text:
Chemical Thermodynamic Functions
I'm a high school student with basic knowledge about chemical thermodynamics. I have three questions regarding the thermodynamic state functions $H, S, G$. I have searched a lot in textbooks but they don't give the complete information as to whether the properties taken are that of the system or surroundings.
Q1):-Enthalpy is given by the formula
$H = U +PV$ .
 What I want to know is that whether $P$ mentioned above is the internal pressure of the system or the externally applied pressure?
Q2):- The change in entropy is given by the formula
$\Delta S= $$\int_1^2$${\frac{dq_{reversible}}{T}}$. Is the temperature mentioned of the system or the surroundings.
Q3)-On the same lines what temperature is taken to define the Gibbs free energy function:-$G=H-TS$.
Thanks in advance.

The three functions you mentioned are applicable only to thermodynamic equilibrium states of the system, for which the pressures and temperatures of the system and its surroundings match one another to within insignificantly small differences.  In the integral to get the entropy change (Q2), the same matching applies; that is, a reversible path consists of a continuous sequence of thermodynamic equilibrium states.
Note that, at the interface between the system and the surroundings, the temperature and pressure of the system always matches the temperature and pressure of the surroundings (irrespective of whether the system is at equilibrium).  It's just that, along an irreversible path of the system (for which the system passes through a sequence of non-equilibrium states), the pressure and temperature within the system vary with spatial position, and their average values (averaged over the volume of the system) do not match those of the surroundings at the interface.

Thanks for your enlightening answer.. Does it mean that I can apply these equations only during initial and final states for an irreversible process (when the system attains equilibrium after passing through a series of non euilibrium states )and not when the system is undergoing that process?

I also came across a statement that change in enthalpy when a gas undergoes free expansion process is equal to zero. How would you account for this statement sir? If I assume that the equation mentioned in q1 is applicable only in equilibrium states and the pressure is that of the gas it would give me a non zero change in enthalpy. Am I going wrong somewhere?

I forgot to mention that the aforementioned process is an adiabatic free expansion.

The answer to your first question is "yes" for the global system undergoing the irreversible change (because the pressure, temperature, and specific volume are varying with position within the system).  But, on a small scale, the answer depends on whether you are willing to approximate the thermodynamic functions locally (per unit mass) using the local conditions of temperature and pressure.  We engineers are, but some physicists, chemists, and other purists are not.

Regarding your questions about Joule Thompson and adiabatic free expansion, I feel it would be better to ask these questions in a new thread.


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Has this way of comparing internal energy got any sense?
Can internal energy between molecules be compared?
For example, $\ce{H2}$, $\ce{O2}$ and $\ce{N2}$ have some internal energy ($U$) but taking infinite distance between atoms as zero. But each reference point is different. So, can $U$ for those molecules be compared?

Yes. This is the binding energy.
In general, the binding energy is defined as the amount of energy released upon bringing infinitely separated particles in to contact. This would then be a negative number. Another sign convention is the binding energy can be viewed as the amount of work required to separate two atoms or molecules (or whatever) which are in contact. This binding energy is the negative of the first one (is a positive number).
For molecules, you might be more familiar with the term bond energy instead of binding energy. For a diatomic, these two things are the same. Notice that when we talk about the bond energy in a molecule, we treat that bond as being a diatomic molecule.
For the diatomics you mentioned, the electronic energy for the heavier diatomics will be artificially larger than the lighter ones simply because there are more electrons. For this reason, when comparisons between systems like this are made, people usually divide by the total number of electrons. So, in this case you can visualize putting in positive work to bring the protons near to each other and then adding the electrons one at a time to the system until it has no net charge. The binding energy per electron is then a measure of the average amount of stabilization the system gained for each new electron. The same thing can be visualized for the nucleus of an atom, but in that context you will read of the binding energy per nucleon.
This same concept is also quite useful for molecular clusters and other non-covalently bound systems as a measure of the strength of intermolecular interactions. In that case it is infinitely separated molecules which define the zero of energy.
This concept of energy relative to infinitely separated components of a system is a very useful one as it gives an unambiguous definition of the strength of interactions. As a fun thing to ponder, consider how to determine the strength of an internal hydrogen bond? As it happens, to the best of my knowledge, there is no general method for determining the strength of internal hydrogen bonds as there is no natural zero of energy to choose. Some people have tried using the red shift in vibrational frequency, but this too is still a bit ambiguous and hard to determine.

An energy relative to what?

Yes, but you can only know this when you bring the atoms into contact and this is exactly why the binding energy is different for the different molecules. They have different electronic and vibrational energies as molecules (and as atoms), but by choosing the zero as the atoms (with all of their electrons), you get a measure of the energy change which comes from just bringing the atoms together. If you wanted, you could choose the zero to be the bare nuclei, but this is a measure of something else entirely, and if you do this you're probably more interested in the atom than the molecule.

What? I'm not sure what is confusing you? If you're confused by choosing the zero of energy as the infinitely separated atoms, consider choosing the zero of energy as the minimum of of the potential energy curve. This is clearly an unambiguous point, and the the potential energy of the system is easily found there. You can choose to think about the kinetic energy of the nuclei if you want.  One of these is $D_e$ and the other is $D_0$ where the two differ only by zero-point energy. These will be approximately described by the Morse Potential

Let us continue this discussion in chat.

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Reactivity of Aluminium Oxide
I know that Aluminium Oxide is an amphoteric substance meaning that it reacts with both acids and bases. I recently learned that Aluminium has very good corrosion resistance due to the "Aluminium spontaneously forms a thin but effective oxide layer that prevents further oxidation."
However, in the test for Nitrate Ions, an Aluminium foil is added into the solution, the Nitrate Ion is reduced by the Aluminium, but this Aluminium foil surely has Aluminium Oxide which should prevent the reaction from happening, but it happens anyway. How can this phenomenon be explained?
Source: http://www.aluminiumdesign.net/design-support/aluminium-corrosion-resistance/

What other chemicals are added in the test for nitrate ions?

Aluminum oxide coating can relatively easily react with either strong base or acid, dissolving in them, exposing naked elemental aluminum to reaction medium. You can read more  here.
To be concise, there are many ways to make aluminum react: you can use a base, you can use an acid, and you can scratch a sample of aluminum and coat it with mercury. Mercury would form an alloy with Al, but that wouldn't affect its reactivity towards the solution of anything, it would just make it harder for oxygen to oxidize the fresh aluminum.
Answering you question: as far as I know, the nitrate test involves strong base, which serves as a reactant for aluminum oxide, dissolving it and letting aluminum to react further.

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Effect of pressure on the freezing temperature of water
A few lines are written in a book by B.H. Mahan regarding the change in freezing point of water on changing the pressure.

The vapor pressure of water at triple point of water is 4.579 mm. What would happen if the external pressure applied to the system by the piston were to be increased above 4.579 mm ? First of all the va pour present would be completely converted into liquid and solid, to restore the equilibrium pressure.

I'm in total agreement of this. But the next lines appear to be odd.

It is found experimentally that as the pressure on the system is further increased the temperature of the system must be decreased, in order for both  ice and water to remain in equilibrium.

I don't understand  the second statement. If the equilibrium is further disturbed why the system should not  behave the same as before? Why do we have to change the equilibrium constant  to restore the equilibrium?
Thanks in advance.

From the triple point (273.16K, 611.657 Pa), if pressure is increased and temperature maintained at 273.16K, all the water being liquid is the equilibrium condition.
To maintain the solid and liquid phases in equilibrium, as pressure is increased above the triple point pressure, temperature must decrease slightly (to 273.15K if the pressure has increased to atmospheric pressure, for example).

Can you please explain this from the perspective of equilibrium and le Charlie's principle?

@gowreeshmago Liquid is the most dense phase of water, so increasing pressure favors the liquid phase with respect to the vapor or solid phase.


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Is it possible to calculate electronegativity of surface atoms?
As the title asks, is it possible to calculate the electronegativity (EN) of specific atoms on extended surfaces (slabs), more specifically using computational methods? There are a few issues here I am running into:
To find the Mulliken EN $\chi$, we can use the relation
$$\chi = \frac{E_\mathrm{i} + E_\mathrm{ea}}{2}$$
where $E_\mathrm{i} = E(\ce{M+}) - E(\ce{M})$ and $E_\mathrm{ea} = E(\ce{M})-E(\ce{M-})$. However, these energies are for the entire system, usually a molecule, and not the individual atoms. I hope I am not misunderstanding the concept here.
Alternatively certain definitions use $\chi = -\mu$, or the work function. Again, this is usually calculated on a surface basis, not per atom of site.
Is there no accurate determination method which is specific to the atom? Or is it conceptually impossible?

Have you looked at the Fermi surface? https://en.wikipedia.org/wiki/Fermi_surface


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Does MOPAC allows calculation of Standard Enthalpy of reaction in this way?
I am using MOPAC2016 to calculate $\Delta H^\circ_\mathrm{r}$ (and then $\Delta G^\circ_\mathrm{r}$) for a reaction in aqueous phase, at $\pu{298 K}$. I will explain how, and then the question. Finally I give an example. Suppose this simple reaction:
\begin{align}
\ce{A(aq)<=>B(aq)}
\end{align}
Then, $\Delta H^\circ_\mathrm{r}$ can be obtained as:
$$\Delta H^\circ_\mathrm{r} = \Delta H_\mathrm{f}^\circ(\ce{B}) - \Delta H_\mathrm{f}^\circ(\ce{A})$$ where each of those values is what we get on a thermodynamic calculation (also in an optimization) with the name of H.O.F. (heat of formation, HOF).
Formally the question should be: Is that correct to calculate $\Delta H^\circ_\mathrm{r}$ (and then $\Delta G^\circ_\mathrm{r}$) by using the value of HOF for each reactant?
Example
Here is an example of MOPAC output for a thermodynamic calculation, suppose it is the one for molecule $\ce{B}$:
                    CALCULATED THERMODYNAMIC PROPERTIES
                                          *
   TEMP. (K)   PARTITION FUNCTION   H.O.F.    ENTHALPY  HEAT CAPACITY ENTROPY
                                   KCAL/MOL   CAL/MOLE    CAL/K/MOL  CAL/K/MOL

 298.00  VIB.     0.2155D+06                  8205.6617    53.3353    51.9400
         ROT.     0.3572D+07                   888.2813     2.9808    32.9651
         INT.     0.7697D+12                  9093.9430    56.3161    84.9051
         TRA.     0.4009D+28                  1480.4688     4.9680    42.5424
         TOT.                     -670.232   10574.4119    61.2841   127.4474

   *: NOTE: HEATS OF FORMATION ARE RELATIVE TO THE
            ELEMENTS IN THEIR STANDARD STATE AT 298K
              (=  Standard Enthalpy of Formation).

            Hvib:  Zero-point energy is not included.
                   frequencies of less than zero cm-1 are not included.
            Hrot = (3/2)RT
            Htra = (3/2)RT + pV = (5/2)RT

            Heat capacity is Cp, not Cv

          For more information, see: HTTP://OpenMOPAC.net/Manual/thermochemistry.html

With that data table $\Delta G^\circ_\mathrm{f}$ of the molecule $\ce{B}$ should be:
$$\Delta G^\circ_\mathrm{f}= \mathtt{H.O.F.} - TS^\circ= -670.232\times1000 - 298\times(127.4474- S_{atoms})$$
$S_{atoms}$ will be canceled when calculating $\Delta G^\circ_r$.
On this mopac site, $\Delta H_\mathrm{f}^\circ$ appears to be obtained in a similar way, for the $\ce{NH4+}$ molecule.
Am I correct?

I have no experience with MOPAC, and I don't completely understand the output, but your method looks correct to me.

great, @Martin-マーチン, and thanks for the comment. For some reasons I don't understand most of people working on it don't use this approach..

Semi-empiric calculations are much too inaccurate for many systems and many people use other methods instead. DFT and other ab initio methods allow to calculate an absolute value for the electronic energy and thermodynamic corrections and using those to calculate the relative energies. (The linked manual section is very worth reading, there is more information on how thermochemistry is treated in MOPAC.)

@Martin-マーチン I mean that they use mopac but calculate gibbs free energy in a different way. I didn't specify how because it is long..; this methods aren't too accurate but are parametrized to work good, so, they are quite accurate for most purposes (and also are fast)..

$\Delta G_f^\circ$, like, $\Delta H_f^\circ$, is relative to elements in their standard state, so the entropy is missing a term, but that will also cancel when you compute $\Delta G_r^\circ$.  So, for 298K, the equation for $\Delta G_f^\circ$ will give the right $\Delta G_r^\circ$.  For other temps you also need $(H_T - 10574.4119)$ where $H_T$ is the corresponding total $H$ for temperature $T$.

@JanJensen right..so gibbs free energy of reaction can be calculated directly with $\Delta G^\circ_f$ of the species in the reaction and don't need to be relative to gas phase species?(I thought you were calculating by this approach)

The free energy change of a reaction occurring in solution can be computed using a thermodynamic cycle:
$$\Delta G^\circ_{r,soln} = \Delta G^\circ_{r,gas} + \Delta\Delta G^\circ_{r,solv} $$
$\Delta G^\circ_{r,soln}$ can also be written as
$$\Delta G^\circ_{r,soln} = G^\circ_{soln}(B)- G^\circ_{soln}(A)$$
where
$$G^\circ_{soln}(X) = G^\circ_{gas}(X) + \Delta G^\circ_{solv} $$
and
$$G^\circ_{gas}(X) = E_{gas}(X) + G^\circ_{gas,RRHO}(X)$$
The solvation energy is usually computed using a continuum as the difference in electronic energy with and without the continuum.
$\Delta G^\circ_{solv}(X) = E^\prime_{soln}(X)-E^\prime_{gas}(X)$
The parameters in the continuum model are usually optimized for a particular level of theory, which may not be appropriate for the computation of $\Delta G^\circ_{r,gas}$, so $E_{gas}(X)$ is not necessarily the same as $E^\prime_{gas}(X)$
However, if $E_{gas}(X) = E^\prime_{gas}(X)$ then
$$G^\circ_{soln}(X) = E_{soln}(X) + G^\circ_{gas,RRHO}(X) $$
The parameters in the continuum model are usually determined using the gas phase geometry $(X)$.  But if $E_{gas}(X) = E^\prime_{gas}(X)$ then many people often choose to use the geometry optimized in solution
$G^\circ_{soln}(X_{soln}) = E_{soln}(X_{soln}) + G^\circ_{soln,RRHO}(X_{soln})$.
For semiempirical calculations, the electronic energy corresponds to the enthalpy of formation at 298K, so if one simply makes the substitution
$$E(X) \rightarrow \Delta H^\circ_{f,298}(X)$$
then (at 298K)
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) -TS^\circ_{gas,RRHO}(X)$$ $G^\circ_{soln}(X)$ doesn't correspond to a measurable free energy because $S^\circ_{gas,RRHO} \ne S^\circ_{f}$ but it can still be used to compute $\Delta G^\circ_{r,soln}$ since the missing terms cancel.
For temperatures other than 298K
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) + (H^\circ_{gas,RRHO}(X,T)-H^\circ_{gas,RRHO}(X,T=298K)) -TS^\circ_{gas,RRHO}(X)$$

I don't understand why the gas-solution cycle is needed, although I understand the procedure you wrote. But if mopac give  us "$G_f$
"for each reactant in solution, why don't we direcly sum those values?(with corresponding index for reactant and products) –

the cycle is needed because the continuum model has parameters that need to be fitted to measured values.  You can't measure $G^\circ_{soln}(X)$, but you can measure $\Delta G^\circ_{solv}(X)$. So you need the cycle to relate $\Delta G^\circ_{solv}(X)$. to $G^\circ_{soln}(X)$


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Why only structures corresponding to valleys on PES are taken into account?
I have had to ask this question again because it was marked as a duplicate but it wasn't. Here it goes..
Suppose we want to estimate $G_f^\circ$ of a molecule. If it has rotational conformers, right way to calculate $G$ should include a Boltzmann average of them.
We will do it as:
$$\large \langle G_f^\circ \rangle = \frac{\sum_{i}G_f(i)e^{-G_f(i)/k_BT}}{\sum_{i}e^{^{-G_f(i)/k_BT}}}$$
where the sum goes over the conformers.
Question:
Why $\langle G_f^\circ \rangle$ is estimated as the sum of $G_f(i)$ of only most stable conformers? (corresponding to minima on PES)
Example
Suppose the following rotational conformers:

Consider the $y-axis$ to be gibbs free energy, we would estimate $\langle G_f^\circ \rangle$ as the sum of structures corresponding to minima (ie 60º, 180º), but there are non-minimum structures around 180º with less energy than the minima in 60º. Why aren't them considered?
EDIT
Someone could argue that a conformer in 150º won't exist because there will be a driving force to 180º, but if we take the relative energies $\frac{G_f(150º)}{G_f(60º)}$ we get that 150º must have almost the same population as 60º. There might be some criteria I am missing or maybe a conceptual error.
Any edit will be welcome.

It could be that the unstable conformers are so low a fraction of the species as to be negligible.

No, the last part of the post tries to explain that point.@a-cyclohexane-molecule

Are you only considering global minima, then? In your question statement, you mention local minima, and I'm not sure to which you refer in the last part of your post.

No, I mean, why estimate mean gibbs free energy as the sum of gibbs free energy of only most stable conformers?@a-cyclohexane-molecule. Thanks for the comments..

I think this is a good question because I would like to see a rigorous explanation of this. However, my intuition from the previous iteration of this question remains the same that the issue in how we define a state. I think the argument at the end of your post is invalid because it implicitly assumes that $150^{\circ}$ is an occupiable state of the system. The only way it would be at that configuration is a brief fluctuation of around the bottom of the $180^{\circ}$ well or en route to the $60^{\circ}$ well.

@Tyberius it is an occupiable state of the system, why not? It is a position of nucleus..

@HernanMiraola by analogy, think of a two state system: you are in an airplane flying at a constant height or you are on a level stretch on the ground. You can pass between the two states and briefly be at a point that doesn't coincide with those two states. But you will rapidly collapse to one or the other. If you jump out of the plane, there is zero probability that you hang in the air somewhere in between the ground and the plane. The same is true if you jump from the ground. The space in between the states is not occupiable in this sense.

When you are using the Gibbs energy, then the vibrational modes within the minimum are already considered, and the nuclei arrangement at angles slightly off the minimum correspond to vibrational distortions.

The vast majority of thermodynamics we consider is at equilibrium, we is the issue here.
At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated.  This this because they have now force acting on them to drive them to another configuration.
An important relationship to remember with potential energy surfaces is:
$$
F = -\frac{dE}{dx}
$$
Force is the negative derivative of potential energy with respect to some coordinate.
This means that the driving force is given by the slope of the PES and is pushing down the PES.
Then a molecule with a $\phi = 150^\circ$ experiences a rotational force pushing it towards the minimum at $\phi = 180^\circ$.
Answer:
The result is that any molecule not at a minimum will be quickly driven towards the nearest minimum, so only the minima will be occupied and need to be summed over.
Side Note - Maxima
Maxima (the peaks) also experience zero force.  However this is only true at the exact peak itself.  Any point infinitesimally far away from the maxima experiences a force pushing it down the hill into a minima.
Thermal fluctuations and collisions with other molecules will continually slightly nudge the molecules around on the potential energy surface.  At maxima this will push any molecules there off the peaks and down into minima.  In minima, the force will bring the molecules back into the minima.

Nice answer. And what about the relation $G_f(150)/G_f(60)$? Why is that reasoning wrong? I mean, we don't consider them because are only a few at equilibria?

That relationship is only valid for comparing the population of different minima, hence not for $G_f(150^\circ)$ as $150^\circ$ is not a minima.  The population of any state that is not a minima will be zero (or at most , briefly, vanishingly small before relaxing back to the minima).

I think there must be a deeper explanation, I mean, I am not sure that because it is not a minima we can't take that fraction of the previous comment. But it is helpful..thanks..

I'm not sure I believe this statement: "At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated." At equilibrium these states should be populated according to the Boltzmann distribution, and certainly higher-energy states have some, if little, population.

@user213305 What about cis/trans isomers when a double is present? Should treat them as rotational conformers?

You are right that this is an approximation. It is, however, a much better approximation than you might think. The assumption is that each of the wells are shaped approximately the same. Furthermore, because the population of a state decreases exponentially with energy, it is really the approximation that each of the wells is shaped the same near the minimum.
It is well known that the smaller the displacements from the minimum, the more harmonic the motion will be. So, a measure of how much error is introduced by this approximation is the difference in harmonic frequency of the torsion in each of these wells. And, in the case that the frequencies in each of these wells is the same, the equation you have written is exact within the harmonic approximation. If the true (experimental) frequencies are identical, then the equation you have written is exact.
To be extra clear, for identically shaped wells, all deviations from the minimum energy structure will cancel out because when you do the Boltzmann averaging in each well, the population above any minimum will follow a Boltzmann distribution. Thus, the contributions from these deviations will be very minimal, and depend on there being different shapes of the well.
Probably the simplest way to include the shapes of the different wells in a calculation like this would be to construct the PES using some kind of ab initio method and then do a dynamics simulation on this surface to get the free energies this way.

The Boltzmann distribution doesn't only take into account the most stable conformers. It can take the energy of any conformer (relative to the energy of the most stable conformer).
I assume that you're calculating the Gibbs energy using $\Delta G = - RT \text{ ln}K$. In this equation, $K$ represents the equilibrium constant between two local minima, i.e. between anti and gauche. So, when you calculate the Gibbs energy in this way, you're actually calculating the Gibbs energy for the rotation between two conformers (you can sort of imagine this as a reaction).

Oh, I am sorry I edit about G, I will add it again.

The standard Gibbs free energy is "the change of Gibbs free energy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states". You only need to consider the elementary components and the minimum energy structure.


text:
Site of first protonation of imidazole and aminoimidazole
I wish to ask where imidazole and aminoimidazole will be protonated on first protonation.
Imidazole:

Aminoimidazole:

I suppose that the nitrogen (without hydrogen) of the ring is the most probable to be protonated, but I am not sure about it. Do you know where it will be protonated first?
Any link to a study where the issue is solved will be good (I need the information for my thesis).

You can predict pKa values with the online webserver Chemicalize https://chemicalize.com/

So you suggest to protonate the different nitrogens and see the result with that tool?

No, just draw the molecule as you did above and Chemicalize will tell you the most probable  de/protonated isomer as a function of pH

@JanJensen I don't know why it doesn't produces a value for -NH2 in 2-aminoimidazole, SMILES is c1cnc([nH]1)N .  Maybe it just can't be protonated because of the resonance.

Yes, Chemicalize doesn't consider all protonation states.  Only the most probable. The $\ce{NH2}$ group won't be protonated for the reason you stated

You already asked about it https://chemistry.stackexchange.com/questions/80504/protonation-of-compounds-with-two-or-three-neighboring-nitrogens

Possible duplicate of Protonation of Guanidine

For imidazole, the lone pair of pyrrole-type nitrogen (the one with hydrogen) is involved in aromaticity of the whole compound, as it contributes two π-electrons in addition to four of those from carbon-carbon or carbon-nitrogen bonds, which makes it 6, which is a classical sign of Huckel aromaticity. In fact, this NH-nitrogen tends more to act as an acidic center than as a basic one, as it loses its hydrogen when treated by strong bases, like $\ce{NaNH2}$.
The pyridine-type nitrogen has its lone pair out of ring conjugation, thus it can act as a acceptor of hydrogen ion, placing its two electrons into a vacant orbital of a $\ce{H+}$.
For reference, you can see any heterocyclic chemistry textbook. I prefer the one authored by Joule, Mills. Or this.

When it comes to 2-aminoimidazole, this publication  (which is unarguably very old) says all three nitrogens are involved in protonation, because the structure is a derivative of guanidine, a strong organic base. Which seems only legit to me if it said that the contribution of pyrrole-type nitrogen is negligible, as aromaticity is ruined in that case. I could also predict that exo-aminogroup gets protonated more likely, as its basicity is somewhat higher: its lone pair is on a sp3-orbital, rather than the lone pair of endo-nitrogen (the one without H), where it is on a sp2-orbital, thus, the contribution of s-orbital is higher and it is closer to the nucleus of nitrogen atom, decreasing its availability to act as a covalent bond acceptor.

(Yes, I know that hybridization theory is false and is not really loved here, but why not use it, when it is a powerful teaching instrument and helps to explain stuff)

Oh no, in the case of organic compounds, hybridisation theory is actually rather well received because it does a good job on explaining this without having to solve the Hamiltonian.

However, the exo-amino group’s lone pair will very likely not be in an $\mathrm{sp^3}$ hybrid orbital until very high $\mathrm{pH}$ values because it takes part in resonance. (Compare with the amino group of aniline.) Also, that third structure you are drawing there in the series of equilibria deserves a downvote.

I do absolutely agree about not-hundred-percent availability of exo-aminogroup lone pair, but still the comparative basicity of those two nitrogens is still a good question, which need more references to be found. But the second scheme, as you can read from my answer, is not my creation, but an illustration for the paper I mentioned.

If the paper you are referring to is the one you linked (B. T. Storey, W. W. Sullivan, C. L. Moyer, J. Am. Chem. Soc. 1964, 29, 3118–3120. DOI: 10.1021/jo01033a537.) I can only suggest you actually read it since it supports none of your claims.

@Jan I think you’re being somewhat unfair. OP already mentioned that the contribution of the pyrrole-type nitrogen is negligible as aromaticity is ruined, which is precisely the third structure you are complaining about (the meaning is that this structure contributes negligibly to the equilibrium mixture). It could perhaps be explained or drawn slightly clearer, but it is not wrong. BTW DMAP is a useful analogy to draw. Where does it get protonated?

Hückel aromaticity in a heterocycle; no.


text:
Any program for conformational searching for sugars?
I want to estimate mean Gibbs free energy of formation of glucose and for that purpose I think conformers of ring flip should be taken into account. Do you know if some free program can find those structures automatically? (for example Balloon).
The following image shows the conformers I need to get. I am trying to avoid the possibility of drawing it.

I use RDKit for conformation searching.  More info here https://proteinsandwavefunctions.blogspot.dk/2016/08/conformer-search-with-rdkit.html

I'm sorry (new to chemistry here, but your question is interesting to me and I would like some clarification.) Are you asking for software that will generate conformers for you? (Just thought it was a cool question and wanted clarity.)

@MelanieShebel yes, some free programs that can do that are: balloon and RDkit, you can download them in the web. Also avogadro can do that, although I never try it :)..

Use RDKit
It sounds like you're only considering free or open source conformer generation tools. In that case, I highly recommend using RDKit, since the distance-geometry (and in turn ETKDG) methods will intrinsically generate all possible ring geometries. That includes not just the two chair forms you indicate, but all possible ring-puckering forms.
There are other free tools which do this, but RDKit has been repeatedly shown to be the best or among the best free conformer generation tools. (Based on current benchmarks.)

"High-Quality Dataset of Protein-Bound Ligand Conformations and Its Application to Benchmarking Conformer Ensemble Generators." J Chem Inf Model 2017 57, 529-539.
"Freely Available Conformer Generation Methods: How Good Are They?" J Chem Inf Model 2012, 52 (5), 1146–1158.

I'd advocate the newer ETKDG method with RDKit, since it seems to produce somewhat better geometries.

Hi, thanks for the reply. As far as I know RDkit doesn't move B-A-O-H diedrals(doesn't see hydrogens), and also doesn't find conformers for ciclohexane. Have you tried that?

With most conformer programs, the hydrogens are added at the end. As for cyclohexane, yes RDKit can generate multiple different conformers.

Is it possible to add hydrogens to oxygen and nitrogen before the conf search for those programs? Otherwise, I think some relevant structures will be missed, mainly in sugars.

I am not aware of a program that does that, unless you want to run a PMF yourself, but, on the side, here is a systematic study of the energetics of monosaccharide sugar puckering: Heather B. Mayes, Linda J. Broadbelt, and Gregg T. Beckham, J. Am. Chem. Soc. 2014, 136 (3), 1008–1022. DOI: 10.1021/ja410264d. You may find here the info you are looking for.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. We prefer to use human-readable references and non-pdf links, I have edited your post accordingly.

great!Thanks Elisa. Would you clarify PMF?

PMF stands for potential of mean force. Here is a review you may find interesting, http://www.sciencedirect.com/science/article/pii/001046559500053I


text:
Why does only a single azeotropic composition exist for a mixture of two given substances?
In the image below, which represents a generic composition vs. temperature plot for a mixture, there is a single intersection point between the vapor and liquid curves. This is due to the fact that the composition corresponding to this intersection point has an extremely low boiling point due to negative deviation from Raoult's law. If this is the case, why is it not possible for the vapor curve to "hug" the liquid curve along an entire section, rather than just at one point? Similarly, why does the vapor curve have to intersect liquid curve at all? Would it be possible for the minimum point of the vapor curve to be located just above the minimum point of the liquid curve?


text:
Why does a carbocation undergo covalent bonding with, say, chloride, rather than ionic bonding, like what occurs between Na+ and Cl-?
I've been trying to look for an answer for a while but I can't find one. Why does a carbocation accept an electron pair from a nucleophile and form a covalent bond rather than just accepting one electron like what happens between a sodium ion and a chloride ion in the formation of sodium chloride?

Why do two chlorine atoms not form an ionic bond with $\ce{Cl+}$ and $\ce{Cl-}$?

check out https://chemistry.stackexchange.com/a/17072/9961 and maybe https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond

Chemistry 001, octet rule?

Some carbocations do form true ionic compounds.  Many are known in which the positive charge is incorporated into an aromatic ring (e.g., cyclopropenyl, tropyl) or conjugated with multiple aromatic rings (e.g., triphenylmethyl).  But in most carbocations you have a compact, relatively low-energy vacant orbital that will overlap to form a bond with an electron pair from, say, a chloride ion.  Alkali metals don't have such compact valence orbitals and thus do not form a strong covalent bond even with the positive charge all on one atom.

You've reminded me that it's possible to make pure hydrocarbons which are actually ionic salts. Amazingly, these substances reversibly dissociate in organic solvents and their solutions conduct electricity!

I remember seeing those.  They are awesome.  Might they be better approaches to true, full-fledged ions than things like sodium chloride?

@OscarLanzi No, if you want as much ionic character as possible then check Non-coordinating anions.

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


text:
Fabrication and/or Selection of a Single Crystal Ionic Conductor
I want to fabricate or buy single crystal ionic conductors for usage as solid state electrolytes in batteries testing. However, I am having trouble finding anyone that can sell appropriate materials.
From my understanding, lanthanium series elements typically have good ionic conductivities. Thus, it seems like Strontium Lanthanum aluminum oxide (SrLaAlO4) or lanthanum aluminum oxide-strontium aluminum tantalum oxide (LSAT) which are both sold in single crystal form, may be suitable. However, I'm having trouble determining if they are compatible with the lithium ion.
Meanwhile, I know that I have seen SiO2 and Al2O3 used as barriers in battery electrodes, and it's possible that single crystals of either of these may be possible to use. However, just because it is a barrier doesn't necessarily render it a solid state electrolyte. Meanwhile, it appears alpha-Al2O3 has previously been used as a separator, I'm unsure if this is truly because the alpha-Al2O3 can transport the lithium versus the pores of the material transporting the lithium. When trying to discover which was the case, I found this paper discusses the transport mechanism of lithium in Al2O3, but I am unsure if this applied to alpha-Al2O3.
Lastly, I can also find MgAl2O4 single crystals, which appear to be good ionic conductors of lithium but only if doped with lithium. However, I do not know a way to create single crystal LiMgAl2O4, starting with MgAl2O4 single crystal.
Does there exist a suitable single crystal material which can be easily fabricated or bought for my purposes?

I'm confused about what you are trying to do. The electrodes in batteries are solids in general. Are you trying to make a completely solid state battery?

At what temperature would this operate? Beta-alumina, for example, provides appreciable current only above 200°C. See http://www.ionotec.com/conductive-ceramics.html for supplier,  https://en.wikipedia.org/wiki/Beta-alumina_solid_electrolyte for info.

To be clear, I'm wanting to use this purely as a solid state electrolyte. That means that it separates the two electrodes, providing a path for lithium ions but having little electrical conductivity.


text:
How to read these graphs and what are they called?
How does one interpret/read the following graph (found here) as well as on wikipedia or

What are these called?

According to your first link, this is from an x-ray diffraction image of human obesity protein, or leptin. It's called a ribbon diagram (and maybe other things, not my area of expertise).

Well, I'd call them coordinate/arbitrary axes...if you're talking about the arrows at the bottom left ;)

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

This depiction is a ribbon diagram of a protein structure. It shows the protein backbone (alpha carbons and the peptide bonds) smoothed out to a continuous line. In certain areas where the protein adopts alpha-helices or beta-sheets, these are indicated by the elaborate spiral in your picture (alpha-helix) or a wide, flat arrow (beta-sheet; not present in your structure). These depictions are supposed to give you an idea of what the protein looks like.
I would assume the two arrows in the bottom to be two of the coordinate axes of the crystal since ribbon diagrams are typically obtained from crystal structures. However, I cannot know without additional information (e.g. in a footnote of a picture).


text:
How can I remove a TiO2 coating from my car windows?
I've tried all the common solvents including hydrogen peroxide, chlorine, acetone. I tried a high speed sander with a rare earth polish. I have spoken to the head of the chemistry dept. at Cornell University and she said she did not have a remedy that would not ruin the glass. From the MDS here's what's in it:
1.Polymeric mixture with nano   5%-10%
2.Pexi Polymer   5%-10%
3.Anti static additive    5%-8%
4.Anti rust additive   3%-6%
5.Complex micro emulsion    2%-5%
6.Organofunctional siloxane blend    4%-8%
7.Synthetic nano driver polymeric fusion   10%-20%
8.Hydrogen hydroxide 40%-65%
Any suggestions would be greatly appreciated. The owner/developer of this coating won't take responsibility and help me get this stuff off my windows of my new car. Thank you.

"Hydrogen hydroxide" Buhahaha :D Nice joke...

The "head of the chemistry dept. at Cornell" is my favourite in this post. Troll?

"high speed sander", "rare earth polish" - nice.

@Mithoron . Hi. This is not a joke. The ingredients I listed, including hydrogen hydroxide, were taken word for word from the data sheet the developer of this product gave me when gave me this product to protect the glass from mineral deposits from my city water.

Of course you didn't know it's babble.  That makes it funnier :D

Well this isn't going like I'd hoped. First off I'm an illustrator so I have very little knowledge of what the ingredients in this coating are or do other than the many hours I've spent researching this on line of late. That's why I posted this question here in hopes someone who knows much more about chemistry might have a suggestion on how to get this stuff off my windows. I did speak with Melisa Hines at Cornell yesterday and I did hire a glass cleaning company to get this off the glass. They said it was a rare earth polish.

From my understanding the purpose of $\ce{TiO2}$ layer is to absorb UV (ultraviolet) part of solar spectrum in order to reduce negative (destructive) impact on the car internals (e.g. UV-light degrades plastics), also you and your passengers. So again, why do you want to get rid of it? Does it affect the visibility? If possible, please add the photo of the glass and/or car model (ideally glass type and manufacturer, if there are visible labels).

@andselisk. Actually the Tio2 in these coatings acts as a photo catalyst  neutralizing pollutants and dirt that are easy washed away by rain etc. So it's used more for it'self cleaning properties than UV protection in auto and building applications.

@Jeffrey Oh, I didn't know that, thank you! So why do you want to get rid of this coating? It sounds like it's been put there for a good reason.

One can feel dumber just by reading these "ingredients" but your coating is probably mostly polymeric, so you could try strongly oxidating mixtures like piranha solution or  CrO3 in H2SO4 Another option is to burn it out with flame or for more gentle approach decompose with less extreme heat.

@andselisk The windows look normal until they get wet then they reveal  runs and spots from rain,water, the car windshield cleaner etc, that have sat on the coating for any length of time. These runs etc once "set in" have to be smoothed out by repolishing. So the windows appear normal again until you get water on them again and the whole cycle repeats. So the coating doesn't go away it just becomes more uniform after polishing. I have a video showing this but I'm not sure how to get it fro m my computer to this site. I also have all the Toyota glass specs.

@Mithoron In the data sheet under heading Reactivity and sub heading Incompatibility it says Strong Oxidizing agents. So I researched that  and tried hydrogen peroxide and pool chlorine. The peroxide had some effect in the sense that the area I working became less slippery temporarily. What is piranha solution or CrO3 inH2SO4? I don't want to try to burn it off  that would probably melt the rubber gaskets around the windows. The data sheet claims it has a boiling point of 180-240F. Also states the PH is 7.2-7.6 if that helps.

Related https://chemistry.stackexchange.com/questions/84001/removing-polyimide-coating-without-damaging-carbon-nor-fused-silica-optical-fibe This "data" of yours is about solution (FYI "hydrogen hydroxide" is water...) which was used to make the coating. Like anything strong enough to remove coating, could also damage rubber, so window has to be taken out from the door before operation.

The head of the chemistry department at Cornell University is right.
The only advice I have is to wait. Eventually the ravages of time shall triumph over the titan's fiendishness.

I have seen photos of sections of buildings that had been treated with this stuff and after 6 years it was still on there and working. I was hoping to find a way to get this coating off my windows a little sooner than that.


text:
Should the mean energy be written in angular brackets or with an overline?
How would you write mean energy? I came up with two different ways:

$\displaystyle\langle G_{\textrm{f}}^\circ(j) \rangle$ with the MathJax code \langle G_{\textrm{f}}^\circ(j) \rangle
$\displaystyle\overline{G}_{\textrm{f}}^\circ(j)$ with the MathJax code \overline{G}_{\textrm{f}}^\circ(j)
Or maybe another way?

I am looking for the proper use of this notation in the scientific literature.

When in doubt, simply indicate what you mean in text.

To the question itself: I believe that the bar notation is most common for point values, while the bracket notation is mainly for functions. In any case I would write $\displaystyle \overline{G_{\textrm{f}}^\circ(j)}$.

Depends on the convention used. Different people use different conventions but both indicate a mean.

I would use \overline for representing arithmetic mean, and \langle \rangle for  expected value (notice that their difference is commonly overlooked in chemistry literature). However, the proper use only requires  stating the meaning in text.

Both the IUPAC 2008 Green Book (section 4.2) as well as the ACS Style Guide (specifically Chapter 11) agree that both methods of representing the mean value are acceptable. There might be some style guides / international standards that state otherwise, but for the average chemist these two are generally sufficient.
If ever in doubt, just state the meaning in text.

The international standard ISO 80000-2 Quantities and units – Part 2: Mathematical signs and symbols to be used in the natural sciences and technology agrees.

Good to know, too. I was lazy to check, hence the disclaimer.


text:
How many atoms can be reduced in a single redox reaction?
I am doing high school chemistry and I am busy working with redox reactions.
The explanation in my text book, of how a redox reaction works, left me a bit confused as to how many atoms can be reduced/oxidized in a single reaction equation.
Can only one atom (species) be reduced/oxidized in a single equation or can there be more than one reduced/oxidized in the same chemical equation?
All of the reaction equations I have come across so far have just one atom reduced and one atom oxidized, for example:
$$ \ce {Mg(s) + 2HCl(aq) -> H_2(g) + MgCl_2(aq)} $$
In this reaction only Mg is oxidized and only H is reduced. Is there a reaction in which more than one species of atom it reduced or oxidized?

Redox reactions can be much more complicated then that.

I know they get much more complex. I am not doing very complex ones yet. I just chose a simple one to try illustrate my question, which is a very simple question.

Yes, there can be more and yes/no questions are suitable for this site.

Thanks. That really helps. I think my text book spoke as if only one could be reduced/oxidized at a time because I would not need to do more complex ones yet. Unfortunately I don't have enough reputation to up-vote you :-(

I'd rather point out that the reactions in which multiple elements get oxidized/reduced are a relative minority (in school chemistry, at least). True, you'd encounter pretty complex reactions further on, but their complexity stems from other reasons; despite it, most of the times there would be only one oxidant and one reducing agent.

Of two species, one would usually be easier to reduce/oxidise than the other. As a reducing agent/oxidiser, I would in most cases just leave out the latter, and work on two of the former. Exemptions, as usual, prove the rule.

I'm not sure if I understand your question correctly, but consider the reaction
$$\ce{CH3-SH + 3 O2 -> SO2 + CO2 + 2 H2O}$$
where methanethiol is oxidized with oxygen to sulfur dioxide, carbon dioxide, and water.
Thereby the oxidation states of sulfur and carbon change from $-2$ to $+4$ and the oxidation state of oxygen from $\pm0$ to $-2$. So sulfur and carbon atoms are both oxidized while oxygen is reduced.
I hope this answers your question.

@adventurin: This did answer the question, thank you. The example equation helped. By the way, any suggestions on how to make the question clearer would be appreciated.


text:
How does the addition of acetone cause a shift from hexaaquacobalt(II) to tetrachloridocobaltate(II)?
For this reaction, at equilibrium,
$\ce{[Co(H_2O)_6]^2+ + 4Cl^- <=> [CoCl_4]^2- + 6H_2O}$
$\ce{[Co(H_2O)_6]^2+}$ and $\ce{[CoCl_4]^2-}$ are red and blue in aqueous solution, respectively.
When acetone is then added, the solution changes from red to blue. What causes this?

I haven't tried this before but it sounds interesting. As the blue chloro-complex is in an equilibrium with the red aqua-complex, I'd suggest the equilibrium is shifted as the water is absorbed by the acetone. As water and acetone are miscible this may take some water from the equilibrium, causing the shift towards the chloro side.
If you use ethanol to dissolve the anhydrous $\ce{CoCl2}$ it will turn blue as well. Yutaka Fukuda mentioned a chloro-ethanol-complex in his book 'Inorganic Chromotropism' for acetone I doubt it will coordinate.
Therefore I'd rather tend to the first one.

Half of this answer is ‘I think’, ‘I assume’ or irrelevant to the question. I was indifferent of it at first but seeing how it gained so many upvotes I have to counter that. Please attempt to find sources and references to back your assumptions.

I think an answer should also include similar cases. Here two possibilities can arise and I know of only this one book mentioned above that considered a coordination of the solvent as well (for this particular equilibrium) so why not give this as an option for someone having a similar question?

Please don’t misunderstand me: I’m all for noting similar or seemingly similar cases once the key case in point has been explained. However, your answer explains the key case with words that suggest you’re just guessing. I could have written a guesswork answer like that but I didn’t because as a scientist I prefer the experimental evidence where I am unsure.

I would rather you consider the problem using concentrations and or activities. I think we need to consider the change in the activity of the water and the chloride in the system.

The great problem is that the ability of a chemical substance to act in some way, is controlled by the activity of the substance and not by the concentration of it. Oftein people consider the concnetration in a first approximation where the activity coefficent of everything is assumed to be one. But for a good prediction or good undertsanding we need the activity coefficents.
for a species X
activity (a) = concentration x activity coefficent (gamma)
We need to understand that as the composition of the solvent changes or the temperture changes the activity coefficents will change.
I think that the answer relates to the relative dielectric constant of the medium, as you replace water with ethanol (or acetone) or increase the temperature then the relative dielectric constant decreases.
By lowering the relative dielectric constant you will make the coordination of the anions to the cations more favourable. I think that this is the reason why a solution of cobalt chloride in ethanol/water is thermochromic. It is pink when cold and blue when hot.
The replacement of the water with a less coordinating solvent will also help to shift the equilibrium in favour of the blue tetrachloro anionic complex.
If you look at the Debye-Huckel equation for the activity coefficient and make sure that you look at what is inside the A term, you will find that when the relative dielectric constant changes that the activity coefficients of ions will change. I think that the change of colour that you see is due to a change in activity coefficient.
The extended Debye-Huckel equation in its full glory is
$$\log (\gamma i) = \left(\frac{-Z_i^2 e^3}{4 \pi \left({\epsilon_0 \epsilon_r k_\mathrm B T}^{1.5}\right)}\right)\frac 1{1 + r_i(8 \pi n e^2/\epsilon_0 \epsilon_r k_\mathrm B T)^{0.5}}(n/2)^{0.5}$$

$Z_i$ is the charge on the ion of interest
$e$ is the electronic charge
$\epsilon_0$ is the vacuum permittivity
$\epsilon_r$ is the relative dielectric constant
$k_\mathrm B$ is Boltzmann’s constant
$T$ is the temperature in kelvin
$n$ is the number of ions in one cubic meter.
$r_i$ is the sum of the radii of the cation and anion which are in contact.

I have done the calculations for the activity coefficient of chloride in mixtures of water and acetone for a 500 μM solution of cobalt(II) chloride at 298 K in mixtures of water and acetone. As the acetone replaces the water the activity coefficient of the chloride goes down, I think that this will be due to the electrostatic interactions of the chloride anions with the cobalt cations (ion pairing):

I repeated the calculation for the activity coefficients of the cobalt ion and the $\ce{CoCl4^2-}$ ion. Bear in mind that the sum of the radii for $\ce{Co^2+}$ and $\ce{CoCl4^2-}$ will be greater than the sum of the radii for $\ce{Co^2+}$ and $\ce{Cl-}$.
Now if we consider the equation
$$K_\mathrm{eq} = a\frac{\ce{CoCl4^2-}}{a(\ce{Co^2+}){a\ce{(Cl-)^4}}}$$
Now if $K_\mathrm{eq}$ is a constant regardless of the solvent system, then as
$$a(\ce{Cl-}) = [\ce{Cl-}] \times \text{activity coefficient of } \ce{Cl-}$$
$$a(\ce{CoCl4^2-}) = [\ce{CoCl4^2-}] \times \text{activity coefficient of } \ce{CoCl4^2-}$$
$$a(\ce{Co^2+}) = [\ce{Co^2+}] \times \text{activity coefficient of } \ce{Co^2+}$$
Then if we have a new equation which is
$$\text{Random greek symbol} = \text{activity coefficient } \ce{CoCl4^2-} / {\text{activity coefficient } \ce{Co^2+} \times (\text{activity coefficient } \ce{Cl-})^4}$$
If we multiply "random greek symbol" with $K_\mathrm{eq}$' we will get a new equilibrium constant for the system.
($K_\mathrm{eq}$' is the $K$ value determined in water where it is an apparent constant - one calculated making the assumption that all activity coefficients are one)
We can get a new equilibrium constant $K$ (solvent composition) which we can plot against the fraction of the solvent mixture which is acetone. Here we have the graph.
Here K' is the true equilibrium constant which would be for a system in water with a zero concentration of cobalt chloride.
We can see as the acetone content increases that the cobalt cations are more likely to react with the chloride anions to form $\ce{CoCl4}$ dianions than they are in the aqueous system.

If you read the papers by people like Bromley[3]The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. who clearly expresses the ideas of Guggenheim you will find a specific ion interaction term in the activity coefficient equation (Bromley equation). I am sure that the value of the specific ion interaction constant for cobalt and chloride will increase as the water is replaced with acetone.
Sadly I very much doubt if the constants for water/acetone mixtures exist, most of the research effort has gone into aqueous only systems. You could make a lot of the measurements by doing UV/vis spectroscopy on solutions of chloride salts in acetone/water mixtures. But it will be a complex problem, I know that while sodium chloride is a strong electrolyte in water it is a weak one in acetone. I know a classic reaction in organic chemistry for the formation of alkyl iodides is to treat an alkyl bromide or alkyl chloride with sodium iodide in acetone. The larger anion size makes sodium iodide more soluble (and a stronger electroyte) in acetone than sodium chloride/sodium bromide.
If you look at the supporting information for [1], you will see at the end an example of thermochromism. Here mixtures of water, cobalt chloride and the lactic acid / choline chloride eutectic are shown both hot and cold.
This effect has also been reported with nickel chloride in deep eutectic solvents made from choline chloride by some chemists in China. Similar effects were seen for nickel by [2]. They have observed thermochromism which I think is due to a change in the coordination chemistry which in turn is induced by the temperature induced change in relative dielectric constant.

References:

Foreman, M. R. S. J.; Holgersson, S.; Mcphee, C.; Tyumentsev, M. S. Activity coefficients in deep eutectic solvents: implications for the solvent extraction of metals. New J. Chem. 2018, 42 (3), 2006–2012. DOI: 10.1039/c7nj03736h.
Gu, C.; Tu, J. Thermochromic behavior of chloro-nickel(II) in deep eutectic solvents and their application in thermochromic composite films. RSC Adv. 2011, 1 (7), 1220. DOI: 10.1039/C1RA00345C.
Bromley, L. A. Approximate individual ion values of β (or B) in extended Debye-Hückel theory for uni-univalent aqueous solutions at 298.15 K. The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. DOI: 10.1016/0021-9614(72)90038-9.

I also tried it experimentally now. I prepared solutions of cobalt chloride in the solvents, added enough water to discolor them and then added more solvent until it turned blue again. For acetone this shift does not take much solvent and the color is quite intense. For THF however, which has a lower dielectricity constant it takes a lot of solvent and only gives a pale blue. DMF on the other hand has no sudden jump from pink to blue but a huge gradient where it turns purple inbetween and the color is intense, too. I don't really understand what happens with the THF though.

THF is a better donor than acetone, I think that in some ways THF is a bit like water. It would be nice if you could post some pictures. Depending on your reputation you could edit my answer to add the pictures. Failing that we will find a way of doing it.

I can try that tomorrow when I'm back at the lab. I know that THF can act as a donor here but it was blue as well when I 'dissolved' the cobalt chloride in pure THF, I only added enough of the salt to fully dissolve it in THF so therefore it should turn blue again at some point.


text:
Reducing the thickness of a layer of elemental sulfur
I want to make a very thin film of elemental sulfur. While mechanical polishing can bring a film to 10 microns in thickness, but I need to bring my thickness lower (hundreds of nanometers). Although I searched, I was unsuccessful in finding any methods for chemical etching elemental sulfur.
Are there any methods for reducing the thickness of elemental sulfur below 10 microns, preferably to lengths on the order of hundreds of nanometers?

Rather than trying to make a thick layer thinner, how about vacuum deposition of sulfur? In other words, just make a thin layer.

As @MaxW states, vacuum (or vapor) deposition is the way to go. I've produced layers of gold just a few atoms thick, so thin that they are completely transparent or a clear, pale, green, that way.

I don't have access to vapor deposition type coating systems for this process, so I was hoping to find some way to do this in a chemical etch method.


text:
How to calculate the concentration of the elements after dissolution of iron(III) chloride?
I was presented with the following problem:

In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

What is the molarity of the solution?
What is the concentration of iron(III) cation in M?
And what is the concentration of chloride anion in M?

I know how to do the first part of the problem, the answer is $\pu{0.406 M}\ \ce{FeCl3}$. But when I go to balance the equation, the computer system tells me the equation is wrong.
My equation:
$$
2\ce{FeCl3 -> 2 Fe + 3Cl2}
$$
The computer system:
$$
\ce{FeCl3 -> Fe^3+ + 3Cl-}
$$
I don't understand the computer systems' answer. Simply because when by itself, chlorine is a diatomic molecule therefore, should be $\ce{Cl2}$ not just $\ce{Cl}$. Am I wrong to think this? Am I correct or the computer system? If the computer system is correct, why?

The key is the difference in chlorine vs chloride.  Any ideas?

Fe has 3+ charge and Chlorine has  a 1- charge. Is that what you are referring to?

The answer needs charges which you seem to have omitted. It's not $\ce{Cl}$, but it's also not $\ce{Cl2}$ for this question.

I've put in an edit with charges in the correct place.  It may make the answer to the question a bit more obvious.

Okay I think I understand. So when you dissolve the FeCl3 it completely dissociates and yields one Fe +3 ion and 3 Cl -1 ions. Elemental Chlorine gas is diatomic, however, Chloride ion is a single chlorine atom with a negative one charge.  When you have an ionic compound, as in the case of FeCl3, the compound is made up of elements in their ionic states, NOT in their elemental states.

@user213305 In mhchem you can (and should) typeset the reaction equation as a whole: \ce{2FeCl3 -> 2Fe + 3Cl2} not \ce{2FeCl3} \rightarrow 3\ce{Fe} + 3\ce{Cl2}.

When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.

In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

\begin{align}
  m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\
  M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\
  V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\
\end{align}

What is the molarity of the solution?

Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.
\begin{align}
  c &= \frac{n}{V}\\
  n &= \frac{m}{M}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\
  c(\ce{FeCl3 (aq)}) &=
    \pu{0.406 mol//L}
\end{align}

What is the concentration of iron(III) cation in M?
And what is the concentration of chloride anion in M?

You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.
For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is
$$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$
From this you can derive the actual relationships between the ion concentrations.
You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore
\begin{align}
  c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)})
  &&=\pu{0.406 mol//L},\\
  c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)})
  &&=\pu{1.22 mol//L}.
\end{align}

The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:
$$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$

The other equation could also be the electrolysis of the molten salt.

you finish question 1 and you need to finish question 2 and 3
but why are you making a equation for that. it seem like that was what stopped you working in the first place.
but to make it balanced... you said dissolved in volumetric flask. if it dissolved its ionic.
and that means there should be a "cation +" and a "anion -"
hence the $$\ce{fe^3+ + 3cl- }$$
and the $\ce{fe^3+}$ is third oxidation state floating around in solution it could had been a $\ce{fe^2+}$ if you started with $\ce{fe(oh)2}$
so the pc is correct....
1 mole of $\ce{fecl3}$ will yield's 1,5 mole of cl2 therefore it has more $cl$ then $fe$ ions.
concentration of $fe$ ion in $M = 0.406$
concentration of $cl$ ion in $M = 1.218$
$final\ equation\ =$ $$molarity\ of\ \ce{fecl3} * number\ of\ fe\ ion's\ in\ the\ molecule$$
$$\ce{fe^3+}=\  0.406M * 1 = 0.406M$$
$$molarity\ of\ \ce{fecl3} * number\ of\ cl\ ion's\ in\ the\ molecule$$
$$\ce{cl^-}=\    0.406M * 3 = 1.218 M$$

Please use capital letters and please don’t misuse MathJax.


text:
Why do chloroform and benzene exhibit negative deviation from Raoult's law?
Chloroform is a polar molecule, and benzene is nonpolar. Shouldn't the chloroform-chloroform and benzene-benzene intermolecular forces be stronger than chloroform-benzene interactions (like dissolves like), which would result in a positive deviation from Raoult's law?

related https://chemistry.stackexchange.com/questions/6955/deviation-from-raoults-law-because-of-adhesive-and-cohesive-forces

This is because when these liquids are mixed, H-bonding type interactions are formed between hydrogen atom of chloroform (partial positive charge due to 3 Cl atoms) and 'pi' electron cloud of benzene ring.
Thus, chloroform-benzene interactions are stronger than chloroform-chloroform and benzene-benzene interactions.

Then, why does acetone and benzene show positive deviation? Doesn't acetone (being a polar molecule) form hydrogen bonds with the "pi" electron cloud of the benzene ring?

@AdityaKumarPanda such an interaction would require availability of a hydrogen bonded to an electronegative atom, which is absent in acetone.

@ShivanshJaiswal Yeah, I've got it. Thanks!


text:
In helium-2, would each electron experience a single positive charge from the nucleus?
If we have an atom of hydrogen-1, we know that there exists one proton with one electron and in helium-2 two protons with two electrons. My question is that do protons get shared equally with electrons? For example, is it true that each electron in helium experiences the same force from the nucleus as each electron in hydrogen, due to the proton to electron ratio being 1:1 in both atoms?

No, it is not true. If it were, the electron energy levels of helium and hydrogen would be the same. However, it is true that in multi-electron atoms (neutral or otherwise) you have to take into account the 'shielding' of the nuclear charge by the other electrons. However, that shielding is not easy to describe, nor is it perfect in the sense that one electron shields one proton. It just doesn't work that way.

Even in a simplistic (non-quantum) model of the atom, the scale of the distance between the nucleus and the "orbit" of the electron would suggest the nucleus looks just like a point to the electrons so there should be no difference based on the "shape" of the distribution of charge in the nucleus. A quantum view is more complex (some orbitals have a positive probability of electrons being "in" the nucleus), but the effects are usually still small.

Sorry for mentioning Helium-2 as I understand that the isotope does not usually exist, but I mentioned it for the simplicity of the situation, although the neutrons would not really make a difference.

To answer this question, I am going to assume that a $\ce{^2He}$ nucleus actually exists long enough for two electrons to surround it — this is probably not a valid assumption since Webelements.com does not list $\ce{^2He}$ while it does list the isotopes $\ce{^6He}$ and $\ce{^8He}$ (although they admittedly have long half-lifes measured in tenths of seconds). But for our case, let us assume so.
The charge radius of a proton — our closest experimental measurement of its diameter, if ‘diameter’ is even a valid term at subatomic scales — is given on Wikipedia as $\pu{8.75e-16 m}$ or $\pu{0.875fm}$. At this point, take a step back to remember that the Bohr radius, approximately the most probable distance between the electron and the proton in a hydrogen atom, is $\pu{5.29e-11 m}$ or $\pu{52.9pm}$. This is a difference in scale by a factor of $10^5$.
Therefore, a reasonable assumption for the $\ce{^2He}$ nucleus — and indeed for practically any nucleus — is that it can be approximated by a dimensionless point of $+2$ charge and not any sphere, ovaloid, double-sphere or other three-dimensional object. Therefore, if we want to use quantum mechanics to calculate the wave function of $\ce{^2He+}$, we assume a spheric potential caused by a positive charge in the origin of a coordinate system; the value of the positive charge is $2e$ with $e$ being the elementary charge.
A two-electron system cannot, to the best of my knowledge, accurately be calculated. However, we may approximate it as just taking the wave function solutions of $\ce{^2He+}$ and adding another spin-opposite electron. (Then, we slightly adjust the energy values to account for electron-electron repulsion.) This way of approaching the problem shows that the helium atom should really better be considered a central positively charged dot with the outer space occupied by the negatively charged electrons, each of which see a $+2$ charge in the centre and an additional negative charge ‘buzzing around them’ (Note: physically incorrect macroscopic picture).
Replacing $\ce{^2He}$ (as is present in your question) with either of the stable isotopes $\ce{^3He}$ or $\ce{^4He}$ would not change the picture; the size of a neutron (see my note in the second paragraph on subatomic scales) is roughtly that of a proton. If anything, adding one or two neutrons to the two protons will make the nucleus smaller since neutrons increase the intranuclear attractive forces.

In many introductory chemistry books, the core and shell model of an atom and the relevant size differences are exemplified by macroscopic comparisons. My chemistry book offered: if you enlargen an atom so that it would fit the Munich Olympic Tower inside it, the nucleus would inflate to the size of a cherry stone, match head or something of that size. This again shows how good the assumption of poiut-sized charges in the nucleus is.

OP surely meant the elements ordinal number when writing H-1, He-2, not the mass.

@Karl I am not willing to make that assumption but I will note that it does not change the general picture if we add one or two neutrons into the equation.

@Jan So when each electron 'sees' a +2 point positive charge with an additional negative charge orbiting it, does the effect of the positive charge for the electron in question in any way get diminished?

@Coder511 no, that’s what I tried to outline in my answer.

Electrostatic interaction is like gravitation: A spherical charge (or body) gives the same attraction as an equal point charge (or mass) at the centre of the sphere. If you are in the sphere, gravity is the same as for a mass with the radius of your current distance to the centre. All the mass or charge closer to the surface than yourself has zero net interaction with you.
The electrons in helium are in the spherical s-orbital. In every instant, one electron feels the full charge of two, while the other one, which is a bit further out, only feels a charge of one, because the other electron shields the core.
This is a bit oversimplistic (hello quantum mechanics), but each electron feels a core charge roughly around 1.5  in average.

You are arguing that electrons in the same orbital shield each other. I'm not sure that is right in either the classical or the quantum view of the situation. Perhaps it is an approximation for electrons in different orbitals with different spatial distributions.

@matt_black It's a rather good approximation for electrons in different (filled towards partially filled) orbitals. For two electrons in the same orbital it is of course a dead alley towards a realistic model, but the general idea is certainly valid. Perhaps it should say "very simplistic" in the conclusion.

First of all, helium-2 is an extremely short-lived species - if you wanted to speak about its existence at all. So I assume that you had a more stable isotope in mind, e.g. helium-3 or helium-4.

... is it true that each electron in helium experiences the same force ...

Both electrons in a helium atom are indistinguishable from each other. If you wanted to apply the concept of a force exerted on an electron at microscopic scale, then both electrons would experience the same force.

... would each electron experience a single positive charge from the nucleus?

No. The charge experienced is the charge of the two protons and the electrons that shield each other.
The second ionization energy of helium is four times the ionization energy of hydrogen, as predicted by quantum mechanics. The first ionization energy of helium is greater than that of hydrogen and smaller than its second ionization energy. This is in accordance with the statement above and suggests an "effective charge" roughly about $1.3$ to $1.4$.

$\pu{1312.0 kJ/mol}$ Hydrogen 1. ionization energy
$\pu{2372.3 kJ/mol}$ Helium 1. ionization energy
$\pu{5250.5 kJ/mol}$ Helium 2. ionization energy

No, the protons are not "shared" and the attractive force from the nucleus on each electron is about 6 times higher for helium than for hydrogen. This is due to the fact that there are now 2 protons pulling on each electron and they are considerably closer, so each one pulls about 3 times as hard. There is a repelling force between both electrons, but there is no simple analytical expression for it.
In a classical description of this system, the force for each particle-pair is given by Coulomb's law. A single electron around a Hydrogen nucleus feels an attractive force that is proportional to $\frac{1}{r^2}$ and each electron around the Helium-nucleus will feel this force for each proton. The smaller radius of the $\ce{He}$ atom (roughly $\frac{1}{\sqrt{3}}$ times the Bohr radius*)  will lead to a ~3 times higher force per proton, so a ~6 times higher force between the entire nucleus and each electron.
If you wanted to relate this to ionization energy, you can effectively assume that the nuclear attraction force on the departing electron is shielded by the remaining electron. As an approximation, this energy is given by the shielded nuclear charge divided by the radius in Rydberg atomic units : $\frac{Z-1}{r} \approx\sqrt{3}$, or $\pu{24 eV}$. The remaining electron of what is now $\ce{He+}$ will have moved closer to the nucleus, at a distance of $\frac{1}{\sqrt{4}}$, i.e. $\frac{1}{2}$. Its ionization potential is $\frac{Z-0}{r} = 4$, or $\pu{54 eV}$. For completeness sake, in these units the radius of $\ce{H}$ is $1$ and its ionization potential is $\frac{Z-0}{r} = 1$, or $\pu{13.6 eV}$.

* the $\frac{1}{\sqrt{3}}$ expression for the radius of $\ce{He}$ is just a shortcut for ease of calculation. It is within 2% of theoretical/experimental values.


text:
Deviation angle for 1st order diffraction

(i) A diffraction grating has 560 lines per mm. Calculate the grating spacing. If the total length of the grating is 2 cm, how many lines are there?
(ii) Light of wavelength 500 nm is incident on the grating with this spacing. Calculate the deviation angle for 1st order diffraction.

I am having difficulty with the (ii) part. For my spacing I got 0.0018 mm. So, then I used
\begin{align}
n \lambda &= d \sin{\Theta} \\
\frac{n \lambda}{d} &= \sin{\Theta} \\
0.27 &= \sin{\Theta} \\
\Theta &= \sin^{-1}{(0.27)} = 15.7^\circ
\end{align}

The grating equation you've used is correct, and in order to find the first order diffraction angle ($n = 1$) you need to use arcsine trigonometric function and carefully place data with respective units:
\begin{align}
n \lambda &= d \sin{\Theta} \tag{1}\\
d &= \frac{\pu{1 mm}}{N}~[\pu{mm}] \tag{2}
\end{align}
where $N$ is number of lines per unit length. Combining (1) and (2) and converting the wavelength from nm to mm:
$$\Theta = \arcsin{\left(\frac{n \lambda}{d}\right)} = \arcsin{\left(\frac{n \lambda N}{\pu{1 mm}}\right)} = \arcsin{\left(\frac{1 \cdot \pu{5e-4 mm} \cdot 560}{\pu{1 mm}}\right)} = 16.26^\circ \tag{3}$$


text:
Finding CML files for small molecules
I'm trying to generate a figure for a physics textbook to show examples of the vector addition of dipole moments in small molecules. For this purpose, I would like to render pictures of a bunch of small molecules, such as CS2, formaldehyde, benzene, SF6, and chloroform. I've got avogadro running on my linux box, but I'm finding it very time-consuming and confusing to construct the molecules using the GUI. Is there anywhere that I could find pre-made CML files for small molecules like these?

Do you have access to chemoffice? Chem3D can produce 3D structures from formulas and if you run the optimization in there it should be good enough for this application. Other than that: how many molecules are we speaking about? With the right software making each of those takes less than a minute each including low level geometry optimization. If it's not too many I'm happy to help, we can speak in chat.
Another option: you can open the 3D model in jmol from their wikipedia page and extract the *.mol file which to my knowledge can be opened by Avogadro.

@DSVA: Thanks for the ideas, and for the offer of help. This pointed me to the info that I'll give in an answer to my own question.

DSVA's comment pointed me to a database that contains the information I need. The wikipedia articles contain links to an app called jsmol, but that app appears to be getting its data from a database called ChEBI. You can search the database using search terms such as "formaldehyde" or "sf6." The entry for each molecule contains a link to a .mol file, which avogadro can read.

Another location, if you're ok with the mol2 format (which Avogadro can open), PQR has links to download structures on every page. For example, on the cadaverine page:

This will trigger a browser download dialog box for a mol2 file.

Consider a blend of Wikimedia Chemical Structure Explorer and OpenBabel.

Wikipedia includes many entries about chemicals, and often, the property box right on top of the page (left, or right, depends on your script) include the condensed structure representation as a SMILES string:

(credit, the Arabic entry about toluene)
With above mentioned freely available, open source structure explorer which you may run without installation from a web browser (further description, e.g., in J Cheminform 7, 10 (2015)) you may query all of these entries by a (sub)structure search, a keyword, or a combination of both to access the English article and find the Wikipedia page on the right hand pane:

(screen photo of the explorer)
As by today, the public index contains 18484 entries.

With the help of OpenBabel, you may convert the SMILES string into the CML format (the the documentation presents many additional input/output formats, too).   After installation of OpenBabel, the program may used from the CLI / cmd.exe, or from a GUI.  Then, you have at least two options:
a) You save the SMILES string from wikipedia in as file, for example toluene.smi.  If you opt for a conversion from the CLI, the command
obabel -ismi toluene.smi -ocml -O toluene.cml
writes the new file toluene.cml.
b) You paste the SMILES string into the command on the CLI when writing
obabel -:CC1=CC=CC=C1 -ocml -O toluene.cml
to generate the .cml file right in the current working directory. (No, do not set a space between the colon and the SMILES string.)
The documentation lists many more examples and showcases the use of the GUI.  OpenBabel may work on multiple input files at once (e.g., obabel *.xyz -opdb -m to convert all .xyz of a folder into .pdb) and interacts well with other programming languages, too (introduction).

If you have Microsoft Word installed on a Windows machine you can use our open source Add In called Chem4Word to search the ChEBI database and import structures.
Once imported they can be optionally edited, then exported as CML.
You can find Chem4Word here https://www.chem4word.co.uk/download


text:
Why does old gasoline seem to have more "kick"?
From my personal experience, old gasoline (leftover in a container from the previous year) has more "kick" (power) when I actually use it. Higher idle RPM, higher throttle sensitivity.
It says here that it's not the gasoline itself that burns, but the vapour. And old gasoline is bad because it has vaporised over time. Thus, old gasoline must be weaker.
However, I read somewhere that gasoline vapour has nothing to do with octane rate, so there's no difference in octane between the gas left out in the open for a while and the gas sealed in a container.
I'm confused. Does old gasoline seem to have more "kick" because it needs more compression to combust? Or what is it?

Gasoline on storage leads to gum separation. Maybe the heavier hydrocarbons separate?

Similar as in the answer by @AJKOER I think that gasoline - a blend of hydrocarbons and other compounds (e.g., MTBE ether) - may alter over time.  Without experimental evidence, assuming it is stored only for about a year in a dry and cool place in the (metal) jerrycan, I speculate its compounds with a lower boiling point evaporate easier than its compounds with a higher boiling point across the cap of the jerrycan.
As stated correctly in the question, it is not the liquid gasoline driving the engine.  It is the compressed aerosol of the gasoline ignited by the spark plug.  With said relative enrichment of the compounds of higher boiling point in comparison to those of lower boiling point, I speculate a higher compression ratio may be attained without premature (auto)ignition of the fuel.  And because of the higher compression ratio accessible, the combustion may offer more mechanical work (or, as mentioned in the question, offer «more kick»).

My candidate to explain the increased performance is the removal of water as part of the corrosion reaction occurring in metal storage containers. To quote a reference:

Since oxygen has different solubilities (higher for the organic and lower for the aqueous phase), anode or cathode regions are formed on the steel or metal surface. This creates favorable conditions for electrochemical corrosion in a given environment.

As such, I would argue that not only does oxygen content impact corrosion rate, but potentially more important, is that the removal of water over time (from corrosion chemistry) also increases O2 solubility (per above). This could be especially favorable when the fuel is vaporized and mixed with air in the ignition chamber.
[EDIT] Further comments, same source:

Oxidation of biodiesel leads to the formation of various products such as peroxides and hydroperoxides. During degradation, these products transform into shorter-chain compounds such as low-molecular weight acids, aldehydes, ketones, and alcohols [9].

Per another source:

In the winter, companies produce a gasoline containing lighter hydrocarbons, making the liquid more volatile and therefore easier to ignite.

This source also notes with more volatile fuels, evaporation can leave behind a heavier fuel mix.
Corrosion induced oxidation of biodiesel, in particular, appears to not only impact potential oxygen content of the fuel, but may add peroxides as well. The creation of lower molecular weight organics should also increase flammability.
An interesting point once presented in Wikipedia on Methanol, to quote:
"As a fuel for mud racers, methanol mixed with gasoline and nitrous oxide produces more power than gasoline and nitrous oxide alone."
Which, in the current context, supports my opinion that fuel mix chemistry (whether corrosion based, or otherwise, involving O2 or N2O) CAN IMPACT PERFORMANCE.
So, more oxygen content, less heavy fuel mix, more complete ignition, may explain some of the so-called 'kick' that has been referenced. On the other hand, significant evaporation is likely detrimental.
[EDIT] Yet another path as a consequence of corrosion chemistry, to quote from Wikipedia on Tetraethyllead:

At the temperatures found in internal combustion engines, (CH3CH2)4Pb decomposes completely into lead and lead oxides as well as combustible, short-lived ethyl radicals. Lead and lead oxide scavenge radical intermediates in combustion reactions. Engine knock is caused by a cool flame, an oscillating low-temperature combustion reaction that occurs before the proper, hot ignition. Lead quenches the pyrolysed radicals and thus kills the radical chain reaction that would sustain a cool flame, preventing it from disturbing the smooth ignition of the hot flame front.

Also on the performance effect of Tetraethyllead (TEL):

Antiknock agents allow the use of higher compression ratios for greater efficiency[17] and peak power.[18] Adding varying amounts of additives such as low percentage TEL or high percentage ethanol to gasoline allowed easy inexpensive control of octane ratings... Aviation spirits with TEL used in WWII reached 150 octane to enable supercharged engines such as the Rolls-Royce Merlin and Griffon to reach high horsepower ratings at altitude.[19] In military aviation, TEL manipulation allowed a range of different fuels to be tailored for particular flight conditions.

So, the corrosion process on even minor alloy metals, found in the metal storage container, may have also introduced an organometal compound with similar properties to TEL in its ability to scavenge radical intermediates.

I'm not sure if I could be served leaded gas at stations in NJ.  (Because I never needed it, I didn't asked the gents yet.)  Even if available (i.e., leaded aviation fuel probably still is in use), lead would damage catalytic converters installed in cars driven by an Otto engine; and I assume a large population of these cars built since the 1990s have them.  Thus I think nowadays (2020s), the section about TEL becomes less and less important to consider.


text:
Why doesn't D block contraction affect the other P block groups 14 15 16 etc
Usually we come across anomalies in the trends of periodic properties for example the size [Atomic radius value from Raymond Chang Chemistry 9th Edition] of Gallium (135 pm) is smaller than Aluminium (143 pm) as opposed to the basic trend that atomic size increases down the group. This is usually attributed to the fact that the poor shielding of nuclear charge by intervening D orbitals leads to a greater $Z_\mathrm{eff}$ hence the radius is less as opposed to the trend.
If that's the reason why don't we observe similar effects for other groups like group 15 having Phosphorus (110 pm) and Arsenic (120 pm) [Here the size doesn't increase much but is still significant as compared to changes taking place in group 13] , where the size actually increases even when there are D orbitals to intervene. Or in short why are the effects so profound for group 13 and not the other members?
There's another example of In (166 pm) and Tl (171 pm). Both belong to group 13. As compared to other groups these two members have the lowest increase in atomic radius.

As a property showing d-block contraction, the question cited atomic radius. However, d-block contraction can manifest itself in other properties as well:

$\ce{GeO}$ is described as much more stable than $\ce{SiO}$. $\ce{Sn(II)}$ compounds are described as reducing. [Riedel]
The pentachlorides of $\ce{P, As, Sb}$ show a kink in their stability: $\ce{PCl5}$ melts at 167°C, $\ce{AsCl5}$ decomposes at $-50\mathrm{°C}$, $\ce{SbCl5}$ melts at 4°C (and decomposes at 140°C) [HoWi]
The redox potentials in acidic solution for the (0)/(IV) oxyanions and oxides of chalkogens show a peak at $\ce{Se}$ [HoWi]:

\begin{align}
\ce{SO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.50 \ \mathrm{V} \\
\ce{SeO_3^{2-}} + \ce{4e-} + \ce{6H+} &\leftrightharpoons \ce{Se} + \ce{3H2O} \quad &E^0 = +0.74 \ \mathrm{V}\\
\ce{TeO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.57 \ \mathrm{V}
\end{align}

the oxidative strength (at both low and high pH) of halogen oxyacids follows the pattern $\ce{BrO4-} > \ce{H_5IO6} > \ce{ClO4-}$. [HoWi]

I think that stability of oxidation states, stability of comparable compounds, and redox potentials of similar compounds are valid examples of d-block contraction. Of course, other chemists may arrive at different conclusions.

Literature:
[Riedel] E Riedel, Anorganische Chemie, Vol. 6, de Gruyter (2004).
[HoWi] AF Holleman, E Wiberg, Lehrbuch der Anorganischen Chemie, Vol. 101, de Gruyter (1995).


text:
What visual observations can confirm that a reactant is the limiting reagent?
I have this lab question for the lab called Copper Collection Stoichiometry, where we choose an amount of the limiting reagent (iron) for a reaction between copper (II) sulfate and iron. We are to dissolve copper (II) sulfate in water, making it blue, then add iron, and after filtrating, we should have copper left.
How do we know if iron was the limiting reagent in terms of the visual observations? We made it so that iron was the limiting reagent, but how do we know visually if it is?
I understand that in order for iron to be a limiting reagent, it should've reacted with the solution and all completely turned into copper after the experiment. However how do we know if all the iron actually turned into copper? There is no way that we can visually count every solid flake to make sure there's no iron. I also understand that if iron was the limiting reagent, then there should be still copper (II) sulfate left in the solution, as the reaction didn't completely occur. Is that the answer?

Throw in more iron and make sure that the reaction starts again. That would verify your last point.

@IvanNeretin I don't have the resources to do that, it is a school lab, and we are meant to confirm visually that iron is actually a limiting reagent

OK, then was the solution at least slightly blue in the end?

@IvanNeretin yes slightly blue, although lighter in colour than the original solution with warm water and copper (II) sulfate. Does that indicate there is still copper (II) sulfate in the filtrate? I searched up what Iron (II) sulfate looked like, however it was also slightly blue, mixed with green...

You got it right. Yes, that counts as a proof. Iron sulfate in comparable concentrations is all but colorless.

@IvanNeretin Ah I see, thank you so much, that makes alot of sense. Maybe you wish to copy that to post as an answer? I can mark it right cause reputation and stuff if you want. Thanks again

You may answer it yourself, this is allowed and even encouraged, and the idea was yours anyway.

Sounds like you do metals transmutation. ..

The colour of the filtrate after filtering out the solid is still lightly blue in colour. This indicates that Copper (II) sulfate still is present in the solution, meaning that there was not enough iron to completely react with all the copper (II) sulfate, visually confirming it to be the limiting reagent.
A solution containing only Iron (II) sulfate is colourless, so if the solution is blue in colour, then it shows that Copper (II) sulfate still is present.
Credit to Ivan Neretin for helping me.


text:
Ester as a side product in the synthesis of bromoalkane from alcohol. Is it correct?
About 5 months ago, I answered a very interesting question albeit a very simple one.

Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

The question was why sulfuric acid is used to convert any alcohol to bromoalkane but not iodoalkane. Jan's answer quite pointed out the explanation:

hydrogen iodide is sufficiently acidic to protonate the corresponding    alcohol and drive the reaction. More acids are not
  needed.
hydrogen bromide is not acidic enough to drive the reaction. Adding    sulphuric acid protonates the alcohol partially and allows the
  reaction to proceed.
in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidize iodide (while sulfuric acid does).

Though I am satisfied with Jan's excellent answer, I further pointed out that sulfuric acid may react with hydrobromic acid to form bromine(aq.) and sulfur dioxide. This bromine may oxidize alcohol to form aldehyde and carboxylic acid and they two will form ester. This was my proposed explanation but I couldn't find any experimental result to back my explanation and heavily relied on theory.
Now recently Nurdrage, a renowned chemist famous for his synthesis of pyrimethamine has uploaded a video on Youtube about the synthesis of bromoethane and bromohexane from ethanol and hexanol using sulfuric acid. He also explained the concern of forming bromine as side product from the reaction of hydrobromic acid and sulfuric acid. He showed a purification technique by adding sodium bicarbonate to react with bromine to form sodium hypobromide.
Now my question is:
Is my explanation of forming ester instead of bromoalkane correct enough? Is there any possibility of forming minute amounts of ester? The reaction may be slow but it can happen before any purification is done (reacting bromine with sodium bicarbonate). If so, what can be done to remove any ester or prevent the reaction of ester formation?
Also related: Why is alcohol boiled with HBr for preparing alkyl bromide?

Technically it could form, but it would be hydrolyzed. If it did form, it would be in negligible quantities. Lots of other stuff could form, and they do, but in miniscule amounts ... bromine oxidized the alcohol to the aldehyde. The aldehyde undergoes aldol condensation. Another alcohol undergoes elimination. The compounds  react in a hetero diels-alder. So dihydropyrans can also form ... Yes, but not in quantities that are observable.

Fair enough but could you back your answer with a reference?


text:
Uv vis and fluorescence spectroscopy: sensitivity
Is there any reason why fluorescence spectroscopy is said to be more sensitive than Uv-Vis?

Fluorescence is a 'zero background' or absolute type of measurement meaning that single photons can be measured against a 'dark' background so the sensitivity is huge, and limited by the fraction of light caught and efficiency of the detector.
Absorption is a relative measurement as it is the difference in two large numbers, one the light intensity with no sample $I_0$ and the other when the sample is present I. In terms of photons these intensities are huge, say $I_0=10^6$ and a 1% change would reduce the number of photons to $10^6-10^4$; still a large number and so the absorption measurement is looking for the small difference in two large numbers. This measurement is limited by noise in detectors and other electronics.   (Its a bit like trying to weigh the captain of a ship by weighing the ship with him standing on the deck and then when he is not there.)

I'll just add that the correct mathematical abstraction for sensitivity is the signal to noise ratio.

yes I should have added that point.

In fluorescence, what is detected is really photons emitted.  The measure is done in absolute term (versus 0 , the dark) and the response can be easily amplified.
Conversely, in absorption is a difference that must be measured. And this can of course be arbitrarily tiny.
It is like massing a sample. Is easy to weight a mg than detecting a mg difference among two " about 10 kg " samples.
This general consideration is exacerbated by other practical details, as listed in the  answer by Dan.
Also your consideration about geometry applies, which again is dictated by the above.
Edit: Porphyrin answers appeared after I submitted this. Else I wouldn't have answered as this is basically a duplicate.

Absorbance may occur from non-analytes, while fluorescence is often quite specific to an analyte or a fluorophore-tagged analyte. Also, instrumental setup could play an effect. UV-Vis specs are often horizontal in terms of incident beams, sample, and the detector. Fluorescence specs are often at 90 degrees to "extract" the fluorescence from the sample to the detector. If you are only "extracting" what you are interested in, I'd imagine the S/N would be better, yielding higher sensitivity.


text:
Is ionic charge just a label?
We say that when an atom loses an electron it becomes positively charged ion, and similarly when an atom gains electrons it becomes a negatively charged ion. My question is that are 'positive' and 'negative' just labels indicating more protons than electrons or vice versa, or do the ions actually display an increased interaction towards other charged particles?

Is the sum printed on your bank account statement just a label indicating more or less dollars, or do you actually display an increased ability to buy goods?

Are you familiar with electroplating ?

Ummm, and here I was thinking for a moment that this is question "If ionic bonding is just a label?" which may be worth these upvotes. And here it looks like joke...

These ionic charges actually do behave as you would expect from electric charges: cations will be drawn towards the negative pole in an electrolytic cell while anions will be drawn towards the positive pole. The same feature can be observed with α and β radiation (which are $\ce{^4_2He^2+}$ and electrons, respectively); these are deflected in an electric field as expected.
On a more philosophical level, the entire concept of one charge being ‘positive’ while the other is ‘negative’ is of course meaningless. We might as well have called them ‘East’ and ‘West’ charges akin to the North and South poles of a compass. The only physically observable is that the two act as opposites; two particles of like charge will repel, two of opposite will attract. Having positive and negative charges makes calculations much easier though.


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Interference and absortion filters
I am learning a bit about Uv-vis spectroscopy filters. As far as I know there are two commonly used: interference and absotion filters.
The first one works with the interference of light principle and the second is a glass with some pigments which absorbs light (of some specific frequency).
Question
Have you got any idea if there is any further classification of absortion filters? Any help will be greatly welcome.

There are a great many of each type of these filters.
Interference filters are optimized for different wavelengths, different bandwidths, transmission vs. reflection etc. by controlling the materials and the thicknesses of the layers. A commercial supplier, Edmund Optics, shows how to specify and order filters.
In a similar manner, absorptive filters depend on the chemicals used to make them. For example, a solution of $\ce{CuSO4}$, copper sulfate, in water removes wavelengths between ~600-750 nm, but letting blue light through..

Possibly I haven't been clear enough. I wonder if there is any further classification of absortion filters...Thanks for the answer anyway!

Sure: you could classify them any number of ways: Solid, liquid or gaseous? Absorption due to chemical formula or due to colloidal properties (e.g. CuSO4 blue vs. colloidal-gold-in-glass red)? Solid filters could be subdivided into single vs. multilayer, glass vs. plastic... What is the point of your question? Classification is arbitrary.