text:
Why do large cations stabilise large anions and small cations stabilise small anions?
I see this trend from enthalpy change of decomposition data, though I struggle to find a decent explanation for why.

Are you asking this with regards to the dissolution of inorganic salts?

related https://chemistry.stackexchange.com/questions/10273/hardness-and-softness-are-terms-used-to-describe-which-aspect-of-acidic-behaviou

For more detailed info, search "cation-anion radius ratio" or "Pauling's rules". Below is my review of this idea.
This ratio is ${R_{C}/R_{A}}$ (C = cation, A=anion).
Small cation (I mean, too small) attracts anions in such a way that they come too close to each other and repulsive force comes into play. This happens when ratio is <0,155. So for larger anion, larger cation suits better. Ratio can be 0.155-1 for stable compounds, and this ratio also determines coordination number and type of void! (Check the picture below, recall that since <0.155 ratio is for unstable compounds, you have empty "example" line there).


text:
Why do we use Δ to mean heating?
To express heating, we usually use $\Delta$ over reaction arrows like
\begin{align}
\ce{CaCO3 ->[\Delta] CaO +CO2}.
\end{align}
Why do we use $\Delta$ to mean heating during chemical reactions? I thought up with one idea that $\Delta$ symbolizes fire. But actually I have no information about the origin of this usage. Does anybody know the origin or reason of this usage of $\Delta$?

Good question, I would like to know how widespread the triangle is as symbol for heating

It looks like a Bunsen burner flame. It used to be below the arrow.

adding heat does not mean an increase in temperature it means to define a temperature and run the reaction at that temperature. Reactions are usually run at constant T and P.

@user55119- Just as $\ce{CaCO_3}$-> $\ce{Ca(HCO_3)}_2$-$\ce{H_2CO_3}$ or $\ce{Ca(HCO_3)}_2$-$\ce{H_2O}$-$\ce{CO_2}$

https://chemistry.stackexchange.com/questions/105105/why-do-we-use-%ce%94-to-mean-heating

@Karsten♦- I want you to put a space between "or" and the symbol of Calcium Bicarbonate in  the second half of the equation where you do break carbonic acid into water and carbon dioxide, no need to give it in the first half of the equation where you do not break carbonic acid into water and carbon dioxide, in one comment which I gave as a reply to @user55119 yesterday, on doing the needful in that one, you will be able to delete this recent comment which I am posting now as well as the comment which I posted 16 minutes ago

The retrieval of this symbol history poses a considerable challenge, as it is scarcely mentioned in textbooks, Google Books and Google Scholar. The current application of the triangle symbol, distinct from the capital Greek letter delta, can be traced back to the mid-1700s in pre-modern chemistry. This is evidenced in Torbern Bergman's 1735-1784 work, "A Dissertation on Elective Attractions," accessible via the Internet Archive. At the end of the book, originally written in Latin, there is long list of symbols with explanations as shown below. In this context, the inverted triangle represented water, while the upright triangle symbolized the "matter of heat," aligning with symbolic alchemy. Wikipedia has a nice page on Alchemical Symbols. For these reasons, it will be impossible to assign a single person behind this notation.

Here are the tables: http://www.chymist.com/Torbern%20Bergman.pdf

And here the entire translated thesis: https://archive.org/details/adissertationon00berggoog/page/n93/mode/1up

Apparently, arrows weren't used until about 100 years later: https://www.chemistryviews.org/details/ezine/2746271/History_and_Usage_of_Arrows_in_Chemistry/

The reference establishes the triangle as a reaction condition (at high temperature, typically in a crucible), distinct from reactions in aqueous solution (designated by a triangle with tip pointing downward).

@Karsten, The history of arrows is also interesting. Thanks for the links

@AChem Thank you for your answer. It's quite intriguing. I'm also intereted in the temporal gap between the usages in alchemy and modern chemistry.

It can be mentioned that the triangle was used as the Alchemical symbol for fire in ancient times. However I don't know if it has been in continuous use as a reaction-symbol that way since, or if it fell out of use and the modern symbol is taken from Delta and is similar only by random coincidence.

This explanation makes much more sense than the link to $\Delta H$, which could mean exothermic or endothermic, and is only loosely connected to heat transfer. I would be nice to find a reference, though.

Another related link, I'll let somebody else judge on its reputability https://www.eoht.info/page/%CE%94

@Karsten: Alchemist's symbol for fire.

@user55119 I have found reputable sources documenting the alchemists symbol for fire. The question is whether this is a coincidence or the symbol with the arrow is meant to symbolize fire.

@Karsten - I wonder if it would be more likely to get a documented answer at History of Science SE? I tried to flag it to be migrated there, but it's too old to migrate. Maybe as a moderator you can override that (if you agree)?

@Andrew I guess I could post a new question there, more focused on when the triangle symbol first appeared in a chemical equation, and referencing this question. Ian Bush mentioned a wacky reference "eoht.info" that makes some specific claims, but I don't have access to the historic sources to check.

@Karsten Another symbol for heating I saw over reaction arrows is a circle with a  diagonal line crossing it

@Karsten Thank you very much for finding the interesting history of the use of Delta in alchemy. I didn't expect my question would see any progress. I'm glad that you post a more specific question about the history of the symbols.

Gosh folks! Way back when the only convenient laboratory heat source was a candle or Bunsen burner the symbol for heating was an isosceles triangle placed below the reaction arrow. It symbolized the shape of a candle or BB flame. With the advent of typewriters and computer fonts a symbol was needed; the equilateral triangle was the best available so the isosceles morphed into an equilateral. That it happens to be a Greek delta and is chemspeak for change is a coincidence.  My only source is that I am old enough to have lived thru all this nonsense.

Thanks for the downvotes. Unfortunately all the high school teachers and college professors who told us this are all dead.

Δ symbolizes difference. For chemical reactions, it's usually difference between energy that reactants already have and energy needed for reaction. Then Δ = $E_a$ - activation energy of the reaction. Activation energy is like a "potential barrier" for reaction to happen. And Δ means that we've added this amount of energy so energy of reactants changed by Δ (difference).

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How to predict molecular geometry of acetylene using VSEPR model?
I need to determine the molecular geometry of acetylene. For this I performed the following procedure.
First, I drew the correct Lewis structure which is a graphical representation that shows the pairs of electrons linking the atoms of a molecule and the pairs of solitary electrons which may exist.
To predict molecular geometry I used the VSEPR model, which is based on the number of electron pairs in the central atom to determine molecular geometry. In this case, there are two central atoms: $\ce{C}$ and $\ce{C'}$ (the prime symbol is used to distinguish between carbons).
$$\large\ce{H-C#C'-H}$$
The first carbon $\ce{C}$ has 4 pairs of electrons, all of which are bonding. Using the VSEPR notation, $\ce{AX4}.$ For the second carbon $\ce{C'}$ it's exactly the same. Therefore, molecular geometry is symmetrical and tetrahedral.
For a complete study, I applied the valence bond model based on the hybridization of the atomic orbitals. This is where I don't know how to see if the link is σ- or π-type.

Also, I know that the molecule can be contained in a plane, but I don't know how to explain it using VSEPR or valence bond theory.

This is one of the examples where VSEPR theory breaks down; it simply cannot be used with ethyne.

How do I know what's wrong with VSEPR theory a priori? For example, in an exam I am asked to determine this structure, how to do it so that it is always the right one. @Martin-マーチン

I would disagree with Martin, the VSEPR theory works for things like benzene, ethylene and acetylene, but it is unable to estimate the number of σ-bonds and lone pairs. The chemist must do this, then the VSEPR can work out the arrangement of the σ-bonds / lone pairs. The problem is that unless you understand pi systems the use of VSEPR might suggest that the torsion angle $\ce{H-C-C-H}$ in ethene is not zero or 180°. In real life the atoms of ethylene are in a single plane.

I don't know how to see if there are pi links (multiple links) or sigma links (single link).

That something doesn't apply is clear. How two atoms can be linked together in a tetrahedral geometry?

What is important is the distribution of electrons about the nuclei. If H2O is considered tetrahedral and the structure of ice suggests that, then what is the arrangement of the electron pairs about the hydroxide ion?

The first carbon $\ce{C}$ has 4 pairs of electrons, all of which are bonding. Using the VSEPR notation, $\ce{AX4}.$ For the second carbon $\ce{C'}$ it's exactly the same. Therefore, molecular geometry is symmetrical and tetrahedral.

Carbon typically has 4 pairs of electrons which are bonding. However, these four pairs of electrons could be in single, double or triple bonds. Electron pairs in a double or triple bond connect the same two atoms, so they have to be grouped together in the VSEPR scheme. For ethylene, the notation would be $\ce{AX2}$, which is expected to be linear.

For a complete study, I applied the valence bond model based on the hybridization of the atomic orbitals. This is where I don't know how to see if the link is σ- or π-type.

The Lewis structures are a way to illustrate the valence bond description as far as possible. So a triple bond in the Lewis structure is one sigma and two pi bonds in the valence bond formalism.

[from the comment] How do I know what's wrong with VSEPR theory a priori? For example, in an exam I am asked to determine this structure, how to do it so that it is always the right one.

You would measure or deduce the bond lengths and angles with an appropriate experiment, and see whether your data fits your prediction. The predictions work pretty well for some compounds made of H, C, N, O atoms, less so for third row main group elements, and they mostly break down for transition metals. Also, the predicted angles mostly will be approximations (except in the case of high symmetry, e.g. methane or acetylene).
In an exam, you just have to hope that the examples given work well for the level of theory (or the rules of thumb) you are applying.

[OP's comment in another answer] I know how to apply the VSEPR model: I determine the Lewis structure and observe how many electron pairs the central atom has. I also understand the concept of hybridization and overlapping of atomic orbitals (simply the algebraic combination of wave functions). But, if I am given a molecule (such as acetylene, ozone or ammonia) I don't know how to apply the [valence bond] Theory correctly, I get confused when it is a sigma link or pi link. So, my doubt was focused on that aspect.

A single bond in a Lewis structure is considered a sigma bond. For multiple bonds, one is a sigma bond and the remainder are pi bonds.

single bond: one sigma bond
double bond: one sigma and one pi bond
triple bond: one sigma and two pi bonds

This assumes that the Lewis structure is a good representation of the molecule. If a Lewis structure shows an "expanded" octet (or for transition metal complexes), you are better off looking at a molecular orbital diagrams to determine what the bond order is and whether the bond in question is sigma or pi (whether there is a node along the bond axis).
The classic example of the Lewis structure breaking down is the dioxygen molecule, which is often written with a double bond as a Lewis structure but experimental evidence indicates that there are two unpaired electrons involved in bonding.

For more detailed info, check "hybridized atomic orbitals". You will need Molecular Orbital Theory.
This is
 . In excited state, one electron from 2 s goes to empty 2p, and you will have 1 electron on each 2-shell orbital.
Triple bond is formed using sp-hybridized atoms (one s orbital is "averaged" with one p orbital). These "averaged" s and p are used for sigma-bonds (2 bonds, since there are 2 orbitals), and two p-orbitals are used for pi-bonds (also 2 bonds). So each carbon has 2 sigma (with C and H) and 2 pi (with C both) bonds. It's clear from your first picture.
I've found a good picture for you, which is below. It explains, why acetylene is linear molecule (hence, it's molecular geometry).

The reasoning here is in reverse, and wrong with that. The molecular structure determines which type of hybridisation (if that model is applicable at all) is used. And hybridisation is not necessary at all to describe the molecular structure of ethyne, it simply is much more convenient to understand the bonding. Also please be more careful with the language you use; atoms cannot be hybridised, only orbitals can. As such, the term 'sp-hybridised atom' is a sloppy and incorrect colloquialism. It will give the wrong impression especially to new students.

David Klein in his 3rd ed. of "Organic Chemistry" uses this term. I also believe that understanding of hybridization can help us imagine geometry of molecule, because it's easy to do with those orbitals, and since question is about geometry, I've started with hybridization and ended with geometry. If something can be explained in easy way - why not? But I agree that my answer is far from ideal, so would you be so kind as to give a proper explanation?

Yes, unfortunately the term gained popularity with many organic chemistry text books. It still is incorrect. The main problem with the interpretation you are presenting is that it is coincidentally predicting the correct molecular structure, but for the wrong reasons. This understanding of hybridisation, is wrong, and will lead to plenty of problems for other molecules. Predicting the linearity of ethyne is nothing for the comment section as there is no easy model that gives this conclusion for the correct reasons.

Usually tasks like "predict molecular geometry of very small molecule" are school|college tasks which can be easily solved with "sp3 is for tetrahedral, sp2 is for triangular planar, sp - linear". I even dare to say that some school|college courses don't even mention that bond angles can differ from 180-120-109.5, so I think that in this case hybridization theory can be applied. Tasks like this are often use to master good old "sp3 is for tetrahedral, sp2 is for triangular planar, sp - linear".

I am well aware of this, and it makes me very, very sad. It is wrong, intrinsically wrong; and just because a plethora of schools refuse to update their curriculum to include newer evidence, we don't have to spread these wrong theories even further. If students don't start with understanding where the limitations of a model system lie, then why bother teaching them anything at all. We are dealing with those misconceptions on a regular basis here, we don't have to add to the pile promoting the falseness.

If you don't mind, could you, please, name a theory or concept that explains molecular geometry better?

I know how to apply the VSEPR model: I determine the Lewis structure and observe how many electron pairs the central atom has. I also understand the concept of hybridization and overlapping of atomic orbitals (simply the algebraic combination of wave functions). But, if I am given a molecule (such as acetylene, ozone or ammonia) I don't know how to apply the Valencia Link Theory correctly, I get confused when it is a sigma link or pi link. So, my doubt was focused on that aspect. @KellyShepphard

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text:
Melting point vs Freezing point
Is it possible to have Melting point different that Freezing point
I mean is there any element/molecule in the liquid status that freeze at point X Centigrade but when it become solid then you need a higher temperature to become liquid again like X + Y degrees in the same conditions?

https://www.thoughtco.com/melting-point-versus-freezing-point-3976093
Short and nice answer to your question. Hope this helps.

https://chemistry.stackexchange.com/questions/53348/difference-between-freezing-point-and-melting-point

In general I believe such an asymmetry in direction of a process is called a hysteresis, you may be interested in reading about it.

That's very common, actually.
https://en.wikipedia.org/wiki/Polymorphism_(materials_science)#Ostwald's_rule
A substance melts, but the recrystallisation from the iostropic melt can be kinetically inhibited. If you cool further down, a less stable form of the same substance, with lower melting point, can crystallise, and re-form over time, into the original phase.
This (according to Ostwalds rule) happens practically every time a suitable, less stable polymorph exists. Or let's say there is a general tendency for it to happen.
Another, very common example (to which Ostwalds rule also loosely applies) are polymers. Because of their chain structure, crystallisation (if possible at all) is much inhibited, and usually never occurs until you cool a good ten K or so below the equillibrium melting point.

Sulfur is such a substance. At room temperature, it is made of stacked up $\ce{S8}$ octogonal molecules. It melts at $119$°C producing a yellow liquid made of free moving octogonal molecules.
When heated in a test tube to $160$°C, the yellow liquid turns reddish : the octogones are suddenly broken : the liquid is then made of open $\ce{S8}$ chains, that are starting polymerization. At $187$°C, the liquid is brown and suddenly so viscous that the test tube can be turned upside down : the liquid will not flow and get out of the tube. The stuff is made of a polymer containing long chains of up to $20'000$ $\ce{S}$ atoms.
When heated to about $240$°C, the polymerized stuff is dark brown and is melting again. The brown liquid can then be dropped directly into a big beaker filled with water. The brown and cold sulfur falls down to the bottom of the beaker. It is soft and can be easily picked up and even kneaded by the fingers without difficulty. At the points of kneading, the polymer chains are broken into 8-atomic segments that are quickly transformed into octogones : the stuff becomes yellow and hard at these points. It is one of the rare cases where chemistry can be modified by hand.
I am a high school teacher. This demonstration has been carried out by more than $3000$ children and students, without any problems. They all love this demonstration. So do I.


text:
Addition reaction of alkene

Correct option is option c but option d also forms  same compound after 1,2hydride shift.So why only option a and b is possible why not d also??

Bro it seems to me that your doubt is correct ..... But actually the shift that should explain your doubt  is a 1,2 hydride shift and not a methyl shift . The only explanation to explain the answer of the module is that we don't consider the hydride and methide shifts but that won't be justified so I suggest you to consult your teacher

@AdityaGarg I just want to know that is there any condition that hydride shift will take place only from primary to secondary carbocation or it also do the same for shift of secondary to tertiary carbocation

No there isn't such a thing

No not necessarily

My answer explains why option D does not produce better yield of 1-chloro-1-methylcyclohexane

As you can see 1-methylcyclohex-2-ene reacts with $\ce{HCl}$ to give two carbocation intermediates one has more hyperconjugation structures and other having less.

Increasing number of hyperconjugating structures meaning more stable carbocation and number of hyperconjucating structure depends on number of alpha hydrogen present in the carbocation.

Therefore, 1-chlorocyclohex-2-ene yield major product 1-chloro-3-methylcyclohexane and minor product is 1-chloro-1-methylcyclohexane

Hence option D is not correct, While the other two reactant yields major product 1-chloro-1-methylcyclohexane see image below

from above discussion we can say that option A and B will both produce good yield of 1-chloro-1-methylcyclohexane so andwer is option C


text:
Addition reaction(Haloalkane)

In these reactions if two reagent is given how to decide who is going to react with the compound.HBr is a strong acid
CH3OH is a weak base.So HBr dissociate easily but CH3OH is reacting with the given compound. How is this possible??

Who said in the reaction ch3oh has to behave as an acid. It is acting as a nucleophile . It donates oxygen's lone pair to the vacant orbital of the carbonation intermediate  forming an oxonium cation which then looses a molecule of H+ to form the product
PS:  to form the carbonation intermediate the H+ was given by HBr which was given back during the last step of the rxn

No doubt that, HBr is strong and it would disassociate into $\text{H}^{+}$ and $\text{Br}^{-}$ but methanol is a solvent in the second reaction and it would attack the Carbocation because of higher concentration and it's a better nucleophile (-ve charge density on Oxygen atom is high) and then the Hydrogen atom would deprotonate.
One more question, why in the first reaction options, $\text{Br}$ is missing, as $\text{HBr}/\text{H}_{2}\text{O}$ is $\text{HOBr}$ in reaction.


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Vanishing integral rule for two-electron integrals
I understand the vanishing integral rule for a one-electron integral to be:
$$\langle i | \hat{O} | j \rangle = 0 \hspace{1cm} \mathrm{if} \hspace{0.2cm} \Gamma_O \neq \Gamma_i \otimes \Gamma_j$$
For electronic repulsion integrals, the Coulomb operator is totally symmetry, so the product must contain the totally symmetric irrep (TSIR). However, I am not sure what form the vanishing integral rule will have. Is it simply
$$\langle ij | kl \rangle = 0 \hspace{1cm} \mathrm{if} \hspace{0.2cm} \Gamma_\mathrm{TSIR} \neq \Gamma_i \otimes \Gamma_j \otimes \Gamma_k \otimes \Gamma_l$$
or does it have a different form? Additionally, will a Schwarz-inequality matrix composed of:
$$Q_{ij} = \sqrt{\langle ii | jj \rangle}$$
have an identical block-diagonal structure to the one-electron matrices constructed in this fashion?

I would agree that the product of irreps should contain the TSIR if it is not to be zero. Will not your Q always be totally symmetric for all i and j.

this should be somewhat more complex, then just the product of all 4 irreps, since there are 2 different electronic coordinates involved. The argument with the vanishing integral always goes over just one coordinate at a time. (I can try for a more detailed answer later)

Ok, it actually seems to be as simple as taking the product of the 4 IRREPs. However, I could not find a decent reference with a detailed derivation. There is though this wikipedia article and a paper mentioned there (although in a slightly different context) DOI: 10.1002/qua.560050606

See this: https://mattermodeling.stackexchange.com/q/1684/5


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Why is the Infrared spectrum usually drawn using transmittance rather than absorbance?
Infrared radiation can be used to identify compounds but I noticed that the spectra are usually drawn using percentage transmittance rather than percentage absorbance . Why is that ?

One reason: you identify weak transitions more easily by looking at transmission (more light transmitted than absorbed), and if you're normalizing spectral intensities, those weak ones get washed out if you went by absorption (more intense lines for strong transitions).

The original experimental measurement is of transmittance, so why mess with it?


text:
Dichlorine monoxide molecular geometry
I need to predict the geometry of Dichlorine monoxide, using the main link theory: Lewis model, VSEPR and hybridization of molecular orbitals.
First, the Lewis structure is a graphical representation of the pairs of electrons bonding between the atoms of a molecule and the pairs of solitary electrons, which may exist.

We proceed to predict the geometry of the electron and molecular pair of the chemical species, using the VSEPR model, which is based on the number of electron pairs of the central atom (oxygen) to determine it.
As we have previously stated, the central atom is oxygen. The number of electron pairs of oxygen is four, of these four: two are shared and two are bonding. Therefore, electronic geometry is tetrahedral and molecular gometry is angular:

To conclude, we will explain geometry by hybridization of atomic orbitals. Orbital hybridization is simply the linear combination of the mathematical expressions of the wave functions of the orbitals. We know that there are two types of overlaps:

Sigma or frontal bond.
Pi or lateral bond.

Here I am asked to study the different overlaps that occur and to explain the consequences on the geometry of that overlap. I suppose there will only be sigma overlap because there are no multiple links. But I can neither say nor explain with certainty what type of overlap and between which electrons or atoms there are and, finally, explain the relevance of this in geometry.
Likewise, ideally the angle should be 120º, but experimentally it is 110.4º. This decrease is due to the presence of two pairs of solitary electrons in the oxygen atom, isn't it? Is there any way to determine such an angle other than in the laboratory, any formula or relationship?

Welcome to chemistry.se! Are you familiar with molecular orbital theory? You have applied VSEPR theory correctly, but unfortunately within this molecule bonding is not as simple. The two lone pairs at the oxygen will not be equal, similarly to the way they are not equal in water. The best explanation for the different angles is Bent's rule. There will also be pi-backbonding from the chlorine to the oxygen.

I understand the theoretical concepts, i.e. I know that a sigma link (occurs in single links) and a pi link (occurs in multiple links). The problem is that I don't know how to apply these concepts to a practical case, like this one. Normally, I am asked for a hybridization scheme (this is to indicate which orbitals overlap, between which electrons the sigma link or pi link is given and to explain how this influences geometry). I've seen several tutorials on the Internet, and none of them get to be very good. I was hoping they could help me here. And finally, what does Bent's rule postulate?

Bent's rule is an empirical observation which covers 'fractional' hybridisation. It is very well explained in the question I linked. The fundamental theory that is a basis to the simplifications you are using is molecular orbital theory. It is more involved and harder to learn, but in the end easier to apply. Hybridisation has no effect on the molecular structure, it is a mathematical tool to describe bonding.

Tomorrow I start giving the theory of molecular orbitals in theoretical classes. @Martin


text:
How to Separate Copper Nanowires in an Alumina Membrane
So, in my Chemistry Lab we had an experiment to use Whatman alumina membranes and GaIn eutectic to create Nickel nanowires through electrodeposition. So essentially Nickel wire electrode and alumina membrane taped to the copper electrode with GaIn between the membrane and the copper to act as a conductive layer. My group wanted to push beyond just the regular protocol with the extra time we had, so we attempted to perform an identical experiment using copper instead of Nickel. So, everything worked perfectly, we currently have copper nanowires confined inside the alumina membranes and GaIn still stuck on the back of the membranes. The issue is that for Nickel we used Nitric Acid to remove GaIn and then KOH to dissolve the membrane, and we could isolate the nickel wires with magnets, but we obviously can't do so with copper wires. So as of now, we have our copper-GaIn eutectic- Alumina Membranes in solutions of 6M KOH and we do not know how to isolate the copper nanowires. We've thrown around ideas for how to resolve this but nothing certain. So any help would be greatly appreciated!


text:
Finding the weighted parameters of the vertex adjacency matrix for a derivative of azulene
In my last question ( Numbering in a derivative of azulene ), I asked about the numbered of a molecule in order to find the Vertex Adjacency Matrix (VAM) of the same molecule. To explain my question properly, I will do my best to show my effort from A to Z as follows:
In general in order to find the VAM we use the formalism described by

To apply it onto the molecule I am studying, I first applied it to azulene and with agreement with the paper Quantum Interference, Graphs, Walks, and Polynomials the VAM represented by A is:

Now by having the correct numbering to my molecule (Numbering in a derivative of azulene), it also should be a 10 x 10 matrix. The difference is instead of hydrogen we now have Oxygen and hydrogen Sulfide involved in positions 1,5 and 2,6 as shown:

To find VAM the only way was to study the VAM of weighted Graphs as shown below:

Here are my questions:
1- Should I include  the parameter h only for the weighted carbons since the numbering of the molecule I am studying includes only carbon?
2- If 1 is no, how should I include the k parameters described before hand?
Any other suggestions to find VAM is appreciated.

As a comment: The matrix-representation you search emerges from graph theory, and mathematics.  The numbers drawn in the structures earlier (1 for the carbonyl carbon, 2 to the carbon adjacent to the -SH group) a synthetic chemist refers to (e.g., "let us introduce an alkyl chain in the 4-position") do not (necessarily) relate to the numbers to denote the vertices in these nets.  These are seasoned descriptions are still reviewed (2nd edition), and apparently  the Open Molecular Generator were one tool to generate them.

I have two questions: 1- Should I not have asked here and asked in the maths section?
2- Would you be able to elaborate more from a chemist perspective on the relationship between numbers and vertices denotation?

@abedhaidar Re #1: It is not wrong to ask at ChemSE, or math; the question is at the interface between the two (like Chemometrics) or even a third (Cheminformatics).  I speculate contacting a not-only-pure-chemistry research group might be beneficial (e.g., Spackman, http://crystalexplorer.scb.uwa.edu.au/index.html; or even Hutchinson, https://hutchison.chem.pitt.edu/, because of his contributions to openbabel).  #2:  The numbering in the chemical structures follows the priniciple to spot a "parental" core structure, oriented on the canvas according to rules, and numbered according to others.

@abedhaidar (Part 2) Then, if there are chains, they are seen like decoration, with their own set of numbering (again a set of rules), eventually merged to the core structure while retaining their (inner) numbering.  Which is why an organic chemist would count 1 to 4 within the square (top most structure 3--6), once to 1 (for the branch 6--7, a methyl group), and once 1 to 2 (for the branch 3-2-1, an ethyl group) and say 1-ethyl-2-methyl-cyclobutane.  The matrix reminded me a bit to the connection tables seen e.g. in MDL's mol2 files (but, CT are not these matrices), that's all I can say here.


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Dilution of Concentrated TiCl4 using deionized water
How can we dilute conc TiCl4 (>99%) to 20 % w/w, using only DI water?

Since $\ce{TiCl4}$ hydrolyzes rapidly in air, releasing $\ce{HCl}$, you need a good hood and protective equipment.
Put the chilled $\ce{TiCl4}$ into a separatory (dropping) funnel or burette with stopcock that can be adjusted to let one drop at a time flow into chilled water. Since the m.p. of $\ce{TiCl4}$ is −24 °C, you can use an ice-water bath to keep things cool.
There is a good demonstration video available.


text:
I need the diagram of "cyclobutanal"
I doubted whether I can say cyclobutanal or not. Im confused and didn't found my answer anywhere

There simply cannot be an aldehyde functional group inside a cyclobutane ring.

Related: https://chemistry.stackexchange.com/questions/19868/when-is-the-carbaldehyde-suffix-correct/19869

Let's have correct names for related alcohols butanol (e.g. butan-1-ol) and cyclobutanol. Here the -ol suffix means $\ce{-OH}$ group attached to the carbon atom.
In butanal ($\ce{CH3CH2CH2CHO}$), suffix -al means not just $\ce{=O}$ group (oxo group), but entire $\ce{-CHO}$ which already contains carbon, which is included in the counting and numbering. (No locant is needed, because the aldehyde group can be only at the terminal carbon, 1.) If you want to name C-1 to C-4 cyclized butanal, its correct name is cyclobutanone.
Now, it's impossible for cyclic compound to contain aldehyde group in the cycle. For $\ce{-CHO}$ group attached (not included) to the parent cycle, suffix -carbaldehyde is used, therefore names “cyclobutanal” or “cyclobutan-1-al” are wrong, correct is cyclobutanecarbaldehyde.

Thankyou mykhal, I've cleared my doubt in nomenclature in both the classroom and here also with this wonderful explanation, thanks alot. Well, is there any more exceptions in naming like this in aldehydes . ?

It's cyclobutanecarboxaldehyde (or cyclobutanecarbaldehyde). I found for you nice little picture of this guy.

Correct name is cyclobutanecarbaldehyde. There's an instance of "…carboxaldehyde" name in latest IUPAC nomenclature book, but it's a mistake fixed in the ERRATA.

Well, here's my real question guys, can't I write it like cyclobutan-1-al. It seems all good with keeping rules in mind , still my teacher got it wrong.

@VishalKumar No, according to the rules, this name does not make sense.


text:
Why does humidifier put measurable PM2.5 in the air?
(I know little science so please help me if I'm in the wrong place or if my question needs editing. I could use help with tagging, for example.)
I use two devices that measure PM2.5:

Kaiterra LaserEgg 2+, an air monitor which uses laser-based light scattering technology
Blueair Classic 280i, an air purifier which measures PM2.5 but doesn't say how

Generally, their PM2.5 measurements are within 20% of each other and the readings go up and down together.
Both devices are in a room measuring about 40sqm.
It's winter so the windows are closed and I keep the Blueair Classic on automatic mode, which varies the fan speed depending on the level of PM2.5 it measures. It is rare for the fan speed to increase above the lowest level.
The level of PM2.5 is usually under 10μg/m3 (safe, according to the WHO), as measured by the LaserEgg and the Blueair Classic.
A few days ago we felt that the air was dry (cracked skin, etc.) and our humidity sensors reports humidity around 25%, so we turned on an ultrasonic humidifier (https://www.amazon.co.uk/Humidifier-Improves-Health-Sleep-Focus/dp/B01C5YHFWO). We use tap water in the humidifier.
Within a few minutes the LaserEgg and the Blueair Classic reported a massive increase in PM2.5, to about 400μg/m3 (very unsafe according to the WHO). The Blueair Classic changed its fan speed to maximum but it seemed to have little effect. Perhaps the output of the humidifier is greater than the cleaning capacity of the purifier.
We have reproduced this situation several times since - every time the humidifier is turned on.
What could be causing the PM2.5 increase?
Could it be harmful?
P.S. This seems relevant: https://www.researchgate.net/publication/51859262_Indoor_particle_concentrations_associated_with_use_of_tap_water_in_portable_humidifiers

Your ultrasonic humidifier does not put out water vapor, it puts out small water droplets (which then evaporate into vapor). The Blueair (which is not a calibrated particle size distribution counter) sees the water droplets as particles.

@JonCuster Thank you. Please can you give me a source for "which is not a calibrated particle size distribution counter"?

@JonCuster Out of interest, does that mean the water droplets are less than 2.5 micrometers in diameter?

The Blueair is not a scientific instrument to measure particle size distributions. It has some little light scattering system to get some gross average measure of particles. Real particle size distribution measurements are done with instruments costing a large multiple of your air cleaner.

I tried boiling some distilled water and my air quality monitor readings shot through the roof. Very likely it is picking up particles of water vapor.

You can also buy a TDS meter, not very expensive, and measure the total dissolved solids in your water.I own a small distiller that I use to make water for my humidifier. My water is quite hard and measures 350 . After distillation it goes down to a 10 or below. My 2.5 meter will indicate an increase of approx. 5 .

All tap water contains some dissolved solids, and "hard water" contains more, particularly calcium salts. There are also microorganisms found even in treated tap water, and some organisms can build up a biofilm inside the humidifier. Ultrasonic humidifiers are effective at putting these particulates into the air. However, as others stated, you may just be detecting fine water droplets from the humidifier before they evaporate.
To find out if it's water droplets or particulates from dissolved material, do a comparison test, using distilled, filtered, low-particulate water such as High Performance Liquid Chromatography (HPLC) grade water.

Clean out the humidifier thoroughly, rinsing repeatedly with a dilute acid safe for that machine (perhaps vinegar, but check the humidifier manual).
Rinse repeatedly with distilled water and finally with HPLC water.
Refill with pure HPLC water.

If this does not greatly increase the particulates above background, but tap water does, then you've found the source of the issue. If not, then you're probably detecting mostly water mist.
Caveat: Buying HPLC water is not inexpensive. If you can't easily make your own on-site, perhaps just leave the dehumidifier off.

But is inhaling the particulates from hard water dangerous?

Possibly, but this site does not address personal medical questions as such. That said, there is information available, e.g. https://www.epa.gov/pm-pollution/health-and-environmental-effects-particulate-matter-pm

Would boiling distilled water also produce this mist you mention (that would be detected by a typical air quality monitor off the shelf?). It appears to be the case when I tried this, so now I am confused as to what is considered particulate matter by my device or simply "humidity".

Folks at Awair conducted an experiment with different water in their ultrasonic humidifier https://blog.getawair.com/awair-investigates-how-your-humidifiers-water-affects-your-health
According to the blogpost, it is not the water that introduces most of the particles to the air but minerals and other things in the water.

Interesting that according to their test, distilled water still adds particles to the air. I suppose it still can contain impurities.

Of course, pure water will not add particles to air from the humidifier, unless the temperature is well be low freezing. It's IOTETMCO that particles are from dissolved solids or particles suspended in the tap water.

It's mostly due to microscopic water droplets as when my dehumidifier is made on it shoots up the PM2.5 reading from 6-7 to 20-21


text:
Is it safe to heat fluorite (CaF2) to 140 °C?
For the manufacture of the optical system, we need to heat the fluorite ($\ce{CaF2}$) lenses to $\pu{140 °C}$. Is it possible:

for some poisonous gas to be emitted? ($\ce{F2, HF}$ or other)
for defects to appear on the glass? (scratches, chips, cracks and the like)

If yes, please supply sources describing the corresponding scenarios.

The melting point is over 1400C. There's no way it can evolve F2 without electrolysis. Think you'll be safe at 140C

@Waylander Thanks, but I have  concerns like operator live saliva (1 mg) which ph <7. Then there will be reaction which  released HF.

Unless your operators drool is highly acidic (pH1 or lower) then they are not going to produce HF.

@Waylander thanks, if you want give me answer below, and I'm ready to accept it.

The melting point of $\ce{CaF2}$ is over $\pu{1400°C}$. There is no way $\ce{F2}$ can be generated without electrolysis of the molten $\ce{CaF2}$. $\ce{HF}$ is produced by reaction of conc sulfuric acid on $\ce{CaF2}$. Concentrated $\ce{HCl}$ does react with fluorite to some extend, but the reaction is very slow so, unless your operators drool like the monster from Alien, there is no chance of $\ce{HF}$ formation.

Some operators, though...

I guess you could link an MSFS showing how safe $\ce{CaF2}$ is …

The MSDS for fluorite is here: http://www.sciencelab.com/msds.php?msdsId=9927120  The major concerns are contact with eyes and skin.


text:
Does a smell imply particles or chemicals in the air?
(I know little science so please help me if I'm in the wrong place or if my question needs editing. I could use help with tagging, for example.)
If I can smell something then does that mean there are particles or chemicals in the air?
My question is motivated by some new window blinds that have been installed in my home. The blinds have a smell, which is most noticeable when the sun is shining on them. I am trying to work out if the blinds are emitting some chemical into the air. This question is my first step.
[This question is not the same as "How does the smell of a compound come about, and is it possible to define a smell?". This is clear from the title.]

Followup question: Why can some dry objects (rocks and metal pieces for example) have a smell and not slowly lose mass and vanish, unless they are catalytic and doing something to materials in the air?

Yes, next time you sniff something in the elevator consider exactly where each of those particles came from--and in order to smell something I don't think we're talking one or two particles--probably thousands?  Millions?

@rackandboneman Who says they aren't slowly losing mass and vanishing? "Slowly" could easily be over the course of centuries or decades.

Note that "chemical" is a term that's often used somewhat pejoratively. Sure, when you smell bleach or petrol fumes, you're smelling chemicals. But everything is a chemical, so the same is true when you're smelling lemons or baking bread.

“Phantosmia, also called an olfactory hallucination, is smelling an odor that is not actually there” - not saying that’s what’s happening here, but it’s a possibility, and an example of a smell that is not caused by airborne particles or gases.

You're detecting some kind of off-gassing, which is also known by the prettier name "new car smell"

If there are no particles or chemicals in the air, then (unless as @HopelessN00b mentioned, you're suffering from olfactory hallucinations) there is nothing to smell.  This also ignores the fact that the air is chemicals.

The question may look different but is just asking the same thing from a different angle. Your’s is: ‘Does smell imply particles?’ The other one is ‘What makes smell?’

Smell works because your nose detects substances in the air.
Everything you can smell is the result of a molecule carried by air interacting with complex proteins in the nose/mouth (smell and taste are related but taste copes with solids and liquids as well as volatile airborne stuff). We broadly understand how smell works but the exact details are often a complete mystery.
The key is that smell-sensing nerves contain some proteins with very specific sensor sites that can detect particular groups of substances when they "fit" into the site. Exactly how the sites translate the fit into nerve signals is often a mystery. But the point that matters for this question is that the stuff being detected has to be carried into the nose via the air (which means it has to be a gas or, possibly, a very small particle that is easily carried by air).
So, yes, smell implies there are chemicals or particles in the air. However, the nose is extremely sensitive to some chemicals and being able to detect the smell does not imply a problem (even nasties like hydrogen sulphide are detectable by smell at below parts per billion concentration, hundreds of times below the level causing harm). Your blinds probably are emitting something but it is most likely both harmless and will probably fade with time like the smell characteristic of a new car.

"Your blinds probably are emitting something but it is is both harmless and will probably fade with time like the smell characteristic of a new car." - How do you know it is harmless?

@cja I can't prove it is harmless. But your blinds are a consumer product and most likely made from a standard substance (maybe PVC) which has been widely used for decades. Consumer regulations are not perfect but limit dangerous substances in consumer products. And, if harmful things were emitted, lots of people would have noticed by now.

Probably you are right, but to be pedantic, if the material emits something moderately dangerous to health and no effects were noticed immediately, why would it ever be thought that the blinds were the cause?

@cja Because we have in the past observed such harms and acted to limit them. smoking, for example, is bad but not in the short term (the late to causing lung cancer, for example, is often 20-30 years. Despite that, epidemiologists noticed the relationship and introduced campaigns against smoking. Many chemicals have much smaller but still notable effects and they have been restricted because the evidence was still clear.

Odor from a material such as PVC is generally not the resin itself, but the plasticizer chemical that is used to make the material less rigid. There are also additives for UV resistance and fire resistance. Some of the additives have been proven to be  harmful and have been eliminated in some or all jurisdictions. "Additive" sounds like it's a small thing but plasticizers can make up a significant percentage of the total material.

So, technically this is not 100% correct. Smell implies that there is a signal in the part of your brain that processes smells. Most commonly, this is due to neural input from the nose, but not always, e.g., when someone has a stroke.

@cja Also, since every smell is caused by some airborne substance, you can't conclude from a smell alone that there is a harmful substance. Otherwise, every smell would be cause for concern. (Neither can you conclude from the absence of smell that there is no harmful substance in the air. E.g. carbon monoxide is completely odorless.) If you know the smell of a particular toxic substance, that's different of course.

Everything you say supports my impression that particles do not smell-if you smell something it is molecular sized and a chemical.  It could be OK or not, but particles are detected and a problem in other ways.

@Zhe Yes, and the words you think you read could just as easily be a hallucination where you fantasised a community based chemistry Q&A site.

@matt_black That's true. :)

Definitively yes, as already described by another user.
I would point out that air conditions, such a temperature and humidity might trigger a kind of smelling sensation in many subjects. The so called "smell of coming snow" is likely an example. It is not related to your windows but I wanted to add this as a curiosity. Very very strictly speaking, smelling something doesn't implies specific molecules or particles are in the air.

Most definitely you are "smelling" a chemical (almost certainly a gas in this case), and many people have over sensitive inflammatory reactions to many "approved" chemicals in the environment (ie perfumes, fabric softeners, solvents, etc.), so you may have a reaction (like bronchial constriction) even though most people would not.


text:
Lead's affinity for oxygen in the cupellation refining process
My inorganic chemistry textbook1 says:

Cupellation is a process, where crude gold is taken in a small bowl called cupell and melted in the presence of air. Due to high oxygen affinity of $\ce{Pb}$, it is converted to $\ce{PbO}$ (volatile) and escapes from the system.

Now, the affinity for sulphur and lead is explained by HSAB (soft acid-soft base). But how to account for the affinity of oxygen (hard base) and lead (soft acid)?
References

Guha, Sudarsan. Concise Inorganic Chemistry for JEE (Main and Advanced). 4th ed., Wiley India Pvt. Ltd., 2017.

More important is general resistance of noble metals to oxidation.

@Mithoron Yes that is true, but noble metals generally resist reacting with most elements, don't they? While I have looked thorough a few books about cupellation,and they all talk about lead's affinity for oxygen, and I'm not quite sure about the rationale for this.

Indeed if you do the cupelation, the lead will form litharge and some parts of the litharge will volatilize.
But usually the cupel is made out of bone matter - this is very porous.
While the molten lead and the gold will stay in the cupel, the litharge will be absorbed by the cupel like a sponge.
If all lead is converted to litharge and all litharge is absorbed by the cupel only the gold shall be left - so far to the theory.


text:
What is the reaction between nitric acid and fluorine?
Nitric acid, $\ce{HNO3}$, is a strong oxidising agent as well as fluorine, $\ce{F2}$. What will be the reaction when pure $\ce{F2}$ gas is bubbled through 100% pure $\ce{HNO3}$. While searching, many products are suggested including $\ce{HF}$, $\ce{O2}$, $\ce{H2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{NF3}$, $\ce{NO2FO}$, etc.. What is correct?

It was first discovered by Cady that when fluorine is bubbled through $\pu{3 N}$ nitric acid, then a remarkable compound, fluorine nitrate ($\ce{FNO3}$) is formed. It boils at $\pu{-42 ^\circ C}$ and is a very strong oxidizing agent and is capable of liberating iodine from iodides and in gaseous states, it explodes around $\pu{200-300 ^\circ C}$.
References

The reaction of Fluorine with Nitric acid and with solid Potassium Nitrate to form NO3F by Don M. Yost and Alan Beerbower,  J. Am. Chem. Soc. April 1935, 57(4) 782. DOI: https://doi.org/10.1021/ja01307a520
Prof. Dr. Otto Ruff, Dr. Walter Kwasnikie,  Die Fluorierung der Salpetersäure. Das Nitroxyfluorid – NO3F, April 1935, DOI: https://doi.org/10.1002/ange.19350481604 [German paper]


text:
Will gamma radiation cause "thermal" decomposition of sodium bicarbonate?
I am conducting research for a new drug that contains powdered sodium bicarbonate, and the drug will need to be sterilized after placement into it's container/closure system. Typically, this is performed with steam, at 122.5 °C for 15 minutes, but in the case of sodium bicarbonate, this would cause decomposition:
$$\ce{2 NaHCO3 → Na2CO3 + H2O + CO2}$$
The alternative is to sterilize with ionizing (gamma) radiation. Will this also cause decomposition? Any help or referral to another authority is much appreciated.

You may already be aware, but the term you're looking for is "radiolysis". A cursory Google search shows some results for bicarbonate radiolysis, but I haven't parsed anything. My instinct is that while it could theoretically cause some decomposition, almost all of the bicarbonate will remain intact in any reasonable sterilising gamma flux. Living matter is far more fragile than an inorganic salt.

Long story short, don't worry: the drug will decompose way sooner than bicarbonate.

I think you are looking in the wrong direction.
If you subject oxoanions such as sulfate or nitrate to intense gamma ray fields then you will see decomposition of the anions to form oxygen and versions of the anions where they have fewer oxygen atoms per anion. So sulfate could be decomposed to sulfite or even sulfide by the radiation.
Carbonate when it is irradated does form formate, which is then transformed into acetate and then oxalate by further irradation. But the amount of formate, acetate and oxalate formed is very small.
A recent study considered this matter and an estimate was made of the formation of these things in dilute (0.2 mM) carbonate solution. In this medium using alpha particles the yields were
400 picomoles per joule of alpha radiation of formate, 1.7 nanomoles of acetate per joule and 1.0 nanomoles of oxalate per joule.
See Carboxylate anion generation in aqueous solution from carbonate radiolysis, a potential route for abiotic organic acid synthesis on Earth and beyond, Johan Vandenborre, Laurent Truche, Amaury Costagliola, Emeline Craff, Guillaume Blain, Véronique Baty, Ferid Haddad and Massoud Fattahi, Earth and Planetary Science Letters, Volume 564, 15 June 2021, 116892.
https://www.sciencedirect.com/science/article/abs/pii/S0012821X21001515
But the doses required to form these substances in large amounts will be very high. One of my academic interests is the radiation chemistry of organic materials, I happen to have access to a research irradator.
I also worry that at large doses of gamma radiation that the organic compound will degrade, the typcial dose required to make food and medical products sterile is 10 kGy (ten thousand joules of radiation per kilo of matter). This is at the bottom end of the dose which changes organic matter. It is hard to see changes in most plastics at 10 kGy.
When we get into the MGy region it is very easy to see changes in materials, I have seen wood and plastic change greatly after MGy doses of radiation. It would be unlikely that anyone would want to deliver MGy doses of gamma rays to medical products or food during the production process.
But back to your question, It is important to understand that gamma rays are "heat rays", they do increase the internal energy of objects. Thus by increasing the internal energy of objects the temperture of the objects increase. I have felt the warmth of the decay heat of a very strong gamma source. The source was in the range of about 1 PBq of cobalt-60, I  touched the outside of a very thick lead cask containing the source. I could feel the warmth of the cask. I also have it on good authority (from the late John Peckett) that kilo lumps of plutonium are warm to the touch. If you pick one up with a gloved hand or keep it in a plastic bag then it feels warm.
It would be hard to imagine a situation where the gamma radiation level is sufficent to heat sodium hydrogen carbonate to the point at which it decomposes to sodium carbonate and carbon dioxide. The atomic bomb works by unleashing a vast flux of gamma and X-rays which then heat the air, rocks, water or whatever near it to an almighty temperture. This is the mechanism by which blast is generated by an atom bomb.
Consider water for a moment, the heat capacity of water is 4181 joules per kilo per kelvin, so if we were to deliver gamma rays to water which was kept in a perfect dewar. Then we would need 334480 joules per kilo to make the water boil. This would be a dose of 344.5 kGy. I predict that at these doses that many organic substances will degrade a lot.
I assume that the irradation machine will not be able to deliver the dose very quickly, so heat losses during the irradation will be significant. This will increase the dose required to boil the water. I am aware of a fatal accident in a commercial sized gamma irradation plant. In the irradation chamber the dose rate was about 10 Gy per minute. If we consider this as a "typcial" food irradation plant then it means that 33448 minutes (557.5 hours, 23.22 days) of irradation will be needed to make water in a dewar flask boil if the water enters the plant at 20 oC.
In real life a lot of the heat energy deposited in the water will be lost to the surroundings which will greatly increase the dose required to boil the water. This will greatly increase the degradation of the drug from the radiation.

There is an equilibrium between decomposition and recombination, i.e. as $\ce{CO2}$ is given off, it joins back to the  $\ce{Na2CO3}$ to form $\ce{NaHCO3}$.

$\ce{2NaHCO3 <-->Na2CO3 + CO2}$

So for moderate doses, given at moderate rates, it's unlikely great pressure would develop, but you'd need to test to see if the drug is degraded by decomposition products.

The likely issue is the presence of any water (which can also be sourced as you noted by the heating of NaHCO3).
Here is a source to quote:

The radiolysis of water due to ionizing radiation results in the production of electrons, H● atoms, ●OH radicals, H3O+ ions and molecules (dihydrogen H2 and hydrogen peroxide H2O2).

The formation of any of such radicals would likely degrade your new drug.
So, does your irradiation, seemingly result in any visible signs of water? If yes, a possible quality issue.

The drug is lidocaine (an injectable weak acid aqueous solution), and will be kept in a separate compartment from the bicarb powder until just prior to injection.  I expect we would find some trace water in the bicarb just from being in the filling hopper.


text:
What is the peak at 3200 cm⁻¹ in the IR spectrum of Benzocaine?
Benzocaine has the following IR spectrum:

and here is it's structure:

From what I understand:
The two leftmost peaks are the antisymmetrical and symmetrical stretchs of the amine group.
Just before the fingerprint region at about 1690 is the carbonyl ester peak and to the right of it (also before the fingerprint region) is the benzene C-C bonding peaks.
The messy broad peaks just before 3000 are the C-H bonds of the non aryl C.
The tiny peak at ~3040 is probably the aryl C-H bonds.
However I can't assign the peak at ~3200. It could be water of crystillisation in the sample (anywhere from 3100-3600cm^1) but surely there wouldn't be any in a proffesional sample? I also considered whether it could be due to -NH3^+ but the sample is made in a procedure that ensures the solution is alkaline before final purification...
Please could someone point me to a resource that will allow me to assign this peak or explain what it is to me?

3200-3600 peak is usually for O-H bond, (2200-3600, if this bond belongs to carboxylic acid) so it's better to check for possible O-H bonds (water, alcohol|acid). It can be even air moisture in the worst case.

@KellyShepphard Thanks. This peak is on all the IR spectrums of benzocaine I have looked at. I looked at quite a few because origionally I thought that the peak was maybe a mistake in my spectrum. So would even the professional IR spectrums have this peak due to water?

Air moisture is omnipresent, but for pro-spectra, sample preparation is way too good to have "water" peak (usually). You have a good spectrum, though! No need to worry (but if you want, you can master sample preparation to make it the best possible in your lab conditions).

I'm betting it's a common primary amine shoulder band. See here: "A shoulder band usually appears on the lower wavenumber side in primary and secondary liquid amines arising from the overtone of the N–H bending band: this can confuse interpretation."

@SendersReagent, I have had a look at the weblink page, however it does not say what causes the shoulder band... I would like to understand what is causing it, not just accept it. Do you know?

@Mirte to be honest, I don't know what causes the overtone from the N-H bend. If I knew more, I'd have posted it as an answer. I was just leaving the information in case it was useful to you or someone who might use it as a starting point for forming a complete answer.


text:
The density of an aqueous solution
Could the density of an aqueous solution be less than 1g/ml ? For an example, can 0.98g/ml be taken as a density for an aqueous sulphuric acid solution ?

Yes. With the exception of negative quantities or ones where the reaction quotient exceeds the solubility product, anything is fair game.

Sol. of H2SO4 tend to be denser then water.

Answer to your question, "could the density of an aqueous solution be less than $\pu{1g/mL}$," is yes. See following examples:
The density of the commercially available 28% ammonia solution ($\ce{NH3}$ 1n $\ce{H2O}$) is listed as $\pu{0.9 g/mL}$ at 25 °C (Sigma-Aldrich Catalog).
MERCK Tabellen für das Labor has listed the density of 35% ammonia solution ($\pu{18 mol/L}$ concentration) as $\pu{0.88 g/mL}$ at 20 °C (Steffen's Chemistry Pages)
As obvious, the densities of various aqueous ethanol solutions are also equal to less than unity since the density of absolute ethanol is $\pu{0.789 g/mL}$ at 20 °C (Wikipedia). By a quick google search, I found the densities of various aqueous ethanol solutions listed in MERCK Tabellen für das Labor  in Steffen's Chemistry Pages, which is pretty interesting.


text:
Long linear molecule with alternating elements?
In the classic freshman physics text Electricity and Magnetism by Purcell, one of his first examples is the computation of the electrical binding energy of a $\ce{NaCl}$ crystal. He models the crystal in terms of perfect ionic bonding and sums up the energy $q_1q_2/r$ pairwise, computing the $r$ values in three dimensions. (Actually he just gives the first few terms explicitly and then says he carried out the computation numerically on a computer.) The result is $U=-0.8738Ne^2/a$, where $N$ is the number of ions, $a$ is the edge length of the cubic lattice, and the units are cgs, so that the Coulomb constant equals 1.
I thought this was a nice example, but it seemed unfortunate that the actual result could only be found through a fairly complicated numerical computation. I thought it would be nice if I could come up with a similar but simpler example for use with my freshman E&M classes, using a linear molecule with charges $\pm e$ arranged like $\cdots +-+-+-\cdots$. A relatively simple calculation gives an energy of
$$-\frac{Ne^2}{4a}\sum_{k=1}^\infty \frac{1}{k(k-1/2)}.$$
I imagine that the sum can be done in closed form, but anyway it's clearly finite and positive, so that the result is a finite binding energy that has the right sign to keep the system bound.
Is there any molecule that actually has this structure, or something close to it? Googling and asking a chemist in the hall resulted in a variety of examples that are not exactly what I had in mind:

There are various 3-ion molecules such as $\ce{BeF2}$.
You can have long carbon molecules, with hydrogens hanging off of them, but they're not exactly linear, and in any case the carbons are all the same element, so there is no ionic bonding.
Acetylene is linear.

Does nature provide anything like the $+-+-$ chain that I'm describing? Do they not occur because they're too unstable with respect to crumpling up? Could they exist on a substrate (the kind of thing people build using scanning electron microscope probes) or as linear impurities in a bulk solid?
For my purposes, I don't think it really matters whether the bonding is perfect ionic bonding, as long as there is some segregation of charge.
Related:

Why does sulfur, but not oxygen, catenate?
Are long oxygen molecules possible?

I believe I read many years ago in a solid state physics textbook that the calculated speed of sound in a 1D crystal is imaginary. If there is no context I am forgetting, wouldn't that suggest they cannot exist?

While bonding in molecule may have more ionic then covalent component, they are rather treated as covalent - any simple ionic model shouldn't be useful.

My main criticism is on line with the comment of @Mithoron. Treating a long chain with alternating (partially) oppositely charged fragments as kept in place by electrostatic force isn't the proper treatment. Perhaps one can estimate a coulombic term but then we enter in the computational and quantum chemistry field, truly.

@Alchimista: By Earnshaw's theorem, no system can be in a stable, static equilibrium due to electrical forces. That isn't what Purcell is claiming. He goes into a lengthy discussion of this.

Therefore I say. It is somewhat the Peierl instability in a  charged version. It is not clear what you want calculate.

related: https://math.stackexchange.com/questions/3026370/evaluating-the-series-sum-k-1-infty-1-kk-1-2

It is even visually clear that a linear configuration of + and - ....can be at best merastable. Your question boiled down to " is there any covalent structure stabilizing such above mentioned linear configuration?". Answers have provided what you were looking for. But for clarity: remind that a system bound is not a system in its more stable configuration. If I orbit earth I'm bound to it. Still I can only decay to a lower orbit. Shortly: the molecules you can use as example are kept as they are not (only) by the electrostatic term you want to calculate.

..... that is why Purcell treat NaCl as it is, a three dimensional array, in spite this requires a tedious computation.

@Alchimista: No, the three-dimensional NaCl is also classically unstable. Earnshaw's theorem doesn't have anything to do with whether the structure has one geometry or another.

I don't get what are you looking for anymore. Do you agree that NaCl does exist? And that is a three dimensional array?

An example of a linear structure is given by cycopentadienyl lithium, whose structure is explored in this question.  There is some covalent interaction in this polymeric structure, as in most organolithium compounds, but the bonding between lithium and cyclopentadienyl groups is highly polar.

Check for polyoxymethylene (POM).

It has partial positive charge on carbons and partial negative on oxygens, and it's "long" enough to serve as plausible example for sum with large maximum k. But I should mention that there are no perfectly linear polymers which are like string of beads.

Not true how about indefinitely long polyynes & cumulenes? https://bit.ly/2Sp81yb I'd imagine any deviations from linearity would be pretty minor! ; )

@ManRow: I'm having trouble parsing your comment. What are you saying is not true? By the way, it's not necessary to give shortened bit.ly links; you can just give a link straight to Wikipedia, which makes it easier for people to understand what the link is going to be about. The link re linear acetylenic carbon is interesting, but it isn't ionic bonding, since every atom in the chain is a carbon...?

Hi! I was commenting on what Kelly Shepphard said about linear polymers "like a string of beads" not existing, and just provided some counterexamples : )

ManRow, they are still not perfectly linear (and I've said about perfect linearity), and  definitely have impurities (as any other polymer). Besides, usually they exist only as rigid part of molecule whis is linear copolymer, dendric structure or, at least, crosslinked polymer. I even bet that such "strings of beads" would crosslink spontaneously.

@KellyShepphard Haha well those polymers I gave are definitely quite unlikely to be stable in any sort of "real world" conditions (aside from some extremely supercooled or weird other  setups). And, while probably quite more "linear" than the other more common polymers that we do see around, they aren't really going to be "perfect" by any stretch, like most things in the real world!

Furthermore, I'm guessing they may get even less "linear" the longer they get too (im guessing), as deviations with the extra length (& having more subunits) increase https://en.wikipedia.org/wiki/Cumulene So propadiene may be the "most linear' of them so far, while butatriene and longer cumulenes even less!

Then it's not a conterexample for now. But maybe one day humanity will master some new reactions and ways to support perfectly linear long structures - this day will be the day when your example becomes counterexample :) For now, it's like example that polymers can't be perfectly linear - even if parts of them are.

Invite both to consider that you are about right and wrong at the same time :)


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What additive can I add to Jelly Wax/Gel Wax to increase its cooling time/setting time?
Jelly Wax. Jelly wax, also commonly known as gel wax, is a clear, dimensionally stable, jelly-like material made from specially processed mineral oils. It can be used to create a variety of special effects and is also suitable for making decorative container candles. It is normally heated until it liquifies and then poured into a mould. It sets quickly back into a jelly.
I want to use it to pour into a transparent PVC pipe for a sculpture. Unfortunately I have found that it cools rather quickly for it to be useful for this purpose.
Anyone knows how I can increase the cooling time, using an additive maybe?

At your own risk, you can try using plasticizers for waxes. Usually, these plasticizers are wax esters of palmitic acid (well-known example is spermaceti, which can be replaced with jojoba oil, or you can buy/obtain those esters). But it's hard to predict exact amount of plasticizer needed. Add too much - and it will be creamy I-will-never-cool substance. I hope we have chemical engineers here who know more about wax plasticizers and can add details to my superficial statement.

Thanks Kelly! The Jojoba seems to be working!

Glad to know that :)

You rock. Saved my ass :)

Sculptures are great, my favourite art form. I wish you good luck and creative successes :)

Thanks! Kelly btw do you know what else I can use besides Jojoba? It is a bit expensive, especially if I want to use this on a large scale

Any plasticizers for waxes. But as far as I know, for example, palmitic acid ester costs at least 5 d. per gramme. You can also try castor oil (such as for massage), it's pretty cheap (30-40 d. per liter),but I don't know whether it will be suitable for your wax.

Depending on the dimensions, the following may help.
1. wrap the PVC with teflon tape.
2. coil nichrome wire around the pipe.
3. connect the ends to a power supply and slowly raise the voltage.
4. monitor the temperature of water in the PVC for testing.
Nichrome wire data will help you design a coil of a given resistance from which you can calculate the power dissipated. You should only try this if you have experience with basic electric, understand Ohm's law, etc.


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Why does a Lower Concentration of SO2 call for an increase of equipment dimensions in a Contact Process Plant?
I'm working on a presentation based on an article about the Contact Process (production of sulfuric acid), and it says that using a low concentration of SO2 at the entrance of the converter (in which the oxidation of SO2 to SO3 takes place as an equilibrium reaction) is better in terms of conversion, but disadvantageous in that it would require using bigger plant equipment dimensions.
Why does using a lower concentration of SO2 require a higher volume of such equipment?
Wouldn't using more gas require more size?

You need higher quality of equipment and larger "chambers" to deal with low concentrations. Besides, in order to meet needs of plant (I mean, each plant has production output) you will need stated amount of raw materials, which you can find in material balance sheet of the applied technological processes. So, when using low-concentrated things as raw materials, even if yield is higher (it's higher usually by less than 10%, not really much) you will need larger volume of raw materials (since needed amount is usually given in mass of pure $SO_2$ in entering gas).


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Why must a buffer solution contain both a weak acid and a salt solution of its conjugate base?
Doesn’t the weak acid dissociate to produce the conjugate base on its own?
I understand that when a strong base is added to the buffer solution, the hydroxide ions will react with the hydrogen ions formed via the dissociation of the weak acid. The equilibrium position will shift rightwards to replenish these hydrogen ions, ensuring that few hydroxide ions remain in the solution.
Conversely, when a strong acid is added to the buffer solution, the hydrogen ions will react with the conjugate base to form the weak acid molecule, thereby ensuring that most of the additional hydrogen ions are removed from the solution, and the pH of the solution remains stable. In the literature I’ve been reading so far, it is mentioned that a salt solution containing the conjugate base ensures that there is a continuous supply of the ions, but can’t the ions simply be replenished by the dissociation of more weak acid molecules, as in the previous case?

You are correct that a weak acid will only partially dissociate in water. Taking for example the weakly acidic salt $\ce{NH4Cl}$ of which the ammonium cation is acidic, in water we get reaction (1).
$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$
To analyse this reaction we need the equilibrium constant which in the case of acids is defined as an acidity constant $K_\mathrm a$ as shown in (2). (Wether we use $\ce{H+}$ or $\ce{H3O+}$ is irrelevant.)
$$K_\mathrm a = \frac{[\ce{NH4+}][\ce{H+}]}{[\ce{NH3}]}\tag{2}$$
Using the tabulated $\mathrm pK_\mathrm{a}$ value of 9.25, we can calculate the amount of $\ce{NH3}$ in a $\pu{1M}$ solution of $\ce{NH4Cl}$:
$$\begin{align}K_\mathrm a &= \frac{[\ce{NH4+}]^2}{[\ce{NH3}]}\\[0.3em]
K_\mathrm a (c_0 - [\ce{NH4+}]) &= [\ce{NH4+}]^2\\[0.5em]
[\ce{NH4+}]^2 + K_\mathrm a [\ce{NH4+}] - K_\mathrm a c_0 &= 0\\
[\ce{NH4+}] &= \frac{-K_\mathrm a \pm \sqrt{K_\mathrm a^2 + 4 K_\mathrm a c_0}}{2}\\
[\ce{NH4+}] &= \pu{2.37e-5 M}\tag{3}\end{align}$$
Whoops. We may have a good concentration of ammmonium chloride but only a very minor amount of it is still ammonium, $99.998~\%$ are conjugate base. That doesn’t look like a good buffer system. If you were to add an acid, that would be fine since the excess base can capture that. But if you add more base the pH will go up rapidly, because there are no ammonium ions in solution to buffer it.
Only if you add both a significant amount of an acid and its conjugate base will you have significant amounts of both compounds in solution which are then able to buffer by capturing excess protons or hydroxides.

Or you can add some strong base to $\ce{NH4Cl(aq)}$ [or some weak acid to $\ce{NH3(aq)}$] to get a better balance of the conjugate weak acid: weak base combination. This is the fastest way of making a buffer (called "adjusting a buffer's pH").


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Why we don't we save water by boiling the contaminated water to get the pure steam?
I see lot of drinkable water wasted in household and industrial activities, water usually gets contaminated with detergents(and other household materials),chemical effluents released from factories .
Why we don't just boil off the contaminated water to get the steam and again condense it to get pure water safe for drinking?
I get this idea from the fact that when we boil out salted water we get all the salt as precipitate thus distilling the water.I don't know much about distillation but I think this process can be economically feasable when done on a large scale.
By this method of treatment,I am considering countries that cannot afford water treatment plants on a large scale and with problem of water scarcity,as this would be quick way to get drinkable water.

All that wasted water does go to a wastewater treatment plant. Evaporating water on a large scale requires a large amount if energy and some mechanism to scrape off the precipitates

@AvnishKabaj I mean to say that by evaporating water we can again use it for drinking thereby saving water.It might be a efficient method for saving water when you live in countries with water scarcity(african countries,latin american countries and espescially India)

You are under-estimating the energy required to boil the water. The specific heat capacity of water is 4.2 J/g per degree C. Try a little calculation to work out how much energy is required to evaporate 1kg of water in the way you suggest.

Boiling a water mixture doesn't always produce pure water.
Boiling a solution of salt and water (or evaporating the water) will separate the salt from the water, but those are compounds with very different properties.
In the making of distilled alcohols (Ethanol), the vapor phase has more alcohol than water.  Repeated distillations can increase the concentration of the alcohol.  But, the leftover undistilled solution is not pure water.  Ethanol and water are very "sticky" to each other.
Ethanol is just an example.  Household 'grey' water (waste water that doesn't include human waste) will have a whole mess of compounds.  Some you can isolate through boiling, some, like ethanol, will 'come along for the ride'.  Wastewater treatment is complicated.  Look at the system that had to be engineered for the International Space Station.  Actually, a version of this system has been tested for use in the types of communities you are describing.
Purifying grey water isn't a simple process.  There are, however, plenty of things that can be done with grey water.  Using grey water to flush toilets is perfectly fine.  With some simple amount of treatment, grey water can be perfectly suitable for watering crops.
So, since it's cheaper and easier to repurpose grey water, and harder to purify it back to drinking-water, reuse is more efficient than trying to completely treat it at home.

We do, and so does the planet
The thing about water is that it is a renewable resource. The weather systems of the Earth already recycle water by evaporating it and them returning it to us as rain. So many countries don't need to take extra measures to have enough potable water as their lakes and rivers already provide enough.
Those countries that don't have enough fresh water already use various means to purify non-drinkable sources (though the most abundant source isn't contaminated water but sea water which is far more abundant than the flows of used water). Also, distillation is nothing like the most efficient way to produce pure water as it is very, very energy intensive. The current favourite tech (widely used in countries with inadequate rainfall but access to the sea) is reverse osmosis.
So distillation is perfectly viable for creating drinkable water, but would be a very poor choice for poor countries as it is very much more expensive than the alternatives.


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The real structure of difluoro(dioxo)xenon (XeO2F2)

I have a problem regarding structure of difluoro(dioxo)xenon ($\ce{XeO2F2}$). I have posted a image above. my teacher is saying that the structure B is correct as there is very less repulsion in structure B that is what I too think but many of the sources and books have written the structure A to be correct and me as well as my teacher is in a doubt that why structure A is correct not structure B as there is an angle of 180° between oxygen atoms. If I am wrong please guide me and also please tell me what are factors should I look at for judging a correct shape for a compound.
Any hint will be very very helpful to me .

Xenon is a noble gas and thus does not need to form any bonds to complete its octet. When such a bond is formed, this is known as a hypercoordinate structure: it contains more bonding partners than the most simple of simple Lewis structure theories would allow. Thanks to Linus Pauling, for many years chemists have been trying to invoke d orbitals to explain this bonding situation when the actual case is far more trivial and can be explained with simple octet considerations.
The two oxygen atoms which are depicted as double-bonded in the d orbital model are actually connected by a single bond. If you want, this bond can be formed by oxygen keeping its six valence electrons in three orbitals and using the forth, empty orbital to bond with a single orbital of xenon. Both electrons can be described as originally donated by xenon and a good way to draw it in the Lewis formalism is to use charge separation:
$$\ce{\overset{-}{O}-\overset{2+}{Xe}-\overset{-}{O}}$$
This leaves the two fluorine atoms. They actually form a slightly more complex bond known as a 4-electron-3-centre bond (thus fluorine only bonds to xenon in neutral compounds in multiples of 2). For a deeper explanation of the 4-electron-3-centre bond, see this previous answer of mine. In a Lewis formalism, this can only described with resonance structures as shown below:
$$\ce{F-\overset{+}{Xe}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Xe}-F}$$
This bonding situation means that the fluoride atoms must occupy opposite positions with $\angle(\ce{F-Xe-F}) \approx 180°$. Thus, A is the correct depiction of  $\ce{XeO2F2}$ which was confirmed in a neutron diffraction experiment by Peterson, Willett and Huston.[1]
The paper determines the $\ce{O-Xe-O}$ angle as approximately $105°$. This is somewhat in-between a pure p orbital and an $\mathrm{sp^2}$ description of xenon, $15°$ from either. At first approximation it would, however, be fine to assume unhybridised xenon and derive the bonding situation from there with each oxygen bonded to one of the p orbitals and the two fluorines bonding to the third.

Reference:
[1]: S. W. Peterson, R. D. Willett, J. L. Huston, J. Chem. Phys. 1973, 59, 453–459. DOI: 10.1063/1.1679827.

Then in $\ce{XeOF5-}$ the axial position should be occupied by two fluorine atoms, since then at least two fluorine atoms out of the 5 will be in a 4-electron-3-centre bond. But that is not the case; the oxygen and the lone pair occupy the axial positions (see 3.10 (i)). So is this something which cannot be generalised?

@AshishAhuja Instead of linking to a problem set, you could also have linked directly to the corresponding research article which the problem set actually references. But that's minor. As for the structure of $\ce{XeOF5-}$, this is complicated both to rationalise and to identify. The authors of the paper note that it has not been possible to crystallise the salt as it violently decomposes. Therefore, the structural analysis is based on inferrals from other analytic data. In addition, the entire picture is more complicated, as we are not dealing with ...

a neutral molecule but a molecular anion. For corresponding halogen-centred anions (e.g. $\ce{IF6-}$), the rule of thumb is to start off with the central atom carrying the necessary overall charge (i.e. $\ce{I-}$) and then adding ligand pairs using 4e3c bonds as needed. This rule already fails for xenon as we would start off with $\ce{[Xe-]^\dagger}$ which requires an electron occupying a higher shell. This already means that $\ce{XeOF5-}$ defies the standard rules of thumb and would require a different, more general set of assumptions.

On the other hand, $\ce{XeOF5-}$ is synthesised from $\ce{XeOF4}$ -- whose structure is easily predictable -- by addition of fluoride and the literature states that it is subject to rapid fluoride exchange and fluctuation (to the point where no fluorine-xenon coupling is observed in the NMR -- but a small side peak corresponding to $\ce{XeOF4}$ is clearly visible in the $\ce{^19F}$ NMR) so I'm not even sure if we can give a good answer to the question on its electronic structure. All things considered, I think it would be a great standalone question!

Thank you for your very detailed response. Could you confirm one last thing: as you've mentioned $\ce{XeOF5-}$ defies the rule of thumb, and I understand; am I correct in saying that the same rule of thumb is also defied for $\ce{IO2F5^2-}$ (because $\ce{I^2-}$ requires an electron in a higher shell) and hence the 4e3c logic cannot be used to place two fluorines in the axial position? (It's 3.42 (j))

@AshishAhuja I would say so.

Ok, thank you. I think I will avoid asking a new question for now since my understanding of the chemistry which is involved here is very limited and I would not be able to phrase one properly with the relevant details, like those which you have mentioned in your third comment. And I probably wouldn't be able to understand the answers either; I'm in high school and I don't even understand NMR properly. Nevertheless, I'm satisfied for now.

More simply, you can see Xe has a double bond with two oxygens, i.e., the bonds between Xe-O are double bonds means typically sp2 hybridized, while the bonds between Xe-F are single bonds means typically sp3 hybridized. More the s-character of the bond, it is more electronegative (or attached with less electronegative elements). Also, form the VSEPR point of view, the axial bonds have more p-character than the equatorial bonds in trigonal bipyramidal geometry. Thus, the axial bonds are less electronegative than equatorial bonds and tend to keep singly bonded F atoms there while the more electronegative equatorial bonds tend to keep lone pair of Sulphur as well as the doubly bonded O atoms.
Hope the explanation will help.
For more information about the comparative s- and p-character of axial and equatorial bonds of trigonal bipyramidal geometry, please go through detailed descriptions.

Think for yourself. The sole objective of the VSEPR Theory is to determine the nearest shape of a molecule such that there is minimum repulsion. In this example as we can see, we have to take into account lp-lp and lp-bp repulsions. Since Fluorine is more electronegative than Oxygen, the electronic cloud density is more in the former. Hence, it is more favorable to put the F-atoms in the equatorial plane with the O-atoms occupying the axial plane. This is to reduce the bp repulsions with the lp : Option B) correct.


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Diffuse functions for Ahlrichs/Karlsruhe basis sets in Gaussian?
As the question states, can one use the basis sets of Ahlrichs and coworkers (e.g. def2-SVP, def2-TZVP, ...) with diffuse functions (e.g. def2-TZVPD) in Gaussian?
The regular basis sets have been implemented, but not the diffuse functions according to their homepage. Is there some other way to access diffuse functions for these basis sets in Gaussian?

If they are not in Gaussian yet, there won't be keyword to include these diffuse functions, but you can add them manually to your jobs.
You can find the basis set specification for def2-TZVPD for the atoms you are interested from the Basis Set Exchange. From there, you can either just insert the diffuse functions after your molecule specifications using the keyword Extrabasis or just insert the whole basis manually using the gen keyword where you would normally put the name of the basis set.

Judging from how long it took them to 'implement' the regular def2 basis sets, I guess this will be the correct answer for quite a while.


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Could Be2- exist?
I understand that Be2 cannot exist, since it has as many electrons in the antibonding as in the bonding orbitals.
But it seems to me that since the next electron would go into the πu orbital, which is a bonding orbital, Be2- would have a bond order of 0.5, and Be22- would have a bond order of 1, since it would have 4 in bonding (2 in σ and 2 in π)and only 2 in the σ*.
Could Be2- or Be22- at least theoretically be produced?  In other words, would the energy released by forming a bond be sufficient to counterbalance the energy needed to stabilize the negative charge gained?

While it is perfectly fine the way you have formatted your question, we also have MathJax available. If you are curious, please have a look here and here. For the matter at hand, I'd say your reasoning is sound; I wouldn't be too sure about whether one electron is already too much. Anyway, you might find this publication interesting: 10.1126/science.1174326 (it is about the neutral dimer though). Judging from 10.1002/qua.20633, it seems to be stable.

related https://chemistry.stackexchange.com/questions/75655/is-be2-diamagnetic https://chemistry.stackexchange.com/questions/55497/what-are-the-molecular-orbitals-shaped-like-in-diberyllium-monocation


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From where does nucleophile attack the ketone in a nucleophilic addition reaction ?
I inserted a picture to help show where my misunderstanding lies.
I know that in the nucleophilic addition of ketones and aldehydes , a racemix mixture is produced . And I know that the reason is that the molecule can be attacked by the nucleophile from “either side of the molecule” or “above and below”
But I can’t seem to picture that .
In the picture I added , I show the one place/direction of attack that makes sense to me , could someone please show me the other “side” ?

Draw in 3D or build a model of  the enantiomers of the cyanohydrin to get you started.

Here's a link to an animation of cyanohydrin formation. You have to click on the reaction arrow and the individual step will play. http://www.chemtube3d.com/Nucleophilic%20substitution%20at%20the%20carbonyl%20group%20-%20Cyanohydrin%20formation.html
Since the carbonyl is completely planar, the nucleophile could either approach from the front or the back. While your diagram seems to indicate the nucleophile attacks in the same plane as the carbonyl (180° angle from O–C–nuc), in reality the angle of approach is roughly 107° (Bürgi–Dunitz angle).


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Oxygen Solubility in Water
This is probably really obvious, but I really don't understand. Why does oxygen dissolve in water when oxygen is a nonpolar molecule and water is a polar molecule? It's obvious that oxygen does dissolve, or else fish could not survive, but I'm confused on how it does so.

Oxygen is relatively insoluble in water, its solubility being only 264 µM at 25 °C. That explains in part why you (and fish) require dedicated oxygen carriers in your blood to transfer sufficient oxygen around your body.
That oxygen is weakly soluble can be explained, as you point out, on the basis of the hydrophobic effect, which amounts to saying that water likes to hydrogen bond, and an oxygen molecule is not very good at that compared to another water molecule.
That oxygen is soluble at all can be explained on two grounds. First, statistics, just the random mixing of matter. Some oxygen molecules will happen to find their way into water and not find their way out for a while. The other explanation is that oxygen is not entirely nonpolar: it lacks a permanent dipole moment, but it is polarizable. Even molecules that we think of as highly water repellent can exist as liquids (oxygen is a gas at room temperature but can be cooled into a liquid state), which means something must be keeping them together, and that is dispersion forces due to the polarizability of all atoms and molecules.


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What is the origin of chirality in the universe?
I am just curious with the fact that our body is made of so many chiral molecules the DNA that we possess is chiral, the amino acids in our body is chiral and there are many other objects other than the molecules which show chirality. So,I want to know what is the origin of chirality in the universe.

For all we know, it is just a coincidence.

Perhaps a distinction should be made. The reason why there are chiral molecules in the first place is that some atoms (notably, carbon) can have stable asymmetric substitution, I'd say. So it's really a fundamental physical property of matter; I'm not sure things could work any other way; maybe some physics experts know. Then the fact that most amino acids making up the proteins in the human body have one specific configuration (S) rather than the opposite one (R) is indeed probably just a coincidence. Very interesting question though.

First consider the comment of @user6376297. Lots of molecules must be chiral by existence! For why in leaving kingdom there is a large excess of one form: digging the web you find scientific hypothesis, of which I link an example https://www.sciencedaily.com/releases/2011/01/110107145634.htm

Interesting article! Reading the conclusion "These findings imply that the selection of a single enantiomer for the molecules of life observed on Earth is not the result of chance but rather of a deterministic physical mechanism." one may then ask the next question: why was there an excess of one enantiomeric form in the place where these extraterrestrial molecules originated? So yes, no problem hypothesizing that what happened on Earth was mechanistic. Different story would be to imply that there was some form of 'choice' at the very origin of matter. Quite a delicate topic.

IT is a big question with a lot of uncertainty about the true answer. But one chemical phenomenon is important whatever the mechanism and that is the ability of certain reactions to increase small differences in the ratio of chiral enantiomers by self-catalytic reactions. In short: start with a small excess and end up with a large one. What caused the initial imbalance could be chance or the result of a small chiral imbalance caused by some nuclear processes some of which are not symmetric in reflection.

I'm voting to close this question as off-topic because it's not about chemistry.

That's pretty much as asking why things can be asymmetrical. Nothing to do with chemistry,  more like philosophy and even that may be a stretch.

But the above mentioned article shows that it is a deterministic physical mechanism.

@Mithoron This question can be understood in two ways. One the one you voted to close for (Why can molecules be chiral in the first place) and one that Ivan commented on (Why are the molecules in our body of one specific enantiomer). Sadly, both are not answerable …

Thus is surely chemistry tough the question looks bombastic. Again the question is not why optically active molecules exist (legit itself) but why there is an asymmetry of enantiomers distribution in living kingdom. There is not an accepted answer: to the best of my knowledge this is ongoing research in chemistry / biochemistry and perhaps we will find out. It might be in principle even due to merely physical mechanisms. I am not sure that a Q is off topic just because we didn't know the answer. Also not sure this comment shall be an answer. But OP should have got it.

So, is there any current update about this research?


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In zero order why we don't need the coefficient of reactant in integration of rate law?
For a reaction: $\ce{aA->product}$
The rate law is $\text{Rate}=-\dfrac{\mathrm d[A]}{a\,\mathrm dt} = k$
Where:

$k$ is the rate constant
$[\ce{A}]$ is the concentration of the reactant $\ce{A}$ in the reaction
$a$ is the stoichiometric coefficient of $\ce{A}$

If we integrate the relation, we get:
$$[\ce{A}]-[\ce{A}]_0=-akt $$
However, my instructor wrote the relation on board as below;
$$[\ce{A}]-[\ce{A}]_0=-kt$$
Why is $a$ not needed in the integrated rate law?

Because what you are referring to (with the inverse of the stoichiometric coefficient multiplied with d[A]/dt)is the reaction rate(r). This rate is related to the concentrations of the reactants via the rate equation(r= f([A]) which can only be determined experimentally. If it is zero order, you already know that d[A]/dt= [A]^(0). Hence there is no role of a here

I think you take the rate  expression of the reaction as $\frac{1}{a}\times(\text{ the rate of disappearance of}\mathrm{A}$), and your instructor take the rate expression of the reaction as equal to (the rate of disappearance $\mathrm{A}$) .
Both cases accurately describe the kinetics, but different in defining the rate and $\mathrm{k}$.

You are right, for consistency the stoichiometric coefficient should be retained.
Keeping track of the stoichiometric coefficients is essential to convert between rates for different species.
If you want to compare the rate of formation of reagent A and product, you write
$$\text{Rate}=-\frac{\mathrm d[A]}{a\,\mathrm dt} = \frac{\mathrm d[product]}{\,\mathrm dt}= k$$
and so:
$$\ce{[A]-[A]_0=-a*k*t} $$
or
$$\ce{\frac{[A]-[A]_0}{a}=-k*t}$$
and
$$\ce{[product]-[product]_0=k*t}$$
However you would say that the reagent A is consumed with an effective rate constant $\ce{k' = a*k}$.


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Difference in leaving group ability due to variation in nucleophiles
Wikipedia writes:

There can still exist contextual differences in leaving group ability in the purest form, that is when the actual group that leaves is not affected by the reaction conditions (by protonation or Lewis acid complexation) and the departure of the leaving group occurs in the rate determining step. In the situation where other variables are held constant (nature of the alkyl electrophile, solvent, etc.), a change in nucleophile can lead to a change in the order of reactivity for leaving groups. In the case below, tosylate is the best leaving group when ethoxide is the nucleophile, but iodide and even bromide become better leaving groups in the case of the thiolate nucleophile.

Why is it so?As far as I know, leaving group ability is basically how stabilized the leaving group is when it exists as an independent species. I can't see why it depends on the nucleophile

You are comparing the leaving groups via the rate of the reaction. and the leaving group and the nucleophile will affect the rate.

The data which you are using to explain a groups tendency to act as a good leaving group is via a nucleophilic substitution reaction.  Nucleophilic sub reactions are affected by 3 things mainly: the nucleophilicity of the nucleophile, the strength of the bond being broken and how easy is it for the leaving group to leave.
This means if you use a better nucleophile the substitution reaction will happen faster. so in the case of your data, you are comparing leaving groups using "relative rates". So the nucleophilicity of the nucleophile will end up impacting the rate. So by looking at two different nucleophiles you are looking at each nucleophiles nucleophilicity and the leaving group.

Ok, granted that the nucleophiles have different hardness and so, they will have different rates of reaction when ccompared with each other.  But why should that affect the leaving group ability for the same set of reactions of one nucleophile? How does 'relative rate' of leaving groups vary for a nucleophile in this case?

I think you are just misinterpreting the data