text:
What is the exact definition of the radial distribution function?
I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.
From Atkins' Physical Chemistry:

Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $\mathrm dr$ at radius $r$. The... volume of the shell... is now $4\pi r^2 \mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|\psi|^2\times 4 \pi r^2 \mathrm dr$. $$P(r) = 4\pi r^2|\psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).

Immediately after this, I find:

It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$

From Housecroft and Sharpe's Inorganic Chemistry:

A useful way of depicting the probability density is to plot the radial distribution function ... $$\text{Radial distribution function} = 4\pi r^2R(r)^2$$

Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4\pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $\bigg(\int \limits _{\theta =0}^{\pi }\int \limits _{\varphi =0}^{2\pi }|Y(\theta ,\varphi)|^2\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi \bigg)$ to get $4\pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).
My confusion is compounded by the fact that there is another function called the probability density which is $|\psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.
So, my questions are:

What is the physical significance of probability distribution and that of radial distribution function?

There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?

What is the correct expression for calculating radial distribution function?

Which begs the question, what is the definition of radial distribution function?

"If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(\mathbf{r})$ and the latter is $p(r) = p(|\mathbf{r}|)$; usually the latter is more relevant.

The atomic orbitals (wavefunctions) $\psi(r,\theta,\phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(\theta,\phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.
As a concrete example, the $2\mathrm p_z$ orbital ($\{n,l,m\} = \{2,1,0\}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:
$$R_{2,1} = \sqrt{\frac{1}{24}}\left(\frac{Z}{a}\right)^{3/2}\rho\mathrm e^{-\rho/2} \qquad Y_{1,0} = \sqrt{\frac{3}{4\pi}}\cos\theta \tag{1}$$
where $a$ is a collection of constants ($a \equiv 4\pi\varepsilon_0\hbar^2/\mu e^2$), and $\rho$ is related to $r$ by another collection of constants ($\rho \equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2\mathrm p_z$ orbital.
The atomic orbitals themselves are normalised in the sense that
$$\int_{\text{all space}} |\psi(r,\theta,\phi)|^2 \mathrm dV = \int_0^{2\pi} \!\!\!\!\int_0^\pi\!\!\!\int_0^\infty |R(r)|^2|Y(\theta,\phi)|^2 r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi = 1  \tag{2}$$
Conventionally, both the radial and angular wavefunctions are separately normalised. This means that
$$\int_0^\infty |R(r)|^2 r^2\mathrm dr = 1; \qquad \int_0^{2\pi}\!\!\!\!\int_0^\pi |Y(\theta,\phi)|^2 \sin\theta\,\mathrm d\theta\,\mathrm d\phi = 1  \tag{3}$$
(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)
Normally, since $\psi = \psi(r,\theta,\phi)$, the probability density $|\psi|^2$ depends on all three coordinates. The probability density (technically, $|\psi|^2\,\mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube frustum-like shape,† located at the point $(r,\theta,\phi)$. By summing these $|\psi|^2\,\mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.
However, often we are not interested in the angular dependency, and so what we can do is to sum up all these bits which have one value of $r$, but different values of $\theta$ and $\phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $\theta$ and $\phi$. By virtue of the angular wavefunction being normalised, the integral over $\theta$ and $\phi$ evaluates to $1$, and hence we have
$$|\psi|^2\,\mathrm dV = (|R|^2r^2\,\mathrm dr)\cdot (|Y|^2\sin\theta\,\mathrm d\theta\,\mathrm d\phi) \overset{\text{integrate over }\theta,\phi}{\longrightarrow} |R|^2r^2\,\mathrm dr \tag{4}$$
So, this quantity $|R|^2r^2\,\mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.
This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $\theta$ and $\phi$, it comes out to be $1$. The widely-cited alternative definition $4\pi r^2 |\psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have $\{l, m\} = \{0,0\}$ and
$$Y_{0,0} = \sqrt{\frac{1}{4\pi}}  \tag{5}$$
(you may check that this is normalised according to equation $(3)$.) Hence
$$|\psi|^2 = |R|^2|Y^2| = \frac{1}{4\pi}|R^2| \tag{6}$$
and it therefore follows that $|R|^2r^2 = 4\pi r^2|\psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4\pi r^2|\psi|^2$ is used (or $4\pi r^2 \rho$, where the probability density $\rho$ is simply $|\psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.
Based on the above arguments this is my take on the matter:

$|R^2|r^2$ is the best definition of the RDF;
$4\pi r^2|\psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);
I still don't fully understand why Housecroft has an additional factor of $4\pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4\pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $\sqrt{1/(4\pi)}$.

† Thanks to Ruslan for pointing this out! If you are curious as to what this looks like, there is a good diagram on LibreTexts, which I have re-uploaded to SE here. Note, however, that the convention in this diagram is $(\rho,\phi,\theta)$ instead of $(r,\theta,\phi)$.

Since you're working in spherical coordinates, $|\psi|^2,\mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($r\approx0$).

@Ruslan Thank you for the correction! I never knew there was a term for it.

Thanks! But I still don't understand why we use Radial Distribution function as a measure of probability of finding electron if $\mathrm{\psi^2 dV}$ gives the same thing? Also, it seems that the volume of the frustum will go to 0 near the centre of the sphere. In that case, why does the plot of $\mathrm{\psi^2 dV}$ show a maximum at 0?

@ShoubhikRajMaiti it is not really the same thing, as the RDF depends only on $r$, whereas $|\psi|^2$ depends on $(r,\theta,\phi)$. They do both describe how electron density is distributed, but from a different perspective. As for why you would use one over the other, if you look in an inorganic textbook, there is usually some discussion of concepts such as shielding and penetration. Although I find these to be somewhat unconvincing, they rely on using and interpreting the RDFs, as trying to directly interpret $|\psi|^2$ would be incredibly complicated due to the angular dependency.

@ShoubhikRajMaiti these are not plots of the probability $|\psi|^2\mathrm dV$, they are plots of the probability density $|\psi|^2$, which (only for some orbitals, mind you) is a maximum at the nucleus. Now, whatever number $|\psi|^2$ may take on at the nucleus, if you multiply that by $\mathrm dV$ you get zero, so the probability ($= |\psi|^2\mathrm dV$) of finding it there is zero.


text:
Molecular orbital explanation of why the protonation of the oxygen atom makes a carbonyl group more electrophilic
I was working on nucleophilic addition reactions to the carbonyl group (namely, hemiacetal formation) when I came across the following explanation for the use of acid catalyst and how it increases reaction rate:

In acid (dilute HCl, say), the mechanism is different in detail. The first step is now protonation of the carbonyl group’s lone pair: the positive charge makes it much more electrophilic so the addition reaction is faster.

The reaction at hand is the following:

Ignoring the presence of the positive formal charge on the oxygen, I am trying to understand on a molecular orbital theory level why this makes sense. I'm suspecting that this somehow lowers carbon's LUMO (${π}^{*}$ orbital), which would enhance the electrophilicity of carbon.
Would someone be able to give an explanation to the enhanced electrophilicity of the carbonyl group based on molecular orbital theory?
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

I think title should reflect you wanna talk about MOs.

Ignoring the presence of the positive formal charge on the oxygen

I think this is exactly what is responsible, even from the MO viewpoint. From the MO perspective, the electrophilicity of the carbonyl group can be viewed as how energetically-accessible is the $\ce {C=O \pi ^*}$ MO to the nucleophile's HOMO. The protonation of the oxygen atom effectively decreases the energies of all MOs of the carbonyl, including the $\ce {C=O \pi ^*}$ MO. Why? The proton is positively-charged. By attaching to the oxygen atom, there is a net decrease in the potential energy of the molecule due to there being more electron-nuclei attractive forces in the resultant entity. Hence, protonation makes the $\pi ^*$ LUMO more accessible to the attacking nucleophile, increasing electrophilicity.

Wouldn't that interaction be only between the oxygen and hydrogen atom? Or does that interaction affect the electrophilicity of the C atom?

That is not true. The positive charge is more delocalised than that. The protonation will cause all the electrons in the molecule to redistribute themselves.

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text:
When an inert gas is added to a system of constant volume, equilibrium is not effected?
But pressure of the system increases. Because of this increase in pressure, wouldn't (according to Le Chatelier's principal) equilibrium shift towards the side with less moles of gas?

It would. Unless there is no such side.

@IvanNeretin not if the volume is constant. At constant volume the addition of inert gases does not influence partial pressures of the products and reactants $p_i$. It's only the total pressure that changes. Since $p_i$ do not change, no shift in equilibrium occurs.

Addition of an inert gas at constant volume:
When an inert gas is added to the system in equilibrium of constant volume though the total pressure exerted by the gases increases there is no shift in the equilibrium.
There is no shift in the equilibrium since there is no change in the molar concentrations(kept at constant volume) of each of the reacting components.
BUT
If an Inert Gas is added to a mixture kept at constant pressure there is a change in volume of overall mixture and hence the molar concentrations of the reacting components also decreases.
Hence according to Le Chatelier’s Principle the reaction shift in direction of lower molar concentration.
Basically - ((Inert/Non Reacting) Gases)) do not affect the original concentrations and hence do not affect the equilibrium in case of constant volume situations

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text:
Explanation for bond lengths in trans-hexatriene
Hexatriene is an unsaturated hydrocarbon with six carbon atoms and five carbon-carbon bonds, three of which are double bonds.
However, the bond lengths of the $\ce{C=C}$ bonds are not the same. The middle $\ce{C=C}$ bond has a length of 137 pm while the $\ce{C=C}$ bonds at the end of the molecule have lengths of 134 pm, the length of a standard $\ce{C=C}$ bond. The two carbon-carbon single bonds are 146 pm long, also off from the standard 154 pm length of carbon-carbon single bonds.

Clayden's organic chemistry hints that the explanation has to do with the molecular orbits formed and the conjugation system in the molecule. However, I do not fully understand this explanation.
Why do these carbon-carbon bonds show this unusual bond length behavior? A thourough explanation using MO theory would be appreciated.
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):

I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.

And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?

Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)

You do not have to make it up.  You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital.  This gives 0.87 pi bond between C-1 and C-2,  0.48 between C-2 and C-3, and 0.78 between C-3 and C-4.  Other bonds are determined by symmetry.


text:
Putting high symmetry points labels to a band structure plot

So far I've got this plot using Quantum Espresso. I want to put gamma, X, L, etc labels to the k-path.
Quantum Espresso's ouput states the following:
 high-symmetry point:  0.0000 0.0000 0.0000   x coordinate   0.0000
 high-symmetry point:  0.0000 0.5136 0.0000   x coordinate   0.5136
 high-symmetry point: -0.5136 0.5136 0.0000   x coordinate   1.0272
 high-symmetry point:  0.0000 0.0000 0.0000   x coordinate   1.7536
 high-symmetry point:  0.0000 0.0000 0.1556   x coordinate   1.9092
 high-symmetry point:  0.0000 0.5136 0.1556   x coordinate   2.4228
 high-symmetry point: -0.5136 0.5136 0.1556   x coordinate   2.9365
 high-symmetry point:  0.0000 0.0000 0.1556   x coordinate   3.6628

However I was expecting these points to be the high-symmetry points:

How should I label the k-path then?
I'd like something like this:

But I guess I'm not understanding QE's high-symmetry points output.

Have you tried to ask the same question at the QE mailing list (which is pretty active and helpful, in general)?

This might be what you are looking for: https://docs.quantumwise.com/tutorials/espresso_bs_dos/espresso_bs_dos.html

Not really. You see they already know in advance those points and their coordinates. Both 3d coord, and the ones in the band structure plot. I've created the path with xcrysden. And I don't remember the exact path I chose. I want to put the high symmetry points labels in the band structure. But there is no map between 3d coordinates and band plot x coordinates in the output. There should be a better way to do it than computing them by hand. The program should tell you where is gamma, X, L point etc in the band structure. But so far I haven't found a way of doing that.

It also seems that they're using a GUI version of quantum espresso. I'm using QE in the terminal.


text:
How to balance this redox reaction taking place in a basic medium?

How to balance this reaction which is taking place in a basic medium?
  $$\ce{Al + MnO4- -> MnO2 + Al(OH)4-}$$

My Attempt

Reduction half reaction:

$$
\begin{align}
\ce{4 H+ + MnO4- + 3 e- &-> MnO2 + 2 H2O}\\
\ce{28 H+ + 7 MnO4- + 21 e- &-> 7 MnO2 + 14 H2O}
\end{align}
$$

Oxidation half reaction:

$$
\begin{align}
\ce{Al + 4 H2O &-> Al(OH)4- + 8 H+ +7 e-}\\
\ce{3 Al + 12 H2O &-> 3 Al(OH)4- + 24 H+ + 21 e-}
\end{align}
$$

Adding half reactions:

$$
\ce{3 Al + 7 MnO4- + 28 H+ + 12 H2O + 21 e- -> 7 MnO2 + 3 Al(OH)4- + 14 H2O + 24 H+ + 21 e-}
$$
Adding $\ce{OH-}$ and eliminating $\ce{e-}$ and formed $\ce{H2O}$.

Final equation:

$$
\ce{3 Al + 7 MnO4- + 2 H2O -> 7 MnO2 + 3 Al(OH)4- + 4(OH)-}
$$
Please tell if I am wrong.

The hydrogens don't seem to be balanced on both the sides of the equation

Writing an answer since I can't comment yet.
You are using the ion-electron method for basic medium, but are balancing using $\ce{H+}$ and later adding $\ce{OH-}$? Sounds suspicious to me, but the outcome must be the same.
Generally, I add $\ce{OH-}$ ions on the oxygen deficient side and then add water to compensate the hydrogen deficiency on the other side.
Note: your first half-reaction balancing the conversion of permanaganate to $\ce{MnO2}$ must be regarded incorrect, as $\ce{MnO2}$, being soluble in acidic conditions (presence of $\ce{H+}$, as the equation indicates!) would yield to further reduction, and end up as $\ce{Mn^2+}$.
Here's how I balanced it:

Oxidation half

$$
\begin{align}
\ce{Al &-> Al(OH)4-}\\
\ce{Al + 4OH- &-> Al(OH)4- + 3e-}
\end{align}
$$

Reduction half:

$$
\begin{align}
\ce{MnO4- &-> MnO2}\\
\ce{MnO4- &-> MnO2 + 2OH-}\\
\ce{MnO4- + 2 H2O &-> MnO2 + 4 OH- + 3 e-}
\end{align}
$$

Adding:

$$
\ce{Al + MnO4- + 2 H2O -> Al(OH)4- + MnO2}
$$
Your method is fine as well. The only mistake is the wrong $\ce{H+}$ coefficient in $\ce{Al}$ oxidation half-reaction, which will be 4 instead of 8.
P. S. (only concern): note the differences in half reactions in basic and acidic media. Conclusion: when products differ in acidic/basic conditions, it is best not to balance in the asker's method, as it cannot be considered entirely correct. Nothing more, nothing less.

It is quite acceptable to first balance the equation in acidic medium and then add hydroxide on both sides to neutralise the protons. This is because it is the same idea as first providing OH- on oxygen deficient side and then adding hydrogen on the opposite side, or doing it vice-versa as the OP (tried) to do.

Nonetheless the Al oxidation half reaction is likely to be technically incorrect, as the product differs in relation to pH.

@YUSUFHASAN thanks for pointing out, i did the necessary edits.

First off, for basic medium there should be no protons in any parts of the half-reactions. Use water and hydroxide-ions if you need to, like it's been done in another answer.
In basic aqueous solution permanganate $\ce{MnO4-}$ is going to be reduced to manganate $\ce{MnO4^2-}$, and not to manganese(IV) oxide $\ce{MnO2}$ ($\ce{MnO2}$ forms in neutral medium). This can be explained by applying to the standard electrode potentials (for example, see this answer to "Stability of MnO4- in basic vs acidic conditions") or simply deduced by applying the Le Chatelier's principle to the following half reaction you'd normally use for neutral medium:
$$
\ce{MnO4- + 2 H2O -> MnO2 + 4 \color{red}{OH-} + 3 e-}
$$
Theoretically, $\ce{MnO4^2-}$ can be hydrolyzed to $\ce{MnO2}$, but for that an excess of a strong reducing agent and only mildly basic solution would be required. None of these conditions are fulfilled here as the strong alkali is requied to remove constantly forming passivating layer of $\ce{Al2O3}$/$\ce{Al(OH)3}$.
That's said, I'd rather propose the following redox scheme (numbers above the elements are the oxidation numbers):
\begin{align}
\ce{3 \overset{+7}{Mn}O4- + 3e- &-> 3 \overset{+6}{Mn}O4^2-} \tag{red}\\
\ce{\overset{0}{Al} + 4 OH- &-> \overset{+3}{Al}(OH)4- + 3e-} \tag{ox}\\
\hline
\ce{3 \overset{+7}{Mn}O4- + \overset{0}{Al} + 4 OH- &-> 3 \overset{+6}{Mn}O4^2- + \overset{+3}{Al}(OH)4-} \tag{redox}
\end{align}


text:
What's the difference between "extent of reaction" and "position of equilibrium"?
My chemistry textbook defines the terms in question as follows:

Extent of reaction: the relative amounts of products compared with reactants. The extent of reaction is indicated by the value of the equilibrium constant.
Position of equilibrium: the relative amounts of products and reactants at equilibrium. The position of equilibrium varies depending on the extent of the reaction.

I don't understand the difference between these two concepts and would appreciate it if someone could explain both how they are different and how they are linked/influence each other.
Thanks.

It seems they say they same (?)

Two differences. Firstly, the extent of reaction $\xi$ can be mathematically defined, whereas the "position of equilibrium" is for the most part qualitative, in the sense that we only say it "moves left" or "moves right", but not "the position of equilibrium is $x$".
More importantly, the extent of reaction does not actually tell you anything about the equilibrium. In this respect, it is more analogous to the reaction quotient $Q$ rather than the equilibrium constant $K$. If you measure the extent of reaction $\xi$ when the system is at equilibrium, then it does provide a measure of the equilibrium position. This is just like how if you measure $Q$ at equilibrium, it is equal to $K$. However, in principle you can measure $\xi$ (or $Q$) at any stage of the reaction, not just when it is at equilibrium.
Both $\xi$ and $Q$ are mathematical ways of describing how much products there are versus how much reactants, but they are not equivalent, they have different forms. See Wikipedia for more information.
So, I don't agree with the statement that "the extent of reaction is indicated by the value of the equilibrium constant" on two counts. Firstly, the extent of reaction is not directly "indicated" by $Q$ – they both measure the same thing, but not in the same way. Secondly, and crucially, $Q$ is not the equilibrium constant $K$: it is only equal to $K$ at equilibrium, which this sentence doesn't capture.


text:
Doesn't Ostwald's law of dilution contradict Le Chatelier's principle?
By Le Chatelier's principle, increasing the concentration of a weak electrolyte brings out more ions.
By Ostwald's dilution law, increasing the concentration of a weak electrolyte brings out less ions ($a = \sqrt{K_\mathrm{a}/C}$).
Where do we stand?

Le Chatelier pr is about keeping equilibrium  Ostwald l. is about defining which value is equilibrium.

You interpret Ostwald's law wrong. Increasing the concentration of a weak electrolyte would produce more ions, just not as much more as it would in the case of strong electrolyte.

Le Chatelier's Principle makes no quantitative predictions.  In this case, all it says is that adding, let us say, more acetic acid to an aqueous acetic acid solution will give more solvated protons and acetate ions in solution.  But it does not guarantee that the amount of additional ions varies in direct proportion to the acetic acid concentration.  We need such a proportional variation to maintain the same value of $a$ in the Ostwald equation, but Le Chatelier's Principle allows the possibility that maybe the increase is not enough to match the proportional variation.  If we suppose that the amount of ions increases only with the square root of added solute, and work out the fractional ionization $a$ from that assumption we get exactly the Ostwald relation.


text:
Infrared Spectroscopy of N2O

Would you be able to identify the stretch and bend in the spectrum? I can not find much information about the $\ce{N2O}$.It's solved, does anyone know the symmetry of these vibrations?

OCO has a different symmetry than NNO, which should be $C_{\infty v}$.

I don't know what flex and link voltage are; however, here's the rest of it as per this site at UCLA and this homework solution key from Illinois State:
The $\ce{N-N}$ stretch is at $\pu{2200 cm-1}$, the $\ce{N-O}$ stretch is at $\pu{1290 cm-1}$, and the doubly-degenerate $\ce{N-N-O}$ bend is at $\pu{590 cm-1}$.

As a general approach there are $3N-6$ normal mode vibrations and $3N-5$ for a linear molecule with N atoms, so for N$_2$O there are 4 modes, but some may be degenerate. Next look for overtones, in your spectrum there are lines at  approx 1300 and approx 2600 so these are likely from the same mode. Finally the asymmetric stretch usually has a higher frequency than a symmetric stretch and  bends the the lowest frequency. This will work with a bit of intuition for small molecules but far less so for bigger ones where there are many modes with similar frequencies.
In your case there are lines at $\approx 600, 1300$ and $2200$.  The highest frequency is the motion where the N-N bond compresses and at the same time the NO stretches, i.e. the asymmetric stretch. The symmetric stretch, 1300, is when the N-N bond stretches and so does the N-O, and the bend is doubly degenerate as it can occur in the xz plane and the yz plane if z is the internuclear axis.


text:
Why are amidines more basic than amines?
Amidines can be thought of as carboxylic acid derivatives with the carbonyl oxygen replaced by a nitrogen and hydrogen atom; amidines exhibit conjugation.

Amidines are more basic (less acidic) than amines and amides by a factor of around ${10^{3}}$ and ${10^{12}}$, respectively (from pKas).
I would naturally think that conjugation would lower the energy level of MO containing the lone pair; thus, the lone pair would be more bound to the nucleus and less available to deprotonate.
This is an argument similar to explaining why pyridine is less basic than piperidine; $\ce{N}$ is $sp^{3}$ hybridized in piperidine but $sp^{2}$ hybridized in pyridine. Since $sp^{2}$ orbitals have less p character than $sp^{3}$ orbitals (2/3 vs 3/4), the lone pair on $\ce{N}$ in pyridine is at a lower energy level than the lone pair on $\ce{N}$ in piperidien, and pyridine is less basic than pyridine.

Although I realize that the positive charge on the amidine is stabilized by conjugation (as explored by a previous post comparing the basicity of amidines and amides), wouldn't my reasoning with MO theory outlined above suggest a lowering of the basicity of amidines?

References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

I feel that the stabilization caused by resonance in the conjugate acid of amidine will dominate here, as the canonical structures being generated are equivalent; so this will correspond to a lot of resonance energy being released

Note that the lone pair that is delocalised in the conjugation is not the lp that is used to accept proton.

Is that lone pair used to accept the proton because of delocalization in the protonated amidine is only exhibited when the lone pair on the nitrogen with only one H atom deprotonates an acid?

Related https://chemistry.stackexchange.com/q/13593/54655

Note that when you are comparing amidines with amines, you are comparing 2 nitrogen atoms with 1 unless you are comparing it with diamines.  Amidines which have 2 ionizable nitrogen atoms are stabilized  by 2 resonant structures, therefore making them more likely to accept protons (that is more basic).  Similarly, guanidine (not to be confused with guanine which is a nucleic base) which has 3 ionizable nitrogen atoms are stabilized by 3 resonant structures thereby making it even more basic.
For additional reading, check out the basicity of the following drugs.
Betrixaban (which has 1 amidine group)
Pentamidine (which has 2 amidine groups)
Guanethidine (which has 1 guanidine group)
Metformin (a biguanide, that is, having 2 guanidine groups fused together)
Altretamine (which has 3 guanidine groups fused together into a cyclic structure)
Famotidine (which has 1 amidine group and 1 guanidine group).


text:
Why does fulminate not minimize formal charge in its structure?
The official structure of fulminate looks like this:

from wikipedia: https://en.wikipedia.org/wiki/Fulminate
However, this structure does not minimize formal charge, since nitrogen has a formal charge of +1, while carbon has a formal charge of -1.
Instead, it would make more sense for fulminate to look like this:

In this structure, the nitrogen and carbon now both have formal charges of zero, and the number of electrons is the same as the official structure.
Carbon seems like it can exist with six electrons, as shown in https://en.wikipedia.org/wiki/Dichlorocarbene.  However, the article says that it is an intermediate.
My question is, why is the first structure of fulminate the official one and not the second one? Does the octet rule take precedence in this case?

The carbon has six electrons, that's a no-no.

Oh... right... I forgot about the octet rule

For now, read "intermediate" as a fancy way to say "does not exist".

https://chemistry.stackexchange.com/questions/34078/the-lewis-structure-for-nco

@orthocresol: Not necessarily; stable/persistent carbenes exist.


text:
How is the hydrolysis of amides done in a lab?
I have seen this video about the hydrolysis of amides in order to convert them to a carboxylic acid. I understood how it works but I want to know how would someone physically do it in a lab.
I need to know the temperature needed, the time required for the reaction to happen, the reactants ratio and the process itself.

The answer, as with so much chemistry, is it depends. What substrate are you hydrolysing, is it a primary, secondary or tertiary amide? What else is in the molecule that you want to preserve?
Both acid and base hydrolysis require concentrated conditions and elevated temperature. There is a lab procedure here that uses sodium hydroxide in water at reflux to hydrolyse benzamide to benzoic acid. The standard cleavage of an acetamide protecting group to give amine plus acetic acid is overnight reflux in 2N HCl or KOH in MeOH at reflux ref here, for a benzamide protecting group it is 48h in 6N HCl or HBr in Acetic Acid  at rt.
In my own experience a particularly resistant acetamide required 3 equivalents of KOH in refluxing ethylene glycol to hydrolyse.

Thank you for your answer!  My amide is a primary amide and the rest of it's structure is formed out of benzenic rings, like benzamide. As in the conversion of benzamide into benzoic acid I need to replace the NH2 bond with an OH bond. Does this mean that the procedure from here is going to work for me as well?

That procedure should work for you. If it does not then try an acid procedure as used for a benzamide protecting group


text:
What should be the IUPAC name of this compound?

I feel that either of 4-bromo-3-chlorocyclohex-1-en-5-yne or 3-bromo-4-chlorocyclohex-5-en-1-yne is correct.

Never mind the name, are you sure it exists? It is going to be horrifically strained.

Yeah, it's very unrealistic example.

You have already correctly identified all the parts of the name and their correct order, only the numbering of locants is wrong. The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have a principal characteristic group that is expressed as a suffix. Therefore, Rule (c) is not relevant in this particular case.
Low locants are assigned to the multiple bonds according to Rule (e). Since there is a choice between the ‘ene’ ending and the ‘yne’ ending, the low locant is assigned first to the double bond, which results in the name cyclohex-1-en-3-yne.
Next, low locants are assigned to the other substituents as a set according to Rule (f). However, the numbering of the ring is already given by the previous step; therefore, the only possible numbering is 5-bromo-6-chlorocyclohex-1-en-3-yne.


text:
What chemical explosives detonate when exposed to alpha particles or nuclear fission products?
Wikipedia's article on nitrogen triiodide $\ce{NI3}$ claims that

Nitrogen triiodide is also notable for being the only known chemical explosive that detonates when exposed to alpha particles and nuclear fission products

referring to the article published 60 years ago [1]. Are there any other examples of chemical explosives capable of that discovered afterwards, or is it still a prerogative of nitrogen triiodide?
References

Bowden F. P.; Young D. A. The Initiation of Explosion by Neutrons, α-Particles and Fission Products. Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences 1958, 246 (1245), 216–219. https://doi.org/10.1098/rspa.1958.0123.

Meh, even IR may lead to explosion http://blogs.sciencemag.org/pipeline/archives/2013/01/09/things_i_wont_work_with_azidoazide_azides_more_or_less

@Mithoron The effect mentioned in this blog is thermal (a Raman laser is rather high-power). Not what the question and the paper cited in it is about.

(not really an answer here, but anyway ...)
I would suspect any highly sensitive explosive is theoretically capable to do that, possibly depending on how close you are to the point where it would just thermally explode. They checked  "Crystals of lead azide, silver azide, cadmium azide, silver acetylyde and nitrogen iodide" in the paper cited above. The four with heavy metal ions don't work, I guess it makes some sense that those can (better) absorb a very local energy surge like from an alpha particle.
My list of suspects would thus include organic ozonides, peroxides, azides, tetrazoles and other compounds that only contain CNOH. Number 1: solid $\ce{H2O2}$.
And basically anything that is known to sometimes explode without an obvious external triggering event.

Or perhaps it requires more than a single particle - a certain minimum radiation density.

@PeterMortensen But then you are close to a thermal effect again, except if it's sth like "two incidents within 10 ns and 2 nm". Or an accumulating effect. But without an actual example, that's pure conjecture. Educated wild guessing. ;-)

@PeterMortensen Not perhaps, but obviously, And I don't see such suggestion in answer

@Mithoron Why shouldn't a single alpha particle be able to trigger a runaway decomposition reaction?


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Nomenclature priority of prefixes consisting of identical letters but containing different locants

I have no idea how to name this compound. Can someone please help give the IUPAC name for this? Also mention the IUPAC nomenclature rules followed while naming.

The compound that is given in the question consists of a cyclohexane ring with two substituent groups: 1-chloroethyl and 2-chloroethyl.
The correct numbering of locants is stipulated by Rule 14.4 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 NUMBERING
  When several structural features appear in cyclic and acyclic compounds, low locants
  are assigned to them in the following decreasing order of seniority:
  (…)

However, the given case is not explicitly mentioned in this rule.
The correct numbering is actually indirectly given by Rule 14.4 (g):

(g) lowest locants for the substituent cited first as a prefix in the name;

Since the correct order of citation as a prefix in the name is given by Rule P-14.5.4:

P-14.5.4 When two or more prefixes consist of identical Roman letters, priority for order of citation is given to the group that contains the lowest locant(s) at the first point of difference.

Thus, ‘1-chloroethyl’ is cited before ‘2-chloroethyl’ in the name, and thus a low locant is assigned first to the 1-chloroethyl group.
Therefore, the complete name for the compound that is given in the question is 1-(1-chloroethyl)-4-(2-chloroethyl)cyclohexane.

Shouldn't one of the aliphatic chains be the main chain because they have the chloro functional group?


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How hydrogen molecule is formed?
How two hydrogen atoms come close to form a bond ? Textbooks refer to a potential energy diagram but what i cant understand is what is this potential energy if the atoms are neutral . How they come close and how when they come close they stay in an specific distance (bond length) ? From classical physics in order to not accelarate NET FORCE must be equal 0 ? is the same with hydrogen bonding and bonding in general ? also if potential energy decreases then kinetic must increase, but where does that kinetic energy goes when a molecules is formed ?

Hydrogen bonds are not formed when two hydrogens attract: they are formed when certain hydrogens attract atoms in other molecules such as oxygen. Water being the archetypal example.

Well the atoms are not exactly neutral. There is a charge seperation between the proton and the electron in a H atom. So both repulsive and attractive forces exist. At a certain point, there is an equilibrium

The potential energy comes from the charge interaction between the negative electrons and the positive nuclei. In a simplistic way, the kinetic energy is a result of the motion of the electrons around the nuclei. The closer the electrons are to the nuclei, the lower the potential energy but the higher the kinetic energy.
From a one-electron quantum mechanical view ($\ce{H_2^+}$), the system is described by Schrodinger's equation $\hat{H}\psi = E\psi$, where $\psi$ is the wavefunction for the electron, $E$ is the energy of the electron, and $\hat{H}$ is an operator that relates the two. The Hamiltonian operator has two parts: $\hat{H}\psi = \hat{T}\psi + \hat{V}\psi = E\psi$. These two parts are attributed to kinetic energy $\hat{T} \propto \frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2}$, and potential energy $\hat{V}$ described by Coulomb potential. In order to be well behaved, the curvature increases as the electron wavefunction is closer to the nucleus, and since the kinetic energy is proportional to curvature, the kinetic energy is higher. The Coulomb potential is smaller when closer to the nucleus. So the equilibrium wave-function is a balance between these two potentials.
This treatment is more complicated when you add a second electron since the potential energy also includes the repulsion between the two electrons. But the concept is the same and the resulting hydrogen molecule is the configuration of nuclei and electrons that minimizes total energy based on the opposing influences of $\hat{T}$ and $\hat{V}$.


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What are the products formed when nitrogen dioxide reacts with water?
I've seen two different equations for the reaction of nitrogen dioxide with water:
$$\ce{2NO2 + H2O -> H+ + NO3- + HNO2}$$
and
$$\ce{3NO2 + H2O -> 2H+ + 2NO3- + NO}$$
Are both of these correct to some degree, and is one more prominent over the other?

Both of the reactions are correct. However, the second reaction is written by 'adding' two reaction equations.
When nitrogen dioxide($\ce{NO2}$) is dissolved in water, it produces a 1:1 mixture of nitric acid($\ce{HNO3)}$ and nitrous acid(HNO2). $$\ce{2NO2(g) +H2O(l)->HNO3(aq) +HNO2(aq)}$$
i.e.$$\ce{2NO2(g) +2H2O(l)->H3O+(aq) +NO3-(aq) + HNO2(aq)}$$
However, since nitrous acid is unstable in any environment except very cold solution, it decomposes slowly into $\ce{NO}$ and $\ce{HNO3}$:
$$\ce{3HNO2(aq)->2NO(g) +H3O+(aq) + NO3-(aq)}$$
Your second reaction equation is obtained by adding these two reaction together.
Reference:

Housecroft, C. E.; Constable, E. C. Chemistry, 4th ed.; Pearson, 2010, p 777.

Note:
I wrote the $\ce{H+}$ ions as $\ce{H3O+}$, so there  is an extra water molecule on the left side of the equation, in each case.

Both can take place, but nitrous acid is unstable.  In warm or concentrated solutions, the nitrous acid will disproportionate, forming nitric oxide and more nitric acid.  So the nitrous acid is essentially only a temporary intermediate.


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How to make rheoscopic fluid using glycol stearate flakes
Rheoscopic fluid can be made by adding mica powder to water. I read that mica powder can settle and condense after sitting around for a long time. So I want to use glycol stearate flakes instead.
What process is required to turn them into a rheoscopic fluid? The manufacturer states the melting point to be $\pu{135-140 °F}$. Would I simply mix them into some water and raise the temp until they melt? Should it stay in suspension after cooling?

Being highly hydrophobic, I think that just like stearic acid (see link, p. 2: https://arxiv.org/pdf/1806.06120.pdf), glycol stearate microcrystals would disperse without completely dissolving.
I would say that vigorous low-temperature stirring and maybe crushing would work. Note that, as reported around the web, high temperature would partially dissolve them, creating disordered aggregates of stearate, which would imply a loss in its "rheoscopic" qualities.
Mica powder in contrast, which is absolutely insoluble in water, would resist high temperatures (eg. useful for showing convection)


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Can an interaction be considered both hard-hard and soft-soft?
The $\ce {B-F}$ bond can be described as having both a large ionic and a large covalent character$\ce {^1}$, with the AIM charges on $\ce {F}$ and $\ce {B}$ being $\ce {-0.81}$ and $\ce {+2.43}$, respectively, and the bond critical point density being only slightly smaller than that for the $\ce {C-C}$ bond. In a similar vein of thought, can interactions between Lewis acids and Lewis bases be both rather hard and soft at the same time?
The Klopman-Salem equation can be used to quantify the degree of hardness and softness in a Lewis acid-base interaction$\ce {^2}$. The second term is a Coulombic attraction term and the third term is a HOMO-LUMO interaction term. Are there cases where both the 2nd and 3rd terms are large in magnitude?
References

Matta, C. F.; Gillespie, R. J. Understanding and Interpreting Molecular Electron Density. J. Chem. Educ. 2002, 79(9), 1141 - 1152
Fleming, I. Molecular Orbitals and Organic Chemical Reactions. John Wiley & Sons, Ltd. United Kingdom, 2009.


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Why is fluoride the most nucleophilic halide in aprotic solvents?
I am confused on the topic of electronegative nucleophiles, and polar aprotic solvents.
From my understanding, the more electronegative a nucleophile is, the weaker of a nucleophile it will be, because atoms will be less likely to share its electron density.
However, in a polar aprotic solvent, because H-bonding cannot occur, it seems as if the trend of nucleophilicity is such that $\ce{F- > Cl- > Br- > I-}$.
Can someone please clarify how this is possible? I would have expected $\ce{F-}$ to be the LEAST nucleophilic nucleophile in a polar aprotic solvent, because it is the most electronegative atom. Therefore according to the first point I learned, I thought it should be the weakest nucleophile.
All this is what I've learned through reading Kaplan's Organic Chemistry Review.

https://chemistry.stackexchange.com/questions/47072/nucleophilicity-with-respect-to-the-solvent https://chemistry.stackexchange.com/questions/95710/order-of-halide-nucleophilicity-in-dmf

When talking about elements of a group, the size of the atom is more effective than the electronegativity factor.
As $\ce{F-}$ size is too short so its electron density will be much higher than other hallogens.
In aprotic solvent the order of nucleophillicity:
$\ce{F- > Cl- > Br- >I-}$(No hydration occurs) In protic solvant(like $\ce{H2O}$) its reversed because of hydration of ions having much electron density.

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{ }$ for chemical formulas/equations, $ $ for math term/equations, and $\pu{ }$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies if you haven't already.


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Is the flint used in disposable lighters toxic?
Especially the sparks that result when using them. I heard this from a friend and can't seem to find an answer on Google.

The "flint" in disposable lighters is made of pyrophoric alloy called "ferrocerium", which consists of approx. 50% cerium, 25% lanthanum and 20% iron; the remaining 5% is magnesium, neodymium and praseodymium. Exact composition may vary depending on the manufacturer.
If you rub such an alloy on a rough surface, tiny metal particles are formed, which are instantly ignited and produce metal oxides. Neither of the expected products are particularly toxic: lanthanoid oxides, particularly $\ce{CeO2}$ and $\ce{La2O3}$, are not toxic and can only cause mild irritation when swallowed, and iron oxide $\ce{Fe3O4}$ is basically rust.
To sum it up, unless you are allergic to transition metals (which people usually discover very early), I cannot think of any health hazards coming from the "flint" or the sparks.


text:
Newman Projection with t-butyl group?
So again in Chapter 4 David Klein book, we have this organic molecule with a t-butyl group (just as an aside, this is t-butyl ethane, right?

Now how is it that the correct Newman projection must be the following?

What I glean from the Lewis structure is that because there are no wedges or dashed lines, the methyl groups ought to be in the same plane. How can there be a methyl group jutting out upwards and to the right, and another methyl group jutting out upwards and to the left in the t-butyl group and not be illustrated as such with dashed lines and wedges? Thanks in advance!

It's not " t-butyl ethane", it's 2,2‐dimethylbutane. Cram bonds are not always drawn, but it doesn't mean that all the atoms belong to the same plane, you should also apply common sense sometimes. Check out the 3D structure to get better understanding of how atoms are arranged.

@andselisk thanks for the corrections! Still trying to wrap my head around proper nomenclature as you can see. Oh gotcha, didn't know about that, I will be sure to keep that in mind going forward. Also, awesome, I didn't know about that webpage, thank you so much! I will be sure to use in it in the future for more Newman projection conundrums I have.

The three front methyl groups identify a plane as always with tree points. This is pedantic but related as for this is exactly how Fisher proj. shall be read or sketched. There will never be any cram bond as for all are tilt pointing to the observer. The presence of a double bond will make the reading obvious as well.

Lewis structures and skeletal structures show which atoms are connected to which other atoms, and the type of bonds (single, double, triple). They do not show the shape (geometry) of the molecule.
For example, the two skeletal structures below refer to the same molecule, butane:

When there is ambiguity (i.e. multiple molecules with the same skeletal structure), you can add information about the stereochemistry by adding dashes bonds and wedges. In your case, the molecule is achiral, so that does not apply.
To figure out the geometry, remember that carbon atoms with four single bonds have a tetrahedral geometry. You get other geometries when there is a double bond (trigonal planar) or a triple bond (linear) or other, more unusual features associated with a carbon atom (lone pair, non-zero formal charge, radicals, two double bonds).


text:
Ketone prefix organic chemistry nomenclature
Is there a difference between the prefixes keto- and oxo-?

keto- $\subset$ oxo-.

oxo- is more used to just describe a c=o while keto- is used when is is specifically part of a ketone

The prefix ‘keto’ doesn’t exist in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). Already in the 1979 Recommendations, the prefix ‘oxo’ is used to indicate $\ce{=O}$ when the group cannot be cited as suffix.
Nevertheless, the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995) still remembers the use of the prefix ‘keto’:

To indicate a double-bonded oxygen that is part of a ketonic structure, the term keto is sometimes used as a prefix, but such use has been abandoned by IUPAC for naming specific compounds. A traditional use of keto is for indicating oxidation of $\ce{CHOH}$ to $\ce{C=O}$ in a parent compound that contains $\ce{OH}$ groups, such as carbohydrates. E.g. 3-ketoglucose.


text:
Why is C2H5 a stronger +I group than CH3?
While reading about the basic character of amines, I came across a note which said that $\ce{CH3-CH2-NH2}$ is more stable than $\ce{CH3-NH2}$ as $\ce{C2H5}$ is a stronger +I group than $\ce{CH3}$.
However, according to what I know, isn't $\ce{CH3}$ a stronger +I group as in a longer chain the +I effect dies out?

C2H5 is stronger than CH3 for the same reason why CH3 is stronger than H. True, the effect should die out along the chain, but what is it that produces the effect in the first place? What is it sitting at the far end of the chain? Nothing. The effect is produced by the chain itself.

@IvanNeretin Thanks! I guess that works. So C3H7 would be an even stronger +I group?

Yeah, a little.

https://chemistry.stackexchange.com/a/47020/102629

@IvanNeretin but why then  methyl has more +I effect than H


text:
Working principle of simple chemical cells
I have just learnt about simple chemical cells but I have no idea how to determine which cells will work and which will not and the working principle of those cells.
I know that a chemical cell with $\ce{Zn}$ and $\ce{Cu}$ as electrodes and $\ce{CuSO4}$ as electrolyte will create a current in the external circuit. But why would $\ce{Zn}$ choose to lose electrons through the external circuit instead of having displacement reaction with the $\ce{CuSO4}$ electrolyte directly? And in the following cases, which of the cells will work and which will not?

    Electrodes   Electrolyte

1   Zn, Cu       ZnSO4

2   Zn, C        CuSO4

3   C, Cu        CuSO4

4   C, Cu        ZnSO4

And why do they (not) give out a current? Does it depend on the location of the elements in the electrochemical series? Does the choice of electrolyte affect the cells in any way?
Thank you.
Update
After reading several related questions, I find some contradictions between the answers. Is there a strict rule to decide which electrolyte to use or is it fine to use anything as long as it has good conductivity and does not react quickly with the electrodes?
Does it matter what electrolyte we use for a Galvanic Cell?
What is the Purpose of an Electrolyte in a Galvanic Cell?

Zn will react with CuSO4 directly all right, because it can. At the same time it would continue losing electrons via the external circuit, for the same reason.

in simple cells the oxidation,reduction reaction happens automatically so it depends on the location of the element in the electrochemical series
......the elements with a higher oxidation number in the top have more ability to lose electrons and and become a positive ion and this ability goes down when you go down the series ......
for ex.   E°(oxidation of Zn ) > E°(oxidation of Cu)
so the Zn loses 2e turning into Zn+2 .....
Cu+2 take these 2e and turn into cu
.....
the electrolyte for sure affect the cell because it contains the ions of the cathode's element (Cu2+) that will be replaced by the onee from anode (Zn+2) as I mentioned before
yet i'm not sure if that's what you want to know :)

Thanks! But for cells that have Zn and C as electrodes, I believe that C cannot take the electrons from Zn. Then would the cell not give out a current or would the electrolyte take part in the redox reaction in some way?

well.. I think that zn will give the 2e directly to the cu2+ ions in the electrolyte so cooper will deposit on the Zn electrode it's like when you but a piece of Zn in the electrolyte CuSO4 and im not sure about the role of C that's why I didn't answer this part of you Question


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Aetiology of the word "fullerene" in Chemistry
I am a mathematician and not a chemist, and I am trying to understand the historical relationship and current usage of the word "fullerene" across Mathematics and Chemistry.
I apologise in advance if I say chemically-ridiculous things, but I am genuinely interested in understanding this.
I will assert some things that loosely describe my understanding of the situation, but which may be chemically wrong. Constructive correction is welcomed.
MY UNDERSTANDING FROM MATHEMATICS
(1) In graph theory (my area of Mathematics) a "fullerene graph" or simply "fullerene" is usually defined to be a 3-regular planar graph, where by "graph" we mean a combinatorial graph with vertices and edges (i.e. abstractions of atoms and bonds).
(2) The standard example of a fullerene is a 60-vertex planar graph, whose faces are arranged as in a standard soccer ball, and of course this is the famous geodesic dome structure or buckyball.
[Actually, mathematicians will also use the dodecahedron as a standard example of a fullerene graph because it has just 12 pentagonal faces and no hexagonal faces. This is fine for a mathematician because 0 hexagons is a perfectly sensible number of hexagons but perhaps it is chemically nonsensical.]
(3) This structure and its name arises from the carbon molecule $C_{60}$ which was first synthesised by Kroto et al., which has this spherical shape, and which they named buckminsterfullerene after Buckminster Fuller.
(4) Mathematics writings on fullerene graphs that attempt to explain the relationship to chemical fullerenes then usually say something like:

an actual physical fullerene must have a molecular structure (atoms and bonds) that is a 3-valent planar graph with pentagonal and hexagonal faces
there are lots of mathematically possible fullerenes, but only a minuscule fraction actually physically exist.
to determine whether or not a mathematical structure might be physically realised, we compute the spectrum of the adjacency matrix and magically believe it relates to a mysterious chemistry thing called the HOMO-LUMO gap.

[In addition, some authors insist that to qualify as a fullerene graph, the pentagons must be separated from each other.]
[I guess that as carbon atoms are 4-valent, there must be a pile of double bonds floating around that us mathematicians have just totally ignored!]
MY QUESTIONS REGARDING THE CHEMISTRY
(1) When did the very specific term "buckminsterfullerene" get relaxed to "fullerene" and by whom?
(2) To exactly what class of carbon molecules does the term "fullerene" apply?
[I have learned that other forms of carbon include diamond where all atoms are 4-valent, then graphite, then fullerenes but I am hazy on the precise mathematical or chemical distinction.  Then I am further confused by the fact that some sources distinguish tube structures (nano-tubes) from spherical structures and only call the latter fullerenes]
Sorry for the ridiculously long first question, but all clarification is gratefully received!

Indeed seeing fulllerene molecules as simply connected graphs ignore a pile of double bonds which are obviously part and strongly dictating the properties of the molecules. The overall ensemble can be or not stable, and so in different extents. The situation is similar in polycyclic aromatic hydrocarbons or simply cyclic  hydrocarbons with different number of C and double bonds. Rules can be emerges and calculations provide insights. Q1 I don't know when and whom Q2 the answer is in your text just skip the recommendation in bracket as I believe there are examples in which two pentagons. ..

......two pentagons have a bond in common.

To question (1): fullerene is shorter, and both are trivial names. They originate from lab-slang. Once it was known that C-60 is not the only one, the whole class (C-70, and many, many more) was called "fullerenes", defined as distinct molecules with a closed structure (to distinguish from carbon nanotubes).

Do you mean etymology?

Aetiology = origin / history (of anything).

Just a note: normally expressions like “magically believing” and so are reserved for mocking others

@Greg - no mockery was intended, merely my own continued amazement at the (numerous) connections between concepts in graph theory and actual observed phenomena in physics and chemistry. "Magically believing" was intended to emphasise the fact that we (by which I mean "mathematicians" or really just myself)  have no understanding of how this works and so — TO ME — it is totally magical. Mathematicians love these unexpected connections because we are constantly asked to justify our existence / funding and genuine physical applications are incredibly valuable.

@GordonRoyle Than maybe it is more about ignorance to the details than magical beliefs.

Fullerenes are distinct (meaning every fullerene type has a well defined structure) carbon-only molecules, where all atoms have three bonds ($sp^2$, or nearly $sp^2$, as they're not exactly planar).
Leaving out very freakish topologies (which are unlikely to be stable and not fullerenes), this leads to a closed shell-like structure where all surface elements are pentagons or hexagons. (Which are not perfect, bond lengths differ slightly if the edge is shared by two pentagons, two hexagons, or one-and-one.)
The simpler, smaller forms are spherical (C60) or prolate (C70, ...), larger ones can probably have a surface that is not completely convex. (anyone seen oblate fullerene structures?)
A carbon nanotube with a well defined, closed end structure can be a fullerene, but they usually haven't got one.

It seems some people also count ordinary CNTs among the fullerenes. However IUPAC https://goldbook.iupac.org/html/F/F02547.html is even stricter: 12 five-rings, the rest six-rings, closed (cage-like) fused-ring polycycle.

So in my research a certain family of graphs has come up - a cluster of 6 pentagons, then an arbitrary number of “layers” each consisting of five hexagons, surrounding the previous structure, then topped off by another “cap” of 6 pentagons. Is this a fullerene? I did some searching (a few months ago) and was pleased to find that it seemed to be a carbon nanotube as well.

The pentagons do not have to be "clusters" of six and six.  They can be anywhere on the surface as long as there are 12 to close the figure.  Wherever a pentagon is inserted, it imparts part of the curvature that closes the surface; in the case of a tube the curvature is in the "caps" at the ends and therefore, so are the pentagons.

The actual word "fullerene" comes from the geodesic dome designed by Buckminster Fuller.  The near-spherical structure of $\ce{C_{60}}$, the simplest and best known of these molecules, is the dual of the triangular-faced geodesic dome.  Mathematical definitions of "fullerenes" come from the same source.


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Optical activity of organic compunds
Why are the molecules which are superimposable on their mirror images optically inactive? For example 1-bromo-2-methylpropane.

Why, that's simple. Imagine an experimental setup in which one such compound rotates light clockwise. Now imagine another setup, where everything (all the way down to molecules) is the mirror image of the first setup. Being a mirror image, it would rotate light counterclockwise. But wait, the molecules are the same! How can they know which way to rotate the light? They can't and don't.

Thanks a lot ivan...but could you explain it in a more scientific way just in case someone asks about this wrt symmetry and chirality

There is no reason to look for a mirror image. The key is intrinsic and is symmetry. Look here https://chemistry.stackexchange.com/questions/83990/what-is-occuring-on-the-quantum-level-when-a-molecule-rotates-plane-polarized-li

In other words the polarization is unaffected even by a single molecule - when the latter has symmetry.

Individual molecules will react with incoming photons depending on their orientation in space. So–at an individual molecule level–light can be rotated in polarity if the molecule has a particular structure. But in the bulk, the molecules will be randomly oriented and the effect of any single interaction will be cancelled out by other, opposite, interactions.
The exceptions to this are crystals where the overall structure is chiral so the orientation of the molecules or components of the crystal don't allow a random mix of orientations so there is a net effect on the polarisation of the light (quartz or calcite are examples).
The other exception is where molecules are chiral. If the image of the molecule in a mirror is different to the original molecule then no amount of randomness in how the molecule is oriented in space will cancel out the individual interactions of polarised light with the molecules and there will be optical activity.
Molecules which are superimposable on their mirror images don't have this property because rotating and translating them in space will provide a random mix of orientations that always cancels out to nothing when interacting with polarised light. Randomly translating or rotating a chiral molecule cannot achieve this as there is no way to create an opposite orientation in space that is the equivalent of a mirror image just by using rotations and translations.
This is basically a geometry issue: molecules that are the same as their mirror image can be rotated and translated to any possible orientation and so show no net interaction with light in the bulk on average whatever the individual molecules does. Molecules where their mirror image isn't the same don't average out to a zero interaction however you rotate or translate them.


text:
Why isn't iron(III) oxide an effective catalyst for decomposing hydrogen peroxide?
I've been working on a project involving the catalyst iron(III) oxide for decomposing hydrogen peroxide.
It barely seemed to produce any oxygen.
This YouTube video also seems to suggest that iron(III) oxide isn't a very effective catalyst.
But why?

Ah but iron(III) does act as a catalyst in peroxide decomposition (YouTube video).
It can be improved by the addition of some copper.
Some hydrogen peroxide is mixed with inhibitors (radical scavengers or chelating agents) to suppress the decomposition.

Perhaps the iron oxide is a solid and so the catalysis involves the reduction to iron(II) then re-oxidation to iron(III). This would probably be more effective if the iron were dissolved.

Because Fe(III) is already the highest oxidation state for iron. Although Fe(IV) is known, it’s only stable in unusual conditions,
Hydrogen peroxide is an oxidiser, and by decomposing it gives away oxygen. It can’t oxidise something that’s already oxidised. Fe(II) or Fe(0) might be a better option. Note that metallic iron powder might ignite in some cases, when exposed to a strong oxidiser like hydrogen peroxide.


text:
Why is helium considered as a non-metal?
How can helium be considered as a non-metal based on its chemical properties? Helium is a noble gas; it does not take part in chemical reactions under ordinary conditions. But non-metals tend to gain electrons to form negative ions. Also, there are many chemical properties that helium does not show as a non-metal. Then how can be helium considered as a non-metal by looking at is chemical properties?

On chemical properties helium is like a noble gas not a metal. It is chemically unreactive like noble gases but unlike metals; it is not electrically conductive even as a solid, unlike metals. Exactly which properties are metal like and not noble gas like?

https://en.wikipedia.org/wiki/Metal

Metal atoms share electrons.
Non-metals don't.
Helium atoms do not share electrons. QED (this ain't quantum electrodynamics).

Of course, there are semimetals, but again, He doesn't qualify, so it's not germane to the answer.


text:
What does it mean that an electron quantum numbers are n1= 4 - l1?
I'm trying to solve some exercises from the 2017 IChO, and the problem 6-A1 states that

in first ionization, an electron with the quantum numbers $n_1 = 4 - l_1$ is removed.

Now, I've never seen this way of expressing quantum numbers, and I don't know what it's supposed to mean, so I would appreciate it if someone could dumb it down to me.

Could you provide a link to the exercises? As is, I can't make out what you writing. If not, it may help if you rewrite the equation in the question using Mathjax so it is more readable. Welcome to the site! @Camila

I've edited your question including the link to the problem I found (please check out whether it is the correct one). Still, to me it's not quite clear what you are trying to find out since $n_i$ and $l_i$ are standard notations for principal and azimuthal Q.N.

@Camila you seem to just be getting confused by the subscripts. Those are just their to label that $n_1$ and $l_1$ are the quantum numbers of the 1st electron ionized.

$n_1 = 4 - l_1$ means the same as $n_1 + l_1 = 4$, so the $n$ and $l$ quantum numbers of the first electron added up should equal 4. This could refer to 3p or 4s.
For the second electron removed, the $n$ and $l$ add up to 5. This could refer to 3d, 4p or 5s.
For chromium, the first electron removed is the 4s, followed by the 3d. For copper, this is the case as well, but its mass is higher than that of iron, so it is not the answer.


text:
As dQ=nCdT, why not dQ is zero in isothermal process?
Explain in detail about the terms used in first law.

Related: If a gas absorbs heat, then how can its temperature remain constant?

Related: https://physics.stackexchange.com/questions/315381/specific-heat-in-isothermal-and-adiabatic-process-for-gas

Your equation only gives the correct relationship between the heat and the temperature change when there is no work done.  Otherwise, it gives the wrong answer (as you have noted).  So, in thermodynamics, where the system can also exchange work with its surroundings, we no longer define heat capacity in terms of the amount of heat transfer.  Instead, $C_v$ is now defined in terms of the internal energy U, such that $dU=nC_vdT$, and $C_p$ is now defined in terms of the enthalpy H, such that $dH=nC_pdT$.  More precisely, $C_v$ is the partial derivative of internal energy per mole with respect to temperature at constant molar volume, and $C_p$ is the partial derivative of the enthalpy per mole with respect to temperature at constant pressure.  Heat, which is a function of path (and not state) is no longer used to define heat capacity.


text:
Calculation of rate constant for bimolecular reaction from Gaussian DFT calculations
I am working with a bimolecular reaction for which I have computed the energy profile. I have output files for TS and reactants which I intend to use for rate constant calculations. From TST we have:
$$k=\frac{k_{\mathrm B}T}{h}\frac{Q^{\ddagger}}{Q_A Q_B}\exp\left(\frac{-E^\ddagger}{RT}\right)$$
But how do I calculate the rate constant from the Gaussian output? I know how to find $E^\ddagger$, but the partition functions are teasing me. In the output file values for $Q$ are shown both for the vibrational motion, the electronic, the translational, and the rotational:

I have found that $Q=q'_\text{trans}q_\text{rot}q_\text{vib}q_e$
I am not too sure about what the prime in $q'_\text{trans}$ means other than it might have something to do with the translational function being per unit volume.
I have stumbled upon a book where it is explained that you have to convert the translational function to being per unit volume to get the correct units of $\pu{m3 mol-1 s-1}$ for bimolecular reactions. I do not quite understand this nor which of the vibrational partition functions to use (Total Bot or Total V=0).
Furthermore, I am uncertain if the unit of $\pu{mol-1}$ is just obtained by multiplication with the Avogadro constant and what the unit of the translational partition function exactly is. I thought the unit was $\pu{m-3}$ based on:
$$\operatorname{dim}q_\text{trans}=\frac{\operatorname{dim} q^\ddagger}{(\operatorname{dim} q_\ce{A})(\operatorname{dim} q_\ce{B})}= \frac{\mathsf{L}^{-3}}{(\mathsf{L}^{-3})(\mathsf{L}^{-3})} = \mathsf{L}^{-3}$$
But with introduction of the relative translational motion per unit volume as explained in the link, I am confused of where the unit comes into play.

Actually, I would prefer to have the rate constant in cm^3molecules^-1s^-1. To obtain this, I would need to multiply with 10^6 and the Avogadro constant. However, I don't see why I am able to multiply with the Avogadro constant to get molecules^-1 as there would need to be a mol unit in the expression. Otherwise, I would obtain mol*molecules^-1, which is undesired.


text:
Is awaruite to nickel sulfate possible?
Much time has passed since my basic chemistry and I am trying to determine if an awaruite mine proposed by FPX Nickel Corp ("FPX") might serve as feed-stock to nickel sulfate for battery production.
Chemically, this is as far as I got and I am not at all certain if the typical counter-current decantation ("CCD") would remove the unwanted iron.  Neither do I know anything about how to hydrate $\ce{NiSO4}$ into battery grade hexahydrate form $\ce{NiSO4 * 6 H2O}$.
Here is the proposition (source: this is a commercial site):

Awaruite comes in 2 forms: $\ce{Ni2Fe}$ and $\ce{Ni3Fe}$. Sulfuric acid is $\ce{H2SO4}$.
  \begin{align}
\ce{Ni2Fe + 3 H2SO4 &-> 2 NiSO4 + FeSO4 + 3 H2} \\
\ce{Ni3Fe + 4 H2SO4 &-> 3 NiSO4 + FeSO4 + 4 H2}
\end{align}
  Then precipitate the iron sulfate to get rid of it and add water to hydrate the nickel sulfate:
  $$\ce{NiSO4 + 6 H2O -> NiSO4(H2O)6}$$

I think this is the form needed for battery production. I think the counter-current decantation (CCD) is the process of precipitating the iron sulfate.

@andeslisk IMO, your edit went too far in reformatting and rearranging the text in the quote. Quotes should remain true to the original. Quoted text should not be changed to improve grammar, unless you explicitly show where such changes are made, usually by having the changes in []. Certainly portions of the quoted text should not be moved out of the quote into non-quote formatting. It's reasonable to apply better formatting to the reactions and compounds, but any quote should remain as true to the original as possible, at least to the extent of actually duplicating what's being quoted.

Here is another option:  Ni/Fe + H2SO4 +O2 --> Ni/FeSO4 + H2O

Please don't worry about this question any longer.  I found the answer I was looking for buried deep in the literature, in a study that was not directly focussed upon awaruite but did contain over 20% awaruite in the matte.  It came from the Lonmin Marikana mine in South Africa: http://www.stockhouse.com/companies/bullboard?symbol=v.fpx&postid=29170221

@Doug If you found your answer, please feel free to self-answer it. It would make this post complete and it will always help future readers who are looking for answers.


text:
Diels Alder stereoselectivity
I'm familiar with the endo rule as well as the regioselectivity of the reaction. However, what I fail to understand is the stereoselectivity of the product in relation to the diene. For example, in the image below, which one is the correct product? I used to consider only 1 but I've seen several examples, in notes handed by my professor and some books, that provide only 2.

Also related, which one is accepted from the approachs below? I've seen only 1 being discussed but why should 2 be excluded? From what I've tried, they lead to different products while the orbital rules are not violated.

I've seen similar questions with no answers however. Also, books like Clayden's or March's do not point out something similar from what I've seen. The same goes for any website I've checked.

The two products you drew are enantiomers. In the absence of any chiral catalyst (or other chiral centres in the starting materials), you can only get a racemic mixture of both enantiomers.

I see. So in both pictures, both approaches are plausible given the absence of chiral entities?

Yes, not just plausible, but exactly equally plausible, such that you get an exactly 1:1 mixture.

When the two achiral entities undergo the DA reaction, two enantiotopic transition states are involved leading to a racemate. The authors presume you are aware of this issue. If they had to draw both enantiomers all the time, your texts would be appreciably more expensive than they already are. ;)

authors: Αντώνιος Κελεσίδης


text:
Does OH- attack carbonyl carbon in aldehyde during aldol reaction?
Why does $\ce {OH^-}$ not attack the carbonyl carbon in an aldehyde during the aldol reaction and instead, deprotonate the $\alpha$ hydrogen to form an enolate intermediate?
Also, why does the enolate intermediate attack the carbonyl carbon of another aldehyde once it is formed?

It does, but the addition is reversible! It is an unproductive step. Welcome to ChemSE!

Thank you.Many reactions are reversible i.e in SN1 reaction carbocation formation is reversible but the reaction still proceeds. could you please provide an alternate explination

Related: https://chemistry.stackexchange.com/questions/64425/why-do-some-nucleophiles-attack-at-the-carbonyl-group-while-others-at-an-alpha-c?rq=1

Could you clarify your question? It isn't clear at the moment. The phrase "But instead enolate intermediate attacks the carbonyl carbon" seems to be rather out of place.

As noted by Jan here, both pathways do proceed, i.e. the hydroxide attacks the carbonyl carbon and also deprotonates the $\ce {\alpha-H}$. To give you a rough gauge of the extent of each reaction, I will now provide some data from Carey & Sundberg (2007).

In general, the hydration of carbonyl compounds (aldehydes, ketones) is unfavourable, i.e. the equilibrium constant has a value that is less than 1. Only in cases where there are inductive (e.g. presence of halogen substituents) or mesomeric effects (e.g. presence of phenyl substituents) operating will the value be much greater than 1. Also, it is noted that the hydration proceeds very quickly, i.e. the rate of reaction is pretty fast. Note that this is reversible.
Having a greater number of steps, amongst other factors, results in the aldol reaction being slower in general. The equilibriums of each individual step in the aldol reaction mechanism are likely to not establish themselves as quickly as that for the hydration of the carbonyl.
From Carey & Sundberg (2007), p. 684:

In general, the reactions in the addition phase of both the base- and acid-catalyzed mechanisms are reversible. The equilibrium constant for addition is usually unfavorable for ketones. The equilibrium constant for the dehydration phase is usually favorable because of the conjugated $\alpha$, $\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall reaction can go to completion, even if the equilibrium constant for the addition step is unfavorable.

So what happens? In the aldol condensation reaction, the dehydration step is what essentially moves the whole equilibrium forward. As the carbonyls react in this aldol pathway and are depleted, the hydrates would convert back to this carbonyl form to react in this aldol pathway. Eventually, if the reaction were to go to completion, the amount of hydrates present would go to zero and all the carbonyls would have been converted to the $\alpha$, $\beta$-unsaturated carbonyl.
Reference
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.


text:
What is the bond in H2O (water) molecule ? And why?
Is it covalent or coordinate covalent bond? Because it can be a bond between $\ce{O}$ and two $\ce{H}$ atoms and so it will be covalent bond. And it can be a bond between  $\ce{H+}$ ion and $\ce{OH-}$ ion and it will be coordinate covalent bond?
If covalent, why not coordinate covalent? And if coordinate covalent, why not covalent?

There is no difference between an 'ordinary' covalent bond and a 'coordinate' covalent bond. As your structures show, they are just two ways of looking at the exact same thing.

The difference is that, in coordinate covalent bond  the 2 shared electrons are from one atom (O) , while in ordinary covalent bond the 2 shared electrons are from the 2 atoms (one from O, and the other from H)

And how is the molecule supposed to know where the electrons are coming from? The electrons don't come with labels attached.

Yeah they don't . But in coordinate covalent ,the bond is between hydroxide ion and hydrogen ion . While in ordinary covalent bond ,the bond is between atoms .

Well it will depend on whether the water is formed from $\ce OH^- + H^+ \rightarrow H_2O$ or from $2H_2+O_2 \rightarrow 2H_2O$. Are hydrogen atoms or hydrogen ions forming the water - this decides whether the hydrogen has an electron to use in the bond forming process.

https://chemistry.stackexchange.com/questions/21709/why-does-coordinate-covalent-bond-form

Thank you for all of you ,I have read this https://chemistry.stackexchange.com/questions/21709/why-does-coordinate-covalent-bond-form# , but still I don't understand which one is right ... or you mean both are possible(it can form coordinate and ordinary covalent bonds)??!  H.Linkhorn , Mithoron , Tyberius.

Nothing change. You can form water even attaching a OH- and a H+. Would you really conclude that you have a covalent bond and an ionic one in water? Probably not.

There's no such thing as dative bond creation between cation and anion, that's it.

@Mithoron thank you so much . Then when do we form dative covalent bond (between what ?).... Do you also mean that , dative bond is not formed if an ordinary covalent bond can be formed ?

@Alchimista thank you so much .. I don't mean ionic bond , I mean ordinary covalent bond and coordinate covalent bond. If they were both possible to be formed in water , then i don't think there's a problem by concluding that I have an ordinary and a coordinate covalent bonds in water ! (Because it sounds strange ,that's why I am asking .I must've understood something wrong )

We are try to say something that otherwise stated might sound rude. Your distinction inside the same and symmetric molecule is senseless! As I said why not to conclude that one bond is ionic then? We are all repeating the same in slightly different way. What you sketch as a dative bond is right the same covalent bond as sketched above. You should see that.

The water molecule is symmetrical. The two bonds in it are the same and cannot be distinguished. It makes little sense to label them as different types of bond.
Since the bonds are symmetric, the electron distributions will be the same. We can't tell, nor do we care, where the electrons "came from" and all we can tell is that electrons are shared between hydrogen and oxygen to create the bonds.
So the distinction between covalent and coordinate-covalent is irrelevant for the water molecules and does not help us understand anything about it.

Thank you so much  .If water has two same and symmetric bonds ( that means it can't have 1 coordinate and 1 ordinary bond ),that means it has 2 ordinary covalent bond ,right ?.....how the distinction between the 2 bonds is irrelevant ?water has covalent bond (covalent bonds can be ordinary or coordinate or polar... ),so I just want to know ,is the bond in H2O actually ordinary or coordinate or both are possible?[because when I used rules , both (ordinary and coordinate) were possible I think (as shown in the picture above )]

@FatimaElzahra My point is that it is a simple covenant bond and any attempt to classify it as some particular type of covalent bond isn't going to help understand it in any way.

ok thanks a lot .. You mean it is not useful to classify it ,but I just think it is strange to have 2 different possible types of bonds in water , and I thought one of them is maybe wrong . That's why I am asking :)

@Fatima Elzahra again in your reasoning there could be even an ionic one. Is the distinction that does make sense / isn't useful.

Perhaps it helps.
Let's take two hydrogen and one oxygen atoms. They come together and react to form water by covalent bonds involving their valence electrons. Fine.
Now let us take a proton and an OH- ion. The ion will "datively" share is lone pair to the proton and water is formed.  Again fine.
How do you resolve now what O-H bond is what? They are the same type of bond and does not help nor does have sense to make a distinction between them.
It is exactly what you have sketched. But the arrowed bond is not meaningful anymore as for the two electrons are one from O and one from H, nothing can keep track of history here.
Or as said in a comment: no concept on having a dative bond between ions.


text:
Why do we need to add hydrogen to amorphous silicon?
I was recently studying about Staebler-Wronski effect observed in amorphous silicon (a-Si). I realised it was due to imperfections in the lattice structure and breaking of Si-Si bond to Si-H bonds.
So if hydrogen atoms will be less the effect will be minimum. Why do we add hydrogen atoms to a-Si?
Is is just to passivate the unsaturated Si atoms at the edges, or am I missing something?

You are essentially correct. You need to passionate any unsatisfied bonds.

@JonCuster xD my goodness...

@Mithoron - Sigh... stupid auto correct. I sense a (niche) market for a technical spell check dictionary...

@JonCuster Oh no!  T'is a thing of beauty :D

Well, one either adds it on purpose: In order to reduce the defect volume - by passivating the dangling bonds. Unsaturated silicon atoms in a-Si is a serial killer of charge carriers. But then the issue is the SWE - you have to handle it somehow. Options are few but research is being done. The common way of handling it right now is to be honest about it. This pv cell will lose some efficiency the first months, but then stabilize. Not a good selling point, but that is where it is at.
Or one adds it accidentally / coincidentally by using a deposition technique or raw material that contains hydrogen. ($\ce{SiH4}$ or $\ce{SiH_xCl_{4-x}}$). But you can usually thermocycle these problems away, as long as your substrate can handle it. If you did it, accidentally, on a plastic surface - it is usually referred to as "scrap" and recycled...


text:
Iodometric Titration - Iodine Crystal formation
During an iodometric titration iodide is oxidised to iodine by an oxidiser, and then titrated using standardised sodium thiosulfate. I have had no problems using iodate as the oxidiser, however, using sodium hypochlorite (bleach) causes trouble for some reason. Instead of the iodine dissolving in the solution upon addition of acid/KI, the iodine forms black crystals which fail to dissolve even upon addition of sufficient sodium thiosulphate (even when excess is added). The solution is still black, but the crystals become readily apparent during titration.
I suspected this to be due to insufficient KI, and have used both solid and aqueous KI, however I have tested this and this is not true.
Does anyone know what is causing the crystals to form?


text:
First Order Chemical Kinetics
I have the following system of reactions:
\begin{align}
M & \xrightarrow{\lambda(M,W)}M+M \\ W & \xrightarrow{\lambda(M,W)}W+W \\ M & \xrightarrow{\mu} \emptyset \\ W & \xrightarrow{\mu} \emptyset
\end{align}
Where $\emptyset$ encodes no species and $\lambda(W,M) = \mu+c(N-(W+\delta M))$ where $\mu , c , N > 0$ and $0 < \delta < 1$.
Both $\lambda(W,M)$ and $\mu$ are the rates per individual.
I come from a mathematical background and I am trying to apply the Law of Mass action to the above system. I am confused since in the right hand side the $M$ and $W$ appears twice. What are the rates of of $M$ and $W$ of the above system?

As it stands, the system looks quite unchemical.

$\ce{M -> 2M}$ makes no sense chemically. Can you explain the proposed chemistry better?

$\ce{M -> \varnothing}$ doesn't make any sense because it is a violation of conservation of matter. It goes to something, even if you don't care what it is.


text:
Is delta S the same in reversible and irreversible processes?
I've read that change in entropy of a system is the same for a system in reversible and irreversible processes since it is a state function and does not depend on the pathway. Does this only apply to the system since I understand change in entropy of the universe is always positive for irreversible processes?

I don't see the contradiction between the two facts. The entropy of anything, including the entropy of the universe, is a state function. Why would that stop it from increasing, though? The state of the universe isn't static, so there's no reason why $\Delta S$ should be 0.

For an irreverislbe process with specified initial and final states of both the system and of the surroundings, calculating the change in entropy  between the same two end states requires you to separate the system from the surroundings, and to devise a separate alternative reversible path for each to determine the integral of dq/T for each.  This will give you the separate changes in entropy of the system, the surroundings, and the sum (i.e., the universe).  So the reversible paths between the initial and final states will not typically bear much resemblance to the actual irreversible path that the combination of system and surroundings had suffered in the irreversible process.


text:
Which of the following compounds will form significant amounts of meta product during mono-nitration reaction?

Hi, I tried to answer this question based wholly on the concepts of inductive effects. But haven't gotten anywhere with it.
I would appreciate any help regarding these questions:

What condition makes a compound to give meta product(instead of ortho or para)?
I know there is some flaw with my approach, so what is the correct way to do it?
Can this condition be applied to other compounds as well, or is it just for aromatic compounds?

Which of these compounds will protonate under nitration conditions? Will this change the directing properties?

@Waylander As far as I know, all these compounds are ortho-para directors. So, I am really confused how do they give more amounts of meta product.

Would aniline protonate under std nitration conditions of sulfuric acid/nitric acid? Would the anilinium cation still be o/p directing?

@Waylander Sorry, but I don't know. Hence the question.

The answer is it is no longer a strong director once it is protonated so you get a fair bit of meta.

@Waylander Okay. But why does it change its character from o-p director to meta director?

The answer to the question is (1) Aniline.
This occurs because under standard strongly acidic nitrating conditions (typically cH2SO4/HNO3) the aniline NH2 is protonated removing the ability of nitrogen lone pair to donate any electrons. Consequently the electron-withdrawing polar effect makes it m-directing. Some p-nitroaniline is formed because there is a small amount of the aniline freebase in equilibrium with the anilinium, but that is much more reactive than the anilinium.
Further reading here
Also discussed in this question  Why does nitration of aromatic amines give meta directed nitro compounds?


text:
Electrolytic cell to remove soft solder from silver?
Hi my chemistry is pretty dire but have vague idea that an electrochemical cell could be used to remove soft solder (lead/tin probably) from a piece of sterling silver in need of a repair.
The correct electro cell arrangement could be a good method of removing contaminants without damaging the silver / copper of the piece since lead and tin are less noble...
Opt 1 (reverse electroplating): Cell with something like a weak sulphuric acid solution as electrolyte, the contaminated silver as cathode, and possibly a stainless/carbon anode. Current should cause oxidation of the soft solder to lead/tin sulphate in solution. I'm guessing lead/tin may build up on the anode where would also be a release of... er, oxygen???
Opt 2 (Conversion to lead/tin chloride): Cell with something like H2O / NaCl solution as electrolyte, the contaminated silver as the anode and carbon / stainless cathode. Current should cause Cl to liberate at the anode and bond with lead / tin which hopefully will dissolve in the solution or can be scraped off easily. (not sure this is any better than just pouring HCl onto piece though!)
Opt 3 (conversion to lead/tin oxide): Cell with pure water or weak NaOH solution as electrolyte, the contaminated silver as the anode, and a carbon/stainless cathode. Current should generate oxygen at anode which might also cause lead/tin to convert to oxides which can hopefully be scraped off easily.
Are any of these methods viable/preferable? Opt 1 seems to be best as i dont think the electrolyte will react with the silver/copper if not too strong. Not sure if the sulphates will go into solution though or just build up / precipitate out from cathode.
Are any of these methods likely to be preferable to just pickling the piece in acid such as HCl?

I know I'm not supposed to give my opinion on here, but you asked..I would go with option 2. The HCl won't do anything (not dissolve copper anyway).This is tough to answer because the way the options are phrased is confusing...


text:
Is there a Material that is High Heat Resistent and soluble?
What i‘m looking for is any sort of material that dissolves, while being able to resist temperatures of up to 1000°C.
And it doesn‘t have to resist during dissolution, only while it‘s solid.
And the solvent should not be corrosive, or at least only be corrosive for the soluble Material, but not for other (or most) materials (mainly metals).
Also it would be good if said material can be reused afterwards.
(Like you can reuse salt if you dissolved it in water and then dried it again.)
And the solvent should be environmentaly friendly.
Now, the first Part is basically what i‘m looking for and the Second is just some bonus, so it‘s good even if it‘s not consistent with the 2nd Part.
If anyone knows such a Material, it‘d be nice if you could tell me.
Thanks in advance

Good quality quartzware does fine at 1000C (but don’t go above 1100C), and dissolves in HF, but that probably isn’t quite what you are looking for...

@JonCuster Thanks for your answer, but sadly no, i forgot to mention that the solvent shouldn‘t be corrosive (or at least not corrosive for metals(or some metals)).

Such high temperatures pretty much rule out any organic material. A very brief search suggests maybe sodium aluminate meets your requirements. It is cheap and easy to make or buy, doesn't decompose, melts at >1600 °C, is apparently highly soluble in water and could easily be recovered by drying the aqueous solution. It's likely possible to also tune the composition somewhat by adding other inorganic salts/oxides, so you can optimise for your applications. I myself have no experience with sodium aluminate, however.

@NicolauSakerNeto Thanks, that is exactly what i was looking for. Though i don‘t know how you found it with a brief search  All I could find the past 2 weeks had at most a resistance to 300°C... also, i had this Same question on Quora for 2 weeks and got 0 answers  glad i joined StackExchange  Thanks again.

I cannot guarantee the materials in this answer will meet all your requirements, so it would still be very helpful if anyone else could pitch in (especially an inorganic chemist or someone with hands-on knowledge of refractory materials). I will describe my thought process, which may ultimately be more useful than the suggestions themselves.
Immediately, the requirement to resist temperatures up to 1000 °C implies organic materials are very unlikely to be suitable. Almost all reasonable organic compounds would boil, decompose or burn (in air) well before then.
So now we focus on inorganic materials. Simple examples of very heat-resistant compounds are quartz ($\ce{SiO2}$) and alumina ($\ce{Al2O3}$), both melting only past 1500 °C and with no decomposition. However, these materials also resist dissolution; quartz dissolves only in concentrated hot aqueous base, whereas alumina dissolves in both of concentrated hot aqueous base or aqueous acid.
That said, it is possible to add other compounds to these materials and change their properties. For example, mixing $\ce{SiO2}$ with a number of other inorganic substances causes its melting point to lower to more manageable temperatures. This is how we make most kinds of glass. Particularly interesting is that the addition of sufficient alkali metal ions can turn the glass water-soluble. For example, there are a variety of sodium silicates, whose exact composition and properties will vary with the sodium content, and at high sodium content they become very water-soluble. However, it appears these silicates melt close to 1000 °C, which may preclude their use.
Fortunately the same approach works with alumina; increasing its alkali metal ion content eventually renders the material highly water-soluble. Sodium aluminate with a 1:1 mol ratio of sodium to aluminium forms $\ce{NaAlO2}$, which appears to be very water soluble and still melts at >1600 °C. In principle, this material could be made into a concentrated, thick aqueous solution or slurry, poured into a mould, dried at low heat to solidify then calcined to obtain a sodium aluminate object of the desired shape. This could then be used for your applications, and redissolved in water when necessary.
In both of the previous examples, it's certainly possible to add further inorganic compounds (e.g. borax, lime, etc.) to tweak properties further. However, much of this kind of research has already been performed, so I would recommend digging into glass fabrication/cement research rather than rediscovering the wheel.
I should also mention there are ways in which my above suggestions may be inadequate. Alkali aluminates/silicates appear to be strongly hygroscopic, which may make their use more cumbersome. More worryingly, at high alkali metal content they become strongly basic, which could corrode some of the metals you plan to use, both during dissolution in water, or by direct reaction with the aluminate/silicate surface at high temperatures. You'll have to see whether this is a potential deal-breaker.

Thanks again, and i‘m not trying to rediscover the wheel, but i need such material for an application i‘m working on, but since i‘m not too versed in chemistry it might just end as a failure  Also, i‘m a bit worried about whether there will be a significant chemical reaction if it were to come into contact with molten aluminum, since that would be one of the materials i‘d like it to have contact with... but that‘s something i should worry about when it happens(or best some Time before it happens, not that it blows up or so...)  Also, did you mention the low-heat drying to avoid cracks?

I had a feeling you wanted to use molten aluminium, which is a somewhat harder case, as it's comparatively quite reactive. Molten aluminium will rip oxygen out of most compounds (I presume you've heard of thermite?), but it might not rip oxygen out of sodium aluminate (which is kind of like ripping oxygen "from itself"), so things may work out. There might be some corrosion of the aluminium metal, especially when water is added to dissolve the sodium aluminate, but I don't know to what extent. I suggested drying in steps to reduce issues with poor setting, yes.

I also want to share another source of information: Lesker. This table shows a number of substances which are heated to high temperatures so they can be evaporated in a vacuum. Look at the "thermal evaporation" columns; they suggest what kind of materials are best able to withstand the hot substance. For example, hot aluminium is best handled by a boron nitride/titanium diboride crucible, or tungsten. However, neither of these are soluble. It also doesn't necessarily mean other crucibles are incompatible.

One more thing: you can look into methods for making aluminium casts. Most people will pour aluminium into clay, but perhaps someone out there has found a casting substrate which is soluble in water.

Well, it will still be good even if i can't use it with aluminium, but i'll work that out when the time comes. But one thing might prove to be problematic, and that is the drying speed... i would need it to go from the dissolved state (with a relatively high viscosity, to maintain it's shape, without a particular strain applied to it) to become completely solid within around 1 minute at most, the faster the better, since it should be a relatively fast process, but without loosing any structural integrity of the resulting shape.

That definitely makes it more complicated. Nevertheless, here is a video on sodium silicate by a good chemistry youtuber NurdRage which can help you get your bearings.