Precipitaiton of aragonite I'm not positive what this questions is asking.  The prompt says "Write balanced equations for the following processes.  Chemical species should be the dominant ones at the given $\mathrm{pH}$. Show your work in deriving these equations."  Then the actual question is:

  Precipitation of aragonite ($\ce{CaCO3}$;  $K_\mathrm{sp} = 4.8 \times 10^{-9}$), at $\mathrm{pH}~8$.

The balanced equation would just be $\ce{CaCO3 -> Ca^{2+} + CO3^{2-}}$ .

I know that in an acid-base reaction I can use the $\mathrm{pH}$ as the $K_\mathrm{sa}$ to find the concentrations of the reactants and products, but I'm not sure where the $\ce{H+}$ would come in here.  I'm also not sure what finding the concentrations has to do with writing a balanced equation.  Does $\ce{Ca(OH)2}$ have something to do with this?

I could change the equation to $\ce{CaCO3 -> Ca(OH)2 + CO3^{2-}}$  and then calculate the concentrations, but is that what this question is even asking?  Thanks!

I think the question aims at the composition of carbonic acid solutions. The concentration of carbonate ions ($\ce{CO3^{2-}}$) depends on the pH of the solution. At low pH values the concentration is low, at high pH values the concentration is high. So the question is if the concentration of $\ce{CO3^{2-}}$ at pH 8 is high enough to precipitate calcium carbonate.

$\ce{H2CO3}$: black, $\ce{HCO3^{-}}$: violet, $\ce{CO3^{2-}}$: turquoise, $\ce{H+}$ and $\ce{OH-}$: grey


Configuration Interaction matrix composition from electronic configurations The six electronic configurations below could all be found in the matrix elements of the CI matrix at CISD level. How do I construct the 6 x 6 CI matrix composed of these six configurations and which of the elements will be zero?
How different would it be if I used MP4(SDTQ) method?

Given those six configurations, the CI matrix with be a six-by-six matrix with rows and columns defined by the respective configuration. The matrix elements themselves will be the Hamiltonian expressed in those configurations.

For example, say column one corresponds to $\psi_0$, or $\left|0\right>$ in bra-ket notation. Similarly, row one corresponds to $\psi_0$, or $\left<0\right|$ in bra-ket notation. The CI matrix element where these coincide, namely $\left<0|H|0\right>$, where H is your Hamiltonian, is simply the ground state energy (the Hartree-Fock energy, if the configurations are the result of HF orbitals). The other matrix elements, like $\left<1|H|0\right>$ or $\left<0|H|2\right>$ have rules for determining them. The best reference is the book Modern Quantum Chemistry, by Szabo and Ostlund. Dr. David Sherrill has a good reference as well, which you can find here: http://vergil.chemistry.gatech.edu/notes/ci/ci.html

Okay, to answer the second part of your question. There are a few rules for determining CI matrix elements. One of the first ones is that singly excited configurations do not mix with the ground state. This is Brillouin's theorem. I don't see any singly excited configurations in the six you gave, so we won't have any of those. The other rule is that configurations that differ by more than two orbital occupations are zero. Since you truncated to doubles, you also won't have this case. The other thing you need to check is spin. If spin is not conserved, you'll have a zero matrix element. I think you'll have this case with the first configuration and the last one, which have different multiplicities.

As to your last question, I'm not real sure what you are asking. MP4(SDQ) is a perturbative method, which is different from CI, which is a variational method. In other words you do a CI by diagonalization, whereas MP4 is not. So I think CISDQ would be a better question...anyway, the point is that you are increasing your configuration space from singles and doubles (SD) to singles, doubles and quadruples (SDQ). The point I made in the preceding paragraph still holds, and expressing your Hamiltonian between ground state and quadruply excited configurations will be zero because their occupation differs by more than two. Same for singly excited configurations and quadruples.

EDIT: Just had the thought, you should also search for "Slater's rules" when trying to construct CI matrices. These are the rules I mentioned earlier.


electrochemistry experiment something strange.
i made an electrochemistry experiment in which i have measured the potential of cell:
Ag, AgCl | (1M)KCl ║ [Fe(CN)6] 3-, [Fe(CN)6] 4-  | Pt

with different concentrations of Fe+2/Fe+3 .
from the results i have found the faraday constant:
96181 Coulon /mol
very close to the published value (99.7% accuracy).

the intercept of the graph of E (v) against ln(Fe+2/Fe+3)   is:

intercept:    0.198985714  V

(the intercept is also the standard potential for the cell)

yet, when i tried to find the standard potential of

Fe3+ +e- -> Fe2+

i got the result of +0.423 V which is almost half of the published value (+0.77 V)

E_cathode =E^° + E_anode

E_cathode=0.198985+0.224= +0.423 V

how can it be that the faraday constant is almost accurate and in the same time the result of the standard potential is far from the real one.? i'm sure i don't have any calculation mistakes, if it's necessary i can describe the method of work here.

thanks!

What you are measuring is not the potential $Fe^{2+}/Fe^{3+}$. It is the potentiel of the couple $[Fe(CN)_6]^{4-}/[Fe(CN)_6]^{3-}$ which is equal to +0.36 V in the table.


What are the physical manifestations of radial nodes? I realize that nodes (both angular and radial) are areas with zero probablility of finding an electron.  I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.

So for the 1s orbital (zero nodes), the graph of the square of the eigenfunction will be one hump.  For the 2s orbital (one node) there will be a small hump followed by a large hump with a zero point in between them (talking about the sqaure of the eigenfunction not the eigenfunction itself).

For the 1s orbital, the peak of the hump will represent the most probable radius the electron will be found from the nucleus.  The probability will lower as the radius increases or decreases from the most probable radius (similar to gaussian distribution), and the electron will have zero probability of being found within the nucleus of the atom (0 radius).

And from what I understand the radial node(for the 2s orbital) is physically manifested at the point where the 1s orbital and 2s orbital meet (my instructor used the analogy of russian nesting dolls), and on the NON squared eigenfunction plot of the 2s orbital, the node is represented where the wave function crosses the x axis going from +y to -y.

So here is what Im really wondering (please correct me on anything stated previously if I am mistaken).  Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)?  This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)

Edit: Perhaps it is incorrect to say an electron from the 2s orbital has the possibility of being in the 1s orbital.  What I really mean is the electron may occupy the same space as the 1s orbital, but since orbitals are just 3d areas where an electron is most likely to be found perhaps I am not incorrect in saying that a 2s electron occupies the 1s orbital since the 1s orbital is technically just a region in space.  My guess is that the probability of a 2s electron occupying the same space as the 1s orbital is lower than it occupying the space of the 2s orbital due to repulsion by the electrons that already occupy the 1s orbital.

Here is a picture representing everything ive tried to explain so far:

I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.

This isn't quite right.  The square of the radial function will give the probability density for s-orbital, but this is not the probability of finding the electron at a given distance from the nucleus, because the number of points a given distance from the nucleus increase with $r^2$.  To get the plots in Fig. c, the radial function squared in multiplied by $r^2$.

For example, for 1s, the probability density is greatest at the nucleus, but by multiplying by $r^2$ when r=0, a radial probability density of zero is obtained.

Furthermore, as you can see from the graphs on pages 187 and 188 here, for 2s and 3s the most probable location (highest probability density) is at the nucleus (r=0). Again, only by multiplying by $r^2$ when r=0, is a radial probability density of zero is obtained.

So "Fig. a" in the OP is misleading, the central region of 2s and 3s should be shaded much more intensely than the outer region(s).

  So here is what Im really wondering (please correct me on anything stated previously if I am mistaken). Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)? This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)

No, the 2s electron is only in the 2s orbital.


Electrochromic properties of ethyl viologen I came across this article on electrochomic displays. It mentions Ethyl Viologen as an electochromic chemical. I would like to attempt an experiment to showcase it's electrochomic properties but don't really know where to start.
First off, where to get some. Googling I came across these, three, pages. Will one of these work? If not where do I get some? Or do I need to synthesize it (if so how)?
Second, what to do with it once I have it? I take it there's more to it than just sticking some electrodes in a beaker of it.
The article states, "If the EV-molecule absorbs a second electron, EV turns pale blue. This reduction however is irreversible and definitely not wanted." What care (if any) do I need to take to avoid this?
Sorry if I ask something obvious, I'm very inexperienced with chemistry.
Thank you.

[…] I'm very inexperienced with chemistry.

Have a look the material safety datasheet (MSDS) of ethyl viologen dibromide and pay attention to the H-statements:

  H302 Harmful if swallowed.
  H312 Harmful in contact with skin.
  H315 Causes skin irritation.
  H319 Causes serious eye irritation.
  H332 Harmful if inhaled.
  H335 May cause respiratory irritation.

Can you handle this?

You asked:

  what to do with it once I have it? I take it there's more to it than just sticking some electrodes in a beaker of it.

It is not that simple.

It is true that 4,4'-dialkyl pyridinium dications undergo a reversible one-electron reduction. This can be achieved by applying a voltage or by electronic excitation in the presence of an electron donor.

However, the resulting radical cations are very sensitive to oxygen and they have been used as probes to monitor oxygen permeation in polymers. With other words, you would have to exclude oxygen from your setup for a lasting colour effect.

In summary, I'd rather advise against the use of ethyl viologen for showcasing, unless the experiments are performed by professionals in a safe environment.


Can physical absorption happen despite the formation of products? As far as I know, the absorption of $\text{SO}_2$ in water goes under physical absorption, meaning the sulfur dioxide bubbles are held in place by Van der Waals forces, in this case Keesom forces, since both water and sulfur dioxide molecules are dipoles. This table classifies the absorption of sulfur dioxide in water as physical absorption, taken from this site.
However, sulfur dioxide does create products (within an equilibrium) with water.
$$\ce{SO2 + H2O ⇌ HSO3^- + H3O+}$$
and
$$\ce{SO2 + OH- ⇌ HSO3^-}$$
Now, with chemical absorption, the absorbate is held within the absorbent due to interfacial chemical bonding. Could it be that these products don't aid with the absorption at all? Or, could it be that they aid with the absorption in conjunction with, but perhaps to a lesser degree than, than the Keesom forces, thus making the absorption more physical than chemical? If so, is this perhaps the norm?
Or, is this absorption actually a chemical one?

When you consider the dissolution of a gas like $\ce{SO3}$ , $\ce{CO2}$, $\ce{SO2}$ etc. there are three equilibria existing at the same time:
$$\ce{SO2(g)\overset{H2O}{<=>}SO2(aq)\overset{H2O}{<=>}H2SO3(aq)<=>H+(aq) +HSO3-(aq)}$$
The first equilibrium is the dissolution equilibrium of the gas, the second is the hydration equilibrium of the dissolved gas, and the third is the dissociation equilibrium of the acid.
The pKa associated with the first deprotonation of $\ce{H2SO3}$ is 1.81 from Wikipedia, so you can pretty much consider the last equilibrium as completely dissociated in dilute solutions.
You can consider $\ce{SO2(aq)}$ as the state where the $\ce{SO2}$ molecules are physically adsorbed in the solution, the rest exist as $\ce{H+(aq)}$ and $\ce{HSO3-(aq)}$.
How much is physcially adsorbed and how much is chemically dissociated would depend on the particular compound. Unfortunately, I cannot find any data on $\ce{SO2}$. However, in case of dissolving $\ce{CO2}$ in water, 99% exist as $\ce{CO2(aq)}$ i.e. the physically absorbed gas, and 1% as $\ce{H2CO3(aq)}$, $\ce{H+(aq)}$ and $\ce{CO3^2-(aq)}$. (reference)
So, the absorption of $\ce{SO2}$ is partly chemical and partly physical, but note that all of these reactions are reversible. Also keep in mind that the distinction between physical and chemical change is not clear-cut, and gets less and less useful the more you delve into chemistry.
The website you mention in the question gets a lot of other things wrong, for example, dissolution of $\ce{SO3}$ in water gets you sulfuric acid, which is quite stable and hygroscopic, so you would be hard pressed to reverse it, but the table lists it as physical absorption. Whereas chlorine does not dissolve that well in water, and you can drive it off by heating, but it's listed as a reversible chemical change. It doesn't make much sense to me, so I would advise you to get a more reliable source of information.


Calculate the molar fraction of a mixture component given only its pressure You have a mixture of $\ce{He, Ne, Ar}$ gases with a total pressure of $8.40$ atmospheres. $\ce{He}$ has a pressure of $1.50$ and $\ce{Ne}$ has $2.00$. What is the molar fraction of $\ce{Ar}$?

The molar fraction is

$$\frac{\text{moles of Ar}}{\text{total moles of mixture}}$$

However, to know the number of moles of any of these components, I need to know their mass in the mixture. How can I know if all I am given is their pressures?

The answer is $0.583$ and I am not sure how was this concluded. All I can tell is that $\ce{Ar}$ has a pressure of $4.9$ atmospheres by Dalton's law but that's it. By the same law I know that all gases there have the same temperature and volume, but since I don't know either value for any of them it isn't useful at all.

Your mole fraction is $$x_\ce{Ar}=\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}$$ and you somehow want to insert your pressures. You can try to convert all moles that you don't know into pressures that you know by using the ideal gas law $$p~V=n~R~T$$ which will make you find that everything you have (1.5, 2.0, 4.9 atm) is all what you need to come up with $x_\ce{Ar}=0.583$.

As your question was answered by midi-chlorian without further explanation, here is why you don't need more than those pressures given to you.

I said that you can use the ideal gas law to find out the amount of substance of each gas. Therefor we need to rearrange it to give $$n=\frac{p~V}{R~T}$$
We know the gas constant is constant ... that's where the name comes from ... and we can assume the temperature to be constant at whatever temperature we want ... it's not important to know the temperature here. On top of that the missing volume of each gas is equal for all three ideal gases, as Avogadro's law says.

This means that not only R and T are constant but that the volume will also not change, which in turn simplifies everything a lot.
$$n=p~\frac{V_m}{R~T}=p~k$$
Let's insert it into the mole fraction (as I told you to do) and see what happens.
$$\begin{align}
x_\ce{Ar}=&~\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}\\
=&~\frac{p_\ce{Ar}~k}{p_\ce{He}~k+p_\ce{Ne}~k+p_\ce{Ar}~k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\cdot\frac{k}{k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\end{align}$$

And this is the simple reason why you do not need more than your pressures.


Can the amount of shielding in proton NMR caused by ring currents be predicted? My textbook (Organic Chemistry, 2nd Ed. Hornback, Joseph) states that if a hydrogen is located near the center of an aromatic ring it will experience an (often quite sizable) upfield shift, as seen in these images (borrowed from aforementioned textbook):

Theoretically it all makes sense - the ring current of the aromatic system shields these protons rather strongly. But is there a way to estimate what the actual upfield shift will be? For example, if provided with structure 1 and asked to estimate the chemical shift of the red hydrogen, how could you do so? Or do you just run the NMR and find out? My book doesn't give any advice on how to figure this out, just that it happens.

Johnson and Bovey first proposed a theory describing the ring current effect,[1] and in 1972 Haigh and Mallion published their often-cited paper on ring current shielding.[2] A later review by these authors in 1979 describes the mathematical prediction of ring currents very well.[3] You may often see this graphic describing ring current contributions to chemical shifts in textbooks.

So, in answer to your question, it is possible in theory to make chemical shift predictions for nuclei within the shielded regions of ring currents, however in practice, it is very difficult and prone to large errors. Calculations require a very thorough knowledge of distances and angles between the aromatic plane and the nucleus of interest. And a reasonable computer.

Very few chemical shift prediction tables in texts or literature give examples of upfield-shifted aromatics, and none that I'm aware of give enough tabulated data to extend an even ballpark prediction. This makes it very, very difficult to make a prediction unless you have experience with the particular type of molecules you are dealing with. For example, I work with one research group who makes quite good estimations of chemical shifts for the inner ring proton within porphyrin systems, but this has evolved over many years of experience of dealing with this particular class of molecule. They would really be at a loss trying to predict any of the shifts for the 4 compounds you have shown.

Most desktop chemical shift predictors use an additive model based on combined chemical shift contributions, and do not do a full calculation of ring current properties. In fact, many desktop predictors give very poor estimations for these types of protons for this very reason, as they do not distinguish shielded and deshielded regions arising from ring current. For small organic molecules the variables dictating the ring current contribution are difficult to predict, especially for those signals experiencing increased shielding, such as overhanging methyl groups. They require knowledge of distance from the centre of the aromatic ring and angle from the plane. This information is often used qualitatively in spectral analysis to construct a 3-dimensional structure of a molecule, in explaining upfielded shifts of certain nuclei.

In proteins, NMR shift prediction models are more inclusive, as the molecule conformations are often very highly constrained and well predicted. Many shift predictors for protein NMR includes a ring current contribution.

References

Johnson, C. E.; Bovey, F. A. Calculation of Nuclear Magnetic Resonance Spectra of Aromatic Hydrocarbons. J. Chem. Phys. 1958, 29 (5), 1012–1014. DOI: 10.1063/1.1744645.
Haigh, C. W.; Mallion, R. B. New tables of ‘ring current’ shielding in proton magnetic resonance. Org. Magn. Reson. 1972, 4 (2), 203–228. DOI: 10.1002/mrc.1270040203.
Haigh, C.; Mallion, R. Ring current theories in nuclear magnetic resonance. Prog. Nucl. Magn. Reson. Spectrosc. 1979, 13 (4), 303–344. DOI: 10.1016/0079-6565(79)80010-2.


What happens when a body of water and a pice of ice are placed in a pricessly controlled zero degree environment My understanding is, if you put ice into a >0 degrees environment, it will be 0 degrees as it melts and if you put water into a <0 degree environment, it will 0 degrees as it freezes. but what happens if you put water and ice in to an environment maintained exactly at 0 degrees. I'm inclined to think it should be said water freezes below 0 and ice melts above 0, I'd go so far, pending a satisfactory answer of course, is to suggest, what happens to water and ice at 0 degrees is nothing at all. Please put me out of my misery on this :-D

My understanding is, if you put ice into a >0 degree environment it
  will be 0 degrees as it melts and if you put water into a <0 degree
  environment it will 0 degrees as it freezes.

Correct. If you place ice in a container where the temperature is above  $0\ \mathrm{^\circ C}$, the ice will first heat up to $0\ \mathrm{^\circ C}$, and then undergo phase transition to a liquid. The same works the other way around, as you stated. Placing water in a sub-zero temperature will cause the water to cool down to $0\ \mathrm{^\circ C}$ and then it will freeze (form a solid). See picture below:

  But want happens if you put water and ice in an environment maintained exactly at 0 degrees.

This really depends on the other qualities of water. If you add salt in to the water, it will remain as a liquid at $0\ \mathrm{^\circ C}$. This is due to freezing-point depression, which is described reasonably well in Wikipedia.

If you are considering pure water (no salt or other solvents mixed), then you will get an equilibrium of water and ice. It might be easier to understand, if you keep in mind that although the overall temperature is $0\ \mathrm{^\circ C}$, there will be some variation between single molecules, some having a lower amount of thermal energy and some having a higher amount.


Solution for storing a glass electrode What kind of storage solution is required for a glass electrode?

Is 3 M $\ce{KCl}$ sufficient for this purpose? If so, what is the approximate pH of that solution (the solution I use presently has a pH $\approx$ 5)?

Can filling solutions be used as storage solutions for glass electrodes?

A glass electrode is usually stored in a 3 or 4M KCl solution, depending on whatever the manufacturer advices you to do.

But the pH that you measure from this solution is absolutely not the pH. It's the measured potential of this concentrated KCl solution that is (in this case wrongly) converted into the pH scale.

The normal pH measurements with glass electrodes cannot be done without the calibration of your device through references solutions, which are common buffer solutions with acidic, neutral and basic pH values like, e.g., pH 4, 7, 9.

In order to calibrate the pH electrode, you or actually your device saves the measured potential for each of these known buffer solutions and calculates a regression function that should give you something near: $$E = 414.12 - 59.16~pH~~~~~(\text{at}~25~^\circ C)$$

(The time when I had to do this in practical courses gave us $E = 401.93 - 56.711~pH$ ... I have no idea what the temperature was :D )

For further informations on all this you can read Mettler-Toledo's pH guide.

Now, as your KCl solution has a certain potential the device converts it through this regression equation into a pH value. But as I said, that is wrong - I guess that pH indicators like the universal indicator would show no big difference from pH 7 (I've never done this actually, so this is seriously a guess).

Your actual question should rather be about what the potential of a 3M or 4M KCl solution should be. And that, as can be found here, is:


Strongest B/L Acid Which Can Exist (Leveling Effect) Got a few issues with this:

Here is the Lewis Structure:

As we can see, we have five lone pairs on which a hydrogen proton can land and form the solvent's conjugate acid. Let's consider them left to right.

1) Oxygen lone pair. Adding a hydrogen proton here would make the oxygen have a formal charge of +1. Not exactly ideal for an atom which is so electronegative (even more EN than nitrogen is).

2) sp3 nitrogen. This seems like a good site; it parallels that the ammonia molecule. However, we would be creating a +1 formal charge.

3) sp2 nitrogen. We again would be creating a +1 formal charge.

4) sp nitrogen. And finally, we would be creating a +1 formal charge.

I know that the more s-character, the more able is the hybridized atom able to stabilize negative charge. It isn't called electronegativity for nothing.

That being said, is electronegativity inversely related to electropositivity?

So should the sp nitrogen be least able to stabilize a positive formal charge?

From a Lewis acid/base consideration, we would say that each of these lone pair sites are pretty nucleophilic. So which is the most nucleophilic. We need the strongest electron donating tendency. And that appears to be the sp3 nitrogen because in an sp3 nitrogen, the electrons are further from the nucleus on a time-average basis than the electrons in an sp2 or sp nitrogen.

Your reasoning is not entirely correct, as you consider that the charge is only located on the atom which bears the proton upon protonation. But nearby atoms, bonds and molecules will also participate in the stabilization of the positive charge, through several factors.

In the gas phase (so considering no solution phase effects), you should at least consider that, in addition to the instrinsic capability to bear a proton of the atom (which goes as the inverse of electronegativity, as you state in your discussion), one has also to consider the following factors:

Resonance effects (delocalization of the charge on other atoms, that may be less electronegative than the atom directly bearing the proton). In the molecule given above, this would lead to a stabilization of the $sp$ and $sp^{2}$ hybridized nitrogens, as there would be a partial positive charge on the neighbouring carbon atom, as: $$\ce{RCH=N+H-R' <-> RC+H-NH-R'} $$ or $$\ce{RC#N+H <-> RC^{+}=NH}$$
Inductive effects (or if you prefer availability of electrons in nearby atoms that can be displaced towards the charged site). In your constrained textbook example, this would be hard to indicate: there are a lot of heteroatoms (nitrogens and oxygens) and electron withdrawing groups (CN) which are likely to remove electrons from the carbon atoms, and in the central part of the molecule, these are likely to contribute to destabilize the molecule.
Intramolecular hydrogen bonds. Although sometimes neglected, these can contribute quite a lot to the stabilization of the charge when mutiple protonation sites involving heteroatoms are present. Just as a reminder, in the gas phase, a hydrogen bond in a protonated molecule can reach up to 120 kJ/mol. In your example they would be hard to predict without modelling the complete structure, but one could consider a possible H-bond between the protonated imine ($sp^{2}$ nitrogen) and the alcohol oxygen.

Then if we were to consider solution phase, the interactions between neighbouring molecules and the propensity to form hydrogen bonds with the protonated molecule will also play a role on the possible protonation sites.

Based on the above considerations, one should consider that the most likely stable forms would be protonation on either the nitrile group ($sp$ nitrogen) or the imine group ($sp^{2}$ nitrogen), as both allow to put a partial positive charge on the neighbouting carbon atom. The protonated nitrile group is a protonation on an $sp$ nitrogen, with charge delocalization on an $sp$ carbon which means that the charge stabilization will be less efficient than for an imonnium ion.

If one considers purely gas phase consideration, one can use the proton affinity scale to rank these various contributions.

If we look for proton affinities in the NIST database we have:

for a secondary amine $\ce{H3C-NH-CH3}$ 929 kJ/mol
for a nitrile $\ce{H3C-C#N}$ 779 kJ/mol

Unfortunately little experimental data is available for imines. But some theoretical data is available in this article by Hammerum et al.. If one considers the methylimine $\ce{Z-CH3-CH=N-CH3}$ proton affinity is calculated at 943 kJ/mol, higher than dimethylamine. But on the other hand if one considers $\ce{H2C=N-CH3}$, its proton affinity is of 880 (exp) or 902 (theo), lower than dimethylamine.

So inductive effects clearly play a very important role in the order between amine protonation and imine protonation in the gas phase. In a really constrained molecule as above, they are likely to play a major role, as is the possibility to form hydrogen bonds in the solution.

If one now puts in the possible formation of intermolecular hydrogen bonds, protonated secondary amines can make two hydrogen bonds with two hydrogen bond acceptors, whereas protonated secondary imines can only make one hydrogen bond. This will likely tip the balance towards amines and would make answer (4) most likely in solution, whereas I would be quite at loss to give an answer in the gas phase.


Aromaticity of 1,1'-bi(cycloprop-2-en-1-ylidene) and 1,4-dioxine Are these Anti aromatic/aromatic or non aromatic?

1.

Here, if we were to consider the resonating structure(s) with charge separation

One ring is aromatic and other is anti aromatic, and secondly there is not continuous delocalization of pi electrons between those rings {if we don't include no resonating structures with no formal charges, as their contribution would not be equalish to normal form} what category would this go in?

2.

Like before, there can be no continuous electron delocalization like stated in previous example , only other ones the diff forms of

If we consider it to have enough delocalization, it would have 4n electrons and hence antiaromatic {although all the orbitals wont be mixed as 3 on each side twice with oxygen's lone pair being common between all of them}

As we can see in pyrene, if we consider central double bond to be non participating, huckle rule works, is there a better (and just by observation by eye) method that we could use for multi ring systems? what is the basic condition for aromaticity besides planar,cyclic,sp2 {nitrogen and oxygen in general change their hybridisation as to make it aromatic with their lone pair(s)}

Antiaromaticity is a fictional state with real consequences,  that any molecule will distort to avoid.

In the case of molecule 1, the primary state will be a triplet (unpaired pi-electrons), with the central bond lengthening to decrease the pairing of the pi-electrons.  This will lead to 3 electrons in each pi-system, and hence preventing antiaromaticity, because there are no longer 4n electrons in the rings.  A fictional molecule, with the constituency shown for molecule 1 that is in a singlet state would be antiaromatic.

In the case of molecule 2, the molecule will distort itself to make the at least one of the oxygens non-coplanaer with the pi-system, and no longer being co-planar, the system will not be antiaormatic.  A fictional molecule with the constituency shown for molecule 2 that is co-planar, would be antiaromatic.

It is unclear what you are asking about for pyrene.


Relative fraction of population of different enzyme conformations, knowing the rate constants Say we have an enzyme that has multiple structural conformations, say A, B, C and D. Additionally, we say that the enzyme must go through all conformations, to get to one end to the other, i.e.

$$\large \ce{A <=>[k_{A-B}][k_{B-A}] B <=>[k_{B-C}][k_{C-B}] C <=>[k_{C-D}][k_{D-C}] D}$$

So there's no skipping any of these steps to go from A to D.

Given all the rate constants derived experimentally, and thus knowing the equilibrium constants for every step, is it possible to derive an overall equilibrium constant that describes which conformation the enzyme is most likely to be found, and perhaps more importantly, the relative fractions for every conformation?

If we use the equilibrium constants:

$$\ce{A <=>[$K_1$] B <=>[$K_2$] C <=>[$K_3$] D}$$

And we call $[\text{E}]$ the total enzyme concentration, we have:

$$ [\text{A}] + [\text{B}] + [\text{C}] + [\text{D}] = [\text{E}]\label{a}\tag{1}$$

It's not hard to see that $\displaystyle K_1 K_2 K_3 = \frac{[\text{D}]}{[\text{A}]}$. From here, we have that $\displaystyle [\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3}$. We also get:

$$[\text{B}] = K_1[\text{A}] = \frac{[\text{D}]}{K_2 K_3}$$
$$[\text{C}] = K_2[\text{B}] = \frac{[\text{D}]}{K_3}$$

Hence, substituting these last three equalities in $\ref{a}$, we get:

$$ \frac{[\text{D}]}{K_1 K_2 K_3} + \frac{[\text{D}]}{K_2 K_3} + \frac{[\text{D}]}{K_3} + [\text{D}] = [\text{E}]$$

So:

$$[\text{D}] = \frac{K_1 K_2 K_3 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

From here we can also express the concentrations of the other species in terms of $[\text{E}]$:

$$[\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3} = \frac{[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{B}] = K_1 [\text{A}] = \frac{K_1 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{C}] = K_2 [\text{B}] = \frac{K_1 K_2[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

With these equations we may compare the concentrations of each species at equilibrium, thus we are able to determine which one is the highest based on the values for the equilibrium constants; and we can also obtain their mole fractions (divide each by $[\text{E}]$, which will eliminate this term from all the numerators).


Why is sodium cyanide an inorganic compound? I have been told that sodium cyanide is an inorganic compound during a lab experiment but I don't know the reason for this.

Shouldn't sodium cyanide be an organic compound because it has a carbon atom bonded to nitrogen?

The Wikipedia article  on cyanide also says that it is an inorganic compound but I fail to find why it is so.

How do we differentiate between an organic compound and an inorganic compound?

Sodium cyanide is classified as an inorganic compound for historic reasons, much like the entire organic-inorganic classification system is historical in reasoning.

The basic initial idea was that forming organic compounds required a vis vitalis, a life force, and could therefore only occur in living systems. On the other side, inorganic compounds were easily synthesised in vitro and have been synthesised even before alchemy was renamed to chemistry. These included most ionic compounds then known.

Naturally, sodium cyanaide is ionic. Most notably, early chemists would have been able to precipitate cyanide from other cyanide sources, redissolve it and add a sodium source (most likely sodium hydroxide) to regain sodium cyanide; thereby classifying it as inorganic.

In 1828, Friedrich Wöhler disproved the vis vitalis assumption and laid the foundation for synthetic organic chemistry. Yet the distinction between inorganic and organic chemistry was still upheld. Rather than defining it by synthesisability, it was said that any molecule containing carbon be organic — or one of many other similar definitions. This could exclude the cyanide salts but does not have to. The only salts and carbon-containing molecules that are always and consistently classified as inorganic are carbon monoxide, carbon dioxide and (hydrogen)carbonates.

By now, the distinction between organic and inorganic chemistry has lost most of its relevance and is only there for historic reasons. In principle, the distinction is that organic chemists are more interested in the carbon-containing compounds while inorganic chemists lean towards other elements more. But then there are numerous in-between fields that don’t really fit anywhere such as organometallic chemistry, bioinorganic chemistry etc.

Therefore, treat the organic/inorganic distinction as a historical residue and don’t read anything into it.


Does the ignition point of flammable substances changes when provided a pure oxygen atmosphere? I'm wondering if autoignition point decreases for substances when provided higher oxygen concentration.

No, but because of a technical point.

From Wikipedia: The autoignition temperature or kindling point of a substance is the lowest temperature at which it will spontaneously ignite in normal atmosphere without an external source of ignition, such as a flame or spark.

So the definition of the autoignition temperature is based on air not pure oxygen.

For the ignition temperature in general the Wikipedia reference goes on: The temperature at which a chemical will ignite decreases as the pressure or oxygen concentration increases.


Van't Hoff factor for citric and phosphoric acids I am trying to determine the van't Hoff Factor for citric and phosphoric acids.

When phosphoric acid is in aqueous solution, it dissociates as follows:

$$\ce{H3PO4 ->H2PO4- + H3O+}$$

Does this mean its van't Hoff factor is 2? For citric acid I cannot find what it dissociates into anywhere, so I am unable to determine the van't Hoff factor.

The Van't Hoff factor isn't an intrinsic property of any species - it depends on the dissociation degree of an electrolyte. Both of those acids are weak and triprotic, meaning they can lose up to three protons in solution (depending on the pH). Therefore, the Van't Hoff factor will likewise change depending on the pH.

1) If the acid is the only species in solution, and since it's a weak acid, you need to use the same formula as for any weak electrolyte:

$i=1+\alpha$

You can figure out the dissociation factor, $\alpha$, from the dissociation constant:

$K_1=\cfrac{[H_2A^-][H^+]}{[H_3A]}=\cfrac{(\alpha C_{H_3A})^2}{(1-\alpha)C_{H_3A}}=\cfrac{\alpha^2}{1-\alpha}C_{H_3A}$

where $C_{H_3A}$ is the formal concentration of the acid.

2) If you have more species in solution, it will depend on the pH of the solution. In general, you'll have this expression:

$i=\cfrac{[H_3A]+[H_2A^-]+[HA^{2-}]+[A^{3-}]+[H^+]}{C_{H_3A}}$

After some tedious substitutions you'll get

$i=1+\cfrac{K_1}{[H^+]+K_1}+\cfrac{K_2}{[H^+]+K_2}+\cfrac{K_3}{[H^+]+K_3}$

Some relevant values:

$pH=0 \rightarrow i \approx 1$

$pH=pK_1 \rightarrow i = 1.5$

$pH=pK_2 \rightarrow i = 2.5$

$pH=pK_3 \rightarrow i = 3.5$

$pH=14 \rightarrow i \approx 4$

and it changes logarithmically in between.


Has anyone encountered the use of Omega to represent equilibrium? One of my lecturers last year (a pharmacologist by training) used the symbol $\Omega ^m$ as a shorthand for equilibrium. He implied it was common practice but I've never come across it elsewhere. I was wondering if anyone else has encountered it?

I would guess $Ω^m$ notation is a personal invention and I couldn't find any sources that would standardize it.

The choice of these symbols for abbreviating "equilibrium" may be justified that the glyph "Libra" ♎ originates from the Greek letter Omega Ω.
Further, from Merriam–Webster Online:

  Equilibrium contains a root from the Latin libra, meaning "weight" or "balance". As a constellation, zodiac symbol, and astrological sign, Libra is usually pictured as a set of balance scales, often held by the blindfolded goddess of justice, which symbolizes fairness, equality, and justice. Equilibrium has special meanings in biology, chemistry, physics, and economics, but in all of them it refers to the balance of competing influences.

Uppercase letter "m" is probably because $Ω^m$ is a shortened form (contruction) of a single word that ends in the same letter: equilibrium.

$$\underset{Ω}{\underbrace{\text{equi}\underset{♎}{\underbrace{\text{libr}}}\text{iu}}}\underset{^m}{\underbrace{\text{m}}}$$


isomerism in coordination complexes Is there any difference between geometrical and cis-trans isomerism in coordination complexes, because in my book it is given that complexes of the form [Ma4bc] does not show cis-trans isomerism but shows geometrical isomerism

Geometrical isomers are compounds where all atoms are connected to the same atoms but their arrangement in space differs. This does include cis-trans isomers but in the context of coordination compounds also fac-mer isomers ($\ce{[MA3B3]}$ cases).

The case of $\ce{[MA4BC]}$ is tricky. In principle, they show the same two possible arrangements that, if $\ce{B}=\ce{C}$ would be termed cis and trans. In lab practice, most chemists would use cis and trans to describe them. In IUPAC theory, that response and terminology may or may not be valid.


Why does H₂Se have a bigger Ka2 than H₂Te Why does $\ce{H2Se}$ have a bigger $K_\text{a2}$ than both $\ce{H2S}$ and $\ce{H2Te}$?

The following data are from the Rubber Book (CRC Handbook of Chemistry and Physics), 89th edition, 2009, p 8-40.

\begin{array}{lll}
\mathrm{acid} & \mathrm{pK_{a1}} & \mathrm{pK_{a2}} \\
\hline
\ce{H2S} & 7.05 & 19\\
\ce{H2Se} & 3.89& 11\\
\ce{H2Te} & 2.6 & 11\\
\end{array}

Note that $\mathrm{pK_{a1}}$ of $\ce{H2Te}$ was determined at 18 °C, all other data at 25 °C


How may copper acetate ligands be manipulated to change colors? I have a solution of copper acetate and I would like to play around with the ligands to get different colors.

Background: The copper acetate was made through mixing vinegar (5% acetic acid), $\ce{NaCl}$, and $\ce{C}_{\textrm{(s)}}$. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room.

[edit] I neglected to mention that the experiment simultaneously produced a $\pu{0.5 inch}$ deposit of what appears to be Copper Carbonate or Verdigris on the bottom of the 1-Liter beaker. Also, I have let the copper acetate solution evaporate for several years now. The former $\pu{1L}$ is now $\pu{0.5 L}$ and has begun precipitating crystals (like the ones on the wiki page). Fun fact: During the Renaissance, glacial acetic acid was made by dry distilling metal acetates and primarily copper(II) acetate.[/edit]

Question(s):

What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color?
Will heating or cooling the solution change the color and/or ligands?

I may have accidentally created copper cloride in addition to the copper acetate. I decided to just let the solution (made form Copper, vinegar, and salt) evaporate with a small fan speeding the evaporation for the last few weeks. The tallest copper acetate crystals started breaking the surface and then the surface of the liquid started to form a skin (pic 1) so I poured off the now green liquid into another jar. Copper acetate is blue, which is why the green solution seems to suggest that there is copper cloride.

Evaporation! Note the blue-green layer of copper carbonate below the dark copper acetate crystals.

Top View

The green liquid on the left is what was left from evaporation (and I think is copper chloride) - next to another jar of copper acetate.

The copper acetate crystals were covered in a thin layer of copper carbonate because last week I disturbed the solution and stirred up a cloud of copper carbonate accidantally. I rinsed them with vinegar and they cleaned nicely.

The largest crystal was just over 1 cm wide. These crystals were grown from a solution that started evaporating 5 years ago.

The copper carbonate goo in the bottom needed a use so I added more vinegar to it and it formed the copper acetate solution three pictures up (next to the green copper chloride).


How to easily distinguish salts physically? Edit I have basically 12 salts to identify in exams. They are ***

  Ammonium Carbonate, Ammonium Sulphate, Ammonium chloride, Aluminium
  Nitrate , Aluminium Sulphate, Barium chloride, Calcium Carbonate,
  Magnesium Carbonate, Magnesium Sulphate, Lead Nitrate, Zinc Sulphide,
  Zinc Sulphate

***. I will be given one of them. I know all the tests I have to undertake to identify these salts (such as ash test, flame test, sodium carbonate extract tests etc). But I would like to know if there is any quick way to easily identify my salt on looking it or some easy tests so that I get a leverage over time and could finish my tests easier. I would like to know ASAP, as I have my exams tomorrow.

Thanks in advance

Draw a matrix of cations vs anions, mark the possible combinations, and calculate the probabilities.

The chances are 4/12 for a sulphate, 3/12 for a carbonate, etc.

I wouldn't start testing for sulphides ;-)


Carbon-13 NMR for chloroform I am slightly confused by what the spectrum would show for carbon-13 NMR of $\ce{CHCl3}$.

My initial guess would be that the peak would be split by coupling to both the proton and the 3 chlorines, as both nuclei have a net spin.

If the peak were split by the chlorine only, then as there are three chlorine atoms we would get a quartet peak. However this cannot be correct because the spectrum for CDCl3 shown here has only three peaks:

I also don't understand here why the areas of the three peaks are the same. Should they not be in some other ratio (for the four peaks I expected it would be $1:3:3:1$ are ratio)?

Then I would expect that the presence of the proton would split every one of the four peaks from $\ce{C-Cl}$ coupling further into a doublet, giving a quartet of doublets.

I didn't find any spectra showing this, only showing very tiny peaks for $\ce{CHCl3}$ compared with $\ce{CDCl3}$ which I don't understand:

To make it more confusing, my lecture notes say that for chloroform, $\ce{CHCl3}$,

  each $\ce{^13C}$ is attached to a spin $1/2$ proton so unless we applied broadband proton decoupling, coupling to the proton would mean that the $\ce{^13C}$ signal would appear as a doublet.

And it makes no mention of the effect of the chlorine atoms on the spectrum.

I would be very grateful is someone could expalin what the spectrum for carbon-13 NMR of chloroform, $\ce{CHCl3}$, would actually look like!

You assumed that coupling to the three chlorides would yield some type of quartet. This is correct in principle. However, chlorine is one of the many quadrupole nuclei that are basically unobservable by NMR due to their rapid relaxation. I hope another answer is around explaining that as I am not good at it.

I can, however, help you interpret the $1:1:1$ triplet of your spectrum. It is not, as you have assumed, the signal for $\ce{CHCl3}$ but that for $\ce{CDCl3}$. The difference between the two in NMR terms is that chloroform contains the spin ½ nucleus protium while deuterochloroform contains the spin 1 nucleus deuterium — and the latter does not relax rapidly as the chlorine isotopes do. Hence, deuterium is well-observable by NMR.

Now your first thought should be something like the following:

  But I am only coupling to one nucleus. That should give me a $1:1$ doublet and not a $1:1:1$ triplet, shouldn’t it?

That simplification is only correct for spin ½ nuclei. The actual formula for calculating the multiplicity of a peak is $2\,n\,I + 1$, where $n$ is the number of chemically equivalent nuclei your observed nucleus is coupling to and $I$ is the spin of the coupling nuclei. Since deuterium is spin 1 ($I = 1$), the formula gives us:

$$2\,n\,I + 1 = 2 \times 1 \times 1 + 1 = 3$$

Or an expected triplet. Since this triplet does not derive from two equivalent couplings to spin ½ nuclei, it is not a $1:2:1$ triplet but a $1:1:1$ one.

Your second spectrum shows a $\ce{^13C}$ NMR of deuterochloroform which still contains residues of chloroform — at least, that is what it is supposed to show. For some reason unknown to me, the protochloroform signal has not been decoupled, i.e. the spectrum is in fact $\ce{^13C}$ and not $\ce{^13C\{^1H\}}$. Thus, you are observing a strong coupling to the single hydrogen nucleus giving a $1:1$ doublet as expected if chlorine is ignored.


In the condensation reaction of an aldehyde with a hydroxyl amine, why is an oxime formed and not an amide? Condensation reactions are those that involve removal of a water molecule. When you analyze the reaction between an aldehyde and hydroxyl amine, I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine, thus leaving behind an amide ($\ce{R-CO-NH2}$) rather than an oxime being formed?

I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine

There is a lot wrong in this sentence and that’s the key to the answer. For one, there is no hydroxide radical (neither in hydroxyl amines nor anywhere else); if anything it is a hydroxyl radical $\ce{HO^{.}}$ — but it is not present in hydroxyl amine. If you check out the mechanism of the condensation, you will realise that no basic hydroxide species occurs:

If any basic species ever exists in here, it is the oxygen anion in the first step. Note especially that the hydroxyl group on nitrogen never loses its proton and thus never becomes basic in this mechanism.

I can hear you asking ‘so why doesn’t the oxyanion in step two abstract the neighbouring hydrogen?’ And this is really the key:

That hydrogen is not acidic!

Never, never, never ever deprotonate this hydrogen on an aldehyde. The only way to make it moderately acidic (but you would still need very strong bases to do so) is something along the lines of the Corey-Seebach umpolung. It generates a cyclic S,S-acetal out of your aldehyde which is stable. Its $\mathrm{p}K_\mathrm{a}$ is low enough that it is actually accessable. That of an aldehyde is not.

Also, to finally generate an amide, you would need some kind of migration of the hydroxyl oxygen to the aldehyde carbon after hypothetic deprotonation. I’m not saying it’s impossible, but rather unlikely (but that is difficult to judge due to the unlikelyness of the previous step). Note that an amide’s carbon has an oxidation state of $\mathrm{+ III}$ while the oxime’s carbon has $\mathrm{+I}$. You need a redox mechanism to get there.


Do isotones share any similarities? A trivial research in atoms and their basic theories led me to this term: "isotones"
Nuclides sharing the same number of neutrons but different atomic numbers.
A simple request or shall I say, quest:
please indicate me to any feasible similarities existing between isotones and please, DO NOT use jargon.

Chemistry is determined entirely by electron interactions. Since the number of electrons in an atom is determined by the number of protons in the nucleus, the atomic number (number of protons) is what defines the chemical behavior of a given element. This means that in terms of chemical properties, isotones would have no similarities unless they happened to be in the same group (column) of the periodic table, but that would be a coincidence and would have nothing to do with the number of neutrons.

As DavePhD and ron said, the nuclear stability could be similar, since nuclear stability has a lot to do with the total number of nucleons (neutrons + protons) as well as the relative proportion of protons to neutrons. The wikipedia article on isotones discusses stability in a little more detail.


Anti matter elements? Quite a self-explanatory question I guess. I wonder whether there is any information or theory on this. I seem to like remember there is such an element with positron in center and electron rotating around but nothing else.

Can there be any use if we somehow get these? Don't think like immediate destruction, the electrostatic force between "normal" elements electron and the other types can still repel each other and prevent immediate annihilation, in my view.

Can there be any spot in universe with mostly antimatter?

There is no theory, nor is there a need for one. To put in in other words: yes, there is a theory, and that's the one you are already familiar with. Antimatter elements are precisely the same as ordinary matter elements, only the other way around. There "are" (read: "could be", though some were actually detected) antihydrogen, antihelium (all with the same energy levels as their real counterparts), also anticarbon, antiwater, anti-$\ce{NaCl}$, antiglucose, and pretty much everything anty-else.

To compose a substance of particles and antiparticles together is an entirely different question. Yes, there is a compound of electron and positron; it is called positronium and has the lifetime of a few microseconds or something. Electrostatic forces attract the opposite particles, rather than repel them (otherwise it would not be a compound, BTW), so it annihilates pretty quickly. There are other similar compounds, equally or more short-lived. Matter and antimatter do not sit well with each other; there is no going around it.

As for the spots of the Universe composed of antimatter, there was once a mainstream view that there must be many of those, about as many as those made of matter. Then the global symmetry broke, and things went... complicated. I'd rather leave it at that. Go ask at Phys.SE for a more comprehensive take.

So it goes.


Quoting from "InChI, the IUPAC International Chemical Identifier" article [Heller et al. Journal of Cheminformatics (2015) 7:23]:

Each atom is described by a number of properties: its
chemical element name; x,y,z- coordinates (all or any of
them may be zero);

When I edit all coordinates to zero, inchi-1 produces a valid InChI identifier:
InChI=1S/C7H6O2/c8-7(9)6-4-2-1-3-5-6/h1-5H,(H,8,9)

and auxiliary info, which reflect no valid atom coordinates available:
AuxInfo=1/1/N:6,3,7,1,4,2,5,8,9/E:(2,3)(4,5)(8,9)/rA:9nCCCCCCCOO/rB:d1;s1;s2;s2;d3;d4s6;s5;d5;/rC:;;;;;;;;;

I did a similar test earlier, but I didn't retain the strict 10 character column width for each coordinate, which produced an error:
Error 34 (no InChI; Cannot interpret atom block line:     0.0000  0.0000    0.0000


Are the electrons also distributed for the alpha decay? Consider this reaction of alpha decay:

$$ \ce{^{222}_{88}Ra -> ^{218}_{86}Rn + ^4_2 He} $$

My question is, whether or not the electrons also get distributed in the reaction like the nucleons?

The equation for the $\alpha$-decay omits the charges.

Since the $\alpha$-particle actually is $\ce{^4_2He^{2+}}$, the resulting nucleus is initially left with a surplus of two electrons. These are emitted too, probably on the same timescale.

So, the answer is yes: the $\alpha$-decay of a neutral nucleus yields another neutral nucleus, a double-positively charged $\alpha$-particle and two electrons.


Difference between Particle in a Box and Harmonic Oscillator approximations Why in the particle in a box model do the values of n begin at 1 but in the harmonic oscillator they begin at 0? I understand what the wave-functions and their corresponding probabilities look like and that the PIB has 0 nodes for n=1 which means the number of nodes is n-1, so for n = 0 it would have -1 nodes which is physically unreasonable. But I do not understand why the PIB cannot have n=0 whereas the Harmonic oscillator can.

Quantum numbers pop up when you solve Schrödingers wave equation for systems whose solutions are discrete due to the presence of a restraining potential. There is some degree of arbitrariness to what the range of quantum numbers might be, because it depends on how you define the quantum number.

You could just as well define the quantum number and associated energy levels for a 1D  pib as $$\begin{align} E_{n'}&=k(n'+1)^2 \\&=k n'^2 +2kn' + k\end{align} $$ where $$k=\frac{\pi^2 \hbar^2}{2mL^2}$$ Here $n'$ defines a "new" quantum number related to the original one as $n'=n-1$ and therefore ranges from 0 to $\infty$.

For the harmonic oscillator, you could similarly rewrite the energy as

$$E_n = \hbar \omega n'$$

where $n'$ ranges in integer increments from $\frac12$ to $\infty$.

Physically speaking, the non-zero ground-state energy of both the pib and harmonic oscillators is usually  attributed to the Heisenberg Uncertainty Principle, because a particle with no energy would have no velocity and therefore would be completely unrestrained in position. There are of course plane waves, but this would be a strange plane wave, with no amplitude.


Why is the base-centered orthorhombic crystal lattice a unique crystal system? I'm having difficulty understanding why the base-centered orthorhombic crystal system is a unique crystal system. When I draw two base-centered orthorhombic unit cells next to each other there appears to be a primitive crystal that is contained within the union of the two lattices as shown in the image below.

This primitive crystal system appears to be monoclinic--though I wouldn't be surprised if I'm wrong on that front--if you take δ to be the typical β angle of a monoclinic crystal system (I've set δ ≠ 120° to avoid the rhombohedral scenario). Why does this not make the base-centered orthorhombic lattice redundant? Am I misguided in trying to understand these crystal systems in terms of their lattice points and unit cell lengths instead of their symmetry?

All the centered Bravais lattices are redundant in the sense that it is possible to use a primitive cell of smaller volume instead. However, this often means that crystal symmetries are no longer aligned with the cell axes. For the orthorhombic system, you either have 2-fold rotation or screw axis parallel to the cell edges, or mirror planes or glide planes perpendicular to the cell edges. If you switch to a primitive lattice for the centered cases, these symmetry elements will no longer be aligned in this simple way (they might be on diagonals, making the symmetry operations harder to figure out).
So it is a convention, and one that is useful.


Does this set of quantum numbers have a unique solution? I received the following question:

  Complete the missing values for the four quantum numbers:
  $n=?$, $\ell = 2 , m_\ell = 0$ , $m_s= ? $ .

The problem is that I think $n$ can be any number greater then 2, and $m_s$ can be $\frac{1}{2} $ or $-\frac{1}{2}$, so there seems to be no unique solution to this problem.

Is this correct?

You haven't gotten anything wrong, you are entirely correct!

A set of quantum numbers can be used to identify the specific electron in the sub-atomic world of an atom, and what path it occupies in three dimension.

$\ell = 2$ refers to the $d$ orbital. Now how many electrons can a d orbital contain? It has 5 "slots" because we can go from -2, -1, 0, 1, 2. These represent the orientations, each one holding up to two electrons at different spins (up or down, -1/2 or +1/2). We therefore know that $\ell = 2$ refers to a possible 10 different unique sets of quantum numbers, because each set could represent a specific electron, of which there could be up to 10 in the d orbitals, only 6 for p orbitals, etc.

The principal quantum number ($n$): Has only integral values (1, 2, 3, etc) starting at 1. Think of this as the power level. An electron in a 3s, 3p, 3d, etc orbital is in the third power level. It represents the numbers before the letter in the electron configuration such as 1s$^2$2s$^2$2p$^6$ so in other words, it represents what level the electron is on.

The angular momentum quantum number ($\ell$): goes only from 0 to $n-1$, also uses integral values like $n$. Think of this as the letter after the number in the electron configuration such has 1s$^2$2s$^2$2p$^6$ so in other words it represents which orbital we are talking about on that power level. 0 represents $s$, 1 for $p$, 2 for $d$, 3 for $f$ etc.

The magnetic quantum number ($m_\ell$): Goes from $-\ell$ to $\ell$ including 0. Think of this value as representing a particular orientation of an orbital. The number of orientations will vary depending on the orbital. s orbitals are spherical and can only have one orientation, s orbitals are referred to by saying $\ell = 0$. If we said $\ell = 1$ we would be talking about the p orbital, which has -1, 0, and 1 orientations (3 total). This can be thought of as the x, y, and z axis. $d$ orbitals have 5 possibilities, $f$ has 7, etc. I like to refer to each of these orientations as a "slot" which can hold two electrons.

The value for $m_s$ is used to identify the spin either up or down, but not both.


Doping of a solid with impurities This question is a continuation of the previous question of mine:press me if you wish

And my question is how can I dope a solid with impurities and if the process can be understood with basic knowledge of solid state physics.

I give a very general answer: it depends on the solid and the dopant.

One of the main issues is best viewed from a thermodynamic point of view: you want to create something similar to a solution. The dopant should randomly occupy some positions in the solid, and this should be comparably stable or more stable than the pure phases themselves. The dopant can substitute an atom (then it is often chosen to be with similar size and chemical properties), but it can go originally unoccupied places, too.
Some examples:

You can introduce dopants with co-crystallization. You have a method to prepare the host solid through crystallization from its melt or from a solution, then you can mix the dopant during the synthesis, and if you are lucky it works. With molecular materials or simple inorganic salt, it can be a relatively easy process.
Chemical vapor deposition techniques are often used to create (doped) thin films.
Some dopants can be introduced by simple immersion. Hydrogen can diffuse inside of the host material, and act as a dopant in metals and other materials. The preparation can be just letting the sample stay in hydrogen gas for a longer period. Similarly, there are many oxides where oxygen acts as a dopant, and its level can be controlled by heat treatment in oxygen (or in a vacuum, to remove the extra oxygen). My understanding is that the boron doping in silica using diborane gas is also based on similar principles.


Why don't Schools follow latest IUPAC for naming groups? In high school I am taught that Group 7 were halogens and Group 8 were noble gases, and Group 3 the boron group, Group 4 was carbon group, etc. Then when I looked up the groups, I found that noble gases were group 18, halogens were group 17, Group 7 was the manganese group, ...

The school said that their numbering is based on counting valence electrons, and the transition metal columns have no business in their lessons.

So schools don't use IUPAC conventions.

How standard is their group name scheme?

The reason for this is that in 1988 (actually more than a few years back), IUPAC changed their naming system for the groups. Originally the transition metals didn't have a group number and weren't counted as a group. However now it is changed so that the transition metals do have a group number. So number instead of carbon being in group 4, it is in group 14. However despite this being changed quite a few years ago, you will still see in several websites and schools like your and mine still following the old IUPAC conventions and say that carbon is in group 4 rather than 14.

The reason that they do this I suppose is that it avoids confusion in students who are just picking up chemistry as they can see there is a clear relationship between valence electrons and group number. However besides this, schools do follow IUPAC conventions.

References:

Pure Appl. Chem. 1988, 60, 431–436, DOI: 10.1351/pac198860030431

(thanks @Loong)


Understanding a passage related to salt buffer Passage from my textbook:

Are reactions possible among any of the solution components; is so, what is their stoichiometry?
Suppose that you are asked to calculate $[\ce{OH-}]$ in a solution that is prepared to be $\pu{0.10 M}~\ce{NaOH}$  and $\pu{0.20 M}~\ce{NH4Cl}$. Before you answer that $[\ce{OH-}]=\pu{0.10M}$, consider whether a solution can be simultaneously $\pu{0.10M}$ in $\ce{OH-}$ and $\pu{0.20M}$ in $\ce{NH4+}$ . It cannot; any solution containing both $\ce{NH4+}$ and $\ce{OH-}$ must also contain $\ce{NH3}$. The $\ce{OH-}$ and $\ce{NH4+}$ react in a $1:1$ mole ratio until $\ce{OH-}$ is almost totally consumed:$$\ce{NH4+ +OH- ->NH3 + H2O}$$
You are now dealing with the buffer solution $\pu{0.10M}~\ce{NH3}-\pu{0.10M}~\ce{NH4+}$

I am having trouble understanding the third last sentence in the paragraph that starts with "2." The statement reads: "It cannot; any solution containing both $\ce{NH4+}$ and $\ce{OH-}$ must also contain $\ce{NH3}$. The $\ce{OH-}$ and $\ce{NH4+}$ react in $1:1$ mole ratio until $\ce{OH-}$ is almost totally consumed"
Can someone please help me out with understanding this passage?

There is no such thing as  $\ce{NH4OH}$ . They would react to form water and ammonia. Thus, to say that both exist in the solution at such high concentrations would be mistaken. They would react. Consider the reaction between ammonium and hydroxide.

$\ce{NH4+ + OH- -> NH3 + H2O}$

However, do note that at low concentrations, they can exist according to dissociation. However that is not what your passage is saying

*On your test, it may be good to assume the reaction goes to completion and then solve for the dissociation of ammonia. That is most likely what is expected of you.

*I thought of a more understandable analogy:

Think about $\ce{OH-}$ and $\ce{H+}$ in solution at 25 celsius. If one increases, the other decreases! According to ${Kw}$. Adding a lot of acid and base together would simply neutralize, they cannot exist at any higher than ${Kw}$


Why does neighbouring group participation increase the rate of reaction? I don't see any reason why the transition state of an SN2 reaction that happens after an internal nucleophilic group attacks(which is known as NGP) would be more stable. In fact, the nucleophilic group that attacks is often a strong nucleophile that is quite reactive (more than the leaving group which is usually a weak base) which would only destabilise the transition state when the external nucleophile does attack.

My question is: Why is the internal attack and the external attack combined faster than simply the external attack since the rate of such reactions where neighboring groups are involved is greater than normal SN2 reactions?

In neighbouring group participation (NGP), an internal nucleophile first displaces a leaving group in an intramolecular sense, forming a ring (usually 3 or 5 membered) which is then opened by the incoming nucleophile intermolecularly.

  General mechanism of neighbouring group participation (taken from March, Advanced Organic Chemistry, 2007

In the scheme above, Z first displaces X to form a 3 membered ring, which is then opened by Y. Your question then becomes why is Z faster than Y at substituting X?

Collision

In order for the substitution to take place, the nucleophile must come into alignment with the leaving group. In a general sense, intramolecular reactions are usually faster than intermolecular since the two reacting groups are tethered together – there is no requirement that they collide in order to react. This is especially evident for intramolecular formation of 3 and 5 reasons, since those ring sizes are especially favourable. (Note that intramolecular reactions isn't always favourable, for example in macrocyclisations of large rings, where high dilution is needed to close the ring rather than getting intermolecular reactivity).

Entropy

The reaction between the substrate and the external nucleophile Y in an intermolecular fashion is accompanied by a decrease in entropy (two molecules are becoming one in the transition state). Whilst this also happens in the intramolecular sense (the molecule is becoming more rigid), it is significantly less than in the bimolecular case.

Its also worth pointing out that its not always 100% clear whether a molecule is reacting via an NGP type pathway, and indeed there are many cases where its been shown that both are operating at the same time, with the intra being faster than the inter, but not sufficiently rapid to prevent any direct intermolecular reaction.


Which procedure(s) will allow a student to differentiate between solid sodium sulfate and solid sodium sulfite? I. Make solutions of each and look for a precipitate when added to 0.10 M Ba(NO3)2.
II. Add crystals of each to 0.10 M HCl and watch for bubbles.
III. Make solutions of each and test with a pH indicator.

The answer is II and III only. I understand how III works, but I don't see how sulfite and sulfate can be differentiated by adding HCl; don't both get reduced to SO2 bubbles?

Sodium sulfite and hydrochloric acid partially react to $\ce{SO2}$ which is gaseous and thus will form bubbles if not dissolved immediately.

When sodium sulfate is added to hydrochloric acid it will dissolve. The solution then contains sulphate and hydrogensulphate ions. No gaseous product will be formed.


Positive electrode of an electrochemical cell? I had the following problem:

Relevant information from Data Book:

  $\ce{Ag+ + e- <=> Ag}$

  $E^0 = +0.80$

  and

  $\ce{Fe^3+ + e^- <=> Fe^2+}$

  $E^0 = +0.77$

This means $E_\mathrm{cell}$ is +0.03. (right)

And (I thought) that the left ($\ce{Fe}$) half cell has the positive electrode/anode as the ions lose electrons to the anode. I thought that the half equation with the more positive $E^0$ value went forwards.

However this is wrong. Where did I go wrong in my reasoning?

As the standard reduction potential of Ag electrode is more than that of Fe electrode, Ag electrode will act as cathode (where reduction or gain of electrons takes place). The Ag electrode will be a positive electrode because it is electron deficient and thats why reduction is taking place at this electrode.

The Fe electrode will act as anode (where oxidation or loss of electrons takes place) and it will be the negative electrode because there is excess of electrons on this electrode.

Ecell = (Ereduction)cathode - (Ereduction)anode

      = +0.80 - (+0.77)

      = 0.03V


Oxidation of glucose By oxidation of 9 g of glucose it is released 150 kJ of heat. Calculate the standard enthalpy of derivation of glucose (in kJ/mol), if the given enthalpies of the reactants are ΔfH (H2O(l)) = -285,8 kJ/mol; ΔfH (CO2(g)) = -393,5 kJ/mol. Mr (glucose) = 180 (The answer is -1075.8)

$$\ce{6 CO2 + 6 H2O -> C6H12O6 + 6O2}$$

$$n(\ce{C6H12O6})=\frac{\pu{9g}}{\pu{180 g mol^-1}}=\pu{0.05 mol}~\ce{C6H12O6}$$

Now, $$\frac{\pu{0.05 mol}}{\pu{150 kJ}}=\frac{\pu{1 mol}}{x}\implies x=\pu{-3000 kJ/mol}$$

$$\Delta_\mathrm rH=\Delta_\mathrm fH(\ce{H2O(l)})+\Delta_\mathrm fH(\ce{CO2(g)})-\Delta_\mathrm fH(\ce{C6H12O6(s)})$$
$$= 6 \times (\pu{-285.8 kJ/mol}) + 6 \times  (\pu{-393.5 kJ/mol}) - (\pu{-3000 kJ/mol}) = \pu{-1075.8 kJ/mol}$$

Did I take the right steps?

I had some difficulties figuring out what the reaction was

The question mentions "oxidation of glucose", that's a very important hint that tells you that the reaction is going to be:

$$\ce{C6H12O6 + O2 -> ?}$$

Now, note that oxidation of any carbohydrate/hydrocarbon/organic compound only yields (in general) carbon dioxide and water.

Thus, your complete reaction should be:

$$\ce{C6H12O6 + 6O2 -> 6 CO2 + 6 H2O}$$

Notice that each of the products is just the "oxide" of the elements of the original compound $\ce{(C, H)}$


Carbocation rearrangement involving three membered rings Question:

Taking into account of various carbocations and, as well as the rules governing mechanisms of carbocation rearrangements, which reaction is most likely to occur during the given reaction?

My take on the question was this mechanism:

I'm torn between this mechanism and option (b). How do I judge between anchymeric assistance  and stabilization of the intermediate through 5 alpha hydrogens. How will the carbocation rearrange?
Source: MS Chouhan, Chapter: Alkenes, Q.17
This inadequate solution was given:

Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.

It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.

In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.

Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]


Atomic emission spectroscopy There is a sentence in my book I don't understand:

  Every atom or ion, which has one valence electron, will have double lines in the atomic emission spectrum. For example $\ce{Na}$ will produce lines at $589.0\:\mathrm{nm}$ and $589.6\:\mathrm{nm}$.

Why does this happen?

When an electron in any element absorbs energy it will be promoted to an excited state.  How much energy we supply determines how many \ which excited states will be created.  As the electron returns from these various excited states to the ground state light can be emitted, this is the basis of atomic emission spectroscopy.

(image source)

In the case of sodium (or any other atom with just one valence electron) many excited states are possible, so many emission lines are observed.  However, a very likely transition will be the one from the s orbital to the next highest energy orbital, the p orbital.

The magnetic energy of an electron spin is split into energy states with different total angular momentum due to the electron spin's interaction with the internal magnetic field set up by the orbital motion of the electron, this is termed spin-orbit coupling. In the case of a p orbital, it is split into two states (the total angular momentum j is given as j = l +/- s, where l is the angular momentum quantum number [l=1 for a p orbital] and s is the electron spin (1/2), therefore j has two values or states for an electron in a p orbital, 1/2 and 3/2). It is the light emission that occurs when the excited electron returns from these two p orbital energy states to the s orbital ground state that produces the doublet seen for all atoms with a ground state involving one electron in an s orbital.

(image source)


Rule to determine the total number of elements in a period If each orbital can take maximum of 3 electrons, then total number of elements in 2nd and 4th periods respectively are = ?

I am not aware of the rule which helps to determine the total number of elements in a period.

The total number of elements formed due to their last electron entering the $n$-th shell will be equal to the total number of electrons that shell can fill. For each electron added or removed (along with a proton), you'll get a new element, since a unique element is itself defined by a unique count of protons.

Thus, this problem is reduced to finding the total number of electrons in $n$-th shell. And, simply enough, the total number of electrons in $n$-th shell = number of orbitals in $n$-th shell $\times$ number of electrons each orbital can hold.

For example, for $n=3$, we have $\ell=0,1,2$. Thus, the number of orbitals here is $=1+3+5=9$ (recall that s-subshell has 1 orbital, p-orbital has 3 orbitals, and so on[1]). Since $s=\pm 1/2$, each orbital can fit two electrons, thus total count of electrons is $=2\times9=18$, which is also the total number of elements in the third shell.

In your question, the only change is that each orbital can now hold three electrons (by some phenomenon we don't know yet!) So, in the final step, just multiply the total number of orbitals by three instead!

To calculate the number of elements in a particular period, you need to note the individual subshells that belong to that period, and then sum up the number of the elements from those subshells.

In my solution above, I have shown how to get the total number of elements formed due to their last electron entering the $n$-th shell, which is not equivalent to the number of elements in $n$-th period.

For example, in the fourth period of the periodic table, you have the $\ce{4s, 3d}$, and the $\ce{4p}$ subshells. This is the calculation that your solution manual has shown in the second row.

[1]: In general, the total number of orbitals $k$ in shell $n$ is:

$$k = \sum_{\ell=0}^{n-1}(2\ell+1)$$

since: a) each $m$ value stands for a unique orbital b) $-\ell\leq m\leq\ell$, thus $m$ has a total of $2\ell+1$ values for a particular $\ell$ value


How do I reduce the smell of store-bought liquid soap? I recently bought a brand of liquid soap I've never used before and it turns out they formulated it with a bit too much of the scent ingredients, to the point where I find it gets uncomfortably close to nauseating during use, and also it persists too long on my hands afterwards, compared to what I find normal.

How do I neutralize some of this scent, what can I try to mix in? All I can think of are baking soda or vinegar (but then vinegar also persists a lot and is unpleasant, plus it's acidic and I'm going to mix it with a probably-alkaline soap, and I don't want any crazy reactions and byproducts if possible).

If the scent reaches out and touches you that much, it is likely volatile. Pour it in a pot, add a volume of water, heat on a hot plate outside your dwelling till the smell goes away. Evaporate to volume and check ingredients. You may want to add back a small amount of a desirable ingredient that boiled off. Raise the temperature slowly. A vigorous boil will make a real mess.


Clarification of IUPAC name Is the IUPAC name of $\ce{CH3CH=CH-C#CH}$ pent-3-en-1-yne or pent-2-en-4-yne?

Subsection P-44.4.1.10.1 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads

  For the endings 'ene' and 'yne' lower locants are assigned first to the endings as a set without regard to type and then to 'ene' endings.

Since the locant set "1,3" is lower than the locant set "2,4" (see here for a discussion of what that means), the appropriate choice of locants for the above compound leads to the name pent-3-en-1-yne.

As an aside, the configuration of the double bond, if unspecified, should be labelled with the symbol Ξ (P-91.2.1.2.1).

  When the configuration is not known or must remain unspecified for lack of configurational homogeneity, the italicized symbol 'ξ' or 'Ξ' is used, preceded by the required locant. The symbol 'ξ' (small Greek letter 'xi') replaces noncapitalized CIP stereodescriptors such as 'r', 's', 'm', 'p'. The symbol 'Ξ' (capital Greek letter 'xi') replaces capitalized CIP stereodescriptors such as 'R', 'S', 'M', 'P', and nonCIP stereodescriptors 'E', 'Z', 'seqCis', and 'seqTrans'.

The preferred IUPAC name (PIN) for the compound with unspecified double bond geometry is therefore (3Ξ)-pent-3-en-1-yne.

Of course, if the configuration is specified, then the (E)/(Z) labels should be used instead.


Why is an electropositive metal like Zinc capable of plating out? I know that electronegative metals like copper or silver can electroplate out of an aqueous solution in electrolysis. However, I've also noticed $\ce{ZnSO4_(aq)}$ can electroplate out when electrolyzed, yet Zinc is more electropositive than hydrogen. This doesn't happen with other electropositive ionic solutions like $\ce{Al(C2H3O2)3_(aq)}$.

Why does this happen, and why is it limited only to Zinc?

It's too simple to think of this situation in terms of electronegativity and it's certainly not limited to zinc. If you look at a table of standard electrode potentials, you can see that the reduction of many metal ions like $\ce{Zn^2+,Pb^2+,Cd^2+,}\textrm{ etc.}$ have standard potentials below that of the reduction of hydronium (0 V) and should be much less favourable than making hydrogen gas if this were the only consideration.

However, in real electrochemical cells, it generally takes a certain amount of extra voltage to get a certain reaction to occur, known as the overpotential. Like all chemical reactions, redox reactions have a certain activation energy separate from the energy change between the reactants and products, e.g. elemental gold is far lower in energy than a gold solution, but a gold solution will not immediately reduce spontaneously because there is an activation barrier to the process, an extra potential must be applied to make it happen at a reasonable rate. Similarly, though the standard potential for reducing hydronium is 0 V, in practice it may not be observed until much more negative applied voltages.

This overpotential for hydrogen evolution is strongly dependant on the type of electrode used. For the reaction to occur, hydronium ion must adsorb to the surface. In electrodes like platinum, this interaction is stabilized so the overpotential is relatively small and it's thus quite difficult to deposit metals like zinc onto platinum. For this reason, other electrodes must be used for working with aqueous solutions at very negative potentials. Glassy carbon, mercury, and bismuth electrodes are often used for this as hydronium is not readily adsorbed to their surfaces, giving large overpotentials for hydrogen evolution. These types of electrodes may be useful down to -1 V or so and more exotic materials like boron-doped diamond electrodes, even more negative.

However, no electrode can prevent hydrogen evolution if the potential is negative enough. Zinc is about the most easily oxidized metal that is practical to electroplate from aqueous solution. Some metals are so easily oxidized that they are basically impossible to plate in aqueous solutions which is why one generally cannot electroplate from aqueous aluminum solutions or solutions of alkali metals (things like sodium and potassium are so easily oxidized that they react violently with water to form hydrogen gas).


How is Electrode Potential Developed on Inert Electrode? How does redox couple develop its potential if electrode isn't the part of the redox system like platinum in SHE or Cl2/Cl- in water solution for example? Usually it is discussed how electrode develops its potential when it isn't inert like Cu/Cu2+ system when copper electrode itself is important for developing potential. What about inert electrodes?

So, let's take a solution of copper sulfate in a beaker and insert an active electrode: a copper wire. OK so far? We connect this wire to a voltmeter.
Then lets connect another wire (we are quite well off, even rich, so we will use a shiny platinum wire) to the other connection of the voltmeter. Platinum is nice, inert, always shiny. How shall we connect to the solution? Just stick the inert wire into the solution! The wire is inert chemically in that it does not replace copper from copper sulfate or hydrogen from water, but it is not an insulator - it does do something: it carries electrons when they are pushed or dragged.
If we had inserted an inert electrode (Pt) into the solution first so we had two inert Pt electrodes, well, nothing would happen. Nothing that we could measure. Those little ions and molecules are hopping all over the place, but to no net effect.
So the copper atoms near the solution can donate electrons which then float thru the voltmeter to the platinum which can accept those electrons, handing them off to water, forming OH$^-$ and H$^.$ (two of which will form H$_2$). This occurs more easily than the other way around. (It's even a bit more complicated: copper doesn't do this so easily, but if O$_2$ is available, it will consume the hydrogen atoms produced at the Pt wire and drive the equilibrium toward copper dissolution. I just didn't want to start off the discussion with an active metal like zinc, and zinc wires, etc., because it's more difficult to paint a picture with unfamiliar objects. I probably could have used aluminum wires...)
The philosophical question then is: Is the inert electrode really inert? At the copper electrode, a copper ion just plops down on the metal (yeah, some activity), while at the platinum electrode, electrons come in, go out into H$_2$O, split it, form OH$^-$ and H$^.$, help that form H$_2$, then bubbles - it's a crazy mess compared to the copper electrode.


Why is Trouton's rule not valid for liquids with boiling points less than 150 K? This is a question from the last year's semester examination.

  Why is Trouton's rule not valid for liquids with boiling  points less
  than 150 K?

I searched the internet but only found that the exceptions mentioned include liquids with strong intermolecular attraction like hydrogen bonding and some similar cases. Can anyone give me a proper answer?

When a liquid vaporises its entropy goes from a modest value to a significantly larger one.  The ratio of $\Delta H_{vap}/T = \Delta S_{vap}$ is found to be approximately constant at the boiling point. Some authors give a value of $80~ \pu{Jmol^{-1}K^{-1}}$ others $90~ \pu{Jmol^{-1}K^{-1}}$.

Experimental values vary rather more than this and for gases such as neon, nitrogen, oxygen and methane whose liquids all boil below $150$ K, have values that are in the range $65 - 75$, benzene, many 'normal' liquids and liquid sodium, lithium and iodine, in the range $80-90$ and ethanol, water, hydrogen fluoride in the range $105 - 115 ~ \pu{Jmol^{-1}K^{-1}}$. Thus there seems to be nothing unusual about $150$ K but rather an influence from  intermolecular interactions.

The value of $ \approx 80~ \pu{Jmol^{-1}K^{-1}}$ corresponds to a cohesive energy of $\approx 9.5 kT$ per molecule and so the boiling point gives an indication of the strength of the cohesive energy holding molecules together in the condensed phase. When the cohesive energy exceeds this value, as in water, then the ratio $\Delta H/T$ is larger and conversely the ratio is smaller when the cohesive energy is less as in Neon or methane.

The $\approx 9.5 kT$ minimum energy per molecule is quite a modest energy; if a molecule has $6$ near neighbours this corresponds to about $3kT/2$ per interaction between two molecules, roughly the average thermal energy.


Scope of Haloform reaction We know that methyl ketones and secondary alcohols oxidizable to methyl ketones, such as isopropanol. The only primary alcohol and aldehyde to undergo this reaction are ethanol and acetaldehyde, respectively. 1,3-Diketones such as acetylacetone also give the haloform reaction.
Do Terminal Alkynes give Haloform Reaction if Yes then how the mechanism will follow?

Terminal alkynes, including phenyl acetylene, react with iodine (in MeOH) to give 1,2-diiodoalkenes according to this JOC paper  here. Presence or absence of base will not alter the outcome. Presence of AgNO3 gives some di-iodo ketone which can then give rise to the iodoform reaction.


What stable arrangements of hydrogen-bonded water exist in the range of 4-20 water molecules? Water clusters made up of various numbers of water molecules have been described.
Are all of these structures theoretical?
What experimental evidence supports any of these structures?

How does the free energy of a particular water cluster (per number of molecules in the cluster) vary from one thermodynamically stable arrangement to another? Is there a general trend, e.g. larger clusters are more stable?

In response to Martin's initial comment:

I am curious about water clusters in the context of receptor-ligand interactions (for example, in protein receptors, enzymes or synthetic host cavities).

In cases where the cavities of these macromolecules is small (for example, a small enzyme active site) what kinds of water clusters are likely to fill this void in the absence of ligand?

Put more broadly, does the structure of the water that is filling the cavity destabilize the receptor in such a way that ligand binding is enhanced?

Considering a hypothetical receptor that is known to bind ligands (whose sizes range between 4 and 20 non-hydrogen atoms) in water, what are likely structures for the encapsulated water cluster that will be displaced by the ligands upon binding?

The Cambridge Cluster Database can give you a start with theoretical binding energies, but only one isomer for each size is shown.

http://www-wales.ch.cam.ac.uk/~wales/CCD/TIP4P-water.html


Chair conformations of cyclohexane with two substituents I have been studying the chair conformations of cyclohexane but I get confused when it comes to two substituents.

I came across an example, in which the one substituent is the ethyl group (Et) and the other one is methyl (Me).  The ethyl is drawn like a solid wedge, whereas the methyl is drawn like a dashed wedge. As far as I  know, I have to draw the chair and then the equatorial positions and the axial ones. I thought that since the bigger substituent is the ethyl group, that it would be better in a equatorial position and the methyl would be better in a axial position. But the example, draws the ethyl group in an axial position and the methyl in an equatorial position

Why is that?I really don't understand.

Thank you.

Here are some structures to help you. You noted that one group was a solid wedge and the other a dashed wedge without mentioning the substitution pattern. I presume that this was a flat, 2-D structure. You also note that in the chair structure, one group is axial, the other equatorial. Only the 1,3-isomer 2 fits these conditions. The conformation 2a is more stable than 2b.
ADDENDUM: The 1,1-disubstituted isomer would also fit the conditions of the question. The questioner did not designate the structure.


Why does a six-membered ring contract in this SN1 reaction? I was watching this video on Khan Academy (YouTube link) regarding SN1 mechanisms and carbocation rearrangements and came across this mechanism.

At around timestamp 4:20 in the video, why does step 2 take place where the ring contracts to 5 carbons? Surely it would be far easier to simply have a methyl shift from the carbon in blue which would also result in a more stable tertiary carbocation.

A cursory look through several organic chemistry texts (Clayden, Carey, March, Vollhardt) as well as SciFinder doesn't reveal any instances of the same reaction used in the Khan Academy video - this would imply that either the reaction is made up to illustrate a point (albeit not very well) or comes from somewhere more obscure (historical literature in a language other than English).

SN1 vs SN2

The first question that it would be useful to answer is whether the reaction proceeds via an SN1 or SN2 pathway, or, to phrase this a more accurate way, to what extent is the halide dissociated at the time when the rearrangement occurs.

In general, secondary halides can undergo both SN1 or SN2 reactions depending on the conditions used (substitutions at secondary centres are always hard to predict, and often go with neither complete inversion nor complete racemisation). In March's Advanced Organic Chemistry,  the chapter on cationic  rearrangements summarises:

  The SN1-type process occurs mostly when B is a tertiary atom or has one aryl group and at least one other alkyl or aryl group. In other cases, the SN2-type process is more likely.

Stereoelectronics

If we accept that the reaction goes via an SN2-like pathway, we must then acknowledge the stereoelectronic requirement that the migrating group must be able to interact with the C-Br σ* orbital.

This requirement immediately rules out the migration of the exocyclic methyl group, leaving only the internal C-C bond to migrate (assuming that bromine adopts the equatorial position and the ring sits on a chair-like conformation - both of which are reasonable assumptions).

Migrating group ability

Even ignoring the steroeelectronic issues, migratory aptitude would favour rearrangement of the internal C-C bond as it is marginally more able to stabilise the developing positive charge during the rearrangement.

tl;dr

In the absence of the original literature, it's hard to say with certainty which group migrates (though it would be an easy experiment to carry out....). That said, following the general trends for rearrangements/migrations found in undergraduate organic chemistry texts, it would seem most likely that the rearrangement takes place as shown rather than the competing 1,2-methyl shift (we're also ignoring the possibility of a proton elimination to give an alkene).


Nature of spontaneous reactions The definition of spontaneous process states that "A spontaneous process is an irreversible process and maybe reversed by some external agency"

So, can we say that all spontaneous reactions are irreversible reactions under the absence of some external force?

A spontaneous reaction, as contrasted with a non-spontaneous reaction, is a reaction quantified by a negative free energy change. A reaction is said to be spontaneous if it occurs without being driven by an outside force. So, we can say:  In any spontaenous process, the path between the reactants and products is irreversible


How do you calculate the absorption spectrum of water? If you look online you can find various graphs of the absorption spectrum of water. The graph below for examples comes from this blog post.

As far as I can tell these graphs come from empirical measurements. Is there a theoretical model that can derive the same or similar results?  Of course if it was easy to implement in software that would be ideal.

The easy answer is no. It is possible to calculate absorption spectra from molecular structures using quantum mechanical calculation programs (for instance DFT using Gaussian) starting from the Schrödinger equation. These programs only work for a relatively small
number of electrons though and therefore usually assume gas phase including a small correction for the surroundings. Liquid phase water, where the molecules are very strongly coupled will be extremely difficult to calculate since it is almost impossible to create a "box" of relevant molecules. I don't have SciFinder or something equivalent, but a search on Google only gave articles calculating X-ray spectra.


Why Bond Angle is different? H2O has two lone pair electron and H2S also have two lone pair electron but Bond Angle in H2O is 104.5° but H2S has 92.12°bond Angle .

Both H2O and H2S has same hybridization ,which is sp3 and i know that the four hybrid orbitals of sp3 hybridization are directed towards the four corner of a regular tetrahedron making Angle of 109.5° .

My teacher taught me that for every lone pair you have to subtract 2.5° from 109.5° so if that's the case then both of them should have same Angle as both of them have 2 lone pair electron ?
If that's not the case then what is the exact way to determine Bond Angle ? And the way my teacher taught me is it ok ?

In H20 it is sp3 hybridisation, but in H2S it is purely p3 orbital , there is no hybridization in H2S, it cannot afford the cost of hybridization you may recall that all the p orbitals I.e. Px,Py, Pz are 90 degree to each other as along x y and z axis respectively and if u compare this with the Bond angle of H2S which is 92.12 degree( almost 90 degree) which clearly shows that it is formed by p orbitals only, hybridization is a costly process , when sulfur can easily bond with p orbitals itself , why will it go for expensive process of hybridization unnecessarily isn't it?


Black Sludge Appearing When Making Iron Oxide Through Electrolysis I am trying to make iron-oxide through electroylsis to make thermite. I'm using nails for the iron on the positive electrode in salt water. While there definitely is a good amount of rust collecting, especially at the surface, after dissolve multiple nails, there are also these black sludge-like solids forming:

I can pick them out of the solution if i'm careful, but they are very brittle
and break up and dissolve almost immediately if I  use too much force. Additionally, if I stir the up or shake the container, more of this stuff comes up from the bottom, most of it dissolved in a black-green watery, solution and you can't even see the rust anymore. If you give it time to settle, though you can see the rust again.

My question is what is this stuff, and will it make the rust impure? Do I need to get rid of it to make the thermite? Will boiling the solution to get rid of the water like I had planned be enough?

Thanks in advance.

The black stuff is your iron oxides. FeO and Fe3O4 are both black, Fe2O3 is red. I would guess you are making some complex iron oxide/ hydroxide/ hydrate. I have seen XRD analyses of many aqueous iron corrosion products , they are typically a complex mix.


what would be p-v diagram for irreversible isothermal expansion We say that area under the curve gives the work done. So for an irreversible iso thermal expansion, we have work done= -P external( V final- V initial) signifying a rectangular graph. But we know it's isothermal so it would be along an isotherm? Or no. I believe if we have external pressure at the y-axis, graph would be rectangular and if we take pressure of gas at y-axis we would get a curve, same as isothermal reversible but would end at a lower volume than the former. So what do we take for the p-v diagram.

If you have constant external pressure $P_{ext}$, the work on the PV diagram will be minus the area of a rectangle that is $P_{ext}$ high and $V_{final} - V_{initial}$ long.

When you want to show work as an area, you must plot external pressure on the y-axis of a PV diagram, not gas pressure. In the special case of a reversible expansion or compression, the gas pressure $P$ is equal to the external pressure $P_{ext}$ at every point during the process, so you are still plotting external pressure if you're relating work to the area under the curve.

Why does the external pressure matter in the calculation of expansion work in irreversible processes, and not the gas pressure?

The reasoning goes like this:

expansion work = opposing force x distance moved

opposing force = external pressure x area

volume change = area x distance moved

so

expansion work  = -(external pressure) x (volume change)

The minus sign comes from our decision to make work negative when energy leaves the system in an expansion, and positive when energy enters the system from the surroundings in a compression. This is the convention used in chemistry; you'll find the opposite convention used in physics!