclopropenium cation. So (b) has a lower bond rotation energy.