text:
Is there a molecule which has only lone pair electrons?
Lone pair electrons are "valence electrons that are not shared with another atom in a covalent bond" (Wikipedia).
Is there a molecule which has only lone pair electrons?

That wouldn't be a molecule. Lone atoms are fine with that, obviously, and so called van der Vaals molecules technically don't have bonding pairs.

Only if atoms can mutually bond and not to bond at the same time.

Thank you guys, I could not understand that part.

An ion pair, if the bonding can be described with good accuracy as fully ionic.

The answer is No ! If a molecule had only lone pairs, it cannot be a molecule. A molecule has bonds, and these bonds are made of paired electrons. Unpaired electrons cannot be used for bonding atoms. So a molecule must have more bonding electrons than non.bonding electrons.

Yeah, but these must be single-atom molecules. As soon as you have multiple atoms, you are going to have a bond, and the bonding electrons are inevitably no longer lone pair electrons.

When copy-pasting links from Wikipedia, please try to avoid using mobile links which go like en.m.wikipedia.org; instead change them to the usual desktop links en.wikipedia.org. If somebody is reading on mobile they will be directed to the correct version anyway. Also, it's better to quote whatever you need from the page, rather than just putting the link without explanation.

Aren't singlet carbenes not examples of this?

@ml in a singlet carbene, there is a lone pair on the carbenoid carbon but also two bonding pairs from that atom to the substituents.

What about nitrogen monoxide and dioxide?

Simple answer: Almost. It is nearly the case that all the valence electrons of the $\ce{CsF(g)}$ monomer can be considered as lone pairs.
At high temperatures you can have gas-phase salt monomers, i.e., single molecules of $\ce{MX(g)}$, where M is a metal ion and X is a non-metal ion (e.g., $\ce{NaCl(g)}$). [See: Gas-Phase Dimerization of Sodium Halides and Formation of Mixed Sodium/Phenyl Iodides] And these would meet IUPAC's definition of a molecule:

An electrically neutral entity consisting of more than one atom (n>1). Rigorously, a molecule, in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

https://goldbook.iupac.org/terms/view/M04002.
If the bond between the ions were purely ionic then, yes, you could have bonding in a molecule where there are only non-bonding electrons, i.e., only lone pairs.  However, even the most ionic bonds have some covalent character, which means there is some electron sharing.
There are many different ways to assess ionic character. For the purposes of answering this question, I'll (somewhat arbitrarily) use Pauling's formula applied to the difference in electronegativity, since it's simple and well-known.  However, note that there are other measures of ionic character besides electronegativity difference; and even when using the latter, there are many formulas besides Pauling's.
Among atoms whose electronegativities have been measured, the most ionic bond possible would be that between F and Cs, which have Pauling electronegativities of $3.98$ and $0.79$, respectively. [Source: Wikipedia: Electronegativities_of_the_elements].
Pauling proposed the following formula to estimate the amount of ionic character in a bond:
$$\text{Amount ionic character} = 1 - e^{-\frac{1}{4}(\chi_A-\chi_B)^2}$$
Plugging those electronegativities into the above formula, we find that the bond in $\ce{CsF(g)}$ has only a small amount (8%) of covalent character:
$$\text{amt. ionic character} = 1 - e^{-\frac{1}{4}(3.98-0.79)^2} = 0.92
\implies \text{amt. covalent character} = 0.08 $$
Thus we can conclude that, in the most ionic molecules, while there is some electron sharing, it is small.  Hence all the valence electrons can nearly be considered purely as lone pairs.
Note:  Using Pauling electronegativities and Pauling's formula gives a rough approximation of ionic character.  A more sophisticated analysis could yield an answer that, while qualitatively similar (it would still show very little covalent character in $\ce{CsF(g)}$), could differ quantitatively.
As an aside (and a caution):  The formula in the IUPAC Gold Book ( https://goldbook.iupac.org/terms/view/IT07058 ) is wildly incorrect, and has apparently been so since 2014.  It looks like some things got lost during the typesetting.  I emailed IUPAC to suggest a correction.

Addendum
It seems we shouldn't be too hard on IUPAC, since Pauling himself appears to have a typo in his own formula! Here's a screenshot from:  Pauling, Linus. The Nature of the Chemical Bond. Third Edition. Ithaca, NY: Cornell university press, 1960, p 98.  [Available from:  https://archive.org/details/natureofthechemicalbondpauling/page/n115/mode/2up ] You can see the electronegativity difference in eqn. 3-15 is properly exponentiated, but it also should be squared, and it's not:

To confirm that there's a typo in Pauling's book, take a look at the table at the bottom of the above page. If we don't square the electronegativity differences, we don't get the right values (calculations on left); if we do square them, we do (calculations on right) (calculations done in Mathematica; these are screenshots of the inputs and outputs):

"Fr should be slightly less electronegative than Cs, but its electronegativity hasn't been measured". Not everyone agrees. See the discussion under "Characteristics" here. Relativistic effects are entering in.

It seems that the original article (https://doi.org/10.1351/pac199971101919) from which the Gold Book entry was excerpted also got the formula wrong.

...Hmm, now I'm pretty intrigued; do you happen to know where Pauling originally proposed this formula? I've been taking a look at https://doi.org/10.1021/ja01348a011 and some other articles in the same series he published, but with no luck. Have to get back to real work now, but I might try looking again later.

With regard to reporting Gold Book errors, please see my comment here: https://chemistry.stackexchange.com/a/115884/79678. Stuart will dig into it!

@OscarLanzi Thanks, I accidentally reversed the statement about francium in the Wikipedia article I cited. It actually said "francium is expected and, to a small extent, observed to be more electronegative than caesium". I'll edit my post. BTW, do you know of more sophisticated assesments of the %ionic character of the Cs–F bond and, if so, what do they say? Also, for salts, do you know if the bond dissoc. energies on which the electronegnativity differences are based are for ion pairs or ions in crystals? I'm wondering about the diff. in %ionic character for ionic bonds in pairs vs. crystals.

@EdV Actually, Stuart was the one I emailed :)!   See also the addendum I added to the end of my post.


text:
What is the correct way to write dissociation reactions?
The variants are as follows:
For a salt (NaCl):
$$\ce{NaCl_{(s)}<=>NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{1}$$
$$\ce{NaCl_{(s)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{2}$$
$$\ce{NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{3}$$
$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{4}$$
$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{5}$$
$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{6}$$
$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{7}$$
$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{8}$$
Which of these are right/wrong and why?

I am afraid you are overthinking about notation. The only problem is that the plus and minus are too far off from the elemental symbol. An chemical equation cannot describe every process. It is just a shorthand notation. You will still need words to explain what is going on.

I am showing the different versions of the equation followed by my comments.

$$\ce{NaCl_{(s)}<=>NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{1}$$

$\ce{NaCl_{(aq)}}$ is the same as $\ce{Na_{(aq)}^{+} + Cl_{(aq)}^{-}}$ so it makes no sense to show them in equilibrium.

$$\ce{NaCl_{(s)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{2}$$

This would be my preference. The physical state is written more conventionally after the charge of the ion, and not as a subscript:
$$\ce{NaCl(s)<=>Na+(aq) + Cl-(aq)}\tag{2a}$$

$$\ce{NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{3}$$

See (1).

$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{4}$$

There is no reason to show a single water molecule. This would suggest a 1:1 stoichiometry but there are multiple water molecules in the solvation shells of the two ions.

$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{5}$$

See (4).

$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{6}$$

See (4).

$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{7}$$

This one is not balanced.

$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{8}$$

See (7).

Which of these are right/wrong and why?

Some of them are not balanced, and some, while balanced, are not describing the process as accurately and succinctly as version (2).

I agree completely.

@user545735 either one is fine. (1) shows an H+ transfer from acetic acid to water (Bronsted acid/base). (2) saves some keystrokes and shows the water implicitly.

@user545735 Yes because there are more than one water molecule involved in hydrating or salvation the hydrogen ion.

source: Stack Exchange

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

sort: votes


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Precipitation of equimolar amounts of chloride and bromide
I was just wondering about the following situation; if we had a solution containing 0.1M bromide and 0.1M chloride ions, and treated it with excess silver nitrate, what would happen and would quantitative precipitation of both silver salts be possible?
If I understood correctly, as AgBr is less soluble, it precipitates first. I know that the solubility product of AgBr is lower than AgCl, but why does it selectively precipitate first?
Anyway, it precipitates until the point where the remaining bromide (about 0.48% of 0.1M) remains; at this point, AgCl is more insoluble and starts precipitating, until AgBr is more insoluble again, that’s how I understood it.
So basically, am I correct in the assumption that quantitative precipitation of both salts in same solution is not possible because at some point both salts precipitate simultaneously although there is still a significant amount of bromide in solution?

In practice, the very first drop of AgNO3 solution will create a huge oversaturation in both AgBr and AgCl, so both will start precipitating chaotically.

It can be said that precipitation of chlorides may be preferred kinetically even before it is preferred thermodynamically.

@KarstenTheis It is even her own question......

All depends on the particular scenario.
Let assume the concentrations of both halogenides and silver nitrate are low enough, so only the solubility product of silver bromide is reached. Then silver bromide is precipitated selectively ( if we neglect coprecipitation effects ).
Similar effect is achieved, if nitrate is being added so slowly the reprecipitation is faster than primary precipitation.
Otherwise, both salts are precipitated simultaneously. Saying this, be aware there will be ongoing fast or slow recrystallization/reprecipitation.
More soluble chloride will gradually dissolve in favour of bromide precipitation until chloride/bromide concentration ratio is equal to ratio of the respective solubility products.

And if we add excess we are basically precipitating both of them at the same time, regardless of relative solubility?

If you think otherwise I am curious about your reasons. // Who asks too much, thinks too little.

Well, in my lab experiment (measuring iodide and chloride) I conducted potentiometry and we measured the potential change of the solution. There were two equivalence points after 1.82 and 3ml of AgNO3 0.1M. That would mean that first the iodide, then the chloride precipitates, we also only gave amounts of 0.1ml of AgNO3 0.1M into it.
In my handbook it says that measuring AgBr with this method wouldn’t be possible because the solubility of AgBr and AgCl are closer together. But I didn’t really understand if they meant giving in AgNO3 as excess or how you put it, as infinitely small additions

Yes, potentiometry is preferred way here. 2 equivalence points do not mean there is no AgCl(s) before the first one. All depends on the scenario details. The point just means I- are done, Generally both nitrate and hologenides are too concentrated.

If some $\ce{AgCl(s)}$ is introduced into a bromide solution, it reacts with $\ce{Br-}$ ions according to : $$\ce{AgCl(s) + Br- -> AgBr(s) + Cl-}$$ so that $\ce{AgCl}$ is quickly transformed into $\ce{AgBr}$ which is nearly insoluble in, for example,  $\ce{NH3}$ solution. So if some $\ce{AgNO3}$ is added to a solution containing both $\ce{Br-}$ and $\ce{Cl-}$ ions, both may react in the beginning with $\ce{Ag+}$ ions, but after some time the precipitate is only made of pure $\ce{AgBr}$, because the $\ce{AgCl}$ precipitate (formed in the beginning) has all reacted with bromide ions. $\ce{AgCl}$ will be obtained only when the $\ce{Br-}$ ions have all been consumed.

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Fugacity vs compressibility
What's the criterion for saying that the forces of attraction are dominant?
f < p or Z < 1
I ask this because there is a situation where
f < p and yet Z > 1 (below Boyle temperature)
if the criterion is fugacity, it is not so intuitive to say that the forces of attraction are dominant and the real gas has a molar volume greater than that of an ideal gas.
f: fugacity;
Z: compressibility factor;
p: pressure

Let $\varphi=\frac{f}{p}$, where f is the fugacity, and p is the pressure.
Then:
$$\ln \varphi = \int_0^p \big( \frac{Z-1}{p} \big)dp $$
[Source for above expression: Wikipedia article on fugacity]
Hence the fugacity coefficient, $\varphi$, at a certain $T$ and $p$, isn't determined by the compressiblity factor, $Z$, at that $T$ and $p$.  Instead, it's determined by integrating $Z$ with respect to $p$, at the given $T$, according to the above expression.
I.e., $\varphi$ is determined by all the values that $Z$ took on in bringing the gas up to the current pressure.  This makes sense, because it's directly related to the free energy of the gas.
Consider these three limiting cases, in which $Z$ has the specified value (or range of values) for the entire process (you should confirm these yourself, for your own understanding, by checking the above integral):
I. $Z=1 \implies \varphi=1$ ($f=p$)
II. $Z>1 \implies \varphi>1$  ($f>p$)
III. $Z<1 \implies \varphi<1$ ($f<p$)
One can explain the result you are seeing as a case in which $Z<1$ for a sufficient part of the integration such that $ \varphi<1$ ($f<p$) but, at the specific pressure you are considering, $Z>1$.
Physically, this means the following:

$Z>1$: Excluded volume effects dominate at the current pressure.

$\varphi<1$ ($f<p$): Attractive forces dominate over most of the range between $p=0$ and $p=\text{current pressure}$.

@ÉdersonD'MartinCosta  I added, at the end, a physical conclusion that corresponds to the mathematical conclusion.


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Alkaline hydrolysis with KOH instead of NaOH. Why?
I really loved the series breaking bad. So I started reading about the science behind. I found this paper:
Harnisch, F., & Salthammer, T. (2013). Die Chemie bei Breaking Bad. In Chemie in unserer Zeit (Vol. 47, Issue 4, pp. 214–221). Wiley. https://doi.org/10.1002/ciuz.201300612
In Breaking Bad they used HF to dissolve a dead corpse. But in the scientific paper they say that it is not a good way to do it. They talk about some other chemicals, I can't remember them. But they also talked about NaOH.
In Dr. Stone they claim that sodium hydroxide is used to melt corpses by the Yakuza.
Reading a little on this topic I found a Wikipedia article. It is called Alkaline hydrolysis (body disposal). The article is about an alterntative method to burial or cremation. But there they use KOH.
I was just wondering: is there a scientific reason why they use KOH instead of NaOH?

KOH and NaOH are pretty much interchangeable.

Have a look at these articles: https://noahchemicals.com/blog/sicomparison-potassium-hydroxide-and-sodium-hydroxide .. http://info.soapwarehouse.biz/blog-0/bid/83765/Sodium-Hydroxide-vs-Potassium-Hydroxide-what-s-the-difference


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Isn't my book doing this problem about ionic product wrongly?
Problem:
After adding a $\pu{150 mL}$ $\pu{6.7\times10^{-5} M}$ $\ce{NaOH}$ solution to a $\pu{100 mL}$ $\pu{2.5\times10^{-5} M}$ $\ce{FeCl2}$ solution, green colored $\ce{Fe(OH)2}$ precipitate is found. What is the ionic product of $\ce{Fe(OH)2}$?
My book's solution:
The concentration of $\ce{Fe^2+}$ will be the same as the concentration of $\ce{FeCl2}$. So, using the formula for dilution we can get the concentration of $\ce{Fe^2+}$:
$$\begin{align}
    M_1V_1 &= M_2V_2 \\
    2.5\times10^5\times100 &= M_2\times250 \\
    M_2 &= \frac{100\times2.5\times10^{-5}}{250} \\
        &= \pu{1\times10^{-5} M}
\end{align}$$
In the same way, the concentration of $\ce{OH-}$ will be the same as the concentration of $\ce{NaOH}$. So, using the formula for dilution we can get the concentration of $\ce{OH-}$:
$$\begin{align}
    M_3V_3 &= M_4V_4 \\
    M_4    &= \frac{150 \times 6.7 \times 10^{-5}}{250} \\
           &= \pu{4.02\times10^-5 M}
\end{align}$$
Now, ionic product of $\ce{Fe(OH)2}$, $K_\mathrm{ip}$,
$$K_\mathrm{ip} = [\ce{Fe^2+}][\ce{OH-}]^2 =1.616\times10^{-14} \quad \text{[Ans.]}$$
My observations:
My book assumed that $\ce{Fe(OH)2}$ remains completely dissociated and thus the concentration of $\ce{Fe^2+}$ will be the same as the concentration of $\ce{FeCl2}$, and the concentration of $\ce{OH-}$ will be the same as the concentration of $\ce{NaOH}$. My book's answer is correct only at the very beginning of the process when any $\ce{Fe(OH)2}$ precipitate hasn't formed yet. However, after the precipitate has formed, isn't my book's answer wrong?

For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ).
See also upright vs italic

The textbook uses a simplified approach with a badly worded task. It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \le  \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive. The constant also does not count with ions trapped in precipitation.

@Poutnik Edited the question using \ce and \pu using the link you kindly provided. Moreover, if you post this comment as an answer, I'll accept it as my answer.

Poutnik is right. The problem should have been better written according to : When mixing small amounts of a $\ce{6.7·10^{-5} M}$ $\ce{NaOH}$ in $ 100$ mL $\ce{2.5·10^{-5}}$ $\ce{ FeCl2}$ solution, nothing happens. But when $150$ mL of this $\ce{NaOH}$ has been added, the first $\ce{Fe(OH)2}$ precipitate becomes visible. Calculate the ionic product of $\ce{Fe{OH}2}$ .

The textbook uses a simplified approach with a badly worded task.
It addresses the point just before precipitation start. In fact, there should be rather inequality $K_\mathrm{sp} \lt \pu{1.616E−14}$, as it does not say when prcipitation started if mixing was progressive.
The constant also does not count with ions trapped in precipitation.


text:
On the definition of eigenframe
While reading the following article,

Elizaveta A. Suturina, Kevin Mason, Carlos F. G. C. Geraldes, Ilya Kuprov, David Parker, Beyond Bleaney's theory: experimental and theoretical analysis of periodic trends in lanthanide-induced chemical shift, Angewandte Chemie, September 25, 2017

I came across the term eigenframe, which I am unfamiliar with. What is an eigenframe?
Here is the context in which it was used:

In particular, pseudo contact chemical shifts (PCS) induced by lanthanide complexes are commonly described by McConnell's long-range relation (SI units),
$$ \delta^{PCS} = \frac{1}{12\pi r^3}[\chi_{ax}(3\cos^2 \theta -1)+3\chi_{rh}\sin^2\theta \cos2\phi]$$
where $\theta, \phi, r$ are the polar coordinates of the nucleus relative to the lanthanide in the eigenframe of the molar magnetic susceptibility tensor, and $\chi_{ax}$ and $\chi_{rh}$ are axiality and rhombicity of the susceptibility tensor.

I can't find anything definitive but both
https://www.thestudentroom.co.uk/showthread.php?t=4172326
and
https://www.physicsforums.com/threads/eigenframe-definition-please.876035/
suggest that an eigenframe is the basis formed by the eigenvectors of the relevant operator, the molar magnetic susceptibility tensor in this case.


text:
Comparing acidity between 4-hydroxybenzaldehyde and 2-hydroxybenzaldehyde

https://hmdb.ca/metabolites/HMDB0011718
https://hmdb.ca/metabolites/HMDB0034170
The above two links provide the pKa values for the 4-hydroxybenzaldehyde (pKa = 7.32) and 2-hydroxybenzaldehyde (pKa = 8.4) ...
(not sure as to how reliable the source is)

How do I compare their acidic nature? In either cases there's an electron withdrawing group (aldehyde) at ortho and para position, conjugate base is stabilised either way.
How would the meta- isomer differ from the acidity comparison?

Consider interaction of groups in ortho position. // Note that questions should be as self-contained as reasonably possible. Links should merely be just data references or source of further study, if the relevant part of link target content can be written as quoted text, eventually formatted by SE Markdown or MathJax/mhchem.


text:
Gibbs energy of electron transfer
The reaction Gibbs free energy of a cell is
$\Delta_\mathrm{r}G^\circ = -nFE^\circ, \tag{1}$
and the Gibbs energy of photoinduced electron transfer according to IUPAC is
$$\Delta_\mathrm{ET}G^\circ = N_\mathrm{A}\Big\{e\big[E^\circ(\ce{D^{+.}/D}) - E^\circ(\ce{A/A^{-.}})\big] + w(\ce{D^{+.}A^{-.}}) - w(\ce{DA})\Big\} - \Delta E_{0,0}. \tag{2}$$
Are both the same except for the vibrational zero electronic energy of the excited partner $\Delta E_{0,0}/\pu{J mol^-1}$ (energy of the excited state)?
The Gibbs energy of photoinduced electron transfer accounts for the effect of Coulombic attraction in the products $w(\ce{D^{+.}A^{-.}})$ and reactants $w(\ce{DA})$. Why does the reaction Gibbs energy of a cell not have these terms?


text:
Inductive Effect of -CH3 group
3° carbocations are the most stable because of the +I effect of -CH3 group. This group gives a slight negative charge to the central carbon atom, which makes it less positively charged and thus, more stable.
But, in the case of -N(CH3)3 and -NH3, the central N atom has a positive charge.
-CH3 group has +I effect on the central N atom, making N atom slightly less positively charged. So, -N(CH3)3 group should have lesser -I effect. Isn't this statement correct?
But, -N(CH3)3 group has greater -I effect than -NH3. Why does this happen?
Why does -N(CH3)3 have greater -I effect than -NH3?

The R3N- group has a full positive charge that you omitted from the structure.

Yeah, but still the +I effect of R will make the N atom less positively charged, similar to the case of carbocations. So, in R3N- group, the central N atom, being slightly less positively charged than that of H3N- group, should give lesser -I effect than H3N- , right?

Sorry if I have made any mistake. I am in high school now. I cannot understand this phenomenon. Why does -N(CH3)3 gives more -I effect than -NH3?

Are you speaking of the positive $\ce{NH3^+}$ ion, or of the negative  (-)$\ce{NH3^-}$ ion, or of the neutral molecule $\ce{NH3}$ ? Or even of the $\ce{R-NH3^+}$ ion ?

NH3+ ion and NR3+ ion. "-NH3" and "-NR3" are common ways of representation of those groups.

The methyl group carbon has a partial negative charge due to the polarity present in the C-H bonds. This partial negative charge is preferentially donated, leading the methyl group to have the +I effect.
An important way to consider the stability of a tertiary carbocation is through the process of hyperconjugation. Neighboring C-H or C-C bonds, parallel to the vacant p-orbital of a carbocation, can donate some of their electron density to the empty p-orbital, thus stabilizing the positive charge. Tertiary carbocations have a greater capability of providing this electron density, and thus are more stable.
The positive charge present on the nitrogen in either the -NR3+ or -NH3+ groups induces the extent of the -I effect exhibited by both groups. Although it may seem that the presence of alkyl groups would lend a greater +I effect to the -NR3+ group, there are other factors that result in this "anomaly" of sorts.
The factor that causes this is the issue of steric hindrance. The alkyl groups surrounding the nitrogen are relatively large and do not want to be in the vicinity of each other. Although the structure of the -NR3+ group would appear to be tetrahedral, as the alkyl groups push away from each other, they move towards what seems to be more of a planar structure. This is an increase in bond angle from the tetrahedral structure.
It is important to note that the greater the bond angle, the greater %s character. S orbitals are typically closer to the nucleus as compared to the p orbitals, and thus, the electrons are held closer -- the electronegativity of the nitrogen atom will be greater than if it was in the -NH3+ group. As such, it exhibits a greater -I effect.


text:
Wave Dispersive X-ray Fluoresence and Bragg's Law
Thermo Fisher website explains that "WDXRF uses crystals to disperse the fluorescence spectrum into individual wavelengths of each element, providing high resolution and low background spectra for accurate determination of elemental concentrations." Additionally, it goes on to say

WDXRF systems are based on Bragg’s law, which states that crystals will reflect x-rays of specific wavelengths and incident angles when the wavelengths of the scattered x-rays interfere constructively. While the sample position is fixed, the angles of the crystal and detector can be changed in compliance with Bragg’s law so that a particular wavelength can be measured. Only x-rays that satisfy Bragg’s law are reflected.

However, I am still confused on how the crystal, in correspondence with Bragg's law, disperse the spectrum to show each residual wavelength. Because there is a large variation of wavelengths coming from the sample, wouldn't this mean that passing it through the crystal would cause some of them to destructively interfere? How does this help with analyzing the spectra?
Hopefully, someone can provide a more in depth explanation on how this technique is used to analyze fluorescence spectra.

Different 'colors' (energies) of x-rays will diffract at different angles from a crystal. That is all. Either have a position sensitive detector, or rotate the crystal to sample the different energies at one fixed detector.

Do you recall optical (visible) diffraction grating? Let us say, wavelegnths from 400-700 nm pass through a transmission grating. What do you see? A rainbow. This means that all wavelengths passed through the diffraction grating, but at different diffraction angles. A grating does not annhilate energy during destructive interference.

"X-ray monochromators are analogous to grating monochromators and spectrometers in the visible portion of the spectrum. If the lattice spacing for a crystal is accurately known, the observed angles of diffraction can be used to measure and identify unknown X-ray wavelengths.

Because of the sensitive wavelength dependence of Bragg reflection exhibited by materials such as silicon, a small portion of a continuous spectrum of radiation can be isolated. Bent single crystals used in X-ray spectroscopy are analogous to the curved line gratings used in optical spectroscopy. The bandwidth of the radiation after it has passed through a high-resolution monochromator can be as narrow as Δλ/λ = 10−4, and, by tilting a pair of crystals with respect to the incident radiation,

the wavelength of the diffracted radiation can be continuously tuned without changing the direction of the selected light. (From Encyclopedia Britannica).


text:
Why are hydrated lithium ions' radii larger than hydrated sodium ions' radii?
Why are hydrated lithium ions' radii larger than hydrated sodium ions' radii i.e. $r_\ce{Li+(aq)}>r_\ce{Na+(aq)}$?
If ionic radii increase down the group i.e. $r_\ce{Li+}<r_\ce{Na+}<r_\ce{K+}$ then the radii of hydrated ions also increase down the group i.e. $r_\ce{Li+(aq)}<r_\ce{Na+(aq)}<r_\ce{K+(aq)}$. But according to my class notes, The radii of hydrated ions are larger than those of gaseous ions. The radii of hydrated ions decrease as the group. I'm not sure I understand. But, according to the class notes and Quora answers, how can the size of an ion alter the hydration radii?
According to a Quora answers,

Hydration energy is the amount of energy liberated when an ion gets hydrated that is , it accepts water molecule. e.g. lithium ion is small in size so it accepts large number of water molecule so it's hydration energy is more whereas caesium ion is bigger in size so it accepts less number of water molecule and has less hydration energy.

Another answer says,

As the size of ion increases its polasiation tendency decreases therefore the tendency to polarise nearby water molecules decreases hence decreasing its hydration radii. This effect can be seen by observing the trend of hydartion radii in alkali metal ions.

Look around the site, it was discussed recently.

Please check you greater/smaller than signs. Are they like you intended?

This current profile is unregistered. Is this your another profile? https://chemistry.stackexchange.com/users/100905/huzaifa-abedeen

Down a group, ionic radii increases. But at the same time, due to decrease in charge density, the ions attract less number of water molecules. The smallest cation (with the same charges) having highest charge/size ratio attracts maximum number of water molecules.

Image credit: https://rwu.pressbooks.pub/bio103/chapter/the-chemical-context-of-life/
Compare the relative increase in ionic radii down a group to effective radii of hydrated ion, i.e., with attached whole water molecules. Clearly, there will be a decrease in radii of hydrated ions.


text:
Normalized absorbance/transmittance definition
In the absorption spectroscopy we can calculate transmittance $T$ of a given sample by comparing how the intensity of the incident beam $I_{0}$ is decreased with the distance (Lambert-Beer Law).
$A = -\ln{\frac{I}{I_{0}}}=\alpha L$
where $A$ is absorbence, $T = 1 -A$, $\alpha$ is an absorption coefficient and $L$ is the distance.
But I encountered something called Normalized transmittence I don't really understand how can I get it from measured data. Normalized how? It appears in scientific articles but I couldn't find any good source explaining it really.
I read that I should divide my measured $I$ by reference intensity (it can fluctuate so it is measured along the way too). So the normalized absorbance would be $A = -\ln{\frac{I}{I_{0}^2}}$ or just by definition $A = -\ln{\frac{I}{I_{0}}} $
Can anyone help me please?
The spectra then look like this:

[https://www.researchgate.net/figure/Hydrogen-cyanide-H-13-C-14-N-2-3-rotational-vibrational-band-spectrum-obtained-by_fig4_242279131]

This was posted by you and got an answer that you accepted: https://physics.stackexchange.com/q/685025/313612. Is there something that is specifically not being addressed?

I’m voting to close this question because it's a cross-posted question from https://physics.stackexchange.com/questions/685025

Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site?

Recall the more familiar definition of light transmittance
$T = \frac{I}{I_o}$, many instruments quote percent transmittance as
$\%T = 100\frac{I}{I_o}$
Normalized transmittance is nothing but the maximum measured transmittance set to unity. This is easy to accomplish. Suppose you have 2000 data points (y-values) in your spectrum collected as a function of wavelength. Locate the maximum value of transmittance, say it was 95% or 0.95 in your data set (depending what was collected experimentally). Now divide the entire y-values with 95.65% or 0.9565. With that your maximum, transmittance will be exactly equal to 1  and the rest of the values will be less than 1. This process is called normalization.
It is just for convenience when you are comparing a lot of spectra.

Upvoted, but it is not clear why the OP first posted at physics SE, accepted an answer and then came here. Modern double beam-in-time instruments are rather good, so double beam-in-space spectrometers are not necessarily preferred for source noise performance.

@EdV, I think the OP was confused how to normalize the experimental data. I did not know it was posted twice.

Yes, that is what I figured. But the HCN IR spectrum was apparently used to calibrate a laser system, based on what I find when searching. In the old days, a pulsed laser might have been used to pump a tunable dye laser and then pulse-to-pulse fluctuations in the pump laser’s output, typically a few percent, had to be normalized out. Missing and runt pulses were especially annoying and experimental data had to be corrected for those. I used to use a window comparator to ensure collection of data from non-defective laser pulses.


text:
Dimerization of aniline to biphenyl
I've been looking for ways to synthesize biphenyl - $\ce{(C6H5)2}$.
The Ullmann coupling is inaccessible, as bromobenzene is inaccessible in my country due to the manufacture of PCP.
I had the idea to use a version of the Sandmeyer coupling between 2 molecules of aniline - $\ce{C6H5NH2}$.
My assumptions that this is possible are because of the synthesis of diphenic acid, wherein anthranilic acid is diazonated and coupled using a copper(I) ammonia hydroxylamine complex, as per this paper: http://orgsyn.org/demo.aspx?prep=cv1p0222.
I believe it would involve dimerization of the phenyl radical, similar to how the Sandmeyer reaction involves radical addition to copper(II), where the phenyl radical comes from copper(I) induced expulsion of nitrogen from diazonium salts. Is this mechanism correct?
Is is possible to do something similar with anilin? Is there literature for this method?
Please be mindful that I'm not too experienced in the mechanisms of these types of reactions.

Looks like you have a good idea. If the route via the diazonium salt fails, you can prepare bromobenzene from it via the Sandmeyer reaction. If you have access to benzene and bromine, you can make bromobenzene.

I see...thanks for the feedback. Would you mind leaving an answer?

As OP commented that they don't want benzene in the synthesis due to its carcinogenicity, here is an alternative route involving halobenzene:

Sandmeyer reaction to form halobenzene

Wurtz reaction or Ullmann reaction

The coupling reaction can be greatly improve (in terms of selectivity and % yield) by using high-end catalyst and improved reaction conditions. Here are some methods:

Fluorobenzene and phenyllithium

Using palladium based catalyst

Palladium on carbon. Here is a paper describing the reaction

Suzuki-Miyaura coupling

Kharasch coupling

Reaction sources:

Continuous Processes for the Synthesis and Isolation of
Functionalized Biphenyls via Suzuki-Miyaura Cross-Coupling
Reactions by Georg Johannes Lichtenegger (pdf)
Zenish J. Jain, Paraag S. Gide, Rani S. Kankate, Biphenyls and their derivatives as synthetically and pharmacologically important aromatic structural moieties, Arabian Journal of Chemistry, Volume 10, Supplement 2,
2017, DOI: 10.1016/j.arabjc.2013.07.035

OP's route through Diphenic acid can also work. Diphenic acid can be decomposed thermally as well as catalytically to form various PAH and its derivative, one of the product being biphenyl. [Beware: PAH are carcinogenic]

Ref.: Oxidative Ring-Opening of Aromatics: Decomposition of Biphenyl
Carboxylic Acids and Zinc Biphenyl Carboxylates by Natalia Montoya Sanchez and Arno de Klerk. (pdf)

Thanks, but I don't want to work with benzene. It's a carcinogen, and can't be stored in a nonvolatile state.

@Jacob Tbh, biphenyl is also carcinogenic and it can cause kidney stone formation and liver tumors. That bein said, do you want an alternative route involving making a biphenyl derivative and then reduce it to biphenyl?

Sure. Benzene is more volatile than biphenyl, hence I don't want to use it.

The thermolysis of diphenic acid works for me.


text:
Why are group 13 compounds in the +1 oxidation state more ionic than those in the +3 oxidation state?
I know that for group 13 metals in the p block, the stability of +1 OS is generally more than the stability of +3 OS as it is energetically not favourable to attain the higher OS of +3 and such ions act as strong oxidising agents.
$$\ce{\overset{+3}{Tl}+2e^-->\overset{+1}{Tl}}$$
This is called the inert-pair effect. Today, however, I learned in group 13 of p-block metals that the metals in +1 OS are more ionic than those in +3 OS for the same metal. Why is metal in its lower OS of +1 more ionic but in its higher OS of +3 more covalent? For example, Thallium(I) chloride $\ce{TlCl}$ (in its +1 OS) is more ionic or less covalent, and Thallium(III) chloride $\ce{TlCl3}$ (in its +3 OS) is more covalent or less ionic. Does this also suggest that ionic is more stable than covalent in this situation?

https://chemistry.stackexchange.com/questions/30854/why-does-thallium-form-tlf-and-tlf3-but-aluminium-only-forms-alf3 ... https://chemistry.stackexchange.com/questions/146851/in-what-form-does-the-respective-halides-of-thallium-exist-as ... https://chemistry.stackexchange.com/questions/49477/why-isnt-thallium-triiodide-stable

It can be explained by fajans' rules. As the positive charge on an atom increases its size decreases as a result polarising power increases. If polarising power increases then it distorts the electron cloud towards itself and hence ionic nature of the bond decreases.   This is the reason why Thallium(I) chloride is more ionic than Thallium(III) chloride.

Does this also suggest that ionic is more stable than covalent in this situation?

It may not be. Because you can't say that ionic is more stable than covalent. For example, consider diamond which is the one of the hardest substance known to us which is a giant covalent polymer. The actual reason for the stability of Tl(I) over Tl(III) is inert pair effect and it is what you have texted first.

There is another effect : relativity. In the 6th line of the periodic table (containing $\ce{Tl}$), relativity effect becomes important. Even though electrons have no velocity, their behavior does change as if outer electrons were moving at nearly the velocity of light. This was shown by Pekka Pyyko, Accounts of Chemical Research, Vol.$12$, No. $8, 1979$, p. $276 - 281$.
As a consequence, electrons of the outer shells are so fast that their dimensions decrease, so that they are "absorbed" by the other inner shells. Their chemical properties look like atoms having "lost" $26$ electrons and protons. This is why Thallium ($Z = 81$) is chemically similar to $\ce{Cs}$ $(Z = 55 = 81 - 26)$: both form soluble and stable $\ce{M^{+}}$ ions.
For the same reason, its neighbours have properties which are also changed in the same way. On its right, lead ($Z = 82$) looks like baryum ($Z = 56 = 82 - 26$), as both $\ce{PbSO4}$ and $\ce{BaSO4}$ are insoluble compounds. On its left, mercury ($Z = 80$} is "nearly" a noble gas like xenon ($Z = 54 = 80 - 26$) : it is liquid, boiling at low temperature ($356°C)$ : so it is "nearly" a gas, at least with respect to its neighbors, boiling at $1457°C˙ (\ce{Tl})$, and  $2600°C˙(\ce{Au})$. On the other hand, mercury is chemically unique : it is the only metal producing an ionic dimer in aqueous solution : the cation $\ce{Hg2^{2+}}$. This property is also known for noble gases, which may produce ions like $\ce{He2^{+}}$ in special circumstances. And left to left, gold ($Z = 79$) is "nearly" a halogen : it looks a little bit like iodine ($Z = 53 = 79 - 26$), as it produces an ionic compound with cesium : $\ce{CsAu}$.

Mercury is not the only metal to produce a dimeric ion in the +1 oxidation state. Zinc, cadmium, and even the alkaline earth metals beryllium, magnesium and calcium have been caught doing the same thing. Mercurous ion is, however, the most common case.


text:
What does n(Na2CO3) mean?
In my book, $n(\ce{Na2CO3})$ was used to indicate the number of moles of $\ce{Na2CO3}$ without any clarification. Is it common knowledge among chemists that $n(\ce{Na2CO3})$ indicates the number of moles of $\ce{Na2CO3}$?

Yes, the standardized quantity symbol according to ISO 80000-9:2009 Quantities and units — Part 9: Physical chemistry and molecular physics as well as the recommended quantity symbol according to IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book) for amount of substance is $n$.
The quantity “amount of substance” shall not be called “number of moles”, just as the quantity “mass” shall not be called “number of kilograms”. In the name “amount of substance”, the words “of substance” can for simplicity be replaced by words to specify the substance concerned in any particular application, so that one may, for example, talk of “amount of hydrogen chloride”, or “amount of benzene”. Although the word “amount” has a more general dictionary definition, this abbreviation of the full name “amount of substance” may be used for brevity.
Thus, $n(\ce{Na2CO3})$ would be called amount of sodium carbonate.

You are unclear if you are confused by $n$ itself, or by the particular notation $n(\ce{Na2CO3})$
$n$ is widely accepted to denote amount of substance, expressed in moles, as @Loong , IUPAC and Wikipedia say.
If $n$ means the amount of the particular substance like $\ce{He}$ or $\ce{H2O}$, it is often written in index form like $n_{\ce{He}}$ or $n_{\ce{H2O}}$.
Multiple levels of sub/superscripts  often bring typographic difficulties. For these reasons, instead of typing $n_{\ce{Na2CO3}}$, some may prefer rather $n(\ce{Na2CO3})$ instead.
It is not official recommendation and less controversial may be substitution by some placeholder.

I often think the term amount of substance has been chosen rather unluckily, as at overlap with less scientific domains, the term can be confused with primary understanding of amount in common language with very  vaque meaning.

Who recommends writing amount of sodium carbonate as $n({\ce{Na2CO3}})$? Is it IUPAC? I don't recall seeing this recommendation in textbooks?

It is not official recommendation, rather practical one, applicable generally on complex indexes, not just on amount of substance. I remember one of mods suggested it.

This is I am wondering, who is making this practical recommendation?

Personally, I think less controversial way would be substitution method, like let A is , then using $X_\mathrm{A}$, where X is general quantity. As $n_{\ce{^{235}_{92}U}}$ or similar may not always look good.

Yes, I have seen subscripted versions but rarely with in-line brackets.

Na2CO3 is fine as subscript yet, but displaying complicated sub+superscripted expressions, when the major quantity has its own superscript, is challenging. AFAIK, the original recommendation was made in context of standard potentials.


text:
What is the general formula for tertiary alcohol?
According to my book, the general formula is $\ce{R-OH}$.
But according to the internet, it is $\ce{R3COH}$.
Are both the general formulas correct?

First, please use the proper term: tertiary alcohol. "3°" is a shorthand notation, fine for taking your own notes, but not appropriate in formal writing. Then, look up the definition of a tertiary alcohol, and you can figure out which one is correct from that. If your book really said that it's ROH, I strongly suggest dumping it and getting a new book.

I agree. You need a better textbook. The internet definition is much more accurate.

The formula $\pu{R-OH}$ could be used for all sorts of alcohols, and even for other substances like phenols.

I wouldn't be so quick to say dump the book. Whether the formula was used correctly depends on the context. For example, "Consider the following reaction involving ROH (a tertiary alcohol) and..."

The formulae are not inconsistent. "R" can represent anything including "R3C" so the second formula is merely a more specific subset of the first.

A tertiary ($\mathrm{3^\circ}$) alcohol is compound in which the hydroxy group ($\ce{-OH}$) is attached to a tertiary carbon; which again indicates a saturated carbon that is attached to three carbons.
So, the general structure would be (stereochemistry not shown, only connectivity):

Your confusion arises from the fact that in organic chemistry, the symbol $\ce{R}$ can be used to represent any carbon chain, and so the precise meaning depends on the context.
The most general formula for tertiary alcohol should be $\ce{R^1R^2R^3COH}$, as shown in the picture, to indicate that all three carbon chains can be different. However, $\ce{R3COH}$ is a reasonable way to represent it, as long as you keep in mind the caveat.
$\ce{R-OH}$ is probably not the best way to represent a tertiary alcohol, because it hides the tertiary nature, but it might be acceptable to draw in a reaction scheme in case there was no change in the carbon chain, so it is okay to hide that complexity.
At any rate, it is better to know what it means, than rely on a specific formula, because representations in chemistry are flexible, and can change depending on context.


text:
When doing free radical substitution, is the free radical a catalyst?
Consider a simple free radical substitution's propagation step for $\ce{CH4}$ to $\ce{CH3Cl}$:
$$
\begin{align}
\ce{CH4 + Cl^. &-> CH3^. + HCl}\tag{R1}\\
\ce{CH3^. + Cl2 &-> CH3Cl + Cl^.}\tag{R2}
\end{align}
$$
My understanding is that since the free radical is regenerated at the end of the reaction, it must therefore be a catalyst of some sort. Is this understanding correct or flawed?

You may find useful these links for text formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

Yes, $\ce{Cl^•}$ acts as a catalyst here.

@Poutnik Thanks for letting me know! I was trying to find the command for the last 5 mins but couldn't find it and thus went for the last resort!

Hmm,  $\ce{Cl^.}$ as $\ce{Cl^.}$ - nice!

@Poutnik No need for \bullet, \cdot or other macro or Unicode chars. Just use ^., as it's stated in manuals for MathJax-mhchem, $\mathrm\LaTeX$-mhchem, or another package chemformula.

See also IUPAC Gold book: Catalyst

@Poutnik thank you for the help with formatting in chemicals. I find it very useful.

@Poutnik Following the gold book, I know realize that this is a case of autocatalysis. Thanks

Your interpretation is interesting. As you state, the free radical $\ce{Cl·}$ is regenerated at the end of the reaction. It behaves like  Vanadium(V) which oxidizes $\ce{SO2}$ into $\ce{SO3}$. And once reduced into Vanadium(IV), it gets reoxidized by $\ce{O2}$ into Vanadium(V), according to the two equations $$\ce{V2O5 + 2 SO2 -> 2 VO2 + 2 SO3}$$
$$\ce{2 VO2 + O2 -> V2O5}$$ The main difference between the free radical substitution and the vanadium oxide catalysis is the fact that both vanadium oxides do exist as stable compounds. And the radical $\ce{Cl·}$ is not stable and cannot be isolated.

The question is, if we should consider transient, intermediate products in nonlinear complex reactions as catalysts. IUPAC Gold book seems to do so, but.....

Well, consider this argument on the undoubtedly actual expanded system, which would commence with the action of strong UV on a Chlorine gas in presence of Methane, and seemingly would proceed based on random kinetics, as follows:
\begin{align}
\ce{Cl2 + UV &-> Cl^.  + Cl^.}\\
\ce{CH4 + Cl^. &-> CH3^. + HCl}\\
\ce{CH3^. + Cl2 &-> CH3Cl + Cl^.}\\
\ce{Cl^.  + Cl^. &-> Cl2}\\
\ce{CH3^. + CH3^. &-> Products}\\
&\vdots
\end{align}
where not all reactions are equally likely again based on kinetics.
In the above more complex system approaching reality, I would claim the Chlorine presence (certainly in the form of the $\ce{Cl^.}$) has catalytic aspects. Note, both the chlorine radical and the methane radical are cycled within the system, albeit, the methane radical self-reaction is a termination step, whereas the chlorine radical self-reaction restarts the system. However, Chlorine is consumed in the system so not by definition a catalyst.

Good point that chlorine is consumed. Only the radicals regenerate each other. I don't see what you mean by "kinetics", likely the wrong word.

On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

@Karl won't the radicals be consumed if a chlorine radical hits a methyl radical?

MaguireFan123: Yes, indeed .CH3 + .Cl = CH3Cl  and the reverse may also be possible under UV light. Note: my use of dots to convene possible continuation of reactions.


text:
Can anyone identify this instrument part and what instrument it comes from?
Can anyone identify this instrument part and what instrument it comes from?

A better picture might help.

Chances to keep a question like this active increase by provision of context.  E.g., found in an analytical lab performing soil analyses; the nozzle-like shape has a fine opening (or not), material is of (magnetic?) metal, and -- like for a crime investigation -- add a scale to the picture.  This could be a ruler (preferred), or (if small) look into your pocket for a well known coin to be put next to the unknown object. If the camera used possesses a macro function (i.e., yields a better focus for objects in proximity), activate it.

It is the inside of a centrifuge where the rotor gets attached (the top spins, and the two pins ensure proper seat of the rotor on the conical piece). Here is a better picture of the inside of a Sorvall RC-5C Plus Floor Centrifuge, with the hands serving as scale bar. This is partially disassembled to show how one would change the motor brushing.

Screenshot from https://youtu.be/rO1q2eHTyJ0?t=434
If this is for a trivia game, I want to be credited. If this is for some crazy exam question, I'm sorry I helped cheating.


text:
Complete mechanism of benzotriazole synthesis from o-phenylenediamine
Does my suggested mechanism for benzotriazole synthesis look correct? Are the arrows for the cyclization in the right place?
Chemdoodle (which I drew this with) makes the curly arrows look a bit weird.

Too many steps, amino could substitute H2O, then deprotonation.

You mean the intermediate with the pendant hydronium would cyclize with expulsion of water?

Your solution is plausible but I am bothered by the use of structure 7 the way you have. Its too far  to reach! If you do an electrocyclization of 7 to form the benzene ring and a 4-membered ring bearing vicinal nitrogens, then you can have the remaining nitrogen as a neutral "nitrene" ready for 1,2-shift of nitrogen and ring expansion. OTOH, acid-catalyzed cyclization of structure 2 followed by dehydration gets you there.

When I was drawing it I figured the electrocyclization through the electronegative N wouldn't work but ended up drawing it anyway. So I was right to be concerned.

The amino-to-diazo electrocyclization and 1,2 shift is interesting, but I think I like Mithoron's answer where structure 5's nitrogen displaces water and then deprotonates better.

https://ibb.co/Lzthh3G

Jacob, you are free to like what you like. I'm not convinced. I'll post an answer.

Species 1 is your seventh structure. The C=N=N atoms are linear and a long reach for the remaining nitrogen. Electrocyclization of 1 to 2 restores aromaticity. Nitrene species 3 allows for a 1,2-shift to an electron-deficient center with concommitant loss of strain in the 4-membered ring to produce 1H-benzo[d][1,2,3]triazole 4.
Cyclization of 5 $\rightarrow$ 6 and dehydration via 7 gets the job done. Perhaps @Mithoron knows something that I am not aware of.


text:
Tetrachlorophosphonium
I am aware that PCl3 and PCl5 both react with water. In these reactions, water breaks the chlorine atoms off of the phosphorus atom and hydroxyl groups take their place:
$\ce{PCl3 + 3H2O -> H3PO3 + 3HCl}$
$\ce{PCl5 + 4H2O -> H3PO4 + 5HCl}$
So, judging by these equations, my prediction is that the tetrachlorophosphonium ($\ce{PCl4+}$) ion should readily react with water as well. This is my guess at how the reaction should go:
$\ce{PCl4+ + 4H2O -> H+ + H3PO4 + 4HCl}$
Is this correct? If so, I predict the tetrahalophosphonium ions (with fluorine, bromine, and iodine) would have the exact same reaction but with HF, HBr, or HI formed instead of HCl.
If this is not correct, what would the resulting hydrolysis products be?

The reaction would form not only phosphoric acid but also phosphorus acid. Not sure about other halogens. Fluorine will most probably only partially hydrolyze, whereas in case of iodine reducing property of HI might be too high and so you might observe I2 vapors.

$\ce{PCl4^+}$ reacts like $\ce{PCl5}$, because of the equilibrium $$\ce{PCl5 <=> PCl4^+ + Cl^-}$$ So the reaction with water will produce $\ce{H3PO4 + HCl}$ as stated in the original post, and no phosphorous acid.


text:
How can I use the MO diagram of hydrogen fluoride to demonstrate that the molecule is polar?

I am trying to explain that HF is a polar molecule using the MO diagram. Normally I say that the F atom has a higher electronegativity than the H atom, which causes a net positive dipole moment.
However now I am trying to use the MO diagram. My rough guess is that because the H atom has one of its electrons moving down to the energy level, it is effectively showing that it is less electronegative?

It's a bit more involved than that; after all, the fluorine atom also has one of its electrons "moving down".
The key is to look at the occupied MOs and to determine whether they are mainly hydrogen-based, or fluorine-based (i.e., which atom contributes more to the MO). This is actually not trivial to determine at all, and in general, the qualitative MO diagram you've attached isn't going to be sufficient.
In this case, however, the bonding MO (labelled $\sigma$) is only constructed from two constituent AOs: one from hydrogen ($\mathrm{1s}$) and one from fluorine ($2\mathrm{p}_z$). The fluorine $2\mathrm{p}_x$ and $2\mathrm{p}_y$ AOs are nonbonding.
Under these circumstances, it's possible to look at the energy levels to indirectly determine the contributions. In this case, the bonding MO is closer in energy to the fluorine AO than it is to the hydrogen AO. This suggests that the contribution of the fluorine AO is larger; often we say that the bonding MO is "centred on fluorine" or "localised on fluorine".* Effectively, the electrons in this MO would belong more to fluorine than they do to hydrogen.
Note that this can't be generalised to more complex molecules and MO diagrams. It only truly works in the specific case where two AOs make up two MOs. Even, for example, carbon monoxide is problematic: it is only formed from two atoms, but its MOs generally have contributions from more than two different orbitals (carbon 2s and 2p, and oxygen 2s and 2p). So this simplification falls apart.

* This can all be formalised properly using quantum mechanics and linear algebra, but I won't do it here. The treatment would be similar to that which I showed here, except that you would need to explicitly calculate the actual eigenvectors. It should follow from this that the lower-energy new eigenvector (bonding MO) will have a larger coefficient of the lower-energy old eigenvector (AO).

Since you've brought up carbon monoxide, I've actually done the qualitative approach you've described here. It's more involved, but still somewhat possible. The whole approach obviously falls apart, right before MO diagrams themselves stop being good visualisations or approximations. https://chemistry.stackexchange.com/a/67001/4945


text:
How do the odd proton-number elements originate?
I have a question about the origin of the odd proton-number elements in the periodic table, please.
As it is generally thought that the big bang produces hydrogen first, and then hydrogen combined to form helium. Helium combines into beryllium, and beryllium into carbon, etc... This means that all fusion atoms have even numbers of protons, does not not?
Then, if radioactive decay involving protons always result in a 2 protons decay, how then can we ever end up with atoms of odd number of protons, such as lithium, boron, nitrogen, etc?

This is question rather for nuclear physics and cosmology. Generally, they are created by the same processes as even-proton nuclei. BTW Li, Be and B are not produced by fusion. See also Nucleosynthesis. Search also for S-process, R-process and Spallation.

@Poutnik  According to the Wikipedia article you link, Li-7 can be produced by the fusion of tritium and He-4; and Be-7 can be produced by the fusion of He-3 and He-4.

@theorist I should say naturally and in majority. You cannot fuse 2 4He to 8Be, unless there is the 3rd 4He to form 12C. 7Be would quickly decay to 7Li.

@Poutnik  The article described these as the "chief nuclear reactions responsible for the relative abundances of light atomic nuclei observed throughout the universe", indicating these are the primary reactions that gave rise to these isotopes.  Yes, the Be-7 would quickly decay (half-life of 52 days), but Li-7 is stable.  I.e., it's saying the main origin of present-day Li is nuclear fusion.

@theorist I agree about the lithium, not about Be and B. The info I wrote may come from other source than the link and may be related  to higher than expected abundance of light elements on Earth, assuming the main their source us due Spallation. It is also supported by the periodic table chart in the link.

Correction: Stable isotopes of Be and B  were not created by nuclear fusion, but by fission by cosmic radiation called cosmic ray Spallation aka X-process.

More generally, it seems the asker may not be aware of beta decay (both minus and plus, but especially the former) as a major process capable of converting nuclei with an even number of protons into a nucleus with an odd number of protons.

@NicolauSakerNeto yes, thank you... i've forgotten about the decay that converts a proton to a neutron. So the odd numbers can be gotten by this beta decay. Thank you!

Even ones too. You cannot obtain stable iron nuclei by just adding alpha particles.

There are several different types of nuclear reaction that can create elements. Small, fresh stars do the one described in the question. But there are different routes to new nuclei that only occur in big stars when they become supernovae (these produce more heavy elements). Even heavier nuclei are thought to come from extreme events like neutron star collisions. Also, many of the elements common on earth are the result of the deaths of the first starts and are not present in the first generation of stars.

Note, in the English language, chemical names (including elements) are capitalized only at the beginning of a phrase.

One of the most significant sources of an odd-numbered element is the CNO cycle that operates to a minor extent in our Sun and more prominently in more massive stars like Sirius. There are actually several CNO cycles, but they all involve carbon, nitrogen and oxygen interchanging through the absorption of protons and the emission of alpha and positron beta particles. The net result is the same as direct hydrogen-to-helium fusion: four protons absorbed into the nuclei during this cycle are converted into one alpha particle and two positrons that are emitted, plus the same energy that we see with the direct fusion. This mechanism is responsible for nitrogen being the second most abundant odd-numbered element in the Universe, behind hydrogen but beating lithium and boron despite the lower atomic numbers of the latter two elements.


text:
How to determine the amount of CO2 sodium hydroxide absorbs?
I first created a solution of $\ce{NaOH}$ and water, 5g of $\ce{NaOH}$, and 100ml of water, with a pH of around 14. I then added $\ce{CO2}$ (mixing baking soda, citric acid, and water) to the solution. The pH then dropped to 11 forming sodium carbonate.
I would like to know how much $\ce{CO2}$ was absorbed by the solution. Can I do this with the data I already have?

I know 1 mole of co2 reacts with 2 moles of NaOH so do I just do that?

You need to do a titration with phenolphthalein and methyl orange.

What is co2 and ph ? Why  Sodium with the capital S ?

I that just a ph indicator? I already found the ph using ph strips

No, you cannot determine precisely with the data you collected (for example, a pH value 'around 14'). Some of the generated carbon dioxide gas also may have escaped or not reacted with the high pH solution.
Remember, pH is a log scale and lose of precision can be significant. However, you may argue, within the measurement error of the pH scale, you are approximately correct. Also, you are assuming correct calibration of the pH device. Mention the nature/degree of possible errors in your write-up.
Also, correct procedure matters, so if you did not add pure carbon dioxide gas directly, meaning you added the citric acid plus Baking Soda into the aqueous $\ce{NaOH}$, some neutralization by reactions with the sodium hydroxide by the citric acid and Baking Soda. The latter will skew the presumed amount of $\ce{CO2}$ actually released.
If appropriate, some added work. For example, test by adding dissolved calcium chloride, in excess, to the solution. Proceed to collect, dry and weigh the recovered calcium carbonate. Moles of the latter equate to moles of consumed carbon dioxide, albeit, expect some $\ce{CaCO3}$ collection loss.

Thank you for your answer, but for clarification, I did add the co2 directly, sorry I didn't say correctly. I'm ok with just getting an average and not an accurate amount of absorbed co2. Since I added 5g of NaOH(along with 100ml of water) and sodium carbonate requires 2 NaOH and 1 co2 molecule, is it safe to assume 2.5g of cpo2 was absorbed?


text:
Will Lithium displace a sodium salt?
This question is based on a discrepancy I have observed in my limited knowledge about the reactivity and electrochemical series.
In the reactivity series it is mentioned that Na>Li in reactivity which is understandable considering the higher ionization energy of Li. This property is shown in various situations such as the more vigorous reaction of Na with water.
On the other hand, in the electrochemical series, Li is given as a stronger reducing agent than Na.
So, my question is whether Li will displace an aqueous solution of a sodium salt, for example, sodium sulphate to form lithium sulphate(typically the outcomes of these displacement reactions are explained on the basis of the reactivity series, not the electrochemical series-at least at my level). Also, will Li displace sodium oxide?
If so, is the higher value of Na in the reactivity series just an indication of the rate of the reaction as opposed to the outcome of the reaction?

Sodium reacts and releases energy faster, lithium releases more energy (in water environment). As the higher released hydration energy of lithium gets the upper hand over higher consumed ionization energy. Lithium will not displace sodium in water solutions, but hydrogen.

@Poutnik Please could you elaborate.

Reactivity serie is matter of kinetics, electrochemical serie is matter of thermodynamics.

On the other hand, I would be pleased to simply know how the redox potentials of alcaline metals have been practically measured. Because they cannot be determined in water, because in electrolysis, water is reduced at the cathode before $\ce{Na^+}$ or $\ce{Li^+}$ . Does anybody which solvant or which technique was used to get $\ce{Li^+}$ at $-3.04$V ?

@Poutnik Then why doesn't lithium displace sodium in an aqueous of sodium sulphate(as per your comment)? How is that linked to kinetics?

Think longer. This is linked to thermodynamics. What are standard redox potentials of Lithium, sodium and hydrogen?

@Poutnik Yes, wouldn't Lithium sulphate be thermodynamically more stable considering the high hydration energy released? So, shoudn't Li displace sodium to form the more thermodynamically stable product?

Hydrogen + lithium hydroxide are thermodynamically much more preferred to lithium sulphate + sodium.

@Poutnik Ok, understood. One final question-Does the greater reducing power of Li compared to Na influence the outcome of any reaction and is there any chemical application/observation to this? If not, can I safely ignore the higher reducing power of Li when dealing with inorganic reaction prediction and work solely on the reactivity series?

@Poutnik Also, if I replace the situation with Li(s) added to molten Na2SO4(l) (not aqueous), will Li now displace sodium?

Consider, in solution, all those alkali metals are not bound, but are ions.


text:
Selectivity in dehydration of isopropyl alcohol to diisopropyl ether
Can isopropyl alcohol be dehydrated by sulfuric acid to diisopropyl ether in a similar way as to how ethanol can? Is there a way to avoid or at least minimize the inevitable dehydration to propylene?

In addition to the necessity of a proper title please refrain from salutations ("hi", "thanks", "hope this helps" etc. are noise) and note that IUPAC discourages use of the name "isopropanol".

Who cares what IUPAC says?  Isopropanol is a universally used name, far more commonly used than whatever IUPAC says it should be @andselisk.  Link https://books.google.com/ngrams/graph?content=isopropanol%2C+propan-2-ol%2C+2-propanol&year_start=1800&year_end=2019&corpus=26&smoothing=3&direct_url=t1%3B%2Cisopropanol%3B%2Cc0%3B.t1%3B%2Cpropan%20-%202%20-%20ol%3B%2Cc0%3B.t1%3B%2C2%20-%20propanol%3B%2Cc0

@CurtF. "Isopropanol is a universally used name, far more commonly used than whatever IUPAC says it should be" is not true. I suggest you do a slightly better background check before posting snarky comments like this solely reflecting personal preferences. Also, please make sure your edits are complete.

@CurtF. I recall an old comment of yours demonstrating another disagreement with IUPAC recommendations regarding units for $m/z$. I'm not sure what your problem with IUPAC is, but please refrain from unsubstantiated criticism of recognized standards and recommendations that should at the very least least be brought to the attention of the questioner.

I will not refrain from criticizing IUPAC when their recommendations are pedantic, unhelpful, or wrong.  Also, my criticisms aren't unsubstantiated.

Also @andselisk, what was incomplete about my edits?

@CurtF. IUPAC (yes, sometimes pedantic to the extreme) at least provides a logical explanation as to why the name "isopropanol" is incorrect. Your argumentation is at the level of "a billion flies can't be wrong" and isn't even backed up by the usage statistics. This type of subjective critics appears quite unsubstantiated to me. I'm afraid I'm not going to continue this debate as I don't see any signs of mutual understanding so far. All I'm asking is to keep snarky comments to yourself or a personal blog. We have enough issues with terminology, formatting and notations to deal with.

Let us continue this discussion in chat.

Note that diisopropylether has reportedly much stronger tendency to form peroxide, similarly as diethylether does so. In both cases, before usage, it is recommended to treat the solvent with $\ce{FeSO4}$ solution to scavenge these peroxides. This  prevents accidental self-ignitions or in worst case explosions, especially if the solvent is treated thermally.

The acid catalyzed dehydration of secondary and tertiary alcohols to give corresponding ethers at room temperature and atmospheric pressure is unsuccessful as elimination competes over substitution and as a consequence, alkenes are easily formed.
However, in the presence of specific conditions such as low temperature and additional catalysts, isopropyl alcohol can get converted to diisopropyl ether.

Secondary alcohols are converted to symmetrical secondary alkyl ethers in high selectivity. The method employs acidic solid metallosilicate catalyst particles to accomplish the etherification in the presence of sulfuric acid by selective intermolecular dehydration of secondary alcohol to form di-secondary alkyl ethers. Preferably, the catalysts are solid shape selective aluminosilicate particles, especially zeolite such as ZSM-5, zeolite HY and zeolite Beta.

Another resource which talks about the preparation of diisopropyl ether is the following research https://www.researchgate.net/publication/251456024_The_mechanism_of_diisopropyl_ether_synthesis_from_a_feed_of_propylene_and_isopropanol_over_ion_exchange_resin

What I have inferred from these is that at standard conditions, a secondary alcohol like isopropyl alcohol will get dehydrated in the presence of $\ce{H2SO4}$ to propylene. But in the presence of an acid along with a metallosilicate catalyst such as zeolite beta or ZSM-5 and at highly specific conditions, we can minimize the inevitable dehydration to propylene.
Also, another method to synthesize diisopropyl ether from excess isopropyl alcohol is by simple dehydration and elimination at high temperatures to obtain propylene(step 1). The propylene is then separated and is subjected to indirect hydration. In the indirect-hydration process, propylene is further reacted with sulfuric acid to produce mono- and diisopropyl sulfates(step 2). The diisopropyl sulfates will then subsequently get hydrated to obtain diisopropyl ether(step 3). Step 2 and 3 occur at high sulfuric acid concentration (> 80% wt) and low temperature (e.g. 20–30 °C). It might appear that in this entire process, we are first dehydrating and and then rehydrating but the reaction conditions in step 1 are different from those in step 2 and 3 which results in the product being different from the compound we began with.
References

"Process for the production of symmetrical ethers from secondary alcohols", author = Brown, Stephen H. (Princeton, NJ), 1995 August

The mechanism of diisopropyl ether synthesis from a feed of propylene and isopropanol over ion exchange resin, Studies in Surface Science and Catalysis, F.P Heese, M.E Dry, K.P Moller


text:
Is intensity in NMR spectra acquired at regular time intervals?
I'm writing a parser that reads fid binary files and ASCII procpar files from FT-NMR experiments, most of them acquired using Varian instruments. This instruments do not keep the time values at which each intensity is acquired. So there is no obvious way you'd plot Intensity vs Time.
How do the standard visual programs get the time values?
What I found though, is that the acquisition time (time they measure intensity) is stored  in an ASCII file, probably in seconds. procpar files store every detail (spectrometer frequency, pulse width (time), pulse width to rotate the magnetisation by 90° etc.)
I could assume intensity is measured at regular intervals, and divide total-time/data-points, but there may be something else I am missing.
I could also test Fourier transforming the data but it would be time consuming.
The 1D FIDs seems to store real data, also some 2Ds that I've found elsewhere. But I am aware that they may store complex data, in which case the data-points are divided by 2, and a different treatment altogether may be needed.
The FIDs may be 1D, 2D etc, only limited by the instrument capabilities. 2D FIDs seem to hold a large set of fid results, all would be linked to the same time axis, or some similar rearrangement afaiu. There can also be '1D arrayed' data, that I am not yet sure how to interpret. May just be a set of 1D FIDs.

In order to apply discrete Fourier transform, the fundamental requirement is to have uniform sampling rate.

@BuckThorn updated, we can chat for more details if you are keen to. I am writing a parser in Typescript for those files. There are parsers in other langs, but I like to do it by myself from scratch. Plots show well w/o time axis bc I plot using array indexes, which are also equidistant.

By intensity, do you mean individual data points in the FIDs?

Yes. It is an electric current induced in a wire by the variable bulk magnetization field, I believe. I have to say that I included the required details but it is going to confuse a reader. The answer is "yes" if I assume the $\Delta t$ are equidistant. Which seems to be correct. I was expecting some further insights maybe. Thanks for the edit @orthocresol.

1D data
Generally, yes, data points in the FID are acquired uniformly. So, the sampling interval will simply be the acquisition time divided by the number of points.
As Buck Thorn pointed out in some (now deleted) comments, however, there is a slight subtlety in that FIDs in NMR are inherently complex-valued, whereas computers only store real numbers (i.e. complex numbers must be stored as a pair of real numbers). I'm not sure what Varian machines do, but on Bruker machines, the data array actually alternates between real and imaginary points. So, data point 1 is actually the real part of the first complex data point, data point 2 is the imaginary component, etc. So instead of $N$ real data points you actually get $N/2$ complex data points, with an effective sampling interval that is twice of what was earlier mentioned.
This is not as big a problem as it might seem. Using $N/2$ complex data points sampled at half the rate for a complex FFT, or $N$ real data points for a real FFT, will actually give you the same spectral window by the Nyquist theorem. The resulting spectrum will also be the same, although you may need to use the fftshift() function to correct the frequencies.
There is more in-depth discussion of this issue in this article: Turner & Hill, J. Magn. Reson. 1986, 66, 410.
I'm aware of examples where 1D data is acquired in a non-uniform manner and the missing points are reconstructed. See e.g. Ndukwe et al., ChemPhysChem 2016, 17, 2799. These are pretty niche, though, and I'd say that virtually all 1D data you will encounter is uniformly sampled.
2D data
2D data is a lot more complicated. Generally, these are functions of two time variables: $S = S(t_1, t_2)$, so they are stored as matrices, built up of many FIDs.
Within each FID, $t_1$ is a constant, and $t_2$ varies. Between each FID, the value of $t_1$ changes. So, it's often the case that each row of the matrix is one FID, and so the columns represent different values of $t_2$, and the rows represent different values of $t_1$. However, you should consult the manufacturer's manuals as to the exact format in which this data is stored (row-major or column-major).
The $t_2$ dimension is entirely analogous to 1D spectra (essentially, 2D NMR is about collecting many 1D spectra with different values of $t_1$). Hence, it's still almost always the case that data is uniformly sampled in the $t_2$ dimension.
However, the $t_1$ dimension is way more complicated, for three reasons:

There is a similar issue to the real/complex difference mentioned previously for 1D data. In the case of 2D data, it's very often the case that the first and second FIDs actually have the same value of $t_1$, but are actually different experiments designed to yield the real and imaginary part of a complex number (the States method). Alternatively, the first and second FIDs have the same value of $t_1$, and are meant to allow you to regain the real and imaginary parts of a complex number through linear combination (echo-antiecho method). Alternatively, the first and second FIDs actually have different values of $t_1$ (TPPI method, or magnitude-mode spectra).
In general, it is not sufficient to directly perform Fourier transformation on a 2D data matrix read from a file without any extra processing.
The entire reason for this is so that positive and negative frequencies can be discriminated in the $t_1$ dimension after Fourier transformation, and is technically known as quadrature detection. This is an extremely important topic in 2D NMR and there is no way I can explain this fully; I instead refer you to a textbook such as Keeler's Understanding NMR Spectroscopy.

The 'complex' signals in $t_2$ are in fact orthogonal to the 'complex' signals in $t_1$, in the sense that the imaginary units used in both dimensions are to be considered separately. That means that, instead of a 2D matrix of complex numbers $a + bi$, you actually have a 2D matrix of hypercomplex numbers (or quaternions) $a + bi + cj + dij$, where $i$ and $j$ represent the imaginary units in the two dimensions. The number $a$ is real in both dimensions, $b$ and $c$ are imaginary in one of the two dimensions each, and $d$ is imaginary in both dimensions. On Bruker machines, each of these four components are stored in separate files. I don't know what Varian data looks like, but I expect it to be broadly similar.

In recent years non-uniform sampling (NUS) in the $t_1$ dimension is becoming more and more common, and there is a non-negligible chance that any random 2D dataset you obtain from someone in fact uses non-uniform sampling. In fact, when people talk about non-uniform sampling in NMR they are almost always referring to the $t_1$ dimension of 2D NMR (or higher-dimensional NMR). A simple Google search for NUS in NMR will likely bring up several reviews.

It is not necessary to acquire NMR data at a uniform sampling rate (regular time intervals), but it is the standard practice in a large majority of cases. Most experiments acquire data with a uniform sampling rate in order to take advantage of numerical methods such as the Fast Fourier Transform (FFT), which is designed for application to regularly sampled data and whose development was a key to widespread adoption of NMR. There are likely other advantages with regular sampling, including simpler electronics, scheduling of acquisition, data processing, and ease of troubleshooting (although tools to process irregularly sampled data can come to the rescue as a troubleshooting tool when regularly acquired data is partly corrupted). Some modern NMR acquisition methods and complementary processing techniques attempt to circumvent shortcomings of regular sampling by applying irregular sampling.
One shortcoming of regular sampling, particularly of a complete FID, is that it requires a lot of time (and memory, but that is secondary), particularly in the case in multidimensional NMR, where experimental acquisition times can become rather lengthy (days, possibly more) and therefore also costly.
One way to shorten acquisition times is to truncate FIDs in either the direct or the indirect dimensions. In the direct observe dimension it is also common to truncate acquisition before completely sampling an FID to avoid problems with application of high-power pulses when suppressing particular NMR interactions during acquisition. Truncation effects can be ameliorated using prediction and apodization algorithms that are included as standard tools in current NMR instrument software.
Irregular sampling is another more complex way to shorten the required number of points acquired in the indirect dimension, with potentially large rewards in saved time. There are a number of techniques available as standard tools on modern NMR instrumentation to schedule irregular sampling and process the data. Older data (maybe more than 15 years old) was most likely not acquired with irregular sampling techniques.
On Varian (now Agilent) instruments the number of points is stored in the procpar file. To determine the acquisition time, sweep width (half the sampling rate), or number of points for your data set I suggest looking at parameters np,sw, and at, which are described in the VnmrJ users manual:

np Number of data points
Description: Sets number of data points
to be acquired. Generally, np is a dependent parameter and is
calculated automatically when sw or at is changed. If a particular
number of data points is desired, np can be entered, in which case at
becomes the dependent parameter and is calculated based on sw and np.
Values: np is constrained to be a multiple of 2 (Acquisition
Controller or Pulse Sequence Controller board) or a multiple of 64
(Output board). (See the acquire statement in the manual User
Programming for a description of these boards.
at Acquisition time
Description: Length of time during which each
FID is acquired. Since the sampling rate is determined by the spectral
width sw, the total number of data points to be acquired (2swat) is
automatically determined and displayed as the parameter np. at can be
entered indirectly by using the parameter np. Values: Number, in
seconds. A value that gives a number of data points that is not a
multiple of 2 is readjusted automatically to be a multiple of 2
sw Spectral width in directly detected dimension
Description: Sets
the total width of the spectrum to be acquired, from one end to the
other. All spectra are acquired using quadrature detection. The
spectral width determines the sampling rate for data, which occurs at
a rate of 2*sw points per second (actually sw pairs of complex points
per second). Note that the sampling rate itself is not entered, either
directly or as its inverse (known on some systems as the dwell time).
If a value of sw is entered whose inverse is not an even multiple of
the time base listed above, sw is automatically adjusted to a slightly
different value to give an acceptable sampling rate. To enter a value
in ppm, append the character p (e.g., sw=200p). If a DSP facility is
present in the system (i.e., dsp='i' or dsp='r') and oversampling in
the experiment has not been turned off by setting oversamp='n', then
the oversampling factor will be recalculated.

Somewhere in the metadata of your measurement, there is a value called "dwell time". That is the length of the interval which is represented by a single point in your FID, or, if you look at an oldschool AD converter, the time interval between triggering the readout of one value and the next.
In the old days (of FTNMR, not the very old days of CW NMR), the real and imaginary FID intensities were sampled alternatingly, using a single ADC.
Later (probably since the eighties?), one would always have two ADCs running at the same acquisition rate, each connected to one output of the quadrature receiver.
Today, the transients are sampled at extremely high rates (oversampling at 100s of MHz) and digitally filtered/downsampled to a reasonable bandwidth/data rate before they are accumulated and stored like a normal FID from the eighties. ;)
That's regular 1D data, and the first dimenson in nD. For all dimension >1, as was explained in the other answers, everything depends on the exact experiment that was carried out. The exact, absolute time in which each single point was measured could be reconstructed, but it is rarely of any importance, unless you're doing reaction monitoring, i.e. measuring just one spectrum after the other.

I guess it's kinda off topic, but do you know where I could read more about the oversampling / digital filter techniques? I've always known they've existed (and the more practical consequences, e.g. the lack of spectral aliasing in direct dimensions), but never found any resource which got into the details. (+1'd it yesterday, naturally.)

@orthocresol There is nothing NMR-specific about it afaik, standard DSP techniques.  Oversampling is simply replacing e.g. 4 consecutive datapoints by a single point having their average value. The simplest filter uses a running average, or you can use filter functions that are not just a boxcar, but e.g. a gaussian. Mathematically that is done via convolution/deconvolution. I've long wanted to have a comprehensive book explaining that in more details. If you find one, let me know.

What you describe as oversampling is actually called downsampling. Oversampling simply means that we are way above the Nyqyuist limit.


text:
Help with mechanism: Isomerization of eugenol
I'm confused about the mechanism for the isomerization of the external alkene eugenol to the internal alkene isoeugenol. I've added an image below of their structures.

The reaction is usually carried out using a strong base (sodium or potassium hydroxide) in a polyol solvent (typically glycerol or ethylene glycol).
I have no idea why this might happen, especially given that there's no carbonyls with acidic alpha carbons in the molecule. Maybe eugenol phenoxide is responsible?

Looks like a sigmatropic rearrangement.

[1,3] sigmatropic shift?

It is thought to be a two-step reaction with cis-iso-eugenol as an intermediate as described here. The direct 1,3 sigmatropic reararrangement to the trans product has a symmetry problem.


text:
What should be done to decrease the pressure of the HPLC column?
After using an HPLC (C18) column several times it doesn't do what it has to do. When we use it the HPLC instrument shows a very high pressure and the process stops. We have washed with methanol, water, and acetonitrile, but none of them have been effective. What are we overlooking? What should we do to decrease the pressure of the column?

Available pump heads Max. flow rate in ml/min is 10
Max back pressure in bar 400
Recommended particle size is 5 μm

What's the pressure, mode (isocratic/gradient), column geometry, analyte/matrix and mobile phase? Was there a downtime before you noticed an issue with pressure? What does the column's manual say regarding washing and storing conditions?

@andselisk with the flow of 1 the pressure reached to 380 bar, column C18 with nanometere column, the analyte is fructose, mobile phase has been water, we worked with that last week and it was good, this time it doesn't work.

You barely answered any of the questions. 1… what? ml/min, μl/s? What's the flow unit? What does "column C18 with nanometere column" mean? Are you doing nanoflow LC with a guard column, or what? What are the dimensions of the column (length, diameter) and particle size? Again, what mode you are working in? What are the specified maximum pressures for the column and for your LC system? What is the matrix and fructose concentration (could it crystallize)? What temperature is your column thermostat is set to? Feel free to [edit] these details into your question instead of posting comments.

@andselisk I'm really sorry if I can't answer your questions properly, I'm just a student and our professor suggested no option other than washing the column with solvent and it wasn't effective, We've been trying to determine the concentration of HMF made from fructose.

It's OK, the problem is that HPLC troubleshooting is a multivariant task and we need as much info as possible. If the washing procedure described in column's manual didn't help, or there is none, as the last resort you could try washing the column in reverse using low flowrate, say, 0.05 to 0.2 ml/min. Modern columns (which model is yours, BTW?) should handle this all right. I'd strongly recommend though first to talk with your supervisor/TA or anyone on campus who has experience with LC and could come by and have a look at your machine.

By the way, are you sure the problem lies with the column? Have you tried to run the solvent with the column disconnected? A bit silly suggestion I should've probably started with, but still…

@andselisk "washing the column in reverse using low flowrate, say, 0.05 to 0.2 ml/min" we have done, but the pressure is still high. Could the high pressure have any reason other than the column? " Have you tried to run the solvent with the column disconnected?" I'm not sure what should I do, but when we do such a thing the pressure is zero and we can't figure out what's wrong. When we use flowrate 0.5 the pressure is good but when the flowrate is 1 the machine stops the process. Our machine is AZURA Compact HPLC.

Well, if you disconnect the column and the pressure drops, then it is indeed the column giving you trouble. Alternatively, you could try to increase thermostat temperature to, say, 60 °C and wash with your MP again. Also, there are several regeneration guides for various columns such as one from Agilent (PDF).

@andselisk Thank you so much, I would be trying to do that again, and read your valuable link, hope it works, thanks again.

If you ran 100% water on a c18 column it possibly suffered phase collapse. Might or might not be recoverable

What your professor has to recognize is that an HPLC column is a consumable. It is not meant to last forever. The back-pressure shows that it has been mistreated in the past with careless handling. Also, the HPLC pump seal could be going bad and generating polymeric microparticles which are clogging the column. At both ends of a column you have very fine sieve called frits (micron sized, typically 2 micron pores).
If your solvent, sample had small particles they can easily clog the frits. Also, if someone were careless enough, a sample can precipitate at the head of the column due to solubility issues. It is always the inlet frit which causes problems.
Look at the schematic of a column:

Some standard ways to check the system:

Prime the pump, which just means that you ensure pump is filled with solvents and there are no air bubbles.

Remove the column and connect the injector to the detector with a long tubing. And start the flow, always start at low flow rates like 0.2 mL/min and then slowly change it to 1 mL/min. How much pressure is observed? There is negligible pressure like 5-6 bars, then the problem is in the column.

HPLC columns have an arrow to show the direction of flow. Intentionally use the opposite direction and start the flow, again slowly. Do not connect the column to the detector. Starting at 0.1 mL/min and then go up to 0.5 mL/min. If it still stays at the same high pressure even after an hour, it is time to either toss it in the garbage.

If your professor is more mechanically oriented, open the column (last resort). Sonicate the inlet frit in various solvents and fill in just bare silica if you accidently scraped a layer of stationary phase and put the cleaned frit back.


text:
Is sodium superoxide more abundant than sodium oxide?
Reference(NCERT book of India);
My professor has taught me that $\ce{Na2O}$ and $\ce{Na2O2}$ are the most common oxides of sodium, but in my reference book it is written that they are $\ce{NaO2}$ and $\ce{Na2O2}$. I have searched the net but couldn't find anything about their relative occurrence (percentage). Can anyone please clarify which of the above is correct?

What is your reference book? $\ce{NaO2}$ sounds farfetched... but exists: https://en.wikipedia.org/wiki/Sodium_superoxide Perhaps the question is, under which circumstances are they "most common"?

@BuckThorn I have edited my answer to include the screenshot.

Sone links for you: https://chemistry.stackexchange.com/questions/50796/why-do-the-alkali-metals-form-different-products-upon-combustion-in-air ... https://chemistry.stackexchange.com/questions/8387/stability-of-superoxides-of-alkali-metals/8813 ... https://chemistry.stackexchange.com/questions/24762/why-does-potassium-form-peroxides-but-sodium-does-not

Sodium superoxide $\ce{NaO2}$ is a rare substance that can only be made by burning sodium at high pressure. Sodium peroxide $\ce{Na2O2}$ is common: it is formed when metallic sodium is burning in oxygen or air at ordinary pressure. The usual (and "logical") monoxide $\ce{Na2O}$ is not common. It is the result of the pyrolysis of different salts, like sodium carbonate, at very high temperature ($\sim2000\ \mathrm{^\circ C}$). Both the Handbook of Chemistry and Physics and the Merck Index give the properties of $\ce{Na2O}$ and $\ce{Na2O2}$: density, melting point. The superoxide $\ce{NaO2}$ is not mentioned, at least in my edition.

@Maurice.Thanks for your answer,but can we comment which among $\ce{NaO2 and Na2O }$ is more common?

"The superoxide NaO2 is not mentioned" is not true. Sodium superoxide is entry 2650 in the CRC Handbook of Chemistry and Physics (96th edition) with mp 552 °C and density 2.2 g/ml. Besides, I'm not sure the missing database entry is a good argument that speaks anything about the abundance. Also, OP indeed asks for comparison between $\ce{NaO2}$ and $\ce{Na2O},$ not $\ce{Na2O2}.$


text:
What is the relationship between the concentration of potassium chloride in water and its specific heat capacity?
I am curious about the effect that adding potassium chloride to water in varying concentrations has on its specific heat capacity. I have been unable to find any online information about this? Theoretically, would one expect a linear relationship or something different? Furthermore, how would this change if we added another soluble material to water? Any insights on the same would be much appreciated.
Note: This question has been previously asked here but received no responses even with a bounty. Hence, I am posting it here.

How exactly I have been unable to find any online information about this looks like? Explicit effort evidence is required, even if no usable info was found.

Generally, solutions are not like a mixture of peas and beans, where the specific heat capacity would be a linear combination of the ones of pure peas/beans.
The relationship is generally nonlinear. Solutes and solvents form bonds and structures not present in pure substances, having different participation in heat capacity. OTOH, the lattice stractures of the solid disappear. All that is furthermore concentration dependent.
For $\ce{KCl}$ and water, there is various degree of $\ce{K+}$ and $\ce{Cl-}$ hydration, affecting the specific heat capacity value. As $\ce{H2O}$ molecules bound to each other by hydrogen bonds have different effective degrees of freedom than if they are partially bound to ions.
This can be approximated by linear relationship, but deviations of that have to be evaluated experimentally, unless there is already empirical handy data for that, in form of a table or empirical formula. As water has, AFAIK, the highest specific heat capacity of all condensed matter, the specific heat capacity would generally decrease with concentration.


text:
Why doesn't XeF6 displace oxygen from water?
Consider the following reactions (unbalanced):
$$
\begin{align}
\ce{XeF2 + H2O &→ Xe + HF + O2} \tag{a} \\
\ce{XeF4 + H2O &→ XeO3 + Xe + O2 + HF} \tag{b} \\
\ce{XeF6 + H2O &→ XeO3 + HF} \tag{c}
\end{align}
$$
Considering that the oxidizing power order of the xenon fluorides is $\ce{XeF6} > \ce{XeF4} > \ce{XeF2},$ I don't understand why $\ce{XeF6}$ is unable to oxidize $\ce{O^2-}$ (in water) to $\ce{O2}.$ $\ce{XeF2}$ and $\ce{XeF4}$ are able to achieve this (rather difficult) oxidation, but $\ce{XeF6}$ is unable despite having greater oxidizing power.
Note: I have a primitive understanding on this topic and a deep and elaborate explanation will be appreciated. Also, an explanation of the reasoning behind each individual reaction ( for example, why only $\ce{XeF4}$ disproportionates?) will be even better.

XeO and XeO2 do not exist. Also you're oversimplifying what happens when XeF6 reacts with water - at least 3 equations would be needed.

XeO2 is known (https://pubs.acs.org/doi/10.1021/ja110618g) and its synthesis is via hydrolysis of XeF4

My guess: Xenon difluoride and xenon tetrafluoride produces xenon(II) oxide as intermediates which is unstable and disproportionates to xenon, xenon(VI) oxide and oxide and that's what you see in the products. Xenon hexafluoride also hydrolyses but as xenon oxide fluorides as intermediates which ultimately converts to xenon(VI) oxide. So, it is necessary for you to break down the reaction and balance them because the reaction are not that simple. See: https://pubs.acs.org/doi/pdf/10.1021/ja01065a009

Section 8.4 and 8.7 of Arrow Pushing in Inorganic Chemistry: A Logical Approach to the Chemistry of the Main‐Group Elements give mechanisms for these reactions. These are not mechanisms which have actually been proven to be true just like most of the book, but nevertheless they convincingly explain why we get the products that we do in these reactions.

@AshishAhuja Could you please convert the content of sections 8.4 and 8.7 into an answer as I can’t access the link you mentioned.

@ShashwatTomar Search for the book on zlib, I had downloaded a pdf from there. I don't think I have much to add to what is given in the book, so any answer I write will basically just be a blatant copy paste, which I think will fall under plagiarism.