Periodic boundary conditions in DFT for cylindrical system As I understand it, a DFT calculation uses periodic boundary conditions, whenever you want to simulate a bulk material. This is convenient because all information can then be extracted from the fundamental unit cell.

Currently I am faced with the problem of calculating the band structure of a carbon nanotube, which has a cylindrical geometry, i.e. only bulk in only one direction, say z, and circular in the x and y directions. As my unit cell I am using a ring of C-atoms surrounded by vacuum and I then employ periodic in the z-direction. Obviously, for large diameters, this approach involves a rather large unit cell, which makes the calculation slow.

My question is: Is there an alternative way to do this, which exploits the cylindrical symmetry of the system? Specifically what I have in mind is to also apply periodic boundary conditions in the other directions, but where the Schrödinger equation is expressed in cylindrical coordinates with the middle of the nanotube as the origin.

Another, less important, question is also: Do you ever NOT use periodic boundary conditions in DFT? For example, when you do a calculation for an atom, it always seems like you have to add vacuum around it, which is essentially, because you are applying pbc, and do not want it to interact with other unit cells. Is there another way to solve this problem, where you assume the density to vanish far away?

Let's dissect this:

  As I understand it, a DFT calculation uses periodic boundary conditions, whenever you want to simulate a bulk material. This is convenient because all information can then be extracted from the fundamental unit cell.

One important point is that you employ periodic boundary conditions (PBC) only when you are treating a periodic system, i.e. something with translational symmetry. And in those cases you use plane waves as a "basis set".

In the ordinary molecular case you use atom centred basis functions, which fulfil the basic requirements of a wave function (in short): orthonormality, continuous, continuous differentiable, vanishing at infinity.
Therefore a quantum chemical calculation electronic energy is always in vacuum at "0 K".

  Currently I am faced with the problem of calculating the band structure of a carbon nanotube, which has a cylindrical geometry, i.e. only bulk in only one direction, say z, and circular in the x and y directions. As my unit cell I am using a ring of C-atoms surrounded by vacuum and I then employ periodic in the z-direction. Obviously, for large diameters, this approach involves a rather large unit cell, which makes the calculation slow.

This approach actually only makes sense if you are treating a very, very long nanotube. I'd believe a molecular approach using appropriate symmetry, Dnh would be far superior.

  My question is: Is there an alternative way to do this, which exploits the cylindrical symmetry of the system? Specifically what I have in mind is to also apply periodic boundary conditions in the other directions, but where the Schrödinger equation is expressed in cylindrical coordinates with the middle of the nanotube as the origin.

A nanotube is discrete in two directions, i.e. the wave function has to vanish in these directions, egro apart from applying vacuum as PBC, there is no other choice in this framework.

An alternative approach is to treat your nanotube as a discrete molecule and cap it off at the end with hydrogen, then increase the length of the tube until you have sufficient convergence, obviously using Gaussian or Slater type (atom centred) orbitals.

  Another, less important, question is also: Do you ever NOT use periodic boundary conditions in DFT? For example, when you do a calculation for an atom, it always seems like you have to add vacuum around it, which is essentially, because you are applying pbc, and do not want it to interact with other unit cells. Is there another way to solve this problem, where you assume the density to vanish far away?

For discrete molecules: always. For periodic systems: only sometimes not (cluster approach).
Prior to the golden 90s, almost every calculation was a molecular calculation. In the early stages of quantum theory nobody ever dared to dream that we would eventually able to treat a huge stone quantum chemically (a priori) with a computer.

A nanotube is typically right in the middle, between a bulk material and a molecule, and you should either rely on the experience of others, maybe you have an advisor, through the literature, or built it yourself.
There are quite a few groups (I suppose) who work with nanotubes, and some of them have surely tried computations. Do a thorough literature review and apply common practices.

I fear that you are getting your knowledge about DFT calculations from working with a calculation suite designed for bulk materials. You should probably brush up on some of the fundamental theory. I recommend Frank Jensen's book for a rough overview, and Attila Szabo & Neil Ostlund's book for some better maths  (See resources for learning chemistry for more details).
A fairly comprehensive "review" of DFT can be found here: DFT Functional Selection Criteria and in the links within.


Which coordination complex does not absorb visible light? A question was asked in one competitive exam (AIPMT 2010). It said:

  Which of the following complex ions is not expected to absorb visible light?

  $\ce{[Ni(CN)4]^2-}$
  $\ce{[Cr(NH3)6]^3+}$
  $\ce{[Fe(H2O)6]^2+}$
  $\ce{[Ni(H2O)6]^2+}$

The exam is notorious for not releasing solutions of their own question papers, hence, several online sites have come up with their own versions of explanations for why the official answer is option A - $\ce{[Ni(CN)4]^2-}$, most of them copying the solution from one another!

I understand CFT and VBT (not LFT!) in terms of coordination theory. I ask this question on this community to get a definite response, as to what should be the correct answer to this question and why?

Absorption of visual light is associated with an energy difference between two orbitals — one occupied, one unoccupied — and electrons must be able to be excited from one to the other.

In coordination complexes, these excitations typically happen within the metal’s d subshell, so it is usually sufficient to examine that and approximately determine which excitations are possible. The main selection rules are:

the spin rule. The electron must be excitable without a spin flip
the Laporte rule. Basically, d to d transitions are forbidden in octahedral complexes

The spin rule is very strongly observed. The colour of manganese(II) whose transition is spin-forbidden is extremely faint. The Laporte rule only holds true as long as the complex is inversion-symmetric, so any asymmetric vibration is enough to make it void; thus, Laporte-forbidden transitions are typically still visible but somewhat faint.

Let’s examine the complexes:

$\ce{[Ni(CN)4]^2-}$ This is expected to be square planar and $\mathrm{d^8}$. The energy difference between the two highest orbitals — $\mathrm d_{xy}$ and $\mathrm d_{x^2 - y^2}$ — is expected to be high. The former is expected to be fully populated, the latter to be unpopulated.
$\ce{[Cr(NH3)6]^3+}$ This is a $\mathrm{d^3}$ system. It is expected to be octahedral with a standard difference between the lower and higher energy levels.
$\ce{[Fe(H2O)6]^2+}$ This is a $\mathrm d^6$ octahedral system. There is no reason to assume a low spin state. The energy difference is expected to be slightly less than in the previous case.
$\ce{[Ni(H2O)6]^2+}$ this is expected to be a high-spin $\mathrm d^8$ system and octahedral. The same expectation regarding energy levels as on the previous case applies.

We realise that the of our complexes are average high spin octahedral complexes. For these, visible light absorption is always expected. Only one case is different. In that different case, the HOMO-LUMO difference is large. While we can still expect absorption, it seems most reasonable to assign this absorption band an ultraviolet wavelength.

Thus, $\ce{[Ni(CN)4]^2-}$ is the answer.


How to calculate mass produced when a given gas is burned? The equation below represents combustion of methane ($\ce{CH4}$, $\pu{16.04 g/mol}$). Balance the equation, and calculate the mass of water ($\pu{18.02 g/mol}$) formed when $\pu{40.0g}$ of methane is burned.
  $$\ce{CH4(g) + O2(g) -> CO2 + H2O(g)}$$

I have started the problem by balancing:

$$\ce{1CH4(g) + 2O2(g) -> 1CO2 + 2H2O}$$

I then continued by calculating the amount of substance of methane:

$$\ce{C ($\pu{12g/mol}$) + H ($\pu{4 \times 1g/mol}$) = $\pu{16 g/mol}$ \implies ($\pu{40 g}$ methane) / ($\pu{16 g/mol}$) = $\pu{2.5 mol}$}$$

From here I wold like to know how to convert or get to amount of substance of $\ce{H2O}$ to get the mass of $\ce{H2O}$ produced in this reaction.

You have already balanced the equation, so we can get the stoichiometric  coefficients. From the equation we can say that 1 mole $\ce{CH4}$ gives 2 moles of $\ce{H2O}$ because their coefficients are 1 and 2, correspondigly. So 2.5 moles of $\ce{CH4}$ would give 5 moles of $\ce{H2O}$, that is $\pu{90 g}$ of $\ce{H2O}$.


What are better acids for removing rust from an iron nail than phosphoric, or oxalic acid? I'm planning to conduct an experiment in which I test how fast rusting occurs with changes to the concentration of NaCl in the solution.

I want to measure that by finding the change in mass of an iron nail over time, but in order to do that I need to clean off all the rust on the nail before weighing. I have a relatively short time to conduct the experiments.

I've heard acids like phosphoric acid, oxalic acid, or citric acid will work. I'm looking for the best acid to separate the rust from the nail preferably without using steel wool or any abrasive method. I can, however, spray down the nail with distilled water if necessary. Any suggestions?

In industry, rust is cleaned using hydrochloric acid, because besides being an acid, it is also a reducing acid.


Arrange in the order of their reactivity of SN2 reaction Here in case of (1) it is more favorable towards SN1 reaction as it will create benzyl carbocation which is very stable.
(2) Here it is more favorable towards SN1 as it is first of all allyl carbocation and in resonance with the benzene and the double bond.
(3) It is allyl carbocation, so it is less favorable to SN1 than (1) and more favorable to SN2.
(4) It is the most favorable towards SN2 reaction.
Hence the order of reactivity towards SN2 reaction should be
(4) > (3) > (1) > (2)
But the answer is
(2) > (1) > (3) > (4)
Why?

The order you've given is absolutely correct with respect to SN2, I believe, the question needs to be authenticated once again, as the order for SN1 would be the exact reverse of this, which is as given in the answer to your question.


Basicity Alteration Due To Ortho Effect In Methoxy Aniline System I wanted to ask something more in reference to this question.

Suppose instead of toluidine, 2-methoxyaniline and 3-methoxyaniline are present. Now, why is 2-methoxyaniline more basic than 3-methoxyaniline (even though 2-methoxyaniline has reduction of basicity due to ortho-effect but 3-methoxyaniline does not)?
Also what would be the correct order of basicity if nitro group was present in place of methoxy group? Would ortho-effect still take place?

Take home message: Just because you have an ortho substituent does not mean you will have an ortho effect.
To have an ortho effect two conditions must be met:

you need an ortho substituent
and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent.

One measure of steric bulk is the cyclohexane $A$-Value, a table of which can be found here.  From the table we see that methoxy $(A=0.6)$ and nitro $(A=1.1)$ groups have a smaller steric size than that of a methyl $(A=1.7)$ group.  Methoxy and nitro groups generally do not produce an ortho effect.
Therefore, the basicities of the various methoxy anilines (the anisidines) are controlled by a combination of resonance and inductive effects.
\begin{array}{l|r} \hline
\text{Substituent} & \mathrm{p}K_\mathrm{b} \\ \hline
\ce{-H} & 9.13 \\
\ce{o-methoxy} & 9.5 \\
\ce{m-methoxy} & 9.8  \\
\ce{p-methoxy} & 8.7  \\ \hline
\end{array}
(data source)
The basicity of p-methoxy aniline is controlled primarily by resonance effects (too far away for inductive effects to be significant).  Drawing resonance structures we see that the p-methoxy group is pushing electron density into the ring, this tends to keep the nitrogen lone pair more localized on nitrogen (if we also delocalize the nitrogen lone pair into the ring like charges repel leading to destabilization). This results in p-methoxy aniline being more basic than aniline.
The basicity of m-methoxy aniline is controlled purely by inductive effects (resonance is not possible from this position).  The inductively electron withdrawing methoxy group removes electron density from nitrogen, making it less basic than aniline.
The basicity of o-methoxy aniline is controlled by a mix of resonance and inductive effects operating in opposite directions.  Apparently the inductive effect is stronger and wins out (this shouldn't be surprising, note that the inductive effect from the meta position had a 0.7 unit effect on the basicity, and now in the ortho position it will be even stronger, while the resonance effect in the para isomer only changed the basicity by 0.4 units) making o-methoxy aniline less basic than aniline.


Change in internal energy for isothermal process A gas expands isothermally against a constant external pressure of $1\ \mathrm{atm}$ from a volume of $10\ \mathrm{dm^3}$ to a volume of $20\ \mathrm{dm^3}$. It absorbs $800\ \mathrm J$ of thermal energy from its surrounding.
The change in internal energy is:

I came across this question and my doubt is, for isothermal process, $\Delta U=0$ ( change in internal energy) for ideal gas is zero right? Then how can we calculate $\Delta U=0$ here, assuming the gas to be ideal?
Is the error with the question or am I missing something?

It must not be an ideal gas.  This is irrespective or whether the process is reversible or irreversible.   That is the only possible conclusion.  Otherwise the amount of heat received would adjust  until it was equal to the work done.


Stability of organophosphorous vs organophosphates? In environmental chemistry, most of the organophosphorous compounds we worry about are actually organophosphate compounds.

Is there something about the $\ce{C-P}$ bond that makes it less stable (and thus less persistent in the environment) than a $\ce{C-O-P}$ bond?

Setting the scene for your question

In Environmental Chemistry there are several pathways a chemical might travel through including:

being subjected to the biochemistry of an animal
photodegradation
chemical degradation (perhaps it slowly reacts with water, etc.)
uptake as a metabolite in plants
degradation by soil microorganisms

and this is ignoring aquatic and air pathways.

Say, within a mouse or a fly, a possible pathway for degradation for some insecticide is cleavage of the P-O bond. How they have determined this in the past is noteworthy; a synthetic chemist would isotope label an atom in the molecule, say $^{33}$P-labeled or $^{14}$C-labeled, and he purified it before finally determining its specific activity. Then a biochemist would dose a mouse with a solution of the compound dissolved in an appropriate solvent, such as olive oil. Urine and feces were collected after administration. An aliquot of the collected urine, for example, would have its radioactivity measured using scintillation counting. The organo-soluble species would be extracted from the rest of the urine and the distribution of radioactivity could then be determined between the aqueous and organic layers. Then one could investigate what were present by TLC if you're old school -- they would actually run, via cochromatography, all the possible metabolite products they could independently think of and synthesize to identify a TLC spot! (That is a lot of work.) These days we can cheat with HPLC-MS. So, you might ask, "are results from this generalizable?" I have to tell you that they generalize in a very rough way. Enzymes are responsible for all the nifty biochemistry and they tend to be rather selective towards a given substrate and, in many cases, chirality even matters! What they are picky about is a tough problem that is tailored to each individual enzyme. One can make all kinds of generalizations about the chemical nature of the bonds and some possible predictions from it, but those won't necessarily hold up in an animal model with a different molecule.

So nitty-gritty mechanisms might not be what we want to look at. Instead we'll step back and take a broader view of this landscape. We need to look at bigger items in terms of physical properties, namely:

1) Persistence: what is its half-life in soil, etc. This is a matter of degradation, whereas the next is all about movement of an intact molecule. A simple chemical way to test this is how they resist hydrolysis.

2) Mobility: over time, does the compound tend to sit or spread itself around?
For example, factors that influence this are:

sorption: how much does our chemical bind to our soil?
water solubility: how easy it is to move in water for our chemical?
vaporizability: how likely is the material to evaporate away from its source?

A famous organophosphate is malathion. The partition coefficient, according to the wiki page, is 2.36. This implies its solubility is greater in organics than water, but some of it will most certainly dissolve in water. I think, based on its size, it is reasonable to assume it has a negligible vapor pressure too. These factors suggest it will have a low mobility: it will tend to stick to organics, won't appreciably evaporate away and will take quite a lot of water to wash it away. Contrast this with a famous organochlorine, DDT, which has an even higher partition coefficient (on the order of 10, so even more sticky and less soluble in water) and an even higher mass (so probably a lower vapor pressure) and thus an even lower mobility. Indeed, organophosphates largely replaced organochlorines as pesticides, in part, because they persist for shorter periods of times in the environment.

Many of these organophosphates were tested by us and their history is relevant. World Wars I and II ushered in a new era of chemical warfare agents and pesticides, notably of the classes of organochlorines, organophosphates and carbamates. Some organophosphates hardly persist at all, such as tetraethyl pyrophosphate, but they are pretty toxic to mammals. Many other organophosphates were screened for low toxicity towards mammals and increased persistence. We actually sought longer-lasting half-lives for these compounds! It turns out that we see alkyl phosphates with shorter lives than the longer-lived aryl species.

We tend to see certain trends within the different organophosphates, such as the water solubility being greater for the less complex and more polar compounds. An ethyl- is less soluble than a methyl- derivative and an aryl-substituted species get partition coefficients approaching those of the organochlorines.

"All this is fine and interesting, but it doesn't have much to do with my direct question," you might say. And most of the above just sets the scene for my answer to your question. Your specific query, I believe, requires us to go back to a nitty-gritty detail which is really only relevant in the case of your comparison between a phosphate (P-O-C) and an organophosphorus compound (P-C).

My answer to your question

In the biochemistry of most life, phosphates are pretty ubiquitous. As a result, many enzymes have evolved to manipulate them in many different forms; one class of these enzymes is called phosphatases and they assist dephosphorylation reactions. They certainly cannot manipulate just any ol' phosphate (hence why more complex phosphates have longer residues). We could say organophosphates are more "natural" than your typical organophosphorus compound. There is, therefore, a greater chance of having a tool -- an enzyme -- that can pick your more "natural" looking compound apart than your harder to find P-C bond.

As you've indicated, these phosphates are actually pretty stable. It's why DNA is built using them! But when you compare the bond lengths between P-C and P-O and O-C, there's a decent chance of finding that a P-C bond length is longer than a P-O or O-C. This leads me to believe that there are many cases in which P-O and O-C are actually stronger than P-C! The same could easily be said in the other way too! This, I think, is why we have to step-back and look at the bigger picture: the exact physical properties vary considerably from one compound to another and paint a landscape that says the issue is complicated. Perhaps, instead of it being a mere thermodynamic issue dealing with a comparison of the general bonding energies, it has more to do with the kinetic advantage of having enzymes equipped to slice the P-O-C rather than P-C. Of course, your mileage may vary depending on your specific examples.


What's the difference between perfect and ideal gas? There are all kinds of different dinstinctions in the internet, and I'd like to see what you guys thought.

Actually, I've been asked if the heat capacity of an ideal gas es independent of temperature. I've said no, even though it's practically invariant under small ranges.

Namely the result $C_V/n = \frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter.

Is this true?

What really is an ideal gas? What's its difference with a perfect gas? Is my answer to the question I was posed correct?

Thanks.

An ideal gas is the same as a perfect gas. Just different naming. The usual name for such gases (for which is assumed that the particles that make up the gas have no interaction with each other) is ideal gas, perfect gas is what such a gas is named in Atkins physical chemistry book. Personally I like the perfect gas naming better as it illustrates the perfect nature of the assumptions made about it.

For the simple systems (like monoatomic gases) when we can assume a perfect/ideal gas, $C_{V,m}$ is independent of temperature. For real gases this is certainly not the case.

Note that $C_{V,m}=\frac{3}{2}R$ only holds for monoatomic gases.


Can someone explain 3‐(propan‐2‐yl)cyclohex‐1‐ene forming a carbocation? Can someone explain this one, especially the formation of the secondary carbocation (in the solution)?

Solution:

After the electrophilic attack by $\ce{H+}$ on the π bond, I had expected the carbocation formation on the other side (the left end of the π bond) because of more hyperconjugation thus ruling out the chance of its rearrangement to the tertiary site. I've seen this happen in other questions too.

You are correct that the $\alpha$-carbon would be protonated... and then deprotonated, and back and forth, until eventually, the $\beta$-carbon is protonated, the hydride shift proceeds, and the molecule is trapped in the product state. Since protonating the $\alpha$-carbon doesn't result in some other reaction, it doesn't really matter which secondary cation is more stable. I think the important thing here is recognizing where and when a hydride shift can occur.


Oxidation of aluminum and manufacturing of electrolytic capacitors My background is in electrical engineering. Recently I saw instructions on how to manufacture electrolytic capacitors at home by mixing sodium bicarbonate (baking soda) to distilled water, submerging two pieces of aluminum foil into the mixture and applying electrical current through the pieces until current dropped to almost zero. The instructions I followed are here:

https://rimstar.org/science_electronics_projects/make_electrolytic_capacitor.htm

The end result was satisfactory.

However, as I have only very basic knowledge in chemistry, I have only a very vague understanding of what happened, and I would like to understand it in more detail. The article did not explain it very well and I haven't been able to obtain good explanation elsewhere either.

I understand that the current somehow oxidizes the aluminium to form an insulating aluminium oxide layer on top of another one of the plates.

So I would like to pose the following questions regarding this experiment:

Why is distilled water required? I understand that distilled water is basically tap water except its impurities such as charged ions have been removed. This, I believe, makes the water much less conductive. But don't we need the water to conduct? I believe the water, as the electrolyte, acts here as the other plate of the capacitor, with the oxide layer insulating it from the other plate in the final product.
What is the role of the sodium bicarbonate? What exactly does it do in the chemical reaction? We are using distilled water, but then we add sodium bicarbonate to it.
What is the reaction formula of the oxidation? Could you show me how the water, sodium bicarbonate and the aluminium react together to form aluminium oxide, and what is the role of the electric current? I understand that this oxidation of metals, such as aluminium, happens in air as well, and this method simply "accelerates" this process to produce a thicker, more insulating layer.

I would greatly appreciate if somebody could help me understand this process!

You are performing an electrolysis.
This means that on the (+) electrode de reaction is Al $\rightarrow$ Al$^{3+}$ and on the (-) electrode the reverse occurs.

Distilled water is required to avoid the presence of chloride ions which will form chlorine at the (+) electrode and attack your aluminum foil.
Sodium bicarbonate is used to raise the pH so that the formed aluminum ions precipitate as the oxide instead of dissolving as Al$^{3+}$ ions.
The oxidation is Al $\rightarrow$ Al$^{3+}$. Then, the basic pH due to the dissolution of bicarbonate:

NaHCO$_3$ $\rightarrow$ Na$^+$ + HCO$_3^-$

HCO$_3^-$ + H$_2$O $\rightarrow$ OH$^-$ + H2CO$_3$

the aluminium oxide precipitates:

2 Al$^{3+}$ + 6 OH$^-$ $\rightarrow$ Al$_2$O$_3$ + 3 H$_2$O

The electric current forces the oxidation reaction to continue.
In air a thin layer of oxide is formed and then the rest of the metal is not in contact with the air anymore.
This is called passivation.

Since the aluminium ions are precipitating as oxide, in the (-) electrode the reduction is of sodium ions:

$\hspace{1cm}$Na$^{+}$ $\rightarrow$ Na

which reacts with water to form hydrogen and OH$^-$ so that the cycle is complete:

$\hspace{1cm}$2 Na + 2 H$_2$O $\rightarrow$ 2 Na$^{+}$ + 2 OH$^-$ + H$_2$


Why do we not see silicon in a structure like graphite? Silicon exists in a structure similar to diamond, with 4 silicon atoms bonded to each other in a tetrahedron.
Why do we not see it exist in a structure like graphite?

Is it even possible to get silicon to exist like this? If so could we make graphene but with silicon atoms?
Not sure why you would, since I understand carbon graphene will still be cheaper. But I would just like to know.

It's possible, and it's already been done! The compound is called Silicene.

However, silicene generally isn't perfectly flat. (Then again, as  Karl Ratzsch points out in the comments, graphene isn't perfectly flat either). The Wikipedia article notes that it usually takes a puckered shape.

I don't know the reasons for the puckered shape, but I recall being quite surprised that silicene could be synthesized at all. The reason why is that I was taught in high school that silicon very rarely forms double bonds due to its large size. For a visualization, please refer to the absolutely awful image below (if anyone could provide an actual graphic I'd be grateful).

The black atoms on the left are carbon. The $p$ orbitals overlap to a significant degree, which means that pi-bonds can easily form. The green atoms to the right are silicon. You can see that, because of the larger atomic radius, the p-orbitals don't overlap as much, meaning that double bonds are generally not favorable in silicon. This explains why compounds like silenes are relatively unstable.

Pi systems are crucial to the chemistry of carbon--they allow carbon to form multiple bonds, flat rings (like in graphene), and generally be stable in a lot of configurations. The lack of pi-bonding in silicon is sometimes given as a rationalization for why silicon-based lifeforms can't exist, and explains why it's so hard to get silicon into the flat sheets that form silicene.


Derivation of the Temperature Dependence of the Boltzmann Distribution In lecture notes, Atkin's Physical Chemistry, etc. the Boltzmann Distribution is derived by first claiming that the system takes on properties of its weightiest configuration. The weight is therefore maximised, utilising Lagrange multipliers.

\begin{align}
\nabla \ln{W} - \alpha \ \nabla \sum_i n_i - \beta \ \nabla \sum_i n_i\epsilon_i &= 0
\\
\to \frac{\partial{\ln{W}}}{\partial{n_i}} - \alpha - \beta \epsilon_i &= 0
\end{align}

The Stirling Approximation transforms the derivative to something we can work with, and the Boltzmann distribution is eventually derived.

\begin{align}
\frac{\partial \ln{W}}{\partial n_i} &= -\ln{\frac{n_i}{N}} \\
&\vdots \\
\frac{n_i}{N} &= \frac{e^{-\beta\epsilon_i}}{\displaystyle \sum_i e^{-\beta\epsilon_i}} \hfill
\end{align}

The derivation is left there with no mention of the value of $\beta$, though it is later claimed $\beta = \frac{1}{kT}$. How is the value of $\beta$ ascertained?

What OP cited is the derivation of the Boltzmann distribution starting from the micro-canonical ensemble, characterized by a fixed energy and a fixed number of constituent particles. In mathematical terms, the problem statement is as follows:

  Maximize the function $\ln W(\{n_i\})$ under the constraints $\sum_i n_i \epsilon_i = E$ and $\sum_i n_i = N$.

The solution is a function of $E$ and $N$, so let's define

  $\ln \Omega(E, N)$ := (constrained maximum of $\ln W$ as a function of $E$ and $N$),

  $\tilde{n}_i (E, N)$ := (value of $n_i$ at the constrained maximum).

Notice that $\ln\Omega = \ln W(\{\tilde{n}_i\})$. Also, this quantity is what appears in Boltzmann's entropy formula $S=k\ln\Omega$ because $\Omega$ is the number of available states for given values of $E$ and $N$. This observation enables us to relate $\ln\Omega$ to the thermodynamic temperature:
\begin{equation}
\frac{\partial \ln \Omega}{\partial E} = \frac{1}{k}\frac{\partial S}{\partial E} = \frac{1}{kT}.
\end{equation}
Now, OP's question can be rephrased as follows:

  When deriving the Boltzmann distribution starting from the micro-canonical ensemble, $\beta$ is introduced as a Lagrange multiplier associated with the constraint $\sum_i n_i \epsilon_i = E$. How can we relate $\beta$ to the thermodynamic temperature, i.e., establish that $\beta = \frac{\partial \ln \Omega}{\partial E}$?

The underlying principle is really the chain rule in multivariate calculus:
\begin{equation}
\begin{split}
\frac{\partial \ln \Omega}{\partial E} &= \sum_i \frac{\partial \ln W}{\partial n_i}\bigg|_{n_i = \tilde{n}_i} \frac{\partial \tilde{n}_i}{\partial E}\\
&= \sum_i (\alpha + \beta\epsilon_i)\frac{\partial \tilde{n}_i}{\partial E}\\
&=  \alpha\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\Big) + \beta\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\epsilon_i\Big)\\
&= \alpha \frac{\partial N}{\partial E} + \beta \frac{\partial E}{\partial E}\\
&= \beta.
\end{split}
\end{equation}


Chemical bonding and hybridization of organic compounds What happens to the lobe of the p-orbital during sp² hybridization?

During the formation of $\ce{C2H4}$ molecule:

We know that both the carbon atoms will have a total of 3 p-orbitals and they will be involved in hybridization.

Let us number the carbon atoms as $\ce{C}1$ and $\ce{C}2$.

Now the p-orbital of $\ce{C}1$ along $x$-axis will form a $\unicode[Times]{x3C3}$ bond with the p-orbital of $\ce{C}2$ and a $\unicode[Times]{x3C3}$ bond with the s-orbital of hydrogen.

And similarly the p-orbital of $\ce{C}2$ along x-axis will form a $\unicode[Times]{x3C3}$ bond with hydrogen.

So both the lobes of the p-orbital of $\ce{C}1$ and $\ce{C}2$ along $x$-axis are involved in sigma bond formation.

Lets say the p-orbital of $\ce{C}1$ and $\ce{C}2$ along the $y$-axis forms a $\unicode[Times]{x3C0}$ bond.

So here also both the lobes of p-orbital of $\ce{C}1$ and $\ce{C}2$ are involved in hybridization.

Let say that the p-orbital of $\ce{C}1$ which is along the $z$-axis forms a $\unicode[Times]{x3C3}$ bond with the s-orbital of hydrogen.

And similarly the p-orbital of $\ce{C}2$ along $z$-axis forms a $\unicode[Times]{x3C3}$ bond with the s-orbital of the other hydrogen.

Now my confusion is that here two p-orbitals along the $x$ and $y$-axes have four lobes and all of them are involved in bond formation.

But the p-orbital which is along $z$-axis has only one of its lobes involved in bonding. What happens to the other lobe of this p-orbital?

You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.

While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only one lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital will not.

This link provides a nice visualisation of the different orbitals involved in ethene bonding.


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

The "flint" in disposable lighters is made of pyrophoric alloy called "ferrocerium", which consists of approx. 50% cerium, 25% lanthanum and 20% iron; the remaining 5% is magnesium, neodymium and praseodymium. Exact composition may vary depending on the manufacturer.

If you rub such an alloy on a rough surface, tiny metal particles are formed, which are instantly ignited and produce metal oxides. Neither of the expected products are particularly toxic: lanthanoid oxides, particularly $\ce{CeO2}$ and $\ce{La2O3}$, are not toxic and can only cause mild irritation when swallowed, and iron oxide $\ce{Fe3O4}$ is basically rust.

To sum it up, unless you are allergic to transition metals (which people usually discover very early), I cannot think of any health hazards coming from the "flint" or the sparks.


chlorine formed in hydrochloric acid Is it plausible that free chlorine could be formed in concentrated hydrochloric acid.

The background to this is that I have bought some 30% HCl to remove calcium carbonate scale from a reverse osmosis water purifier. The latest batch of HCl seems to have a very slight yellowish tint which I had not noticed on previous occasions. Having used some of this acid I find that the performance of the membrane seems to have deteriorated (these membranes are notably sensitive to hypochlorite).

Could some factor (perhaps exposure to direct sunlight) be responsible?

Is there a way that I could detect the presence of dissolved chlorine?

Can anyone suggest an alternative explanation?

Thanks,
Chris.

You could buy the chlorine test strips from a pool supply store to test for hypochlorite. You may need to neutralize the acid before testing, so ask if the test is accurate in a strong acidic solution.


What exactly does it mean for the reaction of superoxide with non-radicals to be spin-forbidden? From Wikipedia:

  The reaction of superoxide with non-radicals is spin forbidden. In biological systems,   this means its main reactions are with itself (dismutation) or with another biological radical such as nitric oxide (NO) or with a transition-series metal.

I know that one of the configuration bands is paired, and one is unpaired. Does the unpaired one have to bond with an unpaired one from another radical? What does this also say about reactions of the excited state of singlet oxygen (shown in the middle below) with non-radicals?

Also, is superoxide a doublet?

What exactly does it mean for the reaction of superoxide with
  non-radicals to be spin-forbidden?

For clarity, let's start with a simpler case.  If we take a ground state singlet molecule (singlet means all electrons are paired) and we irradiate it with photons, photon absorption will excite an electron to an excited state.  Will the excited state be a singlet or triplet immediately after it is produced?  For the transition to the excited state to occur readily it must be spin-allowed.  That means that spin must be conserved.  If we started with a singlet, then our excited state must still be a singlet.  In that excited state the excited electron will still have its spin paired with the electron that was its partner that remains in a lower energy orbital.  Creation of a triplet state, where the excited electron and its ground state partner no longer have their spins paired is spin-forbidden and is therefor much less likely to occur.  Spin must be conserved.

Well rules are made to be broken.  What can we do to increase the probability of a singlet -> triplet conversion that is formally spin-forbidden. Spin-Orbit coupling (the coupling of the electron spin with the orbital angular momentum of non-circular orbits) is our way out. We need to find a spot where the Morse potential curves overlap for both the singlet and triplet. Then, if there is a large spin-orbit coupling interaction we can readily cross over from singlet to triplet, or back.

Large atoms such as bromine and iodine, as well as transition metals with unpaired electrons have large spin-orbit couplings.  If such groups are attached to our molecule of interest, then spin need not be conserved and formally "spin-forbidden" reactions become more likely.

  Does the unpaired one have to bond with an unpaired one from another
  radical?

Yes, if spin is to be conserved.

  What does this also say about reactions of the excited state
  of singlet oxygen (shown in the middle below) with non-radicals?

The singlet oxygen shown in the middle of your figure ia a singlet species and will react readily with other singlet (non-radical) molecules.

  is superoxide a doublet
  Yes superoxide is a doublet.  Ground state $\ce{O2}$ is a triplet - it has two unpaired electrons in energetically degenerate orbitals, if we add one electron to form superoxide radical anion, that electron will pair with one of those two electrons leaving only one unpaired electron.


Why does this formula work? My teacher gave us a formula for determining if a M-O-H like compound will act like a base or not.without explaining much.
Base if:$$|\chi_o-\chi_M|\gt|\chi_o-\chi_H|$$
Flipped inequality if acid.
I am not unable to understand why the formula works.I think its because if the diffrence in electronegativity between Oxygen and element m is high.
It is  more likely for M-O-H to break into M and O-H
.Not certain about that either.

The bond with greater electronegativity difference has greater tendency to break with help of polar $\ce{H2O}$ molecules to hydrated ions, unless it is ionic already. Electrons then stay at the oxygen as the most electronegative atom from all the three.
So if $\ce{M}$ is less electronegative then $\ce{H}$, there will be'
$$\ce{M^{(+)}-O^{(-)}-H(aq) <=> M+(aq) + OH-(aq)}$$
but if $\ce{M}$ is more electronegative than $\ce{H}$ , then there will be:
$$\ce{M-O^{(-)}-H^{(+)}(aq) + H2O <=> MO-(aq) + H3O+(aq)}$$
( The charge in parenthesis is a partial charge.)


Quantum and organic chemistry of vitamin D synthesis I am somewhat confused by certain steps in this mechanism.

To start with, in the conversion of 7-dehydrocholesterol to previtamin D3, how does the UV photon affect the pi-orbital system in such a way as to rearrange the bonds as shown in the diagram? Else, what exactly brings about the change shown?

Similarly, in the thermal isomerisation step from previtamin D3 to Vitamin D3, what quantum effects and other changes cause the rearrangement shown?

Since this is a biological reaction within the human body it also takes place under physiological conditions ($37^\circ C$, 1 atm, pH = 7.4, various intracellular ionic concentrations).

These are pericyclic (i.e. concerted) reactions. They do not need particularly high temperatures, but indeed they can be catalysed by either the right photons or temperature. They are not "reactions" per se, rather just intramolecular rearrangements of electrons over a "chain" of connected atoms.

The first reaction is a photolytically catalysed electrocyclic reaction that results in the opening of the 6-membered ring, they are very common.

The second reaction is a thermally catalysed [1,7] sigmatropic migration of the H. The rate is dependent on the nature of the nucleus that moves, in this case H (D would be slower). Quantum tunneling will indeed enhance the rate of its non-quantum counterpart, but it is not the driving force of the step as you seem to wonder.

As for the reason why these reactions happen in the first place, it's because there is favourable overlap in the transition states between the orbitals of the atoms at the two external ends of the "chain" of atoms interested in each rearrangement.

As you can see below, the HOMO in each case permits favourable overlap (phase-wise and sterics-wise) between the ends of the "chains" in both cases. I couldn't quickly find a picture of a photolytic ring opening but in that case the actual HOMO is a SOMO: one electron from the former HOMO has been promoted (by the photon) to the former LUMO. This will change the overall symmetry properties of the atom chains so that a [1,7] H shift is allowed under photolytic conditions.

Electrocyclic reaction:

Hydride shift:

Reference: https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/pericycl.htm

Suggested reading: "Pericyclic Reactions" by Ian Fleming (OUP, 1998).


Can solvents like 99.9% isopropyl alcohol and/or n-hexane be "carbonated" with CO2? I'm working on a prototype for a new machine, and I'm trying to get solvents like $99.9\%$ isopropyl alcohol or n-hexane to be carbonated like a soda pop at cold temperatures. I have the machines and everything for it I just want to make sure I'm not going to explode or anything.

The following are published[1] solubilities of $\ce{CO2}$ in several organic solvents at 1 atm and $\mathrm{20^oC}$. Solubilities are given in mL/g:

  acetone: 8.2
  ethanol: 3.6
  benzene: 2.7
  methanol: 4.1
  toluene: 3.0
  xylene: 2.3
  heptane: 2.8
  methyl acetate: 7.4
  diethyl ether: 6.3

For reference, the solubility of $\ce{CO2}$ in water is about 0.9 mL/g under the same conditions.

1) Topham S et al; Carbon Dioxide. Ullmann's Encyclopedia of Industrial Chemistry. 7th ed. (1999-2015). New York, NY: John Wiley & Sons


Boyle's Law of pressure and volume Boyle's Law states that pressure is inversely proportional to volume.
However, let's take an example of a balloon: as we fill air in it, its pressure increases, but its volume also increases. Can anyone explain?

(source: ucdavis.edu)

Balloon is not an ideal system to study pressure-volume relationship. Because, on expansion, the elastic skin also expands and there needs to be an additional pressure build-up on the inner side to counter that force, very similar to the extra pressure in a spherical bubble. (where the surface tension acts exactly similar to the elastic balloon skin)

But in your arguments, you are forgetting an important point. Blowing a balloon also involves increasing the amount of air in it. Boyle's law holds for only a fixed volume of gas. If you change the number of moles, Boyle's law no longer stays valid.


How are phenols made? What is the equation and conditions for the synthesis of phenols? Would it work if we added $\ce{NaOH}$ to benzene and then refluxed?

That reaction will definitely not yield a phenol.
Phenols are not usually prepared directly from benzene, but from benzene derivatives, such as toluene or chlorobenzene:

Oxidation of toluene, as developed by Dow Chemical:

$\ce{C6H5CH3 + 2 O2 → C6H5OH + CO2 + H2O}$

Hydrolysis of chlorobenzene, using base or steam (Raschig–Hooker process):

$\ce{C6H5Cl + H2O → C6H5OH + HCl}$

Actually, there are processes that allow us to obtain it directly from benzene, but these are rarely used:

Direct oxidation of benzene with nitrous oxide, a potentially "green" process:

$\ce{C6H6 + N2O → C6H5OH + N2}$

The most widely used process for industrial production (95%) involves the partial oxidation of cumene (isopropylbenzene) via the Hock rearrangement:

$\ce{C6H5CH(CH3)2 + O2 → C6H5OH + (CH3)2CO}$

Source of the reactions: Wikipedia English page for phenol


Why does nucleophilic aromatic substitution occur at C-2 and C-4 of pyridine? In the following reaction of 2,3,4-tribromopyridine with sodium methoxide in methanol,[1] why are only the C-2 and C-4 bromines replaced with methoxy groups?

Why doesn't the bromine at C-3 also get substituted?

Gibson, K. J.; D'Alarcao, M.; Leonard, N. J. Rearrangements of azabiphenylenes. The impact of nitrogen number and position. J. Org. Chem. 1985, 50 (14), 2462–2468 DOI: 10.1021/jo00214a012.

Nucleophilic aromatic substitution on pyridines (or related heterocycles such as quinolines or pyrimidines) regioselectively occurs at the 2- and 4- positions. These can be thought of as being "ortho" or "para" to the nitrogen.

In the mechanism for nucleophilic aromatic substitution, the initial step involves attack of the nucleophile (in this case, methoxide anion) on the π-system of the aromatic ring. Because this step generates a high-energy anionic intermediate which has its aromaticity broken, it is also typically the rate-determining step. In general, the stability of this intermediate is what dictates whether a nucleophilic aromatic substitution is possible or not.

If we consider the resonance forms for this intermediate resulting from attack at each of the three positions, we can understand why attack at C-2 and C-4 is favoured.

When the nucleophile attacks at C-2 or C-4, the resultant intermediate possesses one resonance form where the negative formal charge is located on nitrogen. On the other hand, this is not possible when the nucleophile attacks at C-3.

In general, resonance forms involving negative charges on electronegative atoms such as N and O are much more important contributors than resonance forms with negative charges on carbon. Thus, the anionic intermediate resulting from attack at C-2 or C-4 is better stabilised by resonance, allowing the nucleophilic substitution to take place.


Why doesn't Boron form Diborane with water? Why is it that Boron does not form Diborane with water rather than $\ce{B2O3}$?

I assume the hypothetical reaction you are asking about is:

$\ce{3H2O + 2B -> B2H6 + 3/2O2}$

under standard conditions.

The standard enthalpy of formation of water is −285.8 kJ/mol, whereas that of diborane is 35.4 kJ/mol; this reaction would therefore be very endothermic and hence unfavourable. You're breaking up strong O-H bonds and B-B bonds to form B-H bonds (including two 3c2e bonds, which are very weak) and weak O-O bonds. In contrast the formation of boron trioxide by reaction of boron and water is very exothermic, and has been considered in potential applications of boron as a rocket fuel.


How can iron(III) sulfate be formed? What is the salt formation for iron(III) sulfate?

Wikipedia tells me that the production of iron(III) sulfate is:

$$\ce{2 FeSO4 + H2SO4 + H2O2 → Fe2(SO4)3 + 2 H2O}$$

I was wondering if there was a difference between salt formation equation and production, because I thought it would've went something more like:

$$\ce{2 Fe(OH)3 + 3 H2SO4 → Fe2(SO4)3 + 6 H2O}$$

I was wondering if there was a difference between salt formation equation and production.

Absolutely! In this case, the latter means that you're trying to turn a side product (waste) of a technical product into another marketable product.

In order to understand, why the oxidation of $\ce{FeSO4}$ is interesting, you have to realize that it is a side product in the synthesis of titanium dioxide ($\ce{TiO2}$). Here, ilmenite ($\ce{FeTiO3}$) is processed with concentrated sulfuric acid to yield $\ce{TiOSO4}$ as the valuable intermediate and $\ce{FeSO4}$ as "waste".


Chemdoodle to mestranova Is it possible to copy my chemdoodle displayed structure into mestranovea? I have tried simple dragging (mac) and all the various copy/paste options... any ideas?

ChemDoodle contains many options for communicating chemical data to other applications. The most common way is to use the Edit>Copy as... function and choose a chemical type to be put on the system clipboard as a string.

If you use the copy function in ChemDoodle, a wide range of common data types, from chemical to image, are placed on the system clipboard for a 3rd party application to paste. You can control what is placed on the clipboard in ChemDoodle's advanced preferences.

ChemDoodle also supports round trip editing with applications that will allow it. However, if none of this works with the application you are pasting into, you will have to contact that specific vendor.

More information on working with ChemDoodle data in 3rd party applications can be found in Chapter 11 of the ChemDoodle User Guide: http://www.chemdoodle.com/downloads/UserGuide.pdf

For further help, please open a ticket here: https://support.ichemlabs.com


Nomenclature confusion regarding carboxylate and ester I suddenly thought about these 2 compounds that seem to have the same name. Aren't both of them methyl propanoate? How do we differentiate them apart name wise?

There is not supposed to be a space between the substituent prefixes (e.g. methyl) and the parent structure (e.g. propanoate), so your second structure is methylpropanoate, or more correctly 2-methylpropanoate.

In the ester there is a space between methyl and propanoate, i.e. methyl propanoate.


Reason for the differences in third ionisation energy between elements (magnesium and fluorine) The question is "Why is the third ionisation energy of magnesium higher than that of fluorine?"

Although I understand that differences in ionisation energy have several factors governing it, e.g. the shielding effect, distance from the nucleus, whether the electron is in its own orbital or paired with another electron etc.

The answer to this question is that "Even though the valence electrons of both elements are removed from the same shell, Mg has greater nuclear charge."

I am confused what the answer means about nuclear charge. Doesn't the third ionisation energy of Mg and F have the same nuclear charge as they are both 3+?

The ionisation energy, as a rough guide, can be related to the effective nuclear charge and the principal quantum number:

$$\mathrm{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$

In the ionisation of $\ce{Mg^2+}$ and $\ce{F^2+}$, both electrons are being ionised from the second principal quantum shell, so $n = 2$ in both cases. However, the main reason why the 3rd IE of magnesium is larger is because the nuclear charge $Z$ is larger, and not because of any so-called stability of half-filled or fully filled shells.

Contrary to what you thought, the nuclear charge is not the charge on the species. It is the charge of the nucleus - hence "nuclear charge". The $\ce{Mg}$ nucleus has 12 protons, and the $\ce{F}$ nucleus has 9 protons. So, there is a greater electrostatic force pulling the electron towards the nucleus in $\ce{Mg^2+}$ than in $\ce{F^2+}$, and consequently it is harder to remove the electron from $\ce{Mg^2+}$.

In truth, the raw nuclear charge must be modified by the shielding constant $\sigma$ to obtain the effective nuclear charge, $Z_\mathrm{eff}$. However, the variation in $\sigma$ between the two compounds is really negligible compared to the variation in the actual nuclear charge.


Pauli exclusion principle and resonance I've always been a bit uncomfortable with the concept of more than two electrons in a single orbital-like region(probability-wise) which occurs in resonance.

This seems to disobey Pauli's exclusion principle.

Let's take benzene for example. It has a delocalized cloud of electrons. These all had the "original" (as in, if resonance didn't occur) quantum numbers $n=2,l=1,m=\text{(1,0,or -1; but it must be the same for all)},m_s=\pm\frac12$. But, due to delocalization, they're now all in one extra-large orbital-like region (by orbital-like, I mean that the pi cloud is similar to an orbital in the sense that it is also a probability cloud which defines the probabilities of finding one or more electrons at different coordinates).

Now, it's obvious that the original quantum numbers are insufficient here due to Pauli's exclusion principle. They may not even exist for these electrons.

The easiest thing to do would be to "propose" (not exactly propose, since I guess the Dirac equation can prove their existence) an extra quantum number, which can take twelve values. Or a combination of a few extra quantum numbers that have the same effect.

But this raises the question: do the electrons belonging to the pi cloud "retain" their "original" quantum numbers as well?

Basically I want to know how benzene and other resonating compounds are described quantum mechanically, with focus on quantum numbers. (links to papers are fine, probably necessary here--but please explain the gist of the link)

I'd prefer the answer in terms of simpler quantum mechanics, though I'm not averse to the idea of some extra reading.

To understand why the exclusion principle isn't violated in this system, you really need to shift from valence bond theory to molecular orbital theory. The $p_{z}$ orbitals in benzene combine according to the $D_{6h}$ symmetry of the molecule to generate a set of bonding molecular orbitals (MOs) (which are lower in energy than the isolated $p_{z}$ orbitals) and a set of antibonding molecular orbitals (which are higher in energy). Bonding orbitals are generated from sets of $p_{z}$ orbitals which are mostly or totally in phase, whereas the highest energy antibonding orbital of the set is composed of $p_{z}$ orbitals that are totally out of phase with respect to their adjacent neighbors.

Here, for instance, is the MO that is generated from the totally in-phase set (I calculated this at a spin-restricted RI-BP86/6-311G* level of theory in ORCA(1) and visualised the isosurfaces in VMD(2)):

Now, this MO is shared equally across all of the carbons, i.e. it is not localised. However it only contains 2 electrons and thus satisfies the exclusion principle. (F'x mentions spin-orbitals - this is where the orbitals are split on the basis of their spin, which entails a single electron per orbital. Benzene is a closed-shell singlet so we would expect the $\alpha$ and $\beta$ spin orbitals to be spatially indistinguishable.)

Not to sound like a walking advertisment for Housecroft and Sharpe, but H&S has a great visual introduction to MO theory.

P.S.

It was suggested in chat that I expand upon the generated orbitals - here's one of the two highest occupied molecular orbitals, from the same calculation.

As you can see, there's a big node along one of the mirror planes. This orbital is higher in energy than the earlier example because it's composed of two antiphase sets of 3 $p_{z}$ orbitals, so this orbital has antibonding character for two of the carbon-carbon pairs. It should be easy to extrapolate to the highest lying MO of this type - antibonding all the way around.

(1) Neese, F. ORCA – an ab initio, Density Functional and Semiempirical program package, Version 2.6. University of Bonn, 2008

(2) Humphrey, W., Dalke, A. and Schulten, K., 'VMD - Visual Molecular Dynamics', J. Molec. Graphics 1996, 14.1, 33-38.


Are the values for enthalpy, internal energy and Gibbs free energy the same for a particular process? Internal energy, enthalpy and Gibbs free energy are all units of energy. So, for any particular process where energy is lost from the system the same amount is given off to the surroundings. Am I right in thinking that a change in internal energy/free energy/enthalpy is all a measure of how the energy of the system changes in a process and as a result should be equal to each other for the same process, albeit under different conditions?

Let me expand slightly:
$$dU<TdS-pdV$$ So for a spontaneous change to occur, the internal energy must decrease provided that the entropy and the volume are kept constant. The internal energy will continue to fall until equilibrium is reached. The exact amount of energy lost by the system is gained by the surroundings. As a side question: how exactly is the energy lost to the surroundings?
$$dH<TdS+VdP$$ The enthalpy of the system decreases for a spontaneous change, provided that the entropy and pressure are kept constant. Energy lost by system = energy gained by surroundings.
$$dG<VdP-SdT$$ The Gibbs free energy of the  system decreases for a spontaneous change, provided that the temperature and pressure are kept constant. Energy lost by system = energy gained by surroundings.

For a given process, surely the change in internal energy (at constant S,V) be equal to the change in enthalpy (at constant P,S) and Gibbs free energy (at costant P,V), right? The above argument seems to make sense to me but enthalpy can be defined as $H=U+PV$ which would suggest that the two values would be different wouldn't it?

While enthalpy, internal energy, and Gibbs's energy are all types of energy, they are not all measuring the same quantities. Internal energy is the energy of the molecules in your system. Enthalpy takes into account the internal energy plus whatever work is being done by or on your system. And Gibbs's energy is the maximum amount of non-PV work that you can get from your system. While these three quantities are all in units of energy, they are not just different terms for the total energy. Since enthalpy and Gibbs's energy involve work, and hence exchange of energy with the surroundings, these quantities will be different than just the internal energy of the system alone.


Does the electrochemical series reverse in electrolysis? Are these two statements true?

In a galvanic cell at standard lab conditions (SLC), the strongest reductant is $\ce{Li (s)}$ and the strongest oxidant is $\ce{F2 (g)}$.
In an electrolytic cell at SLC, the strongest oxidant is $\ce{Li+ (aq)}$ and the strongest reductant is $\ce{F- (aq)}$.

No, only the first is. The oxidising or reducing property of a species is tied to the species, not to the process in which the species is used. Therefore regardless of whether you are running a galvanic cell or an electrolytic cell, $\ce{F2}$ will always be a strong oxidant and $\ce{Li}$ a strong reductant.

The only difference about the two types of cells is that in the galvanic cell, you are letting the spontaneous cell reaction happen, which produces energy. In the electrolytic cell, you are using energy to force the non-spontaneous cell reaction to happen. The spontaneous cell reaction refers to the direction of the reaction with $\Delta G < 0$ (or equivalently, $E > 0$).

By this, I mean: you can use a galvanic cell to carry out the reaction $\ce{2 Li + F2 -> 2Li+ + 2F-}$. Since $\ce{Li}$ is a strong reductant and $\ce{F2}$ is a strong oxidant, this reaction will generate a lot of energy.

You can use an electrolytic cell to reverse that reaction (although you'd have to use molten and not aqueous $\ce{LiF}$). But since $\ce{Li+}$ is a weak oxidant and $\ce{F-}$ a weak reductant, you're going to need lots of energy to carry that out. Which makes perfect sense!


What is the correct lewis structure of SCH⁻? I need to find the Lewis structure of the ion $\ce{SCH^-}$.

I think the Lewis structure should be:
$\ce{S=C-H}$, with one lone pair on the carbon and two lone pairs surrounding the sulfur.

Is this correct?

This model seems reasonable, as you can see from the formal charges carbon has the -1 value while the rest are 0. One thing that you have forgot is to add brackets around the structure to show that it is a ion, include its overall charge.

$$\ce{[\overset{..}{\underset{..}{S}}=\overset{..}{C}-H]-}$$

Although, I suppose a structure of $\ce{[\overset{..}{C}#\overset{..}{S}-H]-}$ would work with the same formal charge, except that sulfur has the expanded octet, which is possible but I am not sure if that is more reasonable or not, it probably is not.

It is a very odd ion, to say the least, probably very unstable. Perhaps you have mistaken the hydrogen with nitrogen? It would be more reasonable to be asked to draw the lewis structure for $\ce{[SCN]-}$, thiocyanate ion. I am not even know how to pronounce or name the $\ce{[SCH]-}$ion you are required to draw, or if it even exists. Maybe that is why you need to draw it though.


Is it possible to make micelles to pick up Microplastics from water? I'd like to know if I could get micelles to pick up microplastics in water. I read that microplastics are sometimes charged, so can we make the lyophobic end to pick up microplastics, so that they can then be collected on the water surface and be easily disposed?
This is the idea I'm planning to pitch for a science fair, is this actually feasible?
I read that detergent micelles work well even in hard water, so can we use such detergent micelles to pick up microplastics?

Thanks!

Lyophobic means solvent-hating, and here the solvent is water, so the lyophobic ends (presumably the tails, if we are discussing a normal detergent) will associate to form the hydrophobic core of the micelles. The headgroups meanwhile will preferentially interact with water, forming an interface between solvent and core.

If you add detergent to a solution containing charged plastic particles, the detergent tail is unlikely to interact with the particle unless the particle also displays amphiphilic character. More importantly, headgroups with the same charge as the particle surface will discourage association with the particle. So your proposal lacks a viable aggregation mechanism for at least one reason (charged and hydrophobic regions are not likely to associate) and possibly for a second (oppositely charged regions of detergent and particle will repel). So you need to research carefully:

the expected charge of the particles
the choice of detergent, ideally with a headgroup with charge opposite that of the particle, and which resists precipitation due to the hard water but will encourage flocculation of the particles
scalability, which is a question regarding the volume of water you intend to treat

There is always the possibility that the solution contains ions resulting in charge inversion of the charged particle, another factor to consider.

You'd be well served to look further into the steps used to treat wastewater.


Is possible to figure out if a battery is NiCd or NiMH? Aside from opening it of course. I'm not sure if this is the right place to ask this but I have a rechargeable battery pack for an Xbox One controller and the info on it says that it's a sealed nickel cadmium battery but it also has a recycle symbol with "NI-MH" written underneath. The company's website doesn't seem to have any customer support and I tweeted at them on Twitter but haven't received a response. If it is NiCd, I'm surprised that any batteries of that type are still made. If it were Ni-MH I'm sure they would say it on the packaging. I think I read that NiCd cells don't have a large voltage drop until the end of the charge, so I suppose I could measure that. Does anyone have any suggestions about the battery type or know of a better place I could ask this question?

NiCd cells are much more dense then NiMH, so if you have two cells of the same size (e.g. AA), it's easy to determine by "hefting". However, with a sealed battery, this may not be so obvious. You could determine the density by Archimedes' method, but you probably don't want to get the inside of the battery pack wet (unless you can thoroughly dry it), to prevent electrolytic corrosion.

BTW, NiCd cells are much more resistant to damage from over-charge and from complete discharge than NiMH, and are still widely used in portable power tools and other high-drain applications. NiMH provide more total energy by mass, and more environmentally "friendly'.


possibility of metallic ammonium In chemistry class, we discussed how metallic bonding occurs when metals lose electrons and become cations while the lose electrons become delocalized electrons and bind the metal cations. These electrons are then free to flow around the metal cations. But could the metal be replaced with an ammonium cation instead? (Or any other polyatomic cation, but we only learn ammonium and I'm not aware of any others minus hydronium which doesn't exist out of a solution.) Would the electrons still be free? How would you make it? And how could you prevent the following from occurring?
$$\ce{2NH4+ + 2e- -> 2NH3 + H2}$$

A nomenclature note: conventionally, in chemistry "a metal" is an element (such as a transition metal or alkali metal). If what you're talking about is a substance that behaves as a metal, that's typically referred to as metallic, like metallic glasses.

Therefore, I would rephrase your question as "Can metallic ammonium exist?" - in fact, this has been a question for quite awhile. From what I've read, there's a theoretical basis that it should be possible at given temperatures and pressures, but it hasn't been experimentally observed. The papers I was reading were from the 50s-70s, so I'll keep looking for more recent information.

Solvated electrons are slightly related, in that there is a "free" electron, and this was in fact originally documented in liquid ammonia.


How could you distinguish between an irreversible electron transfer couple and a chemically irreversible couple using cyclic voltammetry? What is the difference between an irreversible electron transfer couple and a chemically irreversible couple?

How could you distinguish between these two processes using cyclic voltammetry?

To this question there are black-and-white as well as grey answer parts.

Chemical reversibility refers to the generation of a species that cannot be reduced or oxidised furthermore. In case you scan at moderate speed (scan rate), you may reduce or oxidise a compound and this intermediate can further react to an electrochemically inert species. If the follow-up reaction occurs fast, this couple is chemically irreversible.

Electrochemical reversibility refers to the rate constant of electron transfer. High rate constants (fast electron transfer) are found in electrochemically reversible couples whereas low rate constants are found in electrochemically irreversible cases.

Okay, you can trick chemical reversibility by scanning so fast, that you oxidise/reduce the intermediate of a redox-reaction at the electrode so fast that it has no time to further react. In this case, you should have only a forward-peak using slow cyclic voltammetry without a backward-peak; but you should start to see a backward-peak as you increase the scan rate.

For electrochemical reversibility, you can use Butler-Volmer kinetics. The main message of Butler-Volmer is that you increase the rate of electron transfer by increasing the potential. Electrochemical reversibility is seen on the peak separation between oxidation and reduction peak. Reversible systems show a peak separation at standard conditions with one electron transferred of 59 mV.

So, how would I dinstinguish them using cyclic voltammetry? By the peak shape and the change of CV by varying the potentials and scan rate.


Nitrogen dioxide fluorescence quenching and lifetime Nitrogen dioxide fluorescence quenching:

$$
\begin{align}
\ce{NO2 + h\nu &->[$\varphi_\mathrm{Ia}$] NO2^{\ast}}\tag{I}\\
\ce{NO2^{\ast} &->[$k_2$] NO2 + h\nu'}\tag{II}\\
\ce{NO2^{\ast} + NO2 &->[$k_3$] 2 NO2}\tag{III}\\
\ce{NO2^{\ast} + Xe &->[$k_4$] NO2 + Xe}\tag{IV}\\
\ce{NO2^{\ast} + NO2 &->[$k_5$] 2 NO + O2}\tag{V}\\
\end{align}
$$

The fluorescence lifetime of $\ce{NO2^{\ast}}$ in the presence of all the reactions, $\tau$, measured at $\pu{298 K}$ at different concentrations (molecules per litre) of the reactants:

$$
\begin{array}{crrr}
\hline
\text{Experiment} & \tau/\pu{μs} & [\ce{Xe}]/\pu{L-1} & [\ce{NO2}]/\pu{L-1}
 \\
\hline
1 & 3.38 & \pu{1.6E19} & \pu{3.2E18} \\
2 & 1.89 & \pu{1.6E19} & \pu{6.4E18} \\
3 & 3.64 & \pu{0.8E19} & \pu{3.2E18} \\
\hline
\end{array}
$$

What is the real fluorescence lifetime of $\ce{NO2^{\ast}}?$

Is this question valid? Isn't the lifetime vs the concentration of the quencher supposed to be linear?

You are looking for the rate constant $1/\tau^0=k_2$. The rate equation for the $\ce{NO2^*}$ is (using $N^*$ for $\ce{[NO2^*]}$ and N for ground state NO2), $\displaystyle \frac{dN^*}{dt}=-k_2N^*$ and when quenched $\displaystyle \frac{dN^*}{dt}=-k_2N^* -N^*((k_3+k_5)N+k_4[Xe])$ which can both be integrated to give exponentials, i.e $N^*=N^*_0e^{-k_2t}$ and similarly for the other equation. Writing the rate constants as $1/\tau^0=k_2$ and $1/\tau_Q=k_2+(k_3+k_5)N+k_4[Xe]$ you can find the values after looking at the pattern of the quencher concentrations in the question.


Image of crystal structure Is there any websites or resources where I can get a good image of a certain crystal structure? In my case I'm looking for lead(II) tungstate $\ce{PbWO4}$ and I can't find a vivid graphic on google (except of the structural formula).

I've seen modeling structure of crystals, but unfortunately the software is commercial.

I recommend to fetch the crystal structures of inorganic compounds via specific Google search pattern site:crystallography.net <keyword(s)>, e.g. in your case

site:crystallography.net PbWO4

yields in this COD page where you can use JSmol to do basic visualization right in your web browser. Alternatively, you can download CIF and work further offline with numerous free tools such as Mercury or Olex2. Both are free and available for Windows, Linux and MacOS.

Mercury

Olex2


Conductivity as a function of acid concentration I have conducted an experiment measuring the conductivity of both hydrochloric acid and sulfuric acid in solution respectively, with varying concentration. Wikipedia as well as a question on the site suggests a $\sqrt{c}$ relationship is viable. However, my data resembles this:

Clearly not a simple $\sqrt{c}$. It reminds me of the inverse gamma distribution. I have tried looking at papers by Onsager on his conductivity theory, but they're unclear, and he doesn't propose a function. Are there any papers that look at accurately modelling conductivity as a function of concentration which exhibit this form? Obviously for low concentrations (e.g. 1-2 molar), a linear or $\sqrt{c}$ would suffice.

At low concentration, conductivity is proportional to concentration (a linear relationship).

Each ion will have its own unique mobility, as discovered by Kohlrausch. $\ce{H+}$ has the highest mobility. As you can see in your graph the acids have higher conductivities than the salts. $\ce{OH-}$ is also highly mobile. As concentration increases, the linear relationship breaks down for two reasons.

Firstly, in infinitely dilute solution, for the strong electrolytes in the graph, there is complete dissociation into separately solvated ions. However, as concentration increases, a portion of the electrolyte exists as ion pairs.  See for example Equations for Densities and Dissociation Constant of NaCl(aq) at 25°C from “Zero to Saturation” Based on Partial Dissociation J. Electrochem. Soc. 1997 vol. 144, pp. 2380-2384.

Secondly, the mobility of the ions that are solvated is decreased by the fact that they are no longer moving through water, but are instead moving past other ions as well.

To have a maximum in the curves of the question, and to account for the above factors, it is necessary to subtract a term from the linear term.

You need a function of the form:

$$\text{Conductivity} = Ac - Bf(c),$$

where $A$ and $B$ are constants, $c$ is concentration, and $f(c)$ is some function of concentration. Historically, the function $Ac - Bc\sqrt{c}$ was the first and most simple to use.

I would start by trying to fit your data to that function.

More advanced treatments replace $\sqrt{c}$ with $\sqrt{I}$, where I is ionic strength. Further advances involve higher order $I$ in addition to $\sqrt{I}$, such as $I \ln (I)$, $I$, and $I^{3/2}$.

The $I \ln (I)$ and $I$ terms arise upon considering the ions as charged spheres of finite diameter, rather than a simple point charges, as explained in Electrolytic Conductance and Conductances of the Halogen Acids in Water and references cited therein. This is the reference to look at if you want the best function to fit the HCl curve in your graph as it includes numerical coefficients.

For more information see The Conductivity of Liquids by Olin Freeman Tower, which although not the most recent work, has the advantage of being understandable, or Calculating the conductivity of natural waters.

For a very advanced consideration, see "Electrical conductance of electrolyte mixtures of any type" Journal of Solution Chemistry, 1978, vol. 7, pp. 533-548.


What is the hybridisation in dimolybdenum? Dimolybdenum's $\mathrm{4s}$ and $\mathrm{4p}$ orbitals are full. It is going to use its $\mathrm{4d}$ and $\mathrm{5s}$ orbitals to form a sextuple bond (2 $\sigma$, 2 $\pi$, 2 $\delta$). I am wondering if the $\mathrm{4s}$ and $\mathrm{4p}$ orbitals are going to hybridize with the $\mathrm{4d}_{z^2}$ orbital (the one forming the first $\sigma$ bond) and the $\mathrm{5s}$, given the differences in energy. The hybridisation would be $\mathrm{ds^2p^3}$, which is certainly uncommon.

If so, do the $\mathrm{s}$ and $\mathrm{p}$ lone pairs always hybridize, even if they don't form any bonds?

The electronic configuration of $\ce{Mo2}$ is given in Bursten, B.E.; Cotton, F.A.; Hall, M.B. Dimolybdenum: nature of the sextuple bond. J. Am. Chem. Soc. 1980, 102 (20), 6348–6349:

  $$\ldots\mathrm{
9\sigma_g^{1.88}
5\pi_u^{3.78}
2\delta_g^{3.42}
10\sigma_g^{1.92}
9\sigma_u^{0.08}
2\delta_u^{0.58}
5\pi_g^{0.22}
10\sigma_u^{0.12}
}$$

Which can be simplified as:

$$\ldots\mathrm{
(9\sigma_g)^2
(5\pi_u)^4
(2\delta_g)^4
(10\sigma_g)^2
(9\sigma_u)^0
(2\delta_u)^0
(5\pi_g)^0
(10\sigma_u)^0
}$$

$9\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{4d_{z^2}+4d_{z^2}}$.
$5\pi_{\mathrm{u}}$ is doubly degenerate; it is $\mathrm{4d_{xz}+4d_{xz}}$ and $\mathrm{4d_{yz}+4d_{yz}}$.
$2\delta_{\mathrm{g}}$ is doubly degenerate; it is $\mathrm{4d_{xy}+d_{xy}}$ and $\mathrm{d_{x^2-y^2}+d_{x^2-y^2}}$.
$10\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{5s+5s}$.

Now, let us have a look at the beautiful 4d orbitals, kindly provided by our user Philipp:

From here, you can see that no hybridization is necessary, because the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals can all form bonds within themselves.


What is the difference between quantities reported as optical rotation and circular birefringence? Optical activity gets complicated quickly and the following question is intended to avoid threshold confusion. The linked lecture note* seems very accessible to me and Table 10.6 in particular is quite clear. On page 2 the author writes, "Optical rotation results from circular birefringence..." and then relates that idea to circular dichroism.

  My question is: How does the angle we measure in a student polarimeter as optical rotation relate to the angle reported as circular birefringence? Do they merely differ by a constant or is there more to it than that?

What I think I get is that circular birefringence measures the ellipticity of light, which involves a difference in the transmission of L and R circularly polarized components of an initially plane-polarized beam. Optical rotation simply measures the change in plane of polarization of light that has passed through a sample. My guess is that the two are different, otherwise one could just translate to degrees of rotation...? But if they are both measures of optical activity it would be also be surprising if they are related in some non-linear way.

*The note is from a lecture at Washington Univ. at St. Louis and I will try to get a better cite.

It gets very complicated, there are many papers and books on the topic and this little table, from Jensen et al. 1, gives a glimpse of the anisotropic effects:

An extremely short answer is this: when linearly polarized light is incident on an optically active medium, it can be thought of as consisting of equal components of right circularly polarized (RCP) and left circularly polarized (LCP) light. The indices of refraction for these are unequal, so RCP and LCP propagate at different speeds through the medium. This is circular birefringence. Upon exiting, it is linearly polarized light that exits, not the RCP and LCP fictive components. But the differential transit speed results in a rotation of the plane of polarization of the exiting linearly polarized light. This is optical activity, so optical activity is circular birefringence. The longer the optical path, the more the rotation and so on.
This is a truly vast topic, so I recommend not jumping directly into the deep end. Shurcliff's wonderful old book 2, maybe still findable, is extremely readable and a delight: the Poincare sphere is genius from a true genius! The book by Kliger et al. 3 is a good starting point and not insanely hard to understand. The paper by Jensen et al. 1 is rewarding, but definitely not easy reading.
There are many more papers, books, animations, and so on from which to learn. But be aware that notations are all over the place and reconciling equations and terminology can be very frustrating.
References:

H.P. Jensen, J.A. Schellman, T. Troxell, “Modulation Techniques in Polarization Spectroscopy”, Applied Spectroscopy 32 (1978) 192-200 (doi.org/10.1366/000370278774331567).

W.A. Shurcliff, Polarized Light, Harvard University Press,
Cambridge, MA, 1962.  For example on archive.org.

Kliger, D. S.; Lewis, J. W.; Randall, C. E. Polarized Light in Optics
and Spectroscopy, 1st ed.; Academic Press: Boston, 1990.  To borrow for  example from archive.org.

EDIT: In response to the OP's comment, I am adding two of my tables.

Circular birefringence (optical activity)
$$
\begin{align}
[φ] &≡ \text{molar optical activity}~(\pu{deg/cm M}) \\
\mathrm{CB} &≡ 2π(n_- - n_+)l/λ ≡ 2[φ]cl × π/180
\end{align}
$$

Circular dichroism
$$
\begin{align}
[θ] &≡ \text{molar ellipticity}~(\pu{deg/cm M}) \\
\mathrm{CD} &≡ \ln 10 (ε_- - ε_+)cl/2 ≡ 2[θ]cl × π/180 \\
g &≡ \text{dissymetry ratio} ≡ Δε/ε ≡ ΔA/A \\
ε &≡ \text{mean molar absorptivity} = (ε_- + ε_+)/2
\end{align}
$$

Complex angle of rotation, $χ$
$$
\begin{align}
χ &≡ χ' + iχ'' ≡ \mathrm{CB}/2 + i\mathrm{CD}/2 = φ + iθ \\
Δε &= 4π[θ]/180\ln 10 ≅ 0.03032[θ]
\end{align}
$$

Definitions of symbols and expressions
$$
\begin{align}
\mathrm{RCP} &≡ \text{right circularly polarized light (‘+’ subscript)} \\
\mathrm{LCP} &≡ \text{left circularly polarized light (‘−’ subscript)} \\
ε_± &≡ \text{molar absorptivities for RCP and LCP}~(\pu{L/mol cm}) \\
ε &≡ (ε_+ + ε_-)/2 ≡ \text{mean molar absorptivity}~(\pu{L/mol cm}) \\
A_± &≡ \text{absorbances for RCP and LCP} \\
A &≡ (A_+ + A_-)/2 ≡ \text{mean absorbance (true)} \\
n_± &≡ \text{(real) indices of refraction for RCP and LCP} \\
c &≡ \text{concentration}~(\pu{M}) \\
l &≡ \text{path length}~(\pu{cm}) \\
λ &≡ \text{wavelength}~(\pu{cm}) = 10^{-7} × \text{wavelength}~(\pu{nm}) \\
n &≡ (n_+ + n_-)/2 ≡ \text{mean index of refraction} \\
A_l &≡ \ln 10 (A_+ + A_-)/2 ≡ \text{mean absorbance} \\
η &≡ 2πnl/λ ≡ \text{phase (radians)} \\
[θ] &≡ \text{molar ellipticity}~(\pu{degree/cm M}) \\
[φ] &≡ \text{molar optical activity}~(\pu{degree/cm M}) \\
\mathrm{CD} &≡ \ln 10 (ε_- - ε_+)cl/2 ≡ 2[θ]cl(π/180) = 2θ \\
\mathrm{CB} &≡ 2π(n_- - n_+)l/λ ≡ 2[φ]cl(π/180) = 2φ \\
a &≡ \mathrm{CD}^2 - \mathrm{CB}^2 + \mathrm{LD}^2 - \mathrm{LB}^2 + \mathrm{LD'}^2 - \mathrm{LB'}^2  \\
b &≡ 2\mathrm{CD} × \mathrm{CB} + 2\mathrm{LD} × \mathrm{LB} + 2\mathrm{LD'} × \mathrm{LB'} \\B &≡ \left[(a^2 + b^2)^{0.5} + a\right]^{0.5}/2\sqrt{2} \\
C &≡ \left[(a^2 + b^2)^{0.5} - a\right]^{0.5}/2\sqrt{2} \\
Q &≡ B + iC
\end{align}
$$

The expressions for a, b, B, C, and Q are used in the general anisotropic medium Jones optical calculus and Mueller optical calculus matrices. These two matrices are denoted by J(GAM) and M(GAM), respectively, and are given in the paper by Jensen et al.


Why doesn't diffusion take place between immiscible liquids? If we take a test-tube of bromine and invert an upside-down test tube of air on top of it, it'll diffuse with the air, until the bromine is thoroughly 'mixed' with the air in both test tubes. But why doesn't the same thing happen with two immiscible liquids? Say oil and water.
When you put a layer of oil on top of water, the two don't mix, but they should. By diffusion.

In gaseous phases, we generally assume that there's very little or no bonding between individual molecules (or atoms in, say, a noble gas). This isn't strictly the case, but it's a pretty good approximation, and many good approximate techniques rely on it. A gas that behaves entirely in this way is known as an 'ideal gas'. (See also: ideal gas law.)

In contrast, liquids have bonding between individual molecules, and their behaviour is strongly affected by molecular properties.

So, in the examples you gave:

1. Bromine and air

In this example, imagine that you have a bunch of ping pong balls bouncing around at a reasonably high speed. (Ping pong balls are often used as examples for ideal gases.) Since the balls are already travelling at high speed (given a sensible temperature), and they only get in each other's way by bouncing off each other, it's relatively easy to get the bromine and the air 'balls' to mix, and there's really nothing apart from gravity to make them separate again.

2. Water and oil

Water molecules form pretty good bonds to other water molecules. (See: hydrogen bond.) Bonding in oils is generally weaker, but also occurs via a very different mechanism to the bonds in water. (See: Van der Waals force.) This means that oil does not form good bonds to water, and any water to diffuse into the bulk oil would have to break the bonds to all the neighbouring water molecules, and any oil to diffuse into the bulk water would have to separate a few water molecules to make a space to do so. Not just that, but since the water and oil molecules can reform their bonds if they manage to connect to other water and oil molecules, if you mix them up, the droplets will tend to recombine into the separate phases.

As a comparison to this example, let's check out two liquids that do mix.

3. Water and ethanol

For the water, we have basically the same situation as before -- water molecules forming good bonds to each other. The ethanol, though, has an -OH group that can form bonds to the water in the same way that the water does (though not as well). This means that ethanol that mixes with water (and vice versa) will tend to stay mixed, and given that the liquids are being mixed around just by random motions, means that you'll get one mixing with the other just as a matter of statistics.


Why are bonds ionic when the electronegativity difference between bonded atoms is greater than 1.7? I'm learning about how to recognise whether a bond is ionic or covalent, based on the difference in electronegativity between the two bonding partners, $\Delta \chi$.

What I have now is a formula:

If $\Delta \chi = 0$, then the bond is nonpolar
If $0 < \Delta\chi \leq 1.7$, then the bond is polar covalent
If $\Delta \chi > 1.7$, then the bond is ionic

But I don't know how scientists determined that formula, the history of it and which experiment indicates that formula.

The first thing to consider is the difference between covalent and ionic bonding, from the UCDavis ChemWiki site Ionic and Covalent Bonds,

  In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms that have the same electronegativity share electrons in covalent bonds since donating or receiving electrons is unfavorable.

The electron donor has a low electronegativity and the electron acceptor has a higher elelctronegativity - so there is a difference in electronegativity $\Delta{EN}$, effectively creating a positive and negative end, an example is below:

Image source: Japan Synchrotron Radiation Research Institute (JASRI)

When the differences in electronegativities of various compounds are graphed against % ionic character, as shown below:

Image source: University of Florida Chemical Bonding page

Values of $\Delta{EN}$ greater than 1.7 correspond to an ionic character of greater than 50%, from the University of Florida website:

  What determines how the electrons are shared is the relative electronegativity (electron greed) of the bonding atoms. The degree of polarity or degree of ionic bonding of any given bond can vary continuosly zero to nearly 100%. We normally say that bonds between atoms with electronegativity difference ($\Delta{EN}$) greater than 1.7 are ionic, although this really means only more than about half ionic in character.

Another resource is from the University of Washington Lecture 23: Ionic to Covalent Bonds.


Required HLB on the Griffin Scale What is the required HLB on the Griffin Scale of 94% water and 6% mineral oil for producing a stable micro-emulsion?

The required HLB doesn't depend on the concentration of water. The required HLB of mineral oil is ca. from 9 to 12. You have to find out the required HLB of your mineral oil at first.


Why is zinc not considered to have a noble gas configuration? I am wondering why we say that, for example, zinc does not have a noble gas configuration?

I would like to say beforehand that this question may has its origin in a confusion about the word "shell" or "outer shell". Wikipedia states that:

  The properties of the noble gases can be well explained by modern theories of atomic structure: their outer shell of valence electrons is considered to be full.

Ok, so argon is a noble gas and its configuration is: $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$. When we speak of a shell we mean the principle quantum number $n$. In this case, the highest principal quantum number is $n=3$. However, for $n=3$ there is also the $\ce{3d}$ subshell available, which is not filled by argon.

So I conclude that "filled shell" does not mean that all available subshells for a given principal quantum number (here $n=3$) have to be filled. Because the $\ce{3d}$ subshell is empty here. Instead, I conclude that "filled shell" means, that whenever we start to fill a subshell (here $\ce{3s}$ or $\ce{3p}$), they must be filled. If there is another subshell with $n=3$ available, it has to be completely empty (like $\ce{3d}$).

But when we look at zinc, we find the configuration: $\ce{[Ar] 3d^{10} 4s^2}$.

Again, for the highest principal quantum number (here $n=4$) we have a filled subshell (here $\ce{4s}$). Additionally, all shells of the $n=3$ are also filled. There is no subshell (neither for $n=3$ nor for $n=4$) which is partially occupied. So, according to the definition from Wikipedia, what is missing for zinc to be in a noble gas configuration?

Again, let me stress, that I am not asking for what a noble gas is (I know that zinc is neither a gas nor a noble gas). I am asking what I am doing wrong, when I conclude that zinc obeys the definition of a noble gas from Wikipedia?

The difference hinges on the word "valence" in "their outer shell of valence electrons is considered to be full".

That's the difference between a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ configuration and a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2}$ one. It's not just filled principle quantum numbers or filled subshells, but (un)filled valence subshells.

The definition of valence is a bit circular here. A valence subshell is one which can react to form a bond. For argon, all the valence subshells are filled because there's a substantial stability gap in adding that last $\ce{3p}$ electron and the first $\ce{4s}$ or $\ce{3d}$ electron. (If you think about potassium, the $\ce{3d}$ orbital is even less easily filled than the $\ce{4s}$ one.) That's why the $\ce{3d}$ orbital doesn't count as a "valence" orbital for Argon -- it takes too much energy to fill. Even though it has a principle quantum number of 3, it (re)acts like it has one of 4.

Contrariwise for zinc, there's a relatively small energy gap between the $\ce{4s}$, $\ce{3d}$ and $\ce{4p}$ orbitals. You probably already know this for $\ce{4s}$ and $\ce{3d}$, as they trade back and forth during the transition metals, but $\ce{4s}$ and $\ce{4d}$ are also relatively close in energy. This is part of the reason why atoms like carbon can "borrow" a $\ce{2s}$ electron to make four one-electron-donated bonds, rather than being forced to make two one-electron-donated bonds and one two-electrons-borrowed one.

So the definition of valence orbitals doesn't depend on their quantum numbers, but on the energy required to fill them. That's why zinc is not a noble gas - the $\ce{4p}$ orbitals count as valence (reactive) orbitals for zinc even while the $\ce{4d}$ don't.


Estimation of calcium EDTA How can i estimate calcium EDTA in a given sample?
We tried estimation by using mercuric nitrate but in vain we cant get a sharp end point.

Something crossed my mind - you could try titrating with Fe(III) standard using SCN- as indicator. First excess of Fe3+ would produce a sharp end-point.


Amine and Amide reactions with ketone Amines react with ketones whereas amides do not. Can anyone please explain why?

Protonation of the carbonyl group of a ketone allow amines to react via nucleophilic addition. This is because the added proton pulls electrons further away from the reaction site, thereby allowing the lone-pairs from the amine to attack the carbonyl carbon.

In the case of an amide, the C=O beside the nitrogen causes electrons to be pulled away because of the positive charge of the carbon. This means that despite the possibility of having a protonated ketone, there will still not be a reaction because the nitrogen is not nucleophilic.


How to balance the following equation? In the following unbalanced chemical equation the oxidation state of nitrogen ($+5$) is both decreased ($+5$ to $+2$) and unchanged ($+5$), so I am unable to balance it .

$$\ce{Cu + HNO3(dil) -> Cu(NO3)2 + 2NO + H2O}$$

My attempt: I tried by making equal decrease in oxidation number and increase in oxidation number but I am  not getting the given answer.

There are two half reactions:

elemental copper to cupric ion
nitrate ion to nitric oxide

The copper half reaction should be easy to balance, but the reduction of nitrate to nitric oxide is more complicated. start with what you know:

Nitrate goes to nitric oxide, one to one.
Then balance oxygen with water since it is aqueous, presumably.
Then balance hydrogen with protons since it is acidic from nitric acid.
Finally, balance the half reaction with free electrons, which should be three electrons for a single nitrogen atom in the reactants to be consistent with nitrogen going from +5 to +2.

There are different numbers of of electrons in the copper oxidation and nitrogen reduction half reactions at this point, so you have to multiply each half reaction to get the numbers of electrons in each equal to cancel out. The free protons should be converted into nitric acid by adding the appropriate amount of nitrate ion to each side. your final balanced equation should have a ratio of elemental copper to nitric acid of 3 to 8, respectively.


How could alcohol be made into a powder? In the news recently is 'Palcohol', powdered alcohol, that you add to water or existing liquids to make it alcohol. Presumably this would dissolve and break apart into alcohol and {some other compound}.

What could bond to ethyl alcohol in order to:
1) Create a solid, and
2) Not be disgusting as a side product?

Most things have some associated taste, $\ce{NaCl}$ is certainly out. If the 'other' components somehow combined to $\ce{H2O}$ or a gas like $\ce{O2}$ or $\ce{CO_2}$ that might remove any residual taste.

What would be in the list of available combinations with ethyl alcohol that are safe to consume, gaseous, or don't otherwise add weird tastes to the liquid?

Cyclodextrins, which comprise a class of cyclic carbohydrates, use their properties of being hydrophobic within their ring(s), yet hydrophilic outside. Alcohol can be put inside these rings. Then, the entire molecule is still capable of dissolving in water, releasing the encapsulated alcohol molecules. (src)

This popsci link has one way of making edible powdered alcohol. A nice casual step-by-step tutorial, actually...

Still you want to maintain a balance between maximizing alcohol (or do you?) and having the powder "feeling wet" due to high absorption of alcohol.

Also this may be a relevant Wikipedia article.


Energy of the Orbital versus Energy of the Electron When introducing the hydrogenic atomic orbitals, it is said that the 2s and 2p orbitals are degenerate. I have also read that for atoms with more than one electron, like He, the 2s and 2p are NOT degenerate.

How does the addition of a second electron in the 1s orbital change the energy of the 2s and 2p orbitals?

Let's imagine triplet $\ce{He}$, hence one electron in the 1s. The other electron could not be in the 1s, too (it would have to have opposite spin, thus, singlet $\ce{He}$). It is in either the 2s or one of the 2p orbitals.$^{1}$ Among other things, the Coulomb interaction with the 1s orbital/electron will affect their orbital energies. Given that the 1s and 2s have the same symmetry, but each of the 2p have a symmetry different from them, it is reasonable to assume that the Coulomb interaction between 1s and 2s would be different from the one of 1s and 2p.$^{2}$ Thus, the orbital energies are different.

$^{1}$ Though I would think that the electron density would always be spherically symmetric for the isolated atom.

$^{2}$ This can also be calculated when assuming some parameters for the helium orbitals (which can be calculated exactly for the one-electron case).


Are polar bond stronger than non-polar bonds? I came across Pauling's electronegativity scale in which he noticed that the experimental bond enthalpy of X-Y is always greater than the average of X-X & Y-Y and proposed that the difference, D, between the experimental bond enthalpy and the average of X-X & Y-Y gave a measure of the ionic character/polarity of the bond/ electronegativity difference
This seems to imply that the greater the electronegativity difference, the higher the bond dissociation enthalpy (as compared to the average of X-X & Y-Y).
I can't think of any explanation for this though. I came across one explanation stating that the partial charges result in an extra electrostatic attraction, but I can't find any reliable source for verification

The argument is circular: electronegativity is a measure of the tendency of atoms to attract electrons. Therefore electrons are attracted more strongly to electronegative atoms. More attractive has no more meaning than "bound more strongly" and a measure of binding strength can be regarded as a mixture of ionization energy and electron affinity, that is, an electronegativity parameter as defined by Mulliken.
Pauling's original definition of electronegativity relied on the idea that the bond dissociation energy contains contributions from a QM interaction plus an additional purely electrostatic interaction (a classical Coulombic interaction not involving a QM description of chemical bonds). In homonuclear diatomic molecules the electrostatic term is not important because the atoms "tug" at the electrons with equal strength, and so only a QM term contributes to bonding. For heteronuclear diatomics one can estimate the magnitude of the contribution of the QM term as an average of the terms in the two homonuclear molecules. Subtracting this average from the dissociation energy for the heteronuclear diatomic then results in an estimate of an electrostatic contribution to bonding. The electrostatic contribution is argued to be due to the different electron attraction in the two atoms, which is encoded in Pauling's electronegativity parameter. This is summarized by the equation for the heat of formation postulated by Pauling (in $\pu{kJ/mol}$):
$$Q_f = 100 (\chi_A-\chi_B)^2$$
The additional electrostatic term enhances the binding energy:

The greater the separation of two elements on the electronegativity scale, the greater is the strength of the bond between them. The extra stability is the resonance energy between the normal covalent structure and the ionic structure.

The excerpt comes "from the horses mouth" (Pauling's General Chemistry textbook). If the explanation is not satisfying I recommend looking into more advanced theories.