What is the maximum extent of a hydride shift to form the most stable carbocation? What is the maximum extent of a hydride shift to form the most stable carbocation?

If we consider the reaction between 2-methylpropan-1-ol and $\ce{HBr}$, then of course, the carbocation formed after a hydride shift will be more stable (and lead to the major product). 3-methylbutan-1-ol will also undergo hydride shift to some extent to form the more stable carbocation, but is there a limit to it?

Like can the number of carbons be so much that there is almost no chance to make the product with hydride shift possible? Like consider the scenario below:

What can be the maximum $n$ possible here to make the product with hydride shift possible?

Advances in Physical Organic Chemistry, volume 19, discusses the relative transition state barriers for 1,2-, 1,3-,1-4-, and 1,5-hydride shifts.

The specific cations considered are 2,3 dimethyl-2-butyl, 2,4 dimethyl-2-pentyl, 2,5 dimethyl-2-hexyl and 2,5 dimethyl-2-heptyl.

Interestingly, they report that whereas the barrier for 1,4 shift in 2,5 dimethyl-2-hexyl is 12 kcal/mole, this decreases dramatically to less than 6 kcal/mole for 1,5 shift in 2,6 dimethyl-2-heptyl

The reference draws the transition states as rings, with the H being transferred as a bridging member of the ring, a five membered ring for the 1,4 shift and a six member ring for the 1,5 shift.

The reference also makes a comparison to intermolecular hydride shift between isobutane and t-butyl cation, which has a barrier of 3.6 kcal/mole.

While the reference does not discuss the possibility of longer distance than 1,5 shifts, it seems that because the ends of a long ion can meet each other, long distance (in terms of number of bonds) shifts are possible because through space the distance is short.


What chemical reactions could ancient people use to propel projectiles? Are there reactions other than burning of gunpowder that ancient people could perform that would produce gas from non gaseous substances, so that it could have been used for propulsion of projectiles had they thought of it? An example would be CaO + H2O reaction, as upon mixing in a closed container this mixture would pressurize the container, and at some pressure the cord tying the projectile to the chamber would rip and projectile would be propelled

The answer depends on what do you mean for ancient people. There are a lot of reactions that could produce gas for ejecting projectiles. The most common was the mixture of sulphur, potassium nitrate $\ce{KNO3}$ and carbon that is known as gun powder and has been reported to be used since the IX century in China.

If you want some less efficient and not used reaction,  you could think of using methane explosion extracted from some swamp. Or some reaction between calcium carbonate and acetic acid. Like this one between sodium hydrogen carbonate:

$$\ce{NaHCO3 + CH3COOH -> NaCH3COO  + CO2 + H2O}$$


can metamers contain N as differing functional group? Are $\ce{Ph - NH - CH3}$ and $\ce{Ph - CH2 - NH3}$ metamers?

This term is so old, I'd never even heard of it, had to look it up. Hooray for wikipedia

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

. As i understand it, this applies to the two compounds you give.

The proper modern term is "isomer", a substance having the same molecular formula as a second one, but in different structural arrangement. (And yes, the two are isomers, although you would only rarely come in a situation where you want to discuss them as such, because they are chemically very different.)


How to synthesize 1,3-pentadiene from butanal? How do you synthesize 1,3-pentadiene from butanal?

I have the (possible) last steps of bromination of 1-pentene at the allylic position and then an E1 reaction to give the final product. I don't understand how to get from butanal to 1-pentene though. Also if you know another way to do the synthesis then feel free to share.

I believe one way would be to start with butanal participating in a Wittig reaction to form 1-pentene.

$\ce{CH3CH2CH2CHO + H2C=PPh3 -> CH3CH2CH2CH=CH2}$

Then allylic bromination, as you said.

$\ce{CH3CH2CH2CH=CH2 ->[{NBS}][{hv}] CH3CH2CHBrCH=CH2}$

Dehydrohalogenation with a hindered base like t-butoxide (to favor the elimination reaction) should give the diene. The only thing I'm not sure about is whether it would be E1 or E2. It isn't very hindered, but it is a secondary halide. With those factors, plus the alkoxide, I think it will be an E2 mechanism.

Let me know if I made a mistake somewhere.


Does a zinc metal coating react with coffee in a coffee grinder? I'm restoring an old coffee grinder and I want to give the grinder mechanism a nice zinc / silver coating.

But I read it is not safe for zinc to be in contact with acidic foods.

When I look at the pH, coffee's is 4.6 – 4.3 which is quite acidic:
https://angelscup.com/blog/taste/coffee-acidity-flavor-ph-acid-reflux-and-low-acid-coffee/

Not sure if the zinc and coffee are likely to react in such a scenario. Should I maybe choose a silver plating?

Before cleaning with citric acid.

After cleaning with citric acid, Al foil and baking soda

If you plate the pieces with zinc or silver, the main source of zinc/silver in the ground coffee will be from abrasion between the parts and the coffee beans and not from chemical attack.

A few milligrams of zinc or silver (or iron for that matter) won't to do any harm to your health, specially if diluted in many coffee cups along several months or years.

In fact, Zinc was used therapeutically in dosages upwards of hundreds of milligrams per day for months, or more, with no permanent health issues.

Silver, on the other hand, has much lower acceptable exposure limits and while not particularly toxic, I'd not put it in parts of the device that are guaranteed to put metallic silver into the ground beans.


Naturally Occurring Vegetable Oil? Is it possible for vegetable oil to occur naturally, without being extracted? Along the same lines, it it possible for Hexane to occur naturally?

Vegetable oil is extracted from plant. According to Wikipedia, the term "vegetable oil" can be narrowly defined as referring only to plant oils that are liquid at room temperature,

So, answer is no.

If it occurs naturally than we will not call it vegetable oil.


Good software to perform simulations on polymer (absorption spectra)? Sorry guys,I'm totally new in this field,for personal interest I'd like to start some calculations on conducting polymers ,could someone suggest a good software to perform simulations on polymer (absorption spectra)?

It depends very significantly on what you want to accomplish.

I'm not going to suggest software, since there are many programs that can (and are) used to predict optical absorption spectra of conjugated polymers.

First, you can either approximate the properties of the polymer by running several calculations on oligomers (e.g., dimer, trimer, tetramer, etc.) and extrapolating the polymer band gap. Because conjugated polymers are similar to the particle-in-a-box model, you can roughly plot the HOMO-LUMO gap or excitation energy vs. 1/(# of repeat units) and get an approximately linear fit.

If you want more information, you could instead run a calculation on a polymer in a periodic unit cell, which will give you the electronic structure of the infinite system (but no information on disorder).

As suggested in the comment above, you can use multiple programs and different computational approaches depending on:

What kind of polymer
What kind of information you want
What computational resources you have
How accurately you want the information

The two most common methods for predicting optical absorption energies are the ZINDO/S semiempirical method, and TDDFT. The latter is more accurate particularly for predicting absorption intensity from the oscillator strength of the excitation energies.


Does unsubstituted benzoic acid not show resonance due to steric effects? My book says that benzoic acid does not show resonance as the carboxylate anion and the benzene ring are not in the same plane due to steric effects.

But there aren't any large groups in the ortho positions to cause steric inhibition of resonance. Then, is the info given in the book wrong? If not, where does the steric effect come from?

I have never heard of this effect occurring with an ordinary, unsubstituted benzoic acid. There is, however, a phenomenon known as the ortho effect that occurs when bulky groups occupy the ortho positions on the ring. In that case, the steric hindrance between the bulky groups and the carboxylic acid moiety forces the carboxylic acid group to rotate out of the plane of the aromatic ring, thereby breaking the conjugation.

One of the consequences of this, which may not be readily apparent, is that this often produces a substantial increase in the acidity of the carboxylic acid, even when the substituents are not electron-withdrawing (viz. 2,6-dimethylbenzoic acid, which is disubstituted at the two ortho positions, having a $\mathrm{p}K_\mathrm{a}$ roughly one unit lower than ordinary benzoic acid, while similar para- and meta-substituted benzoic acids have higher $\mathrm{p}K_\mathrm{a}$ values — that is, are less acidic).

While I've seen references to this effect in multiple sources, I've never actually found a thorough explanation for it. I think it could be argued that the benzene ring is electron-donating within the $\pi$-system (i.e., by resonance), which would only serve to destabilize the carboxylate conjugate base already bearing a negative charge. By rotating out of the plane, the electron-donating effect is removed. Rather than consider the stability of the resulting carboxylate, I suppose one could also consider the stability of the carboxylic acid directly. Here, I'd expect that the reactivity of the carboxylic acid flanked by bulky substituents should be greater, as the loss of $\pi$-conjugation with the carboxylic acid moiety is inherently destabilizing. My reasoning may be flawed, though, so if anybody has some corrections or a better explanation to offer, that would be entirely welcome.

Whether this effect occurs with ordinary benzoic acid or not, I can't be certain. At least one good reason I can think of to doubt that it does is the observed regioselectivity of electrophilic aromatic substitution reactions on benzene rings conjugated to carbonyl groups. They are known to be deactivating (i.e., they withdraw electron density) via their conjugation with the $\pi$ system of the aromatic ring. If they were not coplanar with the ring, the conjugation would be impossible, and the deactivating effect would not be observed.


Molecule with a quadrupole moment in an electric field How does an uncharged non-polar molecule that has a quadrupole moment (such as carbon dioxide) behave in an electric field? I know that in a homogeneous electric field, ions travel while dipoles orient along the field (rotate) and non-polar molecules are not affected.

What kind of electrical field, if any, would exert a force on a molecule with a quadrupole moment?

To determine the force on an arbitrary multipole moment, we first expand the E-field in a Taylor-series around the point $\vec{r}=0$:

$$\vec{E}(\vec{r})=\vec{E}_0+\vec{r}\cdot(\nabla\vec{E})_0+\frac{1}{2}\vec{r}\vec{r}:(\nabla\nabla\vec{E})_0+...$$

where the number of dots in the product denotes how many indices to contract over. We can then determine the force using this expansion, since it is just the charge multiplied by the E-field:

$$\vec{F}=q\vec{E}_0+\vec{\mu}\cdot(\nabla\vec{E})_0+\frac{1}{2}\mathbf{\Theta}:(\nabla\nabla\vec{E})_0+...$$
Typically, people use the traceless quadrupole, which would add an additional factor of $\frac{1}{3}$ in the third term, but I'm going to work with the basic definition throughout. So this shows that a quadrupole subject to a large electric field double gradient (not sure of the correct terminology for this) will experience a force even if it is charge neutral and has no net dipole.

For completeness, we can also write the energy since the force is just the gradient of the energy:

$$W=q\phi-\mu\cdot\vec{E}-\frac{1}{2}\mathbf{\Theta}:(\nabla\vec{E})+...$$
where we see, as TryHard said, that the E-field gradient interacts with the quadrupole.

As an aside, one can also obtain the torque (cross product of force and r) on an arbitrary multipole:
$$\vec{T}=\vec{\mu}\times\vec{E}_0+\frac{1}{2}\mathbf{\Theta}\dot{\times}(\nabla\vec{E})_0+...$$

(I'm using the nonstandard notation $A\dot{\times}B$ to mean $\sum_{ijkl}\epsilon_{jkl}e_iA_{jl}B_{lk}$, where $\epsilon_{jkl}$ is the Levi-Civita operator and the $e_i$ are a set of orthonormal unit vectors). This suggests that a quadrupole will experience a torque when interacting with a field gradient.

For more information on this, I would read chapter 3 of Jeanne McHale's Molecular Spectroscopy. There are some typos and notational quirks(e.g. in the equation for torque, McHale just writes a cross product for the 2nd term, which isn't defined between matrices), but its overall a good book to give background in classical electrostatics and a quantum mechanical description of spectroscopy.


What is the role of acid in keto enol tautomerism I have heard about an acid acting as a catalyst in keto enol tatutomerism.I have yet to understand the mechanism of the reaction though.pictorial depictions are welcome.

In acid-catalysed keto-enol tautomerism, a lone pair of electrons is removed from the oxygen of carbonyl group by the acid.

If enol is the starting compound, markovnikov's addition of $\ce{H+}$ occurs.


Interpreting notation format 1.64E-02 from particulate emission data I'm doing a research project for school on air pollution and part of it is studying pollution caused by aviation and I have found a table of data for the emission indices of various pollutants but I'm not 100% certain, if I'm correctly reading the numbers.

The unit is $\mathrm{g\, kg^{-1}~of~fuel}$ and the numbers are like 1.64E-02 or 2.25E+00. I'm interpreting it as E being for Exa, so in short scale 1.64E would be 1.64 quintillion and that it means 1.64 quintillion particles per g per kg of fuel. Is that the correct meaning? I have no idea what the second bit of the numbers means so please could you explain for me?

The format refers to a quite common scientific notation, where the $\mathrm{E}$ refers to exponent of $10$. It is often used when superscripted values become tedious to read and or write. It is widely known in computational sciences and was introduced for an easier display of huge or tiny numbers.

You can interpret the $\mathrm{E}$ as "times ten raised to the power of", hence the format can be interpreted like this: $$n\mathrm{E}m = n\times10^{m}$$

Therefore $1.64\mathrm{E}-02 = 1.64\times10^{-02}= 0.0164$ or $2.25\mathrm{E}+00=2.25\times10^0=2.25$.


What easy to conduct (at home) reactions could yield solid carbon? I am trying to isolate pure carbon in its elemental form. Are there any decomposition reactions or other easy to carry out reactions to yield solid carbon? and in the case that it is isolated, does it start reacting with other elements in the air?

A classic reaction to yield carbon is the dehydration of sucrose (table sugar) with concentrated sulfuric acid: Put some sugar in a beaker, carefully add sulfuric acid and watch a voluminous "snake" of porous carbon (blow up due to the steam released during the reaction) from the beaker.

You'll find lots of videos on this on youtube or other platforms.

Note that this is performed with concentrated sulfuric acid! If you decide that you must do this at home, protect yourself (googles, gloves, lab coat) and keep the sulfuric acid locked away.

But there's a more simple way to yield carbon:

Hold a spoon (with a long handle) into the flame of a candle. The black stuff that deposits on the spoon is carbon. Play with the distance for optimal results.

As far as the reactivity of carbon is concerned: There's no danger in that! Millions of people stockpile carbon in the form of coal at home in their cellars or sheds for heating.

And actually, this points to the easiest and most safe option: Don't perform any reaction at home and grind some charcoal or anthracite coal instead.


Why HBr has stronger attraction between molecules than CH3NH2? Which of the following compounds exhibits dipole-dipole forces as its strongest attraction between molecules?
a) $\ce{CO2}$
b) $\ce{CH3NH2}$
c) $\ce{Kr}$
d) $\ce{H2}$
e) $\ce{HBr}$

From what I understand, $\ce{CH3NH2}$ has an H-bond and it's the strongest form of dipole-dipole. So I thought that would be the answer.
I also knew that $\ce{HBr}$ shows the greatest electronegativity difference between $\ce{H}$ and $\ce{Br}$ atom, it would be a polar covalent bond (?!).
In that case, $\ce{HBr}$ may have stronger attraction between molecules than $\ce{CH3NH2},$ but it is not dipole-dipole. Isn't that correct? Then shouldn't the answer be $\ce{CH3NH2}$ instead of $\ce{HBr}?$
I must have some concept issues here.

EDIT (after title edited): $\ce{HBr}$ doesn't have stronger interaction than $\ce{CH2NH2}$, but it has dipole-dipole interaction as the strongest forces between it's molecules, which is obviously weaker than H-bonding.

Hydrogen bonding is the strongest intermolecular attraction. It is a type of dipole-dipole interaction1, but it is specific to Hydrogen.
In general, dipole-dipole interactions are considered weaker than H-bonding. The relative strength of forces are as follows,

Hydrogen bonding >  Van der Waals dipole-dipole interactions > Van der Waals dispersion forces

As per your question, (a), (c) and (d) have dispersion forces (as they're non-polar), while (b) has H-bonding as it's strongest interaction (not dipole-dipole).
Finally, as only (e) has dipole-dipole as it's strongest attraction between molecules, hence it's the answer.

Notes:
1: Dipoles are created due to difference in electronegativity. Below image shows that there is sufficient electronegativity difference to create a dipole. H-bonding and dipole-dipole interactions have same origins.

(Source: Master Organic Chemistry)


Sulphonation of benzene Will nitrobenzene and benzene have the same rate towards sulphonation reaction, since the formation of sigma complex is not the RDS of this reaction?

Nitrobenzene and benzene would NOT have the same rate towards any type of electrophilic substitution reaction as strength of nucleophile - (nitro)benzene does matter.

Nitro group as in nitrobenzene is highly deactivating group due its electron withdrawing nature and therefore decreases the nucleophilicity of benzene ring.

IJOART_document


Why is molarity independent of temperature? I can't find a good answer as to why molarity is independent of the temperature. Why is the concentration (in mol/l) independent of the temperature?

First of all, molarity is temperature-dependent.

You might have meant molality, because molality is temperature-independent.

The reason is very simple, if you see molarity definition, molarity is the amount of substance of solute per litre of solvent and that mean it depends on volume.
You know that volume depends on temperature, therefore molarity is also temperature-dependent.

Molality on the other hand is the amount of substance of the solute per mass of the solvent, both are independent of the temperature.


Which chemical reaction to use to remove carbon/peat from sieved earth/gravel? I produce gravel for my aquarium by sieving and rinsing earth from my garden. After boiling and disinfecting it (with a solution against parasites) the rest of the organic material are composted or become invisible duff inside the gravel during the initial cycling of the aquarium. The only thing remaining visible are peat and other carbon particles of the size of the gravel which I currently need to remove mechanically.

Is there a way to remove them /shrink or reduce them to small pieces with a (chain of) chemical reaction(s) which

can be performed with a maximum of chemicals and equipment which can be bought at a supermarket or building supplies store
doesn't harm the fish after introducing them into the aquarium some weeks after the treatment - both through toxines and an instable PH value (more than +/- 0.5)
leaves the color and the size of the gravel untouched because my motivation to use custom gravel are mostly aesthetic and reducing the size would risk to block my undergravel filter (has a filter chamber with a separator for gravel of a certain size)

I thought about:

burning the particles, but that seems wasteful in terms of energy and requires some equipment (burner, fire proof permeable net, etc.)

I'm looking for a solution for some kg of 1mm to 3mm gravel in a 100 l aquarium filled with drinking water with a PH value of 7.2 to 8.

In case it's necessary to have more exact info what the particles in the gravel consist of, I'll ask a separate question how I can figure this out.

It's easier to use flotation or froth flotation to separate out the less-dense peat (silt?) and carbon (I assume from the activated-carbon filter) than to use a chemical reaction..

Simple flotation would depend on making a dense solution (e.g. saturated Epsom salt, $\ce{MgSO4}$) and shaking the gravel mix in it... if the liquid's density is greater than the average density of silt with its entrained air, the gravel will sink and silt float, to be skimmed off. You can reuse the liquid for more separation.

Froth flotation relies on a surfactant wetting the silt and keeping it suspended while the gravel settles out. Try a bit of hand dish-washing
detergent.


Why derived name of nomenclature is used for Alkanes and alcohols only? My chemistry textbook says derived system of nomenclature of organic compounds is only limited to alkanes and alcohols!

What is the reason behind it? Is it chemical or like a ritualistic/cultural?

The derived system is really, really old. Even my textbook from the 20th century has IUPAC naming. My recommendation would be to just find IUPAC naming online, or just get a new textbook.
I can't seem to find much on the derived system, but here's another question about it; LINK
I tried searching about, and there really isn't much about it. I don't see why you'd want to learn about it though, as history of nomenclature isn't very useful or interesting. Besides, if you try to use the derived system, no one else is going to know what compound you're describing.


What effect is on the rate and rate constant when concentration or temperature increases/decreases? I have a few questions regarding rate and rate constant. I am wondering what effect certain concentrations, temperature, etc., has on rate and rate constant. Would be of great help if you can clarify this for me!

If a concentration was halved (and all other factors remained the same), does that mean rate would decrease? If instead the concentration was doubled, I would assume that it would also double.
For rate, if the temperature of the solution decreased, does that mean that it would also decrease since it means it would take less time?

For rate constant, if the concentration was halved, I think there would be no effect (because concentration effects rate only). But, what if the concentration was doubled, does that also mean that there's no effect or would the rate constant decrease?
For rate constant, let's say the ionic strength of the solution increased. Does that mean that the rate constant decreases?

Thanks for clarifying!

Assuming you are speaking about an elementary reaction, and by that I mean a reaction on the form $\ce{A -> B}$ the rate of the reaction is

$$r = \frac{\mathrm d[\ce{B}]}{\mathrm dt} = -\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]$$

Most of the time $k$ can be express following the Arrhenius law:

$$k = A\exp\left(\dfrac{-E_\mathrm{a}}{k_\mathrm{B}T}\right)$$

Where $A$ is the pre-exponential factor (the constant rate when them temperature is really high), and $E_\mathrm{a}$ is the activation energy of the reaction.

In this case the initial concentration has no affect on the rate constant, only the temperature. Looking at the first equation the initial concentration has effect on the rate, this effect will depend on the order of the reaction…

$$r = k[\ce{A}]^2$$

$$r = k[\ce{A}]^3$$

Here a second and third-order reaction. If you study a complex mechanism, it can happen that initial concentration will affect the global rate constant, it can also happen with some approximation:

$$\ce{A + B -> C} \qquad r = k[\ce{A}][\ce{B}]$$

If $\ce{A},$ for example, is in really large quantity compared to $\ce{B},$ you can admit that for every time the concentration will approximately stay the same $[\ce{A}] = [\ce{A}]_0$ and then put it in the constant

$$k' = k[\ce{A}]_0$$

I hope that helped! If not, you need to study more.


Finding overall order of reaction Recently I've been going through an exercise of Chemical Kinetics and Nuclear Chemistry and right now I am stuck on one question that was previously asked in JEE(Advanced)-2013. The question says,

In the reaction,
$\ce{P + Q -> R + S}$
The time taken for $75\%$ reaction of $\ce{P}$ is twice that time taken for $50\%$ reaction of $\ce{P}$. The concentration of $\ce{Q}$ varies with reaction time as shown in the figure. What is the overall order of the reaction?

And here was my approach to the question,
In the question different data have been given for different reactants (i.e. $\ce{P}$ and $\ce{Q}$). For $\ce{P}$ it has been given that it take $\ce{P}$ double the time to react $75\%$ than that from $50\%$. So assuming the time taken for $50\%$ reaction to be $t$ and that of $75\%$ to be $2t$, we can easily find that order of reaction w.r.t P is $1$. As,

$$\frac{t_{\frac12}}{t_{\frac34}}=\frac{\log 2}{\log 4}=\frac12$$

Further w.r.t $Q$ there is a graph given which is something like in form $\ce{[Q]=[Q_o]-mt}$. So comparing with the equation of zero order reaction which is like $\ce{[A]=[A_o]-kt}$ (where $\ce{[A]}$ is the concentration of reactant and $k$ is rate constant), we can conclude that order of reaction w.r.t to $Q$ is $0$.
So I concluded that overall order of reaction will be

order of reaction of $P +$ order of reaction of $Q$
$\implies 1+0=1$

So was this the right approach from my side or there is any kind of error in any step?

As summarized in the IUPAC definition of reaction rate, the rate of change of one reactant must be directly proportional to the rate of change of each other reactant.

If the concentration of one reactant decreases linearly, the concentration of every reactant must decrease linearly (the reaction rate is a constant, and the overall reaction order is zero).

The text of the JEE question contradicts its Figure (which shows linear decrease of [Q]), by saying that the "time taken for 75% reaction of P is twice that time taken for 50% reaction of".

Instead, since [Q] decreases linearly, [P] must decrease linearly, and if 75% of an initial [P] reacts it must take 1.5 times as long as for 50% to react.

It's likely that the author of the question intends that the answer is "first order", but really there is no correct answer.


Nuclear spin isomerism in molecules other than H2 I was reading my teacher's material about ortho- and para- hydrogen, and I found stated that

  "every homonuclear molecule, with nuclides with spin other than zero ($\ce{H2}$, $\ce{D2}$, $\ce{T2}$, $\ce{N2}$, $\ce{^{17}O2}$...) has nuclear spin isomers".

What is not clear to me is: are there any nuclides with spin zero? Is there, hence, any homonuclear molecule without spin isomers?

Nucleons of the same type tend to pair up and cancel out each others' spins, so nuclides with an even number of protons and an even number of neutrons will have zero spin. Some common nuclides with this property include $^4\ce{He}$, $^{12}\ce{C}$, and $^{16}\ce{O}$.


How can difference in neutron number cause a difference in ionisation enthalpies? It’s a well known fact that protium,deuterium and tritium have the same number of electrons and protons but different number of neutrons. It is expected that their ionisation enthalpies should be same as in all the cases only one proton attracts the only electron they have but their enthalpies are different.

What causes this?

There are a handful of additional corrections needed to calculate the exact differences. But the biggest factor is just the mass of the nucleus. Not only is the electron attracted to the nucleus, the nucleus is attracted to the electron. So not only does the electron "move", so does the nucleus. A heavier nucleus moves less. This in turn alters the wave function of the electron.

The difference between protium, deuterium and tritium is reasonably explained by the Bohr model of the atom. The gist is that the electron does not travel around a nucleus of infinite mass, but rather that the electron and nucleus travel around their common center of mass.

For one electron traveling around the nucleus the energy levels of the Bohr system are:

$\text{E}_n = \dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} \cdot \dfrac{\text{Z}^2}{\text{n}^2}\cdot \text{E}_R$

$\text{E}_n$ Energy of the $n$th orbital of the one electron system
$\text{M}_e$ Mass of the electron = $5.489\cdot10^{−4}$ atomic mass units
$\text{M}$ Mass of the nucleus
$\text{Z}$ Charge on the nucleus
$n$ orbital = principal quantum number $n$ = 1, 2, 3, 4, ...
$\text{E}_R$ Rydberg energy = 13.605693 eV

For the $1s^1$ electron of protium, deuterium and tritium, Z=1 and $n$=1.

protium
$\text{M} = 1.007825 - 0.0005489 = 1.007276$ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{1.007276}} = 0.999455$

ionisation enthalpy = $0.999455 \cdot 13.605693 = 13.598$ eV

deuterium
$\text{M} = 2.01410178 - 0.0005489 = 2.0135529 $ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{2.0135529}} = 0.999727$

ionisation enthalpy = $0.999727 \cdot 13.605693 = 13.602$ eV

tritium
$\text{M} = 3.0160492 - 0.0005489 = 3.0155003$ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{3.0155003}} = 0.999818$

ionisation enthalpy = $0.999818 \cdot 13.605693 = 13.603$ eV

Ionization energies From NIST webbook
hydrogen = 13.59844 eV
deuterium = 13.603 eV
tritium = 13.603 eV

Therefore protium < deuterium < tritium

I'll add that when you're doing this sort of thing other factors creep in the calculations.

(1) For a hydrogen atom the electrons aren't relativistic. If you get to uranium though that wouldn't be true.

(2) The 13.6 eV binding energy also reduces the mass of the hydrogen atom. So subtracting the rest mass of the electron isn't absolutely correct, but good enough for the precision used.

13.6 ev $=\quad\quad 0.000 000 014 5$ amu
mass e used $= 0.000 548 9$ amu


Can carboxylic acids like act as nucleophiles in substitution reactions? Can carboxylic acids like formic acid act as nucleophile in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reactions?
For example can formic acid react with an alkyl/aryl halide like benzyl bromide?

I'm asking because I found a very similar question while going through some old organic chemistry test papers. However none of the book's I have read says that carboxylic acids can act as nucleophile in such cases.

Carboxylic acids can act as nucleophiles of course. Consider reactions with isobutene or thionyl chloride (although base certainly helps in this latter case). However, whether it's in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reaction mechanisms is another matter. If you beast something with enough heat it will usually react. I would not be surprised if a carboxylic acid reacted with triflouromethanesulfonic anhydride without base. It's still $\mathrm{S_N2}$ as mechanistic studies have shown the S=O bonds do not open up like acyl electrophiles.
I suspect you would need to resort to a continuous as opposed to a batch approach to push a reaction between the acid and an alkyl halide without using any base. Expose the reaction to high temperature (I'm thinking 250C and higher)  for 5 seconds or less under continuous liquid flow.


What does "reaction rank" in Reaxys stand for? By exporting some search results from Reaxys, I found "Reaction rank" to be one of the columns in the Excel file. In my file, values descend from 107 to 52 and skip some values in-between too. Does anyone know what reaction rank stands for here?

For the Elsevier's Reaxys data base, there is a support page you may consult without subscription, nor log-in to the product.  Actually, this is more like a searchable directory, too, because you search for keywords, like «rank», too.

Doing so here -- time stamped 25/04/2019 -- offers insight about «How do I view my results?».  About half-way through this document, in subsection «Reactions. Order, limit or export your search results», the documentation states:

  "[Reaxys ranking] Defines the method of determining 'relevance'. To
  change the sort type, click 'Reaxys Ranking' and define another sort
  option: 'No of References', 'Reactant Availability', ‘Product
  Availability'', ‘MW of product’, ‘Yield’ or ‘Publication Year’."

And because the content of columns of numeric content (like molar mass, number of reagents, year of publication) may be sorted, this may affect the presentation of the export, too.  It equally may explain the occurrence of gaps in the values you mention.


Should rates and rate constants be expressed in terms of pressure or concentration? If an elementary reaction was conducted in the gaseous phase, say $$\ce{X (g) -> Y(g)},$$ would the formal definition of the rate law be $\mathrm{rate} = kP_\ce{X}$ or $\mathrm{rate} = k[\ce{X}]$?

The ambiguity doesn't lie so much in the definition of the rate, because as long as the constants $a$ and $b$ in $ap_\ce{X}$ and $b[\ce{X}]$ are chosen appropriately, the two expressions are entirely equivalent. So you can define the rate either way, and there is no physical difference, not even a mathematical difference. It is somewhat akin to asking whether you should factorise the number $12$ as $2 \times 6$ or $4 \times 3$. It doesn't matter how you factorise $12$, it's still the same number.
The ambiguity is actually in the definition of the rate coefficient, $k$. The constants $a$ and $b$ above, though directly proportional to each other, are obviously different: and only one of them can be the "true" rate coefficient $k$.
Before launching into the answer proper I should mention that there is no physical difference between using the pressure and the concentration. At the end of the day, it doesn't change the actual rate at which your system is reacting: it's just an arbitrary choice as to what physical constants you want to include in the rate coefficient $k$.
That said, the IUPAC Gold Book seems to come down quite strongly on the side of the concentration, meaning that $k$ is the same as the second constant $b$. They do not make any reference to the phase of the reaction, so it should probably be assumed that they intended it to hold for any general phase. (If anyone knows of a more specialised definition, please point it out!)

order of reaction, $n$ (DOI: 10.1351/goldbook.O04322)
If the macroscopic (observed, empirical or phenomenological) rate of reaction ($nu$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ (expressing in full the dependence of the rate of reaction on the concentrations $[\ce{A}]$, $[\ce{B}], \cdots$)
[...]
The proportionality factor $k$ above is called the ($n$th order) 'rate coefficient'. Rate coefficients referring to (or believed to refer to) elementary reactions are called 'rate constants' or, more appropriately 'microscopic' (hypothetical, mechanistic) rate constants.

This approach is quite consistent throughout. For example, the rate of reaction is also expressed in terms of concentrations, at least under constant-volume conditions:

For the general chemical reaction:
$$\ce{aA + bB -> pP + qQ + \cdots}$$
occurring under constant-volume conditions, without an appreciable build-up of reaction intermediates, the rate of reaction $\nu$ is defined as:
$$\nu = -\frac{1}{a}\frac{\mathrm{d}[\ce{A}]}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}[\ce{B}]}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}[\ce{P}]}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}[\ce{Q}]}{\mathrm{dt}}$$
The quantity $$\dot{\xi} = \frac{\mathrm{d}\xi}{\mathrm{d}t}$$ defined by the equation:
$$\dot{\xi} = -\frac{1}{a}\frac{\mathrm{d}n_\ce{A}}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}n_\ce{B}}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}n_\ce{P}}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}n_\ce{Q}}{\mathrm{dt}}$$
(where $n_\ce{A}$ designates the amount of substance $\ce{A}$, conventionally expressed in units of mole) may be called the 'rate of conversion' and is appropriate when the use of concentrations is inconvenient, e.g. under conditions of varying volume.

One tangential question I can anticipate is: what's the difference between the expressions $r = -(1/a)(\mathrm{d}[\ce{A}]/\mathrm{d}t)$ and $r = k[\ce{A}]$? The answer is: the former is a definition. The rate is literally defined by how fast the reactants disappear, which is a universal truth, applicable to all reactions. The latter is an empirically observed expression, which only holds true for a small subset of reactions (i.e. first-order ones), and it says that the rate at which the reactant disappears is directly proportional to its concentration.


Why bond energy of acetone is more though it is more polar than formaldehyde? I was going through the general properties of carbonyl compounds and then found this (in the pic) data, where it says that dipole moment of acetone (if we assume $\ce{R = -CH3}$) is more than acetaldehyde which in turn is greater than that of formaldehyde.
But I think, as $\ce{-CH3}$ is a $+I$ effecting group, so it will satiate the residing positive pole on carbon slightly and hence will make the $\ce{C=O}$ bond less polar and hence the sequence should be reversed!

Please anyone make me correct where am I mistaking with the concept?

Picture source

EDIT :- I got my answer of the above confusion. Thanks to Physicsapproval.
But with that I came to face another confusion. In the next part of the text (see the paragraph just below the pic in the Picture source) I found that "the larger the dipole moment the greater the polar character of the carbonyl group". So that means, $\ce{C=O}$ bond in acetone is more polar and least in the case of formaldehyde! Now in the next part it is mentioned that the bond energies of $\ce{C=O}$ bond in the mentioned compounds are: $$ \ce{H2C=O} \, ; 170~\mathrm{kcal/mol}$$ $$\ce{RHC=O} ~ ; 175~\mathrm{kcal/mol}$$ $$\ce{R2C=O}~; 180~\mathrm{kcal/mol}$$
Now I know that the more polar the bond is, the less is its bond energy. So, in this case why the trend is reversed, i.e., $\ce{C=O}$ bond of acetone being more polar has the highest bond energy compared to other two?

the more polar the bond is, the less is its bond energy

I'm not sure where you are getting this from.

Take Acid halides as an example series. As the bond get more polar ( HI (0.4 D) < HBr (0.7 D) < HCl (0.9 D) < HF (1.9 D) ) the bond energies follow accordingly ( HI (295 kJ/mol) < HBr (362 kJ/mol) < HCl (428 kJ/mol) < HF (565 kJ/mol) )

As the charge separation gets larger (and therfore more polar) there is a stronger will to not have that large charge separation, as nature wants to naturally come to an equilibrium - and thus the bonds are more energetic.

Does this clear up your confusion?


What chemical regularly transported by road in the UK would cause serious injury/loss of life if it escaped in a road accident? I'm writing a story where a serious multi-vehicle pileup happens in the winter during foggy weather. I need the fallout from this to include the escape of a substance that will cause significant injury/loss of life to those immediately involved in the accident.

What chemical can I use that is regularly transported by road in the UK that could cause this to happen?

Organophosphates in general would be a great idea as you can go from potent insecticides to chemical warfare as it fits the drama of your story.


What is the compound I made from KNO3 and table sugar which subsequently exploded when I exposed it to a flame? I mixed 75% $\ce{KNO3}$ and 25% $\ce{C12H22O11}$ (by volume) in boiling water to make slow burn fuses. Some of the product became a thick brown paste.

After drying, a hard cake of an approximate 1 tablespoon amount exploded when lit. What compound did I make?

It's not a new compound - it's just an intimate mix of fuel (sugar) plus oxidiser (nitrate). It's not the most stable of situations to have...


How to compare the acidity of this multicyclic hydrocarbon to the acidity of cyclopentadiene? This is the question:

  Which one is more acidic?

According to me, cyclopentadiene should have been the answer, as losing a proton would make it aromatic; whereas the first option doesn't need deprotonation to be aromatic, it already is. However the first option is given as the correct one.

I have come up with the following reasonings as to why the first option should be more acidic:

Deprotonation might impart aromaticity to the pentagonal rings too which would act as the driving force to lose the proton.
The negative charge on central carbon (after deprotonation) would be in conjugation with the $\pi$ electrons of the three rings which would make the entire molecule planar and hence reduce the strain on the three bonds that are directly linking the benzene rings. This could be another driving force for deprotonation.

The second reasoning seems much convincing to me but I fail to understand how can I compare that factor with the aromaticity factor involved in cyclopentadiene.

The method presented by Yusuf is a simplified model, which works best for well-behaved molecules. Unfortunately in this case it presents the 'correct' result based on incorrect assumptions. (Even though I did a calculation, I am not 100% convinced this is actually the correct result in the first place.)

In order for the resonance to stabilise a negative charge, the π-systems must be aligned, which is the case for the deprotonation of cyclopentadiene, but it certainly is not true for the other molecule F (Loong said it's called 3a2H-benzo[3,4]pentaleno[2,1,6,5-jklm]fluorene). Here is the calculated structure at the ri-BP86/def2-SVP level of theory:

As you can see, it is domed, and there are three (almost) independent aromatic systems. Through deprotonation this will not change much.

A reason for its relatively high acidity can be found in applying Coulson's Theorem, where the bond angle between two $\mathrm{sp}^x$ hybrid orbitals can be calculated from their angle $\omega_{ij}$ and their hybridisation indexes $\lambda_x$:
$$\cos(\omega_{ij}) = -\frac{1}{\sqrt{\lambda_i\lambda_j}}$$
Since the molecule is C3v symmetric, therefore $\lambda_i = \lambda_j \mathop{:=} x$, this simplifies to:
$$ x = -\sec(x)$$

The inner angle is $\approx\pu{102^\circ}$, therefore $x \approx 5$. The term for hybrid orbitals is actually just an abbreviation:
$$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$

Given that we just form linear combinations, we can see:
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 3\times\mathrm{sp}^5, 1\times\mathrm{sp}
\end{align}
Given the rounding I have done, the remaining orbital is very roughly speaking an $\mathrm{sp}$ orbital; in the same ballpark as acetylene, from which we know that it has a very acidic proton.

The reason for the high acidity is therefore not the stabilisation of the conjugate base, but the destabilisation of the acid. This is the completely reverse reasoning as cor cyclopentadiene, in which the corresponding base is massively stabilised by resonance.

Given the above, how are you supposed to know, and how are you supposed to judge these compounds?

In my opinion, the exercise is something more on the level of quantum chemistry course and belongs into the canon of a masters student. While simply models will accidentally produce the correct result, it is important to see, that they completely fail for F. Structure-property relationships cannot be inferred from a 2D-graph of the molecule. To be honest, the reasoning above was also rather post-hoc, after I have calculated the molecule and analysed its structure.

I believe, (as long as you are not a human calculator), that there is no simple way to tell which of those molecules is more acidic; there is no easy way to weigh the stabilising effect of resonance against the destabilising effect of pyramidalization.

On the ri-BP86/def2-SVP level of theory (which is rather crude, but I don't have a supercomputer at home) it really comes out that F is more acidic. The following isodesmic reaction yields a difference in electronic energy of about $\Delta E_\mathrm{el} = \pu{-92.8 kJ mol-1}$.

Important: Even though the above sound reasonable, it still neglects plenty of things and has other shortcomings; to name the most important: thermal corrections, solvent effects, level of theory.

Addendum:

Upon deprotonation the structure does not change very much (see larger image), it remains very domed, see the overlay of the structures below.

Judging from the three highest occupied molecular orbitals (HOMO, HOMO -1, HOMO -2) we can still presume some delocalisation of the negative charge into the adjacent aromatic rings.

I'm talking more about hybridisation, Bent's rule, and Coulson's theorem in the following: Are the lone pairs in water equivalent?


What will be the final product when the above bridged compound is heated with concentrated sulfuric acid? What will be the final product when the above compound ((1S,2S,4R,7S)-7-methylbicyclo[2.2.1]heptane-2,7-diol) is heated with concentrated sulfuric acid?

I know that the hydroxyl group in the bridgehead position cannot be protonated and removed to form a carbocation. The other hydroxyl group may be protonated and removed to form a carbocation I suppose. But after that what rearrangements will take place ? What will be the final product(s)?

Protonation of both $\ce{OH}$ groups and subsequent formation of the corresponding cations is possible.

The tertiary cation will yield an exomethylene group
The "secondary" non-classical cation leads to a $\ce{C=C}$ double bond in the ring. With other words, a norbornene is formed.

The combined eliminations will give 7-methylene bicyclo[2.2.1]hept-2-ene as the product.

In J. Am. Chem. Soc., 1973, 95, 7801-7813, the same product was obtained by elimination from the tosylate of 7-hydroxy-7-methylnorborn-2-ene.


How to figure out which substance has the highest osmotic pressure? In preparing for an exam and I'm trying to solve the following question. I'm clueless. What should I do?

  Which of the following water solutions has the highest osmotic pressure?

  \begin{align}
\tag1 \pu{0.1 mol/kg} &~ \ce{NaCl}\\
\tag2 \pu{0.1 mol/kg} &~ \ce{KF}\\
\tag3 \pu{0.1 mol/kg} &~ \text{sucrose}\\
\tag4 \pu{0.1 mol/kg} &~ \ce{MgCl2}
\end{align}

$\ce{MgCl2}$ has the highest osmotic pressure because it breaks up into three ions in water, as opposed to all the others, which only break up into two.


Phosphorus does not form pπ-pπ bonds but how could it form dπ-pπ bonds? My textbook says phosphorus cannot form pπ-pπ bonds because their atomic orbitals are so large and diffuse that they cannot have effective overlapping.

My doubt is then how does phosphorus form dπ-pπ bonds? I mean dπ-pπ bond is supposed to be more ineffective overlapping. If pπ-pπ cannot happen then how does dπ-pπ happen?

My doubt is then how does phosphorus form dπ-pπ bonds?

Poorly. Influence of the $d$-orbitals on the chemistry of the $p$-block is minor. The current belief is that 'double' bonds are actually dative bonds in most cases.

See the similar cases of chlorine and sulfur

How does chlorine form more than 1 bond?

What is the chemical structure / hybridization of sulfuric acid?

For phosphorus it isn't that bad, but still bad

http://link.springer.com/article/10.1007%2FPL00010730


Mass Spectrum and Molecular Ion peaks If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?

The major isotopes of chlorine are 35 and 37.  If 4 chlorines are present in a molecule, then the following isotopic combinations of chlorine are possible

35, 35, 35, 35
35, 35, 35, 37
35, 35, 37, 37
35, 37, 37, 37
37, 37, 37, 37

So there will be 5 molecular ion peaks.  You could calculate their relative heights by factoring in the relative isotopic proportions, along with the number of permutations of each of the above combinations.


Gibbs energy derivative vs equilibrium for aqueous reaction I am trying to find the Gibbs-energy equilibrium for the reaction $$\mathrm{A}+\mathrm{B}\rightleftharpoons\mathrm{C}.$$ According to most sources the equilibrium concentrations in an ideal solution should form a ratio $$\Delta G^0=-RT\ln\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}.$$

I am having trouble seeing this when the stoichiometric coefficients on either side of the reaction are unequal, as in this example above.

The Gibbs energy for the solution should be $$G=n_A(\mu_A^0+RT\ln[\mathrm{A}])+n_B(\mu_B^0+RT\ln[\mathrm{B}])+n_C(\mu_C^0+RT\ln[\mathrm{C}]).$$

Assuming we started with a 1 L solution containing 1 mol of A and 1 mol of B, $\xi$ mol of each react to form $\xi$ mol of C, this equation should then be $$G=(1-\xi)(\mu_A^0+RT\ln(1-\xi))+(1-\xi)(\mu_B^0+RT\ln(1-\xi))+\xi(\mu_C^0+RT\ln\xi).$$

My understanding is that if the Gibbs free energy is at its minimum the reaction above will not be spontaneous in either direction and so the solution will be at equilibrium. The derivative of the above equation gives $$0=-(\mu_A^0+RT\ln(1-\xi))-RT-(\mu_B^0+RT\ln(1-\xi))-RT+(\mu_C^0+RT\ln\xi)+RT.$$

Some rearrangements result in $$\Delta G^0=-RT\ln\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}+RT.$$

In other words, my expression is off by $\Delta nRT$. Having been unable to find any help on this matter on the internet, I feel I have either made some silly logical/mathematical mistake or have some fundamental misunderstanding. Please point out the error.

To derive the Gibbs' free energy of an ideal solution you would invoke Raoult's law, which says that
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{p_i}{p*}\right)= \mu_i^*(l)+RT\ln\chi_i$$
Here the asterisk refers to the pure substance. Using the notation in the OP, the total number of moles of reagent and product in solution is

$$n_{tot}=2(n_{ini}-\xi)+\xi+n_{solv}=2n_{ini}-\xi+n_{solv}$$

where $n_{ini}$ is the initial amount of A or B (assumed equal), so that one can write for reagents A and B:
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{n_{ini}-\xi}{2n_{ini}-\xi+n_{solv}}\right)$$
and for product C:
$$\mu_C(l) = \mu_C^*(l)+RT\ln\left(\frac{\xi}{2n_{ini}-\xi+n_{solv}}\right)$$
There is also a term for the solvent:
$$\mu_{solv}(l) = \mu_{solv}^*(l)+RT\ln\left(\frac{n_{solv}}{2n_{ini}-\xi+n_{solv}}\right)$$
which I ignored in earlier edits.
Adding terms to compute the total Gibbs energy and taking the derivative wrt the extent of reaction $\xi$ then results in

$$\begin{align} \Delta G' &= \Delta \mu^* + RT \ln\left(\frac{\xi(2n_{ini}-\xi+n_{solv})}{(n_{ini}-\xi)^2}\right)  \\  &= \Delta \mu^* + RT \ln\left(\frac{n_Cn_{tot}}{n_A n_B}\right)  \\&= \Delta \mu^* + RT \ln \left(\frac{\chi_C}{\chi_A \chi_B}\right) \end{align} $$

($^\dagger$)

Note that here the standard state for each substance refers to the pure liquid in equilibrium with its vapor at the same temperature. To obtain an expression relating a change in standard free energy to the concentration equilibrium constant requires expressing mole fractions in terms of concentrations. This is generally done for dilute solutions where $n_{solv} \approx n_{tot}$ by writing $\chi_i\approx c_iM_{solv}/\rho_{solv}$, which leads to

$$\Delta G' = \Delta \mu^* - RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$
Then by redefining the standard states as
$$\mu_i^\circ = \mu_i^* + RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) $$ leads to $$\Delta G' = \Delta \mu^\circ + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$
If you think the standard state looks strange, consider that if $\chi_i = c_iM_{solv}/\rho_{solv}$ then $$\chi_i(\pu{c^\circ_i= 1 M}) = c_i^\circ \frac{M_{solv}} {\rho_{solv}}$$ and so we can write

$$\mu_i^\circ = \mu_i^* + RT \ln \left( \chi_i(\pu{c^\circ_i= 1 M}) \right) $$ which is more in line with canonical definitions of standard states. The new standard state is defined as a 1 M ideal solution with chemical potential related to that of the pure solute by the above equation.

$^\dagger$/TL;DR :  The mysterious extra $RT$ term in the OP is gone. It's important to keep a term for the solvent. And a lesson from these repeated edits: careful doing math when tired!


Why do people add H(core)+F in Hartree-Fock? I'm trying to implement my own HF code, and it seem usual that, after building the Fock matrix $F$, they add that to the pre-existing $H_{core}$, from the two-point interactions, when computing final energy.

As one example, https://github.com/ipudu/SCFpy/blob/master/SCFpy/scf.py . In lines 135-138, they include Hcore in F. Then in line 156, when they actually return the energy, they add Hcore[i,j]+F[i,j], apparently double-counting Hcore.

As another example, in http://sirius.chem.vt.edu/wiki/doku.php?id=crawdad:programming:project3 , when they compute the energy (see Step #9), they use $$E_{elec}^i = \sum_{\mu\nu}^{AO} D_{\mu\nu}(H^{core}_{\mu\nu} + F_{\mu\nu})$$

again apparently double-counting $H_{core}$. Why? What gives?

My best guess is that this is somehow related to the fact that these are double-filled orbitals, containing both spins. But then, shouldn't $F$ need a factor as well, for all it particles occurring twice as well?

In particular, I have an HF code that agrees with the 'crawdad' link above by copying their steps pretty much verbatim. I'm now trying to extend it into a generalized Hartree-Fock code, but I'm having difficulties, and in trying to understand the problem I became confused by this expression.

Actually, $F$ is double-counting the electron-electron interaction when evaluating the total energey. By adding $H^{\text{core}}$ and using a factor of $\frac{1}{2}$, which I suspect is implicit in $D$, one can obtain the correct energy. This is basically an occurence of the HF energy not being the sum of orbital energies (which are a result of $F$).


What is the product obtained when methyl vinyl ketone is reacted with 1,2-ethanedithiol and the product is further hydrogenated with Raney Nickel? I was solving questions from aldehyde, ketones and ethers and I came across a question that asks:

What is the product obtained when Methyl vinyl ketone is reacted with 1,2-Ethanedithiol and the product is further hydrogenated with Raney Nickel?

My thoughts:

I thought that the reaction could be similar to the one where we react a ketone with ethylene glycol, then on further hydrogenation the double bond breaks. I may be wrong. Any help will be highly appreciated

The answer could be 3 or 4.
1,2-ethanedithiol will react with the carbonyl group to form a dithiane (similar to the ethylene glycol reaction). Reaction of this with Raney Nickel/H2 desulfurises it, so the whole process is effectively a reduction of the carbonyl group to CH2. It depends on the reaction conditions whether the double bond survives. Raney Nickel can reduce alkenes.
You can read more about it here 1 and here 2


Smoke alarm ion neutralisation Smoke alarms are a necessary part of many modern day buildings, especially in laboratories and of course, the family home.

According to the World Nuclear Association page Smoke Detectors and Americium (2014), a major part of the operation of a smoke detector is when the Americium-241 emits alpha particles collide with atmospheric oxygen and nitrogen, ionising them - which form part of an electric current between two electrodes.

When there is smoke (from the article) - bolding mine:

  When smoke enters the space between the electrodes, the smoke particles attach to the charged ions, neutralizing them. This causes the number of ions present – and therefore the electric current – to fall, which sets off an alarm.

What is the chemical process that causes smoke to neutralise the charged oxygen and nitrogen ions in smoke alarms?

From this paper, found the mechanism for neutralisation,

  Smoke particle charge interaction

  When smoke particles enter in the ionization chamber, they are
  subjected to the electric field generated by electrodes but also to the
  one generated by the charges. Close to the charges, the electric field is
  highly divergent resulting in dielectrophoretic force acting on smoke
  particles. The force tends to bind the smoke particles to the charges.
  When smoke particles are considered as isotropic neutral dielectric
  bodies linearly and homogeneously polarisable, dielectrophoretic
  force F can be expressed as

  $ F = \frac{1}{2} \alpha v \overrightarrow{grad} |E^{2} |$

  where $\alpha$ depends on smoke particle polarizability, $v$ is the smoke
  particle volume and $E$ is the electric field produced by the charges
  [11]. Coefficient $\alpha$ is typically of the order of $\epsilon_{0}/3$ [12] where $\epsilon_{0}$ is the
  dielectric constant of air. Volume $v$ is approximately $4 \mu m^{3} $ for a 1-$\mu m $
  diameter smoke particle.
  Amplitude of electric field $E$ produced by charges is given by
  coulomb’s law:

  $ E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^{2}}$

  in which $r$ is the distance between the charge and the smoke particle.
  Due to the divergence of the field, the dielectrophoretic force
  decrease as $\frac{1}{r^{5}}$. If a smoke particle is close to a charge then they can bind. If the smoke particle is relatively far from the charge, the time
  of interaction may be relatively long and it is necessary that the
  charge drifts sufficiently slowly in the chamber to optimize the
  interaction. Indeed if the charge passes too rapidly there will be not
  enough time for the smoke particle to bind and thus nothing would be
  detected. In fact the capture radius reduces with the electric field and
  is of the order of 10 $\mu m $ for $E= 1  V/mm$.

  When the smoke particles attach to the free charges due to
  electrophoresis force, the charges become heavy bodies then the
  charge mobility $\mu$ decreases resulting in a drop of the electric current
  measured between the electrodes of the ionization chamber.
  Even if ionic detectors are very efficient, according to new
  standards they are no longer authorised in almost all of the developed
  countries because of the radioactive sources they include for
  generating the ions. Such radioactive sources are indeed difficult to
  collect and to recycle.


MO diagram for pentaamminechlorocobalt(III)? I was considering the MO diagram for [Co(NH3)5Cl]2+, and figured it’d be as follows:

5 sigma interactions from the NH3 ligands; by reducing a reducible representation one obtains 2A1 + B1 + E

2 pi interactions (pi donation from Cl- px and pz orbitals), which have symmetry E

And the cation atomic orbitals, which we just read off the character table.

Unfortunately this results in a lot of confusing interactions in the MO diagram.

What would the MO diagram look like, and are my group orbitals correct?

Thanks!

Let's start at the bottom. There will be three a1 bonding orbitals. First, we can have the metal $4s$ orbital interact with all six ligands in phase. Let's call this $1a_1$. Next, the metal $d_{z^2}$ can interact with all six ligands as well, but the axial ligands are now opposite phase from the equatorial. Let's call that $2a_1$. Both will of course have antibonding counterparts. The third $a_1$ orbital is an interaction of the metal $p_z$ with out of phase axial ligand orbitals. The equatorial ligands do not contribute to this orbital because they are on the nodal plane. Let's call this $3a_1$.

Mixed in with these a1 orbitals will be the $b_1$ orbital comprised of the equatorial ligand orbitals interacting with $d_{x^2-y^2}$. That will be $1b_1$. It is lower in energy than $3a_1$ but likely higher than $1a_1$ and $2a_1$. (This $1b_1$ and the $2a_1$ correspond to the $1e_g$ orbitals of an $O_h$ complex.)

Near $3a_1$ we have the two $e$ orbitals comprised of the metal $p_x$ and $p_y$, each interacting with a pair of equatorial ligands. Let's call these $1e$. (These along with $3a_1$ correspond to the $1t_{1u}$ orbitals of an $O_h$ complex).

If we are only considering sigma interactions, we next have the $e$ and $b_1$ orbitals representing metal $d_{xy}$, $d_{xz}$, and $d_{yz}$, all of which are nonbonding for sigma. However, the pi interactions with Cl mean that the $e$ orbitals are stabilized and have antibonding destabilized counterparts. Let's call those $2e$. The $b_1$ orbital ($d_{xy}$) remains nonbonding.

Getting the ordering exactly correct is tough in this sort of qualitative analysis, but that should get you a pretty good start.


How to make solution of specific concentration with non-stoichiometric hydrate? I am currently working with a drug molecule known as Amiloride $\ce{HCl}$ and it seems to be a non-stoichiometric hydrate.

How would I go about making a solution with a specific concentration of Amiloride $\ce{HCl}$? I know its anhydrous molecular weight but if I don't know how much water is in the solid crystals how can I get a specific concentration?

Anhydrous Amiloride HCl is commercially available and is stable in air.  This suggests that you should be able to remove the waters of hydration, however many that may be, by repeatedly heating in a drying oven until a constant weight is reached.  If you know the weight before drying you could of course calculate the original degree of hydration after the final drying, if that is of interest.

If you need to further verify dryness, you could have the dried product analyzed by Karl Fischer water anaysis (~$50). Just be sure to seal the sample under a dry atmosphere and label "HYGROSCOPIC" all over the sample and submittal paperwork!

Once you have the constant-weight product, stored under a good desiccator, just use the formula weight of the hydrochloride salt, 266.1, to calculate the concentration of your Amiloride HCl solutions.


How to obtain the radial probability distribution function of a given orbital from a quantum chemical calculation? I know that a very similar question has already been asked here and a very competent and useful answer has been provided (see here). However, that previous question was related to the total electron density whereas I am interested in plotting the radial probability distribution function of a given single orbital. Can I also use MultiWFN tool and Gaussian output to do this?

This is actually a lot simpler than I initially thought. I'll be using the same example as previously explained in How to obtain the radial probability distribution function from a quantum chemical calculation? My test setup has slightly changed since then, i.e. I've tried it with Gaussian 16 B.01 and Multiwfn 3.6.0 and it worked the same.

We start the same way as in the linked question, until you have your ar.fchk and load that into Multiwfn, you should see something like this:

 Multiwfn -- A Multifunctional Wavefunction Analyzer
 Version 3.6, release date: 2019-May-21
 Project leader: Tian Lu (Beijing Kein Research Center for Natural Sciences)
 Below paper *MUST BE CITED* if Multiwfn is utilized in your work:
          Tian Lu, Feiwu Chen, J. Comput. Chem., 33, 580-592 (2012)
 Multiwfn official website: http://sobereva.com/multiwfn
 Multiwfn English forum: http://sobereva.com/wfnbbs
 Multiwfn Chinese forum: http://bbs.keinsci.com/wfn

>>

 Loaded ar.fchk successfully!
 Formula: Ar1
 Molecule weight:        39.94800

We need to modify the wave function now:

                    ************ Main function menu ************
 0 Show molecular structure and view orbitals
 1 Output all properties at a point
 2 Topology analysis
 3 Output and plot specific property in a line
 4 Output and plot specific property in a plane
 5 Output and plot specific property within a spatial region (calc. grid data)
 6 Check & modify wavefunction
 7 Population analysis and atomic charges
 8 Orbital composition analysis
 9 Bond order analysis
 10 Plot total DOS, partial DOS, OPDOS, local DOS and photoelectron spectrum
 11 Plot IR/Raman/UV-Vis/ECD/VCD/ROA spectrum
 12 Quantitative analysis of molecular surface
 13 Process grid data (No grid data is presented currently)
 14 Adaptive natural density partitioning (AdNDP) analysis
 15 Fuzzy atomic space analysis
 16 Charge decomposition analysis (CDA) and extended CDA (ECDA)
 17 Basin analysis
 18 Electron excitation analysis
 19 Orbital localization analysis
 20 Visual study of weak interaction
 21 Energy decomposition analysis
 100 Other functions (Part1)        200 Other functions (Part2)

Chose 6:

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   9.000   9.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 34 Set occupation number of inner orbitals to zero
 36 Invert phase of some orbitals

As you can see there is plenty you can do. For this example I will reset the occupation of some orbitals (26). More precisely, I'll transform the wave function so that only the 2s orbital is doubly occupied. Here is how I did it:

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   9.000   9.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 34 Set occupation number of inner orbitals to zero
 36 Invert phase of some orbitals
26

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
0
 Set to which value? e.g. 1.2
 Note:
 You can also input for example "+1.1" "-1.1" "*1.1" "/1.1" to add, minus, multiply and divide the occupation numbers by 1.1
 To recover the initial occupation numbers, input "i"
 To generate occupation state for calculating odd electron density, input "odd"
0
 Done!

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
2
 Set to which value? e.g. 1.2
 Note:
 You can also input for example "+1.1" "-1.1" "*1.1" "/1.1" to add, minus, multiply and divide the occupation numbers by 1.1
 To recover the initial occupation numbers, input "i"
 To generate occupation state for calculating odd electron density, input "odd"
2
 Done!

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
q

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   1.000   1.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 36 Invert phase of some orbitals
3

 Basic information of all orbitals:
 Orb:     1 Ene(au/eV):  -117.263218   -3190.8944 Occ: 0.000000 Type:A+B
 Orb:     2 Ene(au/eV):   -12.054866    -328.0296 Occ: 2.000000 Type:A+B
 Orb:     3 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     4 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     5 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     6 Ene(au/eV):    -1.154432     -31.4137 Occ: 0.000000 Type:A+B
 Orb:     7 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B
 Orb:     8 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B
 Orb:     9 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   1.000   1.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 36 Invert phase of some orbitals

(The values I have typed in all start in the first column.)

Now leave this screen. You may want to safe with 0 the resulting wave function as new.wfn; but it is not necessary. To get back to the main menu simply type -1.

Now that you have modified the wave function, you can again follow the steps from the linked post.

TL;DR:

200 // other functions
5   // RDF
0   // calc
1   // plot
5   // Export to RDF.txt

I'm sure there  is a lot of fancy stuff you can do with that, I used the same gnuplot file to plot the result:

Shameless self-plug:

I have written some startup scripts for Multiwfn, which make some things easier or extend the functionality to run calculations within a queueing system (e.g. QTAIM charges). Find it on GitHub: https://github.com/polyluxus/runMultiwfn.bash


Can we fill potato chips bags with a gas other than nitrogen? I understand that we fill potato chips bag with nitrogen to prevent oxidation. But why do we use nitrogen, instead of neon or hydrogen or something else?
My first guess is that nitrogen is lighter than neon/argon but what about hydrogen or helium?

As Nilay Ghosh said, nitrogen is cheap. Very cheap. Neon is expensive. Argon is cheaper than neon, but considerably more expensive than nitrogen. Helium is also expensive and needs to be used wisely, for important things, e.g., cryogenics. And hydrogen! I can just see the ads: “Buy our chips: they are lighter than air! But avoid open flames and sparks unless you want to be Hindenburged to a crisp (no pun intended)”
Another fill gas to avoid is sulfur hexafluoride. A tennis ball manufacturer once decided to fill tennis balls with sulfur hexafluoride, assuming this would prevent the balls from going flat as a consequence of the high molar mass of sulfur hexafluoride. But the tennis balls exploded on the shelves because air diffused in.
Thankfully, no one has ever tried using nitrous oxide as the fill gas in potato chip bags!


Conversion factor for chemical to physical scale of atomic masses Before the 1960s chemists and physicists used to have different atomic mass tables. The main difference was that the physicists assigned the $^{16}$O as the O=16 from mass spectrometry. Chemists on the other hand were using the ordinary oxygen which is a mixture of $^{16}$O, $^{17}$O and $^{18}$O as O=16. Of course, this confused everyone before carbon-12 was suggested as a compromise as both parties were not willing to budge. The % abundance of $^{16}$O is 99.76, $^{17}$O is 0.04, $^{18}$O is 0.2.

People were able to develop a conversion factor between a chemist's atomic mass and a physicist's atomic mass tables. A book by Guggenheim "Physicochemical Calculations" shows the example how the conversion factor was obtained. He doesn't explain anything clearly.

Could anyone shed some light, how the author is calculating this factor? I assume by $$\frac{O}{^{16}O}$$ the author mean how heavy is ordinary oxygen with respect to true $^{16}$O . I don't know where 1 is coming from and why there is a 2 besides O-18 isotope.


Sigma bond strengths S-S, S-P, P-P I want to know how the following sigma bond strength values are calculated.
Like S-S = 1.0, S-P=1.71 And P-P=3.0.

How these 1.0, 1.71, 3.0 came?

Just to clarify, these values are relative bond strengths, not absolute bond strength values, which would have units of energy, i.e. Joules, kilocalories, or electron volts. Another detail missed in this question is the fact that these values refer to the bond strength of these types of orbitals directly overlapping along the same axis, as shown in the diagram below.

You'll notice that these relative bond strengths corresponds the amount of overlap occurring in the orbitals. This gives you a qualitative description, but to actually calculate the relative overlaps we can use quantum mechanics. Remember that these orbitals are really just the probability distributions electrons and can be solved analytically by squaring the wave function. We have a mathematical description of both S and P orbitals. By integrating over the volume in which they overlap, we can calculate the relative overlap between S-S, S-P, P-P, and any other orbital overlaps we might be interested in.


What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state? When an electron in a higher energy state jumps to a lower energy state in Bohr’s model of hydrogen atom, it emits a photon which has energy equal to the difference in the energy between the final and initial energy states.
$$h\nu = E_f - E_i$$
where $\nu$ = frequency of photon emitted

Now what I know is each photon emitted corresponds to an emission line in the spectrum.

So, an electron can make the make the transition from a higher energy state to any lower energy state provided it emits a photon given by the above equation. In the question given, if an electron is in higher energy state; the maximum number of spectral lines is when the electron jumps to the next lowest energy state and so on as then photons will be emitted during each transition.

If an electron moves from $n = 6$ to $5$ we get a spectral line. Then from $n = 5$ to $4$ we get one and so on giving us a maximum of $5$ spectral lines. Now if the electron made the transition directly from $n = 6$ to $n = 4$ or any other lower energy state we would get less than $5$ spectral lines.

However, on looking up answer to the question it was given $15$. How can it produce $15$ spectral lines? When an electron from $n = 6$ jumps it has $5$ options and example it makes the jump to $n = 3$ then it has only two options $n = 2$ or $n = 1$. It can’t go back to $n = 6$ unless energy is supplied and it absorbs energy. So how can more than 5 lines be observed in the emission spectrum?

I am missing a key concept in such problems. Can someone tell me what I am missing?

Since comments caused certain level of confusion, I guess I'll try to provide a further illustration.
You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$.
Electron doesn't get stuck forever on any of the levels with $n > 1$.

Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms.
In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$.
The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this.

I put together a rough drawing in Inkscape to illustrate all possible transitions*:

I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots).
To determine $N$, you need to sum the states, as Soumik Das rightfully commented:

$$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$

For $n = 6$:

$$N = \frac{6(6-1)}{2} = 15$$

Obviously the same result is obtained by taking the sum directly.

* Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$.


Potassium citrate in water I take potassium as a supplement in the form of potassium citrate. I dissolve a small amount in tap water and I drink it. Usually, I need to stir a lot, and sometimes I warm the water a little to make it easier to dissolve the potassium.

Recently, I bought some sparkling water (San Pellegrino, but I don't think that makes a difference) and when I put the potassium in it, it reacted making it bubble a little and it dissolved immediately.

Why is that? Did the potassium react with the CO2 or something else in the sparkling water?

Bubbles need nucleation points to form.
The growth of bubbles is limited by the surface tension of the liquid. The surface tension acts stronger on smaller bubbles and less strong on bigger bubbles.
Adding a solid salt to sparkling water means adding a lot of new nucleation points, which subsequently will lead to bubbling.
The small bubbles will combine to bigger bubbles, these bigger bubbles are not so much effected by surface tension and so even more gas is produced.
In other words, adding the salt will trigger a kind of avalanche effect, which weakens either if the dissolved CO2 gets towards used up or the salt gets dissolved completely.

It is not an reaction of dissolved potassium ions with the CO2, nor is it a reaction of dissolved citrate ions with the CO2.
What happens here is a physical reaction and not an chemical reaction.


Oxygen dissolve in water I know that $\ce{CO_2}$ and $\ce{O_2}$ are slightly soluble in water ($\ce{H_2O}$)

In my understanding, when carbon dioxide dissolves in water, it will react to form carbonic acid just like the equation below.

$\ce{H2O + CO2 -> H2CO3}$

So, my problem now is, I can't find the equation for oxygen.
 Will they form hydroxide ions or something else?

Carbon dioxide is not stable in water in its original form, so it undergoes an acid-base reaction to give the carbon dioxide.

On the other hand, oxygen is stable and soluble in water in its $\ce{O_2}$ form, so it will stay like that, no reaction other than solubilization. Besides, it could react with organic matter or other compounds in water, oxidizing them, and getting reduced to $\ce{OH^-}$


Geometries of methyl and silyl radicals The $\ce{.CH3}$ radical is planar but the $\ce{.SiH3}$ radical is a trigonal pyramid. Why are they different? After all, $\ce{C}$ and $\ce{Si}$ are in the same group.

As is currently being discussed in this question, I'm not convinced that the methyl radical is planar, but even if not planar, it is certainly very close to planarity.  It is also certainly less pyramidal than the $\ce{.SiH3}$ radical as you state.

The $\ce{.SiH3}$ radical is more pyramidal than the $\ce{.CH3}$ radical due to differences in electronegativity between $\ce{Si}$ (1.9) and $\ce{C}$ (2.5) and hydrogen (2.1) and how these differences play into Bent's rule.

In a nutshell, Bent's rule states that, "atomic s character concentrates in orbitals directed toward electropositive substituents" (ref).

Since hydrogen is more electronegative than silicon it will draw more of the electron denisty in a $\ce{Si-H}$ bond towards its end of the bond (the reverse of what happens in the case of a $\ce{C-H}$ bond where the carbon is more electronegative than hydrogen) decreasing the electron density in the silicon atomic orbitals involved in the bond.

If there is less electron density in the silicon atomic orbital used in the $\ce{Si-H}$ bond, then according to Bent's rule, the silicon atom will hybridize so that less s-character is used in these atomic orbitals that now have a lower electron density, and will instead use that s-character in the orbital containing the single electron and stabilize it.  Increasing the p-character in the $\ce{Si-H}$ bonds and increasing the s-character in the orbital holding the single electron will tend to push the molecule towards an $\ce{sp^3}$-like, pyramidal geometry.

The reverse is true in the case of carbon.  Due to electronegativities, there will be more electron density towards the carbon side of the $\ce{C-H}$ bond (compared to the $\ce{Si-H}$ bond).  Due to this increased electron density around carbon, it is energetically less favorable to remove s-character from the carbon atomic orbitals involved in the $\ce{C-H}$ bonds (and destabilize this increased electron density) and transfer it to the orbital containing the single electron.  As a result the carbon atomic orbitals involved in the $\ce{C-H}$ bonds will have higher s-character compared to the analogous silicon case and $\ce{.CH3}$ will be less pyramidal.


Determining covalent bond strengths without hybridisation Assuming pure $\ce{2s}$ and $\ce{2p}$ orbitals of carbon are used in forming $\ce{CH4}$ molecule, which of the following statements is false? (single choice question)

  Three $\ce{C-H}$ bonds will be at right angles
  One $\ce{C-H}$ bond will be weaker than the other three $\ce{C-H}$ bonds
  The shape of the molecule will be tetrahedral
  The angle of $\ce{C-H}$ bond formed by $\ce{s-s}$ overlapping will be uncertain with respect to other three bonds.

According to me, the second option is false because $\ce{1s-2s}$ overlapping is stronger than $\ce{1s-2p}$. I just randomly found this question and I don't know the correct answer.

As I already stated, asking this question is wrong on so many levels, but let's just have a look anyway.

We are assuming a set of $3$ rigid $\ce{2p}$ and $1$ rigid $\ce{2s}$ orbitals at carbon. With rigid I mean, hybridisationnote is forbidden - which is wrong, $\text{[very]}_n$ wrong.

As a consequence, a head on overlap between the hydrogen $\ce{s}$ and the carbon $\ce{p}$ results in three bonds, each perpendicular , i.e. $90^\circ$, against each other. This is the feature of the degenerate $\ce{p}$ orbitals. (Answer 1 would be correct.)

As a conclusion from that, the molecule cannot adapt a tetrahedral geometry, because with this geometric arrangement, it is impossible to form a $S_4$ symmetry axis, which is the main feature of the $T_\ce{d}$ or $T$ point groups. Answer 3 would be wrong.

The bond between the carbon $\ce{s}$ and hydrogen $\ce{s}$ also needs to be weaker than the others, since these orbitals are totally symmetric. That means there is a good portion of the carbon $\ce{s}$ orbital already blocked by the other hydrogens. On the other hand, the orbital overlap of the carbon $\ce{p}$ and the hydrogen $\ce{s}$ is quite optimal, since the main contribution of $\ce{p}$ orbitals lie outside of the carbon $\ce{s}$ orbital. (Answer 2 would be correct.)

This $\ce{s-s}$ bond would also be quite wobbly, because neither of the orbitals is directed. Therefore the position of the last hydrogen can easily change without effecting the orbital overlap, and in first approximation the bond strength. (Answer 4 would be correct.)

As a last comment, the molecule in this conformation is not stable, probably not even as an highly excited state. The kinds of thought games actually ruin the more important thing of understanding how orbitals behave. Orbitals are not rigid. They are quite flexible in the end.

Note:
Hybridisation in the generic sense means the linear combination of atomic orbitals of one atom, that contribute to a molecular orbital.


Spectrometer vs Spectrometry vs Spectroscopy How do you differentiate between the three?

I read that "Spectrometry deals with the measurement of a specific spectrum. There are four primary types of spectrometers:

Mass spectrometry  2. Neutron triple axis spectrometry 3. Ion-mobility spectrometry 4. Rutherford backscattering spectrometry"

How about Raman Spectrometers and Infrared Spectrometers. Don't they belong to spectrometry?

Some believe that the main difference is that spectroscopy typically deals with light, mass spectrometry does not.

So do you called it Raman spectroscopy or it is not right to refer it as Raman spectrometry?

It's confusing.

Long time ago, I wrote mass spectroscopy, by mistake, in an undergraduate exam and the instructor told me "You were the only one in my class who wrote mass spectroscopy." The explanation was that MS is mass spectrometry because it does not deal with light or electromagnetic radiation and that's what the textbook said. At that time internet was not that common. Now I searched the term "mass spectroscopy" in papers using Google Scholar and there are 1,100,000 results with this exact phrase. There are 2,330,000 results for mass spectrometry.

It seems, given the usage, and the difference is not that strict. Consulting the controller of all  chemical terminologies, IUPAC says

https://goldbook.iupac.org/terms/view/S05848

  The study of physical systems by the electromagnetic radiation with
  which they interact or that thay [spelling mistake] produce. Spectrometry is the
  measurement of such radiations as a means of obtaining information
  about the systems and their components. In certain types of optical
  spectroscopy, the radiation originates from an external source and is
  modified by the system, whereas in other types, the radiation
  originates within the system itself.

You will see famous books, especially by Griffiths with the title "Fourier Transform Infrared Spectrometry" and nobody raised an eyebrow.


What are the roles of pyridine and DCM in the acylation of an alcohol? What role does pyridine and $\ce{CH2Cl2}$ play in this reaction?

(Prior to this step, the corresponding carboxylic acid was converted to the acyl chloride using DMF, oxalyl chloride, and DCM.)

This reaction is taken from Bear, R. B.; Shea, K. J. Tandem Hetero Diels−Alder Reaction: Synthesis of Oxygenated Macrocycles. Org. Lett. 2001, 3 (5), 723–726. DOI:10.1021/ol007040s..

Dichloromethane (DCM) is a common organic solvent and this is its role in both reactions (and in most cases where it is used). The first reaction is indeed the conversion of a carboxylic acid to an acyl chloride. The DMF is a catalyst and the reactions proceeds through an imidoyl chloride intermediate, $\ce{Me2N+=CHCl}$.

Pyridine is a weak base, with a $\mathrm{p}K_\mathrm{aH}$ of 5.21 and it is often used to remove extra protons during the course of a reaction. It is also a good nucleophile for attacking carbonyl groups to activate them towards attack by other, less good nucleophiles (thanks to @orthocresol and @DGS for pointing this out). In this reaction it serves both these purposes.

An acid anhydride can also be used as the acylating agent. The principle is the same: pyridine acts as a nucleophilic catalyst and generates the same intermediate as above.


Calculating atomic mass of unknown element in given compound (gas) at STP A gas $\ce{X2O5}$ has the density of $\pu{5 g L-1}$ at STP. What is the atomic mass of $\ce{X}$ given $A_\mathrm{r}(\ce{O}) = 16?$

Note: The problem is my homework for school and has been translated from another language. It is probably either missing something or it contains wrong data. Just leaving it as a cautionary tale to anyone who might also have it.
I solved, or at least tried to, by expressing the given density $\rho = \pu{5 g L^-1}$ as
$$\rho = \frac{m}{V}.\tag{1}$$
Also,
$$m = Mn\tag{2}$$
$$V = V_\mathrm{m}n,\tag{3}$$
where $n$ is the amount of substance and $V_\mathrm{m} = \pu{22.4 L mol^-1}.$ So:
$$\rho = \frac{m}{V} = \frac{Mn}{V_\mathrm{m}n} = \frac{M}{V_\mathrm{m}}\tag{4}$$
$$\pu{5 g L^-1} = \frac{M}{\pu{22.4 L mol^-1}} \quad\Rightarrow\quad M = \pu{112 g mol^-1}\tag{5}$$
Then obviously by subtracting the atomic mass of oxygen:
$$M_\mathrm{r} = 5A\mathrm{r}(\ce{O}) + 2A_\mathrm{r}(\ce{X}) \quad\Rightarrow\quad A_\mathrm{r}(\ce{X}) = 16\tag{6}$$
But that is not possible, as $16$ is the atomic mass of oxygen and there is no such thing as $\ce{O2O5}$ as far as I know. What am I doing wrong? Or is the problem wrong altogether?

Assuming it is given that $\ce{X2O5}$ behaves like an ideal gas, the best approach to do this problem is manipulating $pv = nRT$ to get the equation, which fits the given data. Suppose $m$ is the mass of the gas, $M$ is the molar mass of the gas, and $\rho$ is the density of the gas. Then $n =  \frac{m}{M}$:
$$ pv = nRT = \frac{m}{M}RT \ \Rightarrow \ M =  \frac{m}{v} \ \cdot \frac{RT}{p} = \rho \frac{RT}{p} \tag1$$
Now, you can apply $\rho = \pu{5 g L-1}$ at STP to the equation $(1)$ to find $M$. Then apply $2M_\ce{X} + 16 \times 5 = M$ to find $M_\ce{X}$, which is the atomic mass of $\ce{X}$.


Which ligands would be exchanged in this reaction? I have this reaction and need to state what the products would be:

$\ce{(CF3)2PH + [W(CO)5(THF)] -> }$

THF stands for Tetrahydrofuran.

Would this just be a simple ligand exchange, and if so which ones would be exchanged? How would you know?

Complexes like $\ce{M(CO)5(solv)}$ are typically generated by photolysis ($\lambda$ = 350 nm) of $\ce{M(CO)6}$ in the presence of a non-inert solvent. This proceeds through a highly reactive "naked" $\ce{M(CO)5}$ intermediate.

$$\ce{W(CO)6 + solv ->[h\nu] W(CO)5(solv) + CO}$$

When offered another ligand $\mathrm{L}$, the weakest bond  ligand in the $\ce{W(CO)5(solv)}$ complex is replaced. This is THF.
Taking the lone pair on the phosphorous atom into account, is is conceivable that $\ce{(CF3)2PH}$ will do the job:

$$\ce{W(CO)5(thf) + (CF3)2PH -> W(CO)5((CF3)2PH) + THF}$$


Why are amidines so much more basic than amides? Both delocalize the added proton. Is it because the bond lengths in amidines are the same, or perhaps because oxygen is more electronegative than nitrogen and thus is more reluctant at keeping the proton? But how can that explane the huge difference in basicity between amidines and amides (resp $12$/$13$ versus $-1$/$0$ of $\mathrm{p}K_\mathrm{aH}$)?

Here is a diagram comparing what happens when we protonate an amide and an amidine.

Let's compare the protonated products in the two reactions.  In the case of the amidine we can draw two important (the fact that they are equivalent resonance structures is why they both are important) resonance structures to describe the protonated product.  In the case of the protonated amide the resonance structure with the double bond to oxygen places a positive charge on oxygen.  Therefor this resonance structure is a less significant contributor than the corresponding structure for the protonated amidine where the positive charge resides on the less electronegative nitrogen.  Since we can draw more significant resonance structures for the amidine case, the protonated amidine is more stable than the protonated amide.  This difference in protonated product stabilities explains why amidines are more basic than amides.


Reduction of a ketone to an alkane What is the mechanism for the reduction of ketone to alkanes using $\ce{H2}$, $\ce{Pd/C}$ in ethanol? Searching on Google mostly yields the reduction of ketones to secondary alcohols using LAH.

The reduction of a ketone all the way down to an alkane using catalytic hydrogenation over palladium on charcoal is not a general/widely used process.

Quoting from Comprehensive Organic Functional Group Transformations III:

  ...while the reduction of benzylic ketones by hydrogenation over palladium is often an efficient procedure , it ... sees limited use. An isolated report of the reduction of aliphatic ketones by hydrogenation over platinum on Montmorillonite K-10 merits ...attention here and further study

The majority of examples of this reaction occur using benzylic ketones, carrying out the reaction in the presence of aqueous acid in order to facilitate the loss of the hydroxyl caused upon initial reduction of the ketone. The adjacent aromatic ring helps stabilise the developing (+) charge as the C-O bond cleaves, which is what makes benzylic ketones good substrates for this reaction.

  Ref: J. Med. Chem. 2001, 44, 3424

In the context of lab scale chemistry (i.e. not industrial processes), several good methods exist including the Wolff–Kishner (or variants as Jan pointed out in the comments), the Clemmensen, and the Mozingo reduction. Some success is also noted via initial reduction, mesyaltion, and treatment with a hydride source.


Can the enrichment of Uranium be done via manipulating the boiling/melting point difference between differing isotopes of the same element? I have a Physics question for which I could not find the answer on the Internet, so I would like to see if y'all know the answer to it.
I have found that heavy water will have a different boiling point compared to light water, and this trend is present between various isotopes of an element. Based on this research, I have hypothesized that Uranium can be enriched using the procedure below.

Make UF6 (s)

Sublime it in a glass tube

Lower temperature with such precision, that one isotope of Uranium still remains in the gaseous form while another isotope has turned back into a solid.

Separate the gas and solid without disrupting the fragile temperature balance.

Although this link is for Germanium isotopes, it states that there is a 0.15 Celcius change of melting points between isotopes. https://www.sciencedirect.com/science/article/abs/pii/S0921452618300863
So, here are my questions.
Would this apply similarly to Uranium, and how would the use of UF6 affect this boiling point difference?
How exactly does the boiling/melting point change between differing isotopes? And is there a particular formula for finding the melting point for a particular isotope? (I assume it would take in the number of neutrons and some other info)
In addition, is it possible to adjust the temperature of the entire system with decimal grade precision?
Furthermore, would the above setup work at all?
Also, if UF6 is too sensitive, would melting Uranium metal in a kiln with decimal precision temperature adjustability work?
Finally, is there a good, reliable source for melting point differences between isotopes?
I would like to first say that this is all purely theoretical and that I will take no action upon the information obtained; it is all merely a thought experiment to enhance my understanding of Chemistry and Physics. I mean, it's not like I'm gonna make a nuclear bomb; I'm just a normal HS student in the US. If you still think that this information might be dangerous out here, please PM me so we might work something out, like meeting via Zoom/Skype/Phone or something; I am very flexible about that.
I hope y'all have a great day.

Of course isotopes may be separated by distillation, but it's useful, AFAIR, only for hydrogen isotopes.
For uranium, gas centrifuge is proven much cheaper, to some orders of magnitude: https://ru.wikipedia.org/wiki/Газовая_центрифуга
(translate this page, it contains more information than English Wikipedia article).
The essence of this technology is known for decades, and included in basic chemistry textbooks. Only engineering/technological know-hows are sort of classified.
But in the politics, mass media, and perhaps on stackexchange, the whole topic is still considered taboo and is surrounded by fear.